Stepped Circular Shaft Under Two Torques: 04-BS-6, May 2018, Q6

Question 6: Stepped Circular Shaft Under Two Torques (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. $AB$: $500\text{ mm}$ long, hollow, OD $=60\text{ mm}$, ID $=44\text{ mm}$. $BC$: $100\text{ mm}$ long, solid, OD $=60\text{ mm}$ (Check: the source figure gives only ONE outer diameter for AB and states the hole is drilled "in portion AB" only, so BC is taken as the same solid $60\text{ mm}$ shaft continuing from AB — the step to a smaller section occurs only at C, per the given $30\text{ mm}$ CD diameter.). $CD$: $300\text{ mm}$ long, solid, OD $=30\text{ mm}$. $T_C=2{,}000\text{ N}\cdot\text{m}$ CCW, $T_D=200\text{ N}\cdot\text{m}$ CW (opposite sense). $G=80\text{ GPa}$.

Find. Part (a): the maximum shear stress in the shaft and its radial variation there. Part (b): the angle of twist of end D relative to the fixed end A.

Stepped shaft ABCD: AB hollow (OD 60/ID 44 mm), BC solid (OD 60 mm), CD solid (OD 30 mm); torques at C and D act in opposite senses.

Shear-stress variation across the radius of segment AB (hollow): linear from 43.7 MPa at the bore to 59.7 MPa (max) at the outer surface.

Approach. Use the method of sections to find the internal torque carried by each of the three segments (working from the free end D toward the fixed end A), then apply the torsion formula $\tau=T\rho/J$ to each segment to find where the maximum occurs, and sum $T L/(GJ)$ over the three segments for the total twist at D.

Internal torque per segment. Cutting between C and D: only $T_D$ acts beyond the cut, so $T_{CD}=200\text{ N}\cdot\text{m}$ (taking CW as negative and CCW positive at D relative to the sense at C, $T_{CD}=-200\text{ N}\cdot\text{m}$). Cutting between B and C: both $T_C$ and $T_D$ act beyond the cut, so $T_{BC}=T_{AB}=2{,}000-200=\boxed{1{,}800\text{ N}\cdot\text{m}}$ (no torque is applied at B, so AB and BC carry the identical internal torque).
Polar moments. $J_{AB}=\dfrac{\pi}{32}(60^4-44^4)=0.904\times10^{6}\text{ mm}^4$ (hollow); $J_{BC}=\dfrac{\pi}{32}(60)^4=1.272\times10^{6}\text{ mm}^4$ (solid); $J_{CD}=\dfrac{\pi}{32}(30)^4=0.0795\times10^{6}\text{ mm}^4$ (solid).
Part (a) — compare segment shear stresses. $\tau_{AB}=\dfrac{1{,}800{,}000\times30}{0.904\times10^{6}}=59.7\text{ MPa}$; $\tau_{BC}=\dfrac{1{,}800{,}000\times30}{1.272\times10^{6}}=42.4\text{ MPa}$; $\tau_{CD}=\dfrac{200{,}000\times15}{0.0795\times10^{6}}=37.7\text{ MPa}$. The hollow segment AB governs (same torque as BC, but less material resists it): $\tau_{\max}=\boxed{59.7\text{ MPa}}$ at the OUTER surface of AB, well below $\tau_y=160\text{ MPa}$. Across the radius of AB, the shear stress varies linearly from $\tau_i=\tau_{\max}(r_i/r_o)=43.7\text{ MPa}$ at the $22\text{ mm}$ bore surface up to $59.7\text{ MPa}$ at the $30\text{ mm}$ outer surface — note it does NOT start at zero, since there is no material at $r=0$.
Part (b) — angle of twist at D. $\theta_D=\sum\dfrac{T_iL_i}{GJ_i}=\dfrac{1{,}800{,}000(500)}{80{,}000(0.904\times10^6)}+\dfrac{1{,}800{,}000(100)}{80{,}000(1.272\times10^6)}-\dfrac{200{,}000(300)}{80{,}000(0.0795\times10^6)}=0.00478\text{ rad}=\boxed{0.274^{\circ}}$ (the CD segment's opposite-sense torque partially unwinds the twist built up over AB+BC).

Final results
Quantity	Value
T_AB = T_BC	1,800 N·m
T_CD	200 N·m (opposite sense)
τ_max (segment AB, hollow)	59.7 MPa (outer surface)
τ at the bore (r=22mm)	43.7 MPa
θ_D (twist at free end)	0.274°

Topic: Torsion of stepped (multi-segment) circular shafts

Key relations: $\tau=T\rho/J$ (max at outer radius); $J_{solid}=\pi d^4/32$; $J_{hollow}=\pi(d_o^4-d_i^4)/32$; $\theta=\sum T_iL_i/(G_iJ_i)$.

Why this works: The internal torque in each segment is found purely from equilibrium of the piece beyond a cut, independent of the shaft's geometry at that cut; the geometry (solid vs. hollow, diameter) only enters afterward, through $J$, when converting torque to stress or twist.

Common pitfall: Assuming the segment with the largest applied torque, or the smallest diameter, automatically governs the maximum stress; here AB and BC carry the SAME internal torque, so the hollow section (smaller $J$) governs even though its torque is not the largest applied value in the problem.

Source: Hibbeler, Mechanics of Materials, 10th ed., §5.3–5.5 Torsion of Circular Shafts; Angle of Twist

Question 6 of 8: Stepped Circular Shaft Under Two Torques

Question 6: Stepped Circular Shaft Under Two Torques (20 marks)