Distributed Load Resultant

W = \int w (x) d x x ˉ = \frac{\int x w ( x ) d x}{W}

A distributed line load w(x) is equivalent to a concentrated force W equal to the area under the load diagram, acting at the centroid of that diagram. For a uniform w on length L, W = wL at midspan.

$w$ (N/m): Load intensity
$W$ (N): Resultant force
$x ˉ$ (m): Location of the resultant

Step-by-step solved example

Simply supported beam, uniform 4 kN/m over 6 m. Find the equivalent load and its location.

Uniform w is a resultant W = wL at mid-length of the loaded patch.

1. Area
W = 4×6 = 24 kN.
$W = 24 t e x t k N$
2. Centroid
Uniform rectangle → resultant at 3 m from either end.

Answer: 24 kN at midspan

10 practice questions

0/10 correct

1.Uniform 2 kN/m on 5 m → W =

2.Triangular load 0 to w0 over L has W =

3.That triangular resultant sits

4.w has units of

5.A 3 kN/m load over 2 m starting 1 m from the left support of a 6 m beam acts

6.Replacing w(x) by W at the wrong location

7.Hydrostatic on a vertical wall is a

8.Two uniform patches may be

9.w = 8 kN/m, L = 3 m, simply supported. Each reaction =

10.The internal shear just inside the left support equals