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°
\(\alpha = 45^\circ + \frac{{43}}{2} = 66.5^\circ \)