If the bearing capacity of soil is 10 tonnes/m² and the projection of a plain concrete footing from the wall is a cm, the depth D of the footing is: i. D = 0.0775a, ii. D = 0.775a, iii. D = 0.775√a, iv. D = 0.775a²
Try it first: pick an option, then read the explanation.
Asked in SSC JE 27/01/2018 (Evening)
Detailed Explanation
A plain (unreinforced) concrete footing must not crack in bending, because it has no steel to carry tension. The part projecting beyond the wall face behaves like a short cantilever loaded upward by the soil pressure.
Per unit length of wall: \(M = \dfrac{p a^2}{2}\), section modulus \(Z = \dfrac{D^2}{6}\).
\[ f_t = \frac{M}{Z} = \frac{3pa^2}{D^2} \quad\Rightarrow\quad D = a\sqrt{\frac{3p}{f_t}} \]
With \(p = 10\ \text{t/m}^2 \approx 0.1\ \text{N/mm}^2\) and \(f_t \approx 0.5\ \text{N/mm}^2\): \(D = a\sqrt{0.6} = 0.775a\).
Why “Only ii” Is Correct
The depth comes out as a constant times a, so only statement ii (D = 0.775a) has the right form and value. (The printed question says ‘10 tonnes/cm²’, which is clearly a misprint for 10 t/m²; a bearing pressure of 10 t/cm² would be impossibly high for soil.)
Why the Other Options Are Wrong
- A. Only i: 0.0775a is ten times too small.
- C. Only iii: D must be proportional to a, not to √a (check the units: cm vs √cm).
- D. Only iv: D proportional to a² is dimensionally wrong.
Key Point
Plain concrete footing: D = a√(3p/ft). Bigger soil pressure or projection → deeper footing.
Key Concepts for Students
- IS 456 also limits the spread of load in plain concrete footings by a minimum angle.
- Footings for walls are often stepped to save concrete.
- Reinforced footings are used when D would otherwise become uneconomic.
Practice more: this is question 13 in Earth Pressure and Retaining Wall MCQs – Part 1. Try the whole set and check every answer instantly.