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Earth Pressures and Retaining Walls
Muni Budhu - Soil Mechanics and Foundations
2nd
edition
Dr Jie Li
School of Civil, Environmental and Chemical EngineeringRMIT University ( )
Melbourne, Australia
CIVE1129 Geotechnical Engineering 2
Revision Part 2
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Importantly, at K0 state, there are no lateral strains
GL
A
v= 1
h= 3
Lateral Earth Pressure At Rest
Soil is usually deposited in the horizontal layers and consolidation
occurs solely in the vertical direction with no lateral deformation. Such
a soil mass is said to be in the at-rest condition. The ratio between the
horizontal and vertical effective stresses is a constant known as the Ko
coefficient.
' ' '
3' ' '
1
h xo
v z
K
In a homogeneous
natural soil deposit
Lateral Earth PressureCoefficient at Rest:
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smooth wall
Wall movesaway from soil
Wall moves
towards soil
A
B
Lets look at the soil elements A and B during the wall
movement.
Active Earth Pressure in granular soils
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A
z
xz
As the wall moves away from the soil,
Initially, there is no lateral movement.
z= v= z
x= h= K0 z= K0 z
zremains the same; and
xdecreases till failure occurs.
Active state
Active Earth Pressure in granular soils
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Active Earth Pressure in granular soils
z
decreasing x
Initially (K0 state)
Failure (Active state)
As the wall moves away from the soil,
active earth pressure
A
z
x
zWJM Rankine
(1820-1872)
' '( )x za a
K Rankines coefficient of active earth pressure
Can be derived from Geometry of Mohrs circles
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21 sin tan (45 / 2)
1 sin
aK
z(x)a
Az
x45 + /2
90+
Failure plane is at
45 + /2 to horizontal
' '( )x za a
K
Active Earth Pressure in granular soils
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A
zx
z
As the wall moves away from the soil,
xdecreases until failure occurs.
wall movement
x
Active
state
K0 state
Active Earth Pressure in granular soils
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1 and 3 Relation at Failure
1 3
' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c
3 1
' ' 2
tan (45 '/ 2) 2 ' tan(45 '/ 2)c
1 3
1 3
OF 2sinOA
cot '2
c
The failure criterion can also be expressed in terms of principal stresses
3 1
2O
F
A
c
1 3
2
1 3
2
ccot
' ' tan 'f
c
= 45o + /2
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zx
245
active wedge
Active Earth Pressure
2
3 1tan (45 '/ 2) 2x z ztan (45 )
2 a
K
2 'tan (45 )
2
aK
Rankine Theory of Earth Pressure in Sand
Mohr-Coulomb failure criterion:
For sand, c=0. This leads to:
effective active earth pressure
Coefficient of active earth pressure
3 1
' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c
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B
zx
Initially, soil is in K0 state.
As the wall moves towards the soil,
zremains the same, and
xincreases until failure occurs.
Passive state
Passive Earth Pressure in granular soils
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z
Initially (K0 state)
Failure (passive state)
As the wall moves towards the soil,
increasing x
passive earth
pressure
Passive Earth Pressure in granular soils
( ') 'x p P z
K
Rankines coefficient ofpassive earth pressureB
z
x
Again, from Geometry of Mohrs circles
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z (x)p
Azx
90+
Failure plane is at
45 -/2 to horizontal45 -/2
)2/45(tansin1
sin1 2
PK
Passive Earth Pressure in granular soils
( ') 'x p P zK
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B
zx
As the wall moves towards the soil,
xincreases till failure occurs.
wall movement
x
K0 state
Passive state
Passive Earth Pressure in granular soils
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1 3
' ' 2tan (45 '/ 2)
z x
2
x z ztan (45 '/ 2) pK
2tan (45 '/ 2)
pK
245
passive wedge
Passive Earth Pressure
Mohr-Coulomb failure criterion:
For sand, c=0. This leads to:1 3
' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c
Coefficient of passive earth pressure
effective passive earth pressure
Rankine Theory of Earth Pressure in Sand
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' '
1 3' '
1 3
' ' ' '
1 3 1 3
2sin '
2
f f
f f
f f f f
OBOA
Mohr-Coulomb Failure Criterion
'
12
'
3
1 sin ' 'tan 45
1 sin ' 2
f
p
f
k
'
32
'
1
1 sin ' 'tan 45
1 sin ' 2
f
a
f
k
kp = passive earth pressure coefficient ka = active earth pressure coefficient
1 3(1 sin ) (1 sin )
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Normal stress
Shear stress
zoActive state Passive state
The 3 States:
At Rest
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x zx
z
At-Rest
xz
Active pressure
x z
z x
xz
Passive pressure
Rankine Theory of Earth Pressure in Sand
From the at-rest
condition and with thevertical effective stress
kept constant, the soil
can fail in two ways.
First, the horizontal
effective stress may
decrease when the soil
is allowed to have
lateral extension. Thiscreates the active earth
pressure.
Second, the horizontal
stress may increasewhen the soil is
compressed laterally.
This creates the passive
earth pressure.
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2
a x z a
1
2a a aP dz K dz z K dz K H
2
p x z p
1
2p p p
P dz K dz z K dz K H
x
aP
pP
x
PA and PP are the resultant active
and passive thrusts on the wall
Ha
Hp
Rankine Theory of Earth Pressure in Sand
Active thrust:
Passive thrust:
Ka HaKp Hp
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Hv
uaP
h
For the case shown above:z
( )
z z h
h z h z h
)(w hzu
a x
0 0
2
( ( ))
1 1( ) (2 ( ))
2 2
H h H
a a
h
a a
P dz z K dz h z h K dz
K h K H h h H h
x zaK
2w )(
2
1hHdzuU
H
h
a aP P U
1. Calculate the vertical effective stressdistribution.
2. Calculate the effective earth pressure.
3. Calculate the effective earth thrust.
4. Calculate the pore pressure u.
5. Calculate the total force U due to u.
6. Calculate the total earth thrust Pa = Pa + U
Presence of pore pressure:
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The general relation: Kp > Ko > Ka
Kp
Ka
Ko
Kp KaKo> >
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x z 2a a
K c K
zxc
2tan (45 '/ 2)
aK
Active Earth Pressure
active wedge
245
cz
Rankine Theory for Clay
Mohr-Coulomb failure criterion:
S i n c e x= 3a n d z = 1a t f a i l u r e , t h e p a s s i v e e a r t hp r e s s u r e :
3 1
' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c
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Can soil undergo tension?
If z = 0, thenx 0Now if z = z,
At what depth willx = 0?
x
= Ka
z- 2cK
a
This depth is called the depth of cracking, zc, &
defines the potential tension zone
Rankine Theory for Clay
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Depth of Cracking
By definition:
At zc, x = 0
Therefore,
x = 0 = Ka zc - 2cKa
Therefore,
zc = [2cKa][ Ka ]Or
c
a
2cz
K
Rankine Theory for Clay
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x z 2a aK c K
zxc
ac
2K
cz
2tan (45 '/ 2)aK
Active Earth Pressure
active wedge
245
cz
Rankine Theory for Clay
The depth of tensile cracking:,
c
a x
1 2
2
H
a
z a
cP dz K H
K
Active thrust:
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Active earth pressure
aP
x 3 z 1
Passive earth pressure
pP
x 1 z 3
Advantage: Simple and explicit
expressions for earth pressures and
earth thrust.
Solution Steps:
1. Find vertical stress z2. Find horizontal stress x3. Integrate x to find the effective
earth trust.
4. If a water table is present,estimate the forces due to pore
pressures.
5. Estimate the total earth thrust.
Assumptions:
Rankine Earth Pressure: Summary
Horizontal ground surface Vertical retaining wall
Initially at an at-rest state
No friction between the wall and soil
Failure occurs throughout the soil mass behind the wall
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Orientation of Failure PlanesFrom Mohrs circles
Active state:
(45 + /2) to horizontal
Passive state:
(45 -/2) to horizontal
Rankine Earth Pressure: Summary
[ '] ' 2x p P z PK c K
[ '] ' 2x a a z aK c K
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Quiz
1. Which requires more retaining wall rotation to be mobilized?A) Neutral resistance
B) Passive resistance
C) Active resistance
D) None of above
2. For the same soil, what is usually true?
A) Passive pressure > at-rest stress > active pressure
B) Passive pressure > active pressure > at-rest stress
C) Active pressure > passive pressure
D) At-rest stress > passive pressure
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Rankine Active Earth Pressure
Determine the active lateral earth pressure on the frictionlesswall shown below. Calculate the resultant force and its location
from the base of the wall. Neglect seepage effects.
1. The active lateral earth coefficient
2 2tan (45 '/ 2) tan (45 30 / 2) 0.333o o oaK
2. The effective stresses and pore pressure
E l 12 1 R ki A ti E th P ( 495)
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Example 12.1 Rankine Active Earth Pressure (p495)
3. These stresses are sketched as below:
4. The lateral force
Pa = Pa +U = 165 kN/m
Pa = x 17 x 5 = 42.5 kN/m U = Pw = x 49 x 5 = 122.5 kN/m
Since both the distributions of the effective lateral earth
pressure x and the pore pressure u are triangular overthe whole depth, the resultant is at the centroid of the
triangle, that is H/3=1.67m from the base.
5. Location of resultant
S R ki E th P
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x 3 z 1 x 1 z 3
Solution Steps:
1. Find vertical stress z
2. Find horizontal stress x3. Integrate x to find the effective earth trust.
4. If a water table is present, estimate the forces due to pore pressures.
5. Estimate the total earth thrust.
Summary - Rankine Earth Pressure
pPaP
Active earth pressure Passive earth pressure
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1. Vertical backs of walls only
2. Backfill surface must be regular
a solution exists for a sloping backfil l, provided slope
angle, < 3. Backfill loads / surcharge effects approximated
4. Wall friction ignored!
a beneficial effect
Limitations of Rankines earth pressure theory
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Example: Rankine Earth Pressure
A 6 m high smooth retaining wall is to support a soil with unit
weight =17.4 kN/m3, friction angle = 26o and cohesion c =
14.36 kN/m2. The groundwater table is below the base of the wall.
Determine the Rankine active force per unit length of the wall, and
determine the line of action of the resultant.
(1) Active lateral earth pressure coefficient:
2
tan (45 '/ 2) 0.39
o
aK
' ' 2 ' ) 17.95x a z a
K c K kPa
2 '2.64
c
a
cz m
K
At surface: z = 0 kPa
(2) The depth of tensile cracks :
Soil cannot undergo tension!
The depth of tensile cracks
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' ' 2 ' ) 22.77x a z a
K c K kPa At base: z = 17.4 x 6 =104.4 kPa
(3) Active lateral earth pressure:
The active force is the positive area of active earth pressure distribution:' 1 1
( )( 2 ' ) 22.8(6 2.64) 38.3 /2 2
a c a aP H z HK c K kN m
Location of resultant active earth pressure is (H-zc)/3 = (6-2.64)/3 =1.12m
z= 104.4 kPa
2 '2.64
c
a
cz m
K
x= 22.8 kPa
x= -18 kPa
1.12 m
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For the frictionless wall shown below, determine:
1. Active lateral earth pressure distribution
2. Passive lateral earth pressure distribution
3. Magnitudes and locations of active and
passive forces
4. Resultant force and its location
5. Ratio of passive moment to active moment
Earth Pressure in Layered Soils
(1) Lateral earth pressure coefficients
2tan (45 25 / 2) 0.41o oaK
2tan (45 30 / 2) 0.33o oaK
Top layer:
Bottom layer:
2tan (45 30 / 2) 1/ 3o op aK K Passive coefficient:
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(2)Active lateral earth pressure (at the back of the wall)
At surface:
x = Kaz = 0.41 x 20 kPa = 8.2 kPa, u = 0 kPa
At 2- m,z = 20+H = 20+19 x 2 = 58 kPa
z = qs = 20 kPa
x = Kaz = 0.41 x 58 = 23.8 kPa u = 0 kPa
At 2+ m,z = 20+H = 20+19 x 2 = 58 kPa
x = Kaz = 0.33 x 58 = 19.33 kPa u = 0 kPa
At base, z = 58 + H = 58 + (20 9.8) x 4 = 98.8 kPa
x = Kaz = 0.33 x 98.8 = 32.93 kPa
u = w H = 9.8 x 4 = 39.2 kPa
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(3) Passive lateral earth pressure (in the front of the wall)
At surface:
x = Kpz = 3 x 0 = 0 kPa, u = 0 kPa
z = 0 kPa
At base:
z = H2 = (20 9.8) x 4 = 40.8 kPa
x = Kpz = 3 x 40.8 = 122.4 kPa
u = w H = 9.8 x 4 = 39.2 kPa
x = 8.2 kPa
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(4) Active Force
x= 122.4 kPa
U = x 39.2 x 4 = 78.4 kN/m
u = w H = 9.8 x 4 = 39.2 kPa
x
x = 23.8 kPa
x = 19.3 kPa
x= 32.9 kPau = 39.2 kPa u = 39.2 kPa
' 8.2 23.8 19.3 32.92 4 136.4kN/m2 2
aP
The effective active force acting on the back of the wall is the total
area of the active lateral earth pressure distribution:
Pa = Pa + U = 214.8 kN/mTotal
4 m
2 m
x = 8.2 kPa
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(4) Total moment about the base line:
x= 122.4 kPa
Mu = x (39.2 x 4) x 1/3 x 4 = 78.4 kN
u = w H = 9.8 x 4 = 39.2 kPa
x
x = 23.8 kPa
x = 19.3 kPa
x= 32.9 kPau = 39.2 kPa u = 39.2 kPa
' 1 2 1 48.2 2 (1 4) 15.6 2( 4) 19.3 4 2 13.6 4 345.6kN2 3 2 3
aM
Ma = Ma + Mu = 214.8 kNTotal
32.9-19.3 =13.6
Location of active lateral force = Ma/Pa = 450.1/214.8 = 2.1m from the base
4 m
2 m
(5) Passive Force
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Total moment about the base line: Mp
= Pp
x 4/3 = 430.8 kN
Rx = Pp - Pa = 323.2 214.8 =108.4 kN/m
The negative sign indicates the location is below the base.
Ratio of moment = Mp/Ma = 430.8/450.1 = 0.96 < 1
( )
The effective passive force acting on the front face of the wall
is the total area of the passive lateral earth pressure distribution:
Pp = x 122.4 x 4 = 244.8 kN/m
Pp = Pp + U = 323.2 kN/m
U = x 39.2 x 4 = 78.4 kN/m
Location of passive lateral force = 4/3 = 1.33m from the base
Its location is:
The resultant lateral force
(Mp - Ma)/R = (430.8 - 450.1)/108.4 = - 0.18 m
(6) The moment ratio
Wall wil l rotate
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TWO METHODS
to estimate the
earth pressures on structures
1. Rankine
Plausible stress states
2. Coulomb
Plausible failure mechanisms
Relative merits of approaches?
COULOMB APPROACH
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HANDLES: Irregular backfill surfaces
Sloping backs of walls
Sloping backfill Surface surcharges
Wall friction
Interface friction angle, Although failure surfaces were known to be
curved, a planar approximation was adopted
(OK for Active pressures)
Coulombs Earth Pressure Theory
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At failure, the stress on the slip surface through the soil obey the Coulomb
failure criterion Stresses:
ntanc
Slip surface
tanT C N
nT ds N ds C c ds
Passive wedge failure
Slip surface
Active wedge failure
Coulomb s Earth Pressure Theory
Coulomb wedge analysis is a limited equilibrium method. A collapse
mechanism is assumed, which is usually a straight-line slip surface, and theresulting forces are deducted graphically from the principle of equilibrium.
Integrating we obtain the corresponding forces according to
where
n
Coulombs Earth Pressure Theory
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H
Failed soil wedge
T ds
Free-body diagram of failed soil wedge
Pa
W
T
NnN ds
H cot
sand
W = weight of soil wedge - needs support
N = force normal to sliding plane from underlying supporting soil
T = tangential force along sliding plane = N tanPa = Max reaction from wall required for equilibrium for critical
wedge angle,
Coulomb s Earth Pressure Theory
Coulombs Earth Pressure Theory
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T ds H
Failed soil wedgeFree-body diagram of failed soil wedge
Pa
W
T
NnN ds
H cot
Fx = Pa + T cos- N sin= 0
Fy = W - T sin- N cos= 0W = H2 cot
T = N tan
sand
Pa = H2
cot tan(-)Solving for Pa, The critical value of which makes the wall force Pa a maximum is
given by solving Pa/ = 0. This leads to
Coulomb s Earth Pressure Theory
= 45
o
+ /2
Influence of Wall Friction
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Influence of Wall Friction
PaActive
Pp
Passive
angle of frictionbetween wall and soil
For rough wall, = . Forsmooth wall, = 2/3
Coulombs Earth Pressure Theory
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Active Earth Pressure: Sand
H
q
N
T Sand
, ',
tanNT
w wtanT N
21 cot2
W H
cotq
Q qB qH
Bq = H cot
W
Nw
Tw
Consider both Wall friction + Surface surcharges q
Pa
Nw
Tw
Pa
Active Earth Pressure: Sand
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For sand, the analysis is done in terms of effective stress using and and c =
cw= 0. Consider a vertical wall of height of H retaining a sand backfill in level with
its coping. A surcharge q is applied to the ground surface. The water table is
assumed to below the base of the wall.
xF =0
H
q
N
T Sand
, ',
tanNT w wtanT N
21 cot2
W H cotQ qH
Two unknowns: ,wN N
w sin cos sin tan 'cos cos (tan tan ')N N T N N N
y wF =0 cos sin cos (1 tan tan ')Q W T N T N
H cot
W
Nw
Tw
Dividing 1st by 2nd equation, we have tan( ')w
w
N
Q W T
The two equations can be used to solveN and Nw, in terms of the angle .
Active Earth Pressure: Sand
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Active Earth Pressure: Sand
The critical value of which makes the wall force Nw a maximum is
given by solving Nw/= 0. This leads to
2a
12
aP qH H K
2
cos 'tan1 sin ' 1
tan '
aK
2( ) tan( ') 1 cot tan( ')
1 tan tan( ') 2 1 tan tan( ')w
Q WN qH H
tan( ')( ) tan( ')( tan )w w wN Q W T Q W N
with
2
2 ' cos 'tan 452 1 sin '
o
aK
Comparing with Rankine result derived previously
when = 0 (smooth wall) we recover the Rankine expression for Ka
Coulombs Earth Pressure Theory
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Ka and Kp for retaining wall
with sloping back, wall friction
and sloping soil surface.
(12.16)
(12.17)
Page 500
The above type of analysis can
also be extended for the caseof an inclined wall and an
inclined ground surface.
Rankine Theory vs Coulomb Wedge Method
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Rankine Theory
At no point in the soil mass did the stress state exceed the failure
stress state.
In reality, a more efficient distribution of stress could exist
This is a lower bound solution (lower than the true solution)
Coulomb Wedge Method
A planar slip surface is assumed and the solution is obtained by
requiring force equilibrium on the soil mass In reality, a more efficient failure mechanism may be possible
This is upper bound solution (greater than the true solution)
Rankine Theory vs Coulomb Wedge Method
Attention Quiz Time!
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1. Coulombs approach assumes that (click all apply)
A) The soil is rigid-plastic
B) The soil is elastic-plasticC) A slip plane occurs through the soil mass
D) The soil is homogeneous
Attention Quiz Time!
2. Coulombs approach for earth pressure is based on
A) Limit analysis
B) Finite element method
C) Limit equilibrium method
D) Lower bound solution
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1) Erosion
2) Rainfall
3) Earthquakes
4) Geological Features
5) External Loading
6) Construction Activities
Summary - Causes of Slope Failure
SLOPE STABILITY
SUMMARY: KEY POINTS
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Angle of repose for dry granular soils
Influence of seepage on granular soils
Slope stability for homogeneous slopes in saturated clay
- Taylors method/charts
Frictional soils more difficult
- Method of slices
Bishops method for circular slips
Slope stability programs search for failure surface with lowest FS
- Circular or non-circular slips?
Importance of shear strength parameters
- Drained and/or undrained?
- Peak, ultimate or critical state?
Slope Stability Analysis
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Slope Stability Analysis
When the ground surface is sloping, forces exist which tend to cause
the soil to move from high points to low points. The most important of
these forces are the force of gravity and the force of seeping water.
Failure will occur when the resultant shear forces exceeds the shearstrength of the soil mass.
Factor of Safety is the ratio of the available shear strength of the soil to
the minimum shear strength required to maintain equilibrium.
m
fFoS
Where
f is the shear strength of the soil
m
is the minimum shear strength required to maintain stability
Sl St bilit A l i
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Slope Stability Analysis
In stability analysis Factor of Safety is defined as the
ratio of the resisting stress to the disturbing stress
resisting stress
disturbing stressFoS
ESA Effective Stress Analysis
TSA Total Stress Analysis
Infinite slopes
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h
b
Vertical
slice
W
s
Force equilibrium the slice
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W
N = WN = W coss
l = length of sliding surface
s
s
WT = W sins
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Stability IF, WT = T = cl + N tan
where cl = resistance due to cohesion (kN)
and N tan' = resistance due to friction
Factor of Safety (FoS) = restoring force
disturbing force
STABILITY
Restoring forceFoS N = WN = W coss
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CASE 1: c = c = 0 (clean sand)
T
FoSDisturbing force
cL N tan
W
cL cos tan sin
s
s
W aW a
The natural angle of repose
W
s
s s
WT = W sins
cos tan ' tan 'FoS= FS = F
sin tan
s
s s
W a
W a a
At limit equilibrium, FS = 1
s =
CASE 1: c = c = 0 (clean sand)
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The natural angle of repose
s =
the maximum angle of slope (measured fromhorizontal plane) at which loose sand will
come to rest on a pile of sand.
bCase 2: c = 0, Seepage down the slope
Pore force U on sliding base due to pore
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N = WN = b h coss
' cos tan ' ' tan 'FS =
sin tan
f s
m sat s sat s
b h a
bh a a
W
s
Pore force U
h
Seepage force parallel to slope
Js = w b h i = w b h sins
(i =h/l = s)
Available shear strength:
f= N tan = b h coss tan
m = Wsins + Js = b h sins + w b h sins = ( +w)bh sins
sat
Pore force, U, on sliding base due to pore
water pressure
Case 2: c = 0,
Seepage down the slope
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almost only half the FS!
' cos tan ' ' tan 'FS =
sin tan
f s
m sat s sat s
b h a
bh a a
Seepage down the slope
Effective normal force reduced less friction!
At limit equilibrium, FS = 1
' tan ' 1tan = tan '
2s
sat
a
b
Sh t t l t bilit fi d i d
Case 3: TSA, c 0,
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l = b / coss
sin ( )sin= sin cos
/ cos
s ss s
s
W a bh aTh a a
l l b a
2h =z =
sin(2 )u
s
s
a
Ws
h
Short term slope stability, find-grained
soils, Total Stress Analysis (TSA)
Shear strength on l for TSA:
s = 1/2 sin-1(2 su /h )
l
At limit equilibrium, FS = 1
2
FS = =sin cos sin(2 )
u u u
m s s s
s s
h a a h a
zcr= hcr= 2 su /s = 45o min. value of z (or h)
zcr(or hcr) = max. slope under short term loading
Example 13.1 Infinite slope stability considering
seepage
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seepage
s = ?
Example Infinite slope stability considering seepage
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s = cs = 30oStep 2: Determine max. slope under dry condition
For FS =1.25, the safe slope is s
= 24.8o
Step 3: Determine max. slope under saturated condition
For sand, cs not significantly affected by the wet or dry
condition same s
Example Infinite slope stability considering seepage
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' tan ' tan =
s
sat
a
s = 14o
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Q1: A sand has a fiction angle of 24 . Determine the
maximum slope angle impending failure (assuming infiniteslop failure). What is the safe slope for FS =1.25?
s = = 24o s = ?(1)
o
-1 -1 os 24 =tan tan 19 6
FS 1 25tan tan .
.
(2)
Q2: Discharge from a dam cause tidal variation in the
downstream river For a sand of fiction angle of 25o and
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downstream river. For a sand of fiction angle of 25 and
saturated unit weight of 18.8 kN/m3 that is subjected to thistidal variation (rapid drawdown at low tide), determine the
maximum slope angle (infinite slop failure).
' tan ' tan =s
sat
a
318.8 9.8 9 kN m
-1 os
9 =tan tan25 12 618 8
..
Method of Slices
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Bishops Method (1955):
Assume circular slip plane
Consider only moment equilibrium
Neglect seepage forces
Circular Failure Surface
FBD of a slice
Bishops Method (p578)
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FBD of a slice
Fy = 0
Introduce porewater
pressure ratio
Mo = 0
eq (13.18)
Bishops Method (p578)
' tan( ')T N
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FBD of a slice
Substitute
tan( ')FS =
ff j j j
m j j
T N
T T
' tan( ')
FS
j j
j
NT
Into eq 13.18
Solving for Nj
take eq (13.26)
Bishops Method (p578) FBD of a slice
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Ignoring (Xj Xj+1)
1% error, gives
Bishops equation for ESA
(Effective Stress Analysis):
(13.30)
Groundwater below
surface ru = 0(13.31)
Bishops Method (p578)
Total Stress Analysis (TSA)
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FBD of a slice
cosFS =
sin
j
u jf j
m j j
b
s
W
(13.34)
f= (su)j x lj
Bishops equation for TSA:
y ( )
bj = lj cos j
FS =
sin
u jf j
m j j
s l
W
Effect of tension crack:
Modify the failure surface
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Effect of tension crack on
the slip surface
Tension crack may be filled with water (reduce FS) Provide a channel for water to reach underlying soil layers
Bishops method: Moment caused by
hydrostatic pressure
Method of Slices
Procedures to determine FS of a slope using Bishop method
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Procedures to determine FS of a slope using Bishop method.
(1) Draw the slope to scale including the soil layers
R
O
Phreatic
surface
(2) Draw a trial slip surface and identify its point of rotation (locate the
phreatic surface).
(4) Divide soil mass above the slip
surface into a number of slices. Forease of calculations, try to make as
many as possible of the same width.
About 10 slides are satisfactory for
most hand calculations
1 23
45
6
7
8
(3) If soil is clay, calculate the tension crack depth and sketch in a
possible location of tension crack.
Method of Slices
Procedures to determine FS of a slope using Bishop method.
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(5) Measure b, z and etc. Make a table as shown and record b, z,
zw, and for each slice
R
O
Phreatic
surface
1 23
45
6
7
8
b
equipotential
line for slice 4
zW
z
-+
B width,
z mean height,
unit weight,zw vertical projection of the
equipotential line (porewater
pressure head)
Method of Slices
Procedures to determine FS of a slope using Bishop method.
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RO
Phreatic
surface
1 23
45
6
7
8
b
equipotential
line for slice 4
zW
z
-+
(7) Divide the sum of last column by thesum of column 9 to get FS. If FS is not
equal to the assumed value, reiterate
until FS calculated and FS assumed
are same or within a small tolerance(=0.01)
(6) Calculate W = bz, ru = zw w/ z , assume a value of FS anddetermine mj from following equation or the chart shown in Fig 13.13.complete the calculations for the next 3 columns.
Example 13.3 (p585) - Bishops Method
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Homogenous soil
su 30 kPa
' 33 deg.w 9.8 kN/m3
sat 18 kN/m3
zcr 3.33 m
zs 4 m
FS 1.06 assumed
Example 13.3 (p585) - Bishops Method
(1) Draw the slope to scale
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(1) Draw the slope to scale
(2) Find depth of the tension crack
(3) Divide the slide mass into 9 slices
(4) Set up a spreadsheet
bR
(4) Set up a spreadsheet
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ESA TSA
Slice b z W=bz zw ru mj Wsin W (1 - ru)tan' mj sub/cos
m m kN m deg
1 4.9 1 88.2 1 0.54 -23 1.47 -34.5 38.3 159.7
2 2.5 3.6 162.0 3.6 0.54 -10 1.14 -28.1 54.6 76.2
3 2 4.6 165.6 4.6 0.54 0 1.00 0.0 49.0 60.0
4 2 5.6 201.6 5 0.49 9 0.92 31.5 62.1 60.7
5 2 6.5 234.0 5.5 0.46 17 0.88 68.4 72.2 62.7
6 2 6.9 248.4 5.3 0.42 29 0.85 120.4 80.1 68.6
7 2 6.8 244.8 4.5 0.36 39.5 0.86 155.7 87.6 77.88 2.5 5.3 238.5 2.9 0.30 49.5 0.90 181.4 97.5 115.5
9 1.6 1.6 46.1 0.1 0.03 65 1.02 41.8 29.6 113.6
Sum 536.6 570.9 794.8
FS 1.06 1.48
zzw
For ESA, assume a value of FS first
Taylors Method slope stability for undrained shear strength, su (or cu)
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Simple slopes
Homogeneous
- slopes are rarely homogeneous
Failure occurring by rotation
No surcharge/external loading
No open water outside the slope
Relative depth, nd
Stability number, N0
cos
FS =sin
j
u jf j
m j j
bs
W
(13.34)
Based on TSA and the following equation:
Taylors Charts (1) Calculate nd = D0 / H0
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0ou
FSN H
s
Bedrock
D0
H0
sUnit weight of soil =
Shear strength = cu
FS = FoS(2) Calculate
(3) Read s from the Charts
Example H0 = 10 m, D0 = 3 m, F = 1.25
= 18 kN/m3, su = 30 kPa(1)
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nd = D0/H0 = 3/10 = 0.3
0
1.25 (18 10)30
7.5
o
u
FSN Hs
20
(2)
(3) Read s value at theintersection of nd and No
Which of the following is not true for Taylor s method,
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a) Slope failure depends on soil typeb) Slope failure depends on soil colour
c) Slope failure depends on seepage
d) Slope failure depends on slope geometry
In Bishops method, the failure surface is
a) Any shape
b) Planar
c) Circular d) Non-circular
Which of the following is not true for Taylor s method,
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a) used for Homogeneous soilsb) Not consider surcharge
c) Rotation Failure
d) ESA (Effective Stress Analysis)
Which change can lead to slope failure?
a) addition of vegetation cover on slope surface
b) removal of material from the top of a slope
c) addition of material to the base of a sloped) rise of water table in slope material
Bearing Capacity of Soils and
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Settlement of Shallow Foundations
Muni Budhu - Soil Mechanics and Foundations
2nd edition
Dr Jie Li
VERTICAL CENTRIC LOADS
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Short-Term (TSA): qu = 5.14suscdc
Long Term (ESA):
qu = Df(Nq 1) sq dq + 0.5 B'N s d
INCLINED LOADS
Short-Term (TSA): qu = 5.14suic
Long Term (ESA):qu = Df(Nq 1) iq + 0.5 B' N i
Nq
and Nrare all functions of (effective!)
GEOMETRIC FACTORS
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Choice of Strengths
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SAND
Clean sand
c = 0 = c
, ,
CLAY
Saturated, NC
undrained (Short-Term) loading
u = 0
usually more critical
drained (Long-Term) loading
c,
Problem Solving 9.6 A square footing, 3m wide, is located 1.5mbelow the surface of a stiff clay. Determine the allowable bearing
capacity for short term condition if su 100 kPa and sat 20 kN/m3
If
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capacity for short term condition if su= 100 kPa and sat = 20 kN/m Ifthe footing were located on the surface, what would be the allowablebearing capacity? Use FS = 3.
c
B
s 1 0 2 1 0 2 1 2L. . . dc = 1+0.33 tan-1
(Df/B'), take as 1
u u c cq 5 14 s s d 5 14 100 1 2 1 617kPa. . .
Position 1 Df= 1.5 m:
u
a f
q 617q D 20 1 5 236kPa
FS 3.
Position 2 Df= 0 m:
u
a f
q 617q D 20 0 206kPa
FS 3
Problem Solving 9.7 A column carrying a load of 750 kN is to befounded on a square footing at a depth of 2 m below the ground surface
in deep clay stratum. What will be the size of the footing for FS = 3 for
ESA and TSA? The soil properties are = 28o = 18 5 kN/m3 and
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ESA and TSA? The soil properties are p = 28o, sat = 18.5 kN/m3, andsu= 55 kPa. The ground water level is at the base of the footing but it is
expected to rise to the ground surface during rainy seasons.
c
B 2 4s 1 0 2 1 0 2 1 2
L 2 4
.. . .
. dc = 1
Check both short and long-term cases
(a) Short-Term (TSA)
Assume B = 2.4 m
u u c cq 5 14 s s d 5 14 55 1 2 1 339kPa. . .
,
Applied stress = a750
130 2 kPa2 4 2 4 .. .
u
a f
q 339FS 3 0
D 130 2 8 7 2.
' . .
tan 28 2 'tan (45 / 2) 14.7q pN e
(b) Long Term (ESA ) Problem Solving 9.7
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N = 0.1054 exp(9.6p) = 11.5
u 1q 8 7 2 0 13 7 1 53 1 21 8 7 2 4 11 5 0 6 1 0 513kPa2
. . . . . . . . . .
u
a f
q 513FS 4 6
D 130 2 8 7 2
.
' . .
TSA governs, use B = 2.4 m
sq = 1 + B/Ltan(p)=1+2.4/2.4 tan (28) =1.53
Use Davis & Booker expression (p351):
dr= 1
sr= 1 0.4 B/L= 1 0.4 x 2.4/2.4 = 0.6
dq = 1 + 2tan(p)(1-sin(p))2 tan-1(Df/B) =1.21
(Table 9.1 p351)
2nd edition
1) Which analysis should be used to calculate the bearing capacity of
a fine-grained soil?,
A) TSA
B) ESA
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B) ESA
C) Both TSA and ESA
D) Finite Element Method
2) Which analysis should be used to calculate the bearing capacity of
a coarse-grained soil?,
A) TSA
B) ESAC) Both TSA and ESA
D) Finite Element Method
TSA: qu = 5.14 suscdc
ESA: qu = Df(Nq 1) sq dq + 0.5 BN s dN
qand N
rare all functions of (effective!)