--- title: "Brass-steel stepped shaft: max torque, rotation" subject: "Strength of Materials" chapter: 6 key: "semester_3/strength_of_materials:stepped-shaft-brass-steel-max-torque" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/stepped-shaft-brass-steel-max-torque site: BCT Engineering --- # Brass-steel stepped shaft: max torque, rotation Strength of Materials · Chapter 6: Torsion - Asked in: 2066 Bhadra (old course), 8 marks - Video: [Understanding Torsion](https://www.youtube.com/watch?v=1YTKedLQOa0) — The Efficient Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/stepped-shaft-brass-steel-max-torque - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6 ## Question The allowable shear stress in brass is $88\ \text{N/mm}^2$ and steel is $110\ \text{N/mm}^2$. Find the maximum torque 'T' that can be applied at the free end C in the stepped shaft of solid circular section as shown. Find also the total rotation of free end of the shaft with respect to the fixed end, if $G_{brass} = 400\ \text{kN/mm}^2$ and $G_{steel} = 80\ \text{kN/mm}^2$. [Figure: shaft fixed at A; brass segment AB of diameter 80 mm and length 1 m; steel segment BC of diameter 60 mm and length 1.2 m; torque T at free end C.] ## Short answer Torque capacity from τ_allow: brass 8.85 kN·m, steel 4.67 kN·m, so T_max=4.67 kN·m. Twist: φ_brass=0.00290 rad, φ_steel=0.05500 rad; total rotation = 0.0579 rad = 3.32°. ## Answer **Data:** brass AB: $d = 80$ mm, $L = 1000$ mm, $\tau_{allow} = 88\ \text{N/mm}^2$, $G = 400\ \text{kN/mm}^2$; steel BC: $d = 60$ mm, $L = 1200$ mm, $\tau_{allow} = 110\ \text{N/mm}^2$, $G = 80\ \text{kN/mm}^2$. The same torque $T$ acts in both segments (applied at the free end C). #### Maximum torque $$ T_{brass} = \frac{\pi}{16}\tau d^3 = \frac{\pi}{16}(88)(80)^3 = 8.8467\times10^{6}\ \text{N·mm} = 8.847\ \text{kN·m} $$ $$ T_{steel} = \frac{\pi}{16}(110)(60)^3 = 4.6653\times10^{6}\ \text{N·mm} = 4.665\ \text{kN·m} $$ The smaller governs, so $T_{max} = 4.665$ kN·m (the steel segment reaches its limit first). #### Rotation of the free end relative to the fixed end $$ \varphi_{AB} = \frac{TL}{GJ} = \frac{(4.6653\times10^{6})(1000)}{(400\times10^3)\frac{\pi}{32}(80)^4} = 0.00290\ \text{rad} $$ $$ \varphi_{BC} = \frac{(4.6653\times10^{6})(1200)}{(80\times10^3)\frac{\pi}{32}(60)^4} = 0.05500\ \text{rad} $$ $$ \varphi_C = \varphi_{AB} + \varphi_{BC} = 0.05790\ \text{rad} = 3.317^\circ $$ **Answer:** $T_{max} = 4.67$ kN·m; total rotation of C $= 0.0579$ rad $= 3.32^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the same torque T used in both brass and steel segments? A. Both materials have equal G B. Because the stresses are equal C. T is applied at the free end C, so every section transmits T D. The diameters are equal Answer: C. With the load only at the free end, each cross-section carries T. (Key point: Same torque) 2. How is the maximum permissible T chosen? A. Sum of the two B. Larger of the two limiting torques C. Average of the two D. Smaller of the two limiting torques Answer: D. T must not exceed either limit; the smaller (steel, 4.665 kN·m) governs. (Key point: Governing segment) 3. Which expression gives the limiting torque of a segment? A. T = G J phi/L only B. T = (pi/16) tau d^3 C. T = tau d^3 D. T = (pi/32) tau d^4 Answer: B. T = tau J/r = (pi/16) tau d^3. (Key point: Limiting torque formula) 4. How is the rotation of C relative to A found? A. T L/(G J) with average G B. Sum of T L/(G J) for AB and BC (0.0029 + 0.0550 rad) C. Difference of the two twists D. Use only the steel G Answer: B. Each segment is computed with its own G and J and the twists add to 0.0579 rad = 3.32°. 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