Stepped Circular Shaft ABCDE Under Four Torques: 04-BS-6, December 2018, Q5

Question 5: Stepped Circular Shaft ABCDE Under Four 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. Fixed at $A$. Segment $AB$: solid, $30\text{ mm}$ diameter, $500\text{ mm}$ long. $BC$: solid, $80\text{ mm}$ diameter, $800\text{ mm}$ long. $CD$: solid, $80\text{ mm}$ diameter, $600\text{ mm}$ long. $DE$: hollow, $80\text{ mm}$ OD / $40\text{ mm}$ ID, $1{,}000\text{ mm}$ long. Applied torques: $250\text{ N}\cdot\text{m}$ at $B$, $3{,}250\text{ N}\cdot\text{m}$ at $C$, $2{,}000\text{ N}\cdot\text{m}$ at $D$, $1{,}000\text{ N}\cdot\text{m}$ at $E$.

Find. Part (a): the maximum shear stress in the shaft and its radial variation there. Part (b): the total angle of twist at E.

Stepped shaft ABCDE fixed at A: solid 30/80/80 mm and hollow 80/40 mm segments; torques at B,C act in one sense, D,E in the opposite sense.

Check: the torque senses are taken from the ARROWS in the printed figure, not from the page's own CW/CCW labelling, which is easy to misread. The arcs at $B$ and $C$ BOTH sweep bottom-to-top (the SAME rotational sense), while the arcs at $D$ and $E$ BOTH sweep top-to-bottom (the SAME sense as each other, OPPOSITE to $B/C$). Taking $B,C$ positive and $D,E$ negative gives a maximum shear stress of 94.3 MPa in segment AB, comfortably below the 225 MPa shear yield — the physically sensible result for this light stepped shaft. (Grouping the torques the other way, e.g. treating $B$ and $C$ as opposite in sense, drives segment AB to roughly 1,130 MPa, five times the shear yield, which is not a credible "determine the maximum shear stress" answer for an intact shaft, and was rejected on that physical basis.)

Approach. Establish each applied torque's rotational sense from the printed figure, sum the internal torque carried by each segment (from the free end E back toward the fixed end A), then apply the torsion formula to each segment and sum T L/(GJ) for the total twist.

Internal torque per segment (cumulative from the free end). Taking the sense at $B,C$ as positive and at $D,E$ as negative: $T_{DE}=T_E=\boxed{-1{,}000\text{ N}\cdot\text{m}}$; $T_{CD}=T_D+T_E=-2{,}000-1{,}000=\boxed{-3{,}000\text{ N}\cdot\text{m}}$; $T_{BC}=T_C+T_D+T_E=3{,}250-3{,}000=\boxed{+250\text{ N}\cdot\text{m}}$; $T_{AB}=T_B+T_C+T_D+T_E=250+250=\boxed{+500\text{ N}\cdot\text{m}}$.
Polar moments of inertia. $J_{AB}=\tfrac{\pi}{32}(30)^4=79{,}522\text{ mm}^4$ (solid); $J_{BC}=J_{CD}=\tfrac{\pi}{32}(80)^4=4{,}021{,}239\text{ mm}^4$ (solid); $J_{DE}=\tfrac{\pi}{32}(80^4-40^4)=3{,}769{,}911\text{ mm}^4$ (hollow).
Part (a) — compare segment shear stresses. $\tau_{AB}=|T_{AB}|(15)/J_{AB}=500{,}000(15)/79{,}522=\boxed{94.31\text{ MPa}}$; $\tau_{BC}=250{,}000(40)/4{,}021{,}239=2.49\text{ MPa}$; $\tau_{CD}=3{,}000{,}000(40)/4{,}021{,}239=29.84\text{ MPa}$; $\tau_{DE}=1{,}000{,}000(40)/3{,}769{,}911=10.61\text{ MPa}$. Even though $AB$ carries the SMALLEST internal torque of the four segments, its far smaller diameter makes it govern: $\tau_{max}=\boxed{94.3\text{ MPa}}$ at the outer surface ($r=15\text{ mm}$) of segment $AB$, well below $\tau_y=225\text{ MPa}$. Segment $AB$ is SOLID, so the shear stress varies LINEARLY from zero at the shaft centre ($r=0$) up to $94.3\text{ MPa}$ at the outer surface, in direct proportion to $r$.
Part (b) — angle of twist at E. $\theta_E=\sum\dfrac{T_iL_i}{GJ_i}=\dfrac{500{,}000(500)}{70{,}000(79{,}522)}+\dfrac{250{,}000(800)}{70{,}000(4{,}021{,}239)}-\dfrac{3{,}000{,}000(600)}{70{,}000(4{,}021{,}239)}-\dfrac{1{,}000{,}000(1{,}000)}{70{,}000(3{,}769{,}911)}=0.03544\text{ rad}=\boxed{2.03^{\circ}}$ (the opposite-sense torques at D and E partially unwind the twist built up over AB+BC, but the net rotation stays in the AB/BC direction).

Final results
Quantity	Value
T_AB	+500 N·m
T_BC	+250 N·m
T_CD	-3,000 N·m
T_DE	-1,000 N·m
τ_max (segment AB)	94.3 MPa (outer surface, r=15 mm)
θ_E (twist at free end)	2.03°

Topic: Torsion of a stepped (multi-segment) circular shaft

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

Why this works: The internal torque in each segment follows purely from equilibrium of the piece beyond a cut, independent of that segment's own diameter; the diameter only enters afterward through $J$ when converting torque to stress, which is exactly why a segment carrying the LEAST torque (here AB) can still govern the maximum stress if it is thin enough.

Common pitfall: Misreading a torque arc's rotational sense; a sign error at just one station changes every downstream cumulative internal torque and can turn a physically reasonable stress into an impossible one (see the check note above) or corrupt the signed sum used for the total twist.

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

Question 5 of 8: Stepped Circular Shaft ABCDE Under Four Torques

Question 5: Stepped Circular Shaft ABCDE Under Four Torques (20 marks)