| Practice Problem 4-4 |
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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 average shear stress, tS, in the bolts [ksi]? |
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Solution: The average shear stress is simply the force divided by the cross-sectional area. Note that half of the total force is acting on each of the two bolts, thus: tS = (P/2)/(pd2/4) tS = (400 lb)/[p(0.5 in)2/4] = 2037 psi = 2.04 ksi |
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