Question 3.7: Compute the angle of twist of a 10-mm-diameter shaft carryin...

Objective:
Compute the angle of twist in the shaft.
Given:
Torque =T=4.10 \mathrm{~N} \cdot \mathrm{m}; length =L=250 \mathrm{~mm}.
Shaft diameter =D=10 \mathrm{~mm} ; G=80 \mathrm{GPa}.
Analysis:

Use Equation (3-11) (\theta=T L / G J). For consistency, let T=4.10 \times 10^{3} \mathrm{~N} \cdot \mathrm{mm} and G=80 \times 10^{3} \mathrm{~N} / \mathrm{mm}^{2}. From Example Problem 3-6, J=982 \mathrm{~mm}^{4}.

Results:
\theta=\frac{T L}{G J}=\frac{\left(4.10 \times 10^{3} \mathrm{~N} \cdot \mathrm{mm}\right)(250 \mathrm{~mm})}{\left(80 \times 10^{3} \mathrm{~N} / \mathrm{mm}^{2}\right)\left(982 \mathrm{~mm}^{4}\right)}=0.013 \mathrm{rad}
Using \pi \mathrm{rad}=180^{\circ},
\theta=(0.013 \mathrm{rad})(180 \% \pi \mathrm{rad})=0.75^{\circ}
Comment:

Over the length of 250 \mathrm{~mm}, the shaft twists 0.75^{\circ}.