Question 3 of 8: Stepped Circular Shaft — Torsion

Question 3: Stepped Circular Shaft — 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. The printed figure shows the curved torque arrows at C and D sweep the same rotational sense, while the arrow at B sweeps the opposite sense — i.e. T_B opposes T_C and T_D.

T_B = 7 kN·m (CCW), T_C = 4 kN·m (CW), T_D = 1 kN·m (CW); G = 80 GPa; shear yield τ_Y = 250 MPa.

Find. τ_max in the shaft and its distribution over the governing cross-section; the angle of twist at D; the consequence of doubling all three torques.

Approach. Take CW as positive, sum the applied torques on the free-end side of a cut through each segment to get the internal torque, then apply $\tau=Tc/J$ and $\phi=\sum TL/(GJ)$ segment by segment.

1. Internal torque per segment (cut then sum everything toward the free end D): $$T_{AB}=T_B+T_C+T_D = -7+4+1 = -2\ \text{kN}\cdot\text{m}\qquad T_{BC}=T_C+T_D=5\ \text{kN}\cdot\text{m}\qquad T_{CD}=T_D=1\ \text{kN}\cdot\text{m}$$
2. Shear stress in each segment, $\tau = Tc/J$ with $J=\pi d^4/32$ (solid) or $\pi(d_o^4-d_i^4)/32$ (hollow): $$J_{AB}=\frac{\pi(40)^4}{32}=2.513\times10^5\ \text{mm}^4 \;\Rightarrow\; \tau_{AB}=\frac{(2\times10^6)(20)}{2.513\times10^5}=159.2\ \text{MPa}$$ $$J_{BC}=\frac{\pi(60)^4}{32}=1.272\times10^6\ \text{mm}^4 \;\Rightarrow\; \tau_{BC}=\frac{(5\times10^6)(30)}{1.272\times10^6}=117.9\ \text{MPa}$$ $$J_{CD}=\frac{\pi(40^4-30^4)}{32}=1.718\times10^5\ \text{mm}^4 \;\Rightarrow\; \tau_{CD}=\frac{(1\times10^6)(20)}{1.718\times10^5}=116.4\ \text{MPa}$$ $$\boxed{\tau_{max}=159.2\ \text{MPa, in segment AB}\ (<\tau_Y=250\ \text{MPa}\Rightarrow\text{still elastic})}$$ The stress is zero at the shaft centre and increases linearly to this value at the outer radius (part (a) sketch below).
3. Angle of twist at D (segment twists accumulate with sign): $$\phi_D=\sum\frac{T_iL_i}{GJ_i}=\frac{(-2\times10^6)(400)}{80000(2.513\times10^5)}+\frac{(5\times10^6)(1500)}{80000(1.272\times10^6)}+\frac{(1\times10^6)(800)}{80000(1.718\times10^5)}$$ $$\boxed{\phi_D = 0.09210\ \text{rad} = 5.28^{\circ}}$$
4. Part (c): doubling the loads. Every internal torque and hence every stress scales by exactly 2 (linear elasticity), so $$\tau_{AB,\text{doubled}} = 2(159.2)=318.3\ \text{MPa} > \tau_Y = 250\ \text{MPa}$$ Segment AB would yield in shear. The elastic formulas $\tau=Tc/J$ and $\phi=TL/(GJ)$ would no longer hold for AB — the shaft would develop a plastic (yielded) outer annulus there, the twist would grow disproportionately faster than the applied torque, and the design would need to be re-sized (or the AB diameter increased) to remain safe; segments BC and CD (235.8 and 232.8 MPa doubled) would still be elastic but with very little margin left.