Multi-degree-of-freedom systems and modal analysis

Why it matters

A finite-element model of a wing or fuselage has thousands of degrees of freedom, yet its dynamic behaviour in flight is governed by a handful of low modes: first wing bending, first torsion, fuselage bending, control-surface rotation. Modal analysis turns the large coupled set of equations into independent single-degree-of-freedom equations, one per mode, so loads, gust response and flutter can be computed with a small, accurate model. It is also how test data from ground vibration tests are compared with the analysis.

Key ideas

Equations of motion. For n coordinates {x}, [M]·{ẍ} + [C]·{ẋ} + [K]·{x} = {F(t)}. [M] and [K] are n × n, symmetric; [M] is positive definite and [K] is positive definite for a restrained structure (positive semi-definite if rigid-body motion is possible). They come from lumped masses and springs, from influence coefficients (the flexibility matrix [a] = [K]⁻¹) or from finite elements.

Eigenvalue problem. Undamped free vibration gives ([K] − ω²·[M])·{φ} = {0}. There are n eigenvalues ω_r² and n eigenvectors (mode shapes) {φ_r}. Computer methods (subspace iteration, Lanczos) find only the lowest modes needed; by hand you solve det([K] − ω²·[M]) = 0 for 2 or 3 DOF.

Orthogonality and modal properties. For r ≠ s: {φ_r}ᵀ·[M]·{φ_s} = 0 and {φ_r}ᵀ·[K]·{φ_s} = 0. For r = s these products define the modal (generalised) mass M_r = {φ_r}ᵀ·[M]·{φ_r} and modal stiffness K_r = {φ_r}ᵀ·[K]·{φ_r}, with ω_r² = K_r/M_r. Scaling each mode so M_r = 1 gives mass-normalised modes, for which [Φ]ᵀ·[M]·[Φ] = [I] and [Φ]ᵀ·[K]·[Φ] = diag(ω_r²).

Modal coordinates. Write {x} = [Φ]·{q}, where [Φ] has the mode shapes as columns and {q} are modal coordinates. Pre-multiplying by [Φ]ᵀ decouples the equations:

M_r·q̈_r + C_r·q̇_r + K_r·q_r = f_r(t), with f_r = {φ_r}ᵀ·{F}

Each line is an SDOF oscillator, solved with the earlier methods (free vibration, harmonic forcing, Duhamel), and the physical response is rebuilt by superposition. The modal force f_r shows that a force applied at a node of mode r does not excite that mode.

Initial conditions. q_r(0) = {φ_r}ᵀ·[M]·{x(0)}/M_r and similarly for velocities.

Damping. The equations decouple only if [Φ]ᵀ·[C]·[Φ] is diagonal. Proportional (Rayleigh) damping [C] = α·[M] + β·[K] does this, and then ζ_r = α/(2ω_r) + β·ω_r/2. In practice a modal damping ratio for each mode is taken from test (typically 1–3 % for metallic airframes).

Truncation. Usually only the lowest few modes are kept, because loads in the low-frequency range excite them most. Keep enough modes to cover the frequency range of the excitation with margin; for stresses or local loads more modes, or a static correction, may be needed.

Links. This is the generalisation of the 2-DOF topic. Rayleigh–Ritz (next topic) gives approximate low modes cheaply. In aeroelasticity the structural modes are used as generalised coordinates and the aerodynamic forces are projected onto them as generalised aerodynamic forces.

Formulas

[M]·{ẍ} + [C]·{ẋ} + [K]·{x} = {F(t)} — MDOF equations; [M] (kg), [C] (N·s/m), [K] (N/m).

det([K] − ω²·[M]) = 0, ([K] − ω_r²·[M])·{φ_r} = {0} — eigenvalue problem.

M_r = {φ_r}ᵀ·[M]·{φ_r}, K_r = {φ_r}ᵀ·[K]·{φ_r}, ω_r² = K_r/M_r — modal mass, stiffness and frequency.

{x} = [Φ]·{q} — modal superposition; f_r = {φ_r}ᵀ·{F} — modal force (N, for unit-scaled modes).

M_r·q̈_r + 2ζ_r·ω_r·M_r·q̇_r + K_r·q_r = f_r — decoupled modal equation.

q_r(0) = {φ_r}ᵀ·[M]·{x(0)}/M_r — modal initial condition.

[C] = α·[M] + β·[K], ζ_r = α/(2ω_r) + β·ω_r/2 — Rayleigh damping; α (1/s), β (s).

Worked examples

Example 1 (standard). M = diag(2, 1) kg and K = [[6, −2], [−2, 4]] kN/m. Find ω₁, ω₂, the mode shapes and the modal masses and stiffnesses.

1. det([K] − λ[M]) = (6000 − 2λ)(4000 − λ) − 2000² = 0 → 2λ² − 14 000λ + 20 × 10⁶ = 0 → λ² − 7000λ + 10⁷ = 0.
2. λ = 2000 and 5000 (rad/s)², so ω₁ = 44.72 rad/s (7.12 Hz) and ω₂ = 70.71 rad/s (11.25 Hz).
3. From row 1, φ₂/φ₁ = (6000 − 2λ)/2000: mode 1, {1, 1}; mode 2, {1, −2}.
4. Modal masses: M₁ = 2 × 1 + 1 × 1 = 3 kg; M₂ = 2 × 1 + 1 × 4 = 6 kg.
5. Modal stiffnesses: [K]{1, 1} = {4000, 2000}, so K₁ = 6000 N/m; [K]{1, −2} = {10 000, −10 000}, so K₂ = 10 000 + 20 000 = 30 000 N/m.
6. Check: K₁/M₁ = 2000 and K₂/M₂ = 5000, matching λ. Cross term {1, 1}·[M]·{1, −2} = 2 − 2 = 0.

Answer: ω₁ = 44.7 rad/s, {1, 1}, M₁ = 3 kg; ω₂ = 70.7 rad/s, {1, −2}, M₂ = 6 kg.

Example 2 (GATE level). For the same system, mass 1 is displaced 1 mm and both masses are released from rest. Write x(t), and find α and β for 2 % damping in both modes.

1. q_r(0) = {φ_r}ᵀ·[M]·{x(0)}/M_r with x(0) = {1, 0} mm: [M]{x(0)} = {2, 0}. q₁(0) = 2/3 = 0.667 mm; q₂(0) = 2/6 = 0.333 mm.
2. Released from rest, each modal coordinate is q_r(0)·cos(ω_r·t): x(t) = 0.667·{1, 1}·cos(44.72t) + 0.333·{1, −2}·cos(70.71t) mm.
3. Check t = 0: x₁ = 0.667 + 0.333 = 1 mm, x₂ = 0.667 − 0.667 = 0. Correct.
4. Rayleigh damping: solving ζ = α/(2ω) + β·ω/2 at both frequencies gives β = 2ζ/(ω₁ + ω₂) = 0.04/115.43 = 3.47 × 10⁻⁴ s and α = 2ζ·ω₁·ω₂/(ω₁ + ω₂) = 0.04 × 3162.3/115.43 = 1.096 s⁻¹.

Answer: x(t) as above (not periodic, since ω₂/ω₁ = √2.5 is irrational); α ≈ 1.10 s⁻¹, β ≈ 3.47 × 10⁻⁴ s.

Common mistakes

• Using {φ_r}ᵀ·{x(0)} instead of {φ_r}ᵀ·[M]·{x(0)}/M_r for modal initial conditions.
• Forgetting that mode shapes are only defined up to a scale factor; modal mass and modal force depend on the scaling, but the physical response does not.
• Assuming any damping matrix decouples in modal coordinates; only proportional (or classical) damping does.
• Mixing kN/m with kg, which shifts every frequency by √1000.
• Truncating to too few modes for a load with high-frequency content.
• Labelling frequencies out of order; ω₁ is always the lowest.

For GATE AE

Expect 2- or 3-DOF frequency calculations from given [M] and [K], mode shapes and orthogonality checks, modal mass and stiffness, response from initial conditions by modal superposition, and Rayleigh damping coefficients. Conceptual questions test why modal coordinates decouple the equations and when they do not.

Quick check

1. What property of mode shapes makes modal decoupling possible?
2. A mode {1, 2} with M = diag(1, 3) kg: what is its modal mass?
3. If a force acts at a node of mode r, what is f_r?
4. Write ζ_r for Rayleigh damping.
5. For M = diag(2, 1) kg and K = [[4, 0], [0, 3]] N/m, what is ω₁?

Answers: 1. orthogonality with respect to [M] and [K]; 2. 1 + 12 = 13 kg; 3. zero; 4. ζ_r = α/(2ω_r) + β·ω_r/2; 5. √2 = 1.41 rad/s.