When the water table is close to the ground surface, the bearing capacity of a soil is reduced to ______.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 14/11/2022
Detailed Explanation
For a sand (c = 0) the bearing capacity comes from the two terms that contain the unit weight γ:
\[ q_u = \gamma D_f N_q + \tfrac{1}{2}\gamma B N_\gamma \]
Below the water table, the effective unit weight is the submerged unit weight \(\gamma' = \gamma_{sat} - \gamma_w \approx 0.5\gamma\).
\[ q_u' = \gamma' D_f N_q + \tfrac{1}{2}\gamma' B N_\gamma \approx 0.5\,q_u \]
Why “one-half” Is Correct
With the water table at the ground surface, γ is replaced by γ′ ≈ γ/2 in both terms, so the bearing capacity is reduced to about one-half (a 50% reduction).
| Water table position | Reduction factor on the γ terms | Effect on bearing capacity of sand |
|---|---|---|
| At depth ≥ B below the footing base | 1.0 (no reduction) | None |
| At the footing base | Rw2 = 0.5 for the ½BγNγ term | Reduced |
| At the ground surface | 0.5 for both γDfNq and ½BγNγ | About half (50% reduction) |
Rw = 0.5(1 + zw/B or Dw/Df). The submerged unit weight is roughly half the saturated unit weight, which is why the maximum reduction is about 50%.
Why the Other Options Are Wrong
- A. three-seventh: Three-sevenths is not obtained from any standard reduction factor.
- B. one-third: One-third would need γ′ = γ/3, which is too low for real soils.
- D. one-fifth: One-fifth is far too severe.
Key Point
Water table at ground surface → bearing capacity of sand ≈ ½.
Key Concepts for Students
- Water table deeper than B below the base: no reduction.
- Water table at the base: only the Nγ term is halved.
- For φ = 0 clay, the cN_c term is unaffected by the water table.
Practice more: this is question 20 in Shallow and Deep Foundation MCQs – Part 2. Try the whole set and check every answer instantly.