Chapter 3
Q. 3.SP.3
The horizontal shaft AD is attached to a fixed base at D and is subjected to the torques shown. A 44-mm-diameter hole has been drilled into portion CD of the shaft. Knowing that the entire shaft is made of steel for which G = 77 GPa, determine the angle of twist at end A.
Step-by-Step
Verified Solution
Since the shaft consists of three portions AB, BC, and CD, each of uniform cross section and each with a constant internal torque, Eq. (3.17) may be used.
\phi=\sum\limits_{i} \frac{T_{i} L_{i}}{J_{i} G_{i}} (3.17)
Statics. Passing a section through the shaft between A and B and using the free body shown, we find
\Sigma M_{x}=0: \quad(250 N \cdot m )-T_{A B}=0 \quad T_{A B}=250 N \cdot m
Passing now a section between B and C, we have
\sum M_{x}=0:(250 N \cdot m )+(2000 N \cdot m )-T_{B C}=0 \quad T_{B C}=2250 N \cdot m
Since no torque is applied at C,
T_{C D}=T_{B C}=2250 N \cdot m
Polar Moments of Inertia
\begin{aligned}&J_{A B}=\frac{\pi}{2} c^{4}=\frac{\pi}{2}(0.015 m )^{4}=0.0795 \times 10^{-6} m ^{4} \\&J_{B C}=\frac{\pi}{2} c^{4}=\frac{\pi}{2}(0.030 m )^{4}=1.272 \times 10^{-6} m ^{4} \\&J_{C D}=\frac{\pi}{2}\left(c_{2}^{4}-c_{1}^{4}\right)=\frac{\pi}{2}\left[(0.030 m )^{4}-(0.022 m )^{4}\right]=0.904 \times 10^{-6} m ^{4}\end{aligned}
Angle of Twist. Using Eq. (3.17) and recalling that G = 77 GPa for the entire shaft, we have
\begin{aligned}\phi_{A} &=\sum_{i} \frac{T_{i} L_{i}}{J_{i} G}=\frac{1}{G}\left(\frac{T_{A B} L_{A B}}{J_{A B}}+\frac{T_{B C} L_{B C}}{J_{B C}}+\frac{T_{C D} L_{C D}}{J_{C D}}\right) \\\phi_{A} &=\frac{1}{77 GPa }\left[\frac{(250 N \cdot m )(0.4 m )}{0.0795 \times 10^{-6} m ^{4}}+\frac{(2250)(0.2)}{1.272 \times 10^{-6}}+\frac{(2250)(0.6)}{0.904 \times 10^{-6}}\right] \\&=0.01634+0.00459+0.01939=0.0403 rad\end{aligned}
\phi_{A}=(0.0403 rad ) \frac{360^{\circ}}{2 \pi rad } \quad \phi_{A}=2.31^{\circ}
