chapter 10 free vibration of mdof systems...

12 - 1

CHAPTER 10 FREE VIBRATION OF MDOF SYSTEMS

System without Damping

The equation of motion of a two-DOF system in free vibration (no external force) is

mu + ku = 0

The displacements of masses are the solution with an initial condition

( )0=u u and ( )0=u u

If a two-DOF system is let to vibrate with an arbitrary initial displacement, the displacement of each mass will be as shown. They are not a simple harmonic as free vibration of SDF system and deflected shape changes with time as the ratio 1 2/u u varies with time.

12 - 2

An undamped structure would undergo simple harmonic motion without change of deflected shape if free vibration is initiated by appropriate displacement shapes. It will maintain that initial shape and both floors will reach their maximum displacement at the same time.

For the first shape above, both floors move in the same direction.

For the second shape, the two floors move in the opposite directions. The location where displacement remains zero is called “node.” These shapes are called “natural modes of vibration.” The number of modes increases as number of degrees of freedom increases.

12 - 3

For each mode, all floors move like a simple harmonic motion with the same frequency called “natural circular frequency of vibration” ( nω ), or “natural cyclic frequency of vibration” ( nf ) and the time to complete one cycle is called “natural period of vibration.”

2n

n

T πω

= 1n

n

fT

=

The lowest natural frequency is known as the fundamental natural frequency, denoted by 1ω . It corresponds to the longest natural period, which is called fundamental natural period ( 1T ) of vibration.

12 - 4

Natural Vibration Frequencies and Modes Displacement vector ( )tu can be written as a product of mode shape vector nφ and modal coordinate ( )nq t , which is a scalar quantity.

( ) ( )n nt q t φ=u

Modal coordinate varies with time whereas the mode shape vector does not. In free vibration, the modal coordinate varies like a simple harmonic with frequency equal to the natural frequency.

( ) cos sinn n n n nq t A t B tω ω= + Therefore, ( ) [ ]cos sinn n n n nt A t B tφ ω ω= +u

Substitute this in the equation of motion, we get

( )2n n n nq tω φ φ − + = m k 0

This equation can be satisfied by two ways:

(1) ( ) 0nq t = but this is not very useful so it is called trivial solution

(2) 2n n nω φ φ− + =m k 0 or 2

n n nφ ω φ=k m

This is called a matrix eigenvalue problem. The matrices k and m

are known and we want to determine 2nω and nφ

2n n nφ ω φ− =k m 0 2

n nω φ − = k m 0

Again, this equation can be satisfied by either nφ = 0 (trivial) or

2det 0nω − = k m

12 - 5

This determinant is a polynomial of order N in terms of 2nω . This

equation is called “characteristic equation or frequency equation.” It has N real and positive roots for 2

nω because matrices k and m are symmetric and positive definite.

The N roots for 2nω are known as eigenvalues, or characteristic

values, and they correspond to the natural frequencies of N modes of vibration.

For each value of 2

nω , nφ can be solved from the equation 2n nω φ − = k m 0 , but they are not unique because many vectors

obtained from scaling a solution of nφ will also satisfy the equation. The shape of this vector nφ is called natural mode of vibration, or mode shape. It is also known as eigenvector corresponding to the eigenvalue

2nω .

12 - 6

Mode and Spectral Matrices We can put many mode shapes together in a matrix to have them in compact form.

11 12 1

21 22 2

1 2

...

...

...

N

Njn

N N NN

φ φ φφ φ φ

φ

φ φ φ

= =

Φ

The matrix Φ is called modal matrix for the eigenvalue problem. The N eigenvalues 2

nω can be assembled into a diagonal matrix 2Ω , which is known as the spectral matrix of the eigenvalue problem.

21

22 2

2N

ωω

ω

=

Ω

From the eigenvalue problem in vector form

2n n nφ ω φ=k m

It is possible to write them in matrix form 2=kΦ mΦΩ

12 - 7

Orthogonality of Modes The natural modes corresponding to different natural frequencies can be shown to satisfy the following orthogonality conditions. When n rω ω≠ ,

0Tn rφ φ =k 0T

n rφ φ =m

It can be proved as follows: From 2

n n nφ ω φ=k m Premultiply by T

rφ 2T T

r n n r nφ φ ω φ φ=k m

Above is scalar so we can transpose the right hand side 2T T

n r n n rφ φ ω φ φ=k m

Suppose we start from the rth mode, 2

r r rφ ω φ=k m

premultiply by Tnφ , we get

2T Tn r r n rφ φ ω φ φ=k m

Subtract the two equations, we get

( )2 2 0Tn r n rω ω φ φ− =m

When 2 2n rω ω≠ , 0T

n rφ φ =m and also 0Tn rφ φ =k .

12 - 8

The orthogonality condition implies that the matrices T≡K Φ kΦ T≡M Φ mΦ

are diagonal where the diagonal elements are T

n n nK φ φ= k Tn n nM φ φ= m

Because k and m are positive definite so nK and nM are positive are related by

2n n nK Mω=

by ( ) ( )2 2 2T T T

n n n n n n n n n n nK Mφ φ φ ω φ ω φ φ ω= = = =k m m

12 - 9

Interpretation of Modal Orthogonality Consider a structure vibrating in nth node with displacements

( ) ( )n n nt q t φ=u

The corresponding accelerations are ( ) ( )n n nt q t φ=u and inertia forces are

( ) ( ) ( )I n n nnt q tφ= − = −f mu m

Next consider displacements of the structures in its rth mode

( ) ( )r r rt q t φ=u

The work done by inertia force is

( )T TI r n rn

φ φ= −f u m( ) ( ) ( )n rq t q t

which is zero because of modal orthogonality.

Another implication is that the work done by equivalent static

forces associated with displacement in nth mode doing through the rth

mode displacement is zero.

( ) ( ) ( )S n n nnt q tφ= =f ku k

The work done is

( )T TS r n rn

φ φ=f u k( ) ( ) ( )n rq t q t

which is zero because of modal orthogonality.

12 - 10

Normalization of Modes The eigenvector multiplied by a scalar scaling factor is also the eigenvector corresponding to the same eigenvalue. Therefore, only the shape of mode is essential. A mode can be scaled without changing the shape. This process is called normalization. Sometimes the mode shape vector is scaled such that the largest element equal to unity. For a building, mode is often scaled such that the DOF at the roof is unity. Computer program usually normalizes the modes such that

1Tn n nM φ φ= =m or T= =M Φ mΦ I

where I is an identity matrix (diagonal with 1 along the main diagonal). These modes are called a mass orthonormal set. This results in

2 2Tn n n n n nK Mφ φ ω ω= = =k or 2T= =K Φ kΦ Ω

12 - 11

12 - 12

12 - 13

12 - 14

12 - 15

12 - 16

12 - 17

12 - 18

12 - 19

12 - 20

12 - 21

12 - 22

Modal Expansion of Displacements Any set of N independent vectors can be used as a basis for representing any other vectors of order N.

1

N

r rr

qφ=

= =∑u Φq

where rq are scalar multipliers called modal coordinates or normal coordinates.

( )1

NT Tn n r r

r

qφ φ φ=

=∑mu m

Because of modal orthogonality, only the term r n= is nonzero.

( )T Tn n n nqφ φ φ=mu m

Therefore, T Tn n

n Tn n n

qM

φ φφ φ

= =mu mum

This modal expansion is employed to determine the free vibration response and response to excitation of MDF systems.

12 - 23

12 - 24

Free Vibration of Undamped MDF Systems Given ( )0u and ( )0u determine ( )tu Use modal expansion of initial conditions

( ) ( )1

0 0N

n nn

qφ=

=∑u ( ) ( )1

0 0N

n nn

qφ=

=∑u

Therefore,

( ) ( )00

Tn

nn

qM

φ=

mu ( ) ( )0

0Tn

nn

qM

φ=

mu

Then, the free vibration response is

( ) ( )1

N

n nn

t q tφ=

=∑u

( ) ( ) ( )1

00 cos sin

Nn

n n n nn n

qt q t tφ ω ω

ω=

= +

∑u

12 - 25

12 - 26

12 - 27

Free Vibration of Damped MDF Systems If the damping matrix c of the system is such that T=C Φ cΦ is diagonal, we said that the system has classical damping. The equations of motion can be uncoupled and solved by modal analysis. If T=C Φ cΦ is not a diagonal matrix, the system is nonclassically damped and must be solved by numerical method and its eigenvalue will be complex numbers. For classically damped system, the uncoupled equation is

0n n n n n nM q C q K q+ + = where

Tn n nC φ φ= c

The modal damping ratio is

2n

nn n

CM

ζω

=

The solution of ( )nq t will be free vibration of a damped SDF system.

22 0n n n n n nq q qζ ω ω+ + =

( ) ( ) ( ) ( )0 00 cos sinn nt n n n n

n n nD nDnD

q qq t e q t tζ ω ζ ω

ω ωω

− + = +

where 21nD n nω ω ζ= −

The displacement of MDF system is

( ) ( ) ( ) ( )1

0 00 cos sinn n

Nt n n n n

n n nD nDn nD

q qt e q t tζ ω ζ ω

φ ω ωω

−

=

+ = +

∑u

12 - 28