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Earth Pressures on Retaining walls
Rankine’s Method
Modes of failure
F
Modes of failure
F
The force F may be provided by
• friction at the base (gravity retaining walls)
Modes of failure
F
The force F may be provided by
• friction at the base (gravity retaining walls)
•
founding the wall into the ground (sheet retaining walls)
Modes of failure
F
The force F may be provided by
• friction at the base (gravity retaining walls)
•
founding the wall into the ground (sheet retaining walls)
•
anchors and struts
Modes of failure
F
The force F may be provided by
• friction at the base (gravity retaining walls)
•
founding the wall into the ground (sheet retaining walls)
•
anchors and struts
•
external loads
Modes of failure
1. ACTIVE Failure
If the force F is too small failure of the wall will occur with soil pushing
the wall out.
F
direction of soil
movement
Modes of failure
1. ACTIVE Failure
If the force F is too small failure of the wall will occur with soil pushing
the wall out.
F
direction of soil
movement
For most retaining walls active failure is the primary concern.
Modes of failure
1. PASSIVE Failure
If the force F is too large failure of the wall will occur with the wall
pushing into the soil.
F
direction of soil
movement
Modes of failure
1. PASSIVE Failure
If the force F is too large failure of the wall will occur with the wall
pushing into the soil.
F
direction of soil
movement
This mode of failure is usually only relevant when large external
forces are being applied.
Modes of failure
1. PASSIVE Failure
If the force F is too large failure of the wall will occur with the wall
pushing into the soil.
F
direction of soil
movement
This mode of failure is usually only relevant when large external
forces are being applied.
However, local passive conditions may occur if any part of the wall
moves towards the soil.
Rankine’s theory
• Assume that the wall is frictionless
• The normal stress acting on the wall will thus be a principal
stress
• If the wall is vertical and the soil surface horizontal the vertical
and horizontal stresses throughout the retained soil mass will be
the principal stresses
• The vertical stress may then be calculated in the usual way
Rankine’s theory
• Assume that the wall is frictionless
• The normal stress acting on the wall will thus be a principal
stress
• If the wall is vertical and the soil surface horizontal the vertical
and horizontal stresses throughout the retained soil mass will be
the principal stresses
• The vertical stress may then be calculated in the usual way
d1
d2
g1
g2
z
The vertical total stress at depth z
is given by
v  g 1 d1  g 2 ( z  d1 )
Rankine’s theory
• Assume that the horizontal stress can be calculated from the
failure criterion. That is the retained soil is assumed to be
everywhere at failure.
Rankine’s theory
• Assume that the horizontal stress can be calculated from the
failure criterion. That is the retained soil is assumed to be
everywhere at failure.
• From the Mohr-Coulomb failure criterion we have
1  N  3  2 c N
Rankine’s theory
• Assume that the horizontal stress can be calculated from the
failure criterion. That is the retained soil is assumed to be
everywhere at failure.
• From the Mohr-Coulomb failure criterion we have
1  N  3  2 c N
• For active failure the horizontal stress will reduce to its
minimum value. That is h = 3 the minimum principal stress,
and v = 1 .
Rankine’s theory
• Assume that the horizontal stress can be calculated from the
failure criterion. That is the retained soil is assumed to be
everywhere at failure.
• From the Mohr-Coulomb failure criterion we have
1  N  3  2 c N
• For active failure the horizontal stress will reduce to its
minimum value. That is h = 3 the minimum principal stress,
and v = 1 .
• The minimum (Active) horizontal stress is then
 h min 
v  2 c N
N
Rankine’s theory
• For passive failure the horizontal stress will increase to its
maximum value. That is h = 1 the maximum principal
stress, and v = 3.
Rankine’s theory
• For passive failure the horizontal stress will increase to its
maximum value. That is h = 1 the maximum principal
stress, and v = 3.
• The maximum (Passive) horizontal stress is then
 h max  N   v  2 c N 
Rankine’s theory
• For passive failure the horizontal stress will increase to its
maximum value. That is h = 1 the maximum principal
stress, and v = 3.
• The maximum (Passive) horizontal stress is then
 h max  N   v  2 c N 
• If the vertical stress stays constant the horizontal stress is
bounded by the active and passive values.
Rankine’s theory
• For passive failure the horizontal stress will increase to its
maximum value. That is h = 1 the maximum principal
stress, and v = 3.
• The maximum (Passive) horizontal stress is then
 h max  N   v  2 c N 
• If the vertical stress stays constant the horizontal stress is
bounded by the active and passive values.
• In the Rankine method a stress state is found that is in
equilibrium with the applied loads and has the soil at failure. In
plasticity theory this approach is referred to as a lower bound
method, a method which can be shown to produce safe,
conservative solutions.
Rankine’s theory
The relation between the active and passive pressures may be
shown graphically by considering Mohr circles.
t
t  c   tan 
h
v

Rankine’s theory
The relation between the active and passive pressures may be
shown graphically by considering Mohr circles.
t
t  c   tan 
hmin
v

Rankine’s theory
The relation between the active and passive pressures may be
shown graphically by considering Mohr circles.
t
t  c   tan 
hmin
v
hmax

Total Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
total stress, using undrained parameters cu, u.
Total Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
total stress, using undrained parameters cu, u.
A total stress analysis is only appropriate if soil remains undrained. It
can only be used in the short term for soils with low permeabilities.
Total Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
total stress, using undrained parameters cu, u.
A total stress analysis is only appropriate if soil remains undrained. It
can only be used in the short term for soils with low permeabilities.
For the undrained active failure of a wall we have
h 
v  2 cu
N
N
Total Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
total stress, using undrained parameters cu, u.
A total stress analysis is only appropriate if soil remains undrained. It
can only be used in the short term for soils with low permeabilities.
For the undrained active failure of a wall we have
h 
where
N
=
v  2 cu
N
N
1 + sin  u
1 - sin  u
and for a homogeneous soil layer
v = gsat z
Total Horizontal Stress
 2 cu
N
H
z
cu, u, gsat
g sat H  2 c u
N
N
Total Horizontal Stress
 2 cu
N
H
z
cu, u, gsat
g sat H  2 c u
N
N
If u is non-zero this implies that the undrained strength varies with
depth. The soil must be saturated to use a total stress (undrained)
analysis.
Tension cracks
• The analysis indicates negative, tensile, stresses at the surface.
• Soil particles cannot provide tension
• The negative stresses have to come from suctions in the pore
water
• It is difficult to rely on the tensile forces and they are usually
ignored
• The tensile stresses reduce the force required for stability of the
wall. Ignoring the tensile stresses therefore gives a more
conservative solution.
Tension cracks
The horizontal stress distribution becomes
z0
z
H
cu, u
gsat
g sat H  2 c u
N
N
Tension cracks
The horizontal stress distribution becomes
z0
z
H
cu, u
gsat
g sat H  2 c u
N
The depth of the tensile region z0 may be determined from
h = 0
v  2 cu
N
2 cu
N
z0 
g sat
 g sat z 0
N
Tension cracks
In the tensile region a crack can develop.
Tension cracks
In the tensile region a crack can develop.
If water is available it can fill the crack, and reduce the stability of the
wall. The horizontal stresses on the wall become.
Tension cracks
In the tensile region a crack can develop.
If water is available it can fill the crack, and reduce the stability of the
wall. The horizontal stresses on the wall become.
Water
z0
gw z0
Soil
g sat H  2 c u
N
N
Effective Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
effective stress, using effective parameters c’, ’.
Effective Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
effective stress, using effective parameters c’, ’.
An effective stress analysis is always appropriate, irrespective of the
drainage conditions. To perform an effective stress analysis the pore
water pressures must be known. This usually limits effective stress
analysis to investigation of long term stability.
Effective Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
effective stress, using effective parameters c’, ’.
An effective stress analysis is always appropriate, irrespective of the
drainage conditions. To perform an effective stress analysis the pore
water pressures must be known. This usually limits effective stress
analysis to investigation of long term stability.
For active failure of a wall we have
 h 
 v  2 c  N 
N
Effective Stress Analysis
The Mohr-Coulomb failure criterion must be expressed in terms of
effective stress, using effective parameters c’, ’.
An effective stress analysis is always appropriate, irrespective of the
drainage conditions. To perform an effective stress analysis the pore
water pressures must be known. This usually limits effective stress
analysis to investigation of long term stability.
For active failure of a wall we have
 h 
where
and
N
=
 v  2 c  N 
N
1 + sin  
1 - sin  
’v = v - u
Effective Stress Analysis
Consider active failure of a wall retaining dry sand
 2 c
N
H
z
c’, ’, gdry
g dry H  2 c N 
N
Effective Stress Analysis
• The use of values of c’, ’ associated with peak failure leads to
the prediction of impossible tensile stresses.
• It is generally more appropriate and safer to use ultimate or
critical state parameters, c’ = 0, ’ = ’ult
• Using the critical state parameters a larger active force is
required for wall stability, and hence a safer estimate is obtained
Effective Stress Analysis
• The use of values of c’, ’ associated with peak failure leads to
the prediction of impossible tensile stresses.
• It is generally more appropriate and safer to use ultimate or
critical state parameters, c’ = 0, ’ = ’ult
• Using the critical state parameters a larger active force is
required for wall stability, and hence a safer estimate is obtained
c’ , ’
c’ = 0, ’ = ’ult
Effective Stress Analysis
• For passive failure the horizontal stresses on a wall retaining dry
sand are given by
 h  g dry z N   2 c  N 
Effective Stress Analysis
• For passive failure the horizontal stresses on a wall retaining dry
sand are given by
 h  g dry z N   2 c  N 
• In this case the critical state parameters c’ = 0, ’ = ’ult give a
smaller force. However, this is a safe, conservative, estimate of
the maximum force the soil can support.
Effective Stress Analysis
• For passive failure the horizontal stresses on a wall retaining dry
sand are given by
 h  g dry z N   2 c  N 
• In this case the critical state parameters c’ = 0, ’ = ’ult give a
smaller force. However, this is a safe, conservative, estimate of
the maximum force the soil can support.
• It is important to use effective vertical stresses, v’ = v - u to
calculate the effective horizontal stresses, h’. Then the total
horizontal stress is given by h = h’ - u
Effective Stress Analysis
• For passive failure the horizontal stresses on a wall retaining dry
sand are given by
 h  g dry z N   2 c  N 
• In this case the critical state parameters c’ = 0, ’ = ’ult give a
smaller force. However, this is a safe, conservative, estimate of
the maximum force the soil can support.
• It is important to use effective vertical stresses, v’ = v - u to
calculate the effective horizontal stresses, h’. Then the total
horizontal stress is given by h = h’ - u
• If the water level is not the same on both sides of a wall, water
will flow. The pore pressures must then be determined from a
flow net before calculating v’.
Example
A 10 m high retaining wall retains 5 m of clay which overlays 3 m
of sand which overlays 2 m of clay. The water table is at the surface
of the retained soil. Calculate the limiting active pressure
immediately after construction.
5m Clay
cu = 20 kPa
u = 5o
gsat = 15 kN/m3
3m
Sand
c’ = 0
’= 35o
gsat =20kN/m3
Clay
cu = 50 kPa
u = 0o
gsat = 15 kN/m3
2m
Example - Short term analysis
Layer 1: A clay layer so will be undrained in the short term. Will
require a total stress (undrained) analysis
Example - Short term analysis
Layer 1: A clay layer so will be undrained in the short term. Will
require a total stress (undrained) analysis
c  c u  20 kPa
N 
1  sin  u
 119
.
1  sin  u
Example - Short term analysis
Layer 1: A clay layer so will be undrained in the short term. Will
require a total stress (undrained) analysis
c  c u  20 kPa
N 
1  sin  u
 119
.
1  sin  u
Active failure thus 1 = v and 3 = h
Example - Short term analysis
Layer 1: A clay layer so will be undrained in the short term. Will
require a total stress (undrained) analysis
c  c u  20 kPa
N 
1  sin  u
 119
.
1  sin  u
Active failure thus 1 = v and 3 = h
From the Mohr-Coulomb failure criterion
h 
v  2 cu
N
N

 v  43.6
119
.
Example - Short term analysis
Layer 1: A clay layer so will be undrained in the short term. Will
require a total stress (undrained) analysis
c  c u  20 kPa
N 
1  sin  u
 119
.
1  sin  u
Active failure thus 1 = v and 3 = h
From the Mohr-Coulomb failure criterion
h 
v  2 cu
At surface
N
N

 v  43.6
119
.
z = 0, v = 0, h = - 36.6 kPa
At base of layer z = 5 m, v = 5x15, h = 26.4 kPa
Example - Short term analysis
-36.6
z0 = 2.91 m
Analysis predicts tensile stresses between the soil
and the wall. These are not possible, and a
tension crack may develop.
26.4
Example - Short term analysis
-36.6
z0 = 2.91 m
Analysis predicts tensile stresses between the soil
and the wall. These are not possible, and a
tension crack may develop.
26.4
Because the water table is at the surface the crack
will fill with water, and a more pessimistic
pressure distribution will be
Example - Short term analysis
-36.6
z0 = 2.91 m
Analysis predicts tensile stresses between the soil
and the wall. These are not possible, and a
tension crack may develop.
26.4
gw z
9.81 x 2.91
26.4
Because the water table is at the surface the crack
will fill with water, and a more pessimistic
pressure distribution will be
Example - Short term analysis
Layer 2: Sand so excess pore pressures will dissipate rapidly.
Cannot use total stress analysis.
Example - Short term analysis
Layer 2: Sand so excess pore pressures will dissipate rapidly.
Cannot use total stress analysis.
For sand in short term assume fully drained. Must use
effective stress analysis.
c  c  0
N
1  sin  

 3.69
1  sin  
Example - Short term analysis
Layer 2: Sand so excess pore pressures will dissipate rapidly.
Cannot use total stress analysis.
For sand in short term assume fully drained. Must use
effective stress analysis.
c  c  0
N
1  sin  

 3.69
1  sin  
Active failure so ’1 = ’v and ’3 = ’h and from the
Mohr-Coulomb criterion
 h 
 v  2 c  N 
N
 v

3.69
Example - Short term analysis
Layer 2
z
v
u
´v = v - u
´h = ´v/3.69
u
h = ´h + u
5
75
49
26
7
49
56
8
135 78.4
56.6
15.3
78.4
93.7
Example - Short term analysis
Layer 2
z
v
u
´v = v - u
´h = ´v/3.69
u
h = ´h + u
5
75
49
26
7
49
56
8
135 78.4
56.6
15.3
78.4
93.7
Note that most of horizontal pressure is due to water
Example - Short term analysis
Layer 3: Clay, therefore total stress (undrained) analysis for
short term
c  c u  50 kPa
N 
1  sin  u
 1
1  sin  u
Example - Short term analysis
Layer 3: Clay, therefore total stress (undrained) analysis for
short term
c  c u  50 kPa
N 
1  sin  u
 1
1  sin  u
When u = 0 the Mohr-Coulomb criterion reduces to
1 = 3 + 2 c u
Example - Short term analysis
Layer 3: Clay, therefore total stress (undrained) analysis for
short term
c  c u  50 kPa
N 
1  sin  u
 1
1  sin  u
When u = 0 the Mohr-Coulomb criterion reduces to
1 = 3 + 2 c u
z
v
h
8
135
35
10
165
65
Example - Short term analysis
The final pressure distribution is
2.91
28.5
26.4
2.09
56
3
93.7
35
2
65
Example - Short term analysis
The final pressure distribution is
2.91
The force required to prevent active
failure can be determined from the
pressure diagram
2.09
F = 0.5x28.5x2.91
28.5
26.4
56
+ 0.5x26.4x2.09
3
+ 56x3 + 0.5x(93.7-56)x3
93.7
+ 35x2 + 0.5x(65-35)x2
35
2
65
= 393.7 kN/m
Example 2
A 5m high retaining wall retains a clayey soil, which overlies a highly
permeable sandstone. If the water level remains at the surface of the
clay in the retained soil, and is level with the top of the sandstone
determine the minimum force required to maintain the stability of the
wall for short and long term. The soil parameters are:
c u  37 kPa ,  u  5o , c   0,  ult  25o , g sat  19 kN / m 3
Clayey soil
Sandstone
5m
Example 2 - Short term
Short term undrained - total stress analysis
Example 2 - Short term
Short term undrained - total stress analysis
Minimum force for stability - active failure
Example 2 - Short term
Short term undrained - total stress analysis
Minimum force for stability - active failure
h 
v  2 cu
N
N
v

 67.8
119
.
Example 2 - Short term
Short term undrained - total stress analysis
Minimum force for stability - active failure
h 
v  2 cu
N
N
v

 67.8
119
.
At surface h = - 67.8 kPa, at 5 m h = 11.9 kPa
Example 2 - Short term
Short term undrained - total stress analysis
Minimum force for stability - active failure
h 
v  2 cu
N
N
v

 67.8
119
.
At surface h = - 67.8 kPa, at 5 m h = 11.9 kPa
Allowing for tension crack filling with water, pressures acting on the
wall will be
zo = 4.25 m
4.25x 9.81
11.9
Example 2 - Short term
Short term undrained - total stress analysis
Minimum force for stability - active failure
h 
v  2 cu
N
N
v

 67.8
119
.
At surface h = - 67.8 kPa, at 5 m h = 11.9 kPa
Allowing for tension crack filling with water, pressures acting on the
wall will be
zo = 4.25 m
F 
4.25x 9.81
11.9
1
1
 9.81  4.252   11.9  0.75  931
. kN / m
2
2
Example 2 - Long term
Long term - Effective stress analysis
Pore pressures required - To be determined from a flow net
Example 2 - Long term
Long term - Effective stress analysis
Pore pressures required - To be determined from a flow net
5m
Example 2 - Long term
Long term - Effective stress analysis
Pore pressures required - To be determined from a flow net
5m
u  gw (h  z)
Example 2 - Long term
Long term - Effective stress analysis
Pore pressures required - To be determined from a flow net
X
5m
u  gw (h  z)
Take datum at base of clay, then at X
h = ho - Dh = 5 - (5/3)x1 = 10/3
Example 2 - Long term
Long term - Effective stress analysis
Pore pressures required - To be determined from a flow net
X
5m
u  gw (h  z)
Take datum at base of clay, then at X
h = ho - Dh = 5 - (5/3)x1 = 10/3
z = (2/3)x5 = 10/3
u=0
Example 2 - Long term
Effective stress analysis with c’ = 0, ’ = 25o
 h 
 v  2 c  N 
N
Now u = 0, so ’v = v = gsat z
At wall base h = ’h = 38.6 kPa
Hence F = 0.5 x 38.6 x 5 = 96.4 kN/m
 v

2.46
Coulomb’s Method
Failure mechanism
In Coulomb’s method a mechanism of failure has to be assumed
soil movement
wall
movement
Assumed
failure plane
Failure mechanism
In Coulomb’s method a mechanism of failure has to be assumed
soil movement
wall
movement
Assumed
failure plane
If this is the failure mechanism then the Mohr-Coulomb failure
criterion must be satisfied on the assumed failure planes
Limit equilibrium method
• The application of the failure criterion to assumed mechanisms
of failure is widely used in geotechnical engineering. This is
generally known as the limit equilibrium method.
• It is not a rigorous theoretical method but is used because it
gives simple and reasonable estimates of collapse.
• The method has advantages over Rankine’s method
– it can cope with any geometry
– it can cope with applied loads
– friction between soil and retaining walls (and other structural
elements) can be accounted for
Failure criterion
For any point on the failure plane we have
t
=
c +  tan 
Failure criterion
For any point on the failure plane we have
t
=
c +  tan 
If analysis is of undrained stability then the failure criterion must
be expressed in terms of total stress using undrained parameters cu
and u
t
=
c u +  tan  u
Failure criterion
For any point on the failure plane we have
t
=
c +  tan 
If analysis is of undrained stability then the failure criterion must
be expressed in terms of total stress using undrained parameters cu
and u
t
=
c u +  tan  u
If the pore pressures are known or the soil is dry an effective
stress analysis can be conducted and the failure criterion must be
expressed in terms of effective stress and effective strength
parameters c’, ’
t
=
c  +   tan  
Failure criterion
direction of
soil
movement
t

Assumed
failure
plane
Failure criterion
direction of
soil
movement
t

Forces on the failure plane
Shear Force
T =
 t ds
Assumed
failure
plane
Failure criterion
direction of
soil
movement
t

Forces on the failure plane
Shear Force
T =
Normal Force
N =
 t ds
  ds
Assumed
failure
plane
Failure criterion
direction of
soil
movement
t

Forces on the failure plane
Shear Force
T =
Normal Force
N =
Cohesive Force C =
 t ds
  ds
 c ds
Assumed
failure
plane
Failure criterion
If the soil properties are constant
T
=
C + N tan 
Failure criterion
If the soil properties are constant
T
=
C + N tan 
The forces acting on the failure plane are
T
N
Failure criterion
If the soil properties are constant
T
=
C + N tan 
The forces acting on the failure plane are
T
N
which may be more convieniently
represented by
C

R
Failure criterion
For a wedge of soil failing as shown below there has to be relative
movement between the wall and the soil, and the soil must be failing on
this plane.
soil movement
wall
movement
Assumed
failure planes
Failure criterion
For a wedge of soil failing as shown below there has to be relative
movement between the wall and the soil, and the soil must be failing on
this plane.
soil movement
wall
movement
Assumed
failure planes
The failure criterion between the wall and the soil may be written
t
=
c w +  tan  w
Failure criterion
For a wedge of soil failing as shown below there has to be relative
movement between the wall and the soil, and the soil must be failing on
this plane.
soil movement
wall
movement
Assumed
failure planes
The failure criterion between the wall and the soil may be written
t
c w +  tan  w
=
or in terms of forces
Tw
=
C w + N w tan  w
Total Stress Analysis
• A total stress analysis is only valid if the soil is saturated and
does not drain
• In practice this generally means total stress analysis is
limited to assessment of the short term stability of clayey
soils
• Must use total stresses and undrained parameters cu, u
Total Stress Analysis
H tan q
dir. of soil
movement
H
H sec q
q
Soil properties
cu, u
Soil-wall properties
cw, w
Total Stress Analysis - forces consistent with mechanism
C2
W
C1
w
u
R2
q
R1
C 1 = c u H sec q
C2 = c w H
2
W = ½ H tan q g
Total Stress Analysis - polygon of forces
W
q
C2
C1
Total Stress Analysis - polygon of forces
w
90  q  
W
q
C2
C1
u
Total Stress Analysis - polygon of forces
w
R2
R1
90  q  
W
q
C2
C1
u
Total Stress Analysis - polygon of forces
In the polygon of forces
arrows must all be in
the same direction as
you move around the
polygon. A check must
be made that the
indicated directions are
consistent with the
failure mechanism
w
R2
R1
90  q  
W
q
C2
C1
u
Total Stress Analysis - polygon of forces
R2
w
R1
90  q   u
C2
W
q
C1
Total Stress Analysis - polygon of forces
R2
w
R1
90  q   u
Direction of R2 is not
consistent with
assumed mechanism.
Therefore the
mechanism is not
valid.
C2
W
q
C1
Total Stress Analysis - forces on the wall
The polygon gives the forces acting on the soil wedge. Equal and
opposite forces act on the wall. The forces acting on the wall will be
Total Stress Analysis - forces on the wall
The polygon gives the forces acting on the soil wedge. Equal and
opposite forces act on the wall. The forces acting on the wall will be
w
R2
F total
C2
Total Stress Analysis - forces on the wall
The polygon gives the forces acting on the soil wedge. Equal and
opposite forces act on the wall. The forces acting on the wall will be
H
w
R2
F total
F total
C2
V
Total Stress Analysis - forces on the wall
The polygon gives the forces acting on the soil wedge. Equal and
opposite forces act on the wall. The forces acting on the wall will be
H
w
R2
F total
F total
C2
V
For retaining walls the horizontal component of force is usually of
most concern. For the active failure condition different values of q
need to be tried to determine the maximum value of H.
Total Stress Analysis - tension cracks
As with Rankine’s method allowance must be made for tension
cracks, and if water is present the possibility that these cracks may
fill with water.
Total Stress Analysis - tension cracks
As with Rankine’s method allowance must be made for tension
cracks, and if water is present the possibility that these cracks may
fill with water.
The depth, z, of the region affected by tension cracks can be
determined from Rankine’s method. For active failure this gives
h 
g z  2 cu
N
N
 0
Total Stress Analysis - tension cracks
As with Rankine’s method allowance must be made for tension
cracks, and if water is present the possibility that these cracks may
fill with water.
The depth, z, of the region affected by tension cracks can be
determined from Rankine’s method. For active failure this gives
h 
z
g z  2 cu
N
N
=
2 cu
N
g
 0
Total Stress Analysis - tension cracks
z
H
q
Total Stress Analysis - tension cracks
W1
C2
*
W2
C1
w
*
R2
q
*
R1
u
*
Total Stress Analysis - tension cracks
If the tension cracks fill with water this has no influence on the
polygon of forces.
Total Stress Analysis - tension cracks
If the tension cracks fill with water this has no influence on the
polygon of forces.
The water will provide an additional horizontal force on the wall
U
*
R2
*
C2
F total
Total Stress Analysis - Example 1
V
10
o
W
Soil Properties
dir of soil
movement
5m
6.4 m
30
U
o
c u = 10 kPa
o
 u = 10
c w = 2 kPa
o
=
20
w
g = 20 kN/m
3
Total Stress Analysis - Example 1
V
W
Cuw
Cuv
W
o
20
Ruw
o
10
o
30
Ruv
U
Total Stress Analysis - Example 1
20 o
Ruw = 60 kN/m
50 o
160
10
30
o
64
Effective Stress Analysis
An effective stress analysis can be performed whenever the pore
pressures are known.
Effective Stress Analysis
An effective stress analysis can be performed whenever the pore
pressures are known.
In practice this usually means that effective stress analysis can only
be used to assess the long term stability
Effective Stress Analysis
An effective stress analysis can be performed whenever the pore
pressures are known.
In practice this usually means that effective stress analysis can only
be used to assess the long term stability
The failure criterion must be written in terms of effective stress,
that is
t
=
c  +   tan  
Effective Stress Analysis
An effective stress analysis can be performed whenever the pore
pressures are known.
In practice this usually means that effective stress analysis can only
be used to assess the long term stability
The failure criterion must be written in terms of effective stress,
that is
t
=
c  +   tan  
In terms of forces this becomes
T
=
C  + N  tan  
Effective Stress Analysis
An effective stress analysis can be performed whenever the pore
pressures are known.
In practice this usually means that effective stress analysis can only
be used to assess the long term stability
The failure criterion must be written in terms of effective stress,
that is
t
=
c  +   tan  
In terms of forces this becomes
T
=
C  + N  tan  
where N’ = N - U
U = force due to water pressure on failure plane
Effective Stress Analysis - Forces on failure plane
T
Failure plane
N´
U
Effective Stress Analysis - Forces on failure plane
T
Failure plane
N´
U
C´
Failure plane
´
R´
U
Effective Stress Analysis
• When performing effective stress stability calculations the
critical state parameters c’ = 0, ’ = ’ult should be used
• When the soil is dry the pore pressures everywhere will be zero,
and the effective stresses will equal the total stresses. However,
only an effective stress analysis is appropriate.
• If sliding occurs between the soil and a wall appropriate
effective stress failure parameters must be used. The effective
parameters between any interface (eg. a wall) and the soil
should be based on the ultimate conditions so that c´w = 0, ´w=
´wult
Effective Stress Analysis
• In using Coulomb’s method you have to assume a failure
mechanism. However, this may not be the most critical (least
safe) mechanism. Therefore, you need to investigate a number
of mechanisms (values of q) to determine which will be the
most critical.
• For Active failure the Maximum force is needed (Maximum of
Minimum)
• For Passive failure the Minimum force is needed (Minimum of
Maximum)
• The most critical mechanism is unlikely to give an accurate
estimate of the failure load, because observation of real soil
shows failure rarely occurs on planar surfaces.
Effective Stress Analysis
• To select a q value for the assumed failure plane in the soil it is
helpful to remember that the failure plane is inclined at an angle
(p/4 - /2) to the direction of the minor principal stress 3.
Effective Stress Analysis
• To select a q value for the assumed failure plane in the soil it is
helpful to remember that the failure plane is inclined at an angle
(p/4 - /2) to the direction of the minor principal stress 3.
t
F
2

3

Effective Stress Analysis
3
Active

Effective Stress Analysis
3
Active
Passive


3
Effective Stress Analysis
3
Active
Passive


3
• In the presence of steady state seepage it may be necessary
to draw a flow net to determine the pore water Forces U
acting on the soil wedge.
Active failure
W
V
C´uw
C´uv
W
U uw
 ´w
R´uw
q
U
´
R´uv
U uv
Direction of
movement of soil
wedge
Passive failure
V
W
R´ uw
U uw
W
´ w
R´uv
´
C´ uw
q
U
C´ uv
U uv
Effective stress analysis - Example
V
W
X
Soil
Water
Water
m
55 m
6.4 m
30
U
o
10
o
W.T.
Effective stress analysis - Example
V
W
X
Soil
5m
6.4 m
30
U
o
Effective stress analysis - Example
V
W
X
Soil
5m
6.4 m
30
U
o
Soil Properties
c´ = 5 kPa
o
´
=
10

c´ w = 2 kPa
o
´
=
20
 w
gdry = 20 kN/m
gsat = 22 kN/m
3
3
Effective stress analysis - Example
V
W
X
Soil
5m
6.4 m
30
o
U
C´ uv = 5  6.4 = 32 kN/m
C´ uw = 2  5 = 10 kN/m
Soil Properties
c´ = 5 kPa
o
´
=
10

c´ w = 2 kPa
o
´
=
20
 w
gdry = 20 kN/m
gsat = 22 kN/m
3
3
Effective stress analysis - Example
V
W
X
Soil
5m
6.4 m
30
o
Soil Properties
c´ = 5 kPa
o
´
=
10

c´ w = 2 kPa
o
´
=
20
 w
gdry = 20 kN/m
gsat = 22 kN/m
3
3
U
C´ uv = 5  6.4 = 32 kN/m
C´ uw = 2  5 = 10 kN/m
W = 0.5  5  2.89  22 + (8 - 0.5
 5  2.89)  20
= 174.5 kN/m
Example - Water pressures on soil wedge
V
W
X
U
u  5  9.81
Example - Water pressures on soil wedge
V
W
X
U
u  5  9.81
U uv  0.5  5  9.81  5.77  1415
. kN / m
U uw  0.5  5  9.81  5
 122.5 kN / m
Example - Forces on soil wedge
C´uw
Uuw
W
C´uv
R´uw
Uuv
R´uv
Example - Polygon of forces
Cúw
Uuw
R´uw
W
Cúv
Uuv
U uw
R´uv
U uv
60
o
C´ uw
W
30
o
C´ uv
Example - Polygon of forces
Cúw
Uuw
R´uw
W
Cúv
Uuv
U uw
60
o
R´uv
U uv
60
o
C´ uw
W
30
o
C´ uv
Example - Polygon of forces
o
20
Cúw
Uuw
R´uw
W
Cúv
Uuv
Uuw
o
60
R´uv
Uuv
o
60
C´ uw
W
o
30
C´ uv
Example - Forces on the wall
R´uw
U uw
C´uw
Example - Forces on the wall
R´uw
U uw
C´uw
The vertical and horizontal components of the force on the wall are
Vertical
T
uw
= R´ uw sin ´ w + C´ uw = 5.8 + 10 = 15.8 kN/m
Example - Forces on the wall
R´uw
U uw
C´uw
The vertical and horizontal components of the force on the wall are
Vertical
T
uw
= R´ uw sin ´ w + C´ uw = 5.8 + 10 = 15.8 kN/m
Horizontal
N uw = R´ uw cos ´ w + U
uw
= 15.97 + 122.5 = 138.5
kN/m
Example - Forces on the wall
R´uw
U uw
C´uw
The vertical and horizontal components of the force on the wall are
Vertical
T
uw
= R´ uw sin ´ w + C´ uw = 5.8 + 10 = 15.8 kN/m
Horizontal
N uw = R´ uw cos ´ w + U
uw
= 15.97 + 122.5 = 138.5
kN/m
Note that N uw is largely due to water pressure. However, due to water on
the other side of the wall the net resistance required for stability is only
15.97 kN/m
Example 3
Example 3
uD  g w ( hD  zD )
Example 3
uD  g w ( hD  zD )
Choosing a local datum at E: hD = 0
Example 3
x
uD  g w ( hD  zD )
Choosing a local datum at E: hD = 0, zD = - x and uD = + gw x
Example 3
The pore pressure at several points along the failure plane needs to be
determined to evaluate the force due to the water pressures.
Example 3
The pore pressure at several points along the failure plane needs to be
determined to evaluate the force due to the water pressures.
The forces on the assumed failing soil wedge are then
W
2
U w = 0.5 g w H w
 ´ cs
U
From
flow net
´ w
R´
R´ w
K a  tg 2  450   
2

K p  tg 2  450   
2

AÙp löïc chuû ñoäng vaø bò ñoäng cuûa ñaát rôøi
Thay ñoåi tyû soá aùp löïc ñaát ngang vaø ñöùng theo bieán daïng ngang trong thí nghieäm
neùn ba truïc
K a  tg 2  450   
2

K p  tg 2  450   
2

Heä soá aùp löïc ñaát trong tröôøng hôïp maët ñaát nghieâng
Heä soá aùp löïc ñaát trong tröøong hôïp löng töôøng nhaùm
Aùp löïc chuû ñoäng vaø bò ñoäng theo lyù thuyeát Coulomb
Ka 
Kp 
sin 2    

sin    sin     
sin 2  sin    1 





sin



sin





sin 2    
2

sin    sin     
sin 2  sin    1 





sin



sin





2
Phöông phaùp veõ Culmann
Aùp löïc ñaát theo lyù thuyeát Rankine
Toång aùp löïc taùc ñoäng leân töôøng F = ½ (gw +Ka g’)H2
Caùc phöông phaùp thöôøng söû duïng
Tröøong hôïp coù nöôùc chaûy
Töôøng baûn coù doøng nöôùc
Xeùt taûi troïng ñoäng
Ea  [1 / 2]g H 2 (1  kv ) K a
cos 2 (  q   )
Ka 

sin(    ) cos(  q   ) 
cos q cos 2  cos(    q ) 1 

cos(




q
)
cos(



)


 kh 

1

k
v 

q  arctg 
Ñaøo töôøng ñuùc beâ toâng taïi choå
Chuyeån vò töôøng vaø traïng thaùi giôùi haïn cuûa ñaát
Traïng thaùi moment trong caùc giai ñoaïn thi coâng
Aùp löïc ñaát coù xeùt ñeán söï ñaàm chaët ñaát treân maët
HEÄ SOÁ AN TOAØN CHOÁNG TUOÄT
2leg z  tg a 2le  tg a
FS p  

g zK a SV S H
K a SV S H
HEÄ SOÁ AN TOAØN CHOÁNG ÑÖÙT
FSb  
t f y
g HK a SV S H
THÍ DUÏ: moät töôøng chaén cao 8m. Ñaát ñaép sau löng töôøng coù g = 16,6 KN/m2;  = 300. Theùp Galvani ñöôïc
söû duïng ñeå xaây töôøng chaén.Tính heä gia cöôøng vôùi FS9p) = FS(B) = 3
Caùc ñaëc tröng khaùc a = 200 vaø löïc chòu keùo cuûa theùp fy = 2,4105 KN/m2.
GIAÛI: Choïn SV = 0,5m; SH = 1m; maët thanh theùp  = 75mm.
Töø  = 300  Ka = 1/3
Tmax = gHKaSVSH =16,6  8  (1/3) 0,5 1 = 22,14 KN
Beà daày taám theùp
g HK a S v S H  FSb 
22,14  3
t

 0,00369m
5
 fy
0,075  2,4 10
Neáu theùp coù ñoä ró seùt 0,025mm/naêm vaø tuoåi thoï coâng trình laø 50 naêm
Chieàu daày caàn thieát cuûa thanh theùp laø 3,69+0,025 50= 4,94mm choïn t= 5mm
CHIEÀU DAØI CAÀN THIEÁT THANH GIA CÖÔØNG
 0   FS p  K a SV S H
L  H  tg 45   
 13,78m
2
2 tg a

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