Given:
The load varies as width and square of the depth and inversely as the length,
L₁=7127 lb
w₁=5 in=0.42 ft
d₁=7 in=0.58 ft
l₁=13 ft
w₂=3 in=0.25 ft
d₂=3 in=0.25 ft
l₂=18 ft
To find:
The load.
Explanation:
From the question,
[tex]L=\frac{kwd^2}{l}[/tex]
Where k is the proportionality constant.
Thus, substituting the 1st set of values in the above equation,
[tex]\begin{gathered} L_1=\frac{kw_1d_1^2}{l_1} \\ 7172=\frac{k\times0.42\times0.58}{13} \\ \implies k=\frac{7172\times13}{0.42\times0.58} \\ =382742.2\text{ lb/ft}^2 \end{gathered}[/tex]
Substituting the second set of values,
[tex]\begin{gathered} L=\frac{382742.2\times0.25\times0.25}{18} \\ =1329\text{ lb} \end{gathered}[/tex]
Final answer:
A 3 in wide, 3 in deep, and 18 ft long beam can support a load of 1329 lb
