Slope stability analysis

Question 1

Slope stability analysis should satisfy

Answer 1

• equilibrium
• compatibility
• constitutive law
• BC

Question 2

Slope stability analysis applications

Answer 2

-Used for remedial work on existing or theorestical slopes as well as for back analysis of failed slopes

Question 3

Slope stability analysis applications

Answer 3

-Used for remedial work on existing or theoretical slopes as well as for back analysis of failed slopes

Question 4

Analytical method applications

Answer 4

-applied to boundary value problems for linear elastic soils or perfectly plastic ones
-considers movements but not useful for stability

Question 5

Numerical methods

Answer 5

-approximately satisfies all that needs to bee (four from before)
-handles complex constitutive models
-can simulate the complete history of a boundary value problem but limmited meash and integration method applied
-still new and not widely used

Question 6

Usea for analytical methods

Answer 6

Limit equilibrium, limit analysis and stress field all give stability but not movements since they exclude compatibility and displacement BC
- Most common
-Dont give unique solutions
-Doesnt say whether the solution is safe or weak (in general)
-

Question 7

Plain strain

Answer 7

• applies when one dimension is large in relation to the other 2 so its assumed all states // to the x-y plane are same and there are no relative movements
• w = 0 = epsilon_z therfore the shear strains in z are also zero

Question 8

Limit equilibrium assumptions

Answer 8

• Failure surface with failure at all points along it
• can be curved planar or arbitrary
  -Global equilibrium of rigid blocks is considered but not the internal stress within the blocks

Question 9

Why apply FoS

Answer 9

Accounts for uncertainty in
- stratigraphy
-soil strength
-pore pressure distribution

Question 10

FoS on strength
-Planar failure
-Rotational failure

Answer 10

F= shear strength of soil/shear str for equilibrium
shear str for equilibrium = tau_mob

• For drained FoS on str is applied to c’ and tan(phi’)

Planar failure:
F = sum(resisting F)/sum(disturbing F)

Rotational Faiure:
F = sum(resisting M)/sum(disturbing M)

Question 10

FoS on strength
-Planar failure
-Rotational failure

Answer 10

F= shear strength of soil/shear str for equilibrium
shear str for equilibrium = tau_mob

• For drained FoS on str is applied to c’ and tan(phi’)

Planar failure:
F = sum(resisting F)/sum(disturbing F)

Rotational Faiure:
F = sum(resisting M)/sum(disturbing M)

Question 11

Limit equilibrium objective

Answer 11

1. get FoS for a given geometry
2. Get H/angle for a given FoS
3. Determine str from geometry assuming FoS =1, Back analysis

Question 12

Infinite slope assumption

Answer 12

Used to approximate shallow translational slides
-this assumes the toe and the top of the slope have little influence
- failure surface // to the ground
-uniform PWP

Question 13

Limit equilibrium special cases

Answer 13

Dry-cohesionless - Fos is independent of the depth of the slip surface

Saturated cohesionless also independent of the depth but scaled by gamma’/gamma

Question 14

Planar surface analysis

Answer 14

Applied to low strength or this layer overlaying a strong one
- moment equilibrium not used since the distribution of sigma_n’ is unknown and therefore so is the line of action of N’

tau_max = c’ + (N’/L)Tan(phi’)
tau_mob = S/L
- U is average d between the value at the top and bottom of the slope is it varies through it

Question 15

Rotational movements
- getting FoS

Answer 15

Assumes failure surface to be circular and applies the undrained shear str
- Only determinate when phi_u = 0
*Magnitude, direction and line of action of delta P are alll unknown as is F therefore 4 unknowns but only 3 equations: moments =…, delta P = …, delta S = …

To get Fos
- Vary radius with the centre of cirlce constant
- Vary the centre of rotation
- Plot FoS against radii for given COR

Question 16

Method of Slices
-general
-benefits

Answer 16

-Each slice is treated as a distinct sliding block with equilibrium applied to it
* critical surface must be known: geometry, str and PW distribution
* forces, FoS and line of action are unknowns
* apply all 3 equilibrium per slice and failure criterion

Benefits:
-complex geometry can be considered
-variable soil conditions considered
-influence of boundary loads considered

Question 17

Definition of rigorous

Answer 17

the number assumptions matches the number of unknowns (2n - 2 for the method of slices)
- extras can lead to iteration

Question 18

Conventional method of slices

Answer 18

Neglect interslice forces and assume that N’ acts through the centre of the slice leads to 3n -2 assumptions therefore it fails to satisfy constraints
*Forces in one direction are not satisfied
*No iteration
*Gives conservative results
*Uses moments

Question 19

Bishops simplified method of slices

Answer 19

Negelcts inter slice shear force and assume N’ acts through the centre of the slice now 2n -1
assumptions
- 1 too few for horizontal equilibrium so one slice/constraint isnt satisfied and horxontal equilibrium isnt used
- Applies moments but resolves forces perp. to N’ rather than //
-solved iteritavely with the value from convential used as a starting point
-

Question 20

Stability Charts

Answer 20

Assumptions
-single homogeneous slope and a circular slip surface but can be used for non-homogeneous material by using an average strength and slope inclination
-undrained therefore Su = k
-No tension cracks accounted by Su being decreased by 10-15%
-No water pressure on the slope, submerged solpes have gamma replaced by gamma’ but no other water levels applly

Comments
- fast and accurate
- uses dimensionless parameters to give in min FoS so theres no need for the critical slip surface
- D accounts for the location of the hard material

-

Question 21

Stability charts failure types

Answer 21

Slope circle happens when theres a firm base just below the toe so the surface is tangential and daylights above the toe

Toe circle is when the surface is independent of the location pf the base and passes through the toe

Foundation circle is base failure with the centre of the circle being halfway long the slop in x

Question 22

Stability chart for PWP

Answer 22

assume U varries linearly

U = r_u.gamma.z
- since r_u is constant for the slope FoS is a function of it, being linear for simple slopes with a constant slope and no tension cracks

F = m -n.r_u
- m/n are functions of phi’, c’/gamma.H, D and slope angle

Question 23

Compound movements

Answer 23

• Apply the method of slices
  *Leaver arm now off set by distance which effects the line of action of the normal F
• Noncircular slip surface can be taken into account by other solution
• Solutions considered
• Janbu simplified
  *Morgenstein and Price

Question 24

Janbuu Simplifed

Answer 24

- Like simplified bishop but applied horizontal and vertical equilibrium - correction applied for the relative depth of the slide to it =s depth and also soil str *solves issue of negelcting interslice F * is a convergence factor

Question 25

Morgenstein and Price

Answer 25

lambda.f(x) = Xi/Ei with lambda between zero and 1 which introduces n - 1 assumptions with another n coming from the assumption of the line of action of N - since lambda is unknown therefore 2n - 1 unknowns and is rigorous

Question 26

3D limit equilibrium analysis

Answer 26

- applied when the slide has limited lateral extent the shear force at the sides will influence stability , the amount depending on the slides width B -Applies method of slices but as columns instead * the assumptions have a large effect in 3d F = SUM(Rm.B + M1 + M2)/sum(D.B) - M is the moments resisted at the sides - Rm is the resisting moment F - D is the disturbing moment F

Question 27

Other factors effecting stability

Answer 27

Tension cracks -at shallow depths where the tensile strength is exceeded, only occurs in cohesive soils since they have tensile strength -indicate the onset of instability and can start progressive failures Zt = 2Su/gamma -Undrained are usually deeper but shouldnt be more than the cut or half the embankments height Zt = 2c'/gamma . Tan(pi/4 + phi'/2) Vegetation -Roots reinforce soil by bonding -decreases the PWP and WT -most effective in summer when the least drainage is required

Question 28

Seismic effects

Answer 28

Decrease tau_max due to increased power pressure while increasing tau_mob - can be caused by landslide Psuedo static equilibrium techniques are used with limit equlibrium, adding M.a_g force to the static problem, but it wount give deformations - represents a_g as a constant acceleration -no build up of PWP -hydrostatic soil behaviour under loading isnt captured but can be by Dynamic FE which is being researched