Question 3.6: A stepped circular shaft is built in at both ends; the two s...
A stepped circular shaft is built in at both ends; the two shaft segments are the same length, l. If the stepped shaft is made from an elastic–perfectly plastic material with τ_{γ} = 100N/mm², find the magnitude of the torque applied at the step to just cause yield in the smaller shaft.
Stress–strain behaviour
Equilibrium
T = T_{1} + T_{2}The small shaft just reaches yield, so the linear elastic torsion equation can be used for that shaft, i.e.
\frac{T}{J} = \frac{\tau }{r} = \frac{G\theta }{l}\Rightarrow \frac{T_{2}}{\frac{\pi \times 20^{4}}{32} } mm^{4} = \frac{100 N/ mm^{2}}{10 mm}
\Rightarrow T_{2} = 157.1 \times 10^{3} Nmm = 157.1 Nm
For the large shaft, θ and l are the same as the small shaft, and since \gamma = \frac{r\theta }{l} (elastic or plastic), then
\gamma _{1}|_{r = 10 mm} = \gamma _{2}|_{r = 10 mm}Therefore, shaft 1 must also be elastic for r < 10 mm and will be plastic for r > 10 mm.
Thus, for the large shaft,
for r < 10 mm, \tau = \tau _{\gamma }\frac{r}{10} = 10r
for r > 10 mm, \tau = \tau _{\gamma } = 100 N/mm^{2}
Equilibrium
T_{1} = \int_{r = 0}^{r = 10} \underset{\underset{Torque arm \times force}{} }{\underset{|}{r} } \times \underbrace{10r}_{stress} \times \underset{\underset{\underset{\times }{} }{} }{} \underbrace{2\pi rdr}_{area} + \int_{r = 10}^{r = 20} \underset{\underset{Torque arm \times force}{} }{\underset{|}{r} } \times \underbrace{100}_{stress} \times \underset{\underset{\underset{\times }{} }{} }{} \underbrace{2\pi rdr}_{area}
T_{1} = 20\pi \int_{0}^{10}{ r^{3} dr} + 200\pi \int_{10}^{20}{r^{2}dr}
= 20\pi \left[\frac{r^{4}}{4} \right]^{10}_{0} + 200\pi \left[\frac{r^{3}}{3} \right]^{20}_{10}
= 20\pi \times \frac{10^{4}}{4} + 200\pi \left[\frac{20^{3}}{3} – \frac{10^{3}}{3} \right]
i.e. T_{1} = 1623 \times 10^{3} Nmm = 1623 Nm
i.e. total torque = T = T_{1} + T_{2}
= (1623 + 157.1) Nm
∴ T = 1780 Nm
