Numerical Methods for solving Eigenvalues and Eigenvectors

ByyR. K. Verma & G. R. Reddy

Structural and Seismic Engineering SectionReactor Safety DivisionReactor Safety Division

BARC, Mumbai

Equation of Motion for Base Excitation

Considering as a multi degree freedom system the equation ofmotion for the base excitation is written as

[ ]{ } [ ]{ } [ ]{ } [ ] { } 11gxMxKxCxM &&&&& =++In the absence of external excitation

[ ]{ } [ ]{ } [ ]{ } 20=++ xKxCxM &&&Neglecting the structural damping, the free vibration ofthe system is expressed by

Fig. 1: 2DOF System

[ ]{ } [ ]{ } 30 xK xM =+&&{ } { } { } { } t)i (dt)i ( 2 &&L t { } { } { } { } t)sin(-x soand t)sin( x 2 ==Let

{ } [ ]{ } 0 K ][-w 2 =+ Mhavee 2

4 MK =which can also be written as

[ ]{ } { } 5 A =[ ] [ ] and K MA where 2-1 ==

Now the free vibration problem has been reduced to thegeneral Eigenvalue problem. We may have to find out a vector which when multiplied by A yields a scalar multiple of itself Thiswhich, when multiplied by A yields a scalar multiple of itself. Thismultiple is called an Eigenvalue or characteristic value of A andthere are N such Eigenvalues and N such vectors, calledEigenvectors associated with each Eigenvalue i.Examples

Vibration of StructuresNatural frequencies and corresponding mode shapes

Heat Transfer AnalysisLinearized Buckling Analysis

3

Properties of the Eigenvectors

Solution of eq 4 yields n Eigenvalues and corresponding

4 MK =Solution of eq.4 yields n Eigenvalues and correspondingEigenvectors. Each Eigen pairs satisfies eq.4; i.e.; we have ),( ii

6n1,.......,i == ii MK Equation 6 can also be written as

)( )( iii MK =constant nonzeroa is where

Therefore, with being an eigenvector, is also aneigenvector and we say that an eigenvector is defined

i ieigenvector, and we say that an eigenvector is definedonly by its direction in the n-dimensional space.

4

Hence, the eigenvectors obtained need to be normalized withrespect to mass matrix. After normalization, we have 1.=iTi MThis fixes the lengths of the eigenvectors, i.e., the absolutemagnitude of the elements in each eigenvector.

M orthonormality

ijjTi M =

K orthonormalityijij

Ti K =

5

Methods for solution of Eigenvalue problems

There are various numerical methods to obtain the eigenvalues andeigenvectors of the eq.4 which has large degrees of freedom (i.el b f i ) S f h h d d

a Simplified Technique

large number of equations). Some of the methods used are:

a. Simplified Techniqueb. Inverse Iteration Techniquec. Simultaneous Iteration Techniqueqd. Subspace Iteration Techniquee. Lanczos Iteration Technique

The time required for computation of eigenvalues and eigenvectorsreduces from technique a to e and it can be observed in thereduces from technique a to e and it can be observed in thefollowing explanation.

6

Simplified Techniquem11

2 m2

Stiffness Matrix of Each Element

[ ] 1211 EICwhereCK = =3

2

m3

[ ] 311 lCwhereCK e ==Combined stiffness matrix and mass

Fig. 2: Mathematical model of the RCC structure

Column Sizes 100 75 mm10 10 2[ ] [ ]

1

2

1 1 09 1 2 1 ;

mK C M m

= =

matrices are respectively given as follows

E= 2.5 10 10 N/m2 (concrete)Moment of Inertia of the column section I= 3.52 10-6 m4

[ ] [ ] 23

;0 1 2 m

500Kg

1m1

10 1[ ]

=11

mm

MT

500Kg

1000Kg

3

2m2

m3

1.00.67033

2

3

[ ]

=3

2

3

2321

mmM

00.12060

Addedmass

3 0.33 Modeshapevector

Fig. 3: Approximate Eigenvector (mode shape) of the structure

[ ]

=33.067.0

2060256033.067.000.1 MT

7

}{ 0171.0After normalization of mode shape

}{

==0056.00114.0

MT

n

}{ [ ] }{ 9.12652 == nTn K

Fig. 4: Mode shape of the structure bt i d ith 3D FEMobtained with 3D FEM

The frequency is 35.5797 rad /sec (5.96 Hz). The frequency ofthe structure with 3D finite element method considering thegflexibility of floor is about 4.25 Hz.

8

Inverse Iteration Technique

The technique of inverse iteration is very effectively usedto calculate eigenvectors and at the same time the k=5N/m

m1=2Kg

corresponding eigenvalues can also be evaluated.

k=10N/m

m2=2Kg

The free vibration equation of the system shown in Fig. 5f i l i i b i

/

Fig. 5: 2DOF 0)(55)(02 11 =

+txtx&&

for vertical excitation can be written as

System0)(155)(20 22=

+ txtx&&The characteristic equation can be written as

[ ]{ } [ ]{ }XMXK =The above system has two eigenvalues and two eigenvectorsThe above system has two eigenvalues and two eigenvectorsand can be obtained using inverse iteration technique as follows.

9

Evaluation of first eigenvalues and eigenvectorsStep 1- Assume {X} { } = 11X Step 2- Evaluate {R} = [M] {X}

=

22

11

22

Step 3- Evaluate new {X} using [K] {X} ={R}

=

80

22

15555

2

1

X

XX

=

4.08.0

2

1

XX

Step 4 Normalizing the mode shapes 63250Step o a g t e ode s apes{ }{ } [ ]{ }XMX

XT

=

=

3162.06325.0

Step 5- Evaluate eigenvalues

{ } [ ]{ } =KT 22.15.1

==

Consider {X} = {} and repeat the steps 1-4 till required convergenceis reached.

10

Evaluation of eigenvalues and eigenvectors of second mode

One suggested way of checking the convergence is 0001.01

1 =+

+i

ii

g gGram-Schmidt deflation technique is used to make mode shapesindependent to avoid duplication of eigenvalues and eigenvectorswhich is explained as belowwhich is explained as below.

Step 1- New Trial vector {X1} = {X}- {} { }

=

3162.06325.0

11

1 X { } [ ]{ }XMwhere T = [ ] 8974.111223162.06325.0 =

=

N C id i g {X} {X } f ll t 1 4 f b d l tNow Considering {X} ={X1} follow steps 1-4 of above and evaluateeigenvalues and eigenvectors

{ } 2999.0 { }766.26498.7 and6000.02999.0

iteration first After

==

=

The frequencies of the 2 DOF system with finite element method is1.21 and 2.92 Hz. 11

Simultaneous Iteration Technique

{ }

=021120

XIteration-1Step 1- Assume {X}

02

Step 2- Evaluate {R} = [M] {X}

=

1560156031200

1120

015600001560

0312002156000

Step 3- Evaluate new {X} using [K] {X} ={R} 31200011 1211 XX

=

031201560156031200

210121

0119

3231

2221

1211

XXXXXX

C

9573047860XX

=

3590.03590.07180.04786.09573.04786.0

30.1

3231

2221

1211

eXXXXXX

St 4 N r li i g th d hStep 4 Normalizing the mode shapes{ }

[ { } [ ]{ } ]XMXXT=

=

=0182.00119.00035.00158.00172.00158.0

3231

2221

1211

1

XXXXXX

X12

Step 5- Evaluate Eigenvalues{ } [ ]{ } =KT 95.112,149.45

12758,5.2038

21

21

====

Iteration 2 0150.00183.01211 XXEnd of first iteration.

Iteration-2

=

0192.00088.00070.00151.0

0150.00183.030.1

3231

2221

1211

eXXXXXX

122.114,76.40

13024,4.1661

21

21

====

End of second iteration.

Iteration-3

=

01720008400051.00150.0

0179.00186.030.12221

1211

eXXXXXX

0172.00084.03231 XX

12641,3.1655 21 == 43.112,685.40 21 ==

End of third iteration.13

Subspace Iteration Technique

Step 1- Assume {X} { }

=021120

X

Step 2- Evaluate {R} = [M] {X}

=

031201560156031200

021120

156000015600001560

0312002156000

Step 3- Evaluate new {X} using [K] {X} ={R} 31200011 XX

=

031201560156031200

210121

0119

3231

2221

1211

XXXXXX

C

=

7180.04786.09573.04786.0

30.122211211

eXXXX

3590.03590.07180.04786.030.1

3231

2221 eXXXX

14

[M]R = [X]T [M] [X] =0.0009 0.00150 0015 0 0024

Step 4- Reduce the stiffness and mass matrix using above vectors

0.0015 0.0024 [K]R = [X]T [K} [X] =

1.8666 2.24002.2400 4.1066

Step 5- Let new trial vector [X]k,j = 1 00 1

Step 6 Considering load vectors [R] = [M] {X} = 0.0009 0.0015 Step 6- Considering load vectors [R]k,j = [M]R {X}k,j = 0.0015 0.0024

Step 7- Solving [K]R [X] = [R]k,j

[X] = 30.1269 0.2961

1 100.2961 0.4229

Normalizing first vector [X] with [M]R modifies toR

[X]k,1 = 6.9096

16.1224

Fi t i l i i {X} T [K] {X}First eigenvalue is given as {X}k,1T [K]R {X}k,1

11 = 1666.6 15

Step 8- To make the vectors independent use Gram-Schmidtdeflation technique as follows and obtain the second vector

= [X]k,1T [M]R [X]k,2 = [ ] 30.0009 0.0015 0.29616.9096 16.1226 100.0015 0.0024 0.4229 = 2.99610

-5

6.909616.1224

40.9126 100.5507

Modified {X}k,2 = {X}k,2- =St 9 N li i g d t [X] ith [M] difi tStep 9- Normalizing second vector [X]k,2 with [M]R modifies to

{X}k,2 = 141.250485.2352

85.2352 Step 10- Second eigenvalue is given as {X}k,2 T [K]R {X}k,2

22 = 13139.022Considering these as new trial vectors and repeating steps 5-10,converged eigenvalues and vectors can be obtained satisfying thefollowing convergence criteria.

Check convergence, 1 ,

, 1

0.0001j i j ijj i

+

+

= 16

0 4 86 0 9 2 Step 11- Eigen vectors of the original problem can be written as follows

3

0.4786 0.95726.9096 141.2501

10 0.4786 0.717916.1226 85.235

0 3590 0 3590

[] = [X][X]R=0.3590 0.3590

0.0187 0.014 0.0187 0.0140.0149 0.00640.0083 0.0201

[] =

It also satisfied all the orthogonality conditions.

0 1655 0 0039 4 0.1655 0.0039100.0039 1.3142

0 9999 0 0006

T K = 0.9999 0.0006

0.0006 1.0001TM =

17

Lanczos Iteration Technique

The basic steps of the Lanczos iteration technique is to transformgeneralized eigenproblem into a standard form with a tridiagonal

ffi i i Th f llcoefficient matrix. The steps are as follows :

Step 1- Assume {X} { }

=010.10.1

X 0.1

Step 2- Calculate {X1} = {X}/

{ }

= 0146.0

0146.0

1XWhere = [ { } [ ]{ } ]XMX T { } 0146.01 = 68.41Step 3 Let = 0; then calculate for i =1 nXStep 3- Let 0 = 0; then calculate for i =1. . . . . . , n,for i = 1:

{ } 1049.0[ ]{ } [ ]{ }XMXKiX

{ }

=0525.00875.040.11 eX[ ]{ } [ ]{ }ii XMXK =

18

Step 4- 004-5.5841e1 =[ { } [ ]{ }iTii XMX=

Step 5- and if i n,1

0.23321.0e-005 0.0583X

= %

iX% 11 iiiii XXX =0.2915

Step 6-Step 6i [ { } [ ]{ }Ti iX M X% %= 004-1.4924e 1 =

% 01560Step 7-1

ii

i

XX + =% { }

=

0195.00039.00156.0

2X

,2 2Repeat step 3-7 to get0041 1681e 004-1.1681e2 =005-1.480424e 2 =

19

,3 3Repeat step 3-7 to get005-4.2735e3 =3018-5.4912e 3 =

11

= nn

nT

112

221

11

= 5-e1.48044-1.1681e4-1.4924e4-1.4924e 4-5.5841e

T

nn 1

5-4.2735e5-1.4804e

5e1.480441.1681e41.4924e nT

The eigenvalues of T can be obtained using any procedureThe eigenvalues of Tn can be obtained using any procedureexplained above and reciprocals of these will give the eigenvalues oforiginal problem.

The eigenvalues of Tn are 6.041e-4, 7.7e-5and 3.69e-5.Reciprocal of these are 1655.355, 12987.012 and 27100.27. 20

Response of the System by Response Spectrum Method

HfdHf 46501930

Frequencies and mode shapes of the system arek=5N/m

m1=2KgNode1

HzfandHzf 465.0193.0 21 ==

2999.06325.0k=10N/m

m2=2KgNode 2

=

6000.03162.0

Fig. 6: 2DOF System

Participation factor corresponding to each mode can be given as 0.4

0.5

n

(

g

)

Response Spectrum

y

{ } [ ]{ } = 8974.11MT 0.10.2

0.3

A

c

c

e

l

e

r

a

t

i

o

n

21

{ } [ ]{ } 6002.0

Fig. 7: Response Spectrum

0

0 10 20 30 40 50

Frequency (Hz)

The spectral values corresponding to frequencies 0.193 and 0.465 Hz are 0.49and 0.98 m/sec2 respectively.

Acceleration response of node 1 in first mode is0.63251.89740.49=0.588

Acceleration response of node 1 in second mode is-0.29990.60020.98=-0.176

Acceleration response of node 2 in first mode is0 31621 89740 49=0 2940.31621.89740.49=0.294

Acceleration response of node 2 in second mode is0 6000 0 6002 0 98 0 3530.60000.60020.98=0.353

By taking the SRSS combination of modes acceleration responses

22

By taking the SRSS combination of modes, acceleration responsesof node 1 and 2 are 0.614 and 0.459 m/sec2 respectively.

References

1. K. J. Bathe, Finite Element Procedures, 3rd Indianreprint 1996reprint, 1996.

2. Anil K. Chopra, Dynamics of Structures, theory andapplications to earthquake engineering, Secondedition, 2007.

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