A sample of soil failed in a triaxial test under a deviator stress of 200 kN/m² when the confining pressure was 100 kN/m². For the sample, if the confining pressure had been 200 kN/m², what would have been the deviator stress at failure? Assume c = 0.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 05/06/2024 (Morning)
Detailed Explanation
For a cohesionless soil (c = 0) the failure envelope passes through the origin. At failure the ratio of major to minor principal stress is fixed by φ alone:
\[ \frac{\sigma_1}{\sigma_3} = \tan^2\!\left(45^\circ + \frac{\phi}{2}\right) \]
Test 1: \(\sigma_1 = 100 + 200 = 300\ \text{kN/m}^2\), so \(\tan^2(45^\circ+\phi/2) = 300/100 = 3 \Rightarrow \phi = 30^\circ\).
Test 2: \(\sigma_1 = 200 \times 3 = 600\ \text{kN/m}^2\).
\[ \sigma_d = \sigma_1 - \sigma_3 = 600 - 200 = 400\ \text{kN/m}^2 \]
Why “400 kN/m²” Is Correct
Since c = 0, the stress ratio stays at 3. Doubling the cell pressure doubles σ₁ and therefore doubles the deviator stress, from 200 to 400 kN/m².
Why the Other Options Are Wrong
- B. 600 kN/m²: 600 kN/m² is σ₁ in the second test; the deviator stress is σ₁ − σ₃.
- C. 500 kN/m²: 500 kN/m² does not follow from a constant stress ratio of 3.
- D. 100 kN/m²: 100 kN/m² would mean strength fell with more confinement, which is wrong for a frictional soil.
Key Point
c = 0 ⇒ σ₁/σ₃ constant ⇒ deviator stress is proportional to σ₃.
Key Concepts for Students
- tan²(45° + 15°) = tan²60° = 3, which is why φ = 30° gives a ratio of 3.
- Kp = tan²(45° + φ/2) is the same as Rankine's passive earth pressure coefficient.
- Compare with Q260 (φ = 0), where the deviator stress does not change.
Practice more: this is question 7 in Shear Strength of Soils MCQs – Part 1. Try the whole set and check every answer instantly.