Question

A cut of depth 12m is made in a cohesive soil deposit ( c = 40 kn/m , ϕ=0 and γ = 20 kn/m ). If the required factor of safety is 2.0, the stability number is

Answer 1

Answer: The stability number is 0.083.

Explanation:

Given that,

Cohesive soil deposit $c = 40 kN/m^{2}$

Unit weight $\gamma = 20 kN/m^{3}$

Safety factor F = 2.0

Depth of cut H = 12 m

We know that,

The stability number is

$S_{n} = \dfrac{c}{F\gamma H}$

$S_{n} = \dfrac{40 kN/m^{2}}{2.0\times20kN/m^{3}\times 12 m}$

$S_{n} = 0.083$

Hence, the stability number is 0.083.