Critical speed of shafts and whirling

Why it matters

Propeller shafts, turbocharger rotors, pump shafts and crankshafts all carry masses whose centres are never exactly on the axis. At a particular speed, the critical or whirling speed, the shaft bows out violently and can fail or wreck its bearings. A long one-piece propeller shaft in a rear-wheel-drive vehicle is often split into two pieces with a centre bearing precisely to push its critical speed above the maximum driving speed.

Key ideas

Whirling. A rotor of mass m sits on a shaft, with its centre of mass at a small eccentricity e from the shaft axis. As the shaft spins, the centrifugal force m·ω²·(y + e) bends it by y, and the elastic restoring force k·y resists. The bent shaft rotates about the bearing axis; this rotation of the deflected shaft is whirling. In synchronous whirl the shaft whirls at its own spin speed.

Critical speed. Equating the forces: k·y = m·ω²·(y + e), so y = e·r²/(1 − r²) with r = ω/ω_c. When ω reaches ω_c = √(k/m) the deflection becomes unbounded (in practice limited by damping and nonlinearity). The critical speed equals the natural frequency of transverse vibration of the same shaft and rotor, so it can be found from the static deflection: ω_c = √(g/δ).

Behaviour through the critical speed.

• Below ω_c, y and e are in the same direction: the heavy side flies out.
• At ω_c, the deflection is very large, limited only by damping.
• Above ω_c, y = −e·r²/(r² − 1), which tends to −e: the shaft deflects to the opposite side and the centre of mass moves towards the bearing axis (self-centring). Steam and gas turbines run above their first critical speed this way, accelerating quickly through it.

Shaft stiffness for common supports (load W, flexural rigidity E·I, span L):

• Simply supported, load at the centre: k = 48EI/L³.
• Simply supported, load at distances a and b from the supports: k = 3EIL/(a²b²).
• Cantilever, load at the free end: k = 3EI/L³.
• Fixed at both ends (or long rigid bearings), central load: k = 192EI/L³. The support condition matters a lot: fixing the ends raises the central-load critical speed by a factor of 2.

Shaft with its own distributed mass. For a uniform simply supported shaft, ω_c = π²·√(E·I / (μ·L⁴)), μ being the mass per unit length. In terms of the maximum static deflection under its own weight, f_c = 0.5615/√δ_max (Hz, δ in m).

Several loads: Dunkerley's method. 1/ω_c² ≈ 1/ω₁² + 1/ω₂² + … + 1/ω_s², where ω₁, ω₂ are the critical speeds with each load acting alone and ω_s that of the shaft alone. It always underestimates slightly, so it is on the safe side.

Damping and margins. Damping does not change ω_c appreciably but limits the peak deflection. Machines are designed either to run at least about 20 to 30% below the first critical speed (stiff rotors) or well above it with a quick run-up (flexible rotors).

Formulas

ω_c = √(k/m) = √(g/δ) and N_c = 60·ω_c / (2π)

• ω_c: critical speed (rad/s); N_c: in rpm; k: lateral stiffness of the shaft at the rotor (N/m); m: rotor mass (kg); δ: static deflection at the rotor under its weight (m).

y = e·r² / (1 − r²) and, with damping, y/e = r² / √[(1 − r²)² + (2ζr)²]

• y: whirl amplitude of the shaft centre (m); e: eccentricity of the rotor's centre of mass (m); r = ω/ω_c; ζ: damping ratio.

δ = W·L³ / (48·E·I), δ = W·a²·b² / (3·E·I·L), δ = W·L³ / (3·E·I)

• Static deflections under the load for centre-loaded simply supported, off-centre simply supported and cantilever shafts (m); W: load (N); E: Young's modulus (Pa); I = π·d⁴/64 (m⁴).

ω_c = π²·√(E·I / (μ·L⁴)) (uniform simply supported shaft, own mass only)

• μ: mass per unit length (kg/m).

1/ω_c² = 1/ω₁² + 1/ω₂² + … + 1/ω_s² (Dunkerley)

Worked examples

Example 1 (standard: disc on a shaft). A 20 kg disc is mounted at the middle of a steel shaft (E = 200 GPa) of diameter 30 mm, simply supported over a span of 0.8 m. Neglecting the shaft's mass, find the critical speed.

1. I = π × 0.03⁴ / 64 = 3.976 × 10⁻⁸ m⁴.
2. k = 48EI/L³ = 48 × 200 × 10⁹ × 3.976 × 10⁻⁸ / 0.8³ = 381 700 / 0.512 = 745 500 N/m.
3. ω_c = √(745 500 / 20) = √37 275 = 193.1 rad/s, N_c = 193.1 × 60 / 2π = 1844 rpm.
4. Check with static deflection: δ = 20 × 9.81 / 745 500 = 2.63 × 10⁻⁴ m; √(9.81 / 2.63 × 10⁻⁴) = 193.1 rad/s.

Example 2 (GATE level: shaft mass and whirl amplitude). For the same shaft (density 7850 kg/m³), include the shaft's own mass using Dunkerley's method. Then find the whirl amplitude for a disc eccentricity of 0.1 mm at 0.8 and at 2 times the critical speed, neglecting damping.

1. μ = 7850 × π × 0.03² / 4 = 5.549 kg/m.
2. Shaft alone: ω_s = π²·√(EI / (μL⁴)) = 9.870 × √(7952 / (5.549 × 0.4096)) = 9.870 × √3499 = 9.870 × 59.15 = 583.8 rad/s.
3. Dunkerley: 1/ω_c² = 1/193.1² + 1/583.8² = 2.682 × 10⁻⁵ + 0.293 × 10⁻⁵ = 2.975 × 10⁻⁵, so ω_c = 183.3 rad/s (1750 rpm), about 5% below the value without shaft mass.
4. At r = 0.8: y = 0.1 × 0.64 / (1 − 0.64) = 0.178 mm, on the same side as the eccentricity.
5. At r = 2: y = 0.1 × 4 / |1 − 4| = 0.133 mm, on the opposite side; the centre of mass is now only 0.033 mm from the bearing axis.

Common mistakes

• Giving (1/2π)·√(g/δ) and calling it rad/s. That expression is in Hz; ω_c = √(g/δ) is in rad/s.
• Using the deflection formula for the wrong support condition (cantilever vs simply supported vs fixed).
• Forgetting to convert I from mm⁴ to m⁴ or E from GPa to Pa.
• Thinking damping raises the critical speed. It only limits the peak amplitude.
• Using Dunkerley's equation with frequencies added directly instead of their inverse squares.
• Treating any speed below critical as safe. At 0.8 ω_c the deflection is already 1.8 times the eccentricity.

For GATE ME

Expect the critical speed of a shaft with a central or off-centre disc from beam deflection, the critical speed of a uniform shaft, Dunkerley's combination of several loads, and whirl amplitude below or above the critical speed. Practise converting quickly between δ, ω_c, Hz and rpm, and know the stiffness of the four standard support cases.

Quick check

1. Static deflection 1 mm under a disc. Critical speed in rpm?
2. What happens to the rotor's centre of mass well above the critical speed?
3. How does the critical speed change if the span of a centre-loaded simply supported shaft is doubled?
4. Two loads alone give 100 rad/s and 200 rad/s. Dunkerley estimate with both (shaft mass neglected)?
5. Eccentricity 0.05 mm, r = 0.5, no damping. Whirl amplitude?

Answers: 1. √(9.81/0.001) = 99.0 rad/s = 946 rpm. 2. It moves towards the bearing axis (self-centring). 3. k falls by 8, so ω_c falls by √8 = 2.83. 4. 1/√(10⁻⁴ + 0.25 × 10⁻⁴) = 89.4 rad/s. 5. 0.05 × 0.25 / 0.75 = 0.0167 mm.