ME 4213

Vibration of Multi-Degree-of-Freedom (MDOF)

Systems – eigenvalues and eigenvectors

H.P. LEEDepartment of Mechanical Engineering

EA-05-20Email: mpeleehp@nus.edu.sg

Semester 2 2014/2015

ME 4213

Matrices

M M T

A matrix M is defined to be symmetric if

A symmetric matrix M is positive definite if

xTMx 0 for all nonzero vectors x

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Matrices A symmetric positive definite matrix M can be factored

Here L is upper triangular, called a Cholesky matrix

M LLT

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Matrices

The matrix square root is the matrix M 1/2 such that

M 1/2M 1/2 M

If M is diagonal, then the matrix square root is just the root

of the diagonal elements:

L M 1/2 m1 0

0 m2

(4.35)

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Matrices For a symmetric, positive

matrix M

11

2 2

111 1 1/ 2

1 12

1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2

identity symmetric

000, ,

00 0

Let ( ) ( ) and multiply by :

( ) ( ) (4.38)

mm

m m

I K

mM M M

m

t M t M

M MM t M KM t

x q

q q 0

1/ 2 1/ 2or ( ) ( ) where

is called the mass normalized stiffness and is similar to the scalar

used extensively in single degree of freedom analysis. The key here is that

i

t K t K M KM

kK

m

K

q q 0

s a SYMMETRIC matrix allowing the use of many nice properties and

computational tools

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Matrix equation in terms of real symmetric matrix eigenvalue problem

2

2

vibration problem real symmetric eigenvalue problem

(4.40) (4.41)

Assume ( ) in ( ) ( )

, or

j t

j t j t

t e t K t

e K e

K K

q v q q 0

v v 0 v 0

v v v v v 0

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Observations

For a nxn matrix there are n eigenvalues

The eigenvalues are all real and positive

The matrix is similar to a diagonal matrix

The set of eigenvectors are orthogonal (we will

discuss this later)

The set of eigenvectors are independent (we will

discuss this later)

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Observation about the matrices

The original mass matrix is typically symmetric,

positive definite and up to now, diagonal.

The stiffness matrix is typically positive

semidefinite, which means that they may have a

zero eigenvalue.

The stiffness matrix as well as the mass

normalized stiffness matrix are symmetric.

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Orthogonal and Normal Vectors

x

x1

M

xn

, y

y1

M

yn

, inner product is xTy xiyii1

n

x orthogonal to y if xTy 0

x is normal if xTx 1

if a the set of vectores is is both orthogonal and normal it

is called an orthonormal set

The norm of x is x xTx

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Normalization of a vector A vector can be normalized by dividing by its norm.

x

xTx

has norm of 1

x

xTx

xT

xTx

x

xTxxTx

xTx 1

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Recap on the previous example

1 13 31/ 2 1/ 2

2

2 2

1 1 2 2

0 27 3 0

0 1 3 3 0 1

3 1 so which is symmetric.

1 3

3- -1det( ) det 6 8 0

-1 3-

which has roots: 2 and 4

K M KM

K

K I

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Continue ,,,normalized eigenvectors

1 1

11

12

11 12 1

2 11 2

1

( )

3 2 1 0

1 3 2 0

10

1

(1 1) 1

11

12

K I

v

v

v v

v 0

v

v

v

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The second normalized eigenvector

v2 1

2

1

1

, v1

Tv2

1

2(11) 0

v1

Tv1

1

2(11) 1

v2

Tv2

1

2(1 (1)(1)) 1

v i are orthonormal

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Mode u and eigenvector v

u1 v1 and u2 v2

x M 1/2q u M 1/2

v

Note

M 1/2u1

3 0

0 1

13

1

1

1

v1

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Orthonormal set of vectors

1 2

1 1 1 2

2 1 2 2

1 2 1 1 2 2

1 2 21 1 1 2 1 2

1 2

21 2 1 2 2 2

1 0

0 1

0diag( , )

0

T T

T

T T

T T T

T T

T T

P

P P I

P KP P K K P

v v

v v v v

v v v v

v v v v

v v v v

v v v v

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Some terms that you need to know

P is known as the modal matrix

P is a orthogonal matrix

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To continue with the previous example

P v1 v1 1

2

1 1

1 1

PTP 1

2

1

2

1 1

1 1

1 1

1 1

1

2

11 11

11 11

1

2

2 0

0 2

I

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frequency

2

1

2

2

1 1 3 1 1 11 1

1 1 1 3 1 12 2

1 1 2 41

1 1 2 42

4 0 2 0 01

0 8 0 42 0

TP KP

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Conclusion

2diag diag( ) (4.48)T

i iP KP

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Another example

Equations of motion

1 1 1 2 1 2 2

2 2 2 1 2 3 2

( ) 0 (4.49)

( ) 0

m x k k x k x

m x k x k k x

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Matrix form for the equations of motion

1 2 21

2 2 32

00 (4.50)

0

k k km

k k km

x x

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Numerical example

m1 1 kg, m2 4 kg, k1 k3 10 N/m and k2 =2 N/m

1/ 2 1/ 2

2

1 2

1 2

1 0 12 2,

0 4 2 12

12 1

1 12

12 1det det 15 35 0

1 12

2.8902 and 12.1098

1.7 rad/s and 12.1098 ra

M K

K M KM

K I

d/s 3.48

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eigenvector

1

11

21

11 21

1

2 2 2 2 2

1 11 21 11 11

11

For equation (4.41 ) becomes:

12 - 2.8902 1 0

1 3- 2.8902

9.1089

Normalizing yields

1 (9.1089)

0.

v

v

v v

v v v v

v

v

v

21

1 2

1091, and 0.9940

0.1091 0.9940, likewise

0.9940 0.1091

v

v v

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Orthogonal matrix

1 2

0.1091 0.9940

0.9940 0.1091

0.1091 0.9940 12 1 0.1091 0.9940 2.8402 0

0.9940 0.1091 1 3 0.9940 0.1091 0 12.1098

0.1091 0.9940 0.1091 0.9940

0.9940 0.1091 0.9940 0.109

T

T

P

P KP

P P

v v

1 0

1 0 1

It shows that P is an orthogonal matrix

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Another note on v

In the previous section, we could have chosed v2 to be

v2 0.9940

0.1091

instead of v2

-0.9940

0.1091

because one can always multiple an eigenvector by a constant

and if the constant is -1 the result is still a normalized vector.

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Conclusion The procedure for finding the eigenvalues and

eigenvectors have been presented.

We have learnt the concept of modal vectors, normal vectors, modal matrix, normal matrix as well as their properties.

The most important property is that the modal or the normal matrix is an orthogonal matrix.

Or the modal vectors are orthogonal to each others.

The eigenvectors are independent.