Torsional vibration of two- and three-rotor systems

Torsional stiffness, single-, two- and three-rotor natural frequencies, node position and mode shapes, torsionally equivalent stepped shafts and geared systems.

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Why it matters

An engine's crankshaft is twisted by a pulsating gas torque, and the crankshaft with its pulley and flywheel forms a torsional spring-mass system. If a firing harmonic matches a torsional natural frequency, the shaft twists back and forth with large amplitude and can crack, which is why engines carry a torsional damper on the front pulley and why dual-mass flywheels exist. The same models apply to propeller shafts, gearbox shafts and marine drive lines.

Key ideas

Torsional vibration. The rotors twist relative to each other about the shaft axis while the whole assembly may also rotate steadily. Torsional vibration is the oscillating twist superimposed on the steady rotation. Inertia comes from the rotors' mass moments of inertia I (kg·m²), stiffness from the shafts' torsional stiffness k_t = G·J/L (N·m/rad).

Single rotor on a fixed shaft. A disc at the free end of a shaft whose other end is fixed: I·θ̈ + k_t·θ = 0, so ω_n = √(k_t/I).

Two-rotor system (free-free). Two rotors I_A and I_B on a shaft, neither end fixed. There are two solutions:

  • ω = 0: both rotors turn together as a rigid body (this is just steady rotation).
  • One non-zero natural frequency ω_n = √(k_t·(I_A + I_B) / (I_A·I_B)). The rotors swing in opposite directions, and a point on the shaft, the node, does not twist at all. The node divides the shaft so that each part acts as a single-rotor system with the same frequency: I_A·l_A = I_B·l_B, so the node is nearer the larger rotor. The amplitude ratio is θ_A/θ_B = −I_B/I_A.

Three-rotor system. Rotors I_A, I_B, I_C joined by shafts of stiffness k₁ (A–B) and k₂ (B–C). Apart from ω = 0 there are two natural frequencies: a single-node mode (lower frequency; the two end rotors move in opposite directions) and a two-node mode (higher frequency; the middle rotor moves opposite to both ends). They come from the frequency equation below. For a symmetric system (I_A = I_C, k₁ = k₂) the single-node mode has its node at the middle rotor, which stays still, and ω = √(k₁/I_A).

Torsionally equivalent shaft. A stepped shaft (lengths l₁, l₂, l₃ at diameters d₁, d₂, d₃) is replaced by a uniform shaft of diameter d with the same stiffness: l_eq = l₁·(d/d₁)⁴ + l₂·(d/d₂)⁴ + l₃·(d/d₃)⁴. Thin, long sections dominate.

Geared systems. For a rotor or shaft on a second shaft geared at ratio G = (speed of second shaft)/(speed of reference shaft), the equivalent inertia on the reference shaft is G²·I and the equivalent stiffness is G²·k. This reduces a geared drive line to a single-shaft model.

Engine applications. A multi-cylinder crankshaft is modelled as a row of rotors (each crank throw, the pulley and the flywheel). The flywheel is large, so the node of the first mode is near it and the free (pulley) end twists most; that is where the torsional damper goes.

Formulas

k_t = G·J / L and J = π·d⁴ / 32

  • k_t: torsional stiffness (N·m/rad); G: shear modulus (Pa); J: polar second moment of area (m⁴); L: shaft length (m); d: diameter (m).

ω_n = √(k_t / I) (single rotor, other shaft end fixed)

  • I: mass moment of inertia of the rotor (kg·m²); for a solid disc I = m·R²/2.

ω_n = √(k_t·(I_A + I_B) / (I_A·I_B)) (two-rotor, free-free)

l_A = L·I_B / (I_A + I_B) and ω_n = √(G·J / (l_A·I_A)) = √(G·J / (l_B·I_B))

  • l_A, l_B: distances of the node from rotors A and B (m).

I_A·I_B·I_C·ω⁴ − [k₁·I_C·(I_A + I_B) + k₂·I_A·(I_B + I_C)]·ω² + k₁·k₂·(I_A + I_B + I_C) = 0 (three-rotor frequency equation)

  • k₁: stiffness of the A–B shaft; k₂: stiffness of the B–C shaft (N·m/rad). Solve as a quadratic in ω².

l_eq = l₁·(d/d₁)⁴ + l₂·(d/d₂)⁴ + …

  • Equivalent length of a stepped shaft at reference diameter d (m).

I_eq = G²·I and k_eq = G²·k

  • Referring a geared rotor and its shaft to the reference shaft; G = speed of the geared shaft / speed of the reference shaft.

Worked examples

Example 1 (standard: two-rotor system). A steel shaft (G = 80 GPa), 50 mm in diameter and 1 m long, carries rotors of 2 kg·m² and 8 kg·m² at its ends. Find the torsional natural frequency and the node position.

  1. J = π × 0.05⁴ / 32 = 6.136 × 10⁻⁷ m⁴.
  2. k_t = G·J / L = 80 × 10⁹ × 6.136 × 10⁻⁷ / 1 = 49 090 N·m/rad.
  3. ω_n = √(49 090 × (2 + 8) / (2 × 8)) = √30 680 = 175.2 rad/s, f_n = 27.9 Hz.
  4. Node from the 2 kg·m² rotor: l_A = 1 × 8 / 10 = 0.8 m (0.2 m from the larger rotor).
  5. Check: √(G·J / (l_A·I_A)) = √(49 090 / (0.8 × 2)) = √30 680 = 175.2 rad/s.
  6. Amplitude ratio θ_A/θ_B = −8/2 = −4: the small rotor swings four times as far, in the opposite direction.

Example 2 (GATE level: three-rotor system). Three rotors of 5, 10 and 15 kg·m² are joined by two shafts, each of torsional stiffness 800 N·m/rad. Find the two natural frequencies and the mode shape of the first.

  1. Coefficients: a = 5 × 10 × 15 = 750; b = 800 × 15 × (5 + 10) + 800 × 5 × (10 + 15) = 180 000 + 100 000 = 280 000; c = 800 × 800 × 30 = 19 200 000.
  2. 750·ω⁴ − 280 000·ω² + 19 200 000 = 0, so ω² = [280 000 ± √(280 000² − 4 × 750 × 19 200 000)] / 1500 = [280 000 ± √(7.840 × 10¹⁰ − 5.760 × 10¹⁰)] / 1500 = [280 000 ± 144 222] / 1500.
  3. ω₁² = 90.52, ω₁ = 9.51 rad/s; ω₂² = 282.81, ω₂ = 16.82 rad/s.
  4. Mode 1 with θ_A = 1: from rotor A, k₁(θ_A − θ_B) = I_A·ω²·θ_A, so θ_B = 1 − 5 × 90.52 / 800 = 0.434. From rotor C, k₂(θ_C − θ_B) = I_C·ω²·θ_C, so θ_C = k₂·θ_B / (k₂ − I_C·ω²) = 800 × 0.434 / (800 − 1357.8) = −0.623.
  5. Mode shape 1 : 0.434 : −0.623, a single node between B and C.

Example 3 (equivalent shaft). A shaft is 0.4 m long at 60 mm diameter and 0.6 m long at 40 mm diameter. Equivalent length at 40 mm: l_eq = 0.6 + 0.4 × (40/60)⁴ = 0.6 + 0.4 × 0.1975 = 0.679 m.

Common mistakes

  • Calculating √(k/I) for each rotor of a two-rotor system and calling those its natural frequencies. A free-free two-rotor system has one non-zero natural frequency involving both inertias.
  • Putting the node nearer the smaller rotor. It is nearer the larger one.
  • Using the second moment of area I = πd⁴/64 instead of the polar moment J = πd⁴/32 in k_t.
  • Forgetting that the stiffness of a stepped shaft depends on d⁴, so the thin section controls it.
  • Referring a geared inertia with G instead of G².
  • Dropping the zero-frequency (rigid-body) solution and then miscounting the modes of a three-rotor system.

For GATE ME

Expect the natural frequency and node location of a two-rotor system, the frequency equation of a three-rotor system, equivalent length of stepped shafts and referred inertia in geared systems. Practise Example 1 with different inertia ratios and check every answer with √(GJ/(l·I)) from either side of the node.

Quick check

  1. Rotor of 4 kg·m² on a shaft of 400 N·m/rad fixed at the other end. ω_n?
  2. Two-rotor system with I_A = I_B. Where is the node?
  3. How many non-zero natural frequencies does a three-rotor free-free system have?
  4. Doubling the diameter of a shaft changes its torsional stiffness by what factor?
  5. k = 1500 N·m/rad, I_A = 25 and I_B = 35 kg·m². ω_n?

Answers: 1. 10 rad/s. 2. At the middle of the shaft. 3. Two. 4. 16 times. 5. √(1500 × 60 / 875) = 10.14 rad/s.

Interview questions

All Theory of Machines and Vibrations interview questions

Try answering each one aloud before you open it.

  1. 1.What is torsional vibration in mechanical systems?Concept

    Torsional vibration refers to the angular oscillation of an object, typically a shaft, around its axis of rotation. It occurs when the object is subjected to a torque that causes it to twist back and forth. This type of vibration is significant in rotating machinery as it can lead to fatigue failure if not properly controlled.

  2. 2.Explain the concept of a two-rotor system in torsional vibration analysis.Concept

    A two-rotor system is two discs of inertia I_A and I_B on a shaft of torsional stiffness k_t, with neither end fixed. Besides the zero-frequency rigid-body rotation it has one torsional natural frequency, ω_n = √(k_t(I_A + I_B)/(I_A·I_B)), in which the rotors twist in opposite directions about a stationary node. The node lies nearer the larger rotor (I_A·l_A = I_B·l_B), and each side of the node behaves like a single rotor on a fixed shaft with the same frequency. An engine with its flywheel and a driven load is often first modelled this way.

  3. 3.How does a three-rotor system differ from a two-rotor system in terms of torsional vibration?Concept

    A three-rotor system has three inertias joined by two shafts, so besides the zero-frequency rigid-body motion it has two torsional natural frequencies, from a quadratic in ω². The lower is a single-node mode, in which the two end rotors swing in opposite directions; the higher is a two-node mode, in which the middle rotor swings opposite to both ends. A two-rotor system has only one non-zero frequency and one node.

  4. 4.Why is it important to analyze torsional vibrations in automotive drive shafts?Application

    The drive line, from engine flywheel through clutch, gearbox, propeller shaft and axles to the wheels, is a torsional spring-mass system with natural frequencies that engine firing pulses can excite. Resonance causes gear rattle, boom, clunk and high alternating shear stress in shafts and joints, which can lead to fatigue failure. Engineers tune it with clutch damper springs, dual-mass flywheels and shaft stiffness so that the main resonances fall below the normal operating range or are well damped.

  5. 5.What happens if the natural frequency of a rotor system coincides with the operating frequency?Application

    If the natural frequency of a rotor system coincides with the operating frequency, resonance occurs. This can lead to large amplitude vibrations, which may cause excessive stress and potential failure of the system components. It is essential to design the system to avoid such conditions or to implement damping mechanisms to mitigate the effects.

  6. 6.How can damping be introduced in a torsional vibration system?Application

    In engines, torsional damping is added mainly at the free (pulley) end of the crankshaft, where the twist amplitude is largest. Common devices are the rubber-bonded harmonic damper (an inertia ring on a rubber layer, which tunes and damps the crankshaft mode), the viscous damper (a free inertia ring in a casing filled with silicone fluid) and friction dampers. In the drive line, clutch-disc damper springs with friction washers and dual-mass flywheels serve the same purpose.

  7. 7.What is the role of a torsional damper in an engine's crankshaft?Application

    The crankshaft's first torsional mode has its node near the heavy flywheel, so the front (pulley) end twists most, and firing harmonics can excite it within the speed range. A torsional damper on the pulley, usually an inertia ring bonded by rubber or floating in viscous fluid, is tuned near that mode and dissipates energy, cutting the resonant twist amplitude. That reduces alternating shear stress in the crankshaft (preventing fatigue cracks) and protects timing drives and accessories.

  8. 8.Two rotors with mass moments of inertia 10 kg·m² and 15 kg·m² are mounted at the ends of a shaft of torsional stiffness 200 N·m/rad. Calculate the torsional natural frequency.Numerical

    For a free-free two-rotor system, ω_n = √(k(I₁ + I₂)/(I₁·I₂)) = √(200 × 25 / 150) = √33.33 = 5.77 rad/s, or 0.92 Hz. The node is at L·I₂/(I₁ + I₂) = 0.6L from the 10 kg·m² rotor. Torsional problems need mass moments of inertia in kg·m², not masses in kg.

  9. 9.For a three-rotor system, if one rotor is removed, how does it affect the system's natural frequencies?Application

    Removing a rotor leaves a two-rotor system with only one non-zero natural frequency instead of two, and that frequency is generally different from either of the original ones. The remaining shafts may also join into one longer, softer shaft, which lowers stiffness. Because the inertia and stiffness distribution changes, the node position and the speeds at which firing harmonics cause resonance move, so the critical speed map must be recalculated.

  10. 10.Determine the effect of increasing shaft stiffness on the natural frequency of a two-rotor system.Application

    For a two-rotor system ω_n = √(k_t(I_A + I_B)/(I_A·I_B)), so the natural frequency rises with the square root of shaft stiffness: doubling k_t raises it by 41%. Since k_t = GJ/L with J = πd⁴/32, a small increase in diameter has a large effect (10% more diameter gives 46% more stiffness and about 21% higher frequency). The node position does not change, because it depends only on the inertia ratio.

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