Solution of Eigenproblem of Non-Proportional Damping Systems by Lanczos Method

Solution of Eigenproblem of Non-Proportional Damping Systems by Lanczos Method

In-Won Lee, Professor, PEIn-Won Lee, Professor, PE Structural Dynamics & Vibration Control Lab.Structural Dynamics & Vibration Control Lab. Korea Advanced Institute of Science & Technology

The Fourth International Conference on Computational Structures TechnologyThe Fourth International Conference on Computational Structures TechnologyEdinburgh, ScotlandEdinburgh, Scotland18th-20th August 199818th-20th August 1998

Structural Dynamics & Vibration Control Lab., KAIST, Korea 2

OUTLINE

Introduction

Method of analysis

Numerical examples

Conclusions


INTRODUCTION

Free vibration of proportional damping system

where : Mass matrix

: Proportional damping matrix

: Stiffness matrix

: Displacement vector

0)()()( tuKtuCtuM

M

C

K)(tu

(1)


Eigenanalysis of proportional damping system

where : Real eigenvalue

: Natural frequency

: Real eigenvector(mode shape)

Low in cost Straightforward

niMK iii ,,2,1 (2)2ii

ii


Free vibration of non-proportional damping system

(4),0

0

M

KA

where

0M

MCB

tu

tuz

0 tAztzB (3)

(5) tetu Let

tt eetz

(6)

, then

and


Therefore, an efficient eigensolution technique is required.

iii BA (7)

(9): Orthogonality of eigenvector

mjiB ijjTi ,,2,1,

: Eigenvalue(complex conjugate)

: Eigenvector(complex conjugate)

i

(8)

ii

ii

where

Solution of Eq.(7) is very expensive.

nmi 2,,2,1


Current Methods

Transformation method: Kaufman (1974)

Perturbation method: Meirovitch et al (1979)

Vector iteration method: Gupta (1974; 1981)

Subspace iteration method: Leung (1995)

Lanczos method: Chen (1993)

Efficient Methods


Proposed Lanczos algorithm

retains the n order quadratic eigenproblems

is one-sided recursion scheme

extracts the Lanczos vectors in real domain


METHOD OF ANALYSIS Free vibration of non-proportional damping system

where : Mass matrix

: Non-proportional damping matrix

: Stiffness matrix

: Displacement vector

0)()()( tuKtuCtuM

M

C

K)(tu

(11) tetu Let , then

(10)


Quadratic eigenproblem

where : eigenvalue (complex conjugate)

: independent eigenvector (complex conjugate)

02 iiiii KCM

i

i

(12)


where : dependent eigenvector

ji

ji

M

MC

ii

i

T

jj

j

1

0

0

iii *

(13)

ji

jiCMM i

T

ji

T

ji

T

j 1

0**

Orthogonality of the eigenvectors

or

(14)


Proposed Lanczos Algorithm

Assume that m independent

and dependent Lanczos vectors are found

Calculate preliminary vectors and

m ,,, 21

**2

*1 ,,, m

*11

ˆmmm MCK

mm *

1ˆ

1ˆ

m*

1ˆ

m

(15)

(16)


Preliminary vectors can be expressed as

111

~ mmm

*11

*1

~ mmm

2111

~ˆmmmmmmmm

*

2*

1**

1*

1

~ˆmmmmmmmm

are the components of previous Lanczos vectors

(real values)

, and is the pseudo length of and

(17)

(18)

(19)

(20)

1

~m

*1

~m1m

,,, mmm real

where


Orthogonality conditions of Lanczos vectors

miCMM m

T

im

T

im

T

i ,,1for011**

1

1

11111

*1

*11

orCMM mmmmmmm

TTT

111*

1*

111

~~~~~~ m

T

mm

T

mm

T

mm CMMsign

(21)

(22)

(23)

(19)

(20)

111

~ mmm

*11

*1

~ mmm

where


Coefficient m

mm

T

mm

T

mm

T

mm CMM 11**

1ˆˆˆ

mm

T

mmm

T

m

T

mm MMCKCM *1*

the orthogonality conditions Eqs.(21) and (22)

CMT

m

T

m * MT

mEq.(17) + Eq.(18) and Applying

Using Eqs.(15) and (16)

(25)

(24)


Coefficients and

1m 1m

1

2/1

11 mmm 2/1

11 mm

(26)

(27)

(28)

(29)

mm

T

mm

T

mm

T

m

mmm

CMM

111*

1*

11

111

~~~~~~

11 mm sign

Applying the orthogonality conditions Eqs.(21) and (22)

CMT

m

T

m 1*

1 MT

m 1Eq.(17) + Eq.(18) and

where


Coefficients ,m

0

21*

1*

11

mm

T

mm

T

mm

T

mmm CMM (30)


CMT

m

T

m 2*

2 MT

m 2Eq.(17) + Eq.(18) and

0

0


(m+1)th Lanczos vectors and1m*

1m

1

1*1

1

11

~

m

mmmmmm

m

mm

MCK

1

*1

*

1

*1*

1

~

m

mmmmm

m

mm

mm sign

mmm 2/1

,2/1

11 mm

(31)

(32)

mm

T

mmm

T

m

T

mm MMCKCM *1*

mm

T

mm

T

mm

T

mm CMM 111*

1*

111

~~~~~~

where


Reduction to Tri-Diagonal System

Rewriting quadratic eigenproblem

(33)

where (34)

(35)

(36)

where

*iiii MCK

iii *

ii

ii *~

m 21

**2

*1

*m


iiiT (37)


1* KMCTT

MT

Eq.(33) + Eq.(34) and

mm

m

TTTMMCKMCT

4

433

322

21

*1*

Unsymmetric

(38)

where

mmm ,, : Real values

nmi 2,,1for


Eigenvalues and eigenvectors of the system

ii

1

ii

ii **

(39)

(40)

(41)


Physical error norm(Bathe et al 1980)

and : Acceptable eigenpair

6

2

2

2

10][

i

iii

i K

KCMe

ii

Error Estimation

(42)


Comparison of Operations

kk nmnm 23

21 2 nnmnmnm cmk 18482 Proposed method

kk nmnm 23

21 2 nnmnmnm cmk 276104 Rajakumar’s method

mm

kk

nmnm

nmnm

23

21

23

21

2

2

nnmnmnm cmk 216142 Chen’s method

MethodInitial operations(A)

Operations in each row of T(B)

Number of operations = A + p B

p : Number of Lanczos vectors

n : Number of equations

cmk mmm and, : Mean half bandwidths of K, M and C

where


Example : Three-Dimensional Framed Structure

Proposed method

Rajakumar’s method

Chen’s method

Number of total operations Ratio

p = 30

n 1,008

cmk mmmm 81

Method

38.27e+6

53.23e+06

61.38e+06

1.00

1.39

1.60


NUMERICAL EXAMPLES Structures

Cantilever beam with lumped dampers Three-dimensional framed structure with lumped d

ampers Analysis methods

Proposed method Rajakumar’s method (1993) Chen’s method (1988)


Comparisons Solution time(CPU) Physical error norm

Convex with 100 MIPS, 200 MFLOPS


Cantilever Beam with Lumped Dampers

1 2 3 4 99 100 101

C

5

Material PropertiesTangential Damper :c = 0.3

Rayleigh Damping : = = 0.001

Young’s Modulus :1000

Mass Density :1

Cross-section Inertia :1

Cross-section Area :1

System DataNumber of Equations :200

Number of Matrix Elements :696

Maximum Half Bandwidths :4

Mean Half Bandwidths :4


Results of cantilever beam : Physical Error norm(number of Lanczos vectors=30)

ModeNumber

EigenvalueProposedMethod

Rajakumar’sMethod

Chen’sMethod

12345678910

-2.57457+j3.17201-2.57457-j3.17201-1.53800+j18.3566-1.53800-j18.3566-1.69581+j39.6477-1.69581-j39.6477-2.43492+j61.0104-2.43492-j61.0104-3.78360+j82.3222-3.78360-j82.3222

0.32e-070.32e-070.53e-080.53e-080.48e-080.48e-080.51e-070.51e-070.10e-060.10e-06

0.32e-070.32e-070.53e-080.53e-080.48e-080.48e-080.51e-070.51e-070.10e-060.10e-06

0.24e-030.24e-030.19e-030.19e-030.21e-040.21e-040.18e-040.18e-040.16e-040.16e-04

Solution time in second(ratio)

3.86(1.00)

5.21(1.35)

5.67(1.47)


ModeNumber


Rajakumar’sMethod

Chen’sMethod

12345678910

-2.57457+j3.17201-2.57457-j3.17201-1.53800+j18.3566-1.53800-j18.3566-1.69581+j39.6477-1.69581-j39.6477-2.43492+j61.0104-2.43492-j61.0104-3.78360+j82.3222-3.78360-j82.3222

0.25e-070.25e-070.53e-080.53e-080.48e-080.48e-080.51e-070.51e-070.36e-070.36e-07

0.25e-070.25e-070.53e-080.53e-080.48e-080.48e-080.51e-070.51e-070.36e-070.36e-07

0.24e-030.24e-030.19e-030.19e-030.21e-040.21e-040.18e-040.18e-040.13e-040.13e-04


11.63(1.00)

16.09(1.38)

17.10(1.47)

Results of cantilever beam : Physical Error norm(number of Lanczos vectors=60)


Three-Dimensional Framed Structure with Lumped Dampers

Material Properties

Tangential Damper :c = 1,000

Rayleigh Damping : = -0.92

= 0.106

Young’s Modulus: 2.1E+11

Mass Density: 7,850

Cross-section Inertia: 8.3E-06

Cross-section Area: 001

System DataNumber of Equations: 1,008

Number of Matrix Elements :80,784

Maximum Half Bandwidths : 150

Mean Half Bandwidths : 81


Results of three-dimensional framed structure :

Physical Error norm (number of Lanczos vectors=30)

ModeNumber


Rajakumar’sMethod

Chen’sMethod

12345678910

-0.015035+j3.03037-0.015035-j3.03037-0.024784+j3.09011-0.024784-j3.09011-0.243763+j3.65157-0.243763-j3.65157-3.83006+j7.78173-3.83006-j7.78173-3.42807+j8.04792-3.42807-j8.04792

0.13e-060.13e-060.17e-060.17e-060.25e-060.25e-060.93e-050.93e-050.21e-050.21e-05

0.13e-060.13e-060.17e-060.17e-060.25e-060.25e-060.93e-050.93e-050.21e-050.21e-05

0.46e-050.46e-050.46e-050.46e-050.35e-050.35e-050.33e-030.33e-030.14e-040.14e-04


100.27(1.00)

142.38(1.42)

164.44(1.64)


ModeNumber


Rajakumar’sMethod

Chen’sMethod

12345678910

-0.015035+j3.03037-0.015035-j3.03037-0.024784+j3.09011-0.024784-j3.09011-0.243763+j3.65157-0.243763-j3.65157-3.83006+j7.78173-3.83006-j7.78173-3.42807+j8.04792-3.42807-j8.04792

0.13e-060.13e-060.17e-060.17e-060.25e-060.25e-060.43e-050.43e-050.62e-060.62e-06

0.13e-060.13e-060.17e-060.17e-060.25e-060.25e-060.43e-050.43e-050.62e-060.62e-06

0.46e-050.46e-050.46e-050.46e-050.35e-050.35e-050.34e-060.34e-060.71e-050.71e-05


213.57(1.00)

323.56(1.51)

337.44(1.58)

Results of three-dimensional framed structure :

Physical Error norm (number of Lanczos vectors=60)


An efficient solution technique!

CONCLUSIONS The proposed method

needs smaller storage space gives better solutions requires less solution time

than other methods.


Thank you for your attention.


111*

1*

111

~~~~~~ m

T

mm

T

mm

T

mm CMM (A-1)

(A-4)

(A-5)

(A-3)1

11

~

m

mm

(A-2)1

*1*

1

~

m

mm

where If 01 m , 2/1

11 mm

If 01 m , 2/1

11 mm j

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Solution of Eigenproblem of Non-Proportional Damping Systems by Lanczos Method

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