How this instrument works
A shaft under torque does not stress evenly across its cross-section. Shear stress is zero at the centre and climbs in a straight line out to the outer surface, so the material doing the most work is the skin, not the core — which is exactly why a hollow tube resists torsion almost as well as a solid bar of the same outside diameter while using far less steel. This calculator assumes a solid round shaft, the simplest and most common case, and asks how wide that bar must be before its outer-fibre shear stress reaches the allowable limit you set.
The formula falls out of two facts about circles: the polar second moment of area of a solid round section is J = πd⁴ ⁄ 32, and the maximum shear stress at radius d ⁄ 2 is τ = T(d ⁄ 2) ⁄ J. Substitute J and cancel the common terms and the d⁴ in the denominator collapses to a d³, giving τ = 16T ⁄ πd³. Turning that around for the diameter you don't yet know — d = (16T ⁄ πτ)^⅓ — is exactly what this instrument computes, and the cube root is not decoration: it means diameter grows far slower than torque does, which is the whole economic argument for transmitting power through a shaft rather than a cable.
The result is a floor, not a finished drawing. It describes a smooth solid bar under a single steady torque and nothing else — no bending moment from a belt pull or overhung gear, no stress-riser from a keyway, snap-ring groove or shoulder fillet, no fatigue from millions of load reversals. Real shaft design folds those in afterward, usually by choosing a more conservative allowable stress up front or by checking the finished geometry against a combined-loading criterion once the rest of the drivetrain is known.
- Enter the Applied torque the shaft must transmit, in N·m (or switch the field to ft·lb).
- Enter the Allowable shear stress the shaft material may safely carry, in MPa — a fraction of its yield strength, never the full value.
- Read the Minimum shaft diameter in millimetres; switch to centimetres for large industrial shafting.
- Round the result up to the nearest stock bar size before ordering — the figure shown is a minimum, not a size to machine to exactly.
Worked example — sizing a 500 N·m gearbox output shaft
A gearbox output shaft has to transmit 500 N·m of torque, and the engineer has settled on mild steel with an allowable shear stress of 40 MPa — a typical working figure once a safety factor has already been taken off the material's yield strength. Solving d = (16 × 500 ⁄ (π × 40,000,000))^⅓ gives 0.0399294542466 m, so the shaft needs to be at least 39.93 mm across. A standard 40 mm bar clears that minimum with only a sliver of margin, which is normally the point where a designer either accepts the tight fit or steps up to the next stock size.
Double the torque to 1,000 N·m at the same 40 MPa and the minimum diameter rises to only about 50.31 mm, not 79.86 mm — diameter tracks the cube root of torque, so capacity grows far faster than the bar has to grow to carry it. Going the other way, swapping to a stronger alloy rated for 80 MPa instead of touching the torque shrinks that original 500 N·m shaft down to roughly 31.69 mm, the trade an engineer makes when the goal is cutting weight without changing what the shaft has to deliver.
Questions
What counts as a safe allowable shear stress to enter?
It is not the material's full yield strength — it is yield or ultimate strength divided by a safety factor. Shaft-design practice commonly lands near 30 percent of yield strength or 18 percent of ultimate tensile strength for a shaft without a keyway, with a further cut where a keyway, shoulder, or hole concentrates stress locally. Mild steel often works out near the 40 MPa used in the worked example, but always check the specific alloy.
Why doesn't the diameter double when the torque doubles?
Because the formula solves a cubic relationship, d = (16T ⁄ πτ)^⅓, so diameter scales with the cube root of torque rather than one-to-one. Doubling torque from 500 N·m to 1,000 N·m at a fixed 40 MPa raises the minimum diameter from 39.93 mm to only about 50.31 mm — a 26 percent increase, not 100 percent. A modest increase in bar diameter buys a large increase in torque capacity.
Does this formula also cover bending loads on the shaft?
No — it sizes a shaft under pure torsion only. Most real shafts also carry bending from pulleys, gears, or overhung loads, and combined loading needs an equivalent-torque check or a fatigue-based method such as the ASME shaft-design approach that folds bending moment and torque into a single equation. Treat this result as the torsional piece of the sizing problem, not the complete answer where bending is significant.
Does the result already allow for keyways or shoulder fillets?
No. The formula assumes a smooth, solid, round bar under steady torque. A keyway, retaining-ring groove, or shoulder fillet concentrates stress locally and can cut the effective strength of that section by 25 percent or more, so either enter an allowable shear stress conservative enough to absorb it, or check the keyed section separately with a stress-concentration factor once the shaft layout is fixed.
Where does the 16 divided by pi in the formula come from?
It comes from combining the torsion shear-stress equation, τ = Tr ⁄ J, with the polar second moment of area of a solid circle, J = πd⁴ ⁄ 32. Setting r to the outer radius d ⁄ 2 and simplifying leaves τ = 16T ⁄ πd³; rearranging that for d gives the cube-root formula this calculator solves.
Should I machine the shaft to exactly the number shown?
No — round up to the nearest available stock diameter. The formula returns the theoretical minimum with no margin beyond whatever safety factor is already folded into the allowable stress you entered, so a 39.93 mm minimum is normally ordered as 40 mm bar or a comparable inch size, whichever the supplier actually stocks.