Shaft Power, Torque and Speed Calculator

P = T·ω in both unit systems through the exact constants — and then the shaft that has to carry it: torsional stress, polar moment, angle of twist, the diameter each criterion demands, and the published twist limits that disagree with each other by a factor of thirteen.

Last updated 24 Sep 2026

Shaft power, torque and speed

Power, torque, speed → stress and twist

What do you want solvedThe other two fields stay live and the one you are solving for locks and fills itself. All three are the same equation: P = T·ω, with ω = 2πN/60.

Torque (N·m)Torque at the shaft section you are checking. If the shaft steps, the torque is the same all along it but the stress is not — check the smallest section.

Speed (rpm)Shaft speed. A shaft’s own speed matters here only through the power; it matters a great deal for whirling, which is a different page.

Power (kW)Transmitted power. The page also prints it in watts and in mechanical horsepower, and cross-checks the horsepower through the 5252 constant to show the two routes agree.

Shaft outside diameter (mm)The diameter at the section being checked. Torsional stress goes as the cube of it and twist as the fourth power, so this is by far the most sensitive number on the page: ten per cent more diameter is a quarter less stress and a third less twist.

Bore diameter (mm) — 0 for a solid shaftLeave it at nought for a solid shaft. A bore removes metal from where the shear stress is lowest, so it costs far less stiffness than weight: at a bore of 0.75 of the outside diameter a tube keeps 68 per cent of the polar moment for 44 per cent of the section area. The chart below sweeps it.

Length over which the twist is measured (m)The length of shaft carrying the torque, between the driving and the driven feature. Twist is proportional to it, and it is the input most often got wrong — the twist of a transmission shaft is normally quoted per metre precisely so that the length drops out of the comparison.

Material — for G, and this is the least certain number hereThe angle of twist is inversely proportional to G, so an uncertainty in G is the same uncertainty in the answer. Published G for steel spans 75 to 79.3 GPa, about 5.4 per cent; for grey cast iron the two tables read for this page are half as big again apart, because grey iron’s modulus climbs steeply with grade. The table below prints both figures and the value recomputed from E and Poisson’s ratio.

Which published GWhere two sources disagree, both are carried. Switch this and watch the twist change: for grey cast iron it changes by more than half, which is the honest size of the uncertainty rather than a defect in the page.

Allowable shear stress (MPa)The ASME shaft code allowable is 30 per cent of the yield strength but not more than 18 per cent of the ultimate, reduced by a further 25 per cent where there is a keyway. For a C45 shaft at 380 MPa yield and 600 ultimate that is 108 MPa, or 81 with a keyway. The default of 40 here is deliberately conservative, because a real shaft carries bending as well as torsion; the deflection page and the von Mises page cover the combination.

Allowable twist (degrees per metre)And this is where the sources fall out. 0.25°/m is quoted where twist is a timing or positioning error; 0.08°/ft, which is 0.262°/m, is the figure that circulates in machine-design texts; and the one source read for this page that names an application prints 0.8 to 1.0°/ft, which is 2.6 to 3.3°/m — ten times looser. The table below computes the surface shear stress each one implies and shows they are answers to two different questions. A default of 1 sits between them.

Not a circuit: the shaft section on the left with its bore live, and on the right the thing that explains why a bore is nearly free. Torsional shear stress is proportional to the radius — zero on the axis and maximum at the surface — so the triangle is the stress distribution across a radius, drawn to scale. The vertical line inside the triangle is your bore, and the height of the triangle to the left of it is the stress the metal you removed was carrying. At a bore of half the outside diameter that metal was never above half the surface stress, and because the contribution to torque goes as stress times radius times area it was carrying about six per cent of the load while being twenty-five per cent of the section. The figures underneath put numbers on it. Note what the drawing cannot show: a thin wall can buckle in torsion at a stress well below the material's own strength, and that limit depends on the wall thickness and the length rather than on the polar moment, so it is not on this page at all.

8.383MPaExample

150 N·m at 1450 rpm through a 45 mm solid carbon steel shaft, 1.2 m between the driving and driven features

One equation for the power, three for the shaft, and one identity that ties them together

P = Tω · τ = Tr/J · J = π(D⁴ − d⁴)/32 · θ = TL/(GJ) · θ/L = τ/(Gr)

P = Tω: all there is to the first half of the page, with ω = 2πN/60. In practical units it becomes P[kW] = T[N·m]·N/9549.296586 or P[hp] = T[lb·ft]·N/5252.113122, and those two constants are 60000/2π and 33000/2π — the same equation twice. The 33000 is the definition of a horsepower and not a measurement
τ = Tr/J: torsional shear stress, maximum at the surface and zero on the axis. For a solid round shaft it reduces to the familiar 16T/πd³, which is the same statement; this page computes it through J so that the hollow case needs no second formula. Note the cube: ten per cent more diameter is a quarter less stress
J: the polar second moment of area, πD⁴/32 solid and π(D⁴ − d⁴)/32 hollow. The fourth power is why boring a shaft out is nearly free: at a bore of half the outside diameter the shaft loses 6 per cent of its polar moment and 25 per cent of its metal. J is not the same as I — for a round section J = 2I, and for anything that is not round the torsion constant is a different quantity altogether, which is why a square shaft needs its own page
θ = TL/(GJ): the angle of twist, in radians. The handbook form θ[°] = 584 TL/(Gd⁴) is the same thing with the degrees conversion and the 32 folded in: 180×32/π² = 583.610018, and the published 584 is 0.067 per cent high
θ/L = τ/(Gr): THE IDENTITY THAT MAKES THE TWIST LIMITS COMPARABLE, and the most useful line on the page. Twist per unit length is set by the surface shear stress and the modulus of rigidity alone — the torque and the diameter cancel out of it. So any twist limit can be converted into the surface stress it implies, which is what the twist table does, and that is what shows a 0.8°/ft limit to be a strength criterion and a 0.25°/m limit to be a stiffness one
G: the modulus of rigidity, and the least certain number in the calculation: the twist is inversely proportional to it. Published G for steel runs 75 to 79.3 GPa, a 5.4 per cent spread; for grey cast iron the two tables read for this page differ by more than half, because grey iron’s modulus climbs with grade. For an isotropic material it is not independent of the other two constants: G = E/2(1 + ν), and the page prints that value beside the published one
0.30 S_y, 0.18 S_u: the ASME shaft code allowable shear: 30 per cent of yield but not more than 18 per cent of ultimate, less a further 25 per cent with a keyway. For steels in the ordinary range the ultimate always governs, because for engineering steels S_u/S_y runs about 1.2 to 1.6 and the ultimate limb only loses out above 0.30/0.18 = 1.67. It is a torsion-only allowable on a plain section, and a real shaft has bending, a fillet and a reversal every revolution

Worked example

150 N·m at 1450 rpm through a 45 mm solid carbon steel shaft, 1.2 m between the driving and driven features
THE POWER, BOTH WAYS. ω = 2π × 1450/60 = 151.843645 rad/s, so P = 150 × 151.843645 = 22 776.55 W = 22.7765 kW. The shortcut gives the same thing: 150 × 1450/9549.296586 = 22.776547. In imperial, 150 N·m is 110.6343 lb·ft and 110.6343 × 1450/5252.113122 = 30.5439 hp, which is exactly 22 776.55 ÷ 745.69987. The two shortcut constants are 60000/2π and 33000/2π; their ratio is 0.55, because 550 pound-feet per second is one horsepower by definition.
THE SECTION. J = π × 0.045⁴/32 = 4.0258 × 10⁻⁷ m⁴, which is 402 578 mm⁴. The polar section modulus J/r is 17 892 mm³. For a hollow shaft the only change is subtracting the bore's fourth power, which is why one formula does both on this page.
THE STRESS. τ = Tr/J = 150 × 0.0225/4.0258 × 10⁻⁷ = 8.383 MPa, or 1216 psi. The 16T/πd³ form gives the identical figure, as it must. Against an allowable of 40 MPa this shaft is at 21 per cent utilisation, and a 40 MPa allowable would permit a shaft of only 26.73 mm. So torsional strength is NOT what sized this shaft, and that is the normal case: most shafts are sized by bending, by deflection, by a bearing bore or by the nearest standard size.
THE TWIST. θ = TL/(GJ) = 150 × 1.2/(79.3 × 10⁹ × 4.0258 × 10⁻⁷) = 5.6383 × 10⁻³ rad = 0.32305° over the 1.2 m, or 0.26921° per metre. The handbook shortcut 584 TL/(Gd⁴) gives 0.32327, which is 0.067 per cent high because 584 is a rounding of 5760/π² = 583.610018. This page uses the exact value.
WHICH CRITERION ACTUALLY SIZES THE SHAFT. At 1°/m the required diameter is 32.41 mm; at 0.25°/m it is 45.84 mm, which is larger than the shaft we have. So on the tight limit this 45 mm shaft FAILS, and on the loose one it passes with room to spare. The way to see why those two answers are both defensible is the identity θ/L = τ/(G·r): 1°/m on a 45 mm steel shaft corresponds to 31.14 MPa of surface shear, and 0.25°/m corresponds to 7.79 MPa. The first is a stress a steel shaft would be happy at; the second is far below any strength requirement and can only mean that something cares where the shaft is pointing.
AND WHAT G IS WORTH. Everything above used G = 79.3 GPa. The second published figure for carbon steel is 75.0, which is 5.4 per cent lower, and the twist would be 5.7 per cent higher: 0.2846°/m instead of 0.2692. That is the honest precision of any twist calculation, and it is worth keeping in mind before quoting a twist to four significant figures. The selector above lets you switch between the two; for grey cast iron the two published values differ by more than half.

The constants, and where each one comes from

Not one of these is an empirical fit, and every one of them is a rounding of something exact. The 9549.296586 of the metric shortcut is 60000/2π — the 60 turns rpm into revolutions per second, the 2π turns revolutions into radians, and the 1000 turns watts into kilowatts. The famous 5252 of the imperial one is 33000/2π, and the 33000 is not measured either: it is the definition of a mechanical horsepower, 33000 foot-pounds per minute, which James Watt chose. Divide one constant by the other and you get 0.55, which is 550 foot-pounds per second over 1000 — so the two shortcuts are literally the same equation in different clothes, and this page computes the horsepower by both routes and prints both so you can see them agree. The last row is the one worth remembering if you ever read 584 in a handbook: it is 5760/π² = 583.610018, and the rounded version is 0.067 per cent high — small, but there is no reason to carry it when the exact form is one character longer.
Relation	Constant	Really is	Status	Note
P[kW] = T[N·m] × N[rpm] ÷ 9549.296586	9,549.296586	60000 ÷ 2π	exact	the metric shortcut. 2π/60 converts rpm to rad/s and the 1000 is watts to kilowatts
P[W] = 0.10471976 × N[rpm] × T[N·m]	0.10471976	2π ÷ 60	exact	the same constant the other way up, which is how the Engineering ToolBox prints it (as 0.105)
P[hp] = T[lb·ft] × N[rpm] ÷ 5252.113122	5,252.113122	33000 ÷ 2π	exact	the imperial shortcut, and the 33000 is the DEFINITION of horsepower: 33000 foot-pounds per minute
1 hp	745.699872	33000 × lbf × ft ÷ 60	exact, in watts	mechanical horsepower. The pound-force is 0.45359237 kg × 9.80665 m/s², both exact by definition, so this is not a measured number
1 lb·ft	1.3558179483	0.45359237 × 9.80665 × 0.3048	exact, in N·m	which is why 5252.113122 and 9549.296586 are two faces of one constant: their ratio is 0.55, and 550 lb·ft/s is one horsepower
θ[°] = 584 × T × L ÷ (G × d⁴)	583.610018	5760 ÷ π²	exact	the published 584 is a rounding of 583.610018, which is 0.067 per cent high. This page uses the exact value

Four published twist limits, and what each one really asks for

This is the most honest table on the page and the least satisfying. The published allowable twists for a transmission shaft span a factor of thirteen, and there is no reading of the sources that reconciles them as one number. What CAN be done is to work out what each one amounts to, and the fourth column does it, using the exact identity θ/L = τ/(G·r): twist per unit length is set by the surface shear stress and the modulus of rigidity alone, with the torque and the diameter dropping out. Read it and the table resolves itself. The loose limits of 0.8 and 1.0 °/ft correspond to 82 and 102 MPa of surface shear on a 45 mm steel shaft — which is a STRENGTH criterion wearing a stiffness criterion’s clothes, and lands within five per cent of the diameter you would get from a stress allowable in the ordinary range. The tight limits of 0.25 °/m and 0.08 °/ft correspond to 8 and 8.2 MPa, which no strength argument would ever ask for; they can only be stiffness or positioning criteria, and that is exactly what they are quoted for — a camshaft, where twist is a timing error, or a machine-tool feed shaft, where it is a position error. So the question to ask is not “what is the allowable twist” but “does anything on this shaft care where it is pointing”. If nothing does, size it on stress. If something does, the twist criterion will size it and will ask for a shaft nearly twice the diameter. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
As published	In °/m	Diameter it needs for 150 N·m (mm)	Surface shear stress it implies on a 45 mm shaft (MPa)	Where it comes from
0.25 °/m	0.2500	45.84	7.79	the timing-critical figure: quoted for a camshaft, where twist is a phase error rather than a deflection
0.08 °/ft = 0.262 °/m	0.2625	45.29	8.17	the figure that circulates in machine-design texts for machinery shafting. It agrees with the camshaft figure almost exactly, which is the one encouraging thing in this table
0.8 °/ft = 2.625 °/m	2.6247	25.47	81.74	Engineers Edge, for machine shafts
1.0 °/ft = 3.281 °/m	3.2808	24.08	102.17	Engineers Edge, for line shafting

Modulus of rigidity, with the check that makes it usable

The fifth column is why this table was chosen over the several that print G on its own. For an isotropic material the three elastic constants are not independent: G must equal E/2(1 + ν). The source used here prints all three in one row, so the transcription can be CHECKED rather than trusted, and the check was run over every row before this page was written. It holds everywhere to within four per cent, which is as close as a table rounded to two or three significant figures can hold it. It also caught something: the brass figure of 40 GPa that several sites print is seven per cent away from what brass’s own published E and Poisson’s ratio require, so the self-consistent 37 is used as the primary here and the 40 is carried as the alternative. Two rows deserve particular attention. GREY CAST IRON: the two published tables give 41.4 and 27 GPa, which is not a rounding disagreement — grey iron’s modulus climbs steeply with grade, from about 66 GPa of E at the lowest class to 140 at the highest, so the two rows are different irons. Computing twist with the wrong one is out by more than half. STEEL: the spread of 75 to 79.3 GPa across sources is 5.4 per cent, and since twist is inversely proportional to G that is the floor on how well any twist figure can be known. One figure was REJECTED rather than carried: the table that prints rubber at 75.2 GPa, within two per cent of structural steel, where another prints 0.0003. Rubber is not offered here. The same designation can mean different dimensions in different standards families — ANSI against ISO, inch against metric, one national standard against another. The family used here is named beside every figure; check which one your part was made to.
Material	E (GPa)	G published (GPa)	Poisson’s ν	E ÷ 2(1+ν) (GPa)	Difference (%)	Second published G (GPa)	That differs by (%)	What to know
Carbon steel	200.0	79.3	0.29	77.52	-2.2	75.0	-5.4	the default for nearly every transmission shaft, and the one material on this list whose G is worth memorising: 79.3 GPa, or 11.5 million psi
Alloy steel (4140, 4340, EN19)	200.0	79.3	0.29	77.52	-2.2	—	—	three times the yield strength of mild steel and the SAME stiffness, so heat treatment buys stress margin and buys nothing at all on twist
Stainless steel, austenitic (304, 316)	190.3	73.1	0.31	72.63	-0.6	75.0	2.6	a few per cent less rigid than carbon steel, so the same shaft twists a few per cent further. Cold-drawn spring wire is lower again at 69 GPa, which is a property of the drawn condition rather than a different source’s opinion
Grey cast iron	100.0	41.4	0.21	41.32	-0.2	27.0	-34.8	the one row on this list where the two published figures are half as big again apart, because grey iron’s modulus climbs steeply with grade. Both are printed; neither is an average
Ductile or malleable iron	172.0	68.0	0.28	67.19	-1.2	64.0	-5.9	far stiffer than grey iron and much closer to steel, which is most of why a ductile-iron shaft or hub is a usable part where a grey-iron one is not
Aluminium alloy (2024, 6061, 7075)	70.3	26.9	0.33	26.43	-1.8	26.0	-3.3	a third of steel’s rigidity, so an aluminium shaft of the same diameter twists three times as far under the same torque
Titanium alloy (Ti-6Al-4V)	120.0	44.0	0.36	44.12	0.3	41.0	-6.8	steel’s strength at a little over half of steel’s rigidity, which is the whole trade: a titanium shaft sized on stress twists nearly twice as far as the steel one it replaced
Phosphor bronze	111.0	41.4	0.35	41.11	-0.7	38.0	-8.2	a bearing and worm-wheel material rather than a shaft material, carried here because a bronze wheel on a steel worm is the one place two very different moduli appear in the same drive
Brass (red brass C83400)	101.0	37.0	0.35	37.41	1.1	40.0	8.1	about half of steel’s rigidity, and on this list mostly so that a model-engineering or instrument shaft can be computed

Solid against hollow, at the same outside diameter

Columns four and five are the same number at every row, which is the algebra worth seeing: (1 − k⁴)/(1 − k²) factorises to 1 + k² exactly. So a tube’s polar moment per unit of metal is 1 + k² times a solid bar’s — never more than twice, and approaching twice only as the wall thickness goes to zero, which is where buckling and handling damage take over. The practical window is k between 0.5 and 0.8: at k = 0.75 the shaft keeps 68 per cent of its torsional stiffness for 44 per cent of its section, a 56 per cent weight saving for a 32 per cent loss of capacity. The last column is the price at a FIXED outside diameter, which is the usual constraint when a shaft has to fit an existing bearing: boring to k = 0.75 multiplies both the surface stress and the twist by 1.46, because both are proportional to 1/J. If the outside diameter is free, a tube wins easily; if it is fixed, a bore always costs. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Bore ÷ OD, k	J as % of solid	Area as % of solid	J per unit area, relative	1 + k²	Stress and twist multiplier at the same torque
0.00	100.00	100.00	1.0000	1.0000	1.0000
0.25	99.61	93.75	1.0625	1.0625	1.0039
0.40	97.44	84.00	1.1600	1.1600	1.0263
0.50	93.75	75.00	1.2500	1.2500	1.0667
0.60	87.04	64.00	1.3600	1.3600	1.1489
0.70	75.99	51.00	1.4900	1.4900	1.3160
0.75	68.36	43.75	1.5625	1.5625	1.4629
0.80	59.04	36.00	1.6400	1.6400	1.6938
0.85	47.80	27.75	1.7225	1.7225	2.0921
0.90	34.39	19.00	1.8100	1.8100	2.9078

The ASME shaft code allowable shear stress

Two fractions and a reduction, and they are worth knowing because they put a floor under the allowable-stress field in the calculator above. The ASME shaft code gives the allowable shear as 30 per cent of the yield strength but not more than 18 per cent of the ultimate, reduced by a further 25 per cent where the shaft has a keyway. Read the seventh column: at every strength level in this table the ULTIMATE limb governs except in the last row, and the reason is the arithmetic of the two fractions: 0.18 S_u is the smaller of the pair whenever S_u/S_y is below 0.30/0.18 = 1.67, and for engineering steels that ratio runs about 1.2 to 1.6. The yield limb only bites on a material whose yield is unusually low against its ultimate, which an annealed austenitic stainless is — the last row — and which a quenched and tempered steel never is. The strength levels here are illustrative round numbers and only the second row is attributable: 380 and 600 MPa are the C45 figures printed in the shaft-design source read for this page. Use your own material’s certified figures. And note what the code allowable is NOT: it is a torsion-only allowable on a plain section. A real shaft carries bending too, has a fillet or a keyway or a snap-ring groove where the stress concentrates, and sees its load reversed every revolution. The same designation can mean different dimensions in different standards families — ANSI against ISO, inch against metric, one national standard against another. The family used here is named beside every figure; check which one your part was made to. This page sizes a part; it does not certify one. Where the answer carries a consequence — a load path, a lifting duty, a pressure boundary, a fastener holding something that can fall — confirm it against the design code that governs the application, and against the manufacturer’s own rating, before relying on it.
Yield S_y (MPa)	Ultimate S_u (MPa)	0.30 S_y	0.18 S_u	Allowable (MPa)	With a keyway (MPa)	Which governs	What the strength level is
250	400	75.0	72.0	72.0	54.0	ultimate governs	a low-carbon structural steel level
380	600	114.0	108.0	108.0	81.0	ultimate governs	the C45 figures printed in the source read
500	700	150.0	126.0	126.0	94.5	ultimate governs	a quenched and tempered level
700	900	210.0	162.0	162.0	121.5	ultimate governs	a high-tensile alloy level
900	1,100	270.0	198.0	198.0	148.5	ultimate governs	a through-hardened alloy level
215	505	64.5	90.9	64.5	48.4	yield governs	an annealed austenitic stainless level, where the YIELD limb bites instead

Two constants that are one constant, and a twist limit the sources cannot agree on to within a factor of ten

P = Tω, and both of the famous constants are the same constant. In SI, power in kilowatts is torque in newton metres times speed in rpm divided by 9549.296586, and that number is 60000/2π: sixty seconds in a minute, 2π radians in a revolution, a thousand watts in a kilowatt. In imperial, horsepower is torque in pound-feet times rpm divided by 5252.113122, and that number is 33000/2π — where the 33000 is not a measurement at all but the definition of a mechanical horsepower, 33 000 foot-pounds per minute. Divide one constant by the other and you get 0.55, because 550 pound-feet per second is one horsepower. This page computes the horsepower by both routes and prints both figures, so you can see them agree to twelve digits rather than take it on trust.

Then the shaft, and the first thing to notice is that torsional strength probably did not size it. The worked example here is 150 N·m through a 45 mm solid steel shaft, which comes out at 8.4 MPa of surface shear. A 40 MPa allowable — itself conservative; the ASME shaft code would allow 108 MPa for a C45 shaft, or 81 with a keyway — would permit a shaft of 26.7 mm. The 45 mm is doing something else: resisting bending from a pulley or gear overhang, staying inside a deflection or slope limit at a bearing, fitting the bore of a bearing that was chosen for its load rating, or simply being the nearest size of bar in stock. That is the normal case, and it is worth saying because a page that computes only torsional stress will tell you a shaft is four times bigger than it needs to be, which is true and useless.

Twist is where a shaft calculation gets interesting, and where the published limits fall apart. θ = TL/(GJ) is not in dispute. What the allowable value of it should be very much is. One widely quoted figure is 0.25° per metre, given for a camshaft. The figure that circulates in machine-design texts is 0.08° per foot, which is 0.262°/m — the same number, reassuringly. But the one source read for this page that prints a number against a named application gives 0.8 to 1.0° per foot for machine shafts and line shafting, which is 2.6 to 3.3°/m: ten times looser. Both cannot be a general rule, and nothing in either source says which is intended. This page prints them all.

What settles it is an identity worth carrying away: θ/L = τ/(G·r). Twist per unit length is set by the surface shear stress and the modulus of rigidity alone. The torque and the diameter cancel out of it entirely. So any twist limit can be turned into the surface stress it implies, and then the sources stop disagreeing and start answering different questions. On a 45 mm steel shaft, 0.8°/ft corresponds to 82 MPa of surface shear and 1.0°/ft to 102 MPa — which is a strength criterion wearing a stiffness criterion’s clothes, and lands within five per cent of the diameter you would get from an ordinary stress allowable. Meanwhile 0.25°/m corresponds to 7.8 MPa, which no strength argument would ever ask for. It can only be a stiffness or positioning requirement, and that is exactly what it is quoted against. The question to ask about a shaft, therefore, is not what the allowable twist is. It is whether anything on the shaft cares where it is pointing.

G is the least certain number in the whole calculation, and the twist is inversely proportional to it. Published values for steel run from 75 to 79.3 GPa across the three tables read for this page, a spread of 5.4 per cent, which is the floor on how well any twist figure can be known. For grey cast iron the spread is far worse — 41.4 against 27 GPa — and that one is not a measurement disagreement but two different grades of iron, because grey iron’s modulus climbs steeply with grade. Both values are offered as a selector here rather than averaged, and the materials table prints G recomputed from E and Poisson’s ratio beside each published figure. That check is the reason these numbers are usable at all: for an isotropic material G must equal E/2(1 + ν), and the identity holds to within four per cent at every row. It also caught a bad number — the brass figure of 40 GPa that several sites print is seven per cent away from what brass’s own E and ν require — and it caught a worse one, a table that prints rubber’s shear modulus within two per cent of structural steel’s, five orders of magnitude out. That row is not on this page.

And one thing a stronger material will not buy you. Alloy steel has three times the yield strength of mild steel and exactly the same modulus of rigidity. So heat treatment buys stress margin and buys nothing whatsoever against twist: if the twist criterion is governing, the only answers are a bigger diameter, a shorter shaft, or a different geometry. Diameter is the strong lever — twist goes as the fourth power of it — and a bore is nearly free, because the metal near the axis of a shaft in torsion is carrying almost nothing. At a bore of three quarters of the outside diameter a tube keeps 68 per cent of its polar moment for 44 per cent of its section area, and the ratio of those two is exactly 1 + k², which can never exceed two. The chart on this page is that trade.

Frequently asked questions

How do I convert torque and speed to power?

P = Tω with ω = 2πN/60. In practical units, power in kilowatts is torque in newton metres times rpm divided by 9549.296586, and horsepower is torque in pound-feet times rpm divided by 5252.113122. Those two constants are 60000/2π and 33000/2π, so they are the same relation in different units; the 33000 is the definition of a horsepower rather than a measured quantity. 150 N·m at 1450 rpm is 22.7765 kW, which is 30.5439 hp.

Where does the 5252 in the horsepower formula come from?

It is 33000 divided by 2π, and both halves are definitions. A mechanical horsepower is 33 000 foot-pounds per minute, which James Watt chose; a revolution is 2π radians. Put them together and horsepower = torque in pound-feet × rpm ÷ 5252.113122. The same reasoning in SI gives 60000/2π = 9549.296586 for kilowatts from newton metres. The often-noticed fact that an engine’s power and torque curves cross at 5252 rpm is just this constant: at that speed the two numbers are equal by arithmetic, and it means nothing about the engine.

What is the formula for torsional shear stress in a shaft?

τ = Tr/J, maximum at the surface and zero on the axis. For a solid round shaft that becomes 16T/πd³ and for a hollow one 16TD/π(D⁴ − d⁴), which are the same formula with J written out. Note the cube: ten per cent more diameter is a quarter less stress, so a shaft that is marginal on torsion is cheap to fix. 150 N·m in a 45 mm solid shaft gives 8.38 MPa.

How much twist is acceptable in a transmission shaft?

The published figures disagree by a factor of thirteen, so the honest answer is that it depends on what the shaft is for, and the useful way to decide is through the identity θ/L = τ/(G·r). A limit of 0.25°/m corresponds to about 8 MPa of surface shear on a 45 mm steel shaft, which is far below any strength requirement — so it is a positioning criterion, appropriate where twist is a timing or position error, as on a camshaft or a machine-tool feed shaft. A limit of 0.8 to 1.0°/ft, which is 2.6 to 3.3°/m and is what one engineering source gives for machine and line shafting, corresponds to 82 to 102 MPa — which is effectively the strength criterion again. If nothing on your shaft cares where it is pointing, size it on stress and the twist will look after itself.

What value of G should I use for steel?

79.3 GPa (11.5 million psi) is the commonest figure and is what this site uses elsewhere; 75 to 77 GPa also appear in reputable tables, so expect about 5 per cent of uncertainty and do not quote a twist to four significant figures. Alloy steels have the same G as mild steel — heat treatment changes the strength, not the stiffness. Austenitic stainless is a few per cent lower at 73 to 77, and cold-drawn spring wire lower again at about 69, which is a real property of the drawn condition. For an isotropic material G = E/2(1 + ν), and that identity is a good sanity check on any table you are reading.

Is a hollow shaft better than a solid one?

At the same weight, yes, clearly. At the same outside diameter, no, never. The polar second moment of a tube is (1 − k⁴) of a solid bar’s where k is the bore ratio, while its area is only (1 − k²), so its stiffness per unit of metal is 1 + k² times the bar’s — up to twice, approached only as the wall goes to nothing. At a bore of 0.75 the shaft keeps 68 per cent of its torsional capacity for 44 per cent of its section. But if the outside diameter is fixed by a bearing bore, a bore raises both the stress and the twist by 1/(1 − k⁴), which at k = 0.75 is a factor of 1.46. And a thin wall brings in torsional buckling, which this page does not compute.

What allowable shear stress should I use for a shaft?

The ASME shaft code gives 30 per cent of the yield strength but not more than 18 per cent of the ultimate, reduced by a further 25 per cent where the shaft has a keyway. For a C45 shaft at 380 MPa yield and 600 ultimate that works out at 108 MPa, or 81 with a keyway. For steels in the ordinary range the ultimate limb governs for almost every steel, because the ratio of ultimate to yield runs about 1.2 to 1.6 and the ultimate limb only loses out above 0.30/0.18 = 1.67 — an annealed austenitic stainless, with a yield less than half its ultimate, is one of the few places the yield limb bites. Treat all of that as a torsion-only allowable on a plain section: a real shaft carries bending too, has stress raisers at fillets and keyways, and sees its bending stress reversed every revolution, which is a fatigue problem rather than a static one.

Why does the handbook formula for twist have a 584 in it?

Because somebody rounded 5760/π². The angle of twist in radians is 32TL/(Gπd⁴); convert to degrees by multiplying by 180/π and you get 180 × 32/π² = 583.610018, so θ[°] = 583.61 TL/(Gd⁴). The printed 584 is 0.067 per cent high, which is harmless but pointless — the exact form is no harder to use, and this page uses it. It is worth knowing that the constant is exact and unit-free in that form, so the same number works with inches, psi and pound-inches as with metres, pascals and newton metres.

Related calculators

References

The Engineering ToolBox, Torsion of Shafts (read 6 October 2026). The independent statement of the four equations this page computes: τ = Tr/J, J = πD⁴/32 solid and π(D⁴ − d⁴)/32 hollow, α = LT/(JG), and the power relation in the form P = 0.105 n_rpmT — which is 2π/60 = 0.10472 written to three figures. It also prints the shortcut α_degrees ≈ 584 LT/(GD⁴); that 584 is not an empirical constant but exactly 5760/π² = 583.61, and this page uses the exact value rather than the rounded one.
Engineers Edge, Shaft Torsional Deflection and Rigidity (read 6 October 2026). Cited for its allowable-twist figures and for the fact that they do not agree with anybody else’s: it gives the permissible angle of twist as “about 0.8 deg. per foot for machine shafts to about 1.0 deg per foot for line shafting”, which is 2.63 to 3.28 °/m. That is ten times looser than the 0.08 °/ft figure that circulates in machine-design texts. The page prints both and shows, by computing the surface shear stress each one implies, that they are answers to two different questions rather than two answers to one.
Engineers Edge, ASME Shaft Design Allowable Stress and Diameter Equations (read 6 October 2026). The source for the allowable shear stress rule quoted on this page: “S_s(allowable) = 30% of S_y but not over 18% of S_ult for shafts without keyways”, and “these values are to be reduced by 25% if the shafts have keyways”. The ASME shaft code itself is cited by number only and is not reproduced. The same two fractions and the same keyway factor were found independently in a second shaft-design write-up read the same day, which is why they are on the page at all.
Engineers Edge, Average Properties of Structural Materials (read 6 October 2026), which credits its figures to Prof. Mike Ashby, Granta Design, Cambridge. The primary elastic-constant table used here, and the reason it was chosen over the several tables that print G alone: it prints the modulus of elasticity, the modulus of rigidity AND Poisson’s ratio in the same row, so the transcription can be checked against the isotropic identity G = E/2(1 + ν) instead of trusted. That check is run over every row before this batch writes anything, and it holds to within four per cent throughout. Its carbon steel row — E 29.0, G 11.5 Mpsi, ν 0.29 — is the same set this site’s spring pages already use.
R. C. Hibbeler, Average Mechanical Properties of Typical Engineering Materials (SI table), read 6 October 2026 in the University of California, Davis ENG104 handout that reproduces it. The second elastic-constant table, and the one that disagrees: it gives grey cast iron ASTM 20 as E 67.0 GPa and G 27 GPa against the 100 and 41.4 of the table above, which is a real difference between two grades of iron and not a rounding. It also gives red brass C83400 as E 101, G 37, ν 0.35, which satisfies the isotropic identity where the widely printed brass figure of 40 GPa does not.
The Engineering ToolBox, Modulus of Rigidity (read 6 October 2026). The third reading of the elastic constants, used as a tie-breaker and as the check that caught an error in one of the others: Engineers Edge’s separate shear modulus table prints rubber at 10.9 Mpsi = 75.2 GPa, within two per cent of structural steel, where this table prints 0.0003 GPa. The second is the right order of magnitude by five decades. Rubber is therefore not offered on this page, and the episode is the reason none of the figures here rests on a single table.

CalcEngines mechanical calculators are for education and design estimates. Results depend on the values you enter and the assumptions stated on the page, and they do not replace a qualified engineer's check. Standards, material grades and manufacturer data change — verify against the current standard and the supplier's own catalogue before you specify, machine or build anything.