Shear Stress in Torsion
Introduction
Imagine trying to twist a metal shaft, like the axle in a machine or the drive shaft in a car 🚗. When a twisting load is applied, the material inside the shaft resists that twist. That resistance creates shear stress, which is the main stress type studied in torsion. students, in this lesson you will learn how twisting causes shear stress, why the stress changes from the center to the outside of a shaft, and how engineers use these ideas to design safe rotating parts.
Lesson objectives
By the end of this lesson, you should be able to:
- explain the main ideas and terminology behind shear stress in torsion,
- apply solid mechanics reasoning to torsion problems,
- connect shear stress in torsion to the wider topic of torsion,
- summarize how shear stress fits into shaft design,
- use examples and evidence to interpret torsional behavior.
Torsion is important in real life because many machines depend on rotating members to transmit power. If a shaft is too weak, it may twist too much or fail suddenly. Understanding shear stress helps engineers choose the right material and shape for the job ⚙️.
What torsion does to a shaft
Torsion happens when a member is subjected to a turning effect called torque or twisting moment, usually written as $T$. A common example is turning a screwdriver: your hand applies torque, and the shaft of the screwdriver carries that twist to the tip.
When a circular shaft is twisted, one end may rotate relative to the other. The material does not simply spin as a rigid body. Instead, different layers inside the shaft try to slide past each other. This internal resistance creates shear stress, written as $\tau$.
A key idea is that torsion in a circular shaft produces shear stress that varies with distance from the center. The center has very small or even zero shear stress, while the outer surface has the largest shear stress. This is why outer fibers are critical in shaft design.
For an elastic circular shaft, the basic torsion relation is:
$$\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}$$
Here:
- $T$ is the applied torque,
- $J$ is the polar second moment of area,
- $\tau$ is the shear stress at radius $r$,
- $G$ is the shear modulus,
- $\theta$ is the angle of twist,
- $L$ is the length of the shaft.
This equation shows that torsion, shear stress, and angle of twist are closely connected.
Shear stress in torsion
Shear stress is the stress that acts parallel to a surface. In torsion, the internal stresses are not pulling or pushing directly along the shaft axis. Instead, they act tangentially around the shaft’s cross-section. Think of a stack of cards being twisted slightly relative to each other: each layer tries to slide past its neighbor 📚.
For a circular shaft in elastic torsion, shear stress at a radius $r$ is given by:
$$\tau = \frac{Tr}{J}$$
This formula is one of the most important results in torsion. It tells us that:
- if torque $T$ increases, shear stress increases,
- if radius $r$ increases, shear stress increases,
- if $J$ increases, shear stress decreases.
The maximum shear stress occurs at the outer radius $R$:
$$\tau_{\max} = \frac{TR}{J}$$
This is a crucial design result because failure often starts at the outer surface of the shaft, where the stress is highest.
Why stress changes with radius
The linear change of stress with radius is due to the geometry of twisting. The material near the center moves through a smaller arc length than the material near the outside. Since twist causes a relative sliding motion, the outer material experiences a larger tendency to shear.
A useful way to picture this is to imagine a spinning CD. The point near the center moves a short distance in one rotation, while a point near the edge moves much farther. In torsion, the same twisting angle produces more displacement at larger radius, so the shear stress is larger there.
Example 1: finding maximum shear stress
Suppose a solid circular shaft carries a torque of $T = 500\,\text{N·m}$, has radius $R = 0.02\,\text{m}$, and polar moment of area $J = 1.26 \times 10^{-7}\,\text{m}^4$.
The maximum shear stress is
$$\tau_{\max} = \frac{TR}{J}$$
Substituting values gives:
$$\tau_{\max} = \frac{500 \times 0.02}{1.26 \times 10^{-7}}$$
$$\tau_{\max} \approx 7.94 \times 10^{7}\,\text{Pa}$$
So the maximum shear stress is about $79.4\,\text{MPa}$. This example shows how a moderate torque can produce a large internal stress if the shaft is thin.
Polar second moment of area and shaft shape
The term $J$ appears in torsion because it measures how strongly a cross-section resists twisting. For circular sections, the formula for $J$ is well defined and works beautifully in torsion analysis.
For a solid circular shaft:
$$J = \frac{\pi d^4}{32}$$
where $d$ is the diameter.
For a hollow circular shaft:
$$J = \frac{\pi \left(d_o^4 - d_i^4\right)}{32}$$
where $d_o$ is the outside diameter and $d_i$ is the inside diameter.
These formulas show an important design fact: diameter has a very strong effect because it appears to the fourth power. Even a small increase in diameter greatly increases torsional strength and reduces shear stress. That is why engineers often use hollow shafts in cars and aircraft: they can be efficient in weight while still resisting torsion well ✈️.
Example 2: comparing two shafts
If one shaft has diameter doubled from $d$ to $2d$, then for a solid circular shaft:
$$J \propto d^4$$
So the new polar moment becomes:
$$J_{\text{new}} = 2^4 J = 16J$$
This means the larger shaft is much stiffer in torsion and will experience much lower shear stress under the same torque.
Angle of twist and its link to shear stress
Shear stress in torsion is closely related to the angle of twist, which is the amount one end of the shaft rotates relative to the other. The formula is:
$$\theta = \frac{TL}{JG}$$
This tells us that the twist angle increases when:
- torque $T$ increases,
- length $L$ increases,
- polar moment $J$ decreases,
- shear modulus $G$ decreases.
The shear modulus $G$ measures the stiffness of the material in shear. A material with a larger $G$ twists less under the same load.
Since a shaft must often transmit power while staying aligned, designers care about both shear stress and angle of twist. A shaft may have stress below the material limit but still twist too much, which can make a machine inaccurate or noisy.
Real-world example
A drill bit must transmit torque to the cutting tip. If the shaft twists too much, the cutting action becomes less precise. If shear stress is too high, the bit may crack or permanently deform. So engineers must check both stress strength and twist stiffness.
Elastic behavior and the limits of the torsion formula
The formula $\tau = \frac{Tr}{J}$ is valid for a circular shaft under elastic behavior, meaning the material returns to its original shape when the torque is removed. This usually requires the stress to stay below the material’s proportional or yield limit.
If the stress becomes too large, the material may yield. Then the linear relation between stress and radius may no longer be accurate. Non-circular sections also behave differently. For example, rectangular bars twist in a more complicated way and do not have the same simple shear stress distribution as circular shafts.
This is why torsion analysis in Solid Mechanics 2 often focuses on circular shafts. They are common in engineering, and the mathematics is especially clear.
How shear stress in torsion fits into shaft design
Shear stress in torsion is one part of a bigger design process. Engineers usually ask three main questions:
- Will the shaft carry the required torque without failing?
- Will the shaft twist less than the allowable amount?
- Is the chosen shape and material efficient and practical?
To answer these questions, engineers use $\tau = \frac{Tr}{J}$ for strength and $\theta = \frac{TL}{JG}$ for stiffness. They also consider safety factors, material properties, and the way loads vary during operation.
For example, a transmission shaft in an industrial machine may carry changing torque throughout the day. The design must ensure that the maximum shear stress stays within safe limits even when loads increase suddenly. This is a clear example of how theory supports real engineering decisions.
Conclusion
Shear stress in torsion is the internal stress that develops when a shaft is twisted by a torque. students, the most important ideas are that shear stress increases with torque and radius, reaches a maximum at the outer surface, and depends strongly on the shaft’s geometry through $J$. The angle of twist is connected through $\theta = \frac{TL}{JG}$, so torsion problems are always about both strength and stiffness. Understanding these ideas helps explain how shafts, axles, tools, and machine parts work safely in the real world ✅.
Study Notes
- Torsion is twisting caused by torque $T$.
- Shear stress in torsion is written as $\tau$ and acts tangentially on a cross-section.
- For a circular shaft, the basic relation is $\tau = \frac{Tr}{J}$.
- The maximum shear stress occurs at the outer radius: $\tau_{\max} = \frac{TR}{J}$.
- Shear stress increases linearly with radius $r$.
- The polar second moment of area $J$ measures resistance to twisting.
- For a solid circular shaft, $J = \frac{\pi d^4}{32}$.
- For a hollow circular shaft, $J = \frac{\pi \left(d_o^4 - d_i^4\right)}{32}$.
- The angle of twist is $\theta = \frac{TL}{JG}$.
- Larger $J$ means lower stress and smaller twist.
- Larger $G$ means the material is stiffer in shear.
- Torsion formulas are most accurate for circular shafts in the elastic range.
- Engineers design shafts by checking both shear stress and angle of twist.