Question 3.EP.8: A 2.50-in-diameter shaft carrying a chain sprocket has one e...

Objective      Compute the maximum shear stress and the angle of twist in the shaft.

Given      Torque = T= 15 000 lb· in; length = L = 8.00 in. The shaft is square; thus, a = 1.75 in. G=11.5\times 10^{6} psi.

Analysis      Figure 3_10 shows the methods for calculating the values for Q and K for use in Equations (3_12) and (3_13) \tau _{\max }=T/Q

\theta =TL/GK.

Results      Q=0.208 a^{3}=(0.208)(1.75 \mathrm{in})^{3}=1.115 \mathrm{in}^{3}

K=0.141 a^{4}=(0.141)(1.75 \mathrm{in})^{4}=1.322 \mathrm{in}^{4}.  Now the stress and the deflection can be computed \tau_{\max }=\frac{T}{Q}=\frac{15000 \mathrm{lb} \cdot \mathrm{in}}{\left(1.115 \mathrm{in}^{3}\right)}=13460 \mathrm{psi}

\theta=\frac{T L}{G K}=\frac{(15000 \mathrm{lb} \cdot \mathrm{in})(8.00 \mathrm{in})}{\left(11.5 \times 10^{6} \mathrm{lb} / \mathrm{in}^{2}\right)\left(1.322 \mathrm{in}^{4}\right)}=0.0079 \mathrm{rad}  Convert the angle of twist to degrees: \theta=(0.0079 \mathrm{rad})(180 \mathrm{deg} / \pi \mathrm{rad})=0.452 \mathrm{deg}.

Comments      Over the length of 8.00 in, the square part of the shaft twists 0.452 deg. The maximum shear stress is 13 460 psi, and it occurs at the midpoint of each side as shown in Figure 3-10.