Question

For a sandy soil the angle of internal friction is 30 degree if the major principal stress is 50 kilo newton per metre square to failure the corresponding minor principal stress will be

Answer 1

For a sandy soil, the angle of internal friction is 30‚°. If the major principal stress is 50kN/m² at failure, then the corresponding minor principal stress be 16.67 kN/m²

Explanation:

Given

Angle of internal friction

$\phi=30^\circ$

Major principal stress at failure

$\sigma_{1u}'=50 kPa$

The relationship between major $\sigma_{1u}'$ and minor principal stress $\sigma_{3u}'$ at failure is

$\boxed{\sigma_{1u}'=\sigma_{3u}'\tan^2(45^\circ+\frac{\phi}{2})}$

$\implies 50=\sigma_{3u}'\tan^2(45^\circ+\frac{30^\circ}{2})$

$\implies 50=\sigma_{3u}'\tan^2(60)$

$\implies 50=\sigma_{3u}'\times 3$

$\implies \sigma_{3u}'=\frac{50}{3}$

$\implies \sigma_{3u}'=16.67$ kN/m²

Hope this answer is helpful.

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