Lezione 3 Meccanica applicata alle macchine

MDOF system dynamics (part 2)

Linear and natural mechanical system

Kinetic energy can be expressed as a quadratic form of velocities.

Let's introduce a simple M.D.O.F. system (the mass can translate without friction).

The above is a particular 2 D.O.F. system.

m = 2;

Generalised coordinate vector q₁, x₁ absolute displacement q₂, y₁ relative displacement of mass 1 with respect to (w.r.t.) mass 2.

Different sets of coordinates are available, we choose the first one {q₁}.

1. F.B.D. (as many as degree of freedom of the system)
2. F.B.D of mass 1: m₁ẍ₁ + c₁ẋ₁ + k₁(x₁ - x₂) + k₂x₁ = F₁
3. F.B.D of mass 2: m₂ẍ₂ + c₂ẋ₂ + c₃(ẋ₂ - ẋ₁) + k₃(x₂ - x₁) - k₂(x₁ - x₂) = F₂

MDOF system dynamics (part 2)

Linear and natural mechanical system

Kinetic energy can be expressed as a quadratic form of velocities.

Let's introduce a simple M.D.O.F. system (the mass can translate without friction).

(This is a particular 2 D.O.F system).

m = 2;

Generalized coordinate vector

{q₁} = x₁ absolute displacement x₂ {q₂} = γ₁γ₂ relative displacement of mass 1 with respect to (w.r.t) mass 2.

Different sets of coordinates are available, we choose the first one {q₁}.

1. F.B.D. (as many as degree of freedom of the system)

x₁, ẋ₁, ẍ₁

x₂, ẋ₂, ẍ₂

3. EOM dynamic balance of each mass (Newton's second law of motion)

m₁ ẍ₁ + c₁ ẋ₁ + c₂ (ẋ₁-ẋ₂) + k₁ x₁ + k₂ (x₁-x₂) = F₁

m₂ ẍ₂ + c₂ ẋ₂ = c₂ (ẋ₁-ẋ₂) + k₂ x₂ = k₂ (x₁-x₂) = F₂

4 set of seen ...?

Rearrange equations

Rearrange equations considering: Decreasing order of derivatives (Ẋ̇, Ẋ, X) From 1^st to last column of generalized vector (Ẋ̇₁, Ẋ̇₂, Ẋ₁, Ẋ₂, X₁, X₂).

Convention auto à matrix form.

M₁ Ẋ̇₁ + (C₂ + C₁) Ẋ₁ + (K₂ + K₁) X₁ - K₁ X₂ = F₁

M₂ Ẋ̇₂ + C₂ Ẋ₂ + (C₂ + C₁) Ẋ₂ - K₁ X₂ + (K₂ X₁ = -F₂

{q} = {X₁, X₂} → Generalized coordinate vector

{q̇} = {Ẋ₁, Ẋ₂} → Generalized velocity vector

{q̇̇} = ∑ {Ẋ̇₁, Ẋ̇₂} → Generalized acceleration vector

[M] _2x2 [ {Ẋ̇₁} ] + [C] ^2x2 [ {Ẋ₂} ] + [K] ^2x2 [ {X₁} ] = [F] ^2x1 Generated force vector

EOM for m-dof systems

[M] {q̇̇} + [C] {q̇} + [K] {q} = {F}

Particular cases: Unforced system: {F} = 0 → Free undamped free response.

Undamping system: [C] = [ ]→ So we study: Free undamped false response.

[M] {q̇̇} + [C] {q̇} + [K] {q} = 0 → [M] {q̇̇} + [K] {q} = ∅

Properties of [M] and [K] matrices

Before we start let's talk about some properties of [M] and [K] matrices: Constant and the coefficient.

Symmetric matrix (2) [have the same transpose matrix (corresponds to the initial one)].

We can always get symmetrical matrices for a normal mechanical system.

With Lagrange's approach always symmetric.

With Newton's approach we can get any symmetrical matrices.

(P.D.) (P.S.D.) Positive definite and positive semi-definite.

① A square matrix [A] is P.D. if V{ξ}∈ℜⁿ and {ξ}≠{0} and {ξ}^T[A]{ξ} > 0.

② A square matrix [B] is positive semi-definite if V{ξ}^T[ξ} and {ξ}∈ℜⁿ {ξ}^T[B]{ξ} ≥ 0.

So we can say that [M] is P.D. and [K] is P.S.D.

Example (considering a system of N masses connected with N springs).

Kinetic energy of system

What about the kinetic energy of system?

T = ₁/² Σ m_i v_i² and this is a basic equation of the kinetic energy.

T = ₁/² {v}^T [M] {v} ≥ 0 [matrix expression of K-energy; quadratic form of velocity].

{v} = { _{u̇1 u̇2} ... _u̇N }

[M] = diag (m₁, m₂, ..., m_N)

Potential elastic energy

V = 1/2 ∑_i=1^N k_i u_i² - can be written as: V = 1/2 {x}^T [K] {x} > 0.

2-DOF example

u = x₁ - x₂

V = 0 if {x}^T {  } 1/2 {f}{x} - translation of the system without spring elongation (T≠0 V=0) {m₁,   } {m₂,   }   ^{{x₁ } + [  k   -k  ][   ]   {x₂ }         -k ]   {x₁ }  = Ø}

V = 1/2 k (x₁-x₂)² = 1/2 {x}^T [K] {x} = 1/2 {x}^T

Sylvester criterion

Necessary and sufficient condition to have a P.D. matrix is: The determinant of the matrix and all its leading principal minors are positive.

Leading principal submatrices obtained by progressively consider the top-left (1x1, 2x2, ..., n x n) blocks of the original matrix.

Example

[m₁] = _om_o^om_o

1. det([m₁]) = m₁ > 0 ✓
2. det([m₂]) = (m₁) > 0 ✓ this two conditions tells me that [m₁] is P.D.

[K] = _-k1k2k1-k₂k₁k₁k₃

1. det([K]) = _{(k1+k2)(k1k3) - k2²} > 0 ✓ ‽ also [K] is P.D.
2. det(k₁k₂) = k₁k₂ > 0 ✓

The original system is well constrained as it does not allow a rigid body mode.

If k₁=k₂ as can unload from the system [K] become P.S.D. because det([kd]) = 0.

Free response of the undamped MDOF system

EqM _{C(m)x[g] x{+} + [K]x[g] x{-}[g]} Eq _E5

Look for synchronous solutions: all the masses of the system are moving according to a sine function of time.

[_{sin, cos}]_COnkronies {x(t)}=_{x2Qg(t)} ^{Constant non-rod vector describing the displacement pattern}

{ x(t) } = [fx₃ ] ĝ(t)

EOM[m] { x } ĝ(t) + [k] { x } g(t) - f = 0 → pre-multiply by matrix [fx₃ ]

{ fx₃ }ᵀ Cm ] { fx₃ } ĝ(t) + { fx₃ } ᵀ[fx₃ ] { fx₃ } g(t) = { fx₃ }→ = { f }

Strictly positive constant known (fx₃ } ᵀ = 0.

Subtract subtract subtract.

g ĝ = { fx₃ } ᵀ [K] } fx } - ω² ( ξ ){ fx₃ } ᵀ [m] } fx } = 0

ĝ(t) + ω² g(t) = 0

g(t) = cos(ωt+θ) & emsp;⇒ the solution is the harmonic function.

x(t) = { fx₃ } cos (ωt+θ)

{ ˙x(t) } = { fx₃ } sin (ωt+θ)

- ω² [m ] { fx₃ } cos(ωt + θ) + [k ] { fx₃ } cos(ωt + θ) = { f }

([K] - ω² [m])] { ξ } = { φ } _{eigen-value problem (E.V.P )} unknown

The solution of E.V.P. are: ω² = eigenvalues { fx₃ } = eigen vectors.

To have n non-zero solutions: det ( [[K] - ω²[m] ) = 0

_{characteristic determinat of characteristic equation}

Polynomial of the n^th degree in the variable ω² natural frequencies  { i.e. : ω₁ [rad/s]   ∀ : =        2π          H_z Per il pratico equivalente a scrivere quanto a_nω²⁽ⁿ⁺¹⁾ + a_n-1 ω^2(n-1) + … + a₁ω² + a₀ = 0