mechanical vibrations chapter 6 - faculty server...

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 6

Mechanical VibrationsChapter 6

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell


MDOF Equations of Motion

Equation of Motion for 2 DOF

can be written in compact matrix form as

( )

( )

=

−

−++

−

−++

)t(f)t(f

xx

kkkkk

xx

ccccc

xx

mm

2

1

2

1

22

221

2

1

22

221

2

1

2

1

&

&

&&

&&

[ ] [ ] [ ] FxKxCxM =++ &&&



This coupled set of equations can be uncoupled byperforming an eigensolution to obtain ‘eigenpairs’for each mode of the system, that is

‘eigenvalues’ and ‘eigenvectors’

or

‘frequencies’ (poles) and ‘mode shapes’



Equation of Motion[ ] [ ] [ ] )t(FxKxCxM =++ &&&

Eigensolution

Frequencies (eigenvalues) andMode Shapes (eigenvectors)

[ ] [ ][ ] 0xMK =λ−

[ ] [ ]L2122

21

2 uuUand\\

\=

ω

ω=

Ω


MDOF - Orthogonality of Eigenvectors

Normal modes are ‘orthogonal’ with respect to thesystem mass and stiffness matrices.The eigen problem can be written as

For the ith mode of the system, the eigenproblemcan be written as

[ ][ ] [ ][ ][ ]2UMUK Ω=

[ ] [ ] iii uMuK λ= (6.6.1)



Premultiply that equation by the transpose of adifferent vector uj

and write a second equation for uj and premultiplyit by the transpose of vector ui

(6.6.2)

(6.6.3)

[ ] [ ] iTjii

Tj uMuuKu λ=

[ ] [ ] jTijj

Ti uMuuKu λ=



Subtracting 6.6.3 from 6.6.2, yields

Since the eigenvalues for each mode are different,

When i=j, then the values are not zero

(6.6.2)

(6.6.8)

( ) [ ] 0uMu jT

iji =λ−λ

[ ] [ ]

jikuKumuMu

iiiT

i

iiiT

i =

==

[ ] ji0uMu jT

i ≠= (6.6.6)


Modal Matrix and Modal Space Transformation

Define the modal matrix as the collection ofmodal vectors for each mode organized incolumn fashion in the modal matrix

This modal matrix is then used to define themodal transformation equation with a newcoordinate with ‘p’ as the principal coordinate

[ ] [ ]L321 uuuU =

[ ] [ ]

==

M

L 2

1

21 pp

uupUx


Modal Space Transformation

Substitute the modal transformation into theequation of motion

In order to put the equations in normal form, thisequation must be premultiplied by the transpose ofthe projection operator to give

[ ] [ ][ ] [ ] [ ][ ] [ ] [ ][ ] [ ] FUpUKUpUCUpUMU TTTT =++ &&&

[ ][ ] [ ][ ] [ ][ ] FpUKpUCpUM =++ &&&



The first term of the modal acceleration canbe expanded as

[ ] [ ][ ]

[ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( )

=

OMMM

L

L

L

3T

3T

31T

3

3T

22T

21T

2

3T

12T

11T

1

T

uMu2uMuuMuuMuuMuuMuuMuuMuuMu

UMU



Recall from orthogonality that

[ ] ji0uMu jT

i ≠=

so that

[ ] [ ][ ]

[ ] ( ) [ ] ( )

[ ] ( )

=

OMMM

L

L

L

3T

3

2T

2

1T

1

T

uMu000uMu000uMu

UMU


Modal Mass, Modal Damping, Modal Stiffness

The mass becomes

The stiffness becomes

The damping becomes(under special conditions)

[ ] [ ][ ]

=

OMMM

L

L

L

33

22

11

T

m000m000m

UMU

[ ] [ ][ ]

=

OMMM

L

L

L

33

22

11

T

k000k000k

UKU

[ ] [ ][ ]

=

OMMM

L

L

L

33

22

11

T

c000c000c

UCU



The physical set of highly coupled equations aretransformed to modal space through the modaltransformation equation to yield a set of uncoupledequations

Modal equations (uncoupled)

=

+

+

MMM

&

&

M

&&

&&

FuFu

pp

\k

kpp

\c

cpp

\m

mT

2

T1

2

1

2

1

2

1

2

1

2

1

2

1



Diagonal Matrices - Modal Mass Modal Damping Modal Stiffness

Highly coupled system

transformed intosimple system

[ ] FUp\

K\

p\

C\

p\

M\

T=

+

+

&&&

m

k

1

1 c1

mp

3

3

c3

m

k

2

2 c2

f3

k 3

p 2 f2f1p 1

MODE 1 MODE 2 MODE 3



It is very clear to see that these modal spaceequations result in a set of independent SDOFsystemsThe modal transformation equation uncouples thehighly coupled set of equationsThe modal transformation appropriates the forceto each modal oscillator in modal spaceThe modal transformation equation combines theresponse of all the independent SDOF systems toidentify the total physical response


Modal Space Response Analysis

Since the MDOF system is reduced to equivalentSDOF systems with appropriate force, theresponse of each SDOF system can be determinedusing SDOF approaches discussed thus far.The total response due to each of the SDOFsystems can be determined using the modaltransformation equation



[ M ]p + [ C ]p + [ K ]p = [U] F(t)T

m

k

1

1 c1

f1p1

MODE 1

m

k

2

2 c2

p2 f2

MODE 2

mp

3

3

c3

f3

k3

MODE 3

MODAL

[M]x + [C]x + [K]x = F(t)

x = [U]p = [

SPACE

++ u p 33u p 22u p 11

u 3u 2u 1 ]p

.. .

.. .

+

+

Σ=


Modal Space - Modal Matrices

[ ] [ ][ ]

=

\M

\UMU 111

T1

[ ] [ ][ ]

=

\C

\UCU 111

T1

[ ] [ ][ ]

=

\K

\UKU 111

T1Modal Stiffness

Modal Damping

Modal Mass

The mode shapes are linearly independent andorthogonal w.r.t the mass and stiffness matrices

TRUE !!!

???????

TRUE !!!


Proportional Damping

[ ] [ ] [ ][ ][ ]

β+α=β+α

\KM

\UKMU 1

T1

The damping matrix is only uncoupled for a specialcase where the damping is assumed to beproportional to the mass and/or stiffness matrices

Many times proportional damping is assumed sincewe do not know what the actual damping isThis assumption began back when computationalpower was limited and matrix size was of criticalconcern. But even today we still struggle with thedamping matrix !!!


Non-Proportional Damping

If the damping is not proportional then a statespace solution is required (beyond our scope here)

[ ] [ ][ ] [ ]

[ ] [ ][ ] [ ]

=

−

−

F0

xx

K00M

xx

CMM0

1

1

11

1 &

&

&&

[ ] [ ] QYAYB 11 =−&

[ ] [ ] [ ][ ] [ ]

=

11

11 CM

M0B [ ] [ ] [ ]

[ ] [ ]

−

=1

11 K0

0MA

[ ] [ ][ ] 0YBA 11 =λ− [ ] 11 pY Φ=

The equation of motion can be recast as

The eigensolution and modal transformation is then


MATLAB Examples - VTB4_1

VIBRATION TOOLBOX EXAMPLE 4_1

>> clear>> m=[1 0;0 1];k=[2 -1;-1 1];>> [P,w,U]=VTB4_1(m,k)

P =0.5257 0.8507

0.8507 -0.5257w =

0.6180 1.6180U =

0.5257 0.8507 0.8507 -0.5257>> mm =

1 0 0 1>> kk =

2 -1 -1 1

>>


MATLAB Examples - VTB4_3VIBRATION TOOLBOX EXAMPLE 4_3

>> clear>> m=[1 0;0 1];d=[.1 0;0 .1];k=[2 -1;-1 1];>> [v,w,zeta]=VTB4_3(m,d,k)Damping is proportional, eigenvectors are real.

v = -0.5257 -0.8507 -0.8507 0.5257w = 0.6180 1.6180zeta = 0.0809 0.0309>> mm = 1 0 0 1>> dd = 0.1000 0 0 0.1000>> kk =

2 -1 -1 1

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