Torsion moment diagram for a geared shaft

(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):

 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 $= 4.56^\circ$.