A rectangular beam 150 mm × 240 mm deep is simply supported at the ends on a span of 4 m and carries a uniformly distributed load of 4 kN/m on the whole span. What is the point load at the centre it should carry so that the maximum deflection is doubled?
- A24 kN
- B20 kNCorrect
- C16 kN
- D12 kN
Solution
Concept
Central deflection of a simply supported beam
Formula
δudl = 5wL⁴/384EI; δP = PL³/48EI
Step-by-step solution
1. δudl = 5 w L⁴ / 384 EI = 5 × 4 × 4⁴ / 384 EI = 13.33 / EI. 2. A central load P gives δP = P × 4³ / 48 EI = 1.333 P / EI. 3. Taking the deflection due to the point load as twice that of the UDL: 1.333 P = 2 × 13.33 ⇒ P = 20 kN.
Explanation
The deflection of a simply supported beam under a UDL and under a central point load are in the ratio δP/δw = (8/5)(P/wL). With P = 20 kN the central load produces twice the UDL deflection.
Why the other options are wrong
If the total deflection were to be exactly twice that of the UDL alone, the added load would be 10 kN, which is not an option; 24, 16 and 12 kN do not give a clean multiple.
Shortcut
δUDL : δP = 5wL⁴/384 : PL³/48.
Common mistake
Assuming that the section dimensions affect the ratio.
Correct answer: B) 20 kN