A sample of soil failed in a triaxial test under a deviator stress of 200 kN/m² when the confining pressure was 100 kN/m². For the sample, if the confining pressure had been 200 kN/m², what would have been the deviator stress at failure? Assume c = 0.

Concept:

Relation between major and minor principal stress at failure on the basis of Mohr-Coulomb criteria of failure:

\({σ _{1}} = {σ _{3}}{\tan ^2}\left( {45 + \frac{ϕ }{2}} \right) + 2C\tan \left( {45 + \frac{ϕ }{2}} \right)\)

\({σ _{3}} = {σ _{1}}{\tan ^2}\left( {45 - \frac{ϕ }{2}} \right) - 2C\tan \left( {45 - \frac{ϕ }{2}} \right)\)

Calculation:

σ3 = Cell pressure = 100 kPa

σ1= Cell pressure + Deviator stress = 100 + 200 = 300 kPa

c = 0

\({σ _{1}} = {σ _{3}}{\tan ^2}\left( {45 + \frac{ϕ }{2}} \right) + 2C\tan \left( {45 + \frac{ϕ }{2}} \right)\)

300=100(1)+2C(1) 

C=100 kN/m²

For dry sand, C = 0

\(300 = 100 \tan^2\left(45 + \frac{ϕ}{2}\right) + 0 \)

ϕ = 30^o

If confining pressure is 200 kN/m2 then,

\({σ _{3f}} = {σ _{1f}}{\tan ^2}\left( {45 - \frac{ϕ }{2}} \right) - 2C\tan \left( {45 - \frac{ϕ }{2}} \right)\)

\(200 = \sigma_1 f \tan^2\left(45 - \frac{30}{2}\right) - 0 \)

\(\sigma_1 f = 600 \, \text{kN/m}^2 \)

Deviator stress = 600-200 = 400 kN/m²