Torsion

Assumptions

1. The shaft is straight, of uniform circular cross-section, and the material is homogeneous and isotropic.
2. The material obeys Hooke's law; stresses stay within the elastic (proportional) limit.
3. Plane transverse sections remain plane and circular after twisting; radii remain straight.
4. The twist is uniform along the length and the torque acts in a plane perpendicular to the axis.
5. Shear strain varies linearly with the distance from the axis.

Derivation

  fixed end                free end
     |  A ------------- B     line AB before twisting
     |  A ------------- B'    after twisting
     |<------- L ------->|    arc BB' = r*theta

A line AB on the surface of the shaft of radius $r$ and length $L$ moves to AB' when the free end twists through angle $\theta$ (radians). The arc BB' $= r\theta$, and the shear strain (small angle) is

$$\gamma = \frac{BB'}{L} = \frac{r\theta}{L}$$

By Hooke's law in shear, $\tau = G\gamma$, so at the surface

$$\frac{\tau}{r} = \frac{G\theta}{L}$$

At a general radius $\rho$, $\tau_\rho = G\rho\theta/L$, hence shear stress varies linearly with radius: zero at the axis and maximum at the surface.

The torque on a ring of area $dA = 2\pi\rho\,d\rho$ is $dT = \tau_\rho\,\rho\,dA$.

$$T = \int \tau_\rho\,\rho\,dA = \frac{G\theta}{L}\int \rho^2\,dA = \frac{G\theta}{L}J$$

where $J = \int\rho^2\,dA$ is the polar second moment of area. Combining:

$$\boxed{\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}}$$

Section properties

• Polar modulus $Z_p = J/r$ (so $T = \tau_{max}Z_p$). For a solid shaft $Z_p = \pi d^3/16$.
• Torsional rigidity $= GJ$, the torque needed to produce a unit angle of twist per unit length ($T/\theta = GJ/L$).

Power transmitted: $P = T\omega = \dfrac{2\pi N T}{60}$ with $N$ in rpm.

Torque to be transmitted

$$T = \frac{60P}{2\pi N} = \frac{60\times150\times10^3}{2\pi\times300} = 4774.6\ \text{N m} = 4.775\times10^6\ \text{N mm}$$

Solid shaft

$$T = \frac{\pi}{16}\tau d^3 \;\Rightarrow\; d^3 = \frac{16\times4.775\times10^6}{\pi\times60} = 405.3\times10^3$$ $$d = 74.0\ \text{mm} \;\rightarrow\; \text{provide } 75\ \text{mm}$$

Hollow shaft ($k = d_i/d_o = 0.6$)

$$T = \frac{\pi}{16}\tau\,\frac{d_o^4 - d_i^4}{d_o} = \frac{\pi}{16}\tau\,d_o^3(1-k^4)$$ $$1 - k^4 = 1 - 0.6^4 = 0.8704$$ $$d_o^3 = \frac{16\times4.775\times10^6}{\pi\times60\times0.8704} = 465.7\times10^3 \;\Rightarrow\; d_o = 77.5\ \text{mm}$$ $$d_i = 0.6\times77.5 = 46.5\ \text{mm}$$

Provide $d_o = 78$ mm, $d_i = 47$ mm.

Weight comparison

For the same length and material, weight is proportional to cross-sectional area:

$$\frac{A_{hollow}}{A_{solid}} = \frac{d_o^2(1-k^2)}{d^2} = \frac{77.5^2(1-0.36)}{74.0^2} = \frac{3843.6}{5476} = 0.702$$ $$\text{Saving} = 1 - 0.702 = 29.8\ \%$$

The hollow shaft is more economical because the material near the axis, where shear stress is small, is removed.

Answer: hollow shaft $d_o = 77.5$ mm (78 mm), $d_i = 46.5$ mm (47 mm); solid shaft $d = 74.0$ mm (75 mm); weight saving $\approx 29.8\ \%$.

Method

The problem is statically indeterminate: only one equilibrium equation is available for two unknown reactions. Add a compatibility condition.

 //|A-----50mm-----C---40mm---B|//
 //|<---- 0.6 m --->|<-0.4 m->|//
                    | T0 = 2.5 kN m

Equations

1. Equilibrium: $T_A + T_B = T_0 = 2.5\times10^6$ N mm
2. Compatibility: the angle of twist of AC and CB must be equal at C (both ends are fixed and the section C rotates by one angle $\phi_C$)

$$\frac{T_A L_1}{GJ_1} = \frac{T_B L_2}{GJ_2}$$

Polar moments:

$$J_1 = \frac{\pi}{32}(50)^4 = 613.6\times10^3\ \text{mm}^4, \qquad J_2 = \frac{\pi}{32}(40)^4 = 251.3\times10^3\ \text{mm}^4$$

Torsional flexibility of each part, $L/J$: $\dfrac{600}{613.6\times10^3} = 9.778\times10^{-4}$ and $\dfrac{400}{251.3\times10^3} = 15.915\times10^{-4}$.

(a) Reactions

$$T_A = T_0\,\frac{L_2/J_2}{L_1/J_1 + L_2/J_2} = 2.5\times\frac{15.915}{9.778+15.915} = 1.549\ \text{kN m}$$ $$T_B = 2.5 - 1.549 = 0.951\ \text{kN m}$$

(The stiffer portion AC takes the larger share.)

(b) Maximum shear stresses

$$\tau_{AC} = \frac{16T_A}{\pi d_1^3} = \frac{16\times1.549\times10^6}{\pi\times50^3} = 63.1\ \text{MPa}$$ $$\tau_{CB} = \frac{16T_B}{\pi d_2^3} = \frac{16\times0.951\times10^6}{\pi\times40^3} = 75.7\ \text{MPa}$$

(c) Angle of twist of C

$$\phi_C = \frac{T_AL_1}{GJ_1} = \frac{1.549\times10^6\times600}{80000\times613.6\times10^3} = 0.0189\ \text{rad} = 1.08^\circ$$

Check using CB: $\dfrac{0.951\times10^6\times400}{80000\times251.3\times10^3} = 0.0189$ rad.

Answer: $T_A = 1.549$ kN m, $T_B = 0.951$ kN m; $\tau_{AC} = 63.1$ MPa, $\tau_{CB} = 75.7$ MPa; $\phi_C = 0.0189$ rad $= 1.08^\circ$.

(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 an external torque is applied.
3. Find the internal torque from the external torques on one side of the cut.
4. Plot a horizontal line in 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 (outward normal). Positive and negative values only indicate opposite twisting directions.

(b) Numerical

Check equilibrium: $800 = 300+200+300$.

Internal torque in each segment (consider the left side):

 A      B      C      D
 |______|______|______|
 800    500    300   T (N m)
 ■■■■■■|
        ■■■■■■|
               ■■■■■■|

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 $= 4.56^\circ$.