A beam with a central span L between supports and two equal overhangs L₁ carries a UDL w throughout. The maximum positive bending moment equals the maximum negative bending moment. The ratio L₁/L is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 25/09/2019 (Morning)
Detailed Explanation
The overhangs create hogging moments over the supports; the central span sags. The mid-span moment is the simply-supported value wL²/8 reduced by the support moment. Set the two magnitudes equal.
\[ |M_B| = \frac{wL_1^2}{2},\qquad M_E = \frac{wL^2}{8} - \frac{wL_1^2}{2} \]
\[ \frac{wL^2}{8} - \frac{wL_1^2}{2} = \frac{wL_1^2}{2} \Rightarrow \frac{wL^2}{8} = wL_1^2 \Rightarrow \frac{L_1}{L} = \frac{1}{2\sqrt2} \approx 0.354 \]
Why “1/(2√2)” Is Correct
The ratio is 1/(2√2) — option (b). In terms of the total length (L + 2L₁) each overhang is then ≈ 0.207 of the total — the same proportion used for lifting long members at two points.
Why the Other Options Are Wrong
- A. √2: √2 would make the overhangs longer than the span.
- C. 1/√2: 1/√2 makes the hogging moment much larger than the sagging moment.
- D. 1/2: L₁/L = 1/2 gives exactly zero moment at mid-span, not equal moments.
Key Point
Equal + and − moments: L₁ = L/(2√2); each overhang ≈ 0.207 × total length.
Key Concepts for Students
- Equalising + and − moments minimises the largest moment in the beam.
- L₁/L = 1/2 makes the mid-span moment zero.
- Mid-span moment of an overhanging beam = wL²/8 − support moment.
Practice more: this is question 27 in Bending Moment and Shear Force MCQs – Part 2. Try the whole set and check every answer instantly.