A horizontal beam carrying a UDL is supported with equal overhangs a and a central span b. The bending moment at mid-span will be zero if a/b is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
The mid-span moment equals the simply-supported value wb²/8 minus the hogging moment wa²/2 carried in from the overhangs. Set it to zero.
\[ R = \frac{w(2a+b)}{2},\qquad M_{mid} = R\cdot\frac{b}{2} - \frac{w}{2}\left(a + \frac{b}{2}\right)^2 = \frac{wb^2}{8} - \frac{wa^2}{2} \]
\[ M_{mid} = 0 \Rightarrow a^2 = \frac{b^2}{4} \Rightarrow \frac{a}{b} = 0.5 \]
Why “0.5” Is Correct
Each overhang must be half the central span: a/b = 0.5 — option (c). The whole beam is then in hogging, with the BMD just touching zero at the centre.
Why the Other Options Are Wrong
- A. 0.75: a/b = 0.75 makes the mid-span moment hogging (negative).
- B. 0.66: 0.66 also leaves a negative mid-span moment.
- D. 0.33: 0.33 leaves a positive (sagging) moment at mid-span.
Key Point
Equal overhangs with UDL: M_mid = wb²/8 − wa²/2; zero when a = b/2.
Key Concepts for Students
- a/b = 1/(2√2) ≈ 0.354 equalises + and − moments instead.
- Longer overhangs push the whole beam into hogging.
- Overhangs are a way of reducing span moments.
Practice more: this is question 60 in Bending Moment and Shear Force MCQs – Part 4. Try the whole set and check every answer instantly.