A simply supported beam of span L carries a UDL w over the whole span. If the UDL is replaced by a concentrated load W at the centre that produces the same central deflection (all other parameters unchanged), the ratio W/w is:
Try it first: pick an option, then read the explanation.
Correct answer: B. 5L/8
Asked in SSC JE 29/10/2020 (Morning)
Detailed Explanation
Equate the two standard central deflections of a simply supported beam: 5wL⁴/384EI for the UDL and WL³/48EI for a central point load.
\[ \frac{5wL^4}{384EI} = \frac{WL^3}{48EI} \Rightarrow W = \frac{5\times48}{384}wL = \frac{5}{8}wL \Rightarrow \frac{W}{w} = \frac{5L}{8} \]
Why “5L/8” Is Correct
W/w = 5L/8. A central point load is more "effective" at deflecting a beam than the same load spread out, so only 5/8 of the total UDL (wL) is needed.
Why the Other Options Are Wrong
- A. L/8: L/8 would mean W = wL/8, far too small.
- C. 5L³/(48EI): This expression is a deflection-type quantity, not a load ratio.
- D. 8L/5: 8L/5 is the inverse relationship (w/W scaled incorrectly).
Key Point
Equal central deflection: central point load W = (5/8) × total UDL wL.
Key Concepts for Students
- Equal total load: δ(point)/δ(UDL) = 8/5.
- Equal maximum moment: WL/4 = wL²/8 ⇒ W = wL/2.
- Write each deflection with its own standard formula, then equate.
Practice more: this is question 19 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.