Torsion of Shafts

students, imagine turning a screwdriver, tightening a jar lid, or watching a car engine transfer power through a rotating axle 🔧🚗 These are all everyday examples of torsion, the twisting effect caused when a torque is applied to a body. In Solid Mechanics 2, torsion of shafts focuses on how circular members such as drive shafts, axles, and drill bits behave when they are twisted.

In this lesson, you will learn how to:

Explain the key terms used in torsion of shafts.
Understand how torque creates shear stress and angle of twist.
Apply the main torsion formulas to circular shafts.
Connect shaft behavior to the broader topic of torsion.
Use examples to see why shaft design matters in real machines.

By the end, students, you should be able to describe why circular shafts are so important in engineering and how their size, material, and shape affect their strength and stiffness.

What a shaft does in real life

A shaft is a long, usually cylindrical part that transmits rotational motion and power from one place to another. In machines, shafts connect motors to gears, fans, pumps, wheels, and many other components. When a motor turns a shaft, it applies a torque. Torque is the turning effect of a force and is measured in newton-metres, written as $\mathrm{N\,m}$.

A shaft must do two important jobs:

Carry torque without failing.
Limit twisting so the machine still works properly.

If a shaft twists too much, parts may not align correctly. For example, a conveyor belt drive may slip, or gears may no longer mesh smoothly. So torsion is not only about whether a shaft breaks; it is also about whether it stays stiff enough for good performance.

In most basic torsion theory, we study circular shafts because they twist in a simple, uniform way. Non-circular sections behave more complicatedly and are usually avoided for power transmission. This is why round shafts are so common in machines ✅

Main terms and ideas in torsion of shafts

students, before using formulas, it helps to know the language of torsion.

Torque, $T$: the twisting moment applied to the shaft.
Shaft length, $L$: the distance over which twisting occurs.
Radius, $r$: the distance from the center of the shaft to a point inside it.
Polar second moment of area, $J$: a geometric property that shows how strongly a circular cross-section resists twisting.
Shear stress, $\tau$: the stress developed in the material due to twisting.
Angle of twist, $\theta$: how much the shaft rotates relative to one end.
Shear modulus, $G$: a material property that measures rigidity, meaning resistance to shear deformation.

For a circular shaft in elastic torsion, the important idea is that the material near the center experiences less shear stress than the material near the surface. The maximum shear stress occurs at the outer radius.

This happens because the farther a point is from the center, the more it must move around the circle during twisting. So the outer layers of the shaft “work harder” than the inner layers.

Shear stress distribution in a circular shaft

For a circular shaft under torque, the shear stress changes linearly with distance from the center. The basic relationship is

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

where $\tau$ is the shear stress at radius $r$.

This formula shows three important things:

If torque $T$ increases, shear stress increases.
If you move farther from the center, shear stress increases.
If the shaft has a larger $J$, the stress decreases.

At the outer surface, where $r = c$ and $c$ is the outer radius, the maximum shear stress is

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

For a solid circular shaft of diameter $d$,

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

and since $c = \frac{d}{2}$, the maximum shear stress becomes

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

This is a very important result because it lets engineers choose a safe shaft diameter.

Example

Suppose a solid shaft carries a torque of $T = 500\ \mathrm{N\,m}$ and has diameter $d = 40\ \mathrm{mm} = 0.04\ \mathrm{m}$.

Using

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

we get

$$\tau_{\max} = \frac{16(500)}{\pi (0.04)^3}$$

which gives a large shear stress concentrated near the outside of the shaft. This example shows why small changes in diameter make a big difference, because the diameter is raised to the third power. Even a modest increase in size can greatly reduce stress.

Angle of twist and shaft stiffness

A shaft can be strong enough not to break, but still twist too much. That is why angle of twist matters.

For a circular shaft in elastic torsion, the angle of twist is

$$\theta = \frac{TL}{JG}$$

where:

$\theta$ is in radians,
$T$ is torque,
$L$ is shaft length,
$J$ is polar second moment of area,
$G$ is shear modulus.

This formula tells us:

Longer shafts twist more.
Larger torque causes more twist.
A larger $J$ makes the shaft stiffer.
A larger $G$ means the material resists twist better.

This is why engineers care not just about material strength, but also about material stiffness. For example, steel and aluminum may both be used in shafts, but steel usually has a higher shear modulus, so it twists less under the same loading.

Real-world meaning

Think of a bicycle chain drive or a power tool. If the shaft connecting the motor to the working part twists too much, energy is wasted in deformation, and the motion at the output may lag behind. That can reduce efficiency and accuracy. So angle of twist is a serviceability issue as well as a strength issue.

Solid and hollow shafts

students, many shafts are not solid. Some are hollow, especially in vehicles and aircraft. A hollow shaft can be lighter while still resisting torsion effectively.

For a hollow circular shaft with outer diameter $d_o$ and inner diameter $d_i$,

$$J = \frac{\pi}{32}\left(d_o^4 - d_i^4\right)$$

A hollow shaft often gives better torsional efficiency because material is placed farther from the center, where it contributes more to resisting twist. This is useful in design because the outer regions do most of the work in torsion.

Why hollow shafts are useful

They reduce weight.
They can save material.
They often provide a strong torsional resistance for their mass.

For example, a drive shaft in a vehicle may be hollow to lower rotating mass. Less mass can improve performance and reduce energy use, while still allowing the shaft to carry torque safely.

How torsion fits into the bigger topic of solid mechanics

Torsion is one part of Solid Mechanics 2, alongside topics such as shear stress, bending, and deflection. In torsion of shafts, the focus is on how a member behaves under a twisting load rather than a pulling or bending load.

The broader purpose is to connect loads, internal stresses, and deformation. The shaft must satisfy both:

Strength requirement: stress must stay below the material’s allowable limit.
Stiffness requirement: twist must stay within acceptable limits.

These two checks are central in engineering design. A shaft that is too weak may fail by yielding or fracture. A shaft that is too flexible may still work poorly even if it does not break.

A complete design-style example

Suppose a machine shaft must transmit torque smoothly. The engineer may know the required torque $T$, the shaft length $L$, and a maximum acceptable twist $\theta_{\text{allow}}$.

The design process often looks like this:

Choose a material and obtain its shear modulus $G$.
Select a circular section and calculate $J$.
Check the twist using

$$\theta = \frac{TL}{JG}$$

Check the maximum shear stress using

$$\tau_{\max} = \frac{Tc}{J}$$

Adjust the diameter or choose a hollow section if needed.

This is a practical engineering balance. A larger shaft reduces stress and twist, but it also increases weight and cost. A smaller shaft is cheaper and lighter, but may not be safe or stiff enough. Good design means finding the right compromise.

Conclusion

Torsion of shafts is a key part of Solid Mechanics 2 because it explains how circular members transmit power while resisting twist. students, the main ideas are that torque creates shear stress, the maximum stress occurs at the outer surface, and the angle of twist depends on the shaft’s geometry, length, and material. Solid and hollow shafts both have important uses, and each must be checked for strength and stiffness.

Understanding torsion helps engineers design safer cars, machines, tools, and power systems ⚙️ The formulas are useful, but the bigger idea is simple: a shaft must not only carry the load, it must also keep its shape and function.

Study Notes

Torsion is the twisting effect caused by a torque $T$.
Shafts are used to transmit rotational power in machines.
In a circular shaft, shear stress increases linearly with radius $r$.
The torsion formula for shear stress is $\tau = \frac{Tr}{J}$.
Maximum shear stress occurs at the outer surface: $\tau_{\max} = \frac{Tc}{J}$.
For a solid circular shaft, $J = \frac{\pi d^4}{32}$.
For a hollow circular shaft, $J = \frac{\pi}{32}\left(d_o^4 - d_i^4\right)$.
Angle of twist is given by $\theta = \frac{TL}{JG}$.
Larger $J$ means greater resistance to twisting.
Larger $G$ means the material is stiffer in shear.
A shaft must satisfy both strength and stiffness requirements.
Hollow shafts are often efficient because material is placed farther from the center.
Torsion of shafts is a major application of the broader torsion topic in Solid Mechanics 2.