limit equlibrium method. limit equilibrium method failure mechanisms are often complex and cannot be...
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Limit Equlibrium Method
Limit Equilibrium Method
• Failure mechanisms are often complex and cannot be modelled by single wedges with plane surfaces.
• Analysis is straightforward if mechanism consists of
• Multiple wedges• Circular failures
• Most situations can be approximated by one of these mechanisms
• Mechanism appropriate when soil stratigraphy contains weak layers.
These can occur due to
• Thin clay layers in sedimentary deposits
• Pre-existing slips in clayey soils
• Fissures and joints in stiff clays
• Joints in rocks and other cemented soil materials
Multiple wedge mechanisms
Multiple wedge mechanisms
Weak Clay layer( cu , u ) Short term
Long term(c´, ´ )
Multiple wedge mechanisms
12
x
Weak Clay layer( cu , u ) Short term
Long term
Sandy Fill(c´, ´)
(c´, ´ )
Multiple wedge mechanismsOnce the mechanism has been specified, the directions of the forces must be determined. These can be determined from common sense or by drawing a velocity diagram to get the relative velocities.
Multiple wedge mechanismsOnce the mechanism has been specified, the directions of the forces must be determined. These can be determined from common sense or by drawing a velocity diagram to get the relative velocities.
12
Multiple wedge mechanismsOnce the mechanism has been specified, the directions of the forces must be determined. These can be determined from common sense or by drawing a velocity diagram to get the relative velocities.
12
v1
v2
v2 - v1
Multiple wedge mechanismsOnce the mechanism has been specified, the directions of the forces must be determined. These can be determined from common sense or by drawing a velocity diagram to get the relative velocities.
12
v1
v2
v2 - v1 v1 - v2
v1
v2
Multiple wedge mechanisms
W1
P
X
C12
C1
R1
´
u
Multiple wedge mechanisms
L
W2
W1
P
X
XC12
C12
R2
C2
C1
R1
´
´
´
u
Multiple wedge mechanisms
´
L + W2
X
R2
Multiple wedge mechanisms
´
L + W2
X
R2W1
X
C1
P
R1
u
Multiple wedge mechanisms
Weak clay layer
Failure planes
Multiple wedge mechanisms
• When performing stability analyses we are often interested in determining a factor of safety
• The factor of safety can be determined from a limit equilibrium analysis using factored strength parameters
• At failure the stresses are related by the Mohr-Coulomb criterion
c + tan
• At states remote from failure the stresses are assumed to be given by
Factor of Safety
• When performing stability analyses we are often interested in determining a factor of safety
• The factor of safety can be determined from a limit equilibrium analysis using factored strength parameters
• At failure the stresses are related by the Mohr-Coulomb criterion
c + tan
• At states remote from failure the stresses are assumed to be given by
Factor of Safety
mobcF F
tan
Factor of Safety
mobcF F
tan
Factor of Safety
mobcF F
tan
mob m mc tan
This is usually written as
Factor of Safety
mobcF F
tan
mob m mc tan
This is usually written as
where cm (=cF ) is known as the mobilised cohesion
Factor of Safety
mobcF F
tan
mob m mc tan
This is usually written as
where cm (=cF ) is known as the mobilised cohesion
m (= tantan
1 F ) is known as the mobilised friction angle
Failure plane between wedges
Factor of Safety
Consider the 2 wedge mechanism
Failure plane between wedges
Factor of Safety
Consider the 2 wedge mechanism
Between the wedges it is often assumed that the mobilised cohesion, c* and mobilised friction angle, * are given by
ccm m* * 2 2
Failure plane between wedges
Factor of Safety
Consider the 2 wedge mechanism
Between the wedges it is often assumed that the mobilised cohesion, c* and mobilised friction angle, * are given by
ccm m* * 2 2
It is more convienient to take c* = cm and * = m as this must be the case when F=1
Factor of Safetyv1
v2v1 - v2
C12
W1
R1
C1
mc
X 1
Factor of Safetyv1
v2v1 - v2
C12
C12
X 2
W2
C2
R 2
W1
R1
C1
mc
X 1
Factor of Safetyv1
v2v1 - v2
Factor of Safety
W1
C1C12
X1
R1
Factor of Safety
W1
C1C12
X1
R1
W2
C2
C12
R2
X2
• Equilibrium requires that the forces between the two wedges are equal and opposite.
• In the analysis we have assumed a factor of safety which enables X1 and X2 to be determined.
• The values of X1 and X2 depend on F and will not in general be equal
• We need to determine the value of F that gives X1 = X2
• This can be determined by a trial and error process, followed by interpolation
Factor of Safety
• the solution is not necessarily the factor of safety of the slope.
• To determine the actual factor of safety all the possible mechanisms must be considered to determine the mechanism giving the lowest factor of safety.
Factor of SafetyX1 - X2
F
Required solution
• For the slope analysis a unique factor of safety can be determined.
• In the analysis of the retaining wall considered above the force on the wall is related to the factor of safety.
• When analysing retaining walls we are often concerned with the limiting stability, that is when the factor of safety is 1.
• For an active failure mechanism increasing the factor of safety results in a greater horizontal force being required
• For a gravity retaining wall, the wall can be analysed as a single block or wedge
Factor of Safety
• In any analysis the appropriate parameters must be used for c and . In an undrained analysis (short term in clays) the parameters are cu, u with total stresses, and in an effective stress analysis (valid any time if pore pressures known) the parameters are c’, ’ used with the effective stresses.
• In an effective stress analysis if pore pressures are present the forces due to the water must be considered, and if necessary included in the inter-wedge forces.
Factor of Safety
Example
15 m
20 m
12
60o50o
1 2
Example
15 m
20 m
12
60o50o
1 2
1. Calculate areas:
A1 = 86.6 m2 A2 = 118.8 m2
Example
15 m
20 m
12
60o50o
1 2
1. Calculate areas:
A1 = 86.6 m2 A2 = 118.8 m2
2. Assume Factor of Safety
F = 2
Example
15 m
20 m
12
60o50o
1 2
1. Calculate areas:
A1 = 86.6 m2 A2 = 118.8 m2
2. Assume Factor of Safety
F = 2
3. Calculate c, parameters
Weak layer cm = cu/F = 40/2 = 20 kPa, m = 0
Clayey sand cm = 0, ’m =
Example
4. Calculate known forces
Example
4. Calculate known forces
60o
X1
W1
C1
R1
16.1
Example
4. Calculate known forces
60o50o60o
X1
X2
W1
W2
C1
R1
R2
16.1
16.1
16.1
Example
60o
X 1
W1
C1R1
16.1
Example
60o
X 1
W1
C1R1
16.1
60o50o
X2
W2
R216.1
16.1
Example
60o
X 1
W1
C1R1
16.1
60o50o
X2
W2
R216.1
16.1
Example
Example
X2 - X1
F1.0 1.5 2.0
610
238
F = 1.18