# Rectangular Beam

• Dec 11th 2009, 03:50 PM
sologuitar
Rectangular Beam
The maximum safe load that a rectangular beam can support varies directly with the width w and the square of the height h and inversely with the length L of the beam. A 6-foot-long beam that is 2 inches high and 4 inches wide has a maximum safe load of 1000 pounds.

(a) What is the maximum safe load for a 10-foot-long beam that is 4 inches high and 4 inches wide?

(b) How long should a 4 inch beam be to safely support a 6000-pound load?
• Dec 11th 2009, 04:22 PM
Haversine
Quote:

Originally Posted by sologuitar
The maximum safe load that a rectangular beam can support varies directly with the width w and the square of the height h and inversely with the length L of the beam.

Translating this directly into an equation, I get

$\displaystyle M = k \frac{wh^2}{L}$

where $\displaystyle k$ is a proportionality constant that you have to determine from the data you've been given. That is, they've given you M, w, h, and L in one case, and you have to figure out $\displaystyle k$ for that case.

Once you know $\displaystyle k$, you can use it to solve for any of the unknown situations.
• Dec 12th 2009, 05:17 AM
sologuitar
Good...
Quote:

Originally Posted by Haversine
Translating this directly into an equation, I get

$\displaystyle M = k \frac{wh^2}{L}$

where $\displaystyle k$ is a proportionality constant that you have to determine from the data you've been given. That is, they've given you M, w, h, and L in one case, and you have to figure out $\displaystyle k$ for that case.

Once you know $\displaystyle k$, you can use it to solve for any of the unknown situations.

Thanks for the tips. I can take it from here.