Question

Calculate the torque $N$ twisting a steel tube of length $l=3.0m$ through an angle $\phi ={2.0}^{\circ }$ about its axis, if the inside and outside diameters of the tube are equal to $d_1=30mm$ and $d_2=50mm$.

Answer 1

Clearly $N={\int }_{\frac{d_1}{2}}^{\frac{d_2}{2}}\frac{2\pi r^{3}dr\phi G}{l}=\frac{\pi }{32l}G\phi (d_2^4-d_1^4)$
using $G=81GPa=8.1\times {10}^{10}N/m^2$
$d_2=5\times {10}^{-2}m$, $d_2=3\times {10}^{-2}m$
$\phi ={2.0}^{\circ }=\frac{\pi }{90}radians$, $l=3m$
$N=\frac{\pi \times 8.1\times \pi }{32\times 3\times 90}(625-81)\times {10}^{2}Nm$
$=0.5033\times {10}^{3}Nm\approx 0.5kNm$