Correct Answer - Option 3 : 25 kN/m
2
Concept:
We know,
As per triaxial stress, the relationship between principal stress and cohesion -
σ1 = σ3tan2 (45° + ϕ/2) + 2 × c × tan(45° + ϕ/2) ...(1)
For clayey soil subjected unconfined triaxial test, ϕ = 0° and σ3 = 0 kN/m2
From (1)
σ1 = 2c = qu
⇒ c = \(\frac{q_u}{2}\) ...(2)
Here,
c - Cohesion (kN/m2)
qu - Unconfined compressive strength (kN/m2)
σ3 - Confining stress (kN/m2)
σ1 - Major principal stress (kN/m2)
Calculation:
Given,
qu = 50 kN/m2
From (2)
c = \(\frac{50}{2}\) = 25 kN/m2
We know,
As per columb's theory, shear stress is given by -
τ = c + \(\bar{σ}\) tanϕ
For clayey soil, ϕ = 0°
⇒ τ = c = 25 kN/m2