Geotechnical Engineering

DHARM

STABILITY OF EARTH SLOPES 345

\ The factor of safety for the slope = 1.35.

Example 9.6: A cutting is to be made in clay for which the cohesion is 35 kN/m^2 and φ = 0°. The

density of the soil is 20 kN/m^3. Find the maximum depth for a cutting of side slope 112 to 1 if

the factor of safety is to be 1.5. Take the stability number for a 112 to 1 slope and φ = 0° as 0.17.

(S.V.U.—B.E., (N.R.)—Apr., 1966)

c = 35 kN/m^2 φ = 0°

γ = 20 kN/m^3 N = 0.17

Fc = 1.5

∴ cm = c/Fc = 35/1.5 =

70

3

kN/m^2

But N =

c

H

m

γ

∴ 0.17 =

c

HH

m×

×

=

×

××

100

20

70 100

320

∴ H =

70 100

60 017

×

×.

cm

= 6.86 m.

Example 9.7: At cut 9 m deep is to be made in a clay with a unit weight of 18 kN/m^3 and a

cohesion of 27 kN/m^2. A hard stratum exists at a depth of 18 m below the ground surface.

Determine from Taylor’s charts if a 30° slope is safe. If a factor of safety of 1.50 is desired, what

is a safe angle of slope?

Depth factor D = 18/9 = 2

From Taylor’s charts,

for D = 2; b = 30°

N = 0.172

0.172 =

cm

18 9×

cm = 0.172 × 18 × 9 = 27.86 kN/m^2

c = 27 kN/m^2

Fc = c/cm = 27/27.86 = 0.97

The proposed slope is therefore not safe.

For Fc = 1.50

cm = c/Fc = 27/1.5 = 18 kN/m^2

N = 18/18 × 9 = 1/9 = 0.11

For D = 2.0, and N = 0.11

from Taylor’s charts,

we have b = 8 °