Question 5 of 8: Stepped Shaft with Distributed Torque

Question 5: Stepped Shaft with Distributed Torque (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.

G = 80 GPa, f_y = 250 MPa (shear); T_B = 12 kN·m CW; T_C = 7 kN·m CCW; distributed torque t = 2.5 kN·m/m CCW over CD; shaft fixed at A.

Find. (a) Maximum shear stress and its radial variation. (b) Angle of twist at D relative to the fixed end A.

Approach. Take CCW as positive. The shaft is fixed only at A (free at D), so the internal torque at any section equals the sum of every applied torque between that section and the FREE end D; within CD the distributed torque makes the internal torque vary linearly, so its contribution to the twist needs an integral rather than a single $TL/GJ$ term.

1. Internal torque in CD (from D back to C). Measuring ξ from C ($0\le\xi\le1200$ mm), the torque remaining downstream of a cut at ξ is the distributed torque still ahead of it: $$T_{CD}(\xi) = 2.5\frac{(1200-\xi)}{1000}\ \text{kN}\cdot \text{m (CCW)}$$ so $T_{CD}$ is maximum right at C, $T_{CD}(0)=2.5(1.2)=+3.0$ kN·m, and zero at D.
2. Internal torque in BC and AB (constant). $$T_{BC} = T_C + T_{CD}(0) = 7+3.0 = +10.0\ \text{kN}\cdot\text{m (CCW)}$$ $$T_{AB} = T_B(\text{CW}) + T_{BC} = -12+10.0 = -2.0\ \text{kN}\cdot\text{m (i.e. 2.0 kN}\cdot\text{m CW)}$$
3. (a) Shear stress in every segment ($\tau=T r/J$, $J=\pi d^4/32$). $$\tau_{AB}=\frac{16(2.0\times10^6)}{\pi(80)^3} = 19.9\ \text{MPa}, \qquad \tau_{BC}=\frac{16(10.0\times10^6)}{\pi(60)^3} = 235.8\ \text{MPa}$$ $$\tau_{CD,\max}=\frac{16(3.0\times10^6)}{\pi(40)^3} = \boxed{238.7\ \text{MPa}}$$ Even though $T_{BC}>T_{CD}$, the SMALLER 40 mm diameter of segment CD makes it govern (just) over the 60 mm segment BC ($238.7>235.8$ MPa); both stay below the 250 MPa yield. Shear stress varies LINEARLY from zero at the shaft centre to this 238.7 MPa maximum at the outer radius of segment CD, immediately adjacent to C.
4. (b) Angle of twist at D relative to A. $J_{AB}=\pi(80)^4/32= 4.021\times10^6$ mm⁴, $J_{BC}=\pi(60)^4/32=1.272\times10^6$ mm⁴, $J_{CD}=\pi(40)^4/32=0.2513\times10^6$ mm⁴. $$\phi_{AB}=\frac{T_{AB}L_{AB}}{GJ_{AB}} = -0.356^\circ, \qquad \phi_{BC}=\frac{T_{BC}L_{BC}}{GJ_{BC}} = +3.940^\circ$$ For CD, integrate the linearly varying torque over the segment: $$\phi_{CD}=\frac{1}{GJ_{CD}}\int_0^{1200} 2.5\frac{(1200-\xi)}{1000}\,d\xi = +5.129^\circ$$ $$\boxed{\phi_{D/A} = \phi_{AB}+\phi_{BC}+\phi_{CD} = -0.356+3.940+5.129 = 8.71^\circ}\ \text{(net CCW, viewed from D)}$$

Final Results.