The maximum safe load of a rectangle beam varies jointly as its width and the square of its depth and inversely as the length of the beam. If a beam 8.00 in. wide, 11.5 in. Deep, and 16.0 ft long can safely support 15.000 lb, find the safe load for a beam 6.50 in. Wide, 13.4 in. Deep, and 21.0 ft long made of the same material.
x = k *w*d^2/L
15,000 = k *8*11.5^2/16
so
k = 226.8
x = 226.8 * 6.5 * 13.4^2 / 21
= 12,608 pounds
To find the safe load for the second beam, we need to use the joint variation equation and the given information.
The joint variation equation is given as:
Load = k * (width) * (depth^2) / length
where k is the constant of variation.
First, let's find the value of k using the given information for the first beam:
Load1 = k * (width1) * (depth1^2) / length1
15,000 = k * (8.00) * (11.5^2) / 16.0
Now, let's solve for k:
k = (15,000 * 16.0) / (8.00 * (11.5^2))
k = 17.3913
Now we can use this value of k to find the safe load for the second beam:
Load2 = k * (width2) * (depth2^2) / length2
Load2 = 17.3913 * (6.50) * (13.4^2) / 21.0
Load2 ≈ 20,378.53 lb
Therefore, the safe load for the second beam is approximately 20,378.53 lbs.