Torsion of Cylindrical Bars

Abstract

Torsion of cylindrical bars refers to the application of torque (twist) of equal magnitude at the opposite ends of a bar of cylindrical shape.

Appendix: Solutions

• 7.6.1 Determine the polar moment of inertia of a circular bar radius R

Solution

$$J = \int\limits_{0}^{2\pi } {\int\limits_{0}^{R} {r^{2} rdrd\theta = 2\pi \int\limits_{0}^{R} {r^{3} dr = \frac{2\pi R}{4}^{4} = \frac{{\pi R^{4} }}{2}} } }$$

• 7.6.2 What torque is required to twist an aluminum bar through an angle of 10° if the bar is 2 cm in diameter and 1.5 m long?

Solution

$$\begin{aligned} \theta & = \frac{TL}{JG}\quad \Rightarrow \\ T & = \frac{\theta JG}{L} = \frac{{10*\left( {\pi /180} \right)\left( {\pi R^{4} /2} \right)\left( {45.9*10^{6} \,{\text{N}}/{\text{m}}^{2} } \right)}}{1.5\,m} \\ T & = \frac{{10*(\pi )\left( {\pi \left( {1\,{\text{cm}}*1\,{\text{m}}/100\,{\text{cm}}} \right)^{4} } \right)\left( {45.9*10^{6} \,{\text{N}}} \right)}}{{180*2*1.5\,{\text{m}}^{3} }} \\ T & = \frac{{10*(\pi )\left( {\pi \left( {1*1\,{\text{m}}^{4} } \right)} \right)\left( {45.9*10^{6} \,{\text{N}}} \right)}}{{180*2*1.5\,m^{3} *100^{4} }} = \frac{{10*\pi^{2} *45.9*10^{6} }}{{180*2*1.5*10^{6} }}\,{\text{N}} - {\text{m}} \\ T & = \frac{{\pi^{2} *459.0}}{180*3} = 2.097\,{\text{N}} - {\text{m}} \\ \end{aligned}$$

• 7.6.3 What is the maximum shear stress in Pa for the bar in Exercise 7.6.2?

Solution

$$\tau^{\hbox{max} } = \frac{TR}{J} = \frac{2.097*R}{{\pi R^{4} /2}} = \frac{4.195}{{\pi R^{3} }} = \frac{4.195}{{\pi \left( {1*10^{ - 2} } \right)^{3} }} = 1.3352*10^{6} \,{\text{Pa}} = 1.3352\;{\text{MPa}}$$

• 7.6.4 What is the midlength shear strain at a radius r = 0.5 cm for the bar in Exercise 7.6.2?

Solution

$$\gamma = \frac{r\theta }{L}\quad {\text{and}}\quad \theta = \frac{TL}{JG} \Rightarrow \gamma = \frac{r}{L}\theta = \frac{r}{L}10*\frac{\pi }{180} = \frac{0.5}{18*100*0.75} = 0.0004$$

• 7.6.5 Compare the maximum stress in steel, titanium, and aluminum bars for the parameters of Exercise 7.6.2

Solution

$$\tau^{\hbox{max} } = \frac{TR}{J}\quad {\text{and}}\quad T = \frac{\theta JG}{L} \Rightarrow \tau^{\hbox{max} } = \frac{\theta JG}{L}\frac{R}{J} = \frac{\theta R}{L}G$$

Now, $\frac{\theta R}{L}$ is the same for all three bars. Thus,

$$\tau_{Al}^{\hbox{max} } = \frac{\theta R}{L}G_{Al} \quad \tau_{St}^{\hbox{max} } = \frac{\theta R}{L}G_{St} \quad \tau_{Ti}^{\hbox{max} } = \frac{\theta R}{L}G_{Ti}$$