Two identical simply supported beams of span l carry equal total loads W: one as a central point load, the other as a UDL over the entire span. The ratio of their mid-span bending moments is:
Try it first: pick an option, then read the explanation.
Correct answer: B. 2
Asked in SSC JE 2013
Detailed Explanation
Write both mid-span moments in terms of the same total load W. For the UDL, the intensity is w = W/l.
\[ M_1 = \frac{Wl}{4},\qquad M_2 = \frac{wl^2}{8} = \frac{(W/l)\,l^2}{8} = \frac{Wl}{8},\qquad \frac{M_1}{M_2} = 2 \]
Why “2” Is Correct
Concentrating the load at the centre doubles the moment: ratio 2 — option (b).
| Load (total W) | Max shear | Mid-span BM | BMD shape |
|---|---|---|---|
| Central point load W | W/2 | Wl/4 | Triangle |
| UDL w = W/l over span | W/2 | Wl/8 | Parabola |
Same total load, same reactions — but spreading it over the span halves the bending moment.
Why the Other Options Are Wrong
- A. 1/2: 1/2 is the inverse ratio (UDL : point load).
- C. 4: 4 is the ratio of the denominators 8/2 — a slip.
- D. 8: 8 has no basis.
Key Point
Same total W on SS beam: point load at centre (Wl/4) = 2 × UDL (Wl/8).
Key Concepts for Students
- Max shear is W/2 in both cases.
- Central deflection ratio is 8 : 5 (Wl³/48EI vs 5Wl³/384EI).
- Spreading load reduces bending.
Practice more: this is question 55 in Bending Moment and Shear Force MCQs – Part 4. Try the whole set and check every answer instantly.