Question

For a sandy soil, the angle of internal friction is 30‚°. If the major principal stress is 50kpa at failure, then the corresponding minor principal stress (in kpa) will be

Answer 1

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Answer 2

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

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

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

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

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

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

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

$\implies \sigma_{3u}'=16.67$ kPa

Hope this answer is helpful.

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