10.2 Torsion of Circular Shafts, Polar Moment, and Angle of Twist

10.2 Torsion of Circular Shafts, Polar Moment, and Angle of Twist

Torsion is one of the four loading modes named in the NCEES Strength of Materials sub-topic on stress and strain, and it is the mode that governs every rotating shaft. Both a strength check and a stiffness check are required, and for long shafts the stiffness limit usually controls.

Pure Torsion of Circular and Tubular Shafts

When a circular shaft of radius $c$ (or outer radius $r_o$ and inner radius $r_i$) is subjected to an applied torque $T$, internal shear stress $\tau$ acts on the cross-section perpendicular to the longitudinal axis.

Torsional Shear Stress Formula

The torsional shear stress at a distance $r$ from the longitudinal center axis is given by:

$\tau = \frac{T r}{J}$

The maximum torsional shear stress $\tau_{max}$ occurs at the outer boundary ($r = c$):

$\tau_{max} = \frac{T c}{J}$

where $J$ is the Polar Moment of Inertia of the cross-section.

Polar Moment of Inertia ($J$)

• Solid Circular Cross-Section (diameter $d$, radius $c = d/2$): $J = \frac{\pi d^4}{32} = \frac{\pi c^4}{2}$

• Hollow Circular (Tubular) Cross-Section (outer diameter $d_o$, inner diameter $d_i$): $J = \frac{\pi (d_o^4 - d_i^4)}{32} = \frac{\pi (r_o^4 - r_i^4)}{2}$

Angle of Twist ($\phi$)

The angle of twist $\phi$ (in radians) over a shaft length $L$ subjected to constant torque $T$ and shear modulus $G$ is:

$\phi = \frac{T L}{G J}$

For shafts with stepped sections or varying torques, the total angle of twist is $\phi = \sum \frac{T_i L_i}{G_i J_i}$.

Power Transmission in Rotating Shafts

Rotating shafts transmit power $P$ at rotational speed $\omega$ (in rad/s) or frequency $f$ (in Hz / rev/s):

$P = T \omega = 2 \pi f T$

FE Exam Tip: When power is given in horsepower ($1\text{ hp} = 550\text{ ft}\cdot\text{lb/s} = 746\text{ W}$) and rotational speed in RPM ($N$), convert RPM to rad/s using $\omega = \frac{2\pi N}{60}$ before calculating torque.

Why Hollow Shafts Win

Because $J$ depends on the fourth power of radius, material near the axis contributes almost nothing to torsional capacity — and it carries almost no stress either, since $\tau = Tr/J$ vanishes at $r = 0$.

Compare a solid 100 mm shaft with a hollow shaft of 100 mm outside and 60 mm inside diameter:

$J_{\text{solid}} = \frac{\pi(100)^4}{32} = 9.817\times10^6\ \text{mm}^4$

$J_{\text{hollow}} = \frac{\pi\left[(100)^4-(60)^4\right]}{32} = \frac{\pi\left[10^8 - 1.296\times10^7\right]}{32} = 8.545\times10^6\ \text{mm}^4$

The hollow shaft retains 87% of the torsional capacity while removing $(60/100)^2 = 36%$ of the cross-sectional area — so it is 36% lighter for a 13% strength penalty. Per unit mass it is about 36% more efficient, which is why drive shafts, axles, and torque tubes are tubular.

Trap: $J$ for a hollow section is $\frac{\pi}{32}(D_o^4 - D_i^4)$ — subtract the fourth powers, not the diameters. Computing $\frac{\pi}{32}(D_o - D_i)^4$ gives $8.0\times10^4$ mm⁴, low by a factor of 100.

Power Transmission

Shaft problems usually supply power and speed rather than torque:

$P = T\omega \qquad\Longrightarrow\qquad T = \frac{P}{\omega}, \qquad \omega = \frac{2\pi N}{60}\ \text{[rad/s]}$

In US Customary units the standard shortcut is:

$T\ [\text{ft}\cdot\text{lbf}] = \frac{5{,}252 \times \text{hp}}{N\ [\text{rpm}]}$

Worked Example: Sizing a Shaft from Transmitted Power

A solid steel shaft ($G = 79$ GPa) transmits 45 kW at 600 rpm. The allowable shear stress is 55 MPa and the allowable twist is 1.5° over a 2.0 m length. Find the required diameter.

Torque: $\omega = \frac{2\pi(600)}{60} = 62.83\ \text{rad/s}, \qquad T = \frac{45{,}000}{62.83} = 716.2\ \text{N}\cdot\text{m}$

Criterion 1 — shear stress. With $\tau_{\max} = \dfrac{16T}{\pi D^3}$:

$D^3 = \frac{16T}{\pi\tau_{\text{allow}}} = \frac{16(716.2)}{\pi(55\times10^6)} = \frac{11{,}459}{1.728\times10^8} = 6.631\times10^{-5}\ \text{m}^3$

$D = (6.631\times10^{-5})^{1/3} = 0.04049\ \text{m} = 40.5\ \text{mm}$

Criterion 2 — angle of twist. Convert the limit: $1.5° = 0.02618$ rad. With $\phi = \dfrac{TL}{JG}$ and $J = \dfrac{\pi D^4}{32}$:

$J_{\text{required}} = \frac{TL}{\phi G} = \frac{716.2(2.0)}{0.02618(79\times10^9)} = \frac{1{,}432.4}{2.068\times10^9} = 6.926\times10^{-7}\ \text{m}^4$

$D^4 = \frac{32 J}{\pi} = \frac{32(6.926\times10^{-7})}{\pi} = 7.055\times10^{-6}\ \text{m}^4$

$D = (7.055\times10^{-6})^{1/4} = 0.05155\ \text{m} = 51.6\ \text{mm}$

The twist criterion governs. Select $D = 55$ mm (next standard size), and verify:

$\tau = \frac{16(716.2)}{\pi(0.055)^3} = \frac{11{,}459}{5.226\times10^{-4}} = 21.9\ \text{MPa} < 55\ \text{MPa}\ ✓$

The design lesson: both criteria must be checked, and for long slender shafts the stiffness (twist) limit usually controls, not strength. Candidates who check only stress specify a 45 mm shaft that twists 2.7° — nearly double the allowable — and would be offered exactly that as a distractor.

Worked Engineering Problem: Hollow Shaft Torsion

Problem: Hollow Circular Shaft Torsion and Angle of Twist

Scenario: A hollow steel shaft ($G = 77\text{ GPa}$) with outer diameter $d_o = 80\text{ mm}$ ($0.08\text{ m}$) and inner diameter $d_i = 60\text{ mm}$ ($0.06\text{ m}$) has length $L = 2.5\text{ m}$. The shaft transmits a constant torque $T = 6.0\text{ kN}\cdot\text{m} = 6000\text{ N}\cdot\text{m}$. Calculate (a) the maximum torsional shear stress $\tau_{max}$ and (b) the total angle of twist $\phi$ in degrees.

Solution:

1. Calculate the polar moment of inertia $J$: $J = \frac{\pi}{32} (d_o^4 - d_i^4) = \frac{\pi}{32} \left( (0.08)^4 - (0.06)^4 \right)$ $J = \frac{\pi}{32} (4.096 \times 10^{-5} - 1.296 \times 10^{-5}) = \frac{\pi}{32} (2.80 \times 10^{-5}) = 2.7489 \times 10^{-6}\text{ m}^4$

2. Calculate maximum shear stress at outer radius $c = d_o / 2 = 0.04\text{ m}$: $\tau_{max} = \frac{T c}{J} = \frac{(6000\text{ N}\cdot\text{m})(0.04\text{ m})}{2.7489 \times 10^{-6}\text{ m}^4} = \frac{240}{2.7489 \times 10^{-6}} = 8.7308 \times 10^7\text{ Pa} = 87.31\text{ MPa}$

3. Calculate angle of twist $\phi$ in radians: $\phi = \frac{T L}{G J} = \frac{(6000\text{ N}\cdot\text{m})(2.5\text{ m})}{(77 \times 10^9\text{ Pa})(2.7489 \times 10^{-6}\text{ m}^4)} = \frac{15,000}{211,665.3} = 0.070868\text{ rad}$

4. Convert radians to degrees: $\phi_{deg} = 0.070868 \times \left( \frac{180^\circ}{\pi} \right) = 4.06^\circ$