A long rectangular wall is 2·5 m wide. The maximum wind pressure on the face of wall is 1·1 kN/m² and the specific weight of masonry is 22 kN/m³. If the length of the wall is 1 m, the maximum height of the wall so that there is no tension in the base of the wall will be nearly
- A52 m
- B42 mCorrect
- C32 m
- D22 m
Solution
Concept
Wall under wind pressure: no tension at base
Formula
Mwind ≤ W b / 6 (resultant within the middle third)
Step-by-step solution
1. For a 1 m length: wind force = p h = 1.1 h; its moment about the base = 1.1 h × h/2 = 0.55 h². 2. Weight W = γ b h = 22 × 2.5 × h = 55 h. 3. No tension: M ≤ W b/6 ⇒ 0.55 h² ≤ 55 h × 2.5/6 = 22.92 h. 4. h ≤ 22.92 / 0.55 = 41.7 m ≈ 42 m.
Explanation
The resultant of weight and wind must stay in the middle third of the base. The limiting height is where the extreme fibre stress at the windward side becomes zero.
Why the other options are wrong
22, 32 and 52 m do not satisfy h = γ b² / (3p) = 41.7 m.
Shortcut
h = γ b² / (3 p).
Common mistake
Taking the wall thickness as 1 m (the "length").
Correct answer: B) 42 m