Baøi giaûng A. Prof. Dr Chaâu Ngoïc AÅn

Baøi giaûng A. Prof. Dr Chaâu Ngoïc AÅn

ÖÙNG SUAÁT TRONG NEÀN ÑAÁT

ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG

Ñaát baõo hoøa nöôùc

Haït theùp

nöôùc

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn

ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG

Ñaát baõo hoøa nöôùc

Haït theùp

nöôùc

* bình beân traùi theâm vaøo treân maët lôùp ñaát caùc haït theùp taïo moät aùp löïc p, maãu ñaát bò luùn xuoáng. AÙp löïc p coù aûnh höôûng leân öùng suaát khung neân laø öùng suaát höõu hieäu, kyù hieäu laø ’* bình beân phaûi theâm nöôùc treân maët ñeå taïo aùp löïc p, maãu ñaát khoâng luùn xuoáng vì nöôùc theâm vaøo thoâng vôùi nöôùc trong loã roãng taùc ñoäng leân ñaùy bình chöùa, p do coät nöôùc khoâng aûnh höôûng leân khung haït (trung hoøa).


ÖÙng suaát taïi moät ñieåm trong neàn ñaát goàm “öùng suaát giöõa caùc haït” hay öùng suaát höõu hieäu ’ vaø aùp löïc nöôùc trong loã roãng u, theo ñònh ñeà Terzaghi

= ’ + u



TÍNH ÖÙNG SUAÁT TRONG ÑAÁT NEÀN DO TROÏNG LÖÔÏNG BAÛN THAÂN

ÖÙng suaát toång do troïng löôïng baûn thaân ñaát theo phöông thaúng ñöùng kyù hieäu laø bt hay v taïi moät ñieåm baát kyø trong ñaát caùch maët ñaát moät chieàu saâu baèng H, coù theå tính nhö laø troïng löôïng khoái ñaát beân treân truyeàn xuoáng.

H

zbtv dzz0

, )(

n

iizbt h1

,


vôùi K0 laø heä soá aùp löïc ngang ôû traïng thaùi tónh cuûa ñaát coá keát thöôøng (ñaëc ñieåm coá keát thöôøng vaø coá keát tröôùc seõ ñöôïc phaân tích trong caùc chöông sau). Khoaûng nöûa theá kyû tröôùc, heä soá aùp löïc ngang ñöôïc vay möôïn töø lyù thuyeát ñaøn hoài vôùi kyù hieäu laø vaø coù daïng : trong ñoù laø heä soá Poisson.Vôùi toång keát töø raát nhieàu keát quaû thí nghieäm vaø ño ñaïc giaùn tieáp, Jaky ñaõ ñöa ra moät coâng thöùc ñeå tính heä soá aùp löïc ngang ôû traïng thaùi tónh (cuûa ñaát coá keát thöôøng) nhö sau :K0 = 1 - sin’Vôùi ’ laø goùc ma saùt trong ñieàu kieän caét thoaùt nöôùc (seõ phaân tích roõ trong chöông choáng caét). Coâng thöùc cuûa Jaky phuø hôïp cho ñaát rôøi hoaëc ñaát loaïi caùt.Neáu goùc ma saùt ’= 350 thì K0 = 1 – sin350 = 0,426Ñoái vôùi ñaát dính hoaëc ñaát loaïi seùt coá keát thöôøng, Alpan ñeà nghò moät coâng thöùc thöïc nghieäm.

K0 = 0,19 + 0,233logIP Neáu moät maãu seùt coù chæ soá deûo IP = 20, thì heä soá aùp löïc ngang ôû traïng thaùi tónh K0 theo coâng thöùc Alpan laø :

K0 = 0,19 + 0,233log20 = 0,493

10K

'0

'vh K


Seùt, c2

Seùt, c1

Caùùt, s2

Caùùt, s1

V

hWhS1

hC1

hS2

hC2 ’Vu


Seùt, C1

Seùt, C2

Caùùt, S2

Caùùt, S1

hW1

hW2

hS1

hC1

hS2

hC2

BS

BC

z

’u


Nguyeân lyù ño aùp löïc nöôùc loã roãngtrong ñaát

W

W

Soil loaded by an applied weight W

Soil loaded by water weighing W

W

W

Soil loaded by an applied weight W

Soil loaded by water weighing W

Compression No deformation

v Vertical StressVertical Force

Cross Sectional Area (1)

Definition of Total and Effective Stress


Cross Sectional Area

v v wu'

(1)

(2)


Effective vertical stress



v v wu'

(1)

(2)



vuwv́

Case (a) W

A0 W

A



v v wu'

(1)

(2)



vuwv́

Case (a) W

A0 W

A

Case (b) W

A

W

A0

Fig 2 Two Pieces of Rock in Contact

N

T

Water pressure uw

Effective Stress

N

T

Water pressure uw

Effective Force

N N U (3a)

Effective Stress

U = uw ( A - Ac )

N

T

Water pressure uw

Effective Force

N N U (3a)

Effective Stress

U = uw ( A - Ac )

Frictional Failure

T N

N

T

Water pressure uw

Effective Force

N N U (3a)

Effective Stress

U = uw ( A - Ac )

Frictional Failure

T N

Failure in terms of stress

v (3b)

T

A

N

A

U

A

Layer 1

Layer 2

Layer 3

d1

d2

d3

bulk 1

bulk 2

bulk 3

Fig 3 Soil Profile

Surcharge q

v

z

Calculation of Effective Stress

d1

d2

q

v

z

A

Plan

Elevation

Calculation of Total Vertical Stress

z

Force on base = Force on top + Weight of soil

d1

d2

q

v

z

A

Plan

Elevation


z


A v = A q + A 1 d1 + A 2 d2 +

A 3 ( z - d1 - d2 )

d1

d2

q

v

z

A

Plan

Elevation

(4)


z


A v = A q + A 1 d1 + A 2 d2 +

A 3 ( z - d1 - d2 )

v = q + 1 d1 + 2 d2 +3 ( z - d1 - d2 )

Fig 4 Soil with a static water table

Water table

H

P u P Hw w( ) (5)

Calculation of pore water pressure


Water table

H

P u P Hw w( ) (5)

• The water table is the level of the water surface in a borehole.



Water table

H

P u P Hw w( ) (5)

• The water table is the level of the water surface in a borehole.

• It is the level at which the pore water pressure uw = 0


Dry

Saturated

2 m

3m

Fig 5 Soil Stratigraphy

b u l k d r y

b u lk s a t

Step 1: Draw ground profile showing soil stratigraphy and water table

Example: determining the effective stress

Distribution by Volume

Solid

VoidsVv=e Vs

= 0.7m3

Vs= 1m3

Step 2: Calculation of relevant bulk unit weights

Example


Solid

VoidsVv=e Vs

= 0.7m3

Distribution by weight for the dry soil

Vs= 1m3

W V G

k N

k N

s s s w

1 2 7 9 8

2 6 4 6

. .

.

W = 0


Example


Solid

VoidsVv=e Vs

= 0.7m3


Distribution by weightfor the saturated soil

Vs= 1m3

W V kN

kN

kN

w v w

0 7 9 8

6 86

. .

.

W V G

k N

k N

s s s w

1 2 7 9 8

2 6 4 6

. .

.

W V G

kN

kN

s s s w

1 2 7 9 8

26 46

. .

.

W = 0


Example


Solid

VoidsVv=e Vs

= 0.7m3



Vs= 1m3

W V kN

kN

kN

w v w

0 7 9 8

6 86

. .

.

W V G

k N

k N

s s s w

1 2 7 9 8

2 6 4 6

. .

.

W V G

kN

kN

s s s w

1 2 7 9 8

26 46

. .

.

Ww=0


Example

drys wkN

mkN m

G

e

26 46

1701556

133.

.. /


Solid

VoidsVv=e Vs

= 0.7m3



Vs= 1m3

W V kN

kN

kN

w v w

0 7 9 8

6 86

. .

.

W V G

k N

k N

s s s w

1 2 7 9 8

2 6 4 6

. .

.

W V G

kN

kN

s s s w

1 2 7 9 8

26 46

. .

.

Ww=0

drys w

satw s

kN

mkN m

G

e

kN

mkN m

G e

e

26 46

1701556

1

26 46 686

17019 60

1

33

33

.

.. /

( . . )

.. /

( )


Example

2 m

3m

v kPa kN m 15 56 2 19 60 3 89 92 2. . . ( / )

Step 3 Calculate total stress

Example

2 m

3m

v kPa kN m 15 56 2 19 60 3 89 92 2. . . ( / )


Example

u kPaw 3 9 8 29 40. .

Step 4 Calculate pore water pressure

2 m

3m

v kPa kN m 15 56 2 19 60 3 89 92 2. . . ( / )


Example

u kPaw 3 9 8 29 40. .

Step 4 Calculate pore water pressure

Step 5 Calculate effective stress

v v wu kPa89 92 29 40 60 52. . .

0 50 100 150

0m

2m

4m

6m

8m

kPa

pore waterpressure Effective

stress

TotalStress (5m)

Depth

Vertical stress and pore pressure variation

Fig 7 Definition of Stress Components

x

y

z

xz

yz

zz

yy

xy

zy

xx

yx

zx

z

x

Stresses acting on a soil element

Effective stress relations for general stress states

xx xx w yz yz

yy yy w zx zx

zz zz w xy xy

u

u

u

;

;

;(10)

Principle of Effective Stress

Example

Clay

Rock aquifer

Initial GWL

z

Lowered GWL3 m1 m

Example

Clay

Rock aquifer

Initial GWL Lowered GWL

v bulk z bulk z

u w (z - 1) w (z - 3)

v´ bulk - w )z + w bulk - w )z + 3 w

Initial GWL

z

Lowered GWL3 m1 m

• Effective stress increases - soil compresses - ground surface settles

• Effective stress decreases- soil swells - ground surface heaves. The following problems may then occur

• surface flooding

• flooding of basements built when GWL lowered

• uplift of buildings

• failure of retaining structures

• failures due to reductions in bearing capacity

Example


ÖÙNG SUAÁT TRONG NEÀN ÑAÁT DO TAÛI NGOAØI

of 36

http://www.eng.fsu.edu/~tawfiq/soilmech/sld001.html




http://www.eng.fsu.edu/~tawfiq/soilmech/index.html


c1

P

a1 a

c

R

d R

Döôùi taùc duïng cuûa P ñieåm M chuyeån vò moät ñoaïn S theo phöông baùn kính R. M caøng xa O thì S caøng nhoû. Maët khaùc, vôùi R = const, goùc caøng lôùn thì S cuõng caøng nhoû. Xuaát phaùt töø nhaän xeùt ñoù, ta coù theå vieát bieåu thöùc S coù daïng : R

AScos

Tröông töï, taïi M1 caùch M moät ñoaïn dR, coù chuyeån vò S1

dRRAS

cos1


Bieán daïng töông ñoái R cuûa ñoaïn dR laø:

cos.

cos

dR

S - S2

1

dRRR

A

dRR

A

R

A

dRR

Boû qua R.dR vì raát nhoû so vôùi R2

cos2R

AR

Theo giaû thuyeát quan heä giöõa öùng suaát vaø bieán daïng laø tuyeán tính do ñoù öùng suaát xuyeân taâm R gaây neân bieán daïng R ñöôïc xaùc ñònh nhö sau

Trò soá A.B coù theå xaùc ñònh döïa theo ñieàu caân baèng tónh hoïc. Xeùt ñieàu kieän caân baèng tónh hoïc cuûa baùn caàu (O; R)

cos2R

ABR


2

0

.cos

dFP R

Trong ñoù: dF – dieän tích maët ñai troøn caa1c1

dF = 2(Rsin)(Rd)

2

0

22

2

0

sin2coscos.cos

dRR

ABdFP R

dBAP sincos2..2

0

2 BAP ...3

2 P

AB .2

3

cos.2

32R

PR


5

3

2

3

R

zPz

23

2

3

22

5

2

)(

)2(

)(

.

3

21

2

3

zRR

zRx

zRR

zzRR

R

zxPx

23

2

3

22

5

2

)(

)2(

)(

.

3

21

2

3

zRR

zRy

zRR

zzRR

R

zyPy

5

2.

.2

.3

R

zyPzy

5

2.

.2

.3

R

zxPzx

235 )(

).2(..

3

.21..

.2

.3

zRR

zRyx

R

zyxPxy


RR

z

E

Pw

1)1(2

..2

)1(3

2

)(

).21(.

..2

)1(3 zRR

x

R

zx

E

Pu

)(

).21(.

..2

)1(3 zRR

y

R

zy

E

Pv

Chuyeån vò theo chieàu caùc truïc :


yA

x

R1

O

z

M(x,y,z)r

h

h

z

R2

P

72

3**

52

**2**

51

3*

32

*

31

*

30)5(3)43(3

3)21()21(

)1(8

R

Zhz

R

hzZhZz

R

Z

R

Z

R

ZPz

52

2**

32

2**

31

2*

2

2

1

)(62))(43(

)43()1(843

)1(16

R

Zhz

R

zhZ

R

Z

RRG

Pw

Baøi toaùn Mindlin

of 36







ÑÖÔØNG LÖÏC HAY LÖÏC ÑÖÔØNG THAÚNGÖÙng suaát do aûnh höôûng cuûa löïc daïng ñöôøng thaúng phaân boá ñeàu (kM/m) nhö: ñöôøng raây; töôøng chòu löïc trong neàn ñaát, … ñöôïc Flamant phaùt trieån töø baøi toaùn Boussinesq (1892) baèng caùch chia ñöôøng löïc thaønh voâ soá löïc taùc ñoäng pdy leân moät ñoaïn thaät ngaén dy, aùp duïng coâng thöùc Boussinesq cho löïc nhoû naøy roài tích phaân leân caû chieàu daøi taùc ñoäng löïc ñeå coù ñöôïc caùc coâng thöùc sau:

x

y

O

z y

z

p

dy

pdy

M

R

dy

zyx

zpz

25

222

3

.2

.3

222

3.2

zx

zpz

222

2 ..2

zx

zxpx

222

2..2

zx

zxpzx

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Taûi phaân boá ñeàu treân dieän tích baêng (Flamant)

Taûi phaân boá treân dieän tích baêng laø daïng raát thöôøng gaëp trong neàn moùng coâng trình nhö: moùng baêng, ñöôøng, ñeâ,… Khaûo saùt moät ñoaïn dx trong phaïm vi töø -b/2 ñeán +b/2, giaù trò taûi töông öùng laø pdx töông töï moät ñöôøng löïc, tính öùng suaát dz do ñöôøng löïc pdx gaây ra vaø ñoåi bieán soá sang goùc nhìn töø M veà ñaùy moùng.

o

M

1

2

b

d

z

r

x

dx

P

z

x

p

M

3

222

3

cos..

..2..2

r

dxp

zx

zdxpd z

cos

zr

tgzx .

2cos

.dzdx

dp

d z .cos..2 2


2211z 2sin2

1)(2sin

2

1P

1

2

1

2

.cos.21.cos.2 2

dP

dP

z

2211 2sin

2

1)(2sin

2

1

Px

12xzzx 2cos2cos.2

P

Trò soá 2 laáy vôùi daáu döông khi ñieåm M naèm ngoaøi phaïm vi hai ñöôøng thaúng ñi qua hai meùp cuûa taûi troïng


Tröôøng hôïp ñôn giaûn nhaát laø ñoái vôùi caùc ñieåm naèm treân maët chöùa Oz (ñi qua truïc taâm taûi troïng). Vì tính ñoái xöùng cho neân :

1 = 2 =

2sin21 p

z

2sin2P

3x

02cos2cos.2

P12xzzx

2

of 36






Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Taûi phaân boá ñeàu treân dieän tích chöõ nhaät

b

l x

y d

d

p

O

ÖÙng suaát thaúng ñöùng Z cuûa ñieåm naèm treân truïc thaúng ñöùng ñi qua taâm dieän chòu taûi, ôû ñoä saâu z:

pk

zlbzlzb

zlbzlb

zlbz

lbarctg

pz 022

12

122

122

1

221

2111

221

21

11 .22

1

1

1

12

5222

3

)(.2

...3b

b

l

l

z

zyx

zddp

2

ll1

21

bb


ÖÙng suaát thaúng ñöùng z do taûi phaân boá ñeàu treân dieän chòu taûi chöõ nhaät, doïc truïc thaúng ñöùng beân döôùi ñieåm goùc dieän chòu taûi.

pklbzlbz

zlbblz

zlb

zlb

lbzlbz

zlbblzpgz

222222

2/12221

222

222

222222

2/3222

)(

)(2tan

2.

)(

)(2

4

pInmnm

nmmn

nm

nm

nmnm

nmmnpz

1

12tan

1

2

1

12

4 2222

221

22

22

2222

22

z

bm

Hoaëc

z

ln

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn PHÖÔNG PHAÙP ÑIEÅM GOÙC.

A B

CD

M

A B

CD

M

A1

B1

C1

D1

öùng suaát (z) cuûa dieän chöùa taûi ABCD döôùi ñieåm M laø toång caùc öùng suaát goùc M cuûa caùc dieän MA1AD1; MB1BA1; MC1CB1; MC1DD1.Z,M = p[kg(MA1AD1) + kg(MB1BA1) + kg(MC1CB1) + kg(MC1DD1)]p laø aùp löïc phaân boá ñeàu treân dieän chòu taûi ABCD.


A B

CD

M

A B

CD

M

B2

C2

D2 A2

öùng suaát (z) cuûa taïi M do dieän chöùa taûi ABCD taùc ñoäng laø toång ñaïi soá caùc öùng suaát goùc M cuûa caùc dieän MA2CC2; MD2DC2; MD2AB2; MA2BB2. Trong ñoù chæ coù dieän tích ABCD laø chöùa taûi phaân boá ñeàu p, phaàn dieän tích coøn laïi khoâng coù taûi. Z,M = p[kg(MD2AB2) - kg(MD2DC2) - kg(MA2BB2) + kg(MA2CC2)]p laø aùp löïc phaân boá ñeàu treân dieän chòu taûi ABCD.


TÍNH ÖÙNG SUAÁT THEO PHÖÔNG PHAÙP THAÙP LAN TOÛA

)2)(2(

ztglztgb

Qz

))(( zlzb

Qz

Ñôn giaûn hôn thaùp lan toûa coù ñoä doác 2:1 öùng suaát z ôû ñaùy thaùp lan toûa coù daïng:


2b 2b

p

z

M

2 1

A B C x

2b 2b

p

z

M

A B C x

212122

xbb

pz

p

z

2

Ñoái vôùi ñaát caùt thöôøng coù caáu truùc haït vaø ôû ba traïng thaùi rôøi, chaët trung bình vaø chaët. Duø ôû traïng thaùi naøo, dieän tích tieáp xuùc giöõa caùc haït heát söùc nhoû so vôùi dieän tích xung quanh cuûa caùc haït.

Caùc dieåm tieáp xuùc haït trong caáu truùc haït (ñaát caùt)Nhö tyû leä dieän tích tieáp xuùc thöôøng gaëp cuûa caùt laø 0,03%, khi taùc duïng moät öùng suaát phaùp 100kPa thöøng öùng suaát thöïc taùc ñoäng leân dieän tích tieáp xuùc laø 330 Mpa. Vaø vôùi aùp löïc naøy thì nöôùc lieân keát taïi caùc ñieåm tieáp xuùc seõ bò ñaåy khoûi voû nöôùc, do vaäy coù ñieåm tieáp xuùc raát toát giöõa caùc haït.

Stresses in a Soil Mass

How are stresses produced?

• Geostatic stresses– total stress– effective stress– pore water pressure

• Additional stresses– surface loads

• foundation

• embankment

• vehicle

Effects of stresses

• Geostatic stresses– soil compresses or consolidates based on

geostatic stress levels

• Additional stresses– produces additional strain in soil which causes

settlement under point of load.

Normal and Shear Stresseson a Plane

• Use methods learned in Mechanics of Materials

• The normal stress on any plane is

• The shear stress on any plane is

n y x

2 y x

2cos2 xy sin2

n y x

2sin 2 xy cos 2


• Use Mohr’s circle method to graphically depict stresses: orientation, and magnitude.


• The major principal stress is

• The minor principal stress is

n 1 y x

2

y x

2

2

xy2

n 3 y x

2

y x

2

2

xy2

Pole Method of Finding Stresses on a Plane

• Also called “Origin of Planes”

• Draw a line from a known point on the Mohr’s circle parallel to the plane on which the state of stress acts.

• The point of intersection of this line with the Mohr’s circle is called the pole.

• To find the state of stress on any other plane, draw a line parallel to the plane of interest through the pole. State of stress is the intersection of this line with Mohr’s circle.

Basics of Surface Loads

• Categorize into groups based on areal extent– infinite extent (fills, surface surcharge)– finite extent

• point load• line load• strip load• linearly increasing load• uniformly loaded circular area• rectangularly loaded area

Basics of Surface Loads

• Load produces stress and strain– Stress and strain occur in all directions– Commonly focus only on vertical stress

increase

• Analysis based on elastic theory– isotropic, homogeneous material– linear elastic behavior (spring-like)– material comprises a half-space

Stress Caused by a Point Load

• First solved by Boussinesq (1883)

• Stress is maximum nearest applied load and diminishes at distances away.

2/522

1/

1

2

3

ff

zzrz

P

Stress Caused by Line Load

• Line load of magnitude q/unit length acts on half-space surface

• Vertical stress increase is:

222

32

ff

f

zzx

zbP

Superposition

• This principle works because this system is linear elastic, isotropic, and homogeneous.

• This procedure greatly simplifies subsurface stress analysis.

• Merely add up each separate component.

• Tham khảo thêm trong sách Cơ học đất

Baøi giaûng A. Prof. Dr Chaâu Ngoïc AÅn

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