In a drained triaxial compression test conducted on dry sand, failure occurred when the deviator stress was 218 kN/m2 at a confining pressure of 61 kN

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In a drained triaxial compression test conducted on dry sand, failure occurred when the deviator stress was 218 kN/m2 at a confining pressure of 61 kN/m2. The pore pressure at failure was observed to be 10 kPa. The effective angle of shearing resistance and the inclination of the failure plane to the major principal plane will be:
1. 34°, 62°
2. 34°, 28°
3. 40°, 25°
4. 43°, 66.5°

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Correct Answer - Option 4 : 43°, 66.5°

Concept:

In a triaxial test,

${\bar σ _1} = \;{\bar σ _3}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right) + 2c\tan \left( {45 + \frac{\phi }{2}} \right)$

Where σ1 = Major Principal stress, σ 3 = Minor principal stress

σ̅3 = confining pressure

σ̅1 = σ̅3 + σ̅d

σ̅d = deviator stress

c = Effective cohesion

ϕ = angle of internal friction.

α = inclination of failure plane to major principal plane.

$\alpha = 45^\circ + \frac{\phi }{2}$

Calculation:

Given,

Dry sand is given, so c = 0. u = 10 kPa

σd = 218 kN/m2, σc = 61 kN/m2

σ 1 = 61 + 218 = 279 kN/m2

σ 3 = 61 kN/m2

σ̅3 = 61 - 10 = 51 kN/m2, σ̅1 = 279 - 10 = 269 kN/m2.

${\bar σ _1} = {\bar σ _3}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right)$

$269 = 51 \times {\tan ^2}\left( {45 + \frac{\phi }{2}} \right)$

$45^\circ + \frac{\phi }{2} = 66.41^\circ$

$\frac{\phi }{2} = 21.41^\circ$

ϕ = 42.82° ≈  43°

$\alpha = 45^\circ + \frac{{43}}{2} = 66.5^\circ$