Question 8.SP.4: A horizontal 500-lb force acts at point D of crankshaft AB h......
A horizontal 500-lb force acts at point D of crankshaft AB held in static equilibrium by a twisting couple T and 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.
STRATEGY: Begin by determining the internal forces and couples acting on the transverse section containing the points of interest, and then evaluate the stresses at these points due to each internal action. Combining these results will provide the total state of stress at each point.
MODELING: Draw the free-body diagram of the crankshaft (Fig. 1). Find A = B = 250 lb.
+⤻ \Sigma M_x=0: \quad-(500 lb )(1.8 \text{in} .)+T=0 \quad T=900 \text{lb} \cdot \text{in} .
ANALYSIS:
Internal Forces in Transverse Section. Replace 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 (Fig. 2).
\begin{aligned} V & =B=250 \text{lb} \quad T=900 \text{lb} \cdot \text{in} . \\ M_y & =(250 \text{lb} )(2.5 \text { in. })=625 \text{lb} \cdot \text{in} . \end{aligned}
The geometric properties of the 0.9-in.-diameter section are
\begin{aligned} A & =\pi(0.45 \text { in. })^2=0.636 \text { in }^2 \quad I=\frac{1}{4} \pi(0.45 \text { in. })^4=32.2 \times 10^{-3} in ^4 \\ J & =\frac{1}{2} \pi(0.45 \text { in. })^4=64.4 \times 10^{-3} in ^4 \end{aligned}
Stresses Produced by Twisting Couple T. Using Eq. (3.10), determine the shearing stresses at points H, J, K, and L and show them in Fig. 3.
\tau=\frac{T \rho}{J} (3.10)
\tau=\frac{T c}{J}=\frac{(900 lb \cdot in .)(0.45 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, 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. 4.
\begin{aligned} Q & =\left(\frac{1}{2} \pi c^2\right)\left(\frac{4 c}{3 \pi}\right)=\frac{2}{3} c^3=\frac{2}{3}(0.45 \text { in. })^3=60.7 \times 10^{-3} in ^3 \\ \tau & =\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 \end{aligned}
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. Use Eq. (4.15) to determine the normal stresses at points J and L and show them in Fig. 5.
\sigma_m=\frac{M c}{I} (4.15)
\sigma=\frac{\left|M_y\right| c}{I}=\frac{(625 lb \cdot in .)(0.45 in .)}{32.2 \times 10^{-3} in ^4}=8730 psi
Summary. Add the stresses shown to obtain the total normal and shearing stresses at points H, J, K, and L (Fig. 6).
