# SLOPE STABILITY SLOPE STABILITY. Types of Slope Failure. Types of Slope Failure. Assistant Professor Berrak Teymur

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## SLOPE STABILITY

Slope failure

z

### .

z The slope can be natural or man-made.





### strength of the soil.

called slope stability, most likely rupture surface is the

critical plane that has the minimum factor of safety.

Rotational slips

### Types of Slope Failure

Failure surface is circular arc Homogeneous soil conditions

Failure surface is non-circular arc Non-homogeneous soil conditions

Translational slip

(common in coarse grained soils)

### Types of Slope Failure

Compound slip

Adjacent stratum is at greater depth

Adjacent stratum is at shallow depth

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### Slope Failure)

¾ Erosion ¾ Rainfall ¾ Earthquakes ¾ Geological Factors ¾ External Loading ¾ Construction Activities Excavated Slopes Fill Slopes ¾ Rapid Drawdown

¾ Increment of pore water pressure ¾ The change in topography

z

z

z

z

z

z

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z

### Mass Procedure

: The mass of the soil above the surface of sliding is taken as a unit.

z Soil is assumed to be homogeneous.

z

### Method of Slices

: The soil above the surface of sliding is divided into a number of vertical parallel slices.

z The stability of each slice is calculated separately. z Nonhomogeneity of the soils and pore water pressure

can be taken into consideration.

z It also accounts for the variation of the normal stress

along the potential failure surface.

d f s

### =

Factor of Safety:

Fs: Factor of safety w.r.t strength

### τ

f: Average shear strength of soil

### τ

d: Average shear stress developed along the potential failure surface

The shear strength of soil: τf=c+σtanφand τd=cd+σtanφd Therefore;

When Fs=1, the slope is close to failure.

Generally Fs=1.5 is acceptable for the design of a stable slope. Fs>1.5 is safe. d d tan c tan c φ σ φ σ + + = s F

### Seepage (u=0)

L W W=LHγ Na Ta β φ β β γ φ β γ φ β γ τ τ tan tan tan cos tan cos tan cos 2 2 2 + = + + = = H c H c H c F d d d f s

' 2 sat sat s

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### Culmann’s Method

Failure of a slope occurs along a plane when τavtending to cause the slip is more than the shear strength of the soil.

H W

z

### considered.

O

d W

cu La: length of arc

For short term stability of saturated clay slopes.

z

d

f

s

u

s

z

d

u

s

a

s

u

a

### r)/Wd

Disturbing storing

Re Fs=

u

### =0)

z Taylor (1937) published stability coefficients for the

analysis of homogeneous slopes in terms of total stress.

z H= height of slope,

z Ns= stability coefficient (depends on α(slope angle) and D (Depth)).

z Fs= minimum factor of safety

## γ

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u

### Rotational Slips

O x1 cu2 L2 L1 W2 W1 cu1 x2 R

2 2 1 1 2 2 1 1

u u s

w u u s

### =

2 2 1 1 2 2 1 1 O x W cu L Pw y Add Pwy to disturbing moments

### The Method of Slices

The soil mass above a trial failure surface is divided by vertical planes into a series of slices of width b.

### ∑

+ − = ' ( cos )tan' sin 1 α φ α cl W ul W Fs

For total stress, cu and φu used and u=0.

Swedish Solution: Assumes that the resultant of the interslice forces is zero.

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### Procedure for the Method of Slices

1. Draw the slope to scale and note the positions and magnitudes of any external loads.

2. Draw a trial slip surface and identify its point of rotation.

3. Divide the soil mass above the slip surface into a convenient number of slices. More than five slices are needed for most problems.

4. For each slice:

a) Measure the width of slice (b)

b) Determine W (weight of a slice)

c) Measure αjfor each slice d) Calculate ru.

5. Calculate Fs.

O W

z

### ∑

      + − + = α φ α α φ sin ) / tan (tan 1 sec tan ) ( ' ' ' W F ub W b c F s s

u

### Summary:

z For temporary slopes and cuts in fine-grained soils with low permeability choose the undrained strength cu

and carry out an analysis in total stresses.

z For permanent slopes, choose an effective stress strength and calculate pore pressures separately. z Bishop (1955) assumed a circular slip plane and

considered only moment equilibrium.

z He neglected seepage forces and assumed that the lateral normal forces are collinear.

z In Bishop’s simplified method, the resultant interface shear is assumed to be zero.

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## References

Related subjects :