The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and inversely as the length, l. A wooden

Question

The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and inversely as the length, l. A wooden beam 6 in. wide, 5 in. deep, and 12 ft long holds up 7666 lb. What load would a beam 3 in. wide, 7 in. deep and 15 ft long of the same material support? (Round off your answer to the nearest pound.)

Answer 1

 

joint variation means L (safe load) is directly proportional to both w and d²

 

inverse proportionality means L = k/l

So the equation is:

 

L = k(wd²/l)

 

You are given:

w = 6 in

d = 5 in

l = 12 ft

L = 7666 lbs

 

7666 = k(6*5²/12)

 

Solve for the constant, k.

7666 = k (6*25/12)

7666 = k (150/12)

7666 = k (12.5)

613.28 lbs/(ft.in³) = k

 

Use this k-value to solve for L in the last part of the question.

 

What safe load, L would a beam 3 in. wide, 7 in. deep and 15 ft long of the same material support? (Round off your answer to the nearest pound.)

 

w = 3 in

d =7 in

l = 15 ft

k = 613.28 lbs/(ft.in³)

 

Final Safe load = 6010.144 lbs *** Edited for clarity and to fix a multiplication error