Question 4 of 8: Stepped Shaft in Torsion

Question 4: Stepped 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.

[Figure not reproduced: Stepped shaft, fixed at A: T B =12 kN·m and T D =3 kN·m act in one rotational sense, T C =7 kN·m in the opposite sense (confirmed from the arrow curvature/arrowhead position on the source drawing); linear shear-stress distribution shown for the governing segment CD. See the official exam paper.]

Given.

G = 80 GPa, f_y = 250 MPa (shear); T_B=12 kN·m, T_D=3 kN·m (same rotational sense); T_C=7 kN·m (opposite sense); shaft fixed at A.

Find. (a) Maximum shear stress and its radial variation. (b) Angle of twist at D relative to A. (c) The effect of doubling every torque.

Approach. Take the sense of T_B and T_D as positive (T_C is then negative). Cut each segment and sum the applied torques to one side of the cut to get the internal torque; apply $\tau=Tr/J$ in each segment (the governing one is whichever gives the largest τ, not necessarily the largest T) and sum $\phi=TL/(GJ)$ segment by segment for the total twist at D.

1. Internal torque in each segment (cutting from D back to A, signs per the arrows). $$T_{CD}=T_D=+3\ \text{kN}\cdot\text{m}$$ $$T_{BC}=T_C+T_D=-7+3=-4\ \text{kN}\cdot\text{m}\ (4\ \text{kN}\cdot\text{m in the sense of }T_C)$$ $$T_{AB}=T_B+T_C+T_D=12-7+3=+8\ \text{kN}\cdot\text{m}$$ Check: the fixed-end reaction must equal the sum of every applied torque, $12-7+3=8$ kN·m $=T_{AB}$ — confirms the segment signs.
2. (a) Shear stress in every segment, $\tau=16T/(\pi d^3)$. $$\tau_{AB}=\frac{16(8\times10^6)}{\pi(80)^3}=79.6\ \text{MPa}, \qquad \tau_{BC}=\frac{16(4\times10^6)}{\pi(60)^3}=94.3\ \text{MPa}$$ $$\tau_{CD}=\frac{16(3\times10^6)}{\pi(40)^3}=\boxed{238.7\ \text{MPa}}$$ Even though $T_{AB}$ is the largest internal torque, the much smaller 40 mm diameter of segment CD makes it govern: $$\boxed{\tau_{max}=238.7\ \text{MPa, in segment CD}}$$ — below the 250 MPa shear yield, so the shaft remains elastic. The distribution is LINEAR across the cross-section: zero at the shaft centre, rising to 238.7 MPa at the outer radius (r = 20 mm).
3. (b) Angle of twist at D relative to A. $$J_{AB}=\frac{\pi(80)^4}{32}=4.021\times10^6\ \text{mm}^4, \quad J_{BC}=\frac{\pi(60)^4}{32}=1.272\times10^6\ \text{mm}^4, \quad J_{CD}=\frac{\pi(40)^4}{32}=0.2513\times10^6\ \text{mm}^4$$ $$\phi_{AB}=\frac{T_{AB}L_{AB}}{GJ_{AB}}=+1.710^\circ, \qquad \phi_{BC}=\frac{T_{BC}L_{BC}}{GJ_{BC}}=-1.801^\circ, \qquad \phi_{CD}=\frac{T_{CD}L_{CD}}{GJ_{CD}}=+11.969^\circ$$ $$\boxed{\phi_{D/A}=\phi_{AB}+\phi_{BC}+\phi_{CD}=1.710-1.801+11.969=11.88^\circ}\ \text{(net, in the sense of }T_B\text{ and }T_D\text{)}$$
4. (c) Effect of doubling every torque. Doubling every applied torque doubles every internal torque and hence every stress (elastic, linear in T): $$\tau_{CD,doubled}=2(238.7)=477.5\ \text{MPa} \gg f_y=250\ \text{MPa}$$ $$\boxed{\text{Segment CD would YIELD} \text{ — the outer fibres exceed 250 MPa shear stress and the torsion formula }\tau=Tr/J\text{ no longer applies; the shaft develops a plastic core in CD and the angle of twist grows disproportionately faster than the (now doubled) torque.}}$$

Final Results.