CivilStructural AnalysiseasyFE ExamCoursework

Maximum Shear in a Simply Supported Beam

A simply supported beam spans $L = 8\ \text{m}$ and carries a uniformly distributed load of $w = 20\ \text{kN/m}$ over its full length.

Find the maximum shear force.

Given

w	20 kN/m	Uniformly distributed load
L	8 m	Span

Hint 1

Where along a simply supported beam under a UDL is the shear greatest?

Hint 2

Shear is maximum at the supports and zero at mid-span — the opposite of the moment diagram.

Hint 3

Maximum shear equals the support reaction, wL/2.

Worked solution — try the problem first

By symmetry each support carries half the total load:

$R = \frac{wL}{2} = \frac{20 \times 8}{2} = 80\ \text{kN}$

Maximum shear occurs at the supports, immediately inside them, and equals the reaction:

$\boxed{V_{max} = 80\ \text{kN}}$

Shear varies linearly from $+80$ kN at the left support to $-80$ kN at the right, passing through zero at mid-span — which is exactly where the bending moment peaks, since $V = dM/dx$ .

Concepts:Shear force diagramsSupport reactionsDistributed loads

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