Practice Problem 7-5 (a) | Back | Index | Next |
Given: A force of P = 40 lb is applied (along the –z-axis) to the handle of a pipe wrench to tighten a solid shaft into its socket. Let, d = 15 in., R = 1.0 in., and L = 20 in. Req'd: Determine the normal stress, s_{x} at Point B (x=L, y=0, z=R). |
Consider Point B:
- the is no axial force, so no axial stress occurs at B.
- bending moment M_{y}, causes a tensile bending stress at B.
- the other stresses that act in the x-y-z coordinate system are shear stresses.
The normal stress at B is only caused by the bending moment::
s_{x} = M_{y}R/I
The moment of inertia for a solid circle is I = pR^{4}/4, so:
s_{x} = 4PL/[pR^{3}] = 4(40 lb)(20 in.)/[p(0.5 in.)^{3}] = 1020 psi
s_{x} = 1020 psi
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