Soil Mechanics B

1
Coulombs Method
2
Failure mechanism
In Coulombs method a mechanism of failure has to
be assumed
3
Failure mechanism
In Coulombs method a mechanism of failure has to
be assumed
If this is the failure mechanism then the
Mohr-Coulomb failure criterion must be satisfied
on the assumed failure planes
4
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 Rankines 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

5
Failure criterion
For any point on the failure plane we have
6
Failure criterion
For any point on the failure plane we have
If analysis is of undrained stability then the
failure criterion must be expressed in terms of
total stress using undrained parameters cu and fu
7
Failure criterion
For any point on the failure plane we have
If analysis is of undrained stability then the
failure criterion must be expressed in terms of
total stress using undrained parameters cu and fu
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, f
8
Failure criterion
direction of
soil
Assumed
movement
failure
plane
t
s
9
Failure criterion
direction of
soil
Assumed
movement
failure
plane
t
s
Forces on the failure plane Shear Force T

10
Failure criterion
direction of
soil
Assumed
movement
failure
plane
t
s
Forces on the failure plane Shear Force T
Normal Force N
11
Failure criterion
direction of
soil
Assumed
movement
failure
plane
t
s
Forces on the failure plane Shear Force T
Normal Force N Cohesive Force
C
12
Failure criterion
If the soil properties are constant
13
Failure criterion
If the soil properties are constant
The forces acting on the failure plane are
T
N
14
Failure criterion
If the soil properties are constant
The forces acting on the failure plane are
T
N
which may be more convieniently represented by
C
f
R
15
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
16
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
17
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
or in terms of forces
18
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, fu

19
Total Stress Analysis
H tan
q
dir. of soil
movement
H
H sec
q
q
Soil properties cu, fu Soil-wall properties cw,
fw
20
Total Stress Analysis - forces consistent with
mechanism
C
2
W
C
1
f
w
f
R
2
u
q
R
1

C
c
H sec
q
1
u
C
c
H
2
w
2
W ½ H
tan

q
g
21
Total Stress Analysis - polygon of forces
C
2
W
C
1
q
22
Total Stress Analysis - polygon of forces
f
w
90 - q - f
u
C
2
W
C
1
q
23
Total Stress Analysis - polygon of forces
f
w
R
2
R
1
90 - q - f
u
C
2
W
C
1
q
24
Total Stress Analysis - polygon of forces
f
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
R
2
R
1
90 - q - f
u
C
2
W
C
1
q
25
Total Stress Analysis - polygon of forces
R
2
f
w
R
1
90 - q - f
u
C
2
W
C
1
q
26
Total Stress Analysis - polygon of forces
R
2
f
w
R
1
90 - q - f
u
Direction of R2 is not consistent with assumed
mechanism. Therefore the mechanism is not valid.
C
2
W
C
1
q
27
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
28
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
f
w
R
2
F
total
C
2
29
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
f
w
R
2
F
F
total
total
C
V
2
30
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
f
w
R
2
F
F
total
total
C
V
2
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.
31
Total Stress Analysis - tension cracks
As with Rankines method allowance must be made
for tension cracks, and if water is present the
possibility that these cracks may fill with
water.
32
Total Stress Analysis - tension cracks
As with Rankines 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 Rankines
method. For active failure this gives
33
Total Stress Analysis - tension cracks
z
H
q
34
Total Stress Analysis - tension cracks
W
1

C
2
W

C
2
1
f
w
q

f
R
2
u

R
1
35
Total Stress Analysis - Example 1
o
10
V
W
Soil Properties
dir of soil
movement
c
10 kPa
u
o
10
f
5 m
u
6.4 m
c
2 kPa
w
o
20
f
w
3
20 kN/m
g
o
30
U
36
Total Stress Analysis - Example 1
37
Total Stress Analysis - Example 1
o
Ruw 60 kN/m
20
o
50
160
10
o
30
64
38
Effective Stress Analysis - Forces on failure
plane
Failure plane
39
Effective Stress Analysis - Forces on failure
plane
Failure plane
C
Failure plane
f

U
R
40
Effective Stress Analysis

• When performing effective stress stability
  calculations the critical state parameters c
  0, f fult 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 cw 0,
  fw fwult

41
Effective Stress Analysis

• In using Coulombs 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.

42
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 - f/2)
  to the direction of the minor principal stress s3.

43
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 - f/2)
  to the direction of the minor principal stress s3.

44
Effective Stress Analysis
Active
s3
a
45
Effective Stress Analysis
Active
s3
Passive
a
a
s3
46
Effective Stress Analysis
Active
s3
Passive
a
a
s3

• 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.

47
Effective stress analysis - Example
o
10
V
W
W.T.
X
Soil
Water
Water
5 m
6.4 m
5 m
o
30
U
48
Effective stress analysis - Example
V
W
X
Soil
5 m
6.4 m
o
30
U
49
Effective stress analysis - Example
V
W
X
Soil Properties
Soil
c 5 kPa
5 m
o
10
f
6.4 m
c
2 kPa
w
o

20
f
w
3
o
20 kN/m
g
30
dry
3
22 kN/m
g
sat
U
50
Effective stress analysis - Example
V
W
X
Soil Properties
Soil
c 5 kPa
5 m
o
10
f
6.4 m
c
2 kPa
w
o

20
f
w
3
o
20 kN/m
g
30
dry
3
22 kN/m
g
sat
U
C
5

6.4 32 kN/m
uv

C
2
5 10 kN/m
uw
51
Effective stress analysis - Example
V
W
X
Soil Properties
Soil
c 5 kPa
5 m
o
10
f
6.4 m
c
2 kPa
w
o

20
f
w
3
o
20 kN/m
g
30
dry
3
22 kN/m
g
sat
U
C
5

6.4 32 kN/m
uv

C
2
5 10 kN/m
uw




W 0.5
5
2.89
22 (8 - 0.5
5
2.89)
20
174.5 kN/m


52
Example - Water pressures on soil wedge
V
W
X
U
53
Example - Water pressures on soil wedge
V
W
X
U
54
Example - Forces on soil wedge
C
uw
C
W
uv
Uuw
Ruw
Uuv
Ruv
55
Example - Polygon of forces
U
uw
U
uv
o
60
C
W
uw
C
o
30
uv
56
Example - Polygon of forces
U
o
uw
60
U
uv
o
60
C
W
uw
C
o
30
uv
57
Example - Polygon of forces
58
Example - Forces on the wall
R
uw
U
uw
C
uw
59
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
R
sin

C
5.8 10 15.8 kN/m
f
uw
uw
w
uw
60
Example - Forces on the wall
R
uw
U
uw
C
uw
The vertical and horizontal components of the
force on the wall are Vertical Horizontal
T
R
sin

C
5.8 10 15.8 kN/m
f
uw
uw
w
uw
kN/m
N
R
cos

U
15.97 122.5 138.5
f
uw
uw
w
uw
61
Example - Forces on the wall
R
uw
U
uw
C
uw
The vertical and horizontal components of the
force on the wall are Vertical Horizontal Note
that N 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
T
R
sin

C
5.8 10 15.8 kN/m
f
uw
uw
w
uw
kN/m
N
R
cos

U
15.97 122.5 138.5
f
uw
uw
w
uw
uw
62
Example 3
63
Example 3
64
Example 3
Choosing a local datum at E hD 0
65
Example 3
x
Choosing a local datum at E hD 0, zD - x and
uD gw x
66
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.
67
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
0.5
g
H
w
w
w

f
cs

f
w
U
From
R
R
flow net
w
68
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