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ME 563

Mechanical VibrationsLecture #12

Multiple Degree of Freedom

Free Response + MATLAB

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Free Response1

We can solve for the homogeneous solution to a coupled set of

equations in a multiple degree of freedom linear system by:

-Identifying the initial conditions on all the states-Assuming a solution of the form {x(t)}={A}estWhat does this last assumption imply about the response?

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Two DOF System2

Consider the two degree of freedom system of equations:

If we make a solution of the form, , as we did forthe single DOF case, we obtain:

Non-trivial solutions satisfy:

Ms2+ 2Cs+ 2K( ) Ms2 +Cs+K( ) Cs+K( )

2

= 0

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Characteristic Equation3

There are four solutions that satisfy the characteristic

equation and these solutions are expressed as follows when the

modal frequencies are underdamped:

The solution to the homogeneous equation is then written asfollows (for the first two complex conjugate roots):

s1=1+j1, s2=1-j1, s3=2+j2, and s4=2-j2

Real part: decay rate Imaginary part: oscillation frequency

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Free Response Form4

The free response is usually written in the following form for

a multiple degree of freedom system:

Four constants Four initial conditions are required.

Modal vector

(can be scaled)

Decaying co-sinusoid

(common to both degrees of freedom)

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Example Solution5

For K=1 N/m, C=0.1 Ns/m, and M=1 kg, the solution is given

by:

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MATLAB Solution6

To obtain solutions for the free response in MATLAB, the

procedure explained before for a single DOF system is used:

Or

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MATLAB Solution7

We use the following formulation to define the outputs, {y}, if

we are only interested in the displacement variables:

Or:

y{ }21

= C[ ]24

q{ }41

+ D[ ]22

f(t){ }21

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Free Response (Eigen-Analysis)8

We can also solve the homogeneous equations of motion by:

-Identifying the initial conditions on all the states-Identifying the modal frequencies, s, and vectors, {X}, usingeigen-analysis.

Do you remember what an eigenvalue problem looks like?

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Eigenvalue Problem9

Consider the case when we have no damping:

We can write this set of equations of motion as follows:

Now assume a solution of the form {x(t)}={A}est

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Eigenvalue Problem10

Consider the case when we have no damping:

This last equation is the standard eigenvalue problem.Matlab solves this equation with the command eig(A):

[v,d]= eig(A);

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Matlab eig11

How do we interpret what Matlab gives us with this

command?

is a diagonal matrix of -s1,22

is a matrix of modal vectors

How do we find the modal frequencies and vectors?

d

v

d =

2.6180 00 0.3820

v =

-0.8507 -0.52570.5257 -0.8507

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Matlab eig12

First, we take the negative square root of the diagonal entries

of the dmatrix providing the modal frequencies in rad/s:

Second, we can scale the columns of v to produce easy to

interpret modal vectors:

-d =

0+1.6180i 00 0+0.6181i

v =

1.0000 0.6180-0.6180 1.0000

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Matlab Form of Response13

The form of the free response can then be written in the form:

In summary, the eigenvalue problem for an undamped system

is given by:

Where does the damping fit?

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Damped Eigenvalue Problem14

To obtain solutions for the free response in a damped system,

the state variable form of the equations of motion are used:

and then the eigenvalues and eigenvectors of the state matrix

are calculated using eig. This approach works because theassumed solution {q}estis also used for the 1st order system:

[v,d]= eig(A);

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Matlab eig15

How do we interpret what Matlab gives us with this

command in the first order form of the state variable model?

is a diagonal matrix of modal frequencies s1,2and s3,4

(1:2,1:4) is a matrix of modal vectors {X}1 and {X*}1

d

v

diag(d) =

-0.1309 + 1.6127i-0.1309 - 1.6127i

-0.0191 + 0.6177i

-0.0191 - 0.6177i

s3,4

s1,2

{X}2 and {X*}2

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Matlab eig15

Eigenvectors are not the modal vectors of the 2nd order system

eigenvectors involve both position and velocity states. We

look to the position states to obtain the scaled modal vectors.

v = [

-0.0362 - 0.4457i -0.0362 + 0.4457i

0.0224 + 0.2755i 0.0224 - 0.2755i

0.7236 0.7236

-0.4472 + 0.0000i -0.4472 - 0.0000i

-0.4472 + 0.0000i -0.4472 - 0.0000i

-0.7236 -0.7236

0.0085 - 0.2763i 0.0085 + 0.2763i

0.0138 - 0.4470i 0.0138 + 0.4470i

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Damping in MDOF Systems16

Note that for the previous example, the damping matrix[C]

corresponded to a proportionally viscously damped system:

For this type of damping, note that the modal vectors were

entirely real, i.e., the phase between DOFs was either 0 or 180o.

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Damping in MDOF Systems16

What if the system is not proportionally damped? The modal

vectors in this case are complex but the response is not:

For this type of damping, note that the modal vectors are

complex, so the phase between the DOFs is not 0 or 180o.

What does this mean?

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Damping in MDOF Systems17

What if the system is not proportionally damped? The modal

vectors in this case are complex but the response is not:

Real

Imag

Mode #2

Mode #1

2nd order EOMS

cannot be uncoupled

using real modal

vectors!