Geotechnical Engineering

DHARM

344 GEOTECHNICAL ENGINEERING

Cohesive strength c = 200 kN/m^2

∴ Factor of safety with respect to cohesion:

Fc =

c

cm

=^200

16 2.

≈ 12.3.

Example 9.5: An embankment has a slope of 30° to the horizontal. The properties of the soil
are: c = 30 kN/m^2 , φ = 20°, γ = 18 kN/m^3. The height of the embankment is 27 m. Using Taylor’s
charts, determine the factor of safety of the slope.

From Taylor’s charts, it will be seen that a slope with θ = 20° and β = 30° has a stability
number 0.025. That is to say, if the factor of safety with respect to friction were to be unity
(implying full mobilisation of friction), the mobilised cohesion required will be found from

N =

c

H

m

γ

0.025 =

cm

18 27×

∴ cm = 18 × 27 × 0.025 = 12.15 kN/m^2

∴ Factor of safety with respect to cohesion

Fc = c/cm = 30/12.15 = 2.47

But the factor of safety Fs against shearing strength is more appropriate:

Fs =

c+σ φ

τ

tan

F may be found be successive approximations as follows:

Let us try F = 1.5

tanφ

F

= 0.364/1.5 = 0.24267 ≈ tangent of angle 13

2

3

°

With this value of φ, the new value N from charts is found to be 0.055.

c = 0.055 × 18 × 27 = 26.73

∴ F with respect to c = 30/26.73 = 1.12

Try F = 1.3

tanφ

F

= 0.364/1.3 = 0.280 = Tangent of 15

2

3

°

From the charts, new value of N = 0.045

c = 0.045 × 18 × 27 = 21.87

∴ Fc = 30/21.87 = 1.37

Let us try F = 1.35

tanφ

F

= 0.364/1.35 = 0.27 = tangent of angle 15°.1

From the charts, the new value of N = 0.046

c = 0.046 × 18 × 27 = 22.356

∴ Fc = 30/22.356 = 1.342 (= Fφ)

This is not very different from the assumed value.