Problem 570

A uniformly distributed load of 200 lb/ft is carried on a simply supported beam span. If the cross-section is as shown in Fig. P-570, determine the maximum length of the beam if the shearing stress is limited to 80 psi. Assume the load acts over the entire length of the beam.

Solution 570

f_v = \dfrac{VQ}{Ib}

Where:
f_v = 80 \, \text{psi}
V = 100L
Q = 8(5)(2.5) – 6(4)(2) = 52 \, \text{in^3}
I = \frac{1}{12}(8)(10^3) - \frac{1}{12}(6)(8^3) = \frac{1232}{3} \, \text{in}^4
b = 8 – 6 = 2 \, \text{in}

Thus,
80 = \dfrac{100L(52)}{\frac{1232}{3}(2)}
L = 12.64 \, \text{ft } \,\, answer