Shaft Size Calculator
Enter the torque or power and speed your shaft must transmit, choose a material and design basis, and this calculator returns the minimum required diameter. It handles solid and hollow shafts, five design modes (torque only, bending only, combined, fluctuating loads, and torsional rigidity), shock and fatigue factors, and both metric and imperial units. Results include the calculated shear stress and margin of safety.
How shaft diameter is calculated from torque
The fundamental relationship comes from the polar section modulus. For a solid circular shaft carrying a pure torque T, the maximum shear stress tau at the outer surface is tau = 16T / (pi x d^3). Rearranging for diameter gives d = (16 x T / (pi x tau_allow))^(1/3), where tau_allow is the material's allowable shear stress divided by the safety factor. The cube-root dependence means doubling the torque increases the required diameter by only about 26%, while halving the allowable stress (choosing a weaker material or larger safety factor) increases diameter by roughly 26% as well. For hollow shafts, the denominator gains a (1 - k^4) term, where k is the ratio of inner to outer diameter, which accounts for the missing core material.
Design modes: torque, bending, combined and rigidity
Pure torsion sizing applies when the shaft transmits torque between two aligned couplings with negligible lateral loads. Bending-only sizing applies to spindles and mandrels loaded transversely but not twisted. Most real shafts carry both, so the ASME combined method computes an equivalent bending moment Te = sqrt((Km x M)^2 + (Kt x T)^2), then sizes the shaft to Te as though it were a pure torque. The shock and fatigue factors Km and Kt account for the difference between steady and fluctuating loads: a smoothly driven shaft uses Km = Kt = 1.0, while a shaft driven through a jaw coupling or chain drive in a machine tool might use Km = 2.0 and Kt = 1.5. Torsional rigidity sizing is used when the angle of twist matters operationally, such as on precision camshafts or printing-press drives where twist causes phase errors.
Material selection and safety factors
The allowable shear stress you enter should reflect the actual material grade and heat treatment, not just the alloy family. Mild steel in the as-rolled condition has an ultimate tensile strength (UTS) around 400 MPa; the yield shear stress is roughly 58% of tensile yield, and common design practice divides that by a safety factor of 2 to 3 for rotating machinery. Alloy steels such as AISI 4140 quenched and tempered to 900 MPa UTS can sustain considerably higher stress. A safety factor of 1.5 suits well-characterised loads on accurately manufactured parts; 2.0 to 3.0 is more appropriate when loads are uncertain, impact is possible, or failures would be hazardous. Keyways, cross-holes, and press fits introduce stress concentrations that can locally multiply stress by a factor of 1.5 to 3, so the diameter calculated here should be treated as a minimum starting point rather than a finished design.
Solid versus hollow shafts
A hollow shaft is torsionally stiffer per unit mass than a solid shaft of the same outer diameter, because the low-stressed material at the centre contributes little to the polar moment of inertia. A hollow shaft with k = 0.5 (inner diameter half the outer) weighs 75% as much as a solid shaft of the same outer diameter while carrying the same torsional load. At k = 0.7 the weight saving is 51%, at k = 0.8 it is 64%. The trade-off is more complex machining, tighter tolerances on bore concentricity, and reduced fatigue life at the bore surface if the bore is rough. In practice, hollow shafts are chosen for weight-critical applications (propeller shafts on aircraft, marine drives, bicycle bottom brackets) and for shafts that must carry a through-bore for oil feed, electrical cables, or collet mechanisms.
Allowable shear stress by material
| Material | Allowable shear stress (MPa) | Allowable shear stress (psi) | Notes |
|---|---|---|---|
| Mild Steel (e.g. AISI 1020) | 40-60 | 5,800-8,700 | Most common general-purpose shafts |
| Alloy Steel (e.g. AISI 4140) | 80-120 | 11,600-17,400 | Higher-strength power shafts |
| Stainless Steel (304/316) | 60-90 | 8,700-13,000 | Corrosive environments |
| Aluminum 6061-T6 | 30-50 | 4,300-7,300 | Light-weight applications |
| Brass (CuZn) | 35-45 | 5,100-6,500 | Low-load, corrosion-resistant |
| Bronze (CuSn) | 40-55 | 5,800-8,000 | Bearings, marine drives |
| Cast Iron (Grade 20) | 25-40 | 3,600-5,800 | Brittle - avoid shock loading |
Conservative mid-range values for common shaft materials under rotating service. Higher-strength grades and precise heat treatments allow larger values.
Frequently asked questions
What is the formula for minimum shaft diameter?
For a solid shaft under pure torsion, the formula is d = (16 x T x N / (pi x tau))^(1/3), where T is the applied torque in N·m, N is the safety factor, tau is the allowable shear stress in Pa, and d is the diameter in metres. Converting torque from lb·in to N·m, or allowable stress from psi to Pa, lets the same formula work in imperial units. This calculator handles those conversions automatically.
How do I find the torque if I only know power and speed?
Torque (in N·m) equals power (in watts) divided by angular velocity (in radians per second). Angular velocity in rad/s = 2 x pi x rpm / 60. So torque = power (W) x 60 / (2 x pi x rpm). In practice, T (N·m) = P (kW) x 9549 / rpm. Select the "Power and speed" design mode and the calculator performs this step for you.
What safety factor should I use for a shaft?
For steady, well-characterised loads on accurately machined shafts, 1.5 is commonly used. When loads fluctuate or are not precisely known, 2.0 is safer. For impact-loaded shafts, safety-critical drives, or applications where failure could injure people, 2.5 to 3.0 is appropriate. These factors are in addition to any shock factors (Km, Kt) applied in the combined-loading mode.
What is the difference between the Km and Kt shock factors?
The ASME shaft design code introduces Km (the combined shock and fatigue factor for bending) and Kt (the same for torsion). They multiply the bending moment and torque respectively before computing the equivalent moment. For a smooth, gradually applied load Km = Kt = 1.0. For loads applied suddenly with minor shocks (typical of belt drives and gear transmissions) Km = 1.5 and Kt = 1.0. Heavy shocks, as in metal-cutting machine tool drives, require Km = 2.0 to 3.0 and Kt = 1.5 to 3.0.
Should I round up the calculated diameter to a standard size?
Yes. The calculated value is the theoretical minimum. In practice you must round up to the nearest standard shaft diameter in your region - ISO 286 preferred sizes for metric (e.g. 20, 25, 28, 30, 32, 35, 40 mm) or ANSI B4.1 for inch sizes. Rounding up also compensates for keyway stress concentrations, surface roughness, and manufacturing tolerances not captured by the formula.