If the one-way eccentricity along Y-axis is 0.1 m for a 2 m × 2 m square footing, then the effective area of the footing is
- A4 m²
- B3.6 m²Correct
- C3.8 m²
- DNone of these
Solution
Concept
Meyerhof's effective area for eccentric load
Formula
B′ = B − 2e; A′ = B′ × L′
Step-by-step solution
1. Eccentricity e = 0.1 m along one axis of a 2 m × 2 m footing. 2. Effective width B′ = 2 − 2 × 0.1 = 1.8 m; length L′ = 2 m. 3. Effective area A′ = 1.8 × 2 = 3.6 m².
Explanation
With a one-way eccentricity the useful width of the footing is reduced by twice the eccentricity, so the effective area is 3.6 m².
Why the other options are wrong
4 m² is the full area; 3.8 m² would correspond to deducting only e (not 2e).
Shortcut
Effective dimension = B − 2e.
Common mistake
Deducting e instead of 2e (gives 3.8 m²).
Correct answer: B) 3.6 m²