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

Question

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.

Accepted Answer

Free Body. Entire Crankshaft. A = B = 250 lb

[latex]+\circlearrowleft ∑M_x = 0[/latex]: -(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.

[latex]M_{y}=(250 lb )(2.5 ext { in. })=625 lb \cdot ext { in. }[/latex]

The geometric properties of the 0.9-in.-diameter section are

[latex]A= p (0.45 ext { in. })^{2}=0.636 ext { in }^{2}[/latex] [latex]I=\frac{1}{4} p (0.45 ext { in. })^{4}=32.2 imes 10^{-3} ext { in }^{4}[/latex]

[latex]J=\frac{1}{2} p (0.45 ext { in. })^{4}=64.4 imes 10^{-3} ext { in }^{4}[/latex]

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).

[latex]T=\frac{ t _{\max } J}{c}[/latex] (3.8)

[latex]t =\frac{T c}{J}=\frac{(900 lb \cdot in. )(0.45 ext { in. })}{64.4 imes 10^{-3} in ^{4}}=6290 psi[/latex]

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).

[latex]Q=\left(\frac{1}{2} p c^{2} ight)\left(\frac{4 c}{3 p } ight)=\frac{2}{3} c^{3}=\frac{2}{3}(0.45 ext { in. })^{3}=60.7 imes 10^{-3} in ^{3}[/latex]

[latex]t =\frac{V Q}{I t}=\frac{(250 lb )\left(60.7 imes 10^{-3} in ^{3} ight)}{\left(32.2 imes 10^{-3} in ^{4} ight)(0.9 in. )}=524 psi[/latex]

Stresses Produced by the Bending Couple [latex]M_y[/latex]. Since the bending couple [latex]M_y[/latex] 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).

[latex]s _{m}=\frac{M c}{I}[/latex] (4.15)

[latex]s =\frac{\left|M_{y} ight| c}{I}=\frac{(625 lb \cdot ext { in. })(0.45 in .)}{32.2 imes 10^{-3} in ^{4}}=8730 psi[/latex]

Summary. We add the stresses shown and obtain the total normal and shearing stresses at points H, J, K, and L.

Question 8.S-P.4: A horizontal 500-lb force acts at point D of crankshaft AB w...

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