ME 3333 System and VibrationsStep by Step Guide to MDOF system

1 MDOF system

Let’s say that you have a MDOF system, you are trying to find the free response. It comes down to solvinga set of differential equations. For example, your system model may look like

[m1 00 m2

]{x (t)}+

[c1 + c2 −c2−c2 c2 + c3

]{x (t)}+

[k1 + k2 −k2−k2 k2 + k3

]{x (t)} =

[F1F2

](1)

for a mechanical system.The fluid and thermal systems from your last test (test 3) can be written as

[ρA1 00 ρA2

] [h1h2

]+

[ ρgR1

− ρgR1

− ρgR1

ρgR1+ ρg

R2

] [h1h2

]=

[0qin

]

[CA 00 CB

] [TATB

]+

[ 1RA

+ 1RB

− 1RB

− 1RB

1RA+ 1

RB

][TATB

]=

[ 1RATo

1RATo

].

The goal here is to show you how to find the free and the forced response using the Matrix Exponential,eigenvalue problem, and modal analysis.

The matrix exponential approach uses the state-space model. If we let state variables (of the mechanicalsystem) to be

x1 = x1

x2 = x2

x3 = x1

x4 = x2,

where you may recall that x1 and x2 are the displacements of the mass 1 and 2, you will find that thestate-space model is given by

{x}4×1 =

[0 I

M−1K M−1C

]

4×4

{x}4×1 +

[{0}2×1

M−1 {F}2×1

]

4×1

= [A] {x}+ [B] {u} ,

where [M ] , [C] , and [K] are 2× 2 matrices and

[A] =

[0 I

M−1K M−1C

], {u} =

[F1F2

](2)

and [B] is a little harder to figure out. If you have numbers for [M ] , you will be able to figure out [B] matrixis given by

[B] =

[[0]2×2M−1

]

4×2

1.1 Laplace TransformApproach- EOM in terms of 2nd order Differential Equa-

tions

Let’s consider[M ] {x}+ [C] {x (t)}+ [K] {x (t)} = {F (t)} .

After taking the Laplace transform, we have(s2 [M ] + s [C] + [K]

){X (s)} = [M ] {x (0)}+ s [M ] {x (0)}+ [C] {x (0)}+ {F (t)}

{x (t)} = L−1[(s2 [M ] + s [C] + [K]

)−1([M ] {x (0)}+ s [M ] {x (0)}+ [C] {x (0)}+ {F (t)})

].(3)

This looks a bit complicated. However, given enough patience, it is doable.

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1.2 Using Matrix Exponential Approach- EOM in State-Space Model

First, write the system model in the state-space form or

{x} = [A] {x}+ [B] {u} ,

where {x} is a vector of state variables. ALL three system models above can be written in this form.For the free response, we let {u} = 0. Then, the solution is

{x (t)} = e[A]t {x (0)} ,

where e[A]t called matrix exponential and given by

e[A]t = L−1[(s [I]− [A])−1

].

If you have 4 state variables, e[A]t is a 4× 4 matrix, and guess HOW MANY inverse Laplace transforms youneed to take to find {x (t)}????? The answer is 16. If you are interested in x1 (t) , the first state variable,only, HOW MANY inverse Laplace transform do you need to take? The answer is 4.

For the forced response, we start with

{x} − [A] {x} = [B] {u} .

Take a Laplace transform to obtain

(s [I]− [A]) {X (s)} − {x (0)} = [B] {U (s)}

Solve for {X (s)} ,

{X (s)} = (s [I]− [A])−1 ([B] {U (s)}+ {x (0)})

= (s [I]− [A])−1 [B] {U (s)}+ (s [I]− [A])−1 {x (0)} . (4)

Finally, take the inverse Laplace transform to find {x (t)} .Particular solution to Harmonic InputIf you have an harmonic input (only ONE harmonic input), we can find the amplitude of the particular

solution (remember the frequency response plots?) using the transfer function. Start with

{x (t)} − [A] {x (t)} = [B]F cosωf t.

The particular solution is{xp (t)} = Re

([T (iωf )] [B]Fe

iωf),

where[T (s)] = (s [I]− [A])−1 .

The amplitude is{Xp} = |[T (iωf )] [B]F |

Why is the particular solution important? It is important because it is also the steady-state solution in aSTABLE system where the transient response dies out as t → ∞. Engineers are usually interested in thesteady-state response.

Pros: You will ALWAYS get the solution. No imaginary number appears anywhere in the solutionCons: TOO many inverse Laplace transforms to perform.

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1.3 Eigenvalue Problem Approach for the State-Space Model

Free ResponseYou can solve the same problem in the state-space form. SOLVE the eigenvalue problem to find the eigen-

values, λ, and eigenvectors, {u} , of [A] . You do this using your calculator or Matlab ([P lambda]=eig(A)).If there are 4 state variables, there are 4 sets of eigenvalues and eigenvectors. Note that the columns of Pare eigenvectors {u} .

Then, the solution is given by

{x (t)} =n∑

i=1

ci {ui} eλit,

where ci is UNKNOWN constants to be determined from ICs, and n is the number of state variables or thedimension of [A] .

Forced ResponseThere is really no good way to find the forced response

Pros: You have a feel for what the response looks like.Cons: I have many bad things to say about this method. First, the eigenvalue, λ, may be COMPLEX.

Then, you have to rewrite the solution in terms of sin and cos. Second, you have to apply the initialconditions to find ci. If λ is complex, it will be painful unless you rewrote the solution in terms of sin andcos. Third, it is not easy to find the forced response.

1.4 Modal Analysis

For modal analysis, we do NOT start with the state-space model, but a system of 2nd order differentialequation. Equation 1 is a perfect example. One more thing to note is that modal analysis works forUNDAMPED system (c = 0) and a very special case of damped system (proportional damping).

If the system model contains damping and forcing, you set them equal to zero temporarily to find thenatural frequencies and modeshapes. After the damping and forcing terms are stripped away, we are leftwith

[M ] {x (t)}+ {x (t)}+ [K] {x (t)} = {0} .

Free Vibration1. You solve an EIGENVALUE problem, [K] {u} = λ [M ] {u} , using Matlab ([P lambda] =eig(K,M)) or

using your calulator (eig(M−1K)). The eigenvalues are the natural frequency squared and the columnsof the eigenvector matrix are the eigenvectors. The eigenvectors are also called theMODESHAPES.If you are using a calculator, the eigenvalues may not be ordered correctly (must be in ascending order) andyour eigenvectors are most likely NOT normalized properly. You must re-normalize it with respect to [M ] .For example, if your calculator spit out [

0.7070.707

]

as one of the eigenvector, the normalized eigenvector is

{u} = c

[0.7070.707

],

where you find c from{u}T [M ] {u} = 1.

2. The free response is

{x (t)} =n∑

i=1

{ui} (Ai cosωit+Bi sinωit) ,

where Ai and Bi are found using the initial conditions.

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3. Using the orthogonality, you will find that

Ai = {ui}T [M ] {x (0)}

Bi =1

ωi{ui}

T[M ] {x (0)} .

Note that the eigenvalues and eigenvectors of the eigenvalue problem resulting from the state-space model,[A] {u} = λ {u} , where [A] is given in Equaiton 2, are not the natural frequencies and the modeshapes.

Forced Vibration via Modal Analysis - Still undampedLet us consider the case where the EOM is given by

[M ] {x (t)}+ [K] {x (t)} = {F (t)} ,

where there is no damping.1. Repeat step 1 to find the natural frequencies and modeshapes.2. Defind a vector of modal coordinates, {q} , such that

{x (t)} = [P ] {q (t)} , (5)

where [P ] is the matrix of eigenvectors found in step 1.3. Transform the equations into

{q (t)}+[diag ω2

]{q (t)} = [P ]T {F (t)} = {Q (t)} . (6)

4. Find the modal coordinates using the initial conditions,

{q (0)} = [P ]T [M ] {x (0)}

{q (0)} = [P ]T [M ] {x (0)} .

5. Form the final solution,

{x (t)} = [P ] {q (t)}

=N∑

i=1

{ui} qi (t)

Forced Vibration via Modal Analysis - dampedIf [C] is not zero, modal analysis is possible if we have proportional damping. It means that [C] is a

combination of [M ] and [K] or[C] = α [M ] + β [K] ,

where α and β are constant numbers. In that case, [P ]T [C] [P ] is diagonal, and the modal coordinates satisfy

{q (t)}+ [diag 2ζω] {q (t)}+[diag ω2

]{q (t)} = [P ]T {F (t)} = {Q (t)} (7)

instead of Equation 6.

Advantages of Modal Analysis1. Modal analysis is at its best when we need to visualize the motion. The modeshapes show the ‘shape’

of the motion.2. It is EASY to reduce the degree of freedom. The response can be approximated using the first couple

of modes (3 at most is sufficient). That is, even for a 1000-dof system, we can approximate the response as

{x (t)}1000×1 ≈3∑

i=1

{ui}1000×1 qi (t) .

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Note that you can estimate all 1000 coordinates (x1−1000 (t)).This CANNOT be done if you are using the Laplace Transform method or matrix exponential method.

If you have 1000 DOF, you have to take an inverse of 1000×1000 matrix, and take 1,000,000 inverse LaplaceTransforms if you are using the 2nd order DFQ (See Equation 3)1 , or take an inverse of 2000× 2000 matrix,and take 4,000,000 inverse Laplace Transforms if you are using State-Space model (See Equation 4, and yes4 million is correct because [A] is not symmetric). COMPARE THAT to solving just THREE differentialequations, qi + 2ζiωiqi + ω

2i qi = Qi for i = 1, 2, 3. You may say that why don’t we use Matlab? Matlab

will have to solve 2000 first order differential equations. I would say why not solve for the three modalcoordinates q1−3 (t) using Matlab (6 first order DFQ), and now modal analysis is 2000/6 times better.

3. BECAUSE MODAL ANALYSIS gives you a good estimation in short time, you should try to useit even if your damping is not proportional. Estimate your damping matrix with a similar one that isproportional.

1 Actually it is fewer than 1,000,000 inverse LT because(s2 [M ] + s [C] + [K]

)−1

is symmetric. For 1000 dof, you need totake 500,500 inverse Laplace transforms.

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