Question

A soil has an angle of shearing of 30° and cohesion of 35 kN/m². If the specimen of this soil is subjected to a tri-axial compression test, then the value of lateral pressure in the cell for failure to occur at total stress of 300 kN/m² will be
1. 243.21 kN/m²
2. 44.41 kN/m²
3. 103.21 kN/m²
4. 59.59 kN/m²

Answer 1

Correct Answer - Option 4 : 59.59 kN/m²

Concept:

Relationship between C and principal stresses at failure:

Mohr coulomb failure criteria can be expressed in terms of the relationship between the principal stresses σ₁ and σ₃

\({σ _1} = {σ _3}{\tan ^2}\left( {{{45}^o} + \frac{ϕ }{2}} \right) + 2C\tan \left( {{{45}^o} + \frac{ϕ }{2}} \right)\)

Where, σ₁ = Total principal stress, σ₃ = Lateral princiapl stress

C = Cohesion, ϕ = angle of shearing or friction angle

Calculation:

Given,

C = 35 kN/m², σ₁ = 300 kN/m², ϕ = 30° 

∵ We know that, \({σ _1} = {σ _3}{\tan ^2}\left( {{{45}^o} + \frac{ϕ }{2}} \right) + 2C\tan \left( {{{45}^o} + \frac{ϕ }{2}} \right)\)

⇒ \(300 = {σ _3}{\tan ^2}\left( {{{45}^o} + \frac{{{{30}^o}}}{2}} \right) + 2 \times 35 \times \tan \left( {{{45}^o} + \frac{{{{30}^o}}}{2}} \right)\)

σ₃ = 59.585 kN/m²