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In a direct shear test, the shear stress and normla stress on a dry sand sample at failureare 0.6Kg/cm2 and 1Kg/cm2 respectively. The angle of internal friction of the sand will benearby
  • a)
    25°
  • b)
    31°
  • c)
    37°
  • d)
    43°
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
In a direct shear test, the shear stress and normla stress on a dry sa...
Given:
Shear stress = 0.6 Kg/cm²
Normal stress = 1 Kg/cm²

To find: Angle of internal friction

Formula:
Tan φ = Shear stress / Normal stress

Calculation:
Tan φ = 0.6 / 1
φ = tan⁻¹(0.6/1)

φ = 31.01 ≈ 31°

Therefore, the angle of internal friction of the sand is approximately 31°.
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In a direct shear test, the shear stress and normla stress on a dry sa...
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In a direct shear test, the shear stress and normla stress on a dry sand sample at failureare 0.6Kg/cm2 and 1Kg/cm2 respectively. The angle of internal friction of the sand will benearbya)25°b)31°c)37°d)43°Correct answer is option 'B'. Can you explain this answer?
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