Question

In saturated clay a cut slope has been made where slope is making angle B=55° with
horizontal. Taking circular critical sliding surface and the given parameters are
C, (undrained shear strength)=28 kN/m², y=20 kN/m stability number is 0.187.
Determine the maximum depth to which the cut could be possible ?
a) 7.48
b) 7.50
c) 7.52
d) 7.54

Answer 1

Answer: C

Explanation:The equation for stability number is given as:

N = tan(Φ) * (cot(Ψ) + c / (γ * H))

Where Φ is the angle of internal friction, Ψ is the angle of slope, c is the undrained shear strength, γ is the unit weight of soil, and H is the depth of the cut.

For this problem, Φ = 55°, Ψ = 55°, c = 28 kN/m², γ = 20 kN/m³, and N = 0.187.

Rearranging the equation, we can solve for H:

H = c / (γ * (tan(Φ) * (cot(Ψ) - N)))

Plugging in the values, we get:

H = 28 / (20 * (tan(55°) * (cot(55°) - 0.187))) = 7.52 m