10.1 Torsion of circular shafts

Torsion of circular shafts deals with the stresses and deformations that develop when twisting moments are applied to cylindrical components. This is fundamental to designing anything that transmits rotational power: drive shafts, axles, drill bits, and turbine shafts all depend on torsion analysis. The core challenge is predicting how much a shaft twists and whether the internal shear stresses stay within safe limits.

Circular shafts under torsion

Characteristics and components

Circular shafts exist to transmit power and torque between mechanical components like gears, pulleys, and rotors. When a torsional load (a twisting moment) is applied, the shaft rotates about its longitudinal axis, developing internal shear stresses and angular deformation along its length.

Two categories of properties govern how a shaft responds to torsion:

• Geometric properties: diameter, length, and cross-sectional shape. These determine the polar moment of inertia, which controls how effectively the shaft resists twisting.
• Material properties: the shear modulus $G$ (also called the modulus of rigidity) quantifies stiffness against shear deformation, while the shear yield strength sets the stress limit before permanent deformation begins.

Boundary conditions also matter. How the shaft is supported (fixed at one end, fixed at both ends, etc.) and where torques are applied directly affect the internal torque distribution and resulting stresses.

Torsional behavior and failure

Torsional loading produces shear strains throughout the shaft. Shear strain $\gamma$ relates to shear stress $\tau$ through the shear modulus:

$\gamma = \tau / G$

This means a stiffer material (higher $G$) develops less strain for the same stress level.

Design requirements typically limit both the maximum shear stress and the angle of twist. Excessive twist can cause misalignment between connected components or introduce harmful vibrations. The maximum allowable shear stress is set by the material's shear yield strength, which is typically about 50–60% of the tensile yield strength (this follows from yield criteria like Tresca or von Mises).

Three failure modes to know:

• Yielding: permanent deformation occurs when shear stress exceeds the shear yield strength
• Fatigue: progressive damage under repeated cyclic loading, even at stresses below the yield strength
• Fracture: sudden failure when stress exceeds the ultimate shear strength

Shear stress distribution in torsion

Shear stress variation

One of the key results in torsion theory is that shear stress varies linearly from zero at the center of the shaft to a maximum at the outer surface. This comes from the assumption that plane cross-sections remain plane and radii remain straight during twisting (valid for circular cross-sections within the elastic range).

The torsion formula gives the shear stress at any radial distance $\rho$ from the center:

$\tau = \frac{T \cdot \rho}{J}$

where:

• $T$ = applied internal torque
• $\rho$ = radial distance from the shaft center (use $r$ for the outer radius to find maximum stress)
• $J$ = polar moment of inertia of the cross-section

For a solid circular shaft of radius $r$:

$J = \frac{\pi r^4}{2}$

Or equivalently in terms of diameter $d$:

$J = \frac{\pi d^4}{32}$

For a hollow circular shaft with outer radius $r_o$ and inner radius $r_i$:

$J = \frac{\pi}{2}(r_o^4 - r_i^4)$

The maximum shear stress always occurs at the outer surface where $\rho = r$:

$\tau_{max} = \frac{T \cdot r}{J}$

Factors affecting shear stress

• Applied torque: a larger twisting moment produces proportionally larger shear stress everywhere in the cross-section.
• Shaft geometry: since $J$ depends on the fourth power of the radius, even a modest increase in diameter dramatically increases torsional resistance. Doubling the diameter of a solid shaft increases $J$ by a factor of 16.
• Material properties: the shear modulus $G$ does not appear directly in the stress formula, so for a given torque and geometry, the shear stress is the same regardless of material. However, $G$ determines how much the shaft deforms at that stress level, which influences whether the elastic torsion formula remains valid.

Angle of twist in shafts

Angular deformation

The angle of twist $\theta$ measures the relative rotation between two cross-sections separated by a length $L$ along the shaft. For a shaft with constant torque, constant cross-section, and uniform material:

$\theta = \frac{T \cdot L}{G \cdot J}$

where:

• $T$ = internal torque (constant along the length considered)
• $L$ = length of the shaft segment
• $G$ = shear modulus of the material
• $J$ = polar moment of inertia

The product $GJ$ is called the torsional rigidity (or torsional stiffness per unit length). It captures how both material and geometry resist twisting.

If the torque or cross-section varies along the shaft, you need to either sum contributions from individual segments or integrate:

$\theta = \int_0^L \frac{T(x)}{G \cdot J(x)} \, dx$

For a shaft with multiple segments of constant $T$, $G$, and $J$, you simply add up the twist from each segment:

$\theta_{total} = \sum \frac{T_i \cdot L_i}{G_i \cdot J_i}$

Factors affecting angle of twist

• Applied torque: twist is directly proportional to $T$
• Shaft length: longer shafts twist more for the same torque and cross-section
• Shear modulus: a stiffer material (higher $G$) twists less. Steel ($G \approx 80 \text{ GPa}$) twists roughly three times less than aluminum ($G \approx 26 \text{ GPa}$) for the same shaft geometry and torque.
• Cross-sectional geometry: increasing the polar moment of inertia reduces twist. Because $J$ scales with $r^4$, increasing shaft diameter is the most effective geometric change you can make.

Torque, shear stress, and geometry

Interdependence in torsional loading

The torsion formula $\tau = T\rho / J$ and the twist formula $\theta = TL / GJ$ together capture the full picture of torsional loading. Both depend on the polar moment of inertia $J$, which links geometry to stress and deformation. A design change that reduces stress (like increasing diameter) will simultaneously reduce the angle of twist.

You can also combine the two formulas to relate shear stress directly to angle of twist:

$\tau = G \cdot \rho \cdot \frac{\theta}{L}$

This shows that shear stress at any point depends on how much the shaft is actually twisting per unit length ($\theta / L$), scaled by the material stiffness and the radial position.

Design considerations

Solid vs. hollow shafts: Hollow shafts are more weight-efficient because the material near the center of a solid shaft contributes very little to $J$ (and carries very little stress). Removing that core material barely reduces $J$ but significantly reduces weight. For example, a hollow shaft with an inner radius equal to half the outer radius retains about 94% of the solid shaft's $J$ while using only 75% of the material.

Diameter effects: Since $J$ scales with the fourth power of the radius, small increases in diameter have outsized effects. Increasing the radius by just 20% increases $J$ (and therefore torsional resistance) by about 107%.

Material selection: Common shaft materials include steel (high strength and stiffness), aluminum (lightweight but lower stiffness), titanium (high strength-to-weight ratio), and composites (tailorable properties). The choice depends on the balance between strength, stiffness, weight, cost, and environmental conditions.

Factor of safety: Real designs include a factor of safety (typically 1.5 to 3) to account for uncertainties in loading, material variability, and manufacturing tolerances. The required shaft dimensions are sized so that:

$\tau_{max} \leq \frac{\tau_{yield}}{n}$

where $n$ is the factor of safety. Higher consequences of failure (e.g., aircraft vs. hand tools) call for larger safety factors.