Lezione 2 Meccanica applicata alle macchine

Orthogonality property of eigenvectors w.r.t. [K] & [Cm] matrices

Hypothesis

{Cm}, {K}: symmetric and real ⟹ eigenvalues and eigenvectors are real.

Positive definite or positive semi-definite ⟹ eigenvalues are not negative (⟶0).

Theorem

{t_s}^T [Cm] {t_z} = 0, {t_s}^T [K] {t_z} = 0 (s ≠ z ≠ s).

Proof

Consider EVP: ω²{Cm}{x_z}=[K]{x_z}.

Consider two generic e-v ω_s, ω_z and the corresponding e-vectors {t_s}, {t_z}.

1. ω_z² [Cm] {t_z} = [K] {t_z} — mode z.
2. ω_s² [Cm] {t_s} = [K] {t_s} — mode s.

So we pre-multiply the first eq. by t_s^T and second one {t_z}^T.

{t_s}^T ω_z² [Cm] {t_z} = {t_s}^T [K] {t_z}.

{t_z}^T ω_s² [Cm] {t_s} = {t_z}^T [K] {t_s} (✶).

⟵ Transpose these eq. remembering that (CA)^T {t_s} [K] = [B]^-T [A]^T.

ω_s² {t_s}^T [Cm] {t_z} = {t_s}^T [K] {t_z}.

ω_z² {t_s}^T [Cm] {t_z} = {t_s}^T [K] {t_z}.

(ω_s² - ω_z²) {t_s}^T [Cm] {t_z} = 0.

Since \([cm]\) is P.D. \quad modal mass of the iTH mode.

Repeating a similar procedure we can get the same result for (k) matrix:

\[(\omega_i^2 - \omega_j^2) \left\{ \mathbf{\psi_i}^T \right\} [k] \left\{ \mathbf{\psi_j} \right\} = 0 \]

• If \(\omega_i \ne \omega_j \Rightarrow \left\{ \mathbf{\psi_j}^T \right\} [k] \left\{ \mathbf{\psi_i} \right\} = 0 \quad \Rightarrow \text{eigenvectors are K-orthogonal}\).
• If \(\omega_i = \omega_j \Rightarrow \left\{ \mathbf{\psi_i}^T \right\} [k] \left\{ \mathbf{\psi_i} \right\} \ge 0\) because [k] is P.S.D. \quad modal stiffness of the iTH mode.

\[\omega_i^2 \left\{ \mathbf{\psi_i}^T \right\} [cm] \left\{ \mathbf{\psi_i} \right\} = \left\{ \mathbf{\psi_i}^T \right\} [k] \left\{ \mathbf{\psi_i} \right\} \]

\(M_{m} \quad K_m\).

\[ \Rightarrow K_{r} = M_{r} \omega_i^2 \]

By using the modal matrix \([Y] = [\mathbf{y_1}, \ldots, \mathbf{y_n}]\), the result would be:

\[\left[ \mathbf{y}^T \right] (cm) \left[ \mathbf{y} \right] = \text{diag} (M_n)\]

\(M_{11}, \quad , \quad M_{nn}\).

Modal mass

Physical mass.

Modal mass matrix.

C[Nᵀ][K][N]{Y} = diag (kᵣ).

Mass stiffness matrix.

Theory of expansion

In particular, allows to express any motion of the m-DOF of the system as a linear combination of the modal (eigen) vectors.

It is possible of the mode vectors as linearly independent.

i.e. they are a basis of the configuration space.

Theorem

TH: Thereof, {Y} displacement = Σ Cm{Yₘ} vector m=1.

Proof by contradiction

Proof by contradiction: assume that eigenvectors are linearly dependent:

C₁{Y₁} + C₂{Y₂} + ... + Cₘ{Yₘ} = Σ Cₘ{Yₘ} = 0.

Where C are not all zero.

Pre-multiply by {Y₃ᵀ}[Cₘ].

Σ Cₘ {Y₃ᵀ}[Cₘ]{Yₘ} = 0.

From the this product is always null m=1 m=1 except r=s.

Cs: {Yₛᵀ}[Cₘ]{Yₘ} = 0 → Cs = 0.

m≠s ≠p.

Repeating the procedure for all s ≠ 1:

m ⊘ Cs = 0 ∀s this contradict the initial hypothesis → the eigenvectors are linearly independent and any vector displacement {uₛ} can be written as a linear combination of the e-vectors.

Modal participation factor

{U} = Σ_r=1^m C_r {Φ}_r.

Pre-multiply by {Ψ}_s^T [CM].

{Ψ}_s^T [CM] {U} = Σ_r=1^m C_r {Ψ}_s^T [CM] {Φ}_r.

= Σ_r=1^m C_r {Ψ}_s^T [CM] {Φ}_r.

{Ψ}_s^T [CM] {Φ}_r = 0 if s ≠ r + m_r if s = r → form m - orthogonality.

{C}_r = {{Ψ}_s^T [CM] {U}} /m_s.

s = 1, 2, … , m.

EX if {U} = {Φ}_r = {Φ}_j → C_r: = 0 if i ≠ j.

C_r: ≠ 0 if i = j.

Normalization of eigenvectors

We can do it because eigenvectors are defined apart from a multiplicative constant {Ψ}_r → k {Ψ}_r.

Set each first element equal to 1 ({Ψ}₁₁ = {Ψ}₂₂ = {Ψ}_nn = 1) so →[ 1 1 … 1].

| | |.

Set the maximum element of each eigenvector to 1.

X = [ 1 1 ].

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Set each modal mass equal to 1 [m - normalization].

In general m_s = {Φ}_r^T [CM] {Φ}_r ≠ 1.

So we consider another eigenvector {Θ}_r = {Φ}_r / √m_r.

{Θ}_r^T [CM] {Φ}_s = m_s.

{Θ}_s^T [CM] {Θ}_s = 1 ( = m_s).

{Φ} is the modal matrix m-normalized.

[Φ]ᵀ [Cm] [Φ] = [CI].

The modal mass matrix coincide with identity matrix.

[Φ]ᵀ [Ck] [Φ] = diag (ω²).

Modal stiffness matrix.

Decomposing the EOM of M-DOF system using modal analysis

EOM [Cm] {ẍ} + [K] {x} = {} → modal analysis → [Φᵀ ϒ].

diag (mᵢ), diag (kᵢ), ωᵢ² = kᵢ/mᵢ.

Modal matrix.

Modal frequency squared.

{x(t)} = [Φ] {(ξ) }.

Physical coordinate.

Modal coordinate.

Coordinate transformation matrix.

Modal matrix.

Coordinate transformation: from physical {x} to modal {x} coordinate and viceversa.

We said that [Φᵀ] is const → {ẋ(t)} = [Φᵀ] {η̇(t)}.

So replacing: [Cm] [Φ] {η̈} + [Ck] [Φ] {η} = {}.

We pre-multiply the transpose of modal matrix:

[Φᵀ] [Cm] [Φ] {η̈} + [Φᵀ] [Ck] [Φ] {η} = {Φᵀ} {}.

diag (mᵢ) diag (kᵢ).

Thanks to m-orthogonality and k-orthogonality we get a set of n-independent equations in this type:

mᵢ η̈ᵢ + kᵢ ηᵢ = ᵢ.

Cp η̈₂ + ω₂² η₂ = 0 → where i = 4,...,n.

q̈₁ + ω₁²q₁ = 0.

q̈_n + ω_n²q_n = 0.

q are independent and uncoupled.

Which means that the system is elastically and inertially decoupled by modal stiffness matrix is diagonal to modal mass matrix is diagonal.

Solve in modal coordinate and go back into physical through modal transformation.

[x] = [C] [q].

Free response with linear viscous damping

[C][ẍ] + [C]^T[C][ẋ] + [C]^T[K][ẋ] = [p] → modal transf.

[C]^T[C_m][q̈] + [C]^T[C][C_p][q̇] + [C]^T[C_k][C_p][q] = Φ.

diag(m_n) diag(k_k).

In general is not diagonal for generic damping.

Proportional damping

For the special case of "Proportional Damping" → diagonal modal damping matrix.

[C_p] = α[M] + β[K_k].

Constant coefficients.

[C]^T[C_p][C] = [C]^T(α[M] + β[K_k]) [C] = [C]^T[C_m][C] α + [C]^T[K_k][C]^T β = α diag(m_i) + β diag(k_e).

e.o.m. decoupled e.o.m. in modal coordinates.

m_i q̈ + (αm_i + βk_n) q̇ + k_n q = Φ.

q̈ + 2ζ₁ω₁q̇ + ω₁²q₁ = Φ.

That reminds the EOM of S.D.O.F. response.

Free response with prop. damping

2 methods:

1. Inverse modal transformation.
2. Modal approach.

1) Inverse modal transformation

{x₀} = [Φ] x(t=0)   i.c. defined in physical coordinates.

{v₀} = [x](t=0).

We know that the free resp. of the system in modal coordinates (underdamped system) is:

η(t) = η₂₀ e^-ξ₂ω₂t sin (ω_d2 t + θ₂).

= (A₂ cos(ω_d2 t) + B₂ sin(ω_d2 t)) ⋅ e^-ξ₂ω₂t   (✗) (✗).

Initial condition in modal coordinates are needed.

Direct modal transft. is: {x} = [Φ]{z} ⇒ inverse modal transf.

{y₀}=[Φ]^T{x₀}.

{ẏ₀}=[Φ]^T{ẋ₀}.