Calculate the shear strength in terms of effective stress on a plane within a saturated soil mass at a point where the total normal stress is 295 kPa and the pore water pressure is 120 kPa, when c′ = 12 kPa and the angle of internal friction is 30°. (Take tan 30° = 0.578.)
Try it first: pick an option, then read the explanation.
Correct answer: C. 113.15 kPa
Asked in SSC JE 30/10/2020 (Evening)
Detailed Explanation
Terzaghi's principle says soil strength is controlled by effective stress, the part of the total stress carried by the soil skeleton:
\[ \sigma' = \sigma - u \qquad \tau_f = c' + \sigma'\tan\phi' \]
\[ \sigma' = 295 - 120 = 175\ \text{kPa} \]
\[ \tau_f = 12 + 175 \times 0.578 = 12 + 101.15 = 113.15\ \text{kPa} \]
Why “113.15 kPa” Is Correct
Subtract the pore water pressure first, then apply the Mohr–Coulomb equation with c′ and φ′. The result is 113.15 kPa.
Why the Other Options Are Wrong
- A. 135.64 kPa: 135.64 kPa does not follow from the data; using total stress would give 12 + 295 × 0.578 = 182.5 kPa.
- B. 75.68 kPa: 75.68 kPa is too small; it would correspond to an effective stress of about 110 kPa.
- D. 85.06 kPa: 85.06 kPa is about what you get by omitting c′ and using too low a normal stress.
Key Point
Always compute σ′ = σ − u before using c′ and φ′.
Key Concepts for Students
- Water cannot carry shear stress, so pore pressure contributes nothing to strength.
- Rising pore pressure (e.g. after rain) reduces strength — a common cause of slope failure.
- Compare with Q293, which has the same structure.
Practice more: this is question 23 in Shear Strength of Soils MCQs – Part 2. Try the whole set and check every answer instantly.