determ.of eigenvalue & eigenvector

8/8/2019 Determ.of Eigenvalue & Eigenvector

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Noise and VibrationsNoise and Vibrations(BDC4013)(BDC4013)

DR MUHD HAFEEZ B ZAINULABIDIN

MOHD NO RIHAN B IBRAHIM

Universiti Tun Hussein Onn Malaysia

Determination of Natural Frequencies and Mode ShapesDetermination of Natural Frequencies and Mode Shapes


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2

Necessity to Use Computational MethodNecessity to Use Computational Method

In two degrees of freedom system, solvingthe natural frequencies can be conductedby simply calculating the root of the second

order polynomial.

4 2 0 A B C

By assuming 2 0n n

A B C

Then the natural frequencies can be found


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3

Classical MethodsClassical Methods

Standard Matrix Iteration Method

Dunkerlys Method

Rayleighs Method

Holzers Method


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4

Standard Matrix IterationStandard Matrix Iteration

0 M x K x Considering a generalequation of motion

Assuming harmonicmotion

( ) sin( )i i x t X t

Equation to solve 2 0 M X K X


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Standard Matrix IterationStandard Matrix Iteration(Solution procedures to obtain the lowest(Solution procedures to obtain the lowest natnat freq)freq)

(1) Identify matrix [K] and [M]

(2) Calculate [K] -1

(3) Define the initial trial vector {X} and convergence criteria

(4) Multiply [K] -1 [M] {X} = {Xnew}

(5) Normalized the result {Xnew}/ largest Xnew

(6) Check the convergence , use for a new trial {X}

(7) When it is converged

1n

normalizedX


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Example 1

Find the natural frequenciesand mode shapes of thesystem as shown for

k1=k2=k3=kandm1=m2=m3=mby the matrixiteration method.


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Solution

Flexibility matrix [a]=[k]-1=

Dynamical matrix is

Eigenvalue problem:

321

221

1111

k

321

221

111

1

k

mmk

21and321

221

111

where

mkDXXD


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Solution

1st natural frequency:

Assume Hence

By making the first element equal to unity:

we obtain

1

1

1

1X

6

5

3

12 XDX

m

kX 5773.0,0.3,

0000.2

6667.1

0000.1

0.3 112


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Solution

The various iand are shown below:

The mode shape and natural frequencyconverged in 8 iterations.


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Solution

2nd natural frequency:

Deflated matrix

Let the normalized vector

where must be such that

mXXDD T11112

24698.2

80194.1

00000.11 X

129591.9

24698.2

80194.100000.1

100

010001

24698.2

80194.100000.1

2

211

m

mXmX

T

T


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Solution

=0.32799m-1/2 , hence

73699.0

59102.0

32799.02/11

mX

25768.019921.022048.0

19921.023641.002127.0

22048.002127.045684.0

100

010

001

73699.0

59102.0

32799.0

73699.0

59102.0

32799.0

04892.5

321

221

111

2

T

D


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Solution

Let

By using the iterative scheme, we obtain

1

1

1

1X

25763.0

62885.0

22695.0

00000.1

25763.0

16201.0

05847.0

25763.0

2

2

X


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Solution Continuing the procedure,

Hence 2=0.64307, 2=1.24701

80192.0

44496.0

00000.1

, 2Xm

k


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Solution

3rd natural frequency:

Use a similar procedure as before.

Before computing [D3], need to normalize

59102.0

32794.0

73700.0

giveto 22 XX


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DunkerleysFormula

It gives the approx value of the fundamentalfreq of a composite system.

Consider the following general n DOF system:

For a lumped mass system with diagonalmass matrix, the equation becomes:

01or0 22 maImk

0

0...0

0

0

0...0

...

...

...

10...0

0

10

0...01

1 2

1

21

22221

11211

2

nnnnn

n

n

m

m

m

aaa

aaa

aaa


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DunkerleysFormula

i.e.

Expanding:

0

1...

...1

...1

22211

22222121

12121112

nnnnn

nn

nn

mamama

mamama

mamama

(E.1)0...1

)...

...(

1...

1

2

211,,1212112

11,1313311212211

1

22221112

n

nnnnnn

nnnnnn

n

nnn

n

mmaammaa

mmaammaammaa

mamama


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DunkerleysFormula

Let the roots of this equation be 1/12,1/22,, 1/n2. Thus

Equating coefficients of (1/2)n-1 in (E.1) and(E.2):

In most cases,

(E.2)0...11

...11111

...1111

1

222

2

2

1

2222

2

22

1

2

n

n

n

n

n22211122

2

2

1

m...mm1

...11

nn

n

aaa

n2,3,...,i,11

2

1

2

i


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DunkerleysFormula

Thus

Can also be written as

where in=(1/aiimi)1/2=(kii/mi)1/2

formula)s'(Dunkerley...1 2221112 mamama nni

22

2

2

1

2

1...

111

nnnni


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DunkerlyDunkerly FormulaFormula

(calculation procedures)(calculation procedures)

2 2 2 2

11 22

1 1 1 1

n nn

: fundamental (lowest) natural frequency

n

nn

nn

n

k

m

(1) Identify k11, k22 , knn, m1, m2, mn

(2) Calculate natural frequency of the individual component

(3) Predict the fundamental natural frequency of the system

n nn

: natural frequency of a SDOF system

consisting m and spring of stiffness k

nn


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Example 2

Estimate the fundamental natural frequencyof a simply supported beam carrying 3identical equally spaced masses, as shown

below.


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Problem 1

Estimate the fundamental natural frequenciesof the system shown below. Givenk1=k2=k3=k and m1=m2=m3=m


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Solution

Flexibility matrix

Apply the related equations to solve it.

Does the value of nat. freq. smaller & largerthan the exact valueof nat. frequency?

How many percent ?

321

221111

1

ka


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Rayleigh MethodRayleigh Method

This method predicts the fundamental(lowest) natural frequency

This method based on energy method

21

2T mx

212

V kx

1

2

T

T x M x

1

2

T

V x K x


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Rayleigh QuotientRayleigh Quotient

1

2

T

T x M x 1

2

T

V x K x

sin( )

cos( )

x X t

x X t

2max1

2

T

T X M X max1

2

T

V X K X

max maxT V

2

T

T

X K X

X M X


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Rayleigh MethodRayleigh Method

(Calculation procedures)(Calculation procedures)

Identify [K] and [M]

Select any trial vector mode {X}

Predict the fundamental natural frequency

based on the Rayleigh Quotient

2

T

T

X K X

X M X


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Example 3

Estimate the fundamentalfrequency of vibration of thesystem as shown. Assume thatm

1=m

2=m

3=m, k

1=k

2=k

3=k, and

the mode shape is

3

2

1

X


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Solution

Stiffness matrix

Mass matrix

Substitute the assumed mode shape into

110

121012

kk

100

010

001

mm

XR

m

k

m

k

m

k

XR 4629.02143.0

3

2

1

100

010

001

321

3

2

1

110

121

012

321

2


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HolzersMethod

A trial-and-error method to find naturalfrequencies of systems

Requires several trials

The method also gives mode shapes


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HolzerHolzer MethodMethod

1 1 1 1 2

2 2 1 2 1 2 2 3

3 3 3 3 2

( )

( ) ( )

( )

t

t t

t

I k

I k k

I k

2

1 1 1 1 2

2

2 2 1 2 1 2 2 3

2

3 3 3 3 2

( )

( ) ( )

( )

t

t t

t

I k

I k k

I k

2

1

0n

i i

i

I

)cos( tii

Assume


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(calculation)(calculation)

2

1 12 1

1

2

2 3 2 2 1 2 1 2 2

2

1 2 23 2 2 1

2 2

2 2

1 1 2 23 2

2 2

2

3 2 1 1 2 2

2

( )

( )

t

t t t

t

t t

t t

t

I

k

k k k I

k I

k k

I I

k k

I Ik

2

1 1 1 1 2

2

2 2 1 2 1 2 2 3

( )

( ) ( )

t

t t

I k

I k k


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(summary calculation)(summary calculation)

Torsion Translation

1

2

1 12 1

12

3 2 1 1 2 2

2

2 1

1

11

1

2,3,

t

t

i

i i k k

kti

I

k

I Ik

Ik

i n

1

2

1 12 1

12

3 2 1 1 2 2

2

2 1

1

11

1

2,3,

i

i i k k

ki

X

m XX X

k

X X m X m X k

X X m X k

i n


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(calculation procedures)(calculation procedures)

Set initial =0 and set the sweep increment of with a value

Station 1:

X1=1 (or 1=1), calculate M1=2m1X1 (or

2I11)

Station 2:

Calculate X2 (or 2), calculate M2=M1+ 2m2X2 (or

2I22)

Station 3:

Calculate X3 (or 3), calculate M3=M2+ 2m3X3 (or

2I33)

Station n:

Calculate Xn (or n), calculate Mn=Mn-1+ 2mnXn (or

2Inn)


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Example 5

The arrangement of the compressor, turbineand generator in a thermal power plant isshown below. Find the natural frequenciesand mode shapes of the system.


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Solution This is an unrestrained torsional system.

The table below shows its parameters andthe sequence of computations.


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Torsional Systems

The graph below plots the torque Mtappliedat the last disc against the chosen .

The natural frequencies are the at whichMt=0.

The amplitudes i(i=1,2,,n) are the modesha es of the s stem


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42

Problem 1Problem 1

I1=2 kg m2

I2=4 kg m2

I3=2 kg m2

kt1=3 MNm/ radKt2=2 MNm/ rad

Calculate the naturalfrequencies and modeshapes

determ.of eigenvalue & eigenvector

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