Internal Torque, Maximum Shear Stress, and Angle of Twist in a Stepped Shaft: 04-BS-6, Undated paper, Q6

Question 6: Internal Torque, Maximum Shear Stress, and Angle of Twist in a Stepped Shaft

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.

Segment	Length	Section
$AB$	600 mm	hollow, 30 mm OD / 20 mm ID
$BC$	400 mm	solid, 40 mm dia.
$CD$	600 mm	solid, 20 mm dia.

Applied torques: 500 N·m CCW at B, 1000 N·m CW at C, 200 N·m CCW at D. $G=80$ GPa, $\tau_Y=160$ MPa.

Find. (a) the maximum shear stress in the shaft, with its radial distribution at that section; (b) the angle of twist at D, in degrees.

Stepped shaft A-B-C-D, fixed at A; torques 500 CCW at B, 1000 CW at C, 200 CCW at D.

Approach. Working from the free end D back to the fixed end A (method of sections), find the internal torque carried by each segment; compute $\tau=T\rho/J$ (max at the outer radius) per segment to find the governing one; sum $\theta_i=T_iL_i/(GJ_i)$ for the total twist at D.

Internal torque per segment (taking CCW as $+$, consistent with the free end): cutting successively from D toward A and summing the torques on the freed portion, $$T_{CD}=+200\text{ N}\!\cdot\!\text{m},\qquad T_{BC}=-1000+200=-800\text{ N}\!\cdot\!\text{m},\qquad T_{AB}=500-1000+200=-300\text{ N}\!\cdot\!\text{m}$$ (Check: the fixed-end reaction is $-T_{AB}=+300$ N·m, and $300+500-1000+200=0$ — equilibrium confirmed.)
Polar moments of inertia. $$J_{AB}=\tfrac{\pi}{32}(0.030^4-0.020^4)=6.381\times10^{-8}\text{ m}^4$$ $$J_{BC}=\tfrac{\pi}{32}(0.040)^4=2.513\times10^{-7}\text{ m}^4,\qquad J_{CD}=\tfrac{\pi}{32}(0.020)^4=1.571\times10^{-8}\text{ m}^4$$
Shear stress per segment ($\tau=T\,c/J$, outer radius $c$). $$\tau_{AB}=\frac{300(0.015)}{6.381\times10^{-8}}=70.5\text{ MPa},\quad \tau_{BC}=\frac{800(0.020)}{2.513\times10^{-7}}=63.7\text{ MPa},\quad \tau_{CD}=\frac{200(0.010)}{1.571\times10^{-8}}=127.3\text{ MPa}$$ $$\boxed{\tau_{max}=127.3\text{ MPa, in segment CD}}\quad(<\tau_Y=160\text{ MPa})$$ At that section, $\tau$ varies linearly with radius from $0$ at the shaft centre to $127.3$ MPa at $\rho=10$ mm (solid section — no discontinuity, unlike the hollow $AB$ segment which would jump from $0$ at $\rho=10$ mm to a finite value across the bore).
Angle of twist at D. $$\theta_D=\frac{T_{AB}L_{AB}}{GJ_{AB}}+\frac{T_{BC}L_{BC}}{GJ_{BC}}+\frac{T_{CD}L_{CD}}{GJ_{CD}}$$ $$=\frac{-300(0.6)}{80\times10^9(6.381\times10^{-8})}+\frac{-800(0.4)}{80\times10^9(2.513\times10^{-7})}+\frac{200(0.6)}{80\times10^9(1.571\times10^{-8})}$$ $$=-0.03526-0.01592+0.09549=0.04432\text{ rad}$$ $$\boxed{\theta_D=2.54^{\circ}}$$

Result	Value
$T_{AB},T_{BC},T_{CD}$	−300, −800, +200 N·m
$\tau_{max}$	127.3 MPa, segment CD
$\theta_D$	0.0443 rad = 2.54°

Topic: Torsion of a multi-segment (stepped) circular shaft with multiple applied torques.

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

Why this works: A statically DETERMINATE shaft (one fixed support) needs no compatibility equation — the internal torque in each segment follows directly from equilibrium of the free end inward. Even though segment $CD$ carries the SMALLEST applied torque, its small 20 mm diameter (versus 40 mm for BC) makes $J$ shrink as $d^4$, so it produces the largest shear stress: geometry, not load magnitude, controls in stepped shafts.

Common pitfall: Assuming the segment with the largest torque automatically governs the stress, or mixing sign conventions between segments when summing $\theta$ — a consistent CCW/CW convention must be tracked from one end of the shaft to the other.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 5 §5.1–5.4 (Torsion).

Question 6 of 8: Internal Torque, Maximum Shear Stress, and Angle of Twist in a Stepped Shaft

Question 6: Internal Torque, Maximum Shear Stress, and Angle of Twist in a Stepped Shaft