Problema Solution
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 8in.
8
in.
wide, 8in.
8
in.
deep, and 11ft
11
ft
long holds up 1946lb
1946
lb
. What load would a beam 9in.
9
in.
wide, 6in.
6
in.
deep and 10ft
10
ft
long of the same material support? (Round off your answer to the nearest pound.)
Answer provided by our tutors
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:
L = C*w*d^2/l, where C is constant
w = 8 in
d = 8 in
l = 11 ft
l = 11*12 in (1 ft = 12 in)
l = 132 in
L = 1946 lb
C*8*8^2/132 = 1946
........
click here to see the equation solved for C
........
C = 32109/64
Now we need to find L for:
w = 9 in
d = 6 in
l = 10 ft
l = 10*12 in
l = 120 in
L = (32109/64)*w*d^2/l
L = (32109/64)*9*6^2/120
L = 1,355 lb
A wooden beam 9 in wide, 6 in. deep, and 10 ft long holds up 1,355 lb.