A shear box test was performed to give the following results for a cohesive soil sample: Test 1 — normal stress 150 kN/m², shear stress at failure 110 kN/m²; Test 2 — normal stress 250 kN/m², shear stress at failure 120 kN/m². The values of c and tan φ are:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 30/10/2020 (Evening)
Detailed Explanation
Each direct shear test gives one point (σ, τ) on the failure envelope τ = c + σ tan φ. Two points are enough to fix the straight line: its slope gives tan φ and its intercept gives c.
\[ 110 = c + 150\tan\phi \qquad (1) \]
\[ 120 = c + 250\tan\phi \qquad (2) \]
(2) − (1): \(10 = 100\tan\phi \Rightarrow \tan\phi = 0.1\)
From (1): \(c = 110 - 150 \times 0.1 = 95\ \text{kN/m}^2\)
Why “tan φ = 0.1 and c = 95 kN/m²” Is Correct
The shear strength rises only 10 kN/m² for a 100 kN/m² increase in normal stress, so the envelope is nearly flat (tan φ = 0.1, φ ≈ 5.7°) with a large intercept of 95 kN/m². That is typical of a cohesive soil.
Why the Other Options Are Wrong
- A. tan φ = 1.0 and c = 108 kN/m²: tan φ = 1.0 would mean τ rises 100 kN/m² for the 100 kN/m² increase in σ, but it rose only 10.
- B. tan φ = 0.8 and c = 70 kN/m²: tan φ = 0.8 and c = 70 do not satisfy either data point (70 + 0.8 × 150 = 190 ≠ 110).
- D. tan φ = 1.2 and c = 108 kN/m²: tan φ = 1.2 is far too steep and c = 108 does not fit the data.
Key Point
Two tests → subtract equations → tan φ = Δτ/Δσ; then c = τ − σ tan φ.
Key Concepts for Students
- Check the answer by substituting into the second equation: 95 + 250 × 0.1 = 120 ✓.
- With more than two tests, draw the best-fit line.
- φ = tan⁻¹(0.1) ≈ 5.7°.
Practice more: this is question 21 in Shear Strength of Soils MCQs – Part 2. Try the whole set and check every answer instantly.