A simply supported beam of span l carries a point load W at mid-span. The downward deflection under the load is:
Try it first: pick an option, then read the explanation.
Correct answer: C. Wl³/(48EI)
Asked in SSC JE 29/01/2018 (Evening); also SSC JE 2013 and state AE papers
Detailed Explanation
Each half of the beam behaves like a cantilever of length l/2 fixed at the centre (zero slope by symmetry) with an upward tip load W/2 at the support.
\[ \delta = \frac{(W/2)(l/2)^3}{3EI} = \frac{Wl^3}{48EI} \]
Why “Wl³/(48EI)” Is Correct
The central deflection is Wl³/(48EI), and the slope at each support is Wl²/16EI.
Why the Other Options Are Wrong
- A. Wl³/(3EI): Wl³/3EI is the cantilever tip-load result.
- B. Wl³/(8EI): Wl³/8EI resembles the cantilever UDL result.
- D. (5/384)(Wl³/3EI): This mixes the UDL coefficient 5/384 with the cantilever 1/3.
Key Point
Simply supported, central W: δ = Wl³/48EI, θ = Wl²/16EI.
Key Concepts for Students
- Half-beam as a cantilever of length l/2 with load W/2 — a useful trick.
- UDL: 5wl⁴/384EI.
- Stiffness at mid-span = 48EI/l³.
Practice more: this is question 30 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.