Torsional vibration of rotor systems

Why it matters

Crankshafts, propeller shafts, gearbox and generator drive lines, and servo-driven robot joints all twist back and forth about their axes as well as rotate. Torsional vibration is invisible from outside but shows up as cracked crankshafts, broken gear teeth, coupling failures, noise and poor servo positioning. Engine pulses and gear-mesh forces excite it, so the torsional natural frequencies must be kept away from those excitations or damped.

Key ideas

Single-rotor system. A disc (mass moment of inertia J) on a shaft whose other end is fixed behaves like a spring–mass system in rotation: J·θ̈ + k_t·θ = 0, so ω_n = √(k_t/J). The shaft's torsional stiffness is k_t = G·I_p/L from torsion theory (torque per radian of twist).

Effect of shaft inertia. If the shaft's own inertia J_s is not negligible, add one third of it to the disc: ω_n = √(k_t/(J + J_s/3)) (same reasoning as the spring-mass correction).

Two-rotor system. Two discs J_A and J_B on a free–free shaft (both ends free to turn, as in a motor driving a fan through a long shaft) vibrate in opposite senses. Some section of the shaft does not twist at all: the node. Each side of the node acts as a single-rotor system with the same frequency, so l_A·J_A = l_B·J_B and the node lies nearer the heavier rotor. There is also a zero-frequency "mode" in which the whole system simply rotates as a rigid body.

Three-rotor system. Gives two non-zero natural frequencies: a single-node mode and a two-node mode. The frequency equation is a quadratic in ω²; GATE usually asks only for the single- or two-rotor case.

Stepped shafts (torsionally equivalent shaft). Sections in series act like springs in series. Replace them with an equivalent uniform shaft of diameter d: L_eq = L₁ + L₂·(d/d₂)⁴ + L₃·(d/d₃)⁴ + … (taking d = d₁). A thin short section can dominate the flexibility.

Geared systems. Refer everything to one shaft (say the motor shaft). If the other shaft turns at n = ω₂/ω₁ times the reference speed, its inertias and stiffnesses are multiplied by n² when referred to the reference shaft (energy equivalence). Then solve as a simple two- or three-rotor system. A motor with a high-ratio gearhead therefore "sees" the load inertia divided by the square of the reduction ratio, the basis of inertia matching in servo drives.

Excitation and resonance. In a reciprocating engine the torque contains harmonics at multiples ("orders") of the crank speed: for a four-stroke engine, half-orders as well. A torsional critical speed occurs when an order frequency equals a torsional natural frequency, so one natural frequency can produce several critical speeds in the speed range. Gear mesh frequency (teeth × shaft speed) and motor torque ripple are other excitations.

Control measures. Change stiffness or inertia (shaft diameter, coupling stiffness, flywheel position) to move natural frequencies; add damping with a viscous (Houdaille) damper or a tuned rubber damper on the crankshaft nose; use flexible couplings to lower stiffness and add damping between machines.

Formulas

k_t = G·I_p / L (N·m/rad); I_p = π·d⁴/32 (solid round shaft, m⁴), G = modulus of rigidity (Pa)

ω_n = √(k_t / J) (single rotor, rad/s) ; f_n = ω_n / 2π (Hz) ω_n = √(k_t / (J + J_s/3)) (including shaft inertia J_s)

Two-rotor: l_A·J_A = l_B·J_B ; l_A = L·J_B / (J_A + J_B) ω_n = √(G·I_p / (l_A·J_A)) = √(k_t·(J_A + J_B) / (J_A·J_B))

Equivalent shaft: L_eq = L₁ + L₂·(d₁/d₂)⁴ + L₃·(d₁/d₃)⁴

Geared system, referred to shaft 1 (n = ω₂/ω₁): J₂' = n²·J₂ ; k₂' = n²·k₂

J = m·k² (disc of radius of gyration k) ; J = m·R²/2 (solid disc of radius R)

Worked examples

Example 1 (standard): single rotor. Given: a disc with J = 5 kg·m² at the free end of a steel shaft, d = 50 mm, L = 1 m, other end fixed; G = 80 GPa.

1. I_p = π × 0.05⁴ / 32 = 6.136 × 10⁻⁷ m⁴
2. k_t = G·I_p/L = 80 × 10⁹ × 6.136 × 10⁻⁷ / 1 = 49 090 N·m/rad
3. ω_n = √(49 090 / 5) = 99.1 rad/s → f_n = 99.1 / 2π = 15.8 Hz. Answer: f_n ≈ 15.8 Hz. An excitation at 15.8 Hz (948 cycles/min) must be avoided.

Example 2 (GATE level): two-rotor system and its node. Given: a motor rotor J_A = 15 kg·m² and a fan J_B = 45 kg·m² on the two ends of a steel shaft, d = 60 mm, L = 1.5 m, G = 80 GPa.

1. Node position from A: l_A = L·J_B/(J_A + J_B) = 1.5 × 45/60 = 1.125 m (so l_B = 0.375 m; check 1.125 × 15 = 0.375 × 45 = 16.9).
2. I_p = π × 0.06⁴/32 = 1.272 × 10⁻⁶ m⁴
3. ω_n = √(G·I_p/(l_A·J_A)) = √(80 × 10⁹ × 1.272 × 10⁻⁶ / (1.125 × 15)) = √(101 790 / 16.875) = √6032 = 77.7 rad/s
4. Check: k_t = 80 × 10⁹ × 1.272 × 10⁻⁶ / 1.5 = 67 860 N·m/rad; √(67 860 × 60 / (15 × 45)) = √6032 = 77.7 rad/s.
5. f_n = 77.7 / 2π = 12.4 Hz. Answer: node 1.125 m from the motor rotor; f_n ≈ 12.4 Hz.

Common mistakes

• Using the polar second moment I_p (m⁴) where the mass moment of inertia J (kg·m²) is meant, or the reverse.
• Using d³ or d² instead of d⁴ in I_p and in the equivalent-length formula.
• Placing the node nearer the lighter rotor; it lies nearer the heavier one.
• Forgetting that a free–free two-rotor system has a rigid-body zero frequency besides the vibration mode.
• Referring geared inertias by n instead of n².
• Mixing GPa with mm-based dimensions.

For GATE ME

Typical questions: natural frequency of a single- or two-rotor system; node location; equivalent length of a stepped shaft; geared systems referred to one shaft; effect of shaft inertia. Practise computing I_p and k_t with consistent SI units, and the node condition l_A·J_A = l_B·J_B.

Quick check

1. k_t = 800 N·m/rad, J = 2 kg·m². ω_n?
2. In a two-rotor system J_A = 10, J_B = 30 kg·m², L = 2 m. Node distance from A?
3. If shaft diameter doubles, how does k_t change?
4. A load of 0.4 kg·m² is driven through a 10 : 1 reduction. Inertia referred to the motor?
5. How many non-zero natural frequencies does a three-rotor system have?

Answers: 1. 20 rad/s. 2. 1.5 m. 3. 16 times. 4. 0.004 kg·m². 5. Two.