Shaft Size Calculator

Shaft Sizing for Torsion

A solid circular shaft transmitting torque experiences shear stress that increases toward the outer surface. The minimum diameter depends on applied torque, material shear strength, and the required safety factor. Undersized shafts fail by excessive shear; oversized shafts waste material and add inertia.

Shaft Diameter Formula

For a solid circular shaft in torsion:

$$d = \left(\frac{16T}{\pi \cdot \tau \cdot F_s}\right)^{1/3}$$

$T$ is torque, $\tau$ is the material shear stress limit, and $F_s$ is the safety factor. The allowable stress is $\tau / F_s$. Example: 1000 N·m torque, 50 MPa shear stress, and safety factor 2 give a diameter of about 37 mm.

Design Considerations

Use the yield or ultimate shear strength of the shaft material. Steel shafts commonly use factors of 1.5 to 3 depending on load variability and consequences of failure. Account for keyways, shoulders, and stress concentrations separately.

Frequently Asked Questions

What is the safety factor?

It divides the material stress limit to obtain an allowable stress. A factor of 2 means the shaft is designed to operate at half the material limit.

Does this apply to hollow shafts?

No. This formula is for solid circular shafts. Hollow shafts use a different polar moment of inertia.

What shear stress should I use for steel?

Typical carbon steel yield shear stress is 40–60% of tensile yield. Use material data sheets for your specific alloy.

Why is the exponent 1/3?

Shear stress is proportional to diameter cubed in torsion ($\tau \propto 1/d^3$), so solving for diameter gives a cube root.

Can I enter torque in lb·ft?

Yes. Select the torque unit and the tool converts to SI internally before applying the formula.

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Shaft Sizing for Torsion

Shaft Diameter Formula

Design Considerations

Frequently Asked Questions

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