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