Question 3.5: A solid circular shaft of diameter 50 mm and length 1000 mm ...
A solid circular shaft of diameter 50 mm and length 1000 mm is subjected to a torque, T. The shaft is made from an elastic–perfectly plastic material with τ_{γ} = 100 N/mm², G = 70 GN/m². Determine the magnitude of the torque required to cause yielding to occur at a radius of 15 mm (and greater) and the angle of twist.
Equilibrium (moments about the axis)
T = \int_{r = 0}^{r = 25}{\underset{\underset{Torque arm \times force}{| } }{r} } \times \underset{\underset{stress }{| } }{\tau } \times \underset{\underset{\times }{} }{} \underset{\underset{area }{| } }{2\pi rdr}
T = \int_{0}^{15}{ 2\pi \frac{100}{15}r^{3}dr } + \int_{15}^{25}{200\pi r^{2}dr}
= 200\pi \left\{\left[\frac{r^{4}}{15 \times 4} \right]_{0}^{15}+ \left[\frac{r^{3}}{3} \right]_{15}^{25} \right\}
= 200\pi \left[\frac{15^{4}}{15 \times 4} + \frac{25^{3·}}{3} – \frac{15^{3}}{3} \right]
\therefore T = 3.096 \times 10^{6} Nmm = 3.096 kNm
Relationship between \tau _{\gamma } and \gamma _{\gamma }: at the outermost elastic point, τ = \tau _{\gamma } and \gamma = \gamma _{\gamma } and the elastic relation G = \frac{\tau _{\gamma }}{\gamma _{\gamma }} is applicable. It should be noted that outside this outermost elastic point,
i.e. r = 15 mm, the strain, which will be larger, will be elastic–plastic and consequently will not be governed by τ = G_{\gamma }, which is the elastic relation. Nonetheless, at r = 15 mm, we have
\gamma _{\gamma } = \frac{\tau _{\gamma }}{G} = \frac{100}{70 000} = 1.4286 \times 10^{-3} rad
by invoking the compatibility requirement,
r _{\gamma } \theta = \gamma _{\gamma } l
and hence
\theta = \frac{\gamma _{\gamma }l}{r _{\gamma }}
i.e. \theta = \frac{1.4286 \times 10^{-3} \times 1000 mm}{15 mm} \times \left(\frac{360}{2\pi } \frac{deg}{rad} \right)
\therefore \theta = 5.456°