Geotechnical Engineering-II

Unit # II
Slope Stability

Dr. Awdhesh Kumar Choudhary

Assistant Professor
Department of Civil Engineering
NIT Jamshedpur
Types of slope and it’s failure
Types of Slopes: 1. Infinite Slopes; 2. Finite Slopes
Infinite slopes: They have dimensions that extend over great distances and the soil
mass is inclined to the horizontal.
Finite slopes: A finite slope is one with a base and top surface, the height being limited.
The inclined faces of earth dams, embankments and excavation and the like are all
finite slopes.
1. Translational Failure

 It occurs in the case of infinite slopes and here the

failure surface is parallel to the slope surface.
 A slope is said to be infinite, when the slope has no
definite boundaries and soil under the free surface
contains the similar properties up to identical
depths along the slope.
 As said above, when the soil along the slope has
similar properties up to a certain depth and soil
below this layer is strong or hard stratum, the
weak topsoil will form a parallel slip surface when
Translational Failure
failed.
 This type of failure can be observed in slopes of
layered materials or natural slope formations.
Types of slope failure
2. Rotational Failure

 Such failure occurs by rotation along a slip surface and the

shape thus obtained in slip surface is curved. Failed surface
moves outwards and downwards.
 In homogeneous soils, the shape is circular while in case of
non-homogeneous soils it is non-circular. Slope Failure
 Rotational failure may takes place in three different ways :
1) Face failure or slope failure - It occurs when soil
above the toe contains weak stratum. In
this case the failure plane intersects the slope
above toe.
2) Toe failure - It is the most common failure in which
failure plane passes through toe of slope. Toe Failure
3) Base failure - It occurs when there is a weak
soil strata under the toe and failure plane passes
through base of slope.
 Rotational failure can be seen in finite slopes such as
earthen dams, embankments, man-made slopes etc.

Base Failure
Types of slope failure
3. Wedge Failure

 It is also known as block failure or plane failure. It

generates a failure plane that is inclined.
 This type of failure occurs when there are fissures, joints,
or weak soil layers in slope, or when a slope is made of two
different materials.
 It is more similar to translational failure but the difference
is that translational failure only occurs in case of infinite
slopes but wedge failure can occur in both infinite and finite
slopes. Wedge Failure

4. Compound Failure

 It is a combination of translational and rotational failure.

 In this case, the slip surface is curved at two ends like
rotational slip surface and flat at central portion like in
translational failure.
 The slip surface becomes flat whenever there is a hard soil
layer at a considerable depth from toe. Compound Failure
Stability of infinite slopes
Assumptions:

 Soil is homogenous.
 The stress and soil properties on every vertical
plane are identical.
 On any plane parallel to the slope stresses and
soil properties are identical.
Thus, failure in such slope takes place
due to sliding of soil mass along a planer parallel
to the slope along at a certain depth.

Analysis:

 Weight of segment ABCD

W = γ z b (1) , considering unit length
of the soil mass.
 Tangential stress τ down the slope
Stability of infinite slopes

 Normal stress σ within the segment

 Pore water pressure u on the slip surface

u = (z - hw) γw cos2β
 Normal effective stress σ՛
σ՛ = γ z cos2β - (z - hw) γw cos2β
= (γ z – γw z + γw hw ) cos2β

 Shearing strength τf at the base of segment

τf = c՛ + σ՛tan ϕ՛
 Factor of safety can be defined as

FOS =
 In general,

FOS =
Stability of infinite slopes
Special cases:

Case A - Dry cohesionless soil i.e. c՛ = 0 and hw = z

 FOS of an infinite slope with a cohesion-less soil is independent of the depth

of failure plane.
FOS =

 For the critical case FOS = 1  β = ϕ՛  Slope

is just stable
 For β < ϕ՛  τf < τf  Slope is stable
(Independent of depth of slope)
 For β > ϕ՛  τf > τf  Slope would have
already failed at all depths.
σ
Case B – Saturated cohesionless slope i.e. c՛ = 0 and hw = 0

FOS =

Note - FOS of a saturated cohesion-less slope is about ½ for a slope without

saturation.
Taylor’s stability number
 Gravity forces due to unit weight (γ) causes instability while and cohesive forces (cu)
contribute to stability in a soil mass.
 The above statement is valid at every point within the sliding mass as both being body
forces are distributed throughout the soil mass.
 The maximum height Hc of a slope that can be built without failure is thus directly
proportional to cohesion (cu) and inversely proportional to unit weight (γ)
 Maximum height, Hc is also related to the values of ϕu and β where ϕu = angle of
shearing resistance and β = slope angle
Therefore, Hc= (cu / γ) × f (ϕu , β ), where f (ϕu , β) means function of both ϕu
and β
 The above relation will be dimensionally correct only if f (ϕu , β) is a dimensionless
function.
 Taylor(1937) expressed this as a reciprocal of a dimensionless number which he called
as stability number i.e. f (ϕu , β) = 1/Sn

Therefore, Hc= cu / γSn

or, Sn = cu / γ Hc
 If factor of safety Fc is introduced with respect to cohesion, then
Sn = cm / γ H = Sn = cu / Fc γ H, where H is the height of slope (< Hc)
having a factor of safety Fc and cm is the mobilised unit cohesion for equilibrium of a
slope having height H
Stability analysis – Culmann’s method
 This method is based on the assumption that the failure of a slope occurs along a
plane when the average shearing stress tending to cause the slip is more than the shear
strength of the soil.
 Also, the most critical plane is the one that has a minimum ratio of the average
shearing stress that tends to cause failure to the shear strength of soil.
 The adjoining figure shows a slope of height
H. The slope rises at an angle β with the
horizontal. AC is a trial failure plane. If we
consider a unit length perpendicular to the
section of the slope, we find that the weight of
the wedge ABC is equal to

Finite slope analysis - Culmann’s Method

 The normal and tangential components of W with respect to the plane AC are as
follows
Na = Normal Component = W cosθ =

Ta = Tangential Component = W sinθ =

Stability analysis – Culmann’s method
 The average effective normal stress and the average shear stress on the plane AC are
respectively,

and,

Finite slope analysis - Culmann’s Method

 The average resistive shearing stress developed along the plane AC also may be
expressed as

 Equating the value of average shear stress from the above relations i.e. τ = τd
we have,
Stability analysis – Culmann’s method

-------- Equation XY

 The above expression is derived for the trial failure plane AC.
 In an effort to determine the critical failure plane, we must use the principle of
maxima and minima (for a given value of ϕ՛d ) to find the angle θ where the developed
cohesion would be maximum.
 Thus, the first derivative of cd with respect to θ is set equal to zero, i.e.

 Since γ, H and β are constants in equation XY therefore,

=>

 Solving the above equation gives the critical value of θ, i.e.

 Substituting θ = θcr in Eq. XY yields,

-------- Equation MN
Stability analysis – Swedish circle method

 Also known as ordinary method of slices. arc AC

 In this method, the potential failure
surface is assumed to be a circular arc with
centre O and radius r.
 The soil mass (ABCD) above a trial surface
(AC) is divided by vertical planes into a series
of slices of width b.
 The base of each slice is assumed to be a
Swedish circle method
straight line.
 The factor of safety (FS) is defined as the
ratio of the available shear strength τf to the
shear strength τm which must be mobilized to
maintain a condition of limiting equilibrium.
FS = τf / τm
 The FS is taken to be the same for each
slice, implying that there must be mutual
support between slices i.e. forces must act
between them.
FBD of ith slice
Stability analysis – Swedish circle method
 Considering ith slice , we have
1. Total weight of slice W = γ×b×h
2. Total normal force N = σl ( It includes
N′ = σ′l and U = ul). Here u = PWP at
the centre of the base and l = length of
the base.
3. The shear force on the base, T = τm l
4. Total normal forces on sides E1 and E2
5. The shear forces on the sides, X1 and X2
 Considering moments about O, the sum of the FBD of ith slice
moments of the shear forces T on the failure arc AC
must be equal the moment of the weight of the soil
mass ABCD.

Using T = τm l = , we get

Therefore,
Stability analysis – Swedish circle method
 For an analysis in terms of effective stress

 The above equation is exact but approximations are introduced in determining the
forces N′.
 It is assumed that for each slice the resultant of the interslice forces is zero.
 The solution involves resolving the forces on each slice normal to the base i.e. N′ =
Wcosα – ul
 Therefore rewriting the above equation, we get

 The components of Wcosα and Wsinα can be determined graphically for each
slice.
Stability analysis – Swedish circle method
 For an analysis in terms of total stress, the parameters cu and ϕu are used and the
value of u = 0. Therefore we get

 For ϕu = 0, we get

Method of slices