Torsion — Shear Stress in Shafts

The torsion formula tells you the shear stress inside a circular shaft that is carrying torque. For a circular shaft in linear elastic torsion, the stress at radius $r$ is

$$\tau = \frac{Tr}{J}$$

Here, $T$ is the applied torque and $J$ is the polar moment of area. Under these conditions, the shear stress is zero at the center and largest at the outer surface.

For the maximum shear stress, set $r = R$:

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

This result is used for solid and hollow circular shafts when the circular-shaft torsion model is a good fit.

When The Torsion Formula Applies

Use $\tau = Tr/J$ when the member can be modeled as a circular shaft in elastic torsion. If the cross section is not circular, this stress distribution does not generally hold.

That condition matters. The formula is not a universal rule for every twisted part.

What $T$, $r$, $R$, and $J$ Mean

• $T$: applied torque
• $r$: distance from the shaft center to the point of interest
• $R$: outer radius of the shaft
• $J$: polar moment of area of the cross section

For common circular shafts:

$$J = \frac{\pi R^4}{2} \quad \text{for a solid circular shaft}$$ $$J = \frac{\pi \left(R_o^4 - R_i^4\right)}{2} \quad \text{for a hollow circular shaft}$$

Why The Stress Gets Larger Away From The Center

A shaft in torsion does not shear equally everywhere. Material farther from the center has to move through a larger circular path as the shaft twists, so the shear effect grows with radius.

That is why the formula is proportional to $r$. The center line has $r = 0$, so the shear stress there is zero. The outer surface has the largest $r$, so it carries the largest shear stress.

Worked Example: Maximum Shear Stress In A Solid Shaft

Suppose a solid circular shaft has radius $R = 0.020\ \mathrm{m}$ and carries a torque of $T = 120\ \mathrm{N \cdot m}$. Find the maximum shear stress.

First compute the polar moment of area:

$$J = \frac{\pi R^4}{2} = \frac{\pi (0.020)^4}{2} \approx 2.51 \times 10^{-7}\ \mathrm{m^4}$$

Now use the maximum-stress form:

$$\tau_{\max} = \frac{TR}{J}$$ $$\tau_{\max} = \frac{(120)(0.020)}{2.51 \times 10^{-7}} \approx 9.55 \times 10^6\ \mathrm{Pa}$$

So the maximum shear stress is

$$\tau_{\max} \approx 9.6\ \mathrm{MPa}$$

This example shows the main pattern clearly. For the same shaft type, larger torque raises the stress, while a larger polar moment $J$ lowers it.

Common Torsion Formula Mistakes

Using The Formula For The Wrong Cross Section

$\tau = Tr/J$ is the standard elastic torsion result for circular shafts. If the section is not circular, or the material is outside the assumed elastic range, this formula may not describe the real stress correctly.

Mixing Up $J$ And $I$

$J$ is the polar moment of area, not the area moment of inertia $I$ used in ordinary beam bending. Swapping them gives the wrong answer.

Forgetting That Stress Depends On Radius

The stress is not uniform across the shaft. It changes with $r$, so the value at the center is not the same as the value at the surface.

Losing Unit Consistency

If torque is in $\mathrm{N \cdot m}$, radius in $\mathrm{m}$, and $J$ in $\mathrm{m^4}$, then the stress comes out in $\mathrm{Pa}$. Mixing millimeters and meters is a common source of error.

Where You Use The Torsion Formula

The torsion formula is used when engineers and physics students need to estimate shear stress in rotating shafts, drive axles, drill shafts, motor couplings, and similar parts that transmit torque.

In practice, it helps answer a simple design question: is the shaft geometry large enough to carry the torque without exceeding an acceptable shear stress?

Try A Similar Problem

Keep the same shaft, but double the torque. Because $\tau$ is proportional to $T$, the maximum shear stress doubles as well.

If you want to try your own version with a different radius or a hollow shaft, solve a similar torsion problem in GPAI Solver.