Drawing No. EH–ME–011 // Mechanical Engineering
Shaft Design Calculator
Reviewed September 2026
Size a solid or hollow transmission shaft from the power it carries and the speed it turns — from a quick torsion-only diameter through to a combined bending-and-torsion check against yield, with torsional windup for the diameter you settle on.
Calculation workflow
Inputs are converted internally to SI, the engineering model is solved, and results are converted back to the selected unit system.
Design inputs
Design and check results
Cross-section and stress screening
Worked example
15 kW at 900 rpm gives about 159 N·m torque. A simultaneous 800 N·m bending moment dominates this example.
T = P/ω ≈ 159 N·m.
The calculator combines factored bending stress and torsional shear using von Mises stress.
The calculated requirement is a minimum elastic-yield size; choose a practical diameter and verify fatigue and local features.
Background
A rotating shaft commonly experiences simultaneous bending and torsion.
σb = 32M / [πD³(1-k⁴)] ; τ = 16T / [πD³(1-k⁴)]
For a solid shaft k = 0. The selected service factors multiply nominal M and T for the strength check.
σvm = √(σb² + 3τ²)
The required diameter is solved by setting σvm equal to yield strength divided by the target safety factor.
θ = TL / JG
Twist uses the actual operating torque, not the strength service factor. Local torsional flexibility at couplings, splines and keys is not included.
Frequently asked questions
Practical questions about assumptions, limits and interpretation.
Torsion alone is a reasonable first pass for a shaft carrying little bending — a short coupling shaft, for example. Any shaft with gears, pulleys or sprockets between bearings carries bending as well, and the combined check governs. Size on torsion to get a starting diameter, then confirm with the combined result.
They are the ASME-style shock and fatigue factors applied to the bending and torsional moments. Steady, smoothly applied loading sits near 1.5 and 1.0; sudden or reversing duty with shock pushes both substantially higher. They account for the difference between a nominal load and what the shaft really sees.
Use the material's yield strength in shear, or the yield strength in tension divided by the appropriate factor for the failure theory you are working to. The calculator applies your safety factor to the value you enter, so the result is only as conservative as that input.
A shaft can be strong enough and still twist too much. Excess windup shows up as positioning error in drive trains and as a torsional resonance risk in machinery. A common rule of thumb is about one degree of twist per metre of length for general power transmission, but the acceptable figure depends on the application.
No. It is a static strength and stiffness check. Whirling and critical speed depend on mass distribution and bearing support, and fatigue life needs a stress-life or strain-life assessment against the actual load spectrum. Treat this as sizing before those checks, not instead of them.
Yes — the section properties account for the bore. A hollow shaft gives most of the torsional capacity of a solid one at appreciably lower mass, which is why it is common where inertia matters.