-
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.
3
-
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
10
-
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
11
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
12
-
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.
16
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
n
e
n
t
,
K
a
h
l
c
o
m
p
o
n
H
o
r
i
z
o
n
t
a
H
-
EC7 Annex C: Kph - Horizontal Component of phPassive Earth
Pressure
K
p
h
p
o
n
e
n
t
,
K
o
n
t
a
l
c
o
m
H
o
r
i
z
o
-
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
44
-
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
46
-
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
47
-
Worked Example from Craig
48
-
49
-
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
53
-
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
54
-
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
e
n
t
,
K
a
h
c
o
m
p
o
n
e
o
r
i
z
o
n
t
a
l
H
o
-
EC7 Annex C: Kah - horizontal component of active earth
pressurepressure
n
t
,
K
a
h
c
o
m
p
o
n
e
n
o
r
i
z
o
n
t
a
l
c
H
o
-
EC7 Annex C: Kah - horizontal component of active earth
pressurepressure
n
t
,
K
a
h
c
o
m
p
o
n
e
n
o
r
i
z
o
n
t
a
l
c
H
o
-
EC7 Annex C: Kph - horizontal component of passive earth
pressure
K
p
h
p
o
n
e
n
t
,
K
o
n
t
a
l
c
o
m
H
o
r
i
z
o
-
EC7 Annex C: Kph - horizontal component of passive earth
pressureearth pressure
p
h
p
o
n
e
n
t
,
K
p
n
t
a
l
c
o
m
p
H
o
r
i
z
o
n
-
EC7 Annex C: Kph - horizontal component of passive earth
pressureearth pressure
n
e
n
t
,
K
p
h
l
c
o
m
p
o
n
H
o
r
i
z
o
n
t
a
H