Question

Determine the relationship between the torque $N$ and the torsion angle $\phi$ for the tube whose wall thickness $\Delta r$ is considerably less than the tube radius.

• A. $N=\frac{2\pi r^3\Delta r\phi }{3l}G$
• B. $N=\frac{3\pi r^3\Delta r\phi }{l}G$
• C. $N=\frac{2\pi r^3\Delta r\phi }{l}G$
• D. None of these

Answer 1

Answer: (C)

$N=\frac{2\pi r^3\Delta r\phi }{l}G$

Keeping the lower end of the hollow tube fixed, its upper end is twisted by angle $\varphi$ by applying a force $F$. Due to the twist, a shear stress is generated between the lower end and the upper end of the tube.

Thus point $A$ is displaced to $A^{\prime }$ due the force such that $A{A}^{\prime }=dx$

Now from sector $AO{A}^{\prime }$, $A{A}^{\prime }=r\varphi$

Also from sector $AB{A}^{\prime }$, $A{A}^{\prime }=l\theta$ $⟹\theta =\frac{r\varphi }{l}$ ..........(1)

Tangantial stress $=\frac{Force}{Area}=\frac{F}{dx\Delta r}$

$\therefore$ Shear modulus $G=\frac{Stress}{\theta }=\frac{\frac{F}{\Delta rdx}}{\frac{r\varphi }{l}}$ $⟹F=\frac{G\varphi r}{l}\Delta rdx$

Moment of force $dM=Fr=\frac{G\varphi r^2}{l}\Delta rdx$

$\therefore$ Total restoring torque on the annular surface $N=\int dM=\frac{G{r}^2\varphi }{l}\Delta r\int dx$

$⟹N=\frac{2\pi G{r}^3\varphi }{l}\Delta r$ (As $\int dx=2\pi r$)