| Practice Problem 7-5 (a) |
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Given: A force of P = 40 lb is applied to the handle of a pipe wrench to tightens a solid shaft into its socket. Let, d = 15 in., R = 1 in., and L = 20 in. Req'd: |
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Solution: Since there is no axial force the axial stress is a result of the bending moment only. The bending moment is simply the shear force acting at the origin times the length of the pipe: M = P·L. Therefore the axial stress is: sx = MR/I The moment of inertia for a solid circle is I = pR4, therefore: sx = 4P·L/pR3 = 4(40 lb)(20 in)/p(1 in)3 = 1020 psi |
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