Question 8.S-P.4: A horizontal 500-lb force acts at point D of crankshaft AB w...
A horizontal 500-lb force acts at point D of crankshaft AB which is held in static equilibrium by a twisting couple T and by reactions at A and B. Knowing that the bearings are self-aligning and exert no couples on the shaft, determine the normal and shearing stresses at points H, J, K, and L located at the ends of the vertical and horizontal diameters of a transverse section located 2.5 in. to the left of bearing B.
Free Body. Entire Crankshaft. A = B = 250 lb
+\circlearrowleft ∑M_x = 0: -(500 lb) (1.8 in.) + T = 0 T = 900 lb · in.
Internal Forces in Transverse Section. We replace the reaction B and the twisting couple T by an equivalent force-couple system at the center C of the transverse section containing H, J, K, and L.
V = B = 250 lb T = 900 lb · in.
M_{y}=(250 lb )(2.5 \text { in. })=625 lb \cdot \text { in. }
The geometric properties of the 0.9-in.-diameter section are
A= p (0.45 \text { in. })^{2}=0.636 \text { in }^{2} I=\frac{1}{4} p (0.45 \text { in. })^{4}=32.2 \times 10^{-3} \text { in }^{4}
J=\frac{1}{2} p (0.45 \text { in. })^{4}=64.4 \times 10^{-3} \text { in }^{4}
Stresses Produced by Twisting Couple T. Using Eq. (3.8), we determine the shearing stresses at points H, J, K, and L and show them in Fig. (a).
T=\frac{ t _{\max } J}{c} (3.8)
t =\frac{T c}{J}=\frac{(900 lb \cdot in. )(0.45 \text { in. })}{64.4 \times 10^{-3} in ^{4}}=6290 psi
Stresses Produced by Shearing Force V. The shearing force V produces no shearing stresses at points J and L. At points H and K we first compute Q for a semicircle about a vertical diameter and then determine the shearing stress produced by the shear force V = 250 lb. These stresses are shown in Fig. (b).
Q=\left(\frac{1}{2} p c^{2}\right)\left(\frac{4 c}{3 p }\right)=\frac{2}{3} c^{3}=\frac{2}{3}(0.45 \text { in. })^{3}=60.7 \times 10^{-3} in ^{3}t =\frac{V Q}{I t}=\frac{(250 lb )\left(60.7 \times 10^{-3} in ^{3}\right)}{\left(32.2 \times 10^{-3} in ^{4}\right)(0.9 in. )}=524 psi
Stresses Produced by the Bending Couple M_y. Since the bending couple M_y acts in a horizontal plane, it produces no stresses at H and K. Using Eq. (4.15), we determine the normal stresses at points J and L and show them in Fig. (c).
s _{m}=\frac{M c}{I} (4.15)
s =\frac{\left|M_{y}\right| c}{I}=\frac{(625 lb \cdot \text { in. })(0.45 in .)}{32.2 \times 10^{-3} in ^{4}}=8730 psi
Summary. We add the stresses shown and obtain the total normal and shearing stresses at points H, J, K, and L.
