STABILITY OF SLOPES

Slope
An exposed ground surface that stands at an angle with the
horizontal is called an unrestrained slope.

2
TYPES OF SLOPES
A) w.r.t. Method of Construction
1. Natural Slopes
2. Man-made / Engineered Slopes
Embankments, earthen dams, river dikes, excavation trenches, etc.

Natural Slope Engineered Slope

3
TYPES OF SLOPES
B) w.r.t. Extents
1. Infinite Slopes
Having constant slope of infinite extent, e.g. long slope of a mountain
face.
2. Finite Slopes
Slopes of limited heights and extents, e.g. typical man-made slopes

Infinite Slope Finite Slope 4

TYPES OF SLOPES
C) w.r.t. Slope Material
1. Cohesionless
2. Cohesive

Cohesionless Cohesive
5
STABILITY OF SLOPES
A slope is said to be stable if
it meets a prescribed need for
a fixed period of time with a
suitable safety factor (FOS).
California, USA
January, 1997

Nova Frebergo, Brazil

January 13, 2011
6
CAUSES OF SLOPE FAILURE

7
CAUSES OF SLOPE FAILURE

8
CAUSES OF SLOPE FAILURE

9
TYPES/MODES OF SLOPE FAILURE

Toe Failure
 Failure surface passing through
toe of slope
 Material of slope is
homogeneous
 Relatively steep slope angles
11
 TYPES/MODES OF SLOPE FAILURE

Base Failure
 Failure surface passing through foundation
 Foundation soil somehow weaker than slope material
 Relatively gentle slopes
12
SLOPE STABILIZING MEASURES

13
SLOPE STABILIZING MEASURES

14
SLOPE STABILIZING MEASURES

15
 LIMIT EQUILIBRIUM ANALYSIS
 Most common quantitative measure of slope stability
 Stability of slope expressed in terms of factor of safety just
before failure
Failure Surface
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒
𝐹𝑂𝑆 =
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒

𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑆𝑜𝑖𝑙

𝐹𝑂𝑆 =
𝐴𝑝𝑝𝑙𝑖𝑒𝑑 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠

FOS < 1 → Failure

FOS = 1 → Verge of failure/Just stable
FOS > 1 → Stable

Typical design criterion for stable slope; FOS > 1.5

16
 SLOPE STABILITY ANALYSIS
Infinite Slope
L

d
Assumptions:
a
1. Slope face is planar and
of infinite extent a

2. Failure surface is║to c

slope face H h b
3. Water surface is║to
slope face

𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑆𝑜𝑖𝑙

𝐹𝑂𝑆 =
𝐴𝑝𝑝𝑙𝑖𝑒𝑑 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠
17
 SLOPE STABILITY ANALYSIS
Infinite Slope
Case-A: c-f Soil (c>0; f>0) L

𝑊 = 𝛾𝐻𝐿 d

𝑁 = 𝑊 cos 𝛼 = 𝛾𝐻𝐿 cos 𝛼 a

𝑇 = 𝑊 sin 𝛼 = 𝛾𝐻𝐿 sin 𝛼 a

a
c
Available Shear Strength H b
h
of Soil (tr) tr N
W T
𝜏𝑟 = 𝑐 ′ + 𝜎𝑛 ′ ∙ tan 𝜑′ 𝑨𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆 𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝑺𝒐𝒊𝒍
𝑭𝑶𝑺 =
𝑨𝒑𝒑𝒍𝒊𝒆𝒅 𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒆𝒔𝒔

Normal stress on the

failure plane
𝑁 𝛾𝐻𝐿 cos 𝛼
𝜎𝑛 = = = 𝛾𝐻𝑐𝑜𝑠 2 𝛼
(𝐿 cos 𝛼)(1) (𝐿 cos 𝛼)(1)
18
 SLOPE STABILITY ANALYSIS
Infinite Slope
Case-A: c-f Soil (c>0; f>0) L

Available Shear Strength of Soil (tr) d

𝜏𝑟 = 𝑐 ′ + 𝛾𝐻𝑐𝑜𝑠 2 𝛼 ∙ tan 𝜙′ a

a
Applied Shear Stress (t) a
c
𝑇
𝜏= H b
(𝐿 cos 𝛼)(1) h
tr N
W
𝛾𝐻𝐿 sin 𝛼 T
𝜏= 𝑨𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆 𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝑺𝒐𝒊𝒍
(𝐿 cos 𝛼)(1) 𝑭𝑶𝑺 =
𝑨𝒑𝒑𝒍𝒊𝒆𝒅 𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒆𝒔𝒔
𝜏 = 𝛾𝐻 sin 𝛼 cos 𝛼
𝑐 ′ + 𝛾𝐻𝑐𝑜𝑠 2 𝛼 ∙ tan 𝜙′
Factor of Safety (FOS) 𝐹𝑂𝑆 =
𝛾𝐻 sin 𝛼 cos 𝛼
Without considering the effect of WT 2
 SLOPE STABILITY ANALYSIS
Infinite Slope
Case-A: c-f Soil (c>0; f>0) L

Considering 𝑐 ′ + 𝜎𝑛 ′ ∙ tan 𝜙′ d
presence of WT 𝐹𝑂𝑆 =
𝛾𝐻 sin 𝛼 cos 𝛼
a H
 c’ & f’; effective strength parameters
a
 Obtained through drained triaxial test B a
C c
𝜎𝑛 = 𝛾𝐻𝑐𝑜𝑠 2 𝛼 𝜎𝑛′ = 𝜎𝑛 − 𝑢 D

h b
a tr N
𝐴𝐵 = ℎ W T
A
𝐴𝐶 = ℎ cos 𝛼
𝐴𝐷 = ℎ𝑤 = 𝐴𝐶 cos 𝛼 = ℎ 𝑐𝑜𝑠 2 𝛼
Pore water pressure, u
𝑢 = 𝛾𝑤 ℎ𝑤 𝑐 ′ + (𝛾𝐻 − 𝛾𝑤 ℎ) ∙ 𝑐𝑜𝑠 2 𝛼 tan 𝜙′
𝐹𝑂𝑆 =
ℎ𝑤 = 𝐴𝐷 = ℎ 𝑐𝑜𝑠 2 𝛼 𝛾𝐻 sin 𝛼 cos 𝛼

𝑢 = 𝛾𝑤 ℎ 𝑐𝑜𝑠 2 𝛼 3
 SLOPE STABILITY ANALYSIS
Infinite Slope (For c-f soil)

Case-B: Cohesive soil (f=0) 𝑐 ′ + (𝛾𝐻 − 𝛾𝑤 ℎ) ∙ 𝑐𝑜𝑠 2 𝛼 tan 𝜙′

𝐹𝑂𝑆 =
𝛾𝐻 sin 𝛼 cos 𝛼
𝑐′
𝐹𝑂𝑆 = L
𝛾𝐻 sin 𝛼 cos 𝛼

d
Case-C: Granular soil (c=0)
a

(𝛾𝐻 − 𝛾𝑤 ℎ) ∙ 𝑐𝑜𝑠 2 𝛼 ∙ tan 𝜙′ a

𝐹𝑂𝑆 = a
𝛾𝐻 sin 𝛼 cos 𝛼 c

𝛾𝑤 ℎ 𝑐𝑜𝑠 2 𝛼 ∙ tan 𝜙′ H h b
𝐹𝑂𝑆 = 𝛾𝐻 1 − tr N
𝛾𝐻 𝛾𝐻 sin 𝛼 cos 𝛼 W T

𝛾𝑤 ℎ tan 𝜙′
𝐹𝑂𝑆 = 1 −
𝛾𝐻 tan 𝛼

4
SLOPE STABILITY ANALYSIS
Infinite Slope
Case-C: Granular soil (c=0) L

𝛾𝑤 ℎ tan 𝜙′ d
𝐹𝑂𝑆 = 1 −
𝛾𝐻 tan 𝛼
a

For dry slope, h=0 a

a
c
tan 𝜙
𝐹𝑂𝑆 =
tan 𝛼 H h b
tr N
W T

For critical case; FOS = 1

tan 𝛼 = tan 𝜙′ 𝛼 = 𝜙′

i.e. the slope is only stable up to the angle of friction of soil.

5
SLOPE STABILITY ANALYSIS
Infinite Slope
Case-C: Granular soil (c=0) L

𝛾𝑤 ℎ tan 𝜙′
𝐹𝑂𝑆 = 1 − d
𝛾𝐻 tan 𝛼
a
For fully saturated slope, h=H
a
𝛾 − 𝛾𝑤 tan 𝜙′ a
𝐹𝑂𝑆 = c
𝛾 tan 𝛼
H h b
tan 𝜙′ tr N
𝐹𝑂𝑆 = 0.5 W
tan 𝛼 T

For critical case; FOS = 1

tan 𝜙′
1 = 0.5 tan 𝛼 = 0.5 ∙ tan 𝜙′ 𝛼 = 0.5 ∙ 𝜙’
tan 𝛼

i.e. a fully saturated slope will be stable only up to half of its

friction angle 6
Practice Problem #1
Calculate the maximum permissible slope angle for a sandy soil
having gsat = 20 kN/m3, f’ =30°, to give a FOS of 1.5 when;
a. the slope is dry
b. the slope is fully saturated with water table at the ground
surface

7
Practice Problem #2
An infinite slope is shown in the figure. The position of water
table is 2m below the ground surface. Determine the change in
FOS of the slope if WT is brought to a depth of 4m below the
surface.

gsat = 18 kN/m3
c’ = 10 kPa a = 15°
f’ = 20°

H = 6m

8
SLOPE STABILITY ANALYSIS
Finite Slope (Swedish Slip Circle Method)
Assumptions:
1. Material of the slope is homogeneous.
2. Soil is purely cohesive in nature i.e. f = 0.
3. Failure surface has a curved/circular or spoon like surface.
4. Shear strength of the soil is uniformly distributed along
failure plane. (only possible if f = 0)

2
SLOPE STABILITY ANALYSIS
Swedish Slip Circle Method (Cohesive soils (f=0))

C NSL
B

𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝑅 )

𝐹𝑂𝑆 =
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝐷 )

3
 SLOPE STABILITY ANALYSIS
Swedish Slip Circle Method (Cohesive soils (f=0))
Case-I: No Tension Crack x
q
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝑅 )
𝐹𝑂𝑆 = C NSL
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝐷 )
B
R
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 tr = c + sn tan f
For saturated clay under
𝑀𝐷 = 𝑊 ∙ 𝑥 undrained loading; f=0
𝑊 = (𝐴𝑟𝑒𝑎 𝑜𝑓 𝐴𝐵𝐶𝐴 × 1) × 𝛾  tr = c = su
A W
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
𝑀𝑅 = 𝜏𝑟 ∙ 𝐴𝐵 ∙ 𝑅
𝑀𝑅 = 𝑐 ∙ (𝑅 ∙ 𝜃) ∙ 𝑅
𝑀𝑅 = 𝑐 ∙ 𝜃 ∙ 𝑅 2 𝑀𝑅 𝑐 ∙ 𝜃 ∙ 𝑅2
𝐹𝑂𝑆 = 𝐹𝑂𝑆 =
𝑀𝐷 𝑊∙𝑥
→ q in radians
4
 SLOPE STABILITY ANALYSIS
Swedish Slip Circle Method (Cohesive soils (f=0))
Case-II: Development of Tension Crack x
q
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝑅 ) q2
𝐹𝑂𝑆 = C B NSL
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝐷 ) 2𝑐
ℎ𝑡 =
𝛾 𝐾𝑎
R
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 tr = c + sn tan f
𝑀𝐷 = 𝑊 ∙ 𝑥 For saturated clay under
undrained loading; f=0
 tr = c = su
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
A W
𝑀𝑅 = 𝑐 ∙ 𝜃2 ∙ 𝑅2

𝑀𝑅 𝑐 ∙ 𝜃2 ∙ 𝑅2
𝐹𝑂𝑆 = 𝐹𝑂𝑆 =
𝑀𝐷 𝑊∙𝑥

FOS will reduce after development of tension crack [∵ q2 < q]

5
 SLOPE STABILITY ANALYSIS
Swedish Slip Circle Method (Cohesive soils (f=0))
Case-III: Tension Crack filled with water
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝑅 ) x
𝐹𝑂𝑆 =
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 (𝑀𝐷 ) q2
h
C
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 2𝑐 2
ℎ𝑡 = ℎ
3 𝑡
1 R
𝛾 𝐾𝑎
PW
𝑃𝑊 = 𝛾𝑤 ∙ ℎ𝑡2
2 𝛾wht
1 2
𝑀𝐷 = 𝑊 ∙ 𝑥 + 𝛾𝑤 ∙ ℎ𝑡2 ℎ + ℎ𝑡
2 3 tr = c = su

𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 A W
𝑀𝑅 = 𝑐 ∙ 𝜃2 ∙ 𝑅2
𝑐 ∙ 𝜃2 ∙ 𝑅2
𝐹𝑂𝑆 =
1 2
𝑊 ∙ 𝑥 + 2 𝛾𝑤 ∙ ℎ𝑡2 ℎ + 3 ℎ𝑡

FOS will reduce further when tension crack is filled with water
6
Practice Problem #3
Determine the factor of safety of the cohesive slope shown in
the figure for the following two cases;
A. No tension crack
B. 2m deep tension crack filled with water (q1 = 38°)

3.1m
q1=
q2= 40° 4m
35°
C B NSL
gb = 17.75 kN/m3 2m
R
Cu above line AD = 21.5 kPa
Area of ABCA
Cu below line AD = 33.5 kPa = 90m2

D’
A D
W
7
 SLOPE STABILITY ANALYSIS
Ordinary Method of Slices (OMS) (c-f soils)
Guidelines for Slice Selection
 Slices do not have to be of equal
width
 For convenience, base arc of each
slice should pass through one soil
type only
 Slice width should be limited
(curved base approximated as
straight line)
TR

 For c-f soils, normal stress would change along slip circle
 Different normal stress means, shear resistance would also be
different (∵ M-C equation)
 Failing slope divided into slices
8
 SLOPE STABILITY ANALYSIS
Ordinary Method of Slices (OMS) (c-f soils)
b

𝑏
𝑙=
cos 𝛼
𝑙 = 𝑏 sec 𝛼

Wt

l
TR
TR
a
a
𝑀𝑅
𝐹𝑂𝑆 =
𝑀𝐷 TR = Total shear resistance force acting on slice

𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 𝑇𝑅 = 𝜏𝑅 × (𝑙 ∙ 1)
tR = Shear resistance (stress) offered by soil
𝑀𝑅 = 𝑇𝑅 × 𝑟 … (𝐸𝑞. 1)
𝑇𝑅 = (𝑐 ′ + 𝜎𝑛 ′ tan 𝜙) × 𝑙
9
 SLOPE STABILITY ANALYSIS
Ordinary Method of Slices (OMS) (c-f soils)
b
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡
𝑀𝑅 = 𝑇𝑅 × 𝑟 … (𝐸𝑞. 1)
𝑏
𝑇𝑅 = 𝜏𝑅 × (𝑙 ∙ 1) 𝑙=
cos 𝛼
𝑇𝑅 = (𝑐 ′ + 𝜎𝑛 ′ tan 𝜙) × 𝑙 𝑙 = 𝑏 sec 𝛼
Wt sin a
where, Wt cos a a
𝑊𝑡 cos 𝛼 Wt
𝜎𝑛 ′ =
𝑙×1
l
𝑊𝑡 cos 𝛼
𝑇𝑅 = 𝑐 ′𝑙 + 𝑙 tan 𝜙 TR
𝑙×1 a
a
𝑇𝑅 = 𝑐 ′𝑙 + 𝑊𝑡 cos 𝛼 tan 𝜙

𝐸𝑞. 1 𝑀𝑅 = 𝑟 (𝑐 ′ 𝑏 sec 𝛼 + 𝑊𝑡 cos 𝛼 tan 𝜙)

10
 SLOPE STABILITY ANALYSIS
Ordinary Method of Slices (OMS) (c-f soils)
b
𝐷𝑖𝑠𝑡𝑢𝑟𝑏𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡

𝑏
𝑙=
cos 𝛼
𝑙 = 𝑏 sec 𝛼
Wt sin a

Wt cos a a Wt
TR
l

Components of disturbing force (Wt) TR

a
1. Wt cos a  Passes through center of rotation, i.e. zero a
moment
2. Wt sin a  Tangential component; causing sliding

𝑀𝐷 = 𝑟 (𝑊𝑡 sin 𝛼)

11
SLOPE STABILITY ANALYSIS
Ordinary Method of Slices (OMS) (c-f soils)
b
𝑀𝑅
𝐹𝑂𝑆 =
𝑀𝐷
𝑏
𝑙=
𝑀𝑅 = 𝑟 (𝑐 ′ 𝑏 sec 𝛼 + 𝑊𝑡 cos 𝛼 tan 𝜙) cos 𝛼
𝑙 = 𝑏 sec 𝛼
𝑀𝐷 = 𝑟 (𝑊𝑡 sin 𝛼)
Wt sin a

Wt cos a a Wt

TR
a
(𝑐 ′ 𝑏 sec 𝛼
+ 𝑊𝑡 cos 𝛼 tan 𝜙) a
𝐹𝑂𝑆 =
(𝑊𝑡 sin 𝛼)

12
 CRITICAL SLIP CIRCLE
B NSL

1.79
1.55
1.20

A
FOS = 2.4

Many slip circles are possible on any slope

Slip circle having minimum FOS  Critical Slip Circle /

Critical Failure Plane

Design has to satisfy safety against critical slip circle

13
 LOCATION OF CRITICAL SLIP CIRCLE
In Cohesive soils (f=0) (Fellenius Method)
Case-I: Toe Failure
 Plot the configuration according to r
scale q1 B NSL
 Draw two lines from point A and B
at angles as has been shown in
figure.
q2
 Taking radius equal to OA, draw a
circle passing through the slope A

Slope Slope Angle q1 q2

1V : 0.5H 60° 29° 40°
1:1 45° 28° 38° Empirical values of q1 and q2
1 : 1.5 34° 26° 35°
1:2 27° 25° 35°
1:3 19° 25° 35°
14
 LOCATION OF CRITICAL SLIP CIRCLE
In Cohesive soils (f=0) (Fellenius Method)
Case-II: Base Failure 133.5°

 Plot the configuration according to

scale B NSL
 Draw a vertical line at the mid-point
of slope (the center of critical slip
FOS = 1.65
circle always lies on a vertical line FOS = 1.10
passing through the mid-point of
slope) A
 Determine the center by hit and trial
method by comparing the FOS 𝐻 𝐻
2 2
 Circle with minimum FOS is the
critical circle

 Angle made by critical circle at the

center is about 133.5°. (Fellenius)

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 LOCATION OF CRITICAL SLIP CIRCLE
In c-f soils
O4
O3
Omin = Ocr O2
O1

 Plot the configuration according to q1 B NSL

scale
 Find the intersection point of ‘4.5H’ H
horizontal and ‘H’ distance vertical q2
downward from A
A
 Draw the direction angle q1 and q2
H
 Join the points of intersection O and
C 4.5H
C
 Locate OC by hit and trial. For this
try O1, O2,…… and make circles.
 The circle giving minimum FOS is
the critical circle

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 SHORT TERM AND LONG TERM
STABILITY
Construction of dam core
- Clay material
- Very low permeability
- Construction in layers with compaction Clay Core

at OMC

SHORT TERM STABILITY

Stability of slope immediately after construction
 Undrained conditions
 Undrained parameters (cu and fu) to be used for slope stability analysis
 Obtained from UU or CU triaxial tests
 Total unit weight of soil (gb) to be used
 Called as TOTAL STRESS ANALYSIS
 Change in pore water pressure totally dependent upon stress change
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 SHORT TERM AND LONG TERM
STABILITY
Construction of dam core
- Clay material
- Very low permeability
- Construction in layers with compaction Clay Core

at OMC

LONG TERM STABILITY

Stability of slope long time after construction
 Drained conditions
 Drained parameters (cd (or c’) and fd (f’)) to be used for slope stability analysis
 Obtained from CD triaxial tests, or CU tests with PWP measurements
 Effective unit weight of soil (gsub (or g’)) to be used
 Called as EFFECTIVE STRESS ANALYSIS
 Change in pore water pressure independent of stress change
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REFERENCE MATERIAL
Principles of Geotechnical Engineering – (7th Edition)
Braja M. Das
Chapter #13

Essentials of Soil Mechanics and Foundations (7th Edition)

David F. McCarthy
Chapter #17

Geotechnical Engineering – Principles and Practices – (2nd Edition)

Coduto, Yueng, and Kitch
Chapter #17

THE END
In fact, this is just the beginning!
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