Simple-Beam Maximum Moment

M_{m a x} = \frac{P L}{4} (mid point load), M_{m a x} = \frac{w L ^{2}}{8} (uniform)

These closed-form cases are in the Handbook beam tables. Draw V and M diagrams: V is the slope of M, and jumps in V equal concentrated loads. Supports are found from equilibrium first.

$P$ (kN): Concentrated load
$w$ (kN/m): Uniform load
$L$ (m): Span

Step-by-step solved example

A 6 m simple beam carries 8 kN/m. Find M_max and V_max.

Simple span + uniform w: Mmax = wL²/8 at midspan; Vmax = wL/2 at supports.

1. Reactions
Each support takes wL/2 = 24 kN = V_max.
2. Moment
M_max = wL²/8 = 8×36/8 = 36 kN·m at midspan.
$M_{m a x} = 36 kN\cdotm$

Answer: V_max = 24 kN, M_max = 36 kN·m

10 practice questions

0/10 correct

1.Point load P at midspan of simple L: M_max =

2.Cantilever length L, end load P: M_fixed =

3.Cantilever uniform w: M_fixed =

4.Simple beam, P at midspan: V at a support is

5.The shear diagram under uniform load is

6.M is maximum where V is

7.Two equal spans continuous vs two simple spans: midspan M is

8.w = 10 kN/m, L = 4 m simple: M_max =

9.An internal hinge makes M =

10.Fixed-fixed uniform beam: M at ends is