Correct Answer - Option 1 : 617.5 kPa
Concept:
Using the relation
\({\sigma _1} = {\sigma _3}{\tan ^2}\left( {45^\circ + \frac{\phi }{2}} \right) + 2c\tan \left( {45^\circ + \frac{\phi }{2}} \right)\)
Where,
σ1 = maximum principal stress
σ3 = all round cell pressure
\({\sigma _1} = {\sigma _d} + {\sigma _3}\)
σd = deviator stress at failure
ϕ = Angle of internal friction of the soil
c = cohesion of the soil
Calculation:
ϕ = 36°, C = 12 kPa
σ3 = 200 kN/m2
σ1 = σ3 + σd
σ1 = 200 + σd
\({\sigma _1} = 200{\tan ^2}\left( {45^\circ + \frac{{36^\circ }}{2}} \right) + 2\times 12\ \times \tan \left( {45^\circ + \frac{36^\circ }{2}} \right)\)
σ1 = 817.47 kN/m2
Deviator stress at failure
σd = 817.47 – 200
σd = 617.47 kN/m2