In a direct shear test, the shear stress and normal stress on a dry sand at failure are 0.6 kg/cm² and 1 kg/cm² respectively. The angle of internal friction of the sand will be nearly:
Try it first: pick an option, then read the explanation.
Correct answer: B. 31°
Asked in SSC JE (previous year paper)
Detailed Explanation
A dry sand has no cohesion, so its failure envelope passes through the origin: τ = σ tan φ. One direct shear test is then enough to find φ.
\[ \tan\phi = \frac{\tau_f}{\sigma_n} = \frac{0.6}{1.0} = 0.6 \qquad \phi = \tan^{-1}(0.6) = 30.96^\circ \approx 31^\circ \]
Why “31°” Is Correct
tan⁻¹(0.6) is just under 31°, so the answer is 31°.
Why the Other Options Are Wrong
- A. 25°: tan 25° = 0.47, which would need τ = 0.47 kg/cm².
- C. 37°: tan 37° = 0.75, too high.
- D. 43°: tan 43° = 0.93, much too high.
Key Point
c = 0 ⇒ φ = tan⁻¹(τ/σ) from a single test.
Key Concepts for Students
- Useful values: tan 30° = 0.577, tan 31° = 0.60, tan 35° = 0.70, tan 45° = 1.
- Units cancel in τ/σ, so kg/cm² can be used directly.
- φ ≈ 31° suggests a medium-dense sand.
Practice more: this is question 54 in Shear Strength of Soils MCQs – Part 4. Try the whole set and check every answer instantly.