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 ________.
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

The formula for Factor of Safety (FOS) under dry conditions is:
{FOS} = (cA + (W cos θ) tan φ) / (W sin θ)
where:
c = 196 { kPa} = 196 × 10³ {N/m}²
A = 50 {m}²
W = 2000 {tons} = 2000 × 1000 {kg} = 2 × 10⁶ {kg}
g = 9.81 {m/s}²
θ = 60°
φ = 30°

Step 1: Convert weight to Newtons:
W = 2 × 10⁶ × 9.81 = 1.962 × 10⁷ {N}

Step 2: Compute resisting forces:
cA = 196 × 10³ × 50 = 9.8 × 10⁶ {N}
(W cos θ) tan φ = (1.962 × 10⁷ × cos 60°) × tan 30°
= (1.962 × 10⁷ × 0.5) × 0.577 ≈ 9.81 × 10⁶ × 0.577 = 5.665 × 10⁶ {N}
Total resisting force = 9.8 × 10⁶ + 5.665 × 10⁶ = 1.5465 × 10⁷ {N}

Step 3: Compute driving force:
W sin θ = 1.962 × 10⁷ × sin 60° = 1.962 × 10⁷ × 0.866 ≈ 1.699 × 10⁷ {N}

Step 4: Compute FOS:
{FOS} = (1.5465 × 10⁷) / (1.699 × 10⁷) ≈ 0.91