Gravity Wall Verification Manual En

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Verification Manual

of GEO5 Gravity Wall program

Written by: Ing. Veronika Vaneckova, Ph.D.

Edited by: Ing. Jirı Laurin

Verze: 1.0-en

© 1989 - 2009 Fine Ltd.

www.finesotware.eu



INTRODUCTION

This Gravity Wall program Verification Manual contains hand-made calculation of following problems:

• calculation of earth pressures

• calculation of hydrodynamic pressure

• verification of the wall with the help of theory of limit states

• dimensioning of wall section according to EN 1992-1-1 [1]

• verification of the wall influenced by earthquake according to Mononobe-Okabetheory [2], [3]

In general this Verification Manual has been made to show the ability of the GEO5software to provide a solution consistent with the result of hand-made calculations ortextbook solutions.

Results of hand-made calculations are compared with the results from GEO5 GravityWall program.

For further details please refer to our verification example file (Demo.gtz – downloadhttp://www.finesoftware.eu/references/verification-manuals/) using our program GEO5Gravity Wall (http://www.finesoftware.eu/geotechnical-software/gravity-wall/).



Contents

1 Example 1 1

1.1 Problem description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Verification of the whole wall . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Bearing capacity of the foundation soil . . . . . . . . . . . . . . . . . . . 11

1.4 Dimensioning - analysis of wall section . . . . . . . . . . . . . . . . . . . 11

1.5 Verification of the wall - influence of earthquake . . . . . . . . . . . . . . 16

1.6 Bearing capacity of the foundation soil . . . . . . . . . . . . . . . . . . . 23

Reference 23



Verification Manual GEO5 Gravity Wall

1 Example 1

1.1 Problem description

In Fig. 1 is shown example of gravity wall with inclined footing bottom (inclination

1:10). Earth body is compounded from two soil layers and terrain is inclined (1:10).

Top layer (depth 1.5 m) is formed from sandy silt (MS), lower layer is clayey sand

(SC) at front face and back face of the wall. Groundwater table is in the level of layer

SC, i.e. 1.5 m behind the wall and 3.7 m in front of the wall. Soil properties (effective

parameters) are mentioned in the Tab. 1. Wall is made from plain concrete (bulk weight

γ = 23 kN/m3). Verification analysis of the wall is performed with the help of theory of

limit states.

Figure 1: Wall - geometry

Soil parameters (angle of friction and cohesion) are reduced by coefficients

γ mϕ = 1.1 a γ mc = 1.4. Design values used in calculation are in Tab. 2.

1.2 Verification of the whole wall

Calculation of weight and centroid of the wall. Wall is divided into 5 parts (Fig. 1).

Parts 4 and 5 are under groundwater table, therefore bulk weight of concrete is reduced,

www.finesotware.eu 1 © 1989 - 2009 Fine Ltd.




MS SC

bulk weight γ 18.000 18.500 [kN/m3]bulk weight of satur. soil γ sat 20.000 20.500 [kN/m3]angle of internal friction ϕ 26.500 27.000 [◦]cohesion c 12.000 8.000 [kPa]angle of friction struc-soil δ 15.000 15.000 [◦]Poisson ratio ν 0.350 0.350 -

Table 1: Soil properties - effective values

MS SC

angle of internal friction ϕd 24.091 24.545 [◦]cohesion cd 8.571 5.714 [kPa]angle of friction struc-soil δd 13.636 13.636 [◦]

Table 2: Design parameters of soil

i.e. γ = 23 - 10 = 13 kN/m3. In the Tab. 3 are shown dimensions of the parts of the

wall, their weights and centroids.

height width area b. weight weight force point of actionblock hi[m] wi[m] Ai[m2] γ [kN/m3] W i[kN/m] xi[m] zi[m] W i · xi W i · zi

1 0.350 0.700 2.450 23 56.350 1.950 -2.550 109.883 -143.6932 0.350 0.700 1.225 23 28.175 1.367 -1.967 38.506 -55.4113 0.200 2.300 0.460 23 10.580 1.150 -0.700 12.167 -7.4064 0.600 2.300 1.380 13 17.940 1.150 -0.300 20.631 -5.3825 0.230 2.300 0.265 13 3.439 1.533 0.077 5.272 0.264

TOTAL 116.484 186.459 -211.628

Table 3: Weight of construction and centroids of its parts

Centroid of the construction:

x =

5

i=1W i · xi

W =

186.459

116.484= 1.601 m

z =

5i=1

W i · zi

W =−211.628

116.484= −1.817 m

Calculation of front face resistance. Depth of soil in front of the wall is 0.6 m.

Pressure at rest is considered.

For cohesive soils the Terzaghi formula is used for computing of coefficient of earth

pressure at rest Kr:





K r =ν

1−

ν =

0.35

1−

0.35= 0.538

Hydraulic gradient (hw - water tables difference, dd - seepage path downwards, du

- seepage path upwards):

i =hw

dd + du

=3.7 − 1.5

3.03 + 0.6= 0.606

Unit weight of soil in the area of ascending flow:

γ =

γ sat −

γ w −

i·

γ w = 20

.5 − 10 − 0

.606 · 10 = 4

.439 kN/m

3

Vertical normal effective stress σz in the footing bottom:

σz = γ · h = 4.439 · 0.6 = 2.664 kPa

Pressure at rest σr in the footing bottom:

σr = K r · σz = 0.538 · 2.664 = 1.434 kPa

Resultant force of stress at rest S r:

S r =1

2· σr · h =

1

2· 1.434 · 0.6 = 0.430 kN/m

Resultant force S r is horizontally oriented, therefore:

S rx = S r = 0.430 kN/m S rz = 0 kN/m

Point of action of resultant force S r:

x = 0 m

z = −h

3= −0.6

3= −0.200 m

Calculation of active pressure. Construction is at interface between soils MS and SC

divided into two levels, in each is calculated geostatic pressure σz, active pressure σa

and resultant force S a (Fig. 2).





Figure 2: Geostatic pressure σz and active pressure σa

Coefficients of the active pressure in both levels (back face inclination of the struc-

ture α = 0◦, inclination of terrain β = 0◦):

K a = cos2(ϕd − α)

cos2 α · cos(α + δd) ·

1 +

sin(ϕd+δd)·sin(ϕd−β )cos(α+δd)·cos(α−β )

2

K ac =cos ϕd · cos β · cos(δd − α) · (1 + tan(−α) · tan β )

(1 + sin(ϕd + δd − α − β )) · (cos α + δd)

β 1 = arctan

1

10 = 5.711◦

K a1 =cos2 24.091

cos13.636 ·

1 +

sin(24.091+13.636)·sin(24.091−5.711)cos 13.636·cos(−5.711)

2 = 0.410

K ac1 =cos24.091 · cos5.711 · cos13.636 · (1 + tan 0 · tan5.711)

(1 + sin(24.091 + 13.636

−5.711))

·cos13.636

= 0.594





β 2 = arctanγ 1 · tan β 1

γ 2= arctan

18 · tan5.711

18.5= 5.557◦

K a2 =cos2 24.545

cos13.636 ·

1 +

sin(24.545+13.636)·sin(24.545−5.557)cos 13.636·cos(−5.557)

2 = 0.402

K ac2 =cos24.545 · cos5.557 · cos13.636 · (1 + tan 0 · tan5.557)

(1 + sin(24.545 + 13.636 − 5.557)) · cos13.636= 0.588

Unit weight of soil SC in the area of descending flow:

γ 2 = γ sat − γ w + i · γ w = 20.5 − 10 + 0.606 · 10 = 16.561 kN/m3

Vertical geostatic stress σz in 2 levels:

σz1 = γ 1 · h1 = 18 · 1.5 = 27.000 kPa

σz2 = σz1 + γ 2 · h2 = 27 + 16.561 · 3.03 = 77.179 kPa

Active pressure σa in 2 levels:

σa1 = K a1 · σz1 − 2 · cd1 · K ac1 = 0.410 · 27.000 − 2 · 8.571 · 0.594 = 0.886 kPa

σa2a = K a2 · σz1 − 2 · cd2 · K ac2 = 0.402 · 27.000 − 2 · 5.714 · 0.588 = 4.121 kPa

σa2b = K a2 · σz2 − 2 · cd2 · K ac2 = 0.402 · 77.179 − 2 · 5.714 · 0.588 = 24.274 kPa

Depth h0 in the first layer of soil MS, where is neutral active pressure (σa = 0 kPa):

h0 =2 · cd1 · K ac1

γ 1 · K a1=

2 · 8.571 · 0.594

18 · 0.410= 1.380 m

Resultant forces of the active pressure S a and horizontal, resp. vertical components:





S a1 = 0.5 · σa1 · (h1 − h0) = 0.5 · 0.886 · (1.5 − 1.38) = 0.053 kN/m

S ax1 = S a1 · cos δd1 = 0.053 · cos13.636 = 0.052 kN/m

S az1 = S a1 · sin δd1 = 0.053 · sin13.636 = 0.013 kN/m

S a2 = 0.5 · (σa2b − σa2a) · h2 + σa2a · h2

S a2 = 0.5 · (24.274 − 4.121) · 3.03 + 4.121 · 3.03 = 43.019 kN/m



Points of action of resultant forces S a:

x1 = 2.300 m

z1 = −0.8 − 2 − (1.5 − 1.38)

3= −2.840 m

x2 = 2.300 m

z2 = −1

6· 3.032 · (24.274 − 4.121) +

1

2· 3.032 · 4.121

1

2· 3.03 · (24.274 − 4.121) + 3.03 · 4.121

+ 0.23 = −0.927 m

Total resultant force of the active pressure S a and horizontal, resp. vertical com-

ponent:





S ax =2

i=1

S axi = 0.052 + 41.807 = 41.859 kN/m

S az =2

i=1

S azi = 0.013 + 10.142 = 10.155 kN/m

S a =

S 2ax + S 2az =√

41.8592 + 10.1552 = 43.073 kN/m

Point of action of the resultant force S a:

x =

2i=1

S azi · xi

S az

=0.029 + 23.327

10.155= 2.300 m

z =

2i=1

S axi · zi

S ax

=−0.147 − 38.738

41.859= −0.929 m

Calculation of water pressure. The heel of a structure is sunk into permeable subsoil,which allows free water flow below the structure. Therefore acting of hydrodynamic

pressure is considered and its resultant force is calculated (Fig. 3).

Horizontal water pressure σw at interface of level 1 and 2:

σw1 = γ w · h1 = 10 · 2.2 = 22.000 kPa

Resultant forces of water pressure S w in 2 levels:

S w1 =1

2· σw1 · h1 =

1

2· 22 · 2.2 = 24.200 kN/m

S w2 =1

2· σw1 · h2 =

1

2· 22 · 0.83 = 9.130 kN/m

Points of action of resultant forces:

x1 = 2.300 m z1 =

−0.6

−

2.2

3=

−1.333 m





Figure 3: Hydrodynamic pressure σw

x2 = 2.300 m z2 = −0.6 +0.83

3= −0.323 m

Total resultant force of the water pressure S w:

S w =2

i=1

S wi = 24.2 + 9.130 = 33.330 kN/m

Total point of action of the resultant force S w:

x = 2.300 m

z =

2i=1

S wi · zi

S w=−32.267 − 2.952

33.330= −1.057 m

Checking for overturning stability. The moment rotates around the beginning of

system of coordinates (left bottom corner of the footing).

Resisting moment M r is reduced by coefficient of overall stability of structure

γ s = 0.9:





M r = 0.9 · (116.484 · 1.601 + 10.155 · 2.3) = 188.833 kNm/m

Result from program GEO5 Gravity Wall: M r = 188.83 kNm/m

Driving moment M o:

M o = −0.430 · 0.2 + 41.859 · 0.929 + 33.33 · 1.057 = 74.017 kNm/m

Result from program GEO5 Gravity Wall: M o = 74.02 kNm/m

Usage:

V u =M o

M r· 100 =

74.017188.833

· 100 = 39.2 % O.K.

Result from program GEO5 Gravity Wall: V u = 39.2 % O.K.

Checking for slip. Slip in the inclined footing bottom is checked. (Fig. 4).

Figure 4: Forces acting in the footing bottom - normal and shear soil reaction N and T , totalvertical and horizontal forces acting on the structure F vert and F hor

Total vertical and horizontal forces F vert and F hor:

F vert = 116.484 + 10.155 = 126.638 kN/m





F hor = −0.430 + 41.859 + 33.330 = 74.758 kN/m

Normal force in the footing bottom (inclination of footing bottom αb = 5.711◦, see

Fig. 4):

N = F vert · cos αb + F hor · sin αb

N = 126.638 · cos5.711 + 74.758 · sin5.711 = 133.449 kN/m

Shear reaction in the footing bottom:

T = −F vert · sin αb + F hor · cos αb

T = −126.638 · sin5.711 + 74.758 · cos5.711 = 61.786 kN/m

Eccentricity of the normal force (inclined width of the footing bottom d = 2.311

m):

e =d

2−

M r

0.9− M o

N =

2.311

2−

188.830

0.9− 74.017

133.449= 0.138 m

Maximal allowable eccentricity:

ealw =d

3=

2.311

3= 0.770 m > e = 0.138 m O.K.

Resisting force H r:

H r = γ s · (N · tan ϕd + cd · (d − 2 · e))

H r = 0.9 · (133.449 · tan24.545 + 5.714 · (2.311 − 2 · 0.138)) = 65.316 kN/m

Result from program GEO5 Gravity Wall: H r = 65.38 kN/m

Driving force H d:





H d = T = 61.786 kN/m

Result from program GEO5 Gravity Wall: H d = 61.79 kN/m

Usage:

V u =H d

H r· 100 =

61.786

65.316· 100 = 94.6 % O.K.


1.3 Bearing capacity of the foundation soil

Bearing capacity of the foundation soil is set as Rd = 100 kPa and is compared with

stress in the inclined footing bottom.

Usage - eccentricity:

V u =e

ealw

· 100 =0.138

0.770· 100 = 17.9 % O.K.


Stress in the footing bottom:

σ =N

(d − 2 · e)=

133.449

(2.311 − 2 · 0.138)= 65.570 kPa

Result from program GEO5 Gravity Wall: σ = 65.20 kPa

Usage:

V u =σ

Rd

· 100 =65.570

100· 100 = 65.57 % O.K.


1.4 Dimensioning - analysis of wall section

Cross-section in the level of the x-axis in Fig. 5 is analysed. Rectangular section (width

b = 1 m, height h = 1.4 m) is made from plain concrete C20/25 (characteristic strength

of concrete in tension f ctk = 2200 kPa, characteristic cylindrical strength of concrete in

compression f ck = 20000 kPa). Verification of cross-section made from plain concrete is

realized according to EN 1992 1-1 [1].





Figure 5: Dimensioning of cross-section

Calculation of weight and centroid of the wall. Weight of the construction:

W = 23 · (3.5 · 0.7 + 0.5 · 3.5 · 0.7) = 84.525 m

Centroid of the construction:

x =23 · (3.5 · 0.7 · (0.7 + 0.35) + 0.5 · 3.5 · 0.7 · 2 · 0.7

3)

84.525= 0.856 m

z =23 · (3.5 · 0.7 · −3.5

2+ 0.5 · 3.5 · 0.7 · −3.5

3)

84.525= −1.556 m

Calculation of active pressure. In the first level is same active pressure as in the

verification of the whole wall (part 1.2). Centroids of all forces are recalculated. Vertical geostatic stress σz2 in the second level:

σz2 = 27.0 + 16.561 · 2 = 60.121 kPa

Active pressure σa2b at the end of second level (value at the beginning is same, i.e.

σa2a = 4.121 kPa):

σa2b = 0.402

·60.121

−2

·5.714

·0.588 = 17.424 kPa





Resultant force of the active pressure S a2 and horizontal, resp. vertical component:

S a2 = 0.5

·(17.424

−4.121)

·2 + 4.121

·2 = 21.545 kN/m




x1 = 1.400 m z1 = −2.84 + 0.8 = −2.040 m

x2 = 1.400 m

z2 = −0.5 · 4.121 · 22 +

1

6· (17.424 − 4.121) · 22

4.121 · 2 + 0.5 · (17.424 − 4.121) · 2= −0.794 m

Total resultant force of the active pressure S a and horizontal, resp. vertical com-ponent:

S ax = 20.989 kN/m S az = 5.092 kN/m S a = 21.598 kN/m

Point of action of resultant force S a:

x = 1.400 m z =−2.040 · 0.052 − 0.794 · 20.938

21.598= −0.797 m

Calculation of water pressure. Water pressure grows with height (h = 2 m).

Horizontal water pressure σw in the level of analysed cross-section:

σw = γ w · h = 10 · 2 = 20.000 kPa

Resultant force of the water pressure S w:

S w =1

2 ·σw

·h =

1

2 ·20

·2 = 20.000 kN/m





Point of action of the resultant force:

x = 2.300 m

z = −2

3= −0.667 m

Verification of shear strength. Shear and normal force, bending moment and shear

strength of cross-section are calculated.

Design shear force:

V Ed = 20.989 + 20 = 40.989 kN/m

Result from program GEO5 Gravity Wall: V Ed = 40.94 kN/m

Design normal force:

N Ed = 84.525 + 5.092 = 89.617 kN/m

Result from program GEO5 Gravity Wall: N Ed = 89.57 kN/m

Area of cross-section:

Acc = b · h = 1 · 1.4 = 1.4 m2

Stress in cross-section area:

σcp =N Ed

Acc · 1000=

89.617

1.4 · 1000= 0.064 MPa

Design strength of concrete in compression (coefficients αcc.pl = 0.8, γ c = 1.5):

f cd = αcc.pl · f ck

γ c= 0.8 · 20

1.5= 10.667 MPa

Lower value of characteristic strength of concrete in tension:

f ctk.005 = 0.7 · (0.3 · f 2

3

ck) = 0.7 · (0.3 · 202

3 ) = 1.547 MPa

Design strength of concrete in tension (coefficients αct.pl = 0.8, γ c = 1.5):

f ctd = αct.pl · f ctk.005γ c

= 0.8 · 1.5471.5 = 0.825 MPa





Limit stress:

σc.lim = f cd

−2

· f ctd

·(f cd + f fctd)

σc.lim = 10.667 − 2 ·

0.825 · (10.667 + 0.825) = 4.508 MPa

Shear strength:

f cvd =

f 2ctd + σcp · f ctd −

max(0, σcp − σc.lim)

2

2

f cvd =

0.8252 + 0.064 · 0.825 −

0

2

2

= 0.857 MPa

Design shear strength (k = 1.5):

V Rd =f cvd · Acc

k=

0.857 · 1.4

1.5= 799.523 kN/m

Result from program GEO5 Gravity Wall: V Rd

= 799.37 kN/m

Usage:

V u =V Ed

V Rd

· 100 =40.989

799.523· 100 = 5.1 % O.K.


Verification of cross-section loaded by bending moment and normal force. Mo-

ment turns around the cross-section centroid.

Design bending moment:

M Ed = −84.525 · (0.856− 0.7)− 5.092 · 0.7 + 2 0.989 · 0.797+20 · 0.667 = 13.355 kN/m

Result from program GEO5 Gravity Wall: M Ed = 13.32 kNm/m

Normal force eccentricity:

e =M Ed

N Ed

=13.355

89.617= 0.149 m





0.9 · h

2= 0.9 · 1.4

2= 0.63 m > e = 0.149 m

Effective height:

χ = h − 2 · e = 1.4 − 2 · 0.149 = 1.102 m

Design normal strength (η = 1.0):

N Rd = χ · η · f cd = 1.102 · 1 · 10.667 = 11754.239 kN/m

Result from program GEO5 Gravity Wall: N Rd = 11758.56 kN/m

Usage:

V u =N Ed

N Rd

· 100 =89.617

11754.239· 100 = 0.8 % O.K.


1.5 Verification of the wall - influence of earthquake

Second stage of calculation shows the same wall with the influence of earthquake. Calcu-

lation of earthquake effects is made according to Mononobe-Okabe theory [2], [3]. Factor

of horizontal acceleration is kh = 0.05 (inertia force acts horizontally in unfavorable di-

rection) and factor of vertical acceleration kv = −0.04 (inertia force acts downwards).

Coefficients of reduction of soil parameters and coefficient of overall stability of construc-

tion are equal to one (γ mϕ = γ mc = γ s = 1.0). Therefore design soil properties are same

as characteristic values in Tab. 1.

Calculation of weight of the wall. Weight of the wall W = 116.484 kN/m (see part1.2) is increased by weight force from earthquake.

Horizontal and vertical component of weight force from earthquake:

W eq.x = kh · W = 0.05 · 116.484 = 5.824 kN/m

W eq.z = −kv · W = 0.04 · 116.484 = 4.659 kN/m





Calculation of front face resistance. Pressure at rest on front face of the wall is

considered same as in part 1.2, i.e. resultant force is S r = 0.430 kN/m.

Calculation of active pressure. Active pressure σa and resultant force S a are calcu-

lated similarly as in the calculation without earthquake (part 1.2), geostatic stress σz is

same, i.e. σz1 = 27.000 kPa and σz2 = 77.179 kPa.

Coefficients of the active pressure in both levels:

K a1 =cos2 26.50

cos15 ·

1 +

sin(26.5+15)·sin(26.5−5.711)cos15·cos(−5.711)

2 = 0.371

K ac1 =cos26.5 · cos5.711 · cos15 · (1 + tan 0 · tan5.711)

(1 + sin(26.5 + 15 − 5.711)) · cos15= 0.562

K a2 =cos2 27

cos15 ·

1 +

sin(27+15)·sin(27−5.557)cos15·cos(−5.557)

2 = 0.363

K ac2 =cos27 · cos5.557 · cos15 · (1 + tan 0 · tan5.557)

(1 + sin(27 + 15 − 5.557)) · cos15= 0.556

Depth h0 in the first layer of soil MS, where is neutral active pressure (σa = 0 kPa):

h0 =2 · c1 · K ac1

γ 1 · K a1=

2 · 12 · 0.562

18 · 0.371= 2.019 m > h1 = 1.5 m

Active pressure is in the whole first level equal to zero.

Active pressure σa in the second layer of soil SC:

σa2a = K a2 · σz1 − 2 · c2 · K ac2 = 0.363 · 27 − 2 · 8 · 0.556 = 0.903 kPa

σa2b = K a2 · σz2 − 2 · c2 · K ac2 = 0.363 · 77.179 − 2 · 8 · 0.556 = 19.126 kPa

Total resultant force of the active pressure S a and horizontal, resp. vertical com-

ponent:





S a = 0.5 · (σa2b − σa2a) · h2 + σa2a · h2

S a = 0.5 · (19.126 − 0.903) · 3.03 + 0.903 · 3.03 = 30.344 kN/m

S ax = S a · cos δ2 = 30.344 · cos 15 = 29.310 kN/m

S az = S a · sin δ2 = 30.344 · sin 15 = 7.854 kN/m

Point of action of the resultant force:

x = 2.300 m

z = −1

6· 3.032 · (19.126 − 0.903) +

1

2· 3.032 · 0.903

1

2· 3.03 · (19.126 − 0.903) + 3.03 · 0.903

+ 0.23 = −0.826 m

Increase of active pressure caused by earthquake. Earthquake increases the effect

of active pressure.

Seismic inertia angle in the first layer of soil MS:

ψ1 = arctankh

1 − kv

= arctan0.05

1 + 0.04= 2.752◦

Seismic inertia angle in the second layer of soil SC, where is restricted water:

ψ2 = arctan γ sat · kh

(γ sat − γ w) · (1 − kv)= arctan 20.5 · 0.05

(20.5 − 10) · (1 + 0.04)= 5.362◦

Coefficients K ae for the active pressure in both levels:

K ae =cos2(ϕ − ψ − α)

cos ψ · cos2 α · cos(ψ + α + δ) ·

1 +

sin(ϕ+δ)·sin(ϕ−ψ−β )cos(α+ψ+δ)·cos(β −α)

2





K ae1 =cos2(26.5 − 2.752)

cos2.752 · cos(2.752 + 15) · 1 +

sin(26.5+15)·sin(26.5−2.752−5.711)cos(2.752+15)·cos(5.711)

2 = 0.410

K ae2 =cos2(27 − 5.362)

cos5.362 · cos(5.362 + 15) ·

1 +

sin(27+15)·sin(27−5.362−5.557)cos(5.362+15)·cos(5.557)

2 = 0.443

Normal stress σd for calculation of earthquake effects (stress grows from the bottom

of the wall - σd2 is in the footing bottom level and σd0 is in the terrain level):

σd2 = 0 kPa

σd1 = γ 2 · h2 · (1 − kv) = 16.561 · 3.03 · (1 + 0.04) = 52.186 kPa

σd0 = σd1 + γ 1 · h1 · (1 − kv) = 52.186 + 18 · 1.5 · (1 + 0.04) = 80.266 kPa

Increase of the active pressure from earthquake effects in the first layer of soil MS:

σae0 = σd0 · (K ae1 − K a1)

1 − kv

= 80.266 · (0.410 − 0.371)

1 + 0.04= 3.014 kPa

σae1a = σd1 · (K ae1 − K a1)

1 − kv

= 52.186 · (0.410 − 0.371)

1 + 0.04= 1.959 kPa

Increase of the active pressure from earthquake effects in the second layer of soil

SC:

σae1b = σd1 · (K ae2 − K a2)

1 − kv

= 52.186 · (0.443 − 0.363)

1 + 0.04= 4.003 kPa

σae2 = 0 kPa

Resultant forces of increase of the active pressure S ae and horizontal, resp. vertical

components:





S ae1 = 0.5 · (σae0 − σae1a) · h1 + σae1a · h1

S ae1 = 0.5 · (3.014 − 1.959) · 1.5 + 1.959 · 1.5 = 3.730 kN/m

S aex1 = S ae1 · cos δ1 = 3.730 · cos 15 = 3.603 kN/m

S aez1 = S ae1 · sin δ1 = 3.730 · sin 15 = 0.965 kN/m

S ae2 = 0.5 · σae1b · h2 = 0.5 · 4.003 · 3.03 = 6.064 kN/m

S aex2 = S ae2 · cos δ2 = 6.064 · cos 15 = 5.857 kN/m

S aez2 = S ae2 · sin δ2 = 6.064 · sin 15 = 1.569 kN/m


x1 = 2.300 m

z1 = −2.8 −1

2· 1.52 · 1.959 +

1

3· 1.52 · (3.014 − 1.959)

1.5 · 1.959 + 0.5 · 1.5 · (3.014 − 1.959)= −3.603 m

x2 = 2.300 m

z2 = 0.23 − 2

3· 3.03 = −1.790 m

Total resultant force of increase of the active pressure S ae and horizontal, resp.

vertical component:





S aex =2

i=1

S aexi = 3.603 + 5.857 = 9.460 kN/m

S aez =2

i=1

S aezi = 0.965 + 1.569 = 2.535 kN/m

S ae =

S 2aex + S 2aez =√

9.4602 + 2.5352 = 9.794 kN/m

Point of action of the resultant force S ae:

x = 2.300 m

z =

2i=1

S aexi · zi

S aex

=−12.981 − 10.485

9.460= −2.480 m

Calculation of water pressure. Water pressure is same as in the verification of the

whole wall (part 1.2), i.e. resultant force of the water pressure is S w = 33.330 kN/m.

Checking for overturning stability. The moment rotates around the beginning of

system of coordinates (left bottom corner of the footing).

Resisting moment M r:

M r = (116.484 + 4.659) · 1.601 + (7.854 + 2.535) · 2.3 = 217.810 kNm/m

Result from program GEO5 Gravity Wall: M r = 217.83 kNm/m

Driving moment M o:

M o = −0.43·0.2+29.31·0.826+9.46·2.48+33.33·1.057+5.824·1.817 = 93.377 kNm/m

Result from program GEO5 Gravity Wall: M o = 93.43 kNm/m

Usage:

V u =M o

M r

· 100 =93.377

217.810· 100 = 42.9 % O.K.






Checking for slip. Slip in the inclined footing bottom is checked. (Fig. 4).

Total vertical and horizontal forces F vert and F hor:

F vert = 116.484 + 4.659 + 7.854 + 2.535 = 131.531 kN/m

F hor = −0.430 + 29.310 + 9.460 + 33.330 + 5.824 = 77.494 kN/m

Normal force in the footing bottom (inclination of footing bottom αb = 5.711◦, see

Fig. 4):

N = 131.531 · cos5.711 + 77.494 · sin5.711 = 138.589 kN/m

Shear reaction in the footing bottom:

T = −131.531 · sin5.711 + 77.494 · cos5.711 = 64.022 kN/m

Eccentricity of the normal force (inclined width of the footing bottom d = 2.311

m):

e =d

2− M r − M o

N =

2.311

2− 217.810 − 93.43

138.589= 0.258 m

Maximal allowable eccentricity:

ealw =d

3=

2.311

3= 0.770 m > e = 0.258 m O.K.

Resisting force H r:

H r = 138.589 · tan 27 + 8 · (2.311 − 2 · 0.258) = 84.981 kN/m

Result from program GEO5 Gravity Wall: H r = 85.07 kN/m

Driving force H d:

H d = T = 64.022 kN/m

Result from program GEO5 Gravity Wall: H d = 64.05 kN/m





Usage:

V u =H d

H r ·100 =

64.022

84.981 ·100 = 75.3 % O.K.


1.6 Bearing capacity of the foundation soil

Bearing capacity of the foundation soil is set as Rd = 100 kPa and is compared with

stress in the inclined footing bottom.

Usage - eccentricity:

V u =e

ealw

· 100 =0.258

0.770· 100 = 33.5 % O.K.


Stress in the footing bottom:

σ =N

(d − 2 · e)=

138.589

(2.311 − 2 · 0.258)= 77.178 kPa

Result from program GEO5 Gravity Wall: σ = 76.72 kPa

Usage:

V u =σ

Rd

· 100 =77.178

100· 100 = 77.2 % O.K.


References

[1] Czech normalization institute: EN 1992-1-1, Eurocode 2: Design of concrete

structures - Part 1-1: General rules and rules for buildings. CNI, Prague, 2006

[2] Mononobe N., Matsuo, H.: On the determination of earth pressure during

earthquakes. In Proc. Of the World Engineering Conf., Vol. 9, pg. 176, 1929

[3] Okabe S.: General theory of earth pressure. Journal of the Japanese Society of

Civil Engineers, Tokyo, Japan, 1926

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