meca029 – mechanical vibrations – mechanical vibrations gaëtan kerschen space structures &...

MECA029 – Mechanical Vibrations

Gaëtan Kerschen

Space Structures & Systems Lab (S3L)

MECA029 – Mechanical Vibrations

Lecture 6: Reduction and

Substructuring Methods

3

Outline

Preliminary Remarks

4

Quiz

Results and answers…

5

Outline

Previous Lectures

6

Previous Lectures

L1: Analytical dynamics of discrete systems

L2: Undamped vibrations of n-DOF systems

L3: Damped vibrations of n-DOF systems

L4: Continuous systems: bars, beams, plates

L5: Approximation of continuous systems by Rayleigh-Ritz and finite element methods

L6: Reduction and substructuring methods

L7: Direct time-integration methods

L8: Introduction to nonlinear dynamics

7

Previous Lectures

132 Hz

NUMERICAL METHODS OBJECTIVE OF

THIS COURSE

8

Previous Lectures

L1

ODEs:

L2,3

L4

( ) ( ) ( ) ( )t t t t+ + =Mq Cq Kq p&& &0: 0

:

ijjj

i

ji ij

V u Xx

S n tσ

σρ

σ

∂− + =

∂

=

&&

L4LINEARIZATION

L2,3

132 Hz

L5SPATIAL

DISCRETIZATION

PDEs:

L6EFFICIENT

COMPUTATION

ANALYTIC

9

Previous Lectures

Finite element method: a structure is divided into a finite number of elements of simple geometry (bar, beam, membrane, plate, shell)

10

Previous Lectures

1. Kinematic assumptions

2. Write the resulting strain and kinetic energy

3. Determine the interpolation functions

4. Compute the elementary matrices in local axes

5. Coordinate transformation

6. Compute the elementary matrices in structural axes

7. Assembly process

Finite element method:

11

Previous Lectures

Bar element

⇒ Linear interpolation functions inside the element

⇒ Elementary mass and stiffness matrices

2 11 26eL

m ⎡ ⎤= ⎢ ⎥

⎣ ⎦M l 1 1

1 1eLE A −⎡ ⎤

= ⎢ ⎥−⎣ ⎦K

l

12

Previous Lectures

Beam element

⇒ Cubic interpolation functions inside the element

⇒ Elementary mass and stiffness matrices

2 2

2 2

156 22 54 1322 4 13 354 13 156 2242013 3 22 4

eLm

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥− − −⎣ ⎦

M

l l

l l l ll

l l

l l l l

2 2

3

2 2

12 6 12 66 4 6 212 6 12 66 2 6 4

eLE I

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥− − −⎢ ⎥−⎣ ⎦

K

l l

l l l l

l ll

l l l l

13

Previous Lectures

3D Beam element

⇒ Same reasoning as for the previous beam elements

⇒ 6DOFs/node

⇒ 12x12 elementary matrices

14

Previous Lectures

eSeL qRq =

Local ⇒ structural axes

1

X

YZ

O

local axes

3

2y x

z

ψx1

ψy1

ψz1

U1

V1

W1ψx2

U2

V2W2

ψz2

ψy2

, ,eL eL eS eS⇒K M K M

15

Previous Lectures

Assembly process: use localization vectors

, , 0e eS eSΣ ⇒ ⇒ + =K M K M Mq Kq&&

16

Previous Lectures

Convergence to the exact solution when the number of finite elements tends to infinity

Upper bound convergence for a consistent mass matrix (i.e., constructed using the same process as for the stiffness matrix)

Example of a clamped-free bar:

2

2

1 2 1(2 1) 1 ...2 24 2r

EA rrmL N

π πω⎡ ⎤−⎛ ⎞= − + +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

Finite element method

2(2 1)2

1,...,

kEAkml

k

πω = −

= ∞

Continuous system

17

Previous Lectures

Introduction to reduction methods

⇒ Industrial FE models typically have [105-106] DOFs

⇒ But interest in the lower frequency eigensolutions

)1()()1( ×××

=mmnn

yRx

yMyK 2ω=

Reduced matrices

RMRMRKRK TT == and

m n<?

18

Previous Lectures

Guyan: static condensation

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

C

R

CCCR

RCRR

C

R

CCCR

RCRR

xx

MMMM

xx

KKKK 2ω Retained DOF

Condensed DOF

1CC CR−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K KRCRCCDS

R

C

R xKKI

xxx

xx

x ⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧

+=

⎭⎬⎫

⎩⎨⎧

= −1

19

Previous Lectures

Guyan: static condensation

CRCCRCRRT

RR KKKKRKRK 1−−==

CRCCCCCCRC

CRCCRCCRCCRCRRT

RR

KKMKK

MKKKKMMRMRM11

11

−−

−−

+

−−==

( )2 0RR RR Rω− =K M x

20

Today

L1: Analytical dynamics of discrete systems

L2: Undamped vibrations of n-DOF systems

L3: Damped vibrations of n-DOF systems

L4: Continuous systems: bars, beams, plates

L5: Approximation of continuous systems by Rayleigh-Ritz and finite element methods

L6: Reduction and substructuring methods

L7: Direct time-integration methods

L8: Introduction to nonlinear dynamics

21

Today

1. Guyan: Illustrative example

2. Craig-Bampton

22

1. Guyan: Illustrative Example

1 1 2l

2 4

w2 Ψ2

l

3l

3

Ψ3

w3

1

1 2

l

2 3

w2 Ψ2

l

3994484537414620214053

1.38039.39560.6720255.50067.50492.5161

32

42

−−

exactelementselementsr

IELm

rω

No reduction (straightforward application of the finite

element method)

23

1. Guyan: Illustrative Example

1 1 2l

2 4

w2 Ψ2

l

3l

3

Ψ3

w3

+ Guyan ???

2 2

2 2

156 22 54 1322 4 13 354 13 156 2242013 3 22 4

eLm

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥− − −⎣ ⎦

M

l l

l l l ll

l l

l l l l

2 2

3

2 2

12 6 12 66 4 6 212 6 12 66 2 6 4

eLE I

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥− − −⎢ ⎥−⎣ ⎦

K

l l

l l l l

l ll

l l l l

24

1. Guyan: Illustrative Example

1 1 2l

2 4

w2 Ψ2

l

3l

3

Ψ3

w3

+ Guyan ???

3994484537414620214053

1.38039.39560.6720255.50067.50492.5161

32

42

−−

exactelementselementsr

IELm

rω

Guyan (4DOFs⇒2DOFs)506.5

4002

25

Today

1. Guyan: Illustrative example

2. Substructuring and component mode

synthesis

26

Substructuring ?

Complex structure: assembly of several subsystems

⇒ Very often, the subsystem are analyzed by different teams

⇒ Necessary to reconstruct the full model from separate analyses (definition of appropriate interfaces)

⇒ Keep the number of DOFs of each subsystem as small as possible

27

Substructuring: Examples

Vehicle body, 1373000 DOFsNatsiavas et al., JCND2, 2007

Model combining the vehicle body, the suspensions, the wheels and 4 passengers; response to road irregularities

Passenger-seat model

28

Substructuring: Examples

Stator (1278000 DOFs)

Objective: analysis of the radial freeplay

Casing (789072 DOFs)

Rotor (113176 DOFs)

29

Concept of Mechanical Impedance

internal DOF q1

(no applied loads)

( ) gqMK =− 2ω

applied force amplitudes

Impedance matrix Z(w2)

boundary DOF q2

The impedance matrix relates the force amplitudes to the displacement amplitudes

It is the inverse of the FRF matrix H(ω)

30

Concept of Mechanical Impedance

( )( ) ( )

∑=

−−−−

−

−

−−−

+−−−

−=

1

1224

212

21)()(212

214

121

11111

1121121

1121121

1121222

121

112122*22

)( ~~

~~~~n

i ii

Ti

Tiii

ωωω

ωωω

ω

MKxxMK

KKMKKMKKKKMM

KKKKZ

Expression of the reduced impedance matrix

2( ),i iω x% % are the eigenfrequencies and eigenmodes of the subsystem

* * 2 * 422 22 22 2 2

i

i i

αω ωω ω

= − +−∑Z K M

%

Statically condensed stiffness

Statically condensed mass

Clamped vibration modes

31

Concept of Mechanical Impedance

Necessary ingredients for appropriate representation of the dynamic behavior of a given subsystem

⇒ Static boundary modes resulting from static condensation

⇒ Subsystem eigenmodes in clamped boundary configuration

32

Two Ingredients

The Craig-Bampton method, also called component mode synthesis (CMS), relies on these two ingredients:

⇒ The static modes resulting from unit forces on the boundary DOFs

⇒ The internal vibration modes of the substructure fixed on its boundary

33

Two Ingredients

Static mode Internal vibration mode

But how do we calculate the reduction matrix R ???

34

Problem Statement

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡0g

xx

MMMM

xx

KKKK B

I

B

IIIB

BIBB

I

B

IIIB

BIBB 2ω Boundary DOF

Subsystem DOF

Reaction forces with other parts of the system

2 0RR RC R RR RC R

CR CC C CR CC C

ω⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫

− =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

K K x M M xK K x M M x

Retained DOF

Condensed DOF

Guyan

Craig-Bampton

35

Reduction Matrix

Guyan (static condensation)

Craig-Bampton (static + dynamic information)

1CC CR−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K K

1

0???II IB

−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K K

?

36

Reduction Matrix

Guyan (static condensation)

Craig-Bampton (static + dynamic information)

1CC CR−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K K

1

0

II IB I−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K K Φ

Internal vibration modes2

II I II Iω=K Φ M Φ

37

Reduction Matrix

Guyan (static condensation, dynamic information neglected)

Craig-Bampton (static + dynamic information, some internal modes neglected)

1CC CR−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K K

1

0

II IB m−

⎡ ⎤= ⎢ ⎥−⎣ ⎦

IR

K K Φ

Im n<<

38

Reduced Stiffness and Mass Matrices

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

IMMM

MΩ00K

KmB

BmBB

m

BB and2

( ) TBmIBIIIIIB

TmmB

IBIIIIIIBIIBIIBIIBIIBIBBBB

IBIIBIBBBB

MKKMMΦM

KKMKKMKKKKMMM

KKKKK

=−=

+−−=

−=

−

−−−−

−

1

1111

1

with

39

Reduction Performed

Number of DOFs before reduction:

Size of the reduction matrix R:

Number of DOFs after reduction

I Bn n n= +

Internal Boundary

( )Bn n m× +

Bn m+

40

Superelement

The matrices may be interpreted as elementary matrices of a subsystem

The resulting “element” is called a superelement

The superelement of a subsystem is assembled with the rest of the structure like an ordinary finite element

Assumption of linear dynamics for the superelement

andK M

41

Internal Vibration Mode Selection

For simplicity, the choice of the retained modes is usually carried out by considering the first m modes, but this is not optimal !!!

Effective modal masses

⇒ Select modes that contribute significantly to the response of the structure excited at its interface

( )12

1

ni j

T Sj j

m mμ=

− ≈∑Γ e Effective mass of

mode j in direction i

Due to missing mass the equality does not hold

42

Cantilever Beam Example (400 DOFs)

Guyan

Craig-Bampton

xB

xI

1 2 3 199 200

- 1 node at the interface (2DOFs)- We keep the first m modes.

1 2 3 199 200

- Rotational DOFs are condensed- Some translational DOFs are condensed (depending on the reduction).

For example, if the reduction factor is 40, we keep 10 DOFs in the reduced model: DOFs 1, 21, 41, 61, 81, 101, 121, 141, 161, 181

430 5 10 15

0

2000

4000

6000

8000

10000

12000

14000

16000

Mode number

Freq

uenc

y (H

z)

EXACTCRAIG-BAMPTONGUYAN

Cantilever Beam Example (400 DOFs)

NbDOFs_GUYAN=NbDOFs_CB=20

CraigBamptonGuyan

44

Cantilever Beam Example (400 DOFs)

Time series can also be simulated using the superelement

A half-sine pulse is applied to the beam at DOF 1

1 2 3 199 200

0 0.05 0.10

0.5

1

Time

Forc

e

SuperElementTimeSeries

45

Industrial Example (Automotive)

Vehicle body, 1373000 DOFsNatsiavas et al., JCND2, 2007

Model combining the vehicle body, the suspensions, the wheels and 4 passengers; response to road irregularities

Passenger-seat model

46

Industrial Example (Automotive)

Excitation: left suspension / response: vehicle roof

Solid line: exactDashed line: CMS

FRF

Frequency (Hz)

FRF of the vehicle body

47

Industrial Example (Automotive)

Elapsed time (hh:mm:ss) [0-500Hz]

Elapsed time (hh:mm:ss) [0-900Hz]

03:28:27 23:45:31

CMS Full model

09:20:00 >60:00:00

FRF of the vehicle body

48

Industrial Example (Automotive)

Solid line: exactDashed line: CMS

Frequency (Hz)

Number of modes

Modes of the vehicle body

49

Industrial Example (Automotive)

Response of the full, but reduced, model to road irregularities

50

Industrial Example (Space)

Courtesy of EADS Space Transportation, 282000 DOFsAutomated Transfer Vehicle

51

Industrial Example (Space)

All modes up to 100 Hz retained (375 DOFs)

36 interface nodes (216 DOFs)

Total number of DOFs: 591

Reduction: 591/282000=0.21%

52

Industrial Example (Space)

Cumulative effective modal mass

Bas Fransen, 2005

53

Industrial Example (Space)

Example of a dominant lateral mode (high effective modal mass)

54

Model Reduction of Large Molecules

10-monomer amylose chain: 10 substructures

Harmonic excitation (1 Angstrom, 20 GHz)

Amylose structure

55

Model Reduction of Large Molecules

Dowell et al., 2006

Atom displacement

56

Not Discussed in This Course (Chap. 6)

Numerical solution of eigenvalue problems

⇒ Jacobi, Householder, subspace iteration, Lanczosmethods

Sensitivity analysis

⇒ Calculate the sensitivity of eigenfrequencies and eigenmodes to variations of physical parameters (e.g., elasticity modulus, density, thicknesses)

⇒ Useful for structural optimization and model updating