A simply supported beam of span 3·0 m has a cross-section 120 mm × 180 mm. If the permissible stress in the material of the beam is 10 N/mm², what is the maximum uniformly distributed load (UDL) that it can carry ? (Ignore the moment due to self-weight)
- A9·72 kN/m
- B4·52 kN/m
- C5·76 kN/mCorrect
- D8·80 kN/m
Solution
Concept
Flexure formula
Formula
M = σ Z; Z = b d² / 6; M = w L² / 8
Step-by-step solution
1. Z = 120 × 180² / 6 = 6.48 × 10⁵ mm³. 2. M = σ Z = 10 × 6.48 × 10⁵ = 6.48 × 10⁶ N·mm = 6.48 kN·m. 3. w = 8 M / L² = 8 × 6.48 / 9 = 5.76 kN/m.
Explanation
Take the 180 mm side as depth (the usual orientation). The beam's moment of resistance gives the maximum bending moment, from which the UDL follows.
Why the other options are wrong
With Z = 6.48 × 10⁵ mm³ and L = 3 m, w = 8σZ/L² = 5.76 kN/m; 9.72, 4.52 and 8.80 kN/m do not satisfy this.
Shortcut
w = 8 σ b d² / (6 L²).
Common mistake
Using L² as 3 instead of 9.
Correct answer: C) 5·76 kN/m