Critical speed of shafts

Why it matters

A rotating shaft carrying discs, impellers or rotors bends a little because no rotor is perfectly balanced. At one particular speed this bending grows very large and the shaft whirls: bearings, seals and couplings are damaged within minutes. Pumps, fans, turbines, spindles and motor rotors are designed so that the operating speed stays well clear of this critical speed, or passes through it quickly.

Key ideas

Whirling. Consider a disc of mass m on a light shaft of lateral stiffness k, with its centre of mass a small distance e (eccentricity) from the shaft axis. When the shaft spins at ω, the centrifugal force on the eccentric mass bends the shaft by y, and the bent shaft rotates as a bow. Balancing the spring force k·y against the centrifugal force m·ω²·(y + e) gives the deflection formula below.

• Below the critical speed, y is in phase with e and grows as ω rises (heavy side flies out).
• At ω = ω_c = √(k/m), y becomes unbounded in theory (limited by damping in practice).
• Above it, y becomes negative: the disc's centre of mass moves inside the bow, and at very high speed y → −e, so the mass centre approaches the bearing axis and the rotor runs smoothly (self-centring). Turbines and centrifuges run supercritical this way, accelerating quickly through the critical speed.

Critical speed equals the lateral natural frequency. For a rotor carrying a disc, the critical speed in rad/s equals the natural frequency of transverse vibration of the non-rotating shaft–disc system (gyroscopic effects of large discs neglected). So the static-deflection result ω_c = √(g/δ) from free vibration applies, with δ the static deflection at the disc due to its weight as if the shaft were horizontal.

Shaft stiffness from beam theory. k = load / deflection at the disc:

• simply supported, central disc: 48EI/L³;
• simply supported, disc at distances a and b from the bearings: 3EI·L/(a²·b²);
• fixed (long bearings) both ends, central disc: 192EI/L³;
• overhung (cantilever) disc at length L: 3EI/L³. Bearing type therefore changes the critical speed considerably; long, rigid bearings act nearer to fixed ends.

Shaft with its own distributed mass. A uniform simply supported shaft of mass μ per unit length has first critical speed ω = π²·√(EI/(μ·L⁴)), equivalent to f = 0.5615/√δ (δ = maximum static deflection under self-weight, m).

Several discs: Dunkerley and Rayleigh.

• Dunkerley's empirical formula combines the critical speeds each disc (and the shaft alone) would have by itself: 1/ω² = 1/ω₁² + 1/ω₂² + … + 1/ω_s². It always underestimates the true first critical speed, a safe estimate.
• Rayleigh's method uses the static deflections y_i under all loads together: ω² = g·Σ(W_i·y_i)/Σ(W_i·y_i²), which slightly overestimates it. The true value lies between the two.

Design practice. Keep the operating speed at least about 20–30% away from any critical speed (follow your code or data book), balance the rotor well (amplitude scales with e), and add damping in bearings or supports to limit the amplitude while passing through.

Formulas

y = e·r² / (1 − r²) with r = ω/ω_c (undamped whirl amplitude at the disc, m; e = eccentricity, m) With damping: y = e·r² / √((1 − r²)² + (2ζr)²)

ω_c = √(k/m) = √(g/δ) (rad/s) ; N_c = 60·ω_c/(2π) (rpm) ; f_c = 0.4985/√δ Hz (δ in m)

k = 48EI/L³ (SS, central) ; k = 3EI·L/(a²·b²) (SS, off-centre) ; k = 192EI/L³ (fixed–fixed, central) ; k = 3EI/L³ (cantilever)

• E = Young's modulus (Pa); I = π·d⁴/64 for a solid round shaft (m⁴); L, a, b in m.

Uniform SS shaft: ω_s = π²·√(EI/(μ·L⁴)) ; f_s = 0.5615/√δ_max Hz

Dunkerley: 1/ω_c² = Σ 1/ω_i² + 1/ω_s² Rayleigh: ω_c² = g·Σ(W·y) / Σ(W·y²)

Worked examples

Example 1 (standard): single disc at mid-span. Given: steel shaft d = 25 mm, bearings 0.8 m apart (simply supported), 20 kg disc at the centre, E = 200 GPa, eccentricity 0.1 mm. Neglect shaft mass.

1. I = π × 0.025⁴ / 64 = 1.917 × 10⁻⁸ m⁴
2. k = 48EI/L³ = 48 × 200 × 10⁹ × 1.917 × 10⁻⁸ / 0.512 = 3.595 × 10⁵ N/m
3. ω_c = √(359 500 / 20) = 134.1 rad/s → N_c = 134.1 × 60 / 2π = 1280 rpm (check: δ = 20 × 9.81 / 359 500 = 0.546 mm, √(9.81/0.000546) = 134 rad/s).
4. At r = 0.8: y = 0.1 × 0.64 / 0.36 = 0.178 mm (in phase with e).
5. At r = 1.2: y = 0.1 × 1.44 / (−0.44) = −0.327 mm (the bow is opposite the heavy side). Answer: N_c ≈ 1280 rpm; whirl amplitude ≈ 0.18 mm at 0.8 N_c and 0.33 mm at 1.2 N_c.

Example 2 (GATE level): include the shaft's own mass by Dunkerley. Same shaft and disc; steel density 7850 kg/m³.

1. μ = 7850 × π × 0.025² / 4 = 3.853 kg/m
2. ω_s = π² × √(200 × 10⁹ × 1.917 × 10⁻⁸ / (3.853 × 0.8⁴)) = 9.870 × 49.29 = 486.5 rad/s (4646 rpm).
3. Dunkerley: 1/ω_c² = 1/134.1² + 1/486.5² = 5.561 × 10⁻⁵ + 0.4225 × 10⁻⁵ = 5.984 × 10⁻⁵ → ω_c = 129.3 rad/s.
4. N_c = 129.3 × 60 / 2π = 1234 rpm. Answer: N_c ≈ 1234 rpm (about 4% below the light-shaft value). For a 1000 rpm duty this leaves only about 19% margin, so a stiffer shaft or shorter span would be advisable.

Common mistakes

• Using the critical speed in rad/s as if it were rpm or Hz.
• Using the deflection caused by an unspecified load; δ must be the static deflection under the rotor's own weight.
• Picking the wrong beam formula for the support conditions.
• Mixing mm and m in I or L (I in mm⁴ with E in Pa).
• Believing the amplitude keeps growing above the critical speed; it falls back towards e.
• Treating Dunkerley's result as exact; it is a lower bound.

For GATE ME

Typical questions: critical speed from the static deflection or from shaft stiffness and disc mass; effect of support conditions; whirl amplitude from eccentricity at a given speed ratio; Dunkerley combination of several discs or of disc plus shaft. Practise beam deflection formulas and unit conversion to rpm.

Quick check

1. A disc deflects its shaft by 1 mm statically. Critical speed in rpm?
2. If the shaft diameter is doubled (same span and disc), by what factor does ω_c change?
3. Above the critical speed, where is the disc's centre of mass relative to the bent shaft?
4. Two discs alone give 1000 rpm and 2000 rpm. Dunkerley estimate?
5. Does Dunkerley over- or underestimate?

Answers: 1. ω = 99.0 rad/s, about 946 rpm. 2. I rises 16 times, so ω_c rises 4 times. 3. Inside the bow, between the shaft centre and the bearing axis. 4. 1/N² = 1/10⁶ + 1/(4 × 10⁶) → N ≈ 894 rpm. 5. Underestimates.