4. Lateral Earth Pressure and Retaining Walls

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Geotechnical Engineering SNU Geotechnical and Geoenvironmental Engineering Lab. 160 4. Lateral Earth Pressure and Retaining Walls 1) General Influence factors (a) (b) (c) (d) Lateral Earth Pressures can be divided into 1) 2) 3) depending on wall movements (or deformation modes of soil elements). 2) Lateral Earth Pressure i) Lateral earth pressure at rest ' ' V O h K σ σ = Jaky, ' sin 1 φ = O K for loose sand or NC clays. Brooker and Ireland, 0.40 0.007( ) (0 40) 0.64 0.001( ) (40 80) O O K PI PI K PI PI = + = +

Transcript of 4. Lateral Earth Pressure and Retaining Walls

Page 1: 4. Lateral Earth Pressure and Retaining Walls

Geotechnical Engineering

SNU Geotechnical and Geoenvironmental Engineering Lab.

160

4. Lateral Earth Pressure and Retaining Walls

1) General

� Influence factors

(a)

(b)

(c)

(d)

� Lateral Earth Pressures can be divided into

1)

2)

3)

depending on wall movements (or deformation modes of soil elements).

2) Lateral Earth Pressure

i) Lateral earth pressure at rest

''

VOh K σσ =

� Jaky,

'sin1 φ−=OK for loose sand or NC clays.

� Brooker and Ireland,

0.40 0.007( ) (0 40)

0.64 0.001( ) (40 80)

O

O

K PI PI

K PI PI

= + ≤ ≤

= + ≤ ≤

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� For OC soils,

Fig. Stress history for soil under 0K condition

h'

h

0

OCR))(sin-(1

OCR)(

φ=

= −ncOKK

h = 0.4 - 0.5 (Alpan, 1967; Schmertmann, 1975)

or

'sin'

O OCR))(sin1(K φφ−= by Mayne & Kulhawy (1982)

� For dense sands,

5.5)1/()'sin1((min)

×−+−=ddO

K γγφ

by Sherif and et al (1984)

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ii) Active and passive pressures

* Rankine Approach

* Coulomb Approach

a) Rankine

� Assumptions

(i)

(ii)

(iii)

(iv)

(v)

Wall movement for active state Failure plane for active state

Wall movement for passive state

Failure plane for passive state

γ, Ko (<1)

Z

45° - φ/2

45° + φ/2

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Initial condition Final condition Final condition

(Active state) (Passive state)

� Rankine’s Active Earth Pressure

PA = γ'ztan2(45-φ’/2) - 2c’tan(45-φ’/2)

= γ'zKa - 2c aK

� Rankine’s Passive Earth Pressure

PP = γ'ztan2(45+φ’/2) + 2c’tan(45+φ’/2)

= γ'zKp + 2c pK

C’

Initial Condition

γ’z

K0γ’z σ1f’(=Pp)

(Rankine Passive

Earth Pressure)

σ3f’(=PA)

(Rankine Active

Earth Pressure)

Φ ’

Φ ’

C’

σv’ = γ’z

σh’=Koγ’z

σv’ = σ1f’= γ’z

σh’=σ3f’=PA

σv’ = σ3f’=γ’z

σh’=σ1f’=Pp

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� Some Theoretical Earth Pressure Distributions

(1) φ’, c=0

Sands

PA = γ'ztan2(45-φ/2)

Ex) φ = 30o

KAR = tan230 = 0.33

PA = 0.33γz

If γ = 1.8t/m3, PA = 0.594z → Earth Pressure

Pw =γwz = z → Water pressure,

PA << Pw → Water loads are very important.

(2) c, φ=0

Undrained loading

(short term)

for saturated clay

PA = γz - 2c

EAR = 1/2γH2 - 2cH

Hcrit = Theoretical max. height of unsupported cut in clay

H

Z

H/3

EA

R

4c/γ = Hcrit

2c/γ

PA=0(z=2c/γ)

Theoretical

depth of

tension

crack

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� Rankine Active and Passive Earth Pressure For Inclined Granular

backfill

� Based on assumption that the resultant force, EAR, is parallel to slope of

backfill, Rankine’s active pressure and resultant force can be obtained as

below

PA = γHKAR , EAR = 1/2γH2KA

R

where φββ

φβββ

22

22

coscoscos

coscoscoscos

−+

−−=R

AK

Similarly, Rankine’s passive pressure,

PP = γHKPR , EPR = 1/2γH2KP

R

where

φββ

φβββ

22

22

coscoscos

coscoscoscos

−−

−+=R

PK

z

H

H/3

E A R

β

β

β

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b) Coulomb

� Uses a straight line to approximate failure surface

� C

AE is a function of (a) geometry of wedge (H, β, ω, θ)

and (b) soil properties (c, φ’, δ, cα,γ)

� Known : a)

b)

c)

� Unknown :

N = total normal force on

failure surface

Total shear force along

failure surface

(cohesion(c)+friction(φ’))

Total shear force along

the wall

(cohesion(cα)+friction(δ))

R = Resultant force of N and

shear force (friction only)

R

φδ

W

C

AE

wa Lc ⋅Lc ⋅

β

θ

ω

Nwwww

N

H

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� Trial Wedge Procedure (c, φφφφ’ soil)

(i) Assume θ

(ii) Calculate C

AE from vector addition

� Note : Wall friction reduces load against wall.

WWWW

RRRR

ωδ

C

AE

Lc ⋅

θ

wa Lc ⋅

ω

R

φδ

W

C

AE

wa Lc ⋅Lc ⋅

β

θ

ω

Nwwww

N

N

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(iii) Assume additional values of θ, calculate C

AE (θ) and pick max. value.

This max. value is active earth pressure resultant.

* Analytical Solution for Coulomb’s Method (c=0, φ φ φ φ soils)

� Active earth pressure resultant

EAC = 1/2γH2KA

C

where

2

2

2

C

A

)cos()cos(

)sin()sin(1)cos(cos

)(cosK

β−ωω+δβ−φδ+φ

+ω+δω

ω−φ=

� Passive earth pressure resultant

EPC = 1/2γH2KP

C

where

2

2

2

C

P

)cos()cos(

)sin()sin(1)cos(cos

)(cosK

β−ωω+δβ+φδ+φ

−ω−δω

ω+φ=

and for δ=0

2

2

2

C

P

)cos(cos

)sin(sin1coscos

)(cosK

β−ωωβ+φφ

−ωω

ω+φ=

θ

TW

AE

C

AE

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� Notes for Planar Failure Plane of Coulomb’s Method

(1)

(2)

(3)

tan

0r r eα φ=

r0

r

α

Log Spiral Failure Surface

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3) Movements to mobilize limit state (i.e. active or passive failure state)

US army corps No.4, Fig 3-2.

Relationship of Earth Pressures to Wall Movements

(after Department of the Navy 1982)

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Wall Movements (Rotations) to reach the limit state(y/H)

DM7 USAC Das Coduto

A P A P A P A P

Dense Cohesionless

.0005 .002 .0001 .02 .0005 .001 .02

Loose Cohesionless

.002 .006 .0003 .02 .001 .01 .004 .06

Stiff Cohesive

.01 .002 .01 .01 .02

Soft Cohesive

.02 .04 .04 .05 .02 .04

Movements

passive state > active state

clay > sand

loose or soft > dense or stiff