Question

A hillslope is shown below. If the area over the failure plane is 50 m\(^2\) and the weight of the hillslope material (W) is 2000 tons, the Factor of Safety (FOS) for this hillslope in dry conditions is % Given: Cohesion along failure plane = 196 kPa, Dip of failure plane = 60°, Internal friction angle = 30°, Area over failure plane = 50 m\(^2\), Weight of hillslope material = 2000 tons (Round off to two decimal places)

Hint:

The Factor of Safety (FOS) is a measure of slope stability. A value greater than 1 indicates a stable slope, and a value less than 1 suggests failure.

Answer 1

Correct Answer: 0.85

The formula for Factor of Safety (FOS) under dry conditions is: \[ {FOS} = \frac{cA + (W \cos \theta) \tan \phi}{W \sin \theta} \] where:
$c = 196 { kPa} = 196 \times 10^3 \, {N/m}^2$
$A = 50 \, {m}^2$
$W = 2000 \, {tons} = 2000 \times 1000 \, {kg} = 2 \times 10^6 \, {kg}$
$g = 9.81 \, {m/s}^2$
$\theta = 60^\circ$
$\phi = 30^\circ$
Step 1: Convert weight to Newtons: \[ W = 2 \times 10^6 \times 9.81 = 1.962 \times 10^7 \, {N} \] Step 2: Compute resisting forces: \[ cA = 196 \times 10^3 \times 50 = 9.8 \times 10^6 \, {N} \] \[ (W \cos \theta) \tan \phi = (1.962 \times 10^7 \times \cos 60^\circ) \times \tan 30^\circ \] \[ = (1.962 \times 10^7 \times 0.5) \times 0.577 \approx 9.81 \times 10^6 \times 0.577 = 5.665 \times 10^6 \, {N} \] \[ {Total resisting force} = 9.8 \times 10^6 + 5.665 \times 10^6 = 1.5465 \times 10^7 \, {N} \] Step 3: Compute driving force: \[ W \sin \theta = 1.962 \times 10^7 \times \sin 60^\circ = 1.962 \times 10^7 \times 0.866 \approx 1.699 \times 10^7 \, {N} \] Step 4: Compute FOS: \[ {FOS} = \frac{1.5465 \times 10^7}{1.699 \times 10^7} \approx 0.91 \]