Question 7 of 8: Stepped Circular Shaft in Torsion

Question 7: Stepped Circular Shaft in Torsion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. Maximum shear stress and its radial distribution; angle of twist at D; the effect of doubling all three torques.

Approach. Cut each segment and sum the applied torques between the cut and the free end D (honouring T_D's opposite sense with a minus sign) to get each segment's internal torque, then apply $\tau=Tr/J$ and $\varphi=TL/(GJ)$ segment by segment, summing the twists algebraically for the total at D.

1. Internal torque per segment (positive = same sense as T_B, T_C; T_D enters as $-6$): $$T_{AB}=T_B+T_C+T_D=1+12-6=7\ \text{kN}\cdot\text{m}\qquad T_{BC}=T_C+T_D=12-6=6\ \text{kN}\cdot\text{m}\qquad T_{CD}=T_D=-6\ \text{kN}\cdot\text{m}$$
2. Polar moments of inertia. $$J_{solid}=\frac{\pi(60)^4}{32}=1.272\times10^{6}\ \text{mm}^4\qquad J_{hollow}=\frac{\pi(60^4-40^4)}{32}=1.021\times10^{6}\ \text{mm}^4$$
3. Shear stress in each segment ($\tau=T r_o/J$, $r_o=30$ mm throughout): $$\tau_{AB}=\frac{(7\times10^6)(30)}{1.272\times10^6}=165.1\ \text{MPa}\qquad \tau_{BC}=\frac{(6\times10^6)(30)}{1.272\times10^6}=141.5\ \text{MPa}$$ $$\tau_{CD}=\frac{(6\times10^6)(30)}{1.021\times10^6}=\boxed{176.3\ \text{MPa (governs, despite CD carrying the SAME torque as BC)}}$$ CD governs because removing the central material (the 40 mm bore) reduces $J$ faster than it reduces the outer-fibre stress demand. Within CD's material the shear stress still varies LINEARLY with radius, from $\tau(20\ \text{mm})=176.3(20/30)=117.5$ MPa at the bore up to 176.3 MPa at the outer surface (zero only at a hypothetical $r=0$, which is not part of the hollow section); in the solid segments it runs linearly from 0 at the centre to the outer-surface value.
4. Angle of twist at D (segment twists ADD algebraically, so CD's opposite-sign torque partially UNWINDS the twist built up in AB and BC): $$\varphi_D=\sum\frac{T_iL_i}{GJ_i}=\frac{7\times10^6(1000)}{25\,000(1.272\times10^6)}+\frac{6\times10^6(800)}{25\,000(1.272\times10^6)}+\frac{-6\times10^6(1200)}{25\,000(1.021\times10^6)}$$ $$\varphi_D=0.2201+0.1509-0.2821=0.0889\ \text{rad}=\boxed{5.09^\circ}$$
5. Effect of doubling the loads. Doubling every torque doubles every stress: $$2\tau_{AB}=330.1\ \text{MPa}\quad 2\tau_{BC}=282.9\ \text{MPa}\quad 2\tau_{CD}=352.6\ \text{MPa}$$ All three now exceed the 200 MPa shear yield — the shaft would yield in shear over its ENTIRE length, not just the governing CD segment, since even the least-stressed segment (BC) already sits above yield once doubled. The elastic torsion formulas above would no longer apply once yielding begins.