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. If, for the sample, the confining pressure had been 200 kN/m², what would have been the deviator stress at failure? (Assume φ = 0.)
Try it first: pick an option, then read the explanation.
Asked in SSC JE 06/06/2024 (Morning)
Detailed Explanation
For a soil with φ = 0 (for example a saturated clay sheared undrained), the failure envelope is a horizontal line at τ = c. Every Mohr circle at failure must just touch this line, so every circle has the same radius, c.
\[ \sigma_1 = \sigma_3 \tan^2\!\left(45^\circ + \tfrac{\phi}{2}\right) + 2c\,\tan\!\left(45^\circ + \tfrac{\phi}{2}\right) \]
With \(\phi = 0\): \(\sigma_1 = \sigma_3 + 2c\). From the first test: \(300 = 100 + 2c \Rightarrow c = 100\ \text{kN/m}^2\).
Second test: \(\sigma_1 = 200 + 2(100) = 400\ \text{kN/m}^2\), so \(\sigma_d = 400 - 200 = 200\ \text{kN/m}^2\).
Why “200 kN/m²” Is Correct
Because the deviator stress (the diameter of the Mohr circle) equals 2c, raising the cell pressure only slides the circle to the right; it does not make it bigger. The deviator stress at failure stays 200 kN/m².
Why the Other Options Are Wrong
- A. 500 kN/m²: 500 kN/m² is not obtainable from either φ = 0 or c = 0 assumptions with these data.
- C. 100 kN/m²: 100 kN/m² would mean the soil became weaker with more confinement, which does not happen.
- D. 400 kN/m²: 400 kN/m² is σ₁ in the second test, not the deviator stress (and it would also be the answer if c = 0, φ = 30°).
Key Point
φ = 0 ⇒ σ₁ − σ₃ = 2c = constant, whatever the cell pressure.
Key Concepts for Students
- Unconfined compression: qu = 2cu for φ = 0 clays.
- Compare with Q263 (same numbers, but c = 0): there the deviator stress doubles.
- UU tests on saturated clay give φu ≈ 0 because added cell pressure is carried by pore water.
Practice more: this is question 4 in Shear Strength of Soils MCQs – Part 1. Try the whole set and check every answer instantly.