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Home » For a fixed beam of span L carrying a uniformly distributed load w over the entire span, the bending moment at mid-span is:

Structures·MCQ explained

For a fixed beam of span L carrying a uniformly distributed load w over the entire span, the bending moment at mid-span is:

Try it first: pick an option, then read the explanation.

Correct answer: B. wL²/24 sagging

Asked in SSC JE 29/10/2020 (Morning)

Detailed Explanation

Superpose the simply supported BMD (parabola, wL²/8 at mid-span) and the uniform fixing moment (−wL²/12 everywhere).

\[ M_{mid} = \frac{wL^2}{8} - \frac{wL^2}{12} = \frac{3wL^2 - 2wL^2}{24} = \frac{wL^2}{24}\ \text{(sagging)} \]

Why “wL²/24 sagging” Is Correct

The mid-span moment is wL²/24 sagging, half the end moment.

Fixed beam with a full-span UDL: hogging wL²/12 at the ends, sagging wL²/24 at mid-span; the BMD crosses zero at 0.211L from each end.

Why the Other Options Are Wrong

A. wL²/12 sagging: wL²/12 is the end (hogging) moment, not mid-span.
C. wL²/8 sagging: wL²/8 is the simply supported mid-span moment.
D. wL²/12 hogging: Mid-span is sagging, not hogging.

Key Point

Fixed beam, UDL: ends −wL²/12, mid-span +wL²/24.

Key Concepts for Students

Maximum moment is at the supports.
Mid-span deflection = wL⁴/384EI (one-fifth of simply supported).
Contraflexure at 0.211L from each end.

Practice more: this is question 3 in Fixed Beam MCQs – Part 1. Try the whole set and check every answer instantly.