Simple-Beam Maximum Shear

V_{m a x} = \frac{w L}{2} (uniform) V_{A} = P \frac{b}{L}

For a simply supported beam with uniform w, each reaction is wL/2, which is Vmax just inside the support. For a point load P at distance a from A and b from B, VA = Pb/L and VB = Pa/L.

$V_{m a x}$ (N): Maximum shear
$w$ (N/m): Uniform load
$L$ (m): Span
$P$ (N): Point load

Step-by-step solved example

w = 8 kN/m on L = 6 m simple span. Vmax?

Shear is largest at the supports and crosses zero at midspan if uniform and symmetric.

1. Total
W = 8×6 = 48 kN.
2. Reactions
Each reaction = 24 kN = Vmax.
$V_{m a x} = 24 kN$

Answer: 24 kN

10 practice questions

0/10 correct

1.Uniform simple beam: Vmax is at

2.w = 2 k/ft, L = 10 ft. Vmax =

3.Point load P at midspan. Vmax =

4.Shear diagram for uniform simple beam is

5.P = 30 kN at 2 m from A on L = 10 m. VA =

6.V = 0 under uniform load occurs at

7.Cantilever with end load P: V is

8.Relationship: dM/dx =

9.Units of shear in US customary beams often

10.Two equal end reactions with a central P imply