Question

An infinitely long slope is made up of a c-φ soil having the properties: cohesion, C = 20 kPa, and dry unit weight, γ_d = 16 kN/m³ . The angle of inclination and critical height of the slope are 40° and 5 m, respectively. To maintain the limiting equilibrium, the angle of internal friction of the soil (in degrees) is _________________ 
1. 22.4° 
2. 20.8° 
3. 19.4° 
4. None of the above

Answer 1

Correct Answer - Option 1 : 22.4° 

Concept:

Stability analysis of an infinite slope of cohesive soils

Shear strength of cohesive soils is given by,

S = C + (γ_d H Cos² i) tan ϕ

Shear stress of cohesive soil is given by,

τ = γ_d H cos i sin i

As the normal stress σ depends upon the height H of the slope, an expression for the height can be found by equating the shear stress and shear strength

C + (γ_d H Cos² i) tan ϕ = γ H cos i sin i

\({\rm{\gamma_d H}}{\cos ^2}{\rm{i}}\left( {\frac{{\sin {\rm{i}}}}{{\cos {\rm{i}}}} - \tan ϕ } \right) = {\rm{C}}\)

\({{\rm{H}}_{\rm{C}}} = \frac{{\rm{C}}}{{{\rm{\gamma_d }}\left( {\tan {\rm{i }} - \tan ϕ } \right){{\cos }^2}{\rm{i}}}}\)

This the height at which the slope is just stable and is known as critical height (H_C)

Calculation:

Given,

C = 20 kPa, γ_d = 16 kN/m³

i = 40°, H = 5 m

\({{\rm{H}}_{\rm{C}}} = \frac{{\rm{C}}}{{{\rm{\gamma_d }}\left( {\tan {\rm{i }} - \tan ϕ } \right){{\cos }^2}{\rm{i}}}}\)

\({{\rm{5}}} = \frac{{\rm{20}}}{{{\rm{16 }}\left( {\tan {\rm{40 }} - \tan ϕ } \right){{\cos }^2}{\rm{40}}}}\)

tan ϕ = 0.413

ϕ = 22.44°