Limit Equlibrium Method. Limit Equilibrium Method Failure mechanisms are often complex and cannot be modelled by single wedges with plane surfaces. Analysis.

Apresentação em tema: "Limit Equlibrium Method. Limit Equilibrium Method Failure mechanisms are often complex and cannot be modelled by single wedges with plane surfaces. Analysis."— Transcrição da apresentação:

1 Limit Equlibrium Method

2 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

3 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

4 Weak Clay layer ( c u,  u ) Short term Long term (c´,  ´ )

5 Multiple wedge mechanisms 1 2 x  Weak Clay layer ( c u,  u ) Short term Long term Sandy Fill (c´,  ´) (c´,  ´ )

6 Multiple wedge mechanisms Once 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.

7 Multiple wedge mechanisms Once 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. 1 2

8 Multiple wedge mechanisms Once 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. 1 2 v 1 v 2 v 2 - v 1 

9 Multiple wedge mechanisms Once 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. 1 2 v 1 v 2 v 2 - v 1 v 1 2 v 1 v 2  

10 Multiple wedge mechanisms W 1 P X C 12 C 1 R 1  ´  u

11 Multiple wedge mechanisms L W 2 W 1 P X X C 12 C R 2 C 2 C 1 R 1  ´  ´  ´  u

12 Multiple wedge mechanisms  ´ L + W 2 X R 2 

13 Multiple wedge mechanisms  ´ L + W 2 X R 2 W 1 X C 1 P R 1  u 

14 Multiple wedge mechanisms Weak clay layer Failure planes

15 Multiple wedge mechanisms

16 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

17 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

18

19 This is usually written as

20 Factor of Safety This is usually written as wherec m (= c F ) is known as the mobilised cohesion

21 Factor of Safety This is usually written as wherec m (= c F ) is known as the mobilised cohesion  m (= tan        1  F ) is known as the mobilised friction angle

22 Failure plane between wedges Factor of Safety Consider the 2 wedge mechanism

23 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

24 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 It is more convienient to take c* = c m and  * =  m as this must be the case when F=1

25 Factor of Safety v1v1 v2v2 v 1 - v 2

26 C 12 W 1 R 1 C 1  ´ mc X 1 Factor of Safety v1v1 v2v2 v 1 - v 2

27 C 12 C X 2 W 2 C 2 R 2 W 1 R 1 C 1  ´ mc X 1 Factor of Safety v1v1 v2v2 v 1 - v 2

28 Factor of Safety W 1 C 1 C 12 X 1 R 1

29 Factor of Safety W 1 C 1 C 12 X 1 R 1 W 2 C 2 C R 2 X 2

30 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 X 1 and X 2 to be determined. The values of X 1 and X 2 depend on F and will not in general be equal We need to determine the value of F that gives X 1 = X 2 This can be determined by a trial and error process, followed by interpolation Factor of Safety

31 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 Safety X 1 - X 2 F Required solution

32 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

33 In any analysis the appropriate parameters must be used for c and . In an undrained analysis (short term in clays) the parameters are c u,  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

34 Example 15 m 20 m 1 2 60 o 50 o 1 2

35 Example 15 m 20 m 1 2 60 o 50 o 1 2 1. Calculate areas: A 1 = 86.6 m 2 A 2 = 118.8 m 2

36 Example 15 m 20 m 1 2 60 o 50 o 1 2 1. Calculate areas: A 1 = 86.6 m 2 A 2 = 118.8 m 2 2. Assume Factor of Safety F = 2

37 Example 15 m 20 m 1 2 60 o 50 o 1 2 1. Calculate areas: A 1 = 86.6 m 2 A 2 = 118.8 m 2 2. Assume Factor of Safety F = 2 3. Calculate c,  parameters Weak layer c m = c u /F = 40/2 = 20 kPa,  m = 0 Clayey sand c m = 0,  ’ m =

38 Example 4. Calculate known forces

39 Example 4. Calculate known forces 60 o X 1 W 1 C 1 R 1 16.1

40 Example 4. Calculate known forces 60 o 50 o 60 o X 1 X 2 W 1 W 2 C 1 R 1 R 2 16.1

41 Example 60 o X 1 W 1 C 1 R 1 16.1

42 Example 60 o X 1 W 1 C 1 R 1 16.1 60 o 50 o X 2 W 2 R 2 16.1

43 Example 60 o X 1 W 1 C 1 R 1 16.1 60 o 50 o X 2 W 2 R 2 16.1

44 Example

45 X 2 - X 1 F 1.01.5 2.0 610 238 F = 1.18