A beam carrying a UDL has two equal overhangs. For the maximum bending moment in the beam to be as small as possible, the ratio of each overhang to the total length of the beam is:

Asked in SSC JE 04/03/2017 (Morning)

Detailed Explanation

Longer overhangs increase the hogging moments over the supports but reduce the sagging moment at the centre. The largest moment is smallest when the two are equal.

Let each overhang be b and the span between supports be s, so \( l = s + 2b \).

Equal moments: \( \dfrac{wb^2}{2} = \dfrac{ws^2}{8} - \dfrac{wb^2}{2} \Rightarrow s = 2\sqrt2\,b \)

\[ \frac{b}{l} = \frac{b}{2\sqrt2\,b + 2b} = \frac{1}{2(1+\sqrt2)} = 0.207 \]

Why “0.207” Is Correct

Each overhang should be 0.207 of the total length — option (a). The largest moment is then about wl²/46.6.

Symmetrical beam with equal overhangs L₁ and central span L under UDL w, proportioned so that the hogging moment over the supports (wL₁²/2) equals the sagging moment at mid-span. This happens when L₁ = L/(2√2) ≈ 0.354L, i.e. each overhang ≈ 0.207 of the total length.

Why the Other Options Are Wrong

B. 0.307: 0.307 makes the support (hogging) moments too large.
C. 0.407: 0.407 is far too long an overhang.
D. 0.508: 0.508 is more than half the length — impossible for two overhangs.

Key Point

Equal overhangs, UDL: b/l = 0.207 equalises + and − moments and minimises M_max.

Key Concepts for Students

Same ratio as two-point lifting of piles.
Relative to the span between supports: b/s = 1/(2√2) = 0.354.
Zero mid-span moment needs b = s/2 instead.

Practice more: this is question 46 in Bending Moment and Shear Force MCQs – Part 4. Try the whole set and check every answer instantly.