Stepped circular shaft — torsion: 04-BS-6, May 2013, Q8

Question 8: 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.

Shaft ABCD: AB solid (50 mm), BCD hollow (50/40 mm); torque senses as marked.

Given.

Quantity	Value
AB	solid, 50 mm dia., 1000 mm long
BC, CD	hollow, 50 mm OD / 40 mm ID; 800 mm, 1200 mm long
Applied torques	2800 N·m CW at B; 2000 N·m CCW at C; 1500 N·m/m CW along CD
$G$, $F_y$ (shear)	80 GPa, 150 MPa

Find. (a) $\tau_{max}$ in the shaft, and its variation across the radius at that section; (b) the angle of twist at D.

Approach. Taking CCW (matching the torque at C) as positive, sum the applied torques from each cut to the free end D to build the internal torque diagram $T(x)$, then apply $\tau=Tr/J$ per segment (tracking the change in $J$ at B) and sum $\theta=\int T/(GJ)\,dx$ segment-by-segment for the total twist.

Internal torque diagram. With CCW $=+$: $T_B=-2800$, $T_C=+2000$ N·m, and the distributed torque totals $1500\times1.2=-1800$ N·m (CW) over CD. Summing from each cut to the free end D: $$T_{CD}(x)=-1500(3.0-x)\ \text{N}\!\cdot\!\text{m}\ (x\text{ in m from A});\quad T_{CD}(1.8)=-1800,\ T_{CD}(3.0)=0$$ $$T_{BC}=T_C+(\text{dist. total})=2000-1800=+200\ \text{N}\!\cdot\!\text{m}$$ $$T_{AB}=T_B+T_C+(\text{dist. total})=-2800+2000-1800=\boxed{-2600\ \text{N}\!\cdot\!\text{m}}$$ Reaction at the fixed wall A: $+2600$ N·m (equal and opposite to $T_{AB}$), confirming overall equilibrium.
Polar moments of inertia. $$J_{solid}=\frac{\pi(25)^4}{2}=6.136\times10^{5}\ \text{mm}^4\ (\text{AB}),\qquad J_{hollow}=\frac{\pi(25^4-20^4)}{2}=3.623\times10^{5}\ \text{mm}^4\ (\text{BCD})$$
Shear stress in each segment. $\tau=|T|r_{outer}/J$: $$\tau_{AB}=\frac{(2600\times10^3)(25)}{6.136\times10^5}=105.9\ \text{MPa}$$ $$\tau_{BC}=\frac{(200\times10^3)(25)}{3.623\times10^5}=13.8\ \text{MPa}$$ $$\tau_{CD}\big|_{x=C}=\frac{(1800\times10^3)(25)}{3.623\times10^5}=\boxed{124.2\ \text{MPa}}\quad(\text{maximum, in the hollow shaft just past C})$$ Although AB carries the largest torque (2600 N·m), the hollow section just past C — carrying "only" 1800 N·m through a smaller $J$ — produces the larger stress and governs; all values remain below the 150 MPa yield.
Radial variation at the governing section (just past C). Shear stress is linear in $r$, $\tau(r)=Tr/J$, but exists only for $20\le r\le25$ mm (no material inside the bore): $$\tau(20\ \text{mm})=\frac{(1800\times10^3)(20)}{3.623\times10^5}=99.4\ \text{MPa},\qquad \tau(25\ \text{mm})=124.2\ \text{MPa}$$ i.e. a straight line from 99.4 MPa at the bore to 124.2 MPa at the outer surface — with a discontinuity (zero stress, no material) for $r<20$ mm, unlike a solid shaft's stress line which would run all the way to zero at the centre.
Part (b): angle of twist at D. Summing each segment (constant-torque segments directly, CD by integrating the linear $T(x)$): $$\theta_{AB}=\frac{T_{AB}L_{AB}}{GJ_{solid}}=\frac{(-2600\times10^3)(1000)}{(80\,000)(6.136\times10^5)}=-0.05297\ \text{rad}$$ $$\theta_{BC}=\frac{T_{BC}L_{BC}}{GJ_{hollow}}=\frac{(200\times10^3)(800)}{(80\,000)(3.623\times10^5)}=+0.00552\ \text{rad}$$ $$\theta_{CD}=\int_0^{1200}\frac{T_{CD}(s)}{GJ_{hollow}}\,ds=-0.03727\ \text{rad}$$ $$\theta_D=\theta_{AB}+\theta_{BC}+\theta_{CD}=\boxed{-0.0847\ \text{rad}=-4.85^\circ}$$

Quantity	Result
$T_{AB}$, $T_{BC}$, $T_{CD}(C^+)$	−2600, +200, −1800 N·m
$\tau_{max}$ (location)	124.2 MPa, hollow section just past C
$\tau$ at bore ($r=20$ mm)	99.4 MPa
$\theta_D$	−0.0847 rad = −4.85°

Concept & Theory

Topic. Torsion of a stepped, mixed solid/hollow circular shaft with distributed torque.

Key relations. $\tau=Tr/J$ (linear in $r$, evaluated across the actual material only); $\theta=\int T(x)/(GJ)\,dx$, summed segment-by-segment; $J_{solid}=\pi r^4/2$, $J_{hollow}=\pi(r_o^4-r_i^4)/2$.

Why this works. Each prismatic segment carries a constant (or, on CD, linearly varying) internal torque found purely from statics (the shaft is statically determinate since only one end is fixed); linear-elastic torsion theory then gives stress and twist directly per segment, and total twist is the algebraic sum because rotations are small and additive.

Common pitfall. Assuming the segment with the largest applied/internal torque automatically has the largest stress — here the smaller torque in the hollow segment past C produces the higher stress because $J_{hollow} < J_{solid}$ despite the equal outer radius; both $T$ and $J$ must be compared together, segment by segment.

Source. Hibbeler, Mechanics of Materials, Ch. 5 (Torsion: shear stress and angle of twist in circular shafts, statically determinate stepped shafts).

Question 8 of 8: Stepped circular shaft — torsion

Question 8: Stepped circular shaft — torsion (20 marks)