Method of Finite Elements II Page 1

Solution of Equilibrium Equation in Dynamic Analysis

Mode Superposition

Dominik Hauswirth

21.11.07

Method of Finite Elements II Page 2

Contents

1. Mode Superposition: Idea and Equations

2. Example 9.7

3. Modes

4. Include Damping

5. Response contributions

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Mode Superposition1Dynamic Equilibrium Equation

RKUUCUM =++ &&&

Solving methods:• Central difference

• Houbolt

• Wilson θ

• Newmark

Method of Finite Elements II Page 4

Mode Superposition1Effectivity of this methods

Implicite methods(Houbolt, Wilson, Newmark)

1.) Initial calculations (LDLT) -> Number of operations:

2.) For each time step (Mult.) -> Number of operations

2

21

kmno ⋅⋅=

kmno ⋅⋅= 2

Where: n : matrix size, mk: half bandwidth

Number of operation is growing with bandwidth !

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Mode Superposition1Reduction of bandwidth1.) Optimize mesh topology

-> Limited effect

2.) Transformation

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

⋅=

11001210

01210012

kK

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

⋅=

21011210

01101002

kK1

2

3

4

1

4

3

2

Method of Finite Elements II Page 6

Mode Superposition1Modal generalized displacements

{R

T

K

T

C

T

M

T RPXKPPXCPPXMPPRKPXXCPXMP

RKUUCUM

~~~~=++

=++

=++

321&

321&&

321

&&&

&&&

Transformation with unknown P, such that mk smaller

( ) ( )tXPtU ⋅=Dynamic Equation in modal displacements

In theory many different transformations possible –

in practice only one transformation matrix established !

Method of Finite Elements II Page 7

2D-Elements; linear1Transformation matrix

02 =− φωφ MK

Free-vibration equilibrium solutions with damping neglected:

By inserting U we obtain a eigenproblem:

( )( )0sin0

ttUKUUM

−⋅⋅=

=+

ωφ

&&

Postulated solution

with n solutions for eigenvector

iφ

2iωeigenvalue

Normalization:

⎩⎨⎧

=≠

=jiji

M jTi 1

0φφ 222

21 ...0 nωωω ≤≤≤≤

iφ

2iω

Method of Finite Elements II Page 8

2D-Elements; linear1Transformation matrixDefinitions

],...,,[ 21 nφφφ=Φ

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=Ω

2

21

21

2

...

nω

ωω

Since the eigenvectors are M - orthonormal2Ω=ΦΦ=ΦΦ KIM TT

Eigenproblem for n equations

02 =ΦΩ−Φ MK

M and K diagonalized with TRANSFORMATION MATRIX Φ

],...,,[ 21 nφφφ=Φ )()( tXtU ⋅Φ=

Method of Finite Elements II Page 9

2D-Elements; linear1Transformed equation, Damping neglectedMatrix equation

RXX

RXKXMKUUM

T

TT

I

T

Φ=Ω+

Φ=ΦΦ+ΦΦ

=+

Ω

2

2

0

&&

321&&

321

&&

Single decoupled equation i( ) ( ) ( )( ) ( )tRtr

nitrtxtxTii

iii

⋅=

==⋅+

φ

ω ,....,2,12&&

( ) UMtx Tii 00 φ==Transforming the initial conditions

Method of Finite Elements II Page 10

2D-Elements; linear1Solution of decoupled equation iDuhamel integral

( ) ( ) ( )( ) ( ) ( )ttdtrtx iiit

iiii

i ⋅⋅+⋅⋅+−⋅⋅= ∫ ωβωαττωτω cossinsin1

0

Transforming to real displacement base

( ) ( )∑=

⋅=n

iii txtU

1φ

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Example2Example 9.7; p. 789

Method of Finite Elements II Page 12

Example2Exact solution with mode superposition

Time history curve of exact solution

-2

-1

0

1

2

3

4

5

6

7

0 5 10 15 20 25

time [s]

disp

lace

men

t

u1_exaktu2_exakt

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Example2Newmark methodTwo possibilities, leading to the same result

1.) Integrate straight forward -> SLOW

2.) Transform into modal displacements , integrate the decoupled equations , transforme back -> FAST

RKUUM =+&&

RKUUM =+&& RXX TΦ=Ω+ 2&&

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Example2Newmark in MATLAB

25.05.028.0

===∆

αδ

t

Originally proposed as an unconditionally stablescheme by Newmark

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Example2Newmark method

exact solution vs. Newmark

-2

-1

0

1

2

3

4

5

6

7

0 5 10 15 20 25

time [s]

disp

lace

men

t

u1_Newmarku2_Newmarku1_exaktu2_exakt

Method of Finite Elements II Page 16

Example2Benchmark

Time history curve

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

0 0.5 1 1.5 2 2.5 3 3.5 4

time [s]

disp

lace

men

t

Newmarkexakt

Method of Finite Elements II Page 17

Modes3Number of modes in calculationFor n lumped masses in a system, n modes were found - but for

a good approximation often only a few are needed!

Choosing the right modes for calculationIn general the lowest frequencies and modes are approximated in

the best way -> upper bounds for frequencies are found

How many and which modes are taken for calculation depends on the problem:

• Earthquake: In some cases only the 10 lowest modes

• Shock: Many modes necessary, p > 2/3 n

Method of Finite Elements II Page 18

Modes3Modes in Example 9.7

modes

-2

-1

0

1

2

3

4

0 5 10 15 20 25

time [s]

disp

lace

men

t

1st mode u11st mode u22nd mode u12nd mode u2

Method of Finite Elements II Page 19

Modes3Only first mode in Example 9.7 for displacement u1

u1: first mode - exact solution

-2

-1

0

1

2

3

4

0 5 10 15 20 25

time [s]

disp

lace

men

t

exakt_u11st mode u1

Method of Finite Elements II Page 20

Modes3Error MeasurementThe accuracy of a solution p < n can be measured…

..and made better by the so called static correction

( )( ) ( ) ( )[ ]

( )2

2

tR

tKUtUMtRt

ppp

+−=

&&ε

( ) ( )tRtUK

MrRRp

iii

∆=∆⋅

−=∆ ∑=1

)( φ

Mode Superposition has more advantages then onlythe reduction of number of necessary operations!

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Include damping4Include damping• Modal transformation was derived without damping

• Transformation Matrix Φ diagonalizes M and K ….

• …but not a „free“ chosen damping Matrix C. In this case theequations stay coupled and mode Superposition isn‘t possible

Rayleigh damping• If C is a linear combination of M and K decoupling is possible, this

is called Rayleigh damping

KMC ⋅+⋅= βα

• In this case, there are only two free parameter for fitting thedamping rate

Method of Finite Elements II Page 22

Include damping4Decoupled equations in case of Rayleigh daamping

( ) ( ) ( )

ii

i

iiiiii trtxtxtx

βωωαζ

ωζω

+=

=⋅++

2

)(2 2&&&

If an accurate modelling of damping is necessary:

• Direct integration

• Caughey series

Method of Finite Elements II Page 23

Response contributions5Response contributionsSolving a dynamic loaded, damped system, two response contributions

can be observed:

• Transient, damped out solution part

• A permanent dynamic response, which is the static responsemultiplicated by a

dynamic load factor