Triangular Distributed Load Resultant

W = \frac{1}{2} w_{m a x} L, x ˉ = \frac{2}{3} L from the zero (apex) end

Replace a linearly varying line load with a concentrated resultant equal to the area of the load diagram, acting through its centroid. For a triangle, the centroid is 2/3 of the length from the apex (1/3 from the peak).

$w_{m} a x$ (N/m): Peak intensity
$L$ (m): Loaded length

Step-by-step solved example

0-to-12 kN/m triangle over 6 m simple beam, peak at B. Find RA.

Triangular w: resultant at 2/3 from the apex.

1. Resultant
W=½×12×6=36 kN at 4 m from A (2/3 of 6 from apex A).
2. Moment about B
RA×6=36×2 → RA=12 kN.
$R_{A} = 12 kN$

Answer: RA = 12 kN

10 practice questions

0/10 correct

1.Resultant of 0-to-w over L is

2.From the peak end the resultant sits at

3.A trapezoid splits into

4.w=10 kN/m peak, L=4 m, W =

5.Uniform w is a special case with centroid at

6.Hydrostatic on a vertical wall is triangular with w_max=

7.Units of w are force per

8.If the triangle is inverted (peak at A, zero at B) RA on a simple span

9.Internal hinge still requires

10.Simple 9 m beam, triangle 0 at A to 6 kN/m at B: RB =