3.3 Torsion of Shafts, Power Transmission & Thin-Walled Tubes

3.3 Torsion of Shafts, Power Transmission & Thin-Walled Tubes

A twisting moment applied about a member's longitudinal axis produces torsional shear stress. Transmission shafts, couplings, torque tubes, spring wire and drive axles are all sized with the relations below, and on the exam a torsion check almost always comes paired with a power-and-speed conversion and an angle-of-twist (stiffness) limit.

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1. Torsion of Circular Solid & Hollow Shafts

Torsion refers to the twisting of a structural member when subjected to an applied moment acting about its longitudinal axis. In circular cross-sections (solid or hollow), axisymmetric geometry ensures that plane cross-sections remain plane and do not warp during elastic twisting.

The Elastic Torsion Formula

Torsional shear stress $\tau$ increases linearly with radial distance $r$ from the shaft centerline:

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

Maximum shear stress develops at the outermost boundary radius $c = r_o = d_o / 2$:

$$\tau_{\max} = \frac{T c}{J} = \frac{T d_o / 2}{J}$$

Polar Moment of Inertia ($J$)

• Solid Circular Shaft (Diameter $d$): $J = \frac{\pi d^4}{32} = \frac{\pi r^4}{2}$ $\tau_{\max} = \frac{T (d/2)}{\frac{\pi d^4}{32}} = \frac{16 T}{\pi d^3}$
• Hollow Circular Shaft (Outer Dia $d_o$, Inner Dia $d_i$): $J = \frac{\pi (d_o^4 - d_i^4)}{32} = \frac{\pi (r_o^4 - r_i^4)}{2}$ $\tau_{\max} = \frac{T (d_o / 2)}{\frac{\pi (d_o^4 - d_i^4)}{32}} = \frac{16 T d_o}{\pi (d_o^4 - d_i^4)}$

Angle of Twist ($\phi$)

The total angle of twist $\phi$ in radians over length $L$ for a shaft of uniform shear modulus $G$ is:

$$\phi = \frac{T L}{G J} \quad [\text{radians}]$$

To convert radians to degrees: $\phi_{\text{deg}} = \phi_{\text{rad}} \left(\frac{180^\circ}{\pi}\right)$. Torsional stiffness $k_t$ is:

$$k_t = \frac{T}{\phi} = \frac{G J}{L} \quad [\text{N}\cdot\text{m/rad} \text{ or } \text{in-lb/rad}]$$

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2. Torsional Power Transmission: SI & USCS Units

Rotating shafts are ubiquitous in machine systems for transmitting mechanical power between motors, gearboxes, and driven loads. Mechanical power $P$ is the product of applied torque $T$ and angular velocity $\omega$:

$$P = T \omega = 2 \pi N T / 60$$

where $N$ is rotational shaft speed in revolutions per minute (rpm) and $\omega = \frac{2 \pi N}{60}$ in rad/s.

SI Unit Formulation

• Power $P$ in Watts (W) or kilowatts (kW).
• Torque $T$ in Newton-meters ($\text{N}\cdot\text{m}$).
• Speed $N$ in rpm.

$$T = \frac{P}{\omega} = \frac{60 P}{2 \pi N} \approx 9.5493 \left(\frac{P}{N}\right)$$

(Note: When $P$ is in kW, multiply by $10^3$ to obtain $T$ in $\text{N}\cdot\text{m}$).

USCS Unit Formulation

• Power in horsepower (hp), where $1\text{ hp} = 550\text{ ft-lb/s} = 6600\text{ in-lb/s} = 33,000\text{ ft-lb/min}$.
• Rotational speed $n$ in rpm.
• Torque in inch-pounds ($T$ in $\text{in-lb}$):

$$\text{hp} = \frac{T_{\text{in-lb}} \cdot n}{63,025} \iff T_{\text{in-lb}} = \frac{63,025 \cdot \text{hp}}{n}$$

• Torque in foot-pounds ($T$ in $\text{ft-lb}$):

$$\text{hp} = \frac{T_{\text{ft-lb}} \cdot n}{5252} \iff T_{\text{ft-lb}} = \frac{5252 \cdot \text{hp}}{n}$$

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3. Thin-Walled Non-Circular Closed Tubes (Bredt-Batho Theory)

Non-circular sections (rectangles, ellipses, streamline tubes) undergo significant cross-sectional warping under torsion, invalidating the classic circular torsion formula. However, for thin-walled closed tubes with wall thickness $t \ll$ overall dimensions, shear flow $q = \tau t$ is constant around the entire perimeter.

           +---------------------------+
           |  Mean Centerline Area Am  |
           |   +-------------------+   | 
           | t |                   |   |
           |   +-------------------+   |
           +---------------------------+

Bredt-Batho Shear Stress Formula

$$\tau = \frac{T}{2 A_m t}$$

where $A_m$ is the cross-sectional area enclosed by the median line (mean centerline) of the tube wall, and $t$ is the local wall thickness at the point of interest.

Angle of Twist for Closed Thin-Walled Tubes

The rate of twist $\theta = \phi / L$ is obtained by integrating along the median perimeter $S_m$:

$$\theta = \frac{T}{4 A_m^2 G} \oint \frac{ds}{t}$$

For a tube with constant wall thickness $t$ and total median perimeter length $S_m$:

$$\phi = \frac{T L S_m}{4 A_m^2 G t}$$

Open vs. Closed Sections: A closed tubular section exhibits immense torsional stiffness. If a longitudinal slit is cut along the tube wall, it becomes an open thin-walled section. Its torsional stiffness drops by factors of $100$ to $10,000\times$, and peak shear stress escalates dramatically because shear must circulate within the narrow thickness ($J_{\text{open}} \approx \frac{1}{3} \sum b_i t_i^3$).

4. Worked Step-by-Step Shaft Sizing Example

Problem Statement

A solid drive shaft is required to transmit $P = 45\text{ kW}$ of continuous power from an electric motor operating at a rated speed of $N = 900\text{ rpm}$. The shaft span between bearings is $L = 1.20\text{ m}$.

Design Constraints & Material Properties:

• Material: Cold-rolled AISI 1045 steel.
• Allowable shear stress: $\tau_{\text{allow}} = 50\text{ MPa}$.
• Allowable angle of twist: $\phi_{\text{allow}} = 1.0^\circ$ over the total $1.20\text{ m}$ span.
• Shear modulus: $G = 78\text{ GPa} = 78 \times 10^9\text{ Pa}$.

Objective:

1. Calculate the steady-state operating torque $T$.
2. Determine the minimum required shaft diameter $d_{\text{solid}}$ satisfying both strength and torsional stiffness.
3. Compare a hollow shaft design with $d_i / d_o = 0.75$, evaluating the required outer diameter and percentage mass savings.

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Step 1: Operating Torque Calculation

Convert power to Watts ($P = 45,000\text{ W}$) and calculate angular velocity $\omega$:

$\omega = \frac{2 \pi N}{60} = \frac{2 \pi (900)}{60} = 30 \pi \approx 94.2478\text{ rad/s}$

$T = \frac{P}{\omega} = \frac{45,000\text{ W}}{94.2478\text{ rad/s}} = 477.465\text{ N}\cdot\text{m}$

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Step 2: Sizing Solid Shaft for Strength (Stress Criterion)

Applying the elastic torsion formula for a solid circular shaft:

$\tau_{\max} = \frac{16 T}{\pi d^3} \le \tau_{\text{allow}}$

$d \ge \left( \frac{16 T}{\pi \tau_{\text{allow}}} \right)^{1/3} = \left( \frac{16 \times 477.465\text{ N}\cdot\text{m}}{\pi \times 50 \times 10^6\text{ N/m}^2} \right)^{1/3}$

$d \ge \left( \frac{7639.44}{1.57080 \times 10^8} \right)^{1/3} = (4.86341 \times 10^{-5}\text{ m}^3)^{1/3} = 0.03650\text{ m} = 36.50\text{ mm}$

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Step 3: Sizing Solid Shaft for Stiffness (Twist Criterion)

First, convert allowable twist from degrees to radians:

$\phi_{\text{allow}} = 1.0^\circ \times \left(\frac{\pi}{180^\circ}\right) = 0.0174533\text{ radians}$

Using the angle of twist formula for a solid shaft ($J = \pi d^4 / 32$):

$\phi = \frac{T L}{G J} = \frac{32 T L}{\pi G d^4} \le \phi_{\text{allow}}$

$d \ge \left( \frac{32 T L}{\pi G \phi_{\text{allow}}} \right)^{1/4} = \left( \frac{32 \times 477.465\text{ N}\cdot\text{m} \times 1.20\text{ m}}{\pi \times (78 \times 10^9\text{ Pa}) \times 0.0174533\text{ rad}} \right)^{1/4}$

Numerator: $32 \times 477.465 \times 1.20 = 18,334.66\text{ N}\cdot\text{m}^2$.

Denominator: $\pi \times (78 \times 10^9) \times 0.0174533 = 4.27683 \times 10^9\text{ N/m}^2$.

$d \ge \left( \frac{18,334.66}{4.27683 \times 10^9} \right)^{1/4} = (4.28697 \times 10^{-6}\text{ m}^4)^{1/4} = 0.04550\text{ m} = 45.50\text{ mm}$

Criterion Comparison:

• Strength requirement: $d \ge 36.50\text{ mm}$.
• Stiffness requirement: $d \ge 45.50\text{ mm}$.

Governing Design: Torsional stiffness governs. Sizing requires $d_{\text{min}} = 45.50\text{ mm}$. Sizing to the nearest standard millimetric increment yields $d = 46\text{ mm}$ or $d = 48\text{ mm}$.

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Step 4: Hollow Shaft Design & Mass Reduction Analysis

Let the hollow shaft have an inner-to-outer diameter ratio $c_r = d_i / d_o = 0.75$.

The polar moment of inertia is:

$J_{\text{hollow}} = \frac{\pi d_o^4}{32} (1 - c_r^4) = \frac{\pi d_o^4}{32} [1 - (0.75)^4] = \frac{\pi d_o^4}{32} [1 - 0.316406] = 0.683594 \left(\frac{\pi d_o^4}{32}\right)$

For identical torsional stiffness, $J_{\text{hollow}} = J_{\text{solid}} = \frac{\pi d_{\text{solid}}^4}{32}$:

$d_o^4 (0.683594) = d_{\text{solid}}^4 \implies d_o = \frac{d_{\text{solid}}}{(0.683594)^{1/4}} = \frac{45.50\text{ mm}}{0.90928} = 50.04\text{ mm}$

$d_i = 0.75 \times 50.04\text{ mm} = 37.53\text{ mm}$

Now, compare cross-sectional areas to evaluate mass reduction:

$A_{\text{solid}} = \frac{\pi d_{\text{solid}}^2}{4} = \frac{\pi (45.50\text{ mm})^2}{4} = 1626.0\text{ mm}^2$

$A_{\text{hollow}} = \frac{\pi (d_o^2 - d_i^2)}{4} = \frac{\pi [(50.04)^2 - (37.53)^2]}{4} = \frac{\pi [2504.00 - 1408.50]}{4} = \frac{\pi (1095.50)}{4} = 860.4\text{ mm}^2$

$\text{Mass Savings} = \frac{A_{\text{solid}} - A_{\text{hollow}}}{A_{\text{solid}}} \times 100\% = \frac{1626.0 - 860.4}{1626.0} \times 100\% = 47.08\%$

Conclusion: By replacing the solid shaft with a hollow shaft ($d_o = 50.0\text{ mm}, d_i = 37.5\text{ mm}$), the mass is reduced by 47.1% while providing exactly identical torsional rigidity.