| Practice Problem 4-3 |
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Given: Two plates of width, w = 2 in., and thickness, t = 0.375 in., are subjected to a tensile force of P = 800 lb. The plates are connected using two bolts of diameter, d = 0.5 in. Req'd: What is the bearing stress, sB, in the bolts [ksi]? |
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Solution: The bearing stress is simply the force divided by the projected area. Note that half of the total force is acting on each of the two bolts, thus: sB = (P/2)/(t·d) sB = (400 lb)/[(0.375 in)(0.5 in)] = 2133 psi = 2.13 ksi |
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