Craig's soil mechanics 7th solutions manual

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Craigrsquos Soil Mechanics Seventh EditionSolutions Manual

Craigrsquos Soil Mechanics SeventhEdition Solutions Manual

RF CraigFormerly

Department of Civil Engineering

University of Dundee UK

First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

Spon Press is an imprint of the Taylor amp Francis Group

ordf 1992 1997 2004 RF Craig

All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash33294ndashX

This edition published in the Taylor amp Francis e-Library 2004

(Print edition)

ISBN 0-203-31104-3 Master e-book ISBN

ISBN 0-203-67167-8 (Adobe eReader Format)

collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

Contents

1 Basic characteristics of soils 1

2 Seepage 6

3 Effective stress 14

4 Shear strength 22

5 Stresses and displacements 28

6 Lateral earth pressure 34

7 Consolidation theory 50

8 Bearing capacity 60

9 Stability of slopes 74

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 2: Craig's soil mechanics 7th solutions manual

Craigrsquos Soil Mechanics Seventh EditionSolutions Manual

Craigrsquos Soil Mechanics SeventhEdition Solutions Manual

RF CraigFormerly

Department of Civil Engineering

University of Dundee UK

First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

Spon Press is an imprint of the Taylor amp Francis Group

ordf 1992 1997 2004 RF Craig

All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash33294ndashX

This edition published in the Taylor amp Francis e-Library 2004

(Print edition)

ISBN 0-203-31104-3 Master e-book ISBN

ISBN 0-203-67167-8 (Adobe eReader Format)

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Contents

1 Basic characteristics of soils 1

2 Seepage 6

3 Effective stress 14

4 Shear strength 22

5 Stresses and displacements 28

6 Lateral earth pressure 34

7 Consolidation theory 50

8 Bearing capacity 60

9 Stability of slopes 74

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 3: Craig's soil mechanics 7th solutions manual

Craigrsquos Soil Mechanics SeventhEdition Solutions Manual

RF CraigFormerly

Department of Civil Engineering

University of Dundee UK

First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

Spon Press is an imprint of the Taylor amp Francis Group

ordf 1992 1997 2004 RF Craig

All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash33294ndashX

This edition published in the Taylor amp Francis e-Library 2004

(Print edition)

ISBN 0-203-31104-3 Master e-book ISBN

ISBN 0-203-67167-8 (Adobe eReader Format)

collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

Contents

1 Basic characteristics of soils 1

2 Seepage 6

3 Effective stress 14

4 Shear strength 22

5 Stresses and displacements 28

6 Lateral earth pressure 34

7 Consolidation theory 50

8 Bearing capacity 60

9 Stability of slopes 74

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 4: Craig's soil mechanics 7th solutions manual

First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

Spon Press is an imprint of the Taylor amp Francis Group

ordf 1992 1997 2004 RF Craig

All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash33294ndashX

This edition published in the Taylor amp Francis e-Library 2004

(Print edition)

ISBN 0-203-31104-3 Master e-book ISBN

ISBN 0-203-67167-8 (Adobe eReader Format)

collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

Contents

1 Basic characteristics of soils 1

2 Seepage 6

3 Effective stress 14

4 Shear strength 22

5 Stresses and displacements 28

6 Lateral earth pressure 34

7 Consolidation theory 50

8 Bearing capacity 60

9 Stability of slopes 74

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 5: Craig's soil mechanics 7th solutions manual

Contents

1 Basic characteristics of soils 1

2 Seepage 6

3 Effective stress 14

4 Shear strength 22

5 Stresses and displacements 28

6 Lateral earth pressure 34

7 Consolidation theory 50

8 Bearing capacity 60

9 Stability of slopes 74

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 6: Craig's soil mechanics 7th solutions manual

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 7: Craig's soil mechanics 7th solutions manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 8: Craig's soil mechanics 7th solutions manual

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 9: Craig's soil mechanics 7th solutions manual

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 10: Craig's soil mechanics 7th solutions manual

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 11: Craig's soil mechanics 7th solutions manual

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 12: Craig's soil mechanics 7th solutions manual

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 13: Craig's soil mechanics 7th solutions manual

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

Dear-User
Oval

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 14: Craig's soil mechanics 7th solutions manual

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 15: Craig's soil mechanics 7th solutions manual

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

Dear-User
Oval

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 16: Craig's soil mechanics 7th solutions manual

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 17: Craig's soil mechanics 7th solutions manual

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

Dear-User
Rectangle

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 18: Craig's soil mechanics 7th solutions manual

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 19: Craig's soil mechanics 7th solutions manual

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 20: Craig's soil mechanics 7th solutions manual

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 21: Craig's soil mechanics 7th solutions manual

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 22: Craig's soil mechanics 7th solutions manual

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 23: Craig's soil mechanics 7th solutions manual

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 24: Craig's soil mechanics 7th solutions manual

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 25: Craig's soil mechanics 7th solutions manual

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 26: Craig's soil mechanics 7th solutions manual

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 27: Craig's soil mechanics 7th solutions manual

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 28: Craig's soil mechanics 7th solutions manual

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 29: Craig's soil mechanics 7th solutions manual

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 30: Craig's soil mechanics 7th solutions manual

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 31: Craig's soil mechanics 7th solutions manual

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 32: Craig's soil mechanics 7th solutions manual

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 33: Craig's soil mechanics 7th solutions manual

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 34: Craig's soil mechanics 7th solutions manual

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 35: Craig's soil mechanics 7th solutions manual

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 36: Craig's soil mechanics 7th solutions manual

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 37: Craig's soil mechanics 7th solutions manual

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 38: Craig's soil mechanics 7th solutions manual

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 39: Craig's soil mechanics 7th solutions manual

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 40: Craig's soil mechanics 7th solutions manual

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 41: Craig's soil mechanics 7th solutions manual

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 42: Craig's soil mechanics 7th solutions manual

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 43: Craig's soil mechanics 7th solutions manual

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 44: Craig's soil mechanics 7th solutions manual

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 45: Craig's soil mechanics 7th solutions manual

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 46: Craig's soil mechanics 7th solutions manual

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 47: Craig's soil mechanics 7th solutions manual

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 48: Craig's soil mechanics 7th solutions manual

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 49: Craig's soil mechanics 7th solutions manual

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 50: Craig's soil mechanics 7th solutions manual

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 51: Craig's soil mechanics 7th solutions manual

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 52: Craig's soil mechanics 7th solutions manual

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 53: Craig's soil mechanics 7th solutions manual

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 54: Craig's soil mechanics 7th solutions manual

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 55: Craig's soil mechanics 7th solutions manual

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 56: Craig's soil mechanics 7th solutions manual

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 57: Craig's soil mechanics 7th solutions manual

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 58: Craig's soil mechanics 7th solutions manual

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 59: Craig's soil mechanics 7th solutions manual

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 60: Craig's soil mechanics 7th solutions manual

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 61: Craig's soil mechanics 7th solutions manual

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 62: Craig's soil mechanics 7th solutions manual

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 63: Craig's soil mechanics 7th solutions manual

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 64: Craig's soil mechanics 7th solutions manual

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 65: Craig's soil mechanics 7th solutions manual

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 66: Craig's soil mechanics 7th solutions manual

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 67: Craig's soil mechanics 7th solutions manual

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 68: Craig's soil mechanics 7th solutions manual

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 69: Craig's soil mechanics 7th solutions manual

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 70: Craig's soil mechanics 7th solutions manual

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 71: Craig's soil mechanics 7th solutions manual

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 72: Craig's soil mechanics 7th solutions manual

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 73: Craig's soil mechanics 7th solutions manual

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 74: Craig's soil mechanics 7th solutions manual

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 75: Craig's soil mechanics 7th solutions manual

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 76: Craig's soil mechanics 7th solutions manual

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 77: Craig's soil mechanics 7th solutions manual

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 78: Craig's soil mechanics 7th solutions manual

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 79: Craig's soil mechanics 7th solutions manual

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 80: Craig's soil mechanics 7th solutions manual

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 81: Craig's soil mechanics 7th solutions manual

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 82: Craig's soil mechanics 7th solutions manual

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 83: Craig's soil mechanics 7th solutions manual

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 84: Craig's soil mechanics 7th solutions manual

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 85: Craig's soil mechanics 7th solutions manual

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes
Page 86: Craig's soil mechanics 7th solutions manual

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Ch2 Seepage
    • Pb 26 anisotrop net
    • pb 23 isotropic
    • pb 28 isotropic net
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes

Craig's Soil Mechanics 0415327024

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