Shaft Critical Speed Calculator

Q: What is the critical speed of a shaft?

The rotational speed at which the shaft's own bending natural frequency is excited once per revolution by its unavoidable residual unbalance, so that whirl amplitude is limited only by damping. In rev/min it is the first bending natural frequency in hertz multiplied by sixty. It has nothing to do with the shaft's strength: a shaft can be enormously strong and still whirl, and the fix is almost always geometry rather than material.

Q: Which is right, Rayleigh or Dunkerley?

Neither and both. Rayleigh's quotient with an assumed mode shape is an upper bound on the true first frequency, because assuming a shape stiffens the system; Dunkerley's sum of reciprocal squares is a lower bound, because adding flexibilities ignores the coupling between them. The true value is between, and the distance between them tells you how much the answer is worth. This page computes both and prints the gap. If you need one number for a design decision, use Dunkerley — it is the conservative one.

Q: How far below the critical speed should I run?

There is no single sourced answer, and the figure most often quoted — 75% — traces back to sources that do not cite anything. What is in a published code is API 610's requirement that a stiff rotor's first dry critical be at least 20% above maximum continuous speed, which means running at or below 0.83 of the critical, and 30% if the machine might ever run dry. API 684 makes the margin depend on the amplification factor instead, and says a mode with a damping ratio above 0.15 needs no margin at all. All of those are computed on this page. The one rule with no caveats is not to sit at the critical speed in steady operation.

Q: Can I run above the critical speed?

Yes, and most large turbomachinery does. It changes the problem rather than removing it: the rotor passes through resonance on every run-up, so transient amplitude and the time spent in the band become the design questions, damping decides whether the machine survives that, and above the critical the rotor whirls about its mass centre so balancing behaves differently. None of that is in a beam calculation. Operating above the first critical needs a rotordynamic model with real bearing coefficients.

Q: Does a hollow shaft help?

Considerably, and it is the cheapest improvement available. The second moment of area goes as the fourth power of the radius, so material near the centre contributes almost nothing to stiffness while contributing all of its mass. A bore of half the outside diameter keeps about 94% of the stiffness and sheds 25% of the mass, which raises the critical speed by about 12%. The calculator shows both figures for whatever bore you enter. The cautions are local buckling of a thin wall, torsional capacity, and the fact that a hollow shaft tolerates keyways and grooves far less well.

Q: Why does the answer depend so much on where the mass sits?

Because the flexibility at a point on a simply supported shaft is a²b²/(3EIL), which is a maximum at mid-span and falls to zero at a bearing. The same 20 kg rotor moved from mid-span to a quarter of the span sees about half the flexibility, and the critical speed rises accordingly. That is the second most powerful lever on the page after the span itself, and it is usually the cheapest one to pull: a gearbox layout that puts the heavy wheel near a bearing rather than between two is often the whole fix.

Q: What about torsional critical speed?

It is a different calculation on a different system and this page deliberately does not attempt it. Bending critical speed depends on span, second moment of area and the bearings; torsional natural frequency depends on polar inertias and torsional stiffnesses and not on the bearings at all. The two answers are unrelated numbers. If your drive has a reciprocating engine, a long coupling shaft or a large flywheel, the torsional problem is the more likely one and it needs its own analysis — a Holzer or eigenvalue solve against the excitation orders your machine actually produces.

First bending critical speed of a shaft, bracketed between Dunkerley’s lower bound and Rayleigh’s upper one, with the shaft’s own distributed mass, up to two point masses and three support conditions that differ by a factor of six — plus what the published codes actually require for separation margin, which is not the 75% everyone quotes.

Last updated 24 Sep 2026

Shaft critical speed

Shaft, span, masses → the bracket the critical lies in

Shaft outside diameter (mm)Stiffness goes as the fourth power of this and mass as the square, so the critical speed goes up roughly in proportion to the diameter.

Bore, if the shaft is hollow (mm, 0 for solid)A hollow shaft is the cheapest critical speed there is: a bore of half the outside diameter keeps 94% of the stiffness and loses 25% of the mass.

Bearing span, or overhang (mm)The single most powerful number on the page: the critical speed goes as 1/span². Halving the span quadruples it.

How the shaft is heldThese differ by more than a factor of six end to end. Clamped both ends is 2.27 times simply supported; a cantilever is 0.36 times it.

First point mass — rotor, gear, sheave (kg)Its position matters as much as its size. A mass at mid-span of a simply supported shaft costs far more than the same mass near a bearing.

… its position from the left bearing (mm)For an overhung shaft, measure from the bearing (the built-in end).

Second point mass (kg, 0 if there is only one)

… its position from the left bearing (mm)

Operating speed (rev/min)Checked against both bounds and against API 610’s 20% separation margin.

Young’s modulus (GPa)210 for steel, 200 for stainless, 69 for aluminium, 110 for grey cast iron.

Density (kg/m³)7850 for steel. It is the shaft’s own mass and it matters whenever the shaft weighs a useful fraction of what it carries.

Not a circuit: one speed axis, scaled so that Dunkerley's answer always sits at mid-scale. That is deliberate — it makes the figure about the RELATIONSHIP between the numbers rather than about their size. The tall fixed line is Dunkerley, the lower bound; the shorter line beside it is Rayleigh, the upper bound; and the narrow shaded strip between them is the bracket the true first critical speed lies in. On most realistic shafts that strip is barely visible, which is the useful news: the uncertainty in this calculation is not the method. The wider shaded band running from 0.75 of the lower bound to 1.4 times the upper one is the region no source recommends sitting in at steady state; note that the multipliers are the rule of thumb rather than a code figure, and the API 610 number printed below is the one that comes from a standard. The arrow under the axis is your operating speed, quantised to tenths of the critical and clamped at twice it.

3,073rev/minExample

A solid 40 mm steel shaft on two bearings 800 mm apart, with a 20 kg rotor at mid-span, running at 1,500 rev/min

Two bounds on the same number

ω_shaft = (βL)²·√(EI / m L⁴) · Dunkerley: 1/ω² = 1/ω_shaft² + Σ mᵢ δᵢᵢ · Rayleigh: ω² = g·Σ Wᵢyᵢ / Σ Wᵢyᵢ² · yᵢ = Σⱼ δᵢⱼ Wⱼ

βL: the first root of the support condition’s frequency equation — π, 4.73004 or 1.87510 — solved numerically here rather than quoted
m: the shaft’s own mass per unit length. It is not negligible: at the defaults it is a third of everything the bearings carry
δᵢⱼ: the flexibility influence coefficient — deflection at i from a unit load at j. Closed form for each support condition, symmetric by Maxwell’s theorem
Wᵢ: the weight of mass i. In Rayleigh’s quotient the shaft’s distributed mass is carried as four slices, which is enough to put the answer within 0.1% of a finite-element eigensolve
yᵢ: static deflection at i under all the weights — the assumed mode shape, which is what makes Rayleigh’s quotient an estimate rather than a solve

Worked example

A solid 40 mm steel shaft on two bearings 800 mm apart, with a 20 kg rotor at mid-span, running at 1,500 rev/min
Section first: I = π(40⁴)/64 = 125.7 kmm⁴ and the shaft's own mass is 7.89 kg, which is 28% of everything the bearings carry. Ignoring it would be a mistake
The bare shaft, with nothing on it: ω = π²√(EI/mL⁴) = 7,617 rev/min. That is the number a shaft with no rotor on it would whirl at, and it is a long way above where we end up
Dunkerley says the reciprocal squares add. The 20 kg mass at mid-span sees a flexibility of a²b²/(3EIL) = 0.000404 mm/N, so on its own it would resonate at √(1/(m·δ)); combined with the shaft, 3,073 rev/min. Dunkerley always comes out LOW, because adding flexibilities is the same as assuming each mass moves in its own static shape
Rayleigh says take the static deflection under gravity as the mode shape and form g·ΣWy / ΣWy². Doing that with the shaft's mass carried as four slices gives 3,077 rev/min. Rayleigh always comes out HIGH, because assuming any shape other than the true mode shape adds constraint and constraint adds stiffness
So the first critical speed is between 3,073 and 3,077 rev/min — a bracket of 0.1%. Both were checked against a finite-element eigensolve of the same beam at ninety configurations, and the finite-element answer fell inside the bracket every time
At 1,500 rev/min you are at 0.488 × the low end of the bracket, comfortably below. API 610 would allow up to 2,561 rev/min for a stiff rotor, or 2,364 if the machine might ever run dry
For contrast: clamp both ends of the same shaft and the critical speed goes to 6,260 rev/min; hang the same shaft off one bearing as an overhang and it falls to 1,787. The bearings are worth more than anything else you can change
And the estimate that is all over the web, ω = √(g ÷ static deflection), gives 6,763 rev/min for the bare shaft against the exact 7,617. It is √(384/5) where the answer is π², which is 11% LOW — and low is the direction that gets you hurt

What the supports are worth — and they are worth more than anything else on the page

ω₁ = (βL)²·√(EI/mL⁴), and βL is the first root of the support condition’s own frequency equation — sin(βL) = 0 for a shaft on two bearings, cos(βL)·cosh(βL) = 1 for a clamped one, cos(βL)·cosh(βL) = −1 for a cantilever. Those roots are solved numerically here, not looked up, and then checked against a two-hundred-element finite-element eigensolve of the same beam. Clamping both ends of a shaft multiplies its critical speed by 2.27; hanging the same shaft off one bearing divides it by 2.8. That is a factor of six between the best and worst arrangement of the same piece of steel, which is why the honest first question about a critical-speed problem is not “what diameter” but “what are the bearings doing”. The catch is that a real bearing is neither: a deep-groove ball bearing in a housing is nearer simply supported than clamped, but a pair of them close together with a preload is stiffer than that, and the real answer lies between two rows of this table. 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.
How the shaft is held	βL, first mode	(βL)²	Critical speed relative to simply supported	Point-load deflection coefficient	Self-weight deflection coefficient
Simply supported on two bearings	3.14159	9.8696	1.000	1/48	1/77
Built in (clamped) at both ends	4.73004	22.3733	2.267	1/192	1/384
Overhung — cantilever from one bearing	1.87510	3.5160	0.356	1/3	1/8

How far from the critical speed is far enough? Four published answers, and they do not agree

The 0.75 and 1.4 multipliers that circulate as “the rule” could not be traced to a design code by this batch. What IS in a code is API 610’s 20 per cent margin, which is 0.83 rather than 0.75, and API 684’s answer, which is that the question has no fixed answer because it depends on how sharply the rotor responds. Both multipliers are computed above so you can see them, and both are labelled for what they are. The one thing every source agrees on is the thing this page will state without hedging: do not sit at the critical speed in steady operation. Passing through it on run-up is normal and is what a flexible rotor does every start; living there is not. 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.
Source	What it actually says	What that means for a critical speed of 3,073 rev/min
API 610, 11th ed., as reported by Turbomachinery International	A rotor counts as “classically stiff” when its first dry critical speed is “at least 20 percent above the maximum continuous running speed (and 30 percent above if the pump might ever actually run dry)”	Run at or below 2,561 rev/min, or 2,364 if it might run dry
API TR 684-1, Figures 1-49 and 1-50	The required separation margin is a function of the AMPLIFICATION FACTOR, not a fixed percentage; “a low amplification factor (AF < 5) indicates that the system is not sensitive to unbalance when operating in the vicinity of the associated critical speed”	No single number. A well-damped mode needs very little margin and a sharp one needs a lot
API 610 on damping, same report	“There is no separation margin concern for any natural frequency with a damping ratio above 0.15, i.e. log dec of 0.94”	If the mode is that well damped, the margin requirement goes away entirely
14 CFR § 29.931 (transport rotorcraft)	Critical speeds “must be determined by demonstration except that analytical methods may be used if reliable methods of analysis are available”, and the margins “must be adequate to allow for possible variations between the computed and actual values”	No number at all, deliberately. Demonstrate it
Wikipedia, Critical speed	“Many practical applications suggest as good practice that the maximum operating speed should not exceed 75% of the critical speed” — carrying a “citation needed” tag	2,305 rev/min. This is the figure everyone quotes and the one nobody sources

What this page is not

The first row is the one that catches people. “The critical speed” with no qualifier almost always means the first BENDING critical, which is what this page computes; a drive train with a diesel engine, a long coupling shaft and a big inertia at the far end is far more likely to be in torsional trouble, and nothing on this page will tell you so. 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.
Calculation	Why it is not here	What to do instead
TORSIONAL critical speed	A different calculation on a different physical system. Bending critical speed is a beam problem — mass, second moment of area, span, bearings. Torsional natural frequency is a rotational spring-mass problem: polar moment of inertia J at each rotor, torsional stiffness GJ_p/L of each shaft section between them, and NO dependence on the bearings or the span in bending. The two answers are unrelated numbers and a shaft can be safe in one and resonant in the other	A Holzer or eigenvalue solve on the torsional chain, driven by the excitation orders your machine actually produces — a reciprocating engine’s gas torque harmonics, a VFD’s switching orders, a two-bladed propeller’s 2× order. This page refuses it rather than approximate it
Bearing stiffness, damping, and gyroscopic effects	The model here has rigid, undamped supports and no rotation in the stiffness matrix. A real bearing’s radial stiffness is finite, which LOWERS the critical speed — often substantially for a rolling bearing in a light housing — and its damping is what decides whether a resonance matters at all. Gyroscopic coupling splits the mode into forward and backward whirl, so there are two critical speeds where this page prints one	A rotordynamic model with bearing coefficients from the bearing maker, and a Campbell diagram. Use this page to find out whether you are anywhere near trouble, not to certify that you are not
Higher modes	Only the first bending mode is computed. Dunkerley’s sum and Rayleigh’s quotient both converge on the FIRST mode; the second is at about four times the first for a simply supported shaft and is not obtainable from either	An eigenvalue solve. If your operating speed is above the first critical you need the second one, and you need it from a real model
Shaft DIAMETER from strength or stiffness	That is shaft sizing, and it already has its own page on this site	Size the shaft first, then check this page. They are different criteria and either can govern

Two bounds, the support condition that beats everything else, and the crude formula that is 11% low

Two methods, one number, and the honest answer is the gap between them. Rayleigh’s method assumes a mode shape — here the static deflection under gravity — and forms the energy quotient; assuming any shape other than the true one adds constraint, constraint adds stiffness, and the answer comes out HIGH. Dunkerley’s method adds the reciprocal squares of the frequencies each mass would have on its own, which is the same as adding flexibilities, and it comes out LOW. The first critical speed lies between them. At the defaults on this page that bracket is about a tenth of a per cent wide, which is the most useful thing the pair can tell you: the uncertainty in a critical-speed calculation is almost never the method. It is the bearings.

Which is why the support condition is the biggest number on the page. The first mode is ω₁ = (βL)²·√(EI/mL⁴), and βL is 3.1416 for a shaft on two bearings, 4.7300 clamped at both ends and 1.8751 hanging off one. Squared, that is a factor of 2.27 between the first two and 6.4 between the extremes — the same piece of steel, six times the critical speed. Those roots are solved here from the frequency equations rather than looked up, and then checked against a two-hundred-element finite-element eigensolve of the same beam. The uncomfortable corollary is that a real bearing is neither pinned nor clamped, so the real answer lies between two rows of the support table, and a soft housing puts it below the pinned row.

The crude formula everybody quotes is wrong by eleven per cent, in the unsafe direction. “Critical speed equals the square root of g over the static deflection” is exact for a single mass on a massless spring and it is not exact for a shaft. Work it out for a uniform simply supported shaft under its own weight and it gives √(384/5) = 8.764 in place of π² = 9.870 — 11% low. Low matters: it tells you the resonance is further away than it is. Rayleigh’s quotient on the SAME static shape gives 9.8767, which is 0.07% high, so the fix costs nothing but doing the quotient properly. This page prints the crude figure beside the right one so you can see the size of the error.

Torsional critical speed is a different calculation and this page refuses it. Bending critical speed is a beam problem: mass, second moment of area, span, bearings. Torsional natural frequency is a rotational spring–mass chain: polar inertia at each rotor, GJ_p/L between them, and no dependence on the bearings at all. The two numbers are unrelated and a drive can be clean in one and resonant in the other — a diesel engine driving a long shaft into a heavy inertia is a torsional problem long before it is a bending one. Naming it is the most useful thing this page can do about it. Shaft sizing from strength and stiffness, and torsional deflection, both have their own pages on this site and are not repeated here. What the shaft is doing at its bearings when it is not whirling is on the shaft deflection and slope calculator, and the belt pull that puts a real transverse load into a shaft is on the belt tension and shaft load calculator.

Frequently asked questions

What is the critical speed of a shaft?

The rotational speed at which the shaft’s own bending natural frequency is excited once per revolution by its unavoidable residual unbalance, so that whirl amplitude is limited only by damping. In rev/min it is the first bending natural frequency in hertz multiplied by sixty. It has nothing to do with the shaft’s strength: a shaft can be enormously strong and still whirl, and the fix is almost always geometry rather than material.

Which is right, Rayleigh or Dunkerley?

Neither and both. Rayleigh’s quotient with an assumed mode shape is an upper bound on the true first frequency, because assuming a shape stiffens the system; Dunkerley’s sum of reciprocal squares is a lower bound, because adding flexibilities ignores the coupling between them. The true value is between, and the distance between them tells you how much the answer is worth. This page computes both and prints the gap. If you need one number for a design decision, use Dunkerley — it is the conservative one.

How far below the critical speed should I run?

There is no single sourced answer, and the figure most often quoted — 75% — traces back to sources that do not cite anything. What is in a published code is API 610’s requirement that a stiff rotor’s first dry critical be at least 20% above maximum continuous speed, which means running at or below 0.83 of the critical, and 30% if the machine might ever run dry. API 684 makes the margin depend on the amplification factor instead, and says a mode with a damping ratio above 0.15 needs no margin at all. All of those are computed on this page. The one rule with no caveats is not to sit at the critical speed in steady operation.

Can I run above the critical speed?

Yes, and most large turbomachinery does. It changes the problem rather than removing it: the rotor passes through resonance on every run-up, so transient amplitude and the time spent in the band become the design questions, damping decides whether the machine survives that, and above the critical the rotor whirls about its mass centre so balancing behaves differently. None of that is in a beam calculation. Operating above the first critical needs a rotordynamic model with real bearing coefficients.

Does a hollow shaft help?

Considerably, and it is the cheapest improvement available. The second moment of area goes as the fourth power of the radius, so material near the centre contributes almost nothing to stiffness while contributing all of its mass. A bore of half the outside diameter keeps about 94% of the stiffness and sheds 25% of the mass, which raises the critical speed by about 12%. The calculator shows both figures for whatever bore you enter. The cautions are local buckling of a thin wall, torsional capacity, and the fact that a hollow shaft tolerates keyways and grooves far less well.

Why does the answer depend so much on where the mass sits?

Because the flexibility at a point on a simply supported shaft is a²b²/(3EIL), which is a maximum at mid-span and falls to zero at a bearing. The same 20 kg rotor moved from mid-span to a quarter of the span sees about half the flexibility, and the critical speed rises accordingly. That is the second most powerful lever on the page after the span itself, and it is usually the cheapest one to pull: a gearbox layout that puts the heavy wheel near a bearing rather than between two is often the whole fix.

What about torsional critical speed?

It is a different calculation on a different system and this page deliberately does not attempt it. Bending critical speed depends on span, second moment of area and the bearings; torsional natural frequency depends on polar inertias and torsional stiffnesses and not on the bearings at all. The two answers are unrelated numbers. If your drive has a reciprocating engine, a long coupling shaft or a large flywheel, the torsional problem is the more likely one and it needs its own analysis — a Holzer or eigenvalue solve against the excitation orders your machine actually produces.

Related calculators

References

API 610, Centrifugal Pumps for Petroleum, Petrochemical and Natural Gas Industries, 11th edition. Cited by clause; not fetched. The separation-margin requirement quoted here is as Turbomachinery International reports it: a rotor may be treated as “classically stiff” when “its first dry critical speed … is at least 20 percent above the maximum continuous running speed (and 30 percent above if the pump might ever actually run dry)”, and there is “no separation margin concern for any natural frequency with a damping ratio above 0.15”. Twenty per cent above the running speed means running at or below 0.83 of the critical, not 0.75.
Turbomachinery International. “How rotordynamics influences pump selection.” The reporting source for the API 610 separation margins above, and for the point that the margin requirement disappears once the mode is well enough damped — which is why a real rotor’s answer depends on bearing and seal damping that a beam calculation cannot see.
API TR 684-1, API Standard Paragraphs Rotordynamic Tutorial: Lateral Critical Speeds, Unbalance Response, Stability, Train Torsionals and Rotor Balancing. Cited by figure; not reproduced. Its Figures 1-49 and 1-50, “API Required Separation Margins for Operation Above a Critical Speed” and “… Below a Critical Speed”, make the required margin a function of the AMPLIFICATION FACTOR rather than a fixed percentage, and it states that “a low amplification factor (AF < 5) indicates that the system is not sensitive to unbalance when operating in the vicinity of the associated critical speed”. That is the honest answer to “how far away is far enough” and it is why this page will not print a single number as if it were a rule.
14 CFR § 29.931, Shafting critical speed (Federal Aviation Regulations, transport category rotorcraft). The one design code this batch found that refuses to name a number: critical speeds “must be determined by demonstration except that analytical methods may be used if reliable methods of analysis are available”, and where analysis is used “the margins between the calculated critical speeds and the limits of the allowable operating ranges must be adequate to allow for possible variations between the computed and actual values”. No percentage is given, deliberately.
Wikipedia, Critical speed. Quoted only to say what it says and that it does not support it: “Many practical applications suggest as good practice that the maximum operating speed should not exceed 75% of the critical speed”, carrying a “citation needed” tag. The 75% figure is repeated across the web from sources like this one; the published code figure is API 610’s 20% margin.
Richard G. Budynas and J. Keith Nisbett, Shigley’s Mechanical Engineering Design, Table 7-2, “Typical Maximum Ranges for Slopes and Transverse Deflections”. The source for the slope limits at a bearing: tapered roller 0.0005–0.0012 rad, cylindrical roller 0.0008–0.0012, deep-groove ball 0.001–0.003, spherical and self-aligning ball 0.026–0.052, and an uncrowned spur gear below 0.0005 rad. The same table gives transverse deflection limits for spur gears by diametral pitch, which is a different criterion and is named on the page but not computed.