Some Examples of Static Equilibrium — Question 4

Question 4

A triangular distributed load does not act at the beam midpoint

A weightless beam is supported at $x=0$ and $x=L=4\,\mathrm m$. Downward load density is $w(x)=w_0x/L$, with $w_0=60\,\mathrm{N/m}$.

1. Find the total load and its line of action.

2. Find both vertical support reactions.

Question 4 - Solution

A triangular distributed load does not act at the beam midpoint

(a) Derivation and calculation

Integrate the load and its first moment: $$W=\int_0^L\frac{w_0x}{L}\,dx=\frac{w_0L}{2}=120\ \mathrm N,$$ $$Wx_R=\int_0^Lx\frac{w_0x}{L}\,dx=\frac{w_0L^2}{3}=320\ \mathrm{N\,m}.$$ Hence $$\boxed{x_R=2L/3=8/3\ \mathrm m}.$$

(b) Result and interpretation

Taking moments about the left support, $$R_BL=Wx_R\quad\Longrightarrow\quad\boxed{R_B=80\ \mathrm N}.$$ Vertical balance then gives $$\boxed{R_A=120-80=40\ \mathrm N}.$$ The resultant lies toward the end with larger load intensity.

Physical check

The reactions sum to 120 N, and the right reaction contributes the required 320 N m moment.