A point load at mid-span of a fixed beam of uniform section produces fixed-end moments of 60 kN·m. The same load spread uniformly over the entire span will produce fixed-end moments equal to:
Try it first: pick an option, then read the explanation.
Correct answer: C. 40 kN·m
Asked in SSC JE 01/03/2017 (Morning)
Detailed Explanation
The total load W is the same in both cases: as a point load, FEM = WL/8; spread uniformly (w = W/L), FEM = wL²/12 = WL/12.
\[ \frac{WL}{8} = 60 \Rightarrow WL = 480 \qquad M = \frac{WL}{12} = \frac{480}{12} = 40\ \text{kN·m} \]
Why “40 kN·m” Is Correct
The fixed-end moments become 40 kN·m (two-thirds of 60).
| Load on a fixed beam of span L | Fixed-end moments (hogging) | Max sagging moment |
|---|---|---|
| Central point load W | WL/8 at each end | WL/8 at mid-span |
| Point load W at a from A, b from B | MA = Wab²/L², MB = Wa²b/L² | 2Wa²b²/L³ under the load |
| UDL w over the whole span | wL²/12 at each end | wL²/24 at mid-span |
| Support B sinks by Δ (no load) | MA = MB = 6EIΔ/L² (opposite senses) | — |
Compare the simply supported values: WL/4 and wL²/8. Fixing the ends lowers the maximum moment and the deflection.
Why the Other Options Are Wrong
- A. 20 kN·m: 20 kN·m = 60/3 — wrong ratio.
- B. 30 kN·m: 30 kN·m = 60/2 — confuses with the ratio of simply supported moments.
- D. 45 kN·m: 45 kN·m = 60 × 3/4 — wrong ratio.
Key Point
Same total load: UDL FEM / point-load FEM = (WL/12)/(WL/8) = 2/3.
Key Concepts for Students
- Simply supported: wL²/8 vs WL/4 → ratio 1/2.
- Spreading a load always reduces the moments.
- Mid-span sagging: WL/8 (point) vs WL/24 (UDL).
Practice more: this is question 5 in Fixed Beam MCQs – Part 1. Try the whole set and check every answer instantly.