Question

Determine the relationship between the torque $N$ and the torsion angle $\phi$ for the solid rod of circular cross-section. Their length $l$, radius $r$, and shear modulus $G$ are supposed to be known.

• A. $N=5\frac{\pi r^{4}G\phi }{3l}$
• B. $N=\frac{\pi r^{4}G\phi }{2l}$
• C. $N=3\frac{\pi r^{4}G\phi }{2l}$
• D. None of these

Answer 1

The correct option is B $N=\frac{\pi r^{4}G\phi }{2l}$

Keeping the lower end of the solid rod 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 solid rod.

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

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

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

Tangantial stress $=\frac{Force}{Area}=\frac{F}{dxda}$

$\therefore$ Shear modulus $G=\frac{Stress}{\theta }=\frac{\frac{F}{dadx}}{\frac{a\varphi }{l}}$ $⟹F=\frac{G\varphi a}{l}dadx$

Moment of force $M=Fa=\frac{G\varphi a^2}{l}dadx$

$\therefore$ Restoring torque on the annular surface $\delta N=\frac{G{a}^2\varphi }{l}da\sum dx$

$⟹\delta N=\frac{2\pi G{a}^3\varphi }{l}da$ (As $\sum dx=2\pi a$)

Now total restoring torque $N={\int }_0^r\delta N==\frac{2\pi G\varphi }{l}{\int }_0^r{a}^{3}da$

$\therefore N=\frac{2\pi G\varphi }{l}\times \frac{r^4}{4}$ $⟹N=\frac{\pi r^{4}G\varphi }{2l}$