Question

A shear box test was performed to give the following results for a cohesive soil sample.

Result:	(1)	(2)
Normal stress σ (kN/m2)	150	250
Shear stress at failure τ (kN/m2)	110	120

 

The value of c and tan ϕ are;

1. tan ϕ = 0.1 and c = 95 kN/m²
2. tan ϕ = 0.8 and c = 70 kN/m²
3. tan ϕ = 1.0 and c = 108 kN/m²
4. tan ϕ = 1.2 and c = 108 kN/m²

Answer 1

Correct Answer - Option 1 : tan ϕ = 0.1 and c = 95 kN/m²

Concept:

Coulomb's equation,

τ = C + σ tan ϕ

Where,

τ = Shear strength of soil

C = Apparent cohesion

σ = Normal stress on the plane of rupture

ϕ = Angle of internal friction

Calculation:

Given,

σ₁ = 150 kN/m² ,  σ₂ = 250 kN/m2

τ₁ = 110 kN/m2 ,   τ₂ = 120 kN/m²

We know,

τ = C + σ tan ϕ

When, σ1 = 150 kN/m² and τ1 = 110 kN/m2

110 = C + 150 tan ϕ........(1)

When, σ1 = 250 kN/m2 and τ1 = 120 kN/m2

120 = C + 250 tan ϕ........(2)

Equation (2) - (1)

10 = 100 tan ϕ

tan ϕ = 0.1

Put tan ϕ in equation (1), 

110 = C + 150 × 0.1

C = 95 kN/m².