Lateral Earth Pressure - ReviewLateral Earth Pressure Review

Outline E th P Th i Earth Pressure Theories

RankineC l bCoulomb

At Rest Earth Pressure Distribution of Earth Pressure Distribution of Earth Pressure Water Pressure Effect of Surface Loading and Surcharge Effect of Surface Loading and Surcharge

Lateral Earth PressureLateral Earth PressureIt is often necessary in geotechnical engineering to retain y g g gmasses of soil. Eg:

Retaining unstable soil next to road or railway

Raising a section of soil with minimal land takea s g a sect o o so t a a d ta e

Creation of underground space

Seawalls Seawalls

Bridge abutment

Caissons

The design of all these structures requires a knowledge of the t i d b th il th t t istresses imposed by the soil mass on these structures, in

particular, the lateral pressures and forces.

Functions of Retaining WallsFunctions of Retaining Walls

To provide lateral support to the soil andsupport to the soil and retaining it at the desired position.

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Lateral Earth PressureLateral Earth Pressure

In the absence of external loading, stresses in soil originates primarily from self-weight of soil, .The capacity of a soil to resist shear is governed by its shear strengthThe capacity of a soil to resist shear is governed by its shear strength.

A stronger soil will imposed less lateral pressure on adjacent support structures.structures.

A soil can exhibit shear resistance in 3 ways:Cohesionless soil (c=0): strength due entirely to friction.Cohesionless soil (c 0): strength due entirely to friction.Cohesive soil (=0): strength due entirely to cohesion.Cohesive-frictional soil (c-): strength depend on both cohesion and friction.

The calculated values of lateral pressure generated by the weight of the il b id bl i if l d f tisoil can be considerably in error if wrong values are assumed for operative

values of cohesion, c and angle of friction, .

Lateral Earth Pressure TheoriesLateral Earth Pressure Theories

Most theories are derived based on some idealized assumptions about stress-strain behaviour of soil.

o If both stresses and displacements are i d th l ti i i l i frequired, the solution requires inclusion of

complex stress-strain behaviour of soil. Solution involves complex partial differential equations and boundary conditions.

o May be possible with computer modelling such as finite element analysis

Stress-strain curve of real-soil

as finite element analysis.

o If the failure condition is the primary interest and consideration of displacement is notand consideration of displacement is not required, possible to use concept of plastic collapse by assuming a rigid plastic model.

o Only failure criterion and strength parameters required. Idealized Stress-strain

curve

Limit Theorems of Plastic CollapseLimit Theorems of Plastic Collapse

Lower Bound Theorem (Eg: Rankine Theory)Lower Bound Theorem (Eg: Rankine Theory)If a state of stress can be found, which at no point exceeds the failure criterion and is in equilibrium with a system of external loads (including

lf i ht f th il) th ll t Th t l l dself weight of the soil), then collapse cannot occur. The external load system constitutes a lower bound to the true collapse load Reason: a more efficient stress distribution may exist, which would be in easo a o e e c e t st ess d st but o ay e st, c ou d beequilibrium with higher external loads.

Upper Bound Theorem (Eg: Coulomb Theory)If a mechanism of plastic collapse is postulated and if in an increment of displacement the work done by a system of external loads is equal to thedisplacement the work done by a system of external loads is equal to the dissipation of energy by the internal stresses, then collapse must occur. The external load system thus constitutes an upper bound to the true collapse loadcollapse load. Reason: a more efficient mechanism may exist resulting in collapse under lower external loads

Rankine Lateral Earth Pressure TheoryRankine Lateral Earth Pressure Theory

Scenario:Scenario:

- Vertical wall with infinite depth.

S i i fi i il i h il id- Semi-infinite soil mass with soil to one side of wall, empty on the other

Question: what will happen to the stress in theQuestion: what will happen to the stress in the ground when the wall moves (translate laterally) either away from the soil or into the

?soil?

Answer: A state of plastic equilibrium (failure) will be induced in the whole soil mass will be induced in the whole soil mass behind the wall at some point in both cases.

What are the magnitudes of the stresses at theWhat are the magnitudes of the stresses at the state of plastic equilibrium?

Rankine Lateral Earth Pressure TheoryRankine Lateral Earth Pressure Theory

When wall moves away from soil:When wall moves away from soil:- x will decrease until failure is triggered- At that point, x will become the minor

principal stress- Vertical stress z remains the major

principal stressprincipal stress- Mohr circle will touch the failure envelop

When wall moves into the soil:- x will increase until failure is triggered x

gg- At that point, x will become the major

principal stressV ti l t i th i- Vertical stress z remains the minor principal stress

- Mohr circle will touch the failure envelopo c c e ouc e a u e e e op

Rankine Theory - AssumptionsRankine Theory Assumptions

The soil is homogeneous and isotropicg p Failure surface is a plane Ground surface is a plane Wall is infinitely long: i.e. plane strain condition Sufficient wall movements to develop active or passive state Wall is vertical Wall is smooth no friction The resultant force is parallel to ground surface The resultant force is acting at H/3 above the wall bottom

H/3

H

Rankine Active Earth Pressure in Sand (c=0)Rankine Active Earth Pressure in Sand (c=0)

'x = Ko 'z'

No wall movement No wall movement'x

a z

a = Ka 'zHere we are dealing with effective stress only dry soil or drained analysis.

Wall moved away from soil

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Active Earth Pressure in Sand (Rankine)( )

'A = KA 'z '

45 + '/2

B

Active earth pressure 'z'A O A

Failure plane at active state

45 + '/2

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p

Rankine Passive Earth Pressure in Sand (c=0)Rankine Passive Earth Pressure in Sand (c=0)

'o = Ko 'z

'z'x 'P No wall movement

p = Kp 'z

Here we are dealing withHere we are dealing with effective stress only.

Wall moved towards soil

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Passive Earth Pressure in Sand (Rankine)Passive Earth Pressure in Sand (Rankine)

'P = KP 'z'

P P z

Passive earth pressure

P'z 3 1

45 - '/2

13Failure plane at passive state

Active Earth Pressure in Clay (Rankine)Active Earth Pressure in Clay (Rankine)(under short term undrained condition)

= ' + u

= Ko'z + ux x u u=0cu

No wall movementzxA

(-) c = ( )A = z 2 cu

cu = (z - A)A = - 2c

W ll d f il

(+)

Here we are dealing with total stress

A z 2cu

14Wall moved away from soil Here we are dealing with total stress.

Passive Earth Pressure in Clay (Rankine)Passive Earth Pressure in Clay (Rankine)(Under short-term undrained condition)

x = 'x + u

u=0= Ko'z + u cu

zx P No wall movement

c = ( - )P = z+ 2cu cu (P z)P = z + 2cu

P z cu

H d li ith t t l t

P z 2cuWall moved towards soil

15Here we are dealing with total stress.

Active and Passive Earth Pressure in c- Soils (Rankine)(under long-term drained condition)

'(-)

(+)

'z'A c'

Active earth pressure

Here we are dealing with effective stress only.

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only.Passive earth pressure

Rankine Active Pressure (c- soil)Rankine Active Pressure (c soil)

Mohr Circle at Failure

Failure surfaces in active stateFailure surfaces in active state

Rankine Active Pressure Tension CrackRankine Active Pressure Tension Crack

Negative tensile stresses tension crack

If the crack is filled with water, hydrostatic pressure will act on the wall. Crack will extend to:

Active Thrust (force)

Rankine Passive Pressure (c- soils)Rankine Passive Pressure (c soils)Passive Pressure distributionPassive Pressure distribution

Pressure distribution

Passive Thrust (resultant force)Passive Thrust (resultant force)Passive pressure at depth z

Lateral Earth Pressure Due to SurchargeLateral Earth Pressure Due to Surcharge

Uniform Surcharge) (kN/m2

Active PassiveActive Passive

Orientation of Failure Surfaces Active Case

pole

Orientation of Fail re planesMohr Circle at Failure

Orientation of Failure planes 2 sets

Orientation of Failure Planes Passive CaseOrientation of Failure Planes Passive Case

Failure planes 2 setsFailure planes 2 sets

pole

90-90-

pole

Mohr circle at failure

Rankine Pressure with Sloping Backfill (Active)p g ( )

Active pressure coef:

Failure Plane orientation

Active Thrust (Resultant force)

Pressure at depth z: Direction: parallel to slopep p

Rankine Pressure with sloping backfill (Passive)p g ( )

Failure Plane orientationPassive Coefficient and pressure

Passive Thrust

Passive pressure at depth z:Direction: parallel to slopep p

Earth Pressure At Rest

In a semi-infinite, homogeneous soil mass with weight density , the vertical istress is:

The horizontal stress is expressed as:

zzo '' =The horizontal stress is expressed as:

zKK ozooho ''' ==Ko is called the earth pressure coefficient at rest.

Pressure variation with depth:

Earth Pressure At rest

Ko can be evaluated from triaxial test - lateral strain is kept at zero.o p

For Normally Consolidated soils,(Jaky):

F lid t d il ( M & K lh )For overconsolidated soils ( Mayne & Kulhawy) :

Eurocode 7:

When terrain is sloped at angle to the horizontal

Earth Pressure At RestEarth Pressure At RestVariation of Ko with OCR

NB:For highly OC soils, Ko >1, in-situFor highly OC soils, Ko 1, in situ horizontal stress can be bigger than in-situ vertical stress

> ho > ho

Coulomb (1776) Earth Pressure Theory( ) y

Consider the stability of a wedge Assumptions:y gof soil between a retaining wall and a trial failure plane in a condition of limiting equilibrium

p

o The soil is homogeneous and isotropic.condition of limiting equilibrium.

o Failure surface is planar.o Ground surface is planar.po Wall is infinitely long: i.e. plane strain

condition.

o Sufficient wall movements to develop active or passive state.

o Wall can be inclined ( 90).o Wall can be rough ( 0). dg ( )o The resultant force is acting at an

angle or 'w with the normal to wall.28

Coulomb Active Earth Pressure (c=0, sand)

Force Polygon

The point of action is assumed to be at H/3 from base of wall

Coulomb Active Pressure (c>0)

Assumptions:o c is greater than zero (i.e. cu in the

undrained case, c in the drained case).

ll dh i to wall adhesion parameter = cwo tension cracks depth = z0

th t i l f il t di f th h l fo the trial failure extending from the heel of the wall to the bottom of the tension zone,

Forces on wedge:Forces on wedge:

o weight of the wedge (W).ti (P) b t th ll d th ilo reaction (P) between the wall and the soil

o Adhesion: (Cw = cw x EB).o Reaction (R) on the failure plane, acting at angle below the normal.o The force on the failure plane due to the constant component of shear

strength (C = c x BC)strength (C = c x BC).

Coulomb Active Pressure (c>0)( )

Coulomb Active Earth Pressure (c, soils)( , )For fully drained condition (c,), the active pressure at depth z is

NB: For undrained case when u=0, Kac=1 and

Coulomb Earth Pressure Passive CaseCoulomb Earth Pressure Passive Case

In general the passive pressure at depth z is:In general, the passive pressure at depth z is:

Log Spiral Method on Lateral Earth Pressure

When > /3, curvature of failure surface must be taken into account orsurface must be taken into account or the passive resistance will be significantly overestimated, representing an error on the unsafe side. Log Spiral Method.

Pa= Ka H2 Pp= Kp H2K d K

PaH = Pa cos P P i

PpH = Pp cos P P i

Ka and Kp are obtained from Tables or charts.

34PaV = Pa sin PaV = Pp sin

Earth Pressure Coefficient - Curved Failure Surfaces(Kerisel and Absi)(Kerisel and Absi)

Passive Earth Pressure coefficient Horizontal component, Kph

Active Earth Pressure coefficient: Horizontal component, Kah

Kav = Kah tan()Kpv = Kph tan()

Eurocode 7 Annex C: Kah - Horizontal Component of ahActive Earth pressure

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EC7 Annex C: Kph - Horizontal Component of phPassive Earth Pressure

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Application of Earth Pressure TheoryApplication of Earth Pressure Theory

In Rankine theory, a semi-infinite soil mass is subject to expansion or y j pcompression.

In retaining wall, depth is finite so only a finite zone of soil mass is i l dinvolved.

In both active and passive cases, plastic equilibrium can only occur within a wedge of soil between wall and failure planewithin a wedge of soil between wall and failure plane.

The wall must deformed sufficiently to mobilize the limit conditions.

Wall DeformationWall Deformation

The passive pressure is induced by wall movement into the soil.

The relatively large deformation necessary for the full development of limit passive resistance would be unacceptable from serviceability standpointunacceptable from serviceability standpoint.

Hence, full passive pressure is seldom realized under working conditionworking condition.

The design lateral pressure under working conditions would be between the at-rest and passive valueswould be between the at-rest and passive values.

Influence of Wall Yield on Pressure Distribution

Active Case:

In the case of passive earth pressure, the movement necessary to bili th lti t l f b l ti l l f lmobilize the ultimate value of can be large, particularly for loose

sand. One can rarely expect more than one half the value of ultimate passive pressure will be developed.p p p

Active Earth Pressure:

= 20 kN/m3PAV

Active Earth Pressure:

5 mPAH

PA

PAV

Comparisons of Method of Analysis

Theory ' KA PA PAH

y

Theory KA PA PAHLoose Sand

Rankine 30 0 0 33 83 3 83 3Rankine 30 0 0.33 83.3 83.3Coulomb 30 20 0.30 75 70.5Log Spiral 30 20 0 32 80 75 2Log-Spiral 30 20 0.32 80 75.2

Dense SandRankine 40 0 0 22 55 55Rankine 40 0 0.22 55 55Coulomb 40 30 0.20 50 43L S i l 40 30 0 22 55 48

41Log-Spiral 40 30 0.22 55 48

Passive Earth Pressure:Passive Earth Pressure:

= 20 kN/m35 m

PPH

PP

PPV

Comparisons of Method of Analysis

Theory ' KP PP PPHPPPPV

Loose SandRankine 30 0 3 750 750Coulomb 30 20 6.1 1525 1433Log-Spiral 30 20 5 2 1300 1222Log Spiral 30 20 5.2 1300 1222

Dense SandRankine 40 0 4 6 1150 1150Rankine 40 0 4.6 1150 1150Coulomb 40 30 24.9 6225 5391

42Log-Spiral 40 30 13.1 3275 2836

Summary

Theory Roughness Inclination Application

Summary

Theory Roughness Inclination ApplicationRankine smooth vertical KaCoulomb any any Ka

Log-Spiral any any K & KLog Spiral any any Ka & Kp

PAH

PA

PAV

PPHPAH

PA

PAV

PPPPV

Rankine Coulomb & Log Spiral43

Coulomb & Log Spiral

Total Active Earth Pressure on Wall in Sand

q'A= KAq

Sand, 'A= KA'zH PA1

PA2'A= KA (q + H)

PA2

PA1 = KAq H PA2 = KA H2A1 Aq A2 A 2

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Total Active Earth Pressure on Wall in Sand

q

u'A= KA'

Sand

A KA z

+

Earth pressure &surcharge

Water pressuresurcharge pressure

45

Total Active Earth Pressure on Wall in Clayy

q

z

Clay cu

zA= z 2cuwhere z = z + qz q

Earth pressure

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Total Active Earth Pressure on WallTotal Active Earth Pressure on Wall in Layered Soil

q

u'A= KA'z

+

Sand

Water pressure

+

Clay

A= z 2cuEarth pressure

& Surcharge& Surcharge

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Worked Example from Craig

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Total Active Earth Pressure on Wall in Clayy

Ignore negative pressureDry condition

2

- Tension crackzcz

Clay cu A= z 2cu++

Earth pressure

How do you determine zc?

A= zc 2cu = 0 50

A c u

Total Active Earth Pressure on Wall in ClayTotal Active Earth Pressure on Wall in Clay

After heavy downpourInitial crack

Final crack depth

zwInitial crack

depth

p

Clay cu A= 2cA z 2cu

Earth pressure

How to determine zw? A= zw 2cu = zw w

Simplified Pressure Diagram for Designp g g

designactual

Clay

A= z 2cuA= z 2cu

Pressure on Wall due to Line Load on Surface Spangler (1938)

Resultant PHm 0.4, PH = 0.55QLine load Q Hm > 0.4, PH = 0.64Q/(m2 + 1)

Qx = mH

m R/Hpressure

0.1 0.600 3 0 60

H PHR 0.3 0.60

0.5 0.560 7 0 480.7 0.48

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Load on Wall Due to Point Load on SurfaceLoad on Wall Due to Point Load on Surface

Point load Qpx = mHx = mH

m R/H (PHH)/Qp

H PH

m R/H (PHH)/Qp0.2 0.59 0.78

H

R 0.4 0.59 0.78

0 6 0 48 0 450.6 0.48 0.45

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Example 6.1 (Smith)Example 6.1 (Smith)

Example 6.1 (Smith) - 2Example 6.1 (Smith) 2

Example 6.2 (Smith) - 1Solve Example 6.1 using Coulomb earth pressure theory. Assume that = /2

p ( )

Assume that = /2.

Example 6.1 (Smith) - 2 p ( )

This value is inclined at 17.5o to the normal to the back of the wall. So the total horizontal active thrust (force) is ( )58.43 x Cos(17.5o) = 55.7 kN

N t if d i d t b 0 th C l b ti th tNote: if d is assumed to be 0, the Coulomb active thrust would have been the same as that obtained by RankineTheory in Example 6.1y p

Example 1 (Smith) - 3

Solve Example 6.1 using Coulomb earth pressure theory.

p ( )

p g p yAssume that = /2.

Solution when soil surface sloping at 35o:

Substituting =35o, =35o, =17.5o and =90o into the Coulomb formula gives Ka=0.704. Hence:

Total active thrust =0.5 x 0.704 x 19 x 52 = 167.2 kN

Total Horizontal thrust = 167.2 x cos(17.5o) = 159.5 kN( )

Increase in horizontal thrust = 159.5 55.7 = 104 kN

Example 6.2 (Smith) - 1 p ( )

Example 6.2 (Smith) - 2

Example 6.2 (Smith) - 3 p ( )

Example 6.4 (Smith) - 1 p ( )

Example 6.4 (Smith) - 2 p ( )

Charts on Horizontal Earth PressureCharts on Horizontal Earth Pressure

Coefficient in E rocode 7 Anne esCoefficient in Eurocode 7 Annexes

EC7 Annex C: Kah - horizontal component of active earth pressurepressure

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EC7 Annex C: Kah - horizontal component of active earth pressurepressure

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EC7 Annex C: Kah - horizontal component of active earth pressurepressure

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EC7 Annex C: Kph - horizontal component of passive earth pressure

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EC7 Annex C: Kph - horizontal component of passive earth pressureearth pressure

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EC7 Annex C: Kph - horizontal component of passive earth pressureearth pressure

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