1

Modal Testing and Analysis

Undamped MDOF systems

Saeed Ziaei-Rad

2

Undamped MDOF systems

)}({}]{[}]{[ tfxKxM

[M] and [K] are N*N mass and stiffness matrices.

{f(t)} is N*1 force vector

k1 k2 k3m1 m2

f1 f2

x1x2

3

Undamped 2DOF system

21223222

12212111

)(

)(

fxkxkkxm

fxkxkkxm

or

2

1

2

1

322

221

2

1

2

1

0

0

f

f

x

x

kkk

kkk

x

x

m

m

mMNk

mMNkk

kgmm

/8.0

/4.0

1

2

31

21

4

Undamped 2DOF system

800000800000

8000001200000][

10

01][

K

M

5

MDOF- Free Vibration

0|][][|det

:      

}0{}]){[]([

}{)}({

0)}({

)}({}]{[}]{[

2

2

MK

solutiontrvialnonhaveTo

eXMK

eXtx

tf

tfxKxM

ti

ti

6

MDOF- Free Vibration

2222

21 ,,,,, Nr

Natural Frequencies:

Mode shapes:

Nr ,,,,, 21

Or in matrix form:

,  2 r Modal Model

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2DOF System- Free Vibration

0)108.0()104.2( 12264 Solving the equation:

2622

2521

)/  (102

)/  (104

sRad

sRad

Numerically:

11

11][

1020

01046

52r

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Orthogonality Properties of MDOFThe modal model possesses some very important Properties, stated as:

rT

rT

kK

mM

][

][ Modal mass matrix

Modal stiffness matrix

][12rrr km

Exercise: Prove the orthogonality property of MDOF

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Mass-normalisationThe mass normalized eigenvectores are written asAnd have the following property:

2][

][

rT

T

K

IM

The relationship between mass normalised mode shapeand its more general form is:

2/1][

}]{[}{    }    {1

}{

r

rT

rrr

r

r

m

Mmwherem

10

Mass-normalisation of 2DOF

][1040

08.1011

11

2.18.

8.2.1

11

11][

][20

02

11

11

10

01

11

11][

66r

T

rT

kK

mM

Clearly:

261 1020

04.0][ rrr km

707.707.

707.707.

11

11][ 2/12/1

rr mm

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Multiple modes The situation where two (or more) modes have

the same natural frequency. It occurs in structures with a degree of

symmetry, such as discs, rings, cylinders. Free vibration at such frequency may occur not

only in each of the two modes but also in a linear combination of them.

a

b

c

a Vertical mode

b Horizontal mode

c Oblique mode

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Forced Response of MDOF

titi eFeXMK

tfxKxM }{}]){[]([

)}({}]{[}]{[2

Or by rearranging:

}{])[]([}{ 12 FMKX

})]{([}{ FHX Which may be written as:

Response model

kmexceptmallforFF

XH m

k

jjk             0     )(

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Forced Response of MDOFThe values of matrix [H] can be computed easily at

each frequency point. However, this has several advantages:

It becomes costly for large N. It is inefficient if only a few FRF expression is

required. It provides no insight into the form of various

FRF properties.

Therefore, we make use of modal properties for deriving the FRF parameters instead of spatial properties.

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Forced Response of MDOF12 )]([])[]([ HMK

Premultiply both sides by and postmultiply by T

Tr

Tr

TT

H

H

HMK

122

122

12

)]([

)]([

)]([])[]([

][]][[][][ 11 I

Note that: Diagonal matrix

Equation 1

Inverse both sides

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Forced Response of MDOFAs H is a symmetric matrix then:

kjjk HH or

jkkj FXFX // Principle of reciprocity

Using equation 1:

N

r rr

krjrN

r r

krjrjk m

H1

221

22 )(

or

N

r r

jkrjk

AH

122

Modal constant

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Forced Response of 2DOF

2122322

2

1221212

1

)]([

)]([

FXkXkkm

FXkXkkm

Which gives:

)())(( 3231212

12312214

21

2232

021

1

kkkkkkkmkmkmmmm

mkk

F

X

F

Numerically:

12264

26

021

1

108.0104.2

102.1

FF

X

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Forced Response of 2DOF

222

212

221

211

1

111

)()(

F

XH

Or numerically:

262611102

5.0

104.0

5.0

H

Receptance FRF ( ) for 2dof system

11H