What will be the end moment of a fixed beam of length l carrying a uniformly distributed load of w kN/m?
Try it first: pick an option, then read the explanation.
Correct answer: D. ±wl²/12
Asked in SSC JE 23/09/2019 (Morning)
Detailed Explanation
By symmetry both end moments are equal. Their value makes the area of the fixing-moment diagram equal to the area of the free BMD (zero end slope).
\[ M\,l = \frac23\cdot\frac{wl^2}{8}\cdot l \;\Rightarrow\; M = \frac{wl^2}{12} \]
Why “±wl²/12” Is Correct
The end moments are ±wl²/12 (written with opposite signs in the slope-deflection sign convention).
| 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. ±wl²/2: wl²/2 is a cantilever's fixed-end moment under UDL.
- B. ±wl²/8: wl²/8 is the simply supported mid-span moment.
- C. ±wl/8: wl/8 is dimensionally wrong for a UDL; Wl/8 is the FEM for a central point load W.
Key Point
FEM: UDL wl²/12, central point load Wl/8.
Key Concepts for Students
- Area of free BMD = area of fixing BMD (zero end slopes).
- Moments of the two areas balance for zero relative deflection.
- Used directly in moment distribution.
Practice more: this is question 4 in Fixed Beam MCQs – Part 1. Try the whole set and check every answer instantly.