Question 3.6: Compute the maximum torsional shear stress in a shaft having...

Objective: Compute the torsional shear stress in the shaft.
Given: \quad Torque =T=4.10 \mathrm{~N} \cdot \mathrm{m} ; shaft diameter =D=10 \mathrm{~mm}.
c= radius of the shaft =D / 2=5.0 \mathrm{~mm}
Analysiçs: Use Equation (3-7) to compute the torsional shear stress: \tau_{\max }=T c / J . J is the polar moment of inertia for the shaft: J=\pi D^{4} / 32 (see Appendix 1 ).

APPENDIX 1 Properties of Areas
\begin{array}{ll}A=\text { area } & r=\text { radius of gyration }=\sqrt{I / A} \\I=\text {moment of inertia } & J=\text { polar moment of inertia } \\S=\text { section modulus } &Z_{p}=\text { polar section modulus }\end{array}
	\begin{array}{ll}A=\pi D^{2} / 4 & r=D / 4 \\I=\pi D^{4} / 64 & J=\pi D^{4} / 32 \\S=\pi D^{3} / 32 & Z_{p}=\pi D^{3} / 16\end{array}
	\begin{array}{ll}A=\pi\left(D^{2}-d^{2}\right) / 4 & r=\frac{\sqrt{\left(D^{2}+d^{2}\right)}}{4} \\I=\pi\left(D^{4}-d^{4}\right) / 64 & J=\pi\left(D^{4}-d^{4}\right) / 32 \\S=\pi\left(D^{4}-d^{4}\right) / 32 D & Z_{p}=\pi\left(D^{4}-d^{4}\right) / 16 D\end{array}
	\begin{aligned}&A=H^{2} \quad\quad\quad r=H / \sqrt{12} \\&I=H^{4} / 12 \\&S=H^{3} / 6\end{aligned}
	\begin{array}{ll}A=B H & r_{x}=H / \sqrt{12} \\I_{x}=B H^{3} / 12 & r_{y}=B / \sqrt{12} \\I_{y}=H B^{3} / 12 & \\S_{x}=B H^{2} / 6 & \\S_{y}=H B^{2} / 6 &\end{array}
	\begin{aligned}&A=B H-b h \\&I_{x}=\frac{B H^{3}-b h^{3}}{12} \quad S_{x}=\frac{B H^{3}-b h^{3}}{6 H}\quad r_{x}=0.289 \sqrt{\frac{B H^{3}-b h^{3}}{B H-b h}} \\&I_{y}=\frac{H B^{3}-h b^{3}}{12} \quad S_{y}=\frac{H B^{3}-h b^{3}}{6 B} \quad r_{y}=0.289 \sqrt{\frac{H B^{3}-h b^{3}}{H B-h b}}\end{aligned}
	\begin{aligned}&A=B H / 2 \quad\quad\quad r=H / \sqrt{18} \\&I=B H^{3} / 36 \\&S=B H^{2} /24\end{aligned}
	\begin{aligned}&A=\pi D^{2} / 8 \quad\quad\quad r=0.132 D \\&I=0.007 D^{4} \\&S=0.024D^{3}\end{aligned}
	\begin{aligned}&A=0.866 D^{2} \quad\quad\quad r=0.264 D \\&I=0.06 D^{4} \\&S=0.12 D^{3}\end{aligned}
	\begin{aligned}A &=H(a+B) / 2 \\y &=\frac{H(a+2 B)}{3(a+B)} \quad\quad\quad\quad\quad\quad S=\frac{H^{2}\left(a^{2}+4 a B+B^{2}\right)}{12(a+2 B)} \\I_{x} &=\frac{H^{3}\left(a^{2}+4a B+B^{2}\right)}{36(a+B)} \quad\quad r=\frac{H^{2}\left(a^{2}+4 a B+B^{2}\right)}{18(a+B)^{2}} \\y &=\underset{\text { Maximum distance from } x \text {-axis to }}{\text { outer surface of section }}\end{aligned}
	\begin{aligned}&A=\pi b h \\&I=\frac{\pi h^{3} b}{4} \\&S=\frac{\pi h^{2} b}{4} \\&r=h / 2\end{aligned}

Results
\begin{aligned}&J=\pi D^{4} / 32=\left[(\pi)(10 \mathrm{~mm})^{4}\right] / 32=982 \mathrm{~mm}^{4} \\&\tau_{\max }=\frac{(4.10 \mathrm{~N}\cdot \mathrm{m})(5.0 \mathrm{~mm}) 10^{3} \mathrm{~mm}}{982 \mathrm{~mm}^{4}}=20.9 \mathrm{~N} / \mathrm{mm}^{2}=20.9 \mathrm{MPa}\end{aligned}
Comment The maximum torsional shear stress occurs at the outside surface of the shaft around its entire circumference.