--- title: "Torsion moment diagram for a geared shaft" subject: "Strength of Materials" chapter: 4 key: "semester_4/strength_of_materials_mechanical:torsion-moment-diagram-gear-shaft" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/torsion-moment-diagram-gear-shaft site: BCT Engineering --- # Torsion moment diagram for a geared shaft Strength of Materials · Chapter 4: Torsion - Asked in: year not given, 4+4 marks - Video: [Torsion: Shaft Carrying Multiple Torques](https://www.youtube.com/watch?v=yfXCX9H7tS8) — Abrenian Lecturer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/torsion-moment-diagram-gear-shaft - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-4 ## Question (a) What is a torsion moment (torque) diagram? Explain how it is drawn and the sign convention used. (b) A uniform shaft ABCD, 40 mm in diameter, has gears at A, B, C and D spaced 1.0 m apart. Gear A is the input and supplies 800 N m. The gears at B, C and D take off 300 N m, 200 N m and 300 N m respectively. Draw the torsion moment diagram, and find the maximum shear stress and the total angle of twist of D relative to A. G = 80 GPa. ## Short answer A torsion moment diagram plots internal torque along a shaft; each segment’s torque is found by cutting and using right‑hand‑rule sign. For the shaft: AB=800 N·m, BC=500 N·m, CD=300 N·m. Maximum shear stress τ=max=16T/(πd³)=63.7 MPa. Total twist φ=Σ(TL/GJ)=0.0796 rad (4.56°). ## Answer #### (a) Torsion moment diagram A **torsion moment diagram** plots the internal torque $T$ at each section along the axis of a shaft. It shows where the torque is maximum, which governs the design of the shaft. Steps: 1. Find any unknown support torque from equilibrium ($\sum T = 0$ about the axis). 2. Cut the shaft in each segment between the points where external torques are applied. 3. Find the internal torque from the external torques on one side of the cut. 4. Plot a horizontal line for each segment; the diagram jumps at each external torque by the size of that torque. **Sign convention:** the torque is positive if the torque vector, drawn by the right-hand rule, points away from the cut section (along the outward normal). Positive and negative values only indicate opposite twisting directions. #### (b) Numerical Check equilibrium: $800 = 300+200+300$. Internal torque in each segment (considering the left side): | Segment | Internal torque | |---|---| | AB | $+800$ N m | | BC | $800 - 300 = +500$ N m | | CD | $500 - 200 = +300$ N m (equal to the torque taken off at D) | ```text T (N m) 800 +------+ | | 500 | +------+ | | 300 | +------+ | | 0 +------+------+------+---- x A B C D ``` The maximum torque is $800$ N m, in AB. $$ \tau_{max} = \frac{16T}{\pi d^3} = \frac{16\times800\times10^3}{\pi\times40^3} = 63.7\ \text{MPa} $$ Total angle of twist (sum over the segments, each 1000 mm long): $$ J = \frac{\pi}{32}(40)^4 = 251.3\times10^3\ \text{mm}^4 $$ $$ \phi_{D/A} = \sum\frac{T L}{GJ} = \frac{(800+500+300)\times10^3\times1000}{80000\times251.3\times10^3} = 0.0796\ \text{rad} = 4.56^\circ $$ **Answer: torques $800$, $500$, $300$ N m; $\tau_{max} = 63.7$ MPa in AB; total twist $0.0796$ rad $= 4.56^\circ$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the internal torque in each segment found? A. From the external torques on one side of the cut B. By averaging all torques C. By ignoring the gears D. From the diameter alone Answer: A. Cut in each segment and balance external torques on one side; the diagram jumps at each external torque. (Key point: Drawing torque diagram) 2. What are the internal torques in AB, BC, CD? A. 800, 500, 300 N m B. 800, 300, 200 N m C. 300, 500, 800 N m D. 800, 1100, 1300 N m Answer: A. AB = 800; BC = 800 - 300 = 500; CD = 500 - 200 = 300 N m. (Key point: Segment torques) 3. How is the twist of D relative to A found? A. Using only the maximum torque 800 N m over 3 m B. Summing T L/(G J) for each segment C. Using the average torque only over AB D. Dividing 800 by G Answer: B. Torque changes by segment, so sum TL/GJ: (800+500+300) x 10^3 x 1000/(80000 x 251.3 x 10^3) = 0.0796 rad. 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