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For a sandy soil the angle of internal friction is 30 °. if the major principal stress is 50kn/m2 at failure, the corresponding minor principal stress will be?
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For a sandy soil the angle of internal friction is 30 °. if the major ...
Answer:

Given:
Angle of internal friction (ϕ) = 30°
Major principal stress (σ1) = 50 kN/m2

Formula:
σ1/σ3 = (1 + sinϕ) / (1 – sinϕ)

Calculation:
Let us assume the minor principal stress as σ3.
σ1/σ3 = (1 + sinϕ) / (1 – sinϕ)
50/σ3 = (1 + sin30°) / (1 – sin30°)
50/σ3 = (1 + 0.5) / (1 – 0.5)
50/σ3 = 1.5/0.5
σ3 = 50/3
σ3 = 16.67 kN/m2

Therefore, the minor principal stress is 16.67 kN/m2.

Conclusion:
The minor principal stress for a sandy soil with an angle of internal friction of 30° and a major principal stress of 50 kN/m2 is 16.67 kN/m2.
Community Answer
For a sandy soil the angle of internal friction is 30 °. if the major ...
16.6kn/M^2
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For a sandy soil the angle of internal friction is 30 °. if the major principal stress is 50kn/m2 at failure, the corresponding minor principal stress will be?
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