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In a direct shear test, the shear stress and normal stress on a dry sand sample at failure are 0.6 kg/cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearly:
  • a)
    250
  • b)
    310
  • c)
    370
  • d)
    430
Correct answer is option 'B'. Can you explain this answer?
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Calculation of Angle of Internal Friction:
1. The formula to calculate the angle of internal friction (φ) in a direct shear test is:
φ = tan^(-1)(τ / σn)
2. Given values:
Shear stress (τ) = 0.6 kg/cm2
Normal stress (σn) = 1 kg/cm2
3. Substituting the values in the formula:
φ = tan^(-1)(0.6 / 1)
φ = tan^(-1)(0.6)
4. Calculating the value of angle:
φ = tan^(-1)(0.6)
φ ≈ 31 degrees

Conclusion:
Therefore, the angle of internal friction of the sand sample at failure is approximately 31 degrees. Hence, the correct option is B) 31 degrees.
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In a direct shear test, the shear stress and normalstress on a dry sand sample at failure are 0.6 kg/cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearly:a)250b)310c)370d)430Correct answer is option 'B'. Can you explain this answer?
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In a direct shear test, the shear stress and normalstress on a dry sand sample at failure are 0.6 kg/cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearly:a)250b)310c)370d)430Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a direct shear test, the shear stress and normalstress on a dry sand sample at failure are 0.6 kg/cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearly:a)250b)310c)370d)430Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a direct shear test, the shear stress and normalstress on a dry sand sample at failure are 0.6 kg/cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearly:a)250b)310c)370d)430Correct answer is option 'B'. Can you explain this answer?.
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