Lezione 1 Meccanica applicata alle macchine

20/12

We start the lesson with a diagram.

These are all auto-receptance.

Ideal 1/x cross-receptance.

To sum up:

• For n-DOF undamped MDOF system we will get: α__xx (auto-receptance), we will have M resonances and M-1 anti-reson.
• α__xy (cross-receptance), it is M resonances and M-1 anti-resonance or saddle point.

The number of spikes increases moving the measurement point further from the excitation.

20/12

We start the lesson with a diagram.

These are all auto-receptance.

Ideal 1/x cross-receptance.

To sum up:

1. For n-DOF un-damped MDOF system we will get: d_xx (auto-receptance), we will have M resonances and M-1 anti-reson.
2. d_xx (cross-receptance), "", "", M resonances and M-1 anti-resonance or saddle point.

The number of spikes increases moving the measurement point further from the excitation.

Dynamic absorber of tuned mass damper

Let's start with a simple system [SDOF system excited by an harmonic force F_ext]₊₁ → System response (steady state).

CQ = ₀  if the excitation Ω<ω₁.

CQ = _π  if the excitation Ω>ω₁.

ω₁ = √(k₁/_M1)  (nat. freq.).

FRF: transmissibility.

Force transmitted to the ground through the spring

₁/F_0F0/0^{k₁ X₀}/_{k1 - M1 Ω²} = 1/_{1 - Α²/ω²}Ω.

So the design goal for the tuned mass damper:

• To suppress the resonance of the system.
• To minimize the force transmitted to the ground when Ω = ω₁ to minimize the vibration of the mass.

Dynamic absorber

A solution could be: Dynamic Absorber.

Add a spring & mass system.

1 &half; DOF.

Modify Freq. Resp. Function (FRF) from one Resonance (SDOF) to two Resonances (1+1 Anti-resonance) of SDOF.

[I can set my Anti-resonance when FRF of SDOF has his maximum.]

Set the A-R of the 2 DOF at the resonance freq. of S.D.O.F.

(WAR)_2DOF = (W_N)_SDOF = W₂.

EOH[m₁ 0][m₂][k₁ + k₂ -k₂][-k₂ k₂].

Response to F(t) = F₀cos(wt) = {x(t)} = {X₀}cos(wt).

Evaluate the receptance matrix

[α] = ([K] - ω²[C])^-1.

Note that for 2x2 matrix the matrix inverse is A = [^{d bc a}] A^-1 = 1/det. [^{d -b-c a}].

[α]₂ = 1/det([K_adj]).

det. = (k₁k₂ - ω²₁m₁ - ω²₂).

Impose that det(k - ω²M) = 0 (i.e., mod. freq. of the mov system).

det. = ω_max = (ω²m₂ + ω₂)²d_a.

We can impose.

1. _X1 = 0, _α1 = 0, F₀ = ^{K₂ m₂ω2}⁄_det(Kred) F₀ - Ω_AR = 0, so max. σ = 0! (because of that det is null) The den is 0 → det = 0 (what response of eqy_of f) why? because the den is common to every oc (offenar REF).

→ vibration amplitude of m₂(2) X₂ = 0, _α2 = 0, F₀ = K₁ F₀det(K_red) → many σ = 0 → no ANTIRESONANCE = we expect a Saddle Point for m₂.

So, let's try to solve problem Ⓡ many σ = 0 K₂ = K₂ m₂ ω_AR² = 0 Ω_AR = ^K₂⁄_m2.

• Ω_AR can be tuned freely acting only on the added mass and spring (m₂,k₂).
• Impose that Ω_ARε, ωε → K₂ = m₂ ω_i2² = 0 ^K₂⁄_m2→.

Now let's plot what we have done.

We have to choose the two parameters so ^K₂⁄_m2^K₁⁄_m1. That's the design guideline for dynamic absorber!

Saddle point

i) From ⓑ we fix the ratio _K2/_m1 ^{= m₂} suppression of resonance peak of original system (SDOF).

From ⓑ fix the absolute value of m₂ response.

Choose K₂ (or m₂) to comply with max oscillation amplitude of TMD → Increase ω₂ to lower X₂.

Free response of a MDOF system with generic (non-proportional damping)

[_M] {q̈} + [_C] {q̇} + [_K] {q} = {f(t)} [Eqn of the System].

Conversion of the system equation into the state-space representations

{ẋ(t)}= A{x(t)} + B{u(t)} ^{State-Space Form of a Liner (L.T.I) Dynamic System}.

{y(t)}= C{x(t)} + D{u(t)} x → state vector u → input vector y → output vector.

A, B, C, D → state-space matrixes of the dynamic system...

x_z = {q̇}_z ^{i.e q̇_z = [q̇]}.

{ ẋ₁ } = [ O ] { x₁ } + [ I ] { x₂ }.

{ ẋ₂ } = [ -M^-1][ K ] { x₁ } + [ -M^-1][ C ] { x₂ } + { I^{[sub]F] u̇₁ }}.

x₁ = x x₂ = x x₁ = [ M^-1[_K][C] x₂] x₁ = FA matrix → Singular (System) Matrix n x n, Input n x m, Non Dynamic Matrix.

The eigenvalues of A contains the information of natural frequency and damping factors for the damped system.

Free response: {_x(t)} = [_ф] e_{t λ} λ{_x}.

The solution has the exponential form {_x(t)} = e^λt {_x} {_x}.

Substituting we get the e.v problem: A{_x} = λ{_x}.