The maximum bending moment for a simply supported beam with a uniformly distributed load w per unit length is:
Try it first: pick an option, then read the explanation.
Correct answer: C. wL²/8
Asked in SSC JE 04/03/2017 (Morning)
Detailed Explanation
The maximum moment occurs where the shear is zero — at mid-span for a symmetric UDL. Take moments of the left half about the centre.
\[ M_{L/2} = \frac{wL}{2}\cdot\frac{L}{2} - \frac{wL}{2}\cdot\frac{L}{4} = \frac{wL^2}{4} - \frac{wL^2}{8} = \frac{wL^2}{8} \]
Why “wL²/8” Is Correct
M_max = wL²/8 — option (c).
| Beam and load | Max shear force | Max bending moment | Where max BM occurs |
|---|---|---|---|
| Cantilever, point load W at free end | W | WL | Fixed end |
| Cantilever, UDL w over full length | wL | wL²/2 | Fixed end |
| Simply supported, central point load W | W/2 | WL/4 | Mid-span |
| Simply supported, UDL w over full span | wL/2 | wL²/8 | Mid-span |
| Simply supported, load 0 → w (triangular) | wL/3 | wL²/(9√3) | L/√3 from the zero-load end |
Why the Other Options Are Wrong
- A. wL/2: wL/2 is the maximum shear (reaction), not a moment.
- B. wL²/4: wL²/4 forgets the moment of the load on the left half.
- D. wL²/12: wL²/12 is the fixed-end moment of a fixed beam with UDL.
Key Point
SS + UDL: M_max = wL²/8; SS + central W: WL/4.
Key Concepts for Students
- Fixed beam with UDL: support moment wL²/12, mid-span wL²/24.
- Cantilever with UDL: wL²/2.
- Same total load as a central point load doubles the moment (WL/4 vs WL/8).
Practice more: this is question 45 in Bending Moment and Shear Force MCQs – Part 3. Try the whole set and check every answer instantly.