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

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.

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

$A= p (0.45 \text { in. })^{2}=0.636 \text { in }^{2}$                  $I=\frac{1}{4} p (0.45 \text { in. })^{4}=32.2 \times 10^{-3} \text { in }^{4}$

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)

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).

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)

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