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

[math]f_v = \dfrac{VQ}{Ib}[/math]

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

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