Geotechnical Engineering

DHARM

342 GEOTECHNICAL ENGINEERING

9.4 Illustrative Examples

Example 9.1: Fig. 9. 27 shows the details of an em-
bankment made of cohesive soil with φ = 0 and c = 30
kN/m^2. The unit weight of the soil is 18.9 kN/m^3. De-
termine the factor of safety against sliding along the
trial circle shown. The weight of the sliding mass is
360 kN acting at an eccentricity of 5.0 m from the
centre of rotation. Assume that no tension crack de-
velops. The central angle is 70°.

Sliding moment = 360 × 5 = 1800 kNm

Restoring moment

= c. r^2 θ = 30 9

70

180

×× ×π^2

= 2970 kNm

Factor of safety against sliding,

F = 2970/1800 = 1.65.

Example 9.2: A cutting is made 10 m deep with sides sloping at 8 : 5 in a clay soil having a
mean undrained strength of 50 kN/m^2 and a mean bulk density of 19 kN/m^3. Determine the
factor of safety under immediate (undrained) conditions given the following details of the im-
pending failure circular surface: The centre of rotation lies vertically above the middle of the
slope. Radius of failure arc = 16.5 m. The deepest portion of the failure surface is 2.5 m below
the bottom surface of the cut (i.e., the centre of rotation is 4 m above the top surface of the cut).
Allowance is to be made for tension cracks developing to a depth of 3.5 m from surface. Assume
that there is no external pressure on the face of the slope.
(S.V.U.—B.E., (R.R.)—Sep., 1978)
The data are shown in Fig. 9.28.

10 m

2

1

(^45)
6
2.5 m
W=
3040
kN
3
5
8 3.5 m
Tension crack
4m q¢= 93°
q= 105°
e5m»
r = 16.5 m
Fig. 9.28 Trial failure surface (Ex. 9.2)
Mean undrained strength = 50 kN/m^2 , ∴ c = 50 kN/m^2
Factor of safety Fc = cr^2 θ′/W. e
5m
1
1
9m
6m W
70°
Fig. 9.27 Trial slip circle (Ex. 9.1)