CivilStructural AnalysiseasyFE ExamCoursework

Fixed-End Moment Under a UDL

A beam is built in (fixed) at both ends, spans $L = 6\ \text{m}$ , and carries a uniformly distributed load of $w = 15\ \text{kN/m}$ .

Find the magnitude of the fixed-end moment at a support.

Given

w	15 kN/m	Uniformly distributed load
L	6 m	Span

Hint 1

This is a standard fixed-end moment case worth memorising.

Hint 2

The coefficient is not the same as the simply supported wL²/8.

Hint 3

M_FEM = wL²/12 at each support.

Worked solution — try the problem first

For a fixed-fixed beam under a full-span UDL, the standard fixed-end moment is

$M_{FEM} = \frac{wL^2}{12}$

$= \frac{15 \times 6^2}{12} = \frac{15 \times 36}{12} = 45\ \text{kN·m}$

$\boxed{M_{FEM} = 45\ \text{kN·m}}$

Compare with the simply supported case. There, $M_{max} = wL^2/8 = 67.5$ kN·m at mid-span. Fixing the ends redistributes moment to the supports: $wL^2/12$ at each end and $wL^2/24 = 22.5$ kN·m at mid-span. The peak moment drops by a third — that is the structural benefit of fixity, and why continuity is worth having.

Concepts:Fixed-end momentsIndeterminate beamsMoment distribution

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