A friction pile of diameter 50 cm is embedded 15 m into a homogeneous consolidated clay deposit. If the adhesion factor is 0.7 and the cohesion (c) developed between the clay and pile shaft is 4 t/m², what is the safe load (t) for a factor of safety of 3.0?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 27/01/2018 (Evening)
Detailed Explanation
A friction pile in clay carries its load mainly by adhesion (skin friction) along its shaft. The adhesion is a fraction α of the clay's undrained cohesion:
\[ Q_s = \alpha\,c\,A_s = \alpha\,c\,(\pi D L) \]
\[ A_s = \pi \times 0.5 \times 15 = 23.56\ \text{m}^2 \]
\[ Q_u = 0.7 \times 4 \times 23.56 = 65.97\ \text{t} \qquad Q_{\text{safe}} = \frac{65.97}{3} \approx 22\ \text{t} \]
Why “22” Is Correct
End bearing is ignored for a friction pile, so the ultimate load is the shaft adhesion, 66 t. Dividing by the factor of safety of 3 gives about 22 t.
Why the Other Options Are Wrong
- A. 15: 15 t would need a much smaller adhesion or a larger factor of safety.
- C. 26: 26 t does not follow from these numbers.
- D. 30: 30 t is roughly what you get by forgetting the adhesion factor (4 × 23.56/3 = 31.4).
Key Point
Friction pile in clay: Qu = α·c·π·D·L; safe = Qu/FOS.
Key Concepts for Students
- α is about 1 for soft clay and falls (0.3–0.5) for stiff clay.
- For end-bearing add Qb = 9c × base area.
- Group capacity is checked separately (block failure).
Practice more: this is question 15 in Earth Pressure and Retaining Wall MCQs – Part 1. Try the whole set and check every answer instantly.