Question

A simply supported wood beam with a span of L = 23 ft supports a uniformly distributed load of w₀ = 300 lb/ft. The allowable bending stress of the wood is 1.95 ksi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 2.25, calculate the minimum width b that can be used for the beam.

Wo 2L Answer: b - in. The number of significant digits is set to 3; the tolerance is +/-1%

Answer 1

zy = 0 Ay + Cy = 300 times 15.33 = 4599 Sigma MA 0 Cy times 2.3 - 300 times 15.33 times 15.33/2 = 0 Cy = 1533.33lb Ay = 4599 - 1533.33 = 3065.67 lb Shear for zero had position will be maximum bending moment occurs. x tan A Ay = 300 times 2 = 0 x = 3065.67/300 = 10.22 ft maximum bending moment BM/ma = 3065.67 times 10.22 = 300 times 10.22^2/2 = 15667.26 lbFt = given section h/b = 2.25 Implies h = 2.25 b moment of Interia sectin I = bh^3/12 = b times(2.25b)^3/12 = 11.39 b^4/12 bending stress sigma = 1.95 ksi we equation BM/ma/I = sigma/1/t 15667.26 times 10/1.95 times 10^3 = 11.39b^4/12 h/2 = 11.39/b^4/12 times 2.25b/2 b = 4.85 in