Torsion of Circular Shafts

When you apply a torque to a wrench, you feel the metal shaft twist as it transmits that turning force to a bolt. This fundamental action of torsion, the twisting of a structural member due to an applied torque, is critical in countless engineering systems, from automotive drive shafts and industrial machinery to the simple propeller on a toy boat. Understanding torsion is not just about applying a formula; it’s about predicting how much a component will deform and ensuring it doesn’t fail under load. This analysis centers on calculating the internal shear stress and the resulting angle of twist for both solid and hollow circular shafts, providing the tools to design efficient and reliable power-transmitting components.

The Linear Shear Stress Distribution

When a pure torque $T$ is applied to a shaft, it induces shear stress within the material. A key, often counterintuitive, finding is that this stress is not uniform across the cross-section. Instead, it varies linearly from zero at the very center of the shaft (the longitudinal axis) to a maximum value at the outermost surface. This occurs because the deformation, or shear strain, due to twisting is greatest at the farthest point from the center.

The formula that defines this linear relationship is fundamental: $$\tau =\frac{T\rho }{J}$$ Here, $\tau$ (tau) is the shear stress at a specific point, $T$ is the applied internal torque, and $\rho$ (rho) is the radial distance from the center to that point. The term $J$ is the polar moment of inertia, a geometric property of the cross-section that quantifies its resistance to torsional deformation. The maximum shear stress, ${\tau }_{max}$, always occurs at the outer surface where $\rho$ equals the outer radius $c$. Therefore, the formula for design checks is often written as: $${\tau }_{max}=\frac{Tc}{J}$$

Imagine a solid steel rod. The material at the very core experiences almost no shear stress; it is relatively "idle." As you move outward, each concentric "tube" of material carries more stress, with the outer skin doing the heaviest lifting. This linear distribution is a direct consequence of the shaft remaining elastic, plane cross-sections remaining plane, and the material being homogeneous and isotropic—core assumptions of the classical torsion theory for circular members.

The Polar Moment of Inertia, $J$

The polar moment of inertia, $J$, is the rotational analog of the area moment of inertia used in bending. It depends solely on the geometry of the cross-section. For a solid circular shaft of radius $c$ and diameter $D$, it is calculated by integrating the squares of elemental areas about the axis: $$J_{solid}=\frac{\pi c^4}{2}=\frac{\pi D^4}{32}$$ This $D^4$ relationship is powerful; doubling the diameter increases $J$ by a factor of 16, dramatically reducing stress and deformation for the same applied torque.

For a hollow circular shaft with an outer radius $c_o$ and inner radius $c_i$ (or outer diameter $D_o$ and inner diameter $D_i$), the polar moment of inertia is the difference between the $J$ values for the outer and inner solid circles: $$J_{hollow}=\frac{\pi }{2}(c_o^4-c_i^4)=\frac{\pi }{32}(D_o^4-D_i^4)$$ This property is central to understanding why hollow shafts are so efficient, as removing material from the low-stress center has a minimal impact on $J$ but a significant impact on weight.

Calculating the Angle of Twist

While stress tells us about strength, deformation tells us about stiffness. The angle of twist $\varphi$ quantifies how much one end of a shaft rotates relative to the other when torque is applied. For a prismatic shaft (uniform cross-section and material) under constant torque, the formula is: $$\varphi =\frac{TL}{GJ}$$ In this equation, $T$ is the constant internal torque, $L$ is the length of the shaft segment, $G$ is the shear modulus (a material property describing its stiffness in shear), and $J$ is the polar moment of inertia.

Let's walk through a step-by-step example. A solid steel shaft ($G=80$ GPa) with a diameter of 50 mm ($J=\pi (0.05{)}^4/32=6.136\times 10^{-7}$ m${}^4$) is subjected to a torque of 1500 N·m over a length of 2 m. The angle of twist in radians is: $$\varphi =\frac{(1500\text{Ncdotpm})(2\text{m})}{(80\times 10^9\text{Pa})(6.136\times 10^{-7}{\text{m}}^4)}\approx 0.0611\text{radians}$$ To convert to more intuitive degrees, multiply by $180/\pi$, yielding approximately 3.5 degrees. This equation shows that twist is directly proportional to torque and length, and inversely proportional to the product $GJ$, known as the torsional rigidity. A larger rigidity means a stiffer shaft that twists less.

The Efficiency of Hollow Shafts

Hollow shafts provide efficient material usage for torque transmission. This principle is paramount in aerospace, automotive, and any weight-sensitive design. Since shear stress is zero at the center and maximum at the surface, the material near the shaft's axis is under-utilized. By removing it to create a hollow tube, you redistribute material to the outer radius where it is most effective at resisting torsional stress.

The efficiency gain is revealed by comparing the polar moment of inertia per unit cross-sectional area (a proxy for weight). Consider a solid shaft of diameter $D$. Now, create a hollow shaft with the same outer diameter $D$ but with a carefully chosen inner diameter. While the area (and thus weight) decreases, $J$ decreases at a much slower rate initially because it depends on the difference of diameters to the fourth power. For example, a hollow shaft with $D_i=0.5D_o$ retains about 94% of the polar moment of inertia of a solid shaft with the same outer diameter, but uses only 75% of the material. This results in a higher strength-to-weight and stiffness-to-weight ratio, allowing for lighter, more efficient designs without sacrificing performance, provided stability (like buckling under torsion) is not a concern.

Common Pitfalls

1. Confusing Radius and Diameter in Formulas: The most frequent computational error is substituting a diameter $D$ into the formula $J=\pi c^4/2$, which requires the radius $c$. Always double-check which version of the formula you are using. The safe practice is to consistently use radius ($c$) or diameter ($D$) in a given calculation, not a mix.
2. Misapplying the Angle of Twist Formula: The formula $\varphi =TL/GJ$ assumes a prismatic segment with constant $T$, $G$, and $J$. For shafts with multiple segments, changing materials, or stepped diameters, you must calculate the angle of twist for each segment and sum them algebraically. Applying a single formula to a non-uniform shaft will give an incorrect answer.
3. Overlooking Material Property Dependence: Shear stress depends only on geometry and load ($\tau =T\rho /J$), but the angle of twist critically depends on the shear modulus $G$. Using an incorrect $G$ value (e.g., confusing it with Young's modulus $E$) will lead to large errors in deformation prediction. Always verify the correct material property is used.
4. Ignoring Stress Concentrations: The classical torsion formulas assume smooth, prismatic shafts. Any discontinuity—such as a keyway, sudden change in diameter, or hole—creates a stress concentration where the local shear stress can be several times higher than ${\tau }_{max}$ calculated by $Tc/J$. In practical design, these factors must be accounted for with stress concentration factors.

Summary

• Torsion in circular shafts creates a linear shear stress distribution, from zero at the center to a maximum at the outer surface, calculated by ${\tau }_{max}=Tc/J$.
• The polar moment of inertia $J$ is the key geometric property resisting torsion. For a solid shaft, $J=\pi D^4/32$; for a hollow shaft, $J=\pi (D_o^4-D_i^4)/32$.
• The angle of twist is calculated as $\varphi =TL/GJ$, highlighting that deformation is inversely proportional to the torsional rigidity $GJ$.
• Hollow shafts are materially efficient because they remove under-utilized core material, offering a higher polar moment of inertia per unit weight compared to solid shafts of the same outer diameter.