# In a triaxial shear test conducted on a soil sample it having a cohesion of 12 kPa and the angle of shearing resistance of 36°, if the cell pressure i

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In a triaxial shear test conducted on a soil sample it having a cohesion of 12 kPa and the angle of shearing resistance of 36°, if the cell pressure is 200 kPa, the deviator stress at failure will be :

1. 617.5 kPa
2. 817.5 kPa
3. 770.37 kPa
4. 47.1 kPa
5.

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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

${\sigma _1} = {\sigma _d} + {\sigma _3}$

σ= deviator stress at failure

ϕ = Angle of internal friction of the soil

c = cohesion of the soil

Calculation:

ϕ = 36°, C = 12 kPa

σ= 200 kN/m2

σ= σ3 + σd

σ= 200 + σd

${\sigma _1} = 200{\tan ^2}\left( {45^\circ + \frac{{36^\circ }}{2}} \right) + 2\times 12\ \times \tan \left( {45^\circ + \frac{36^\circ }{2}} \right)$

σ= 817.47 kN/m2

Deviator stress at failure

σd = 817.47 – 200

σd = 617.47 kN/m2