Dynamics Ppt

8/17/2019 Dynamics Ppt

1/52

1

Dynamics of viscously

damped linear MDOF systems

Dr C S ManoharDepartment of Civil Engineering

Professor of Structural Engineering

Indian Institute of Science

Bangalore 560 012 India

[email protected]

mailto:[email protected]:[email protected]


2/52

2

Topics

Nature of equations of motion and uncoupling

Classical and non - classical damping models

Input - output relations in time domainInput - output relations in frequency domain

Forced vibration analysis using modal expansion


3/52

3

A rigid bar supported on two springs

1 1 1 2 2

2

O : Elastic centre

O : Centre of gravity

k L k L

Two DOF-s

1 Translation

1 Rotation

1 2 1 2l l L L L


4/52

4

1 1 2 2

2 2 2 1 1 1

1 2 1 1 2 2

2 2

1 1 2 2 1 1 2 2

0

0

00

0

my k y l k y l

I k y l l k y l l

k k k l k l m y y

k l k l k l k l I

is diagonal and is non-diagonal M K

Static coupling


5/52

5

1 2

2 2

1 1 2 2

00

0

k k m me z z

k L k Lme m

is non-diagonal and is diagonal M K

Inertial coupling


6/52

6

1 1 2 2

2

1 2 2

0m ml k k k L x x

ml m k L k L

is non-diagonal and is non-diagonal M K

Static and inertial coupling


7/52

7

Remarks

•Equations of motion for MDOF systems are generally

coupled

•Coupling between co-ordinates is manifest in the form of

structural matrices being nondiagonal

•Coupling is not an intrinsic property of a vibrating system.

It is dependent upon the choice of the coordinate system.

This choice itself is arbitrary.

•Equations of motion are not unique.

They depend upon the choice of coordinate system.


8/52

8

Remarks (continued)

•The best choice of coordinate system is the one

in which the coupling is absent. That is, the structural

matrices are all diagonal.

•These coordinates are called the natural coordinates

for the system. Determination of these coordinates for

a given system constitutes a major theme in structural

dynamics. Theory of ODEs and linear algebra help us.


9/52

9

0 00 ; 0

, and , in general, are non-diagonal

Equations are coupledSuppose we introduce a new set of dependent

variables ( ) using the transformation

( )where is a tra

MX CX KX F t X X X X

M C K

Z t

X t TZ t T n n

nsformaiton matrix, to be

selected.

How to uncouple equations of motion?


10/52

10

0 00 ; 0

( )( )

( )

( )

, ,& structural matrices in the new coordinate system.

( ) force vector in the new coord

t t t t

MX CX KX F t

X X X X

X t TZ t MTZ t CTZ t KTZ t F t

T MTZ t T CTZ t T KTZ t T F t

MZ t CZ t KZ t F t

M C K

F t

inate system

Can we select such that , ,& are all ?

If yes, equation for ( ) would then represent a set of uncoupled

equations and hence can be solved easily.

T M C K

Z t

Question

DIAGONAL


11/52

11

How to select to achieve this?T

Consider the seemingly unrelated problem of

undamped free vibration analysis

0

Seek a special solution to this set of equations in which

all points on the sturcutre oscillate harmonically at thesa

MX KX

2

2

me frequency.

That is

exp ; 1,2, ,

or, exp where is a 1 vector.

exp & exp

exp 0

k k x t r i t k n

X t R i t R n

X t i R i t X t R i t

MR KR i t


12/52

12

2

2

2

exp 0

0

This is a algebraic eigenvalue problem.

;

is positive semi-definite

is positive definite

Eigensolutions would be real valued

and eigenvalues w

t t

MR KR i t

RM KR

KR MR

K K M M

K

M

Note

ould be non-negative.


13/52

13

2

2

12

12 2

12

12

0

Let exist.

0

0 0

If exists, =0 is the solution.

Condition for existence of nontrivial solution is that

should not exist

KR MR

K M R

K M

K M K M R

IR R

K M R

K M

2

2 2 2

1 2

1 2

.

0

This is called the characteristic equaiton.

This leads to the characteristic values

and associated eigenvectors

, , , .

n

n

K M

R R R


14/52

14

2

2

2

Consider - th and -th eigenpairs.

(1)

(2)

(1)

(3)

(2)

r r r

s s s

t

s

t t

s r r s r

t

r

t

r s s

r s

KR MR

KR MR

R

R KR R MR

R R KR

Orthogonaility property of eigenvectors

2

2

2

2 2

(4)

Transpose both sides of equation (4)

Since & , we get

(5)

Substract (3) and (5)

0

t

r s

t t t t

s r s s r

t t

t t

s r s s r

t

r s s r

R MR

R K R R M R

K K M M

R KR R MR

R MR

0

0

t s r

t

s r

R MR r s

R KR r s

2

Normalization

1t

s s

t s s s

R MR

R KR


15/52

15

21 ( )

2 2 2

1 2

Introduce

Diag

nn n

n

R R R

t

t

M I

K

Orthogonality relations

Select T


16/52

16

0 0

2

0 ; 0

( )( )

( )

( )

; 1, 2, ,

How about initial conditions?(0) 0

(0) 0 0

0 (0) & 0 (0)

t t t

r r r r

t t

t t

MX KX F t

X X X X

X t Z t M Z t K Z t F t

M Z t K Z t F t

IZ t Z t F t

z z f t r n

X Z

MX M Z Z

Z MX Z MX

Consider

Undamped

Forced

Vibration Analysis


17/52

17

0

1

1 0

0 10 cos sin sin

010 cos sin sin

t r

r r r r r r

r r

n

k kr r

r

t n

r kr r r r r r

r r r

z z t z t t t f d

X t Z t

x t z t

z z t t t f d


18/52

18

0 00 ; 0

( )

( )

( )

( )

If is not a diagonal matrix, the

equations

t t t t

t

t

MX CX KX F t

X X X X

X t Z t

M Z t C Z t K Z t F t


IZ t C Z t Z t F t

C

How about damped forced response analysis?

of motion would still remain coupled.


19/52

19

If the damping matrix is such that

is a diagonal matrix, then equaitons would

get uncoupled.

Such matrices are called classical damping matrices.

Rayleigh's proport

t

C

C

C

Classical damping models

Example

ional damping matrix

t t t

C M K

C M K

I


20/52

20

2

2

[ ]

[ ] [ ]

2 2

T T

T T

i

n n

nn

n

C M K

C M K

I K

I Diag

c

•Rayleigh damping model imposes prescribed variations

•on modal damping values as function of the mode count

•There are only two free parameters.

•These parameters can be determined using known values of damping ratios

•

for two modes.


21/52

21

100

101

102

103

10-4

10-3

10-2

10-1

100

frequency rad/s

e t a

mass and stiffness proportional

stiffness proportional

mass proportional


22/52

22

2

0

( )

0 (0) & 0 (0)

2 ; 1,2, ,

with 0 & 0 specified.

exp cos sin

1 exp

t

t t

r r r r r r r

r r

r r r r dr r dr

t

r r r

dr

IZ t C Z t Z t F t

Z MX Z MX

z z z f t r n

z z

z t t a t b t

t f d


23/52

23

0

1

1 0

1exp cos sin exp

1exp cos sin exp

1,2, ,

t

r r r r dr r dr r r r

dr

n

k kr r r

t n

kr r r r dr r dr r r r

r dr

z t t a t b t t f d

X t Z t

x t z t

t a t b t t f d

k n


24/52

24

MDOF system with -th dof driven by an unit harmonic force s

exp0 0 1 0 0

MX CX KX F i t F

th entry s

response of the -th coordinate due tounit harmonic driving at -th coordinate.

rs X t r

s

lim ?rst

X t


25/52

25

0

0

2

0

2

0 0 0

2

0

2

0

exp

0 0 1 0 0

lim exp

exp

exp

exp exp exp exp

exp exp

t

MX CX KX F i t

F

X t X i t

X t X i i t

X t X i t

MX i t CX i i t KX i t F i t

M i C K X i t F i t

M i C K X F


26/52

26

2

0

0 0

2

0

2

0

2

0

2

0

exp exp

&

is classical (Diagonal) with 2

t t

t

nn n n

t t

t t t t

t

M i C K X F

X t X i t Z i t

M I K

C C

M i C K Z F

M i C K Z F

M i C K Z F

I i Z F

Diagonal


27/52

27

2

0

1 1

0 2 2 2 2

0 2 2

0 0

1

2 21

2 2

Recall

0 0 1 0 0 ( -th entry=1; rest=0)

2

lim exp exp

2

t

N N t

nk k kn k

k k

nn n n n n n

sn

n

n n n

N

r rn nt

n

N rn sn

n n n n

I i Z F

F F

Z i i

F s

Z i

X t Z i t X t Z i t

i

2 2

1

exp

exp

2

rs rs

N rn sn

rs

n n n n

i t

X t H i t

H

i

N


28/52

28

2 21

2 21

12

2 21

exp2

2

is symmetric but not Hermitian

2

N rn sn

rs

n n n n

N

rn snrs

n n n n

rs sr

rs sr

rs

N rn sn

n n n n

X t i t i

H i

X t X t H H

H H

H

H M i C k i

Remarks


29/52

29

*

12

2 21

Conceptually simple

Computationally difficult to implement

2

Computationally easier to implement

Not al

N N rn sn

n n n n

H M i C k

H i

l modes need to be included

(Nor it is advisable to include all modes)


30/52

30

MDOF system with -th dof driven by an unit impulse force s

0 0; 0 0

0 0 1 0 0

MX CX KX F t

X X

F

th entry s

response of the -th coordinate due to

unit impulse driving at -th coordinate.

rs X t r

s


31/52

31

0 0 1 0 0

&

is classical (Diagonal) with 2

t t

t

nn n n

t t t t

t

MX CX KX F t

F

X t Z t

M I K

C C



IZ Z Z F t


32/52

32

2

1

1

1

2

0 0; 0 0

exp sin

1exp sin

t

N

n n n n n n jn j ns

j

n n

snn ns n n n dn

dn

N

r rn n

n

N

rs rn sn n n dn

n dn

IZ Z Z F t

z z z F t t

z z

z t h t t t

X Z

X t z t

h t t t

1N


33/52

33

1

1exp sin

Remarks

Matrix of impulse response functions

Not all modes need to be included in the summation

If an arbitrary load

N

r rs rn sn n n dn

n dn

rs sr

rs

t

s

X t h t t t

h t h t

h t h t

h t h t

f

0

10

is applied at the -th dof (instead of

unit impulse excitation)

1exp sin

t

rs rs s

t N

s rn sn n n dn

n dn

s

X t h t f d

f t t d


34/52

34

2

2 1 1

Are & the only two orthogonality relations?

.

Consider

(It may be noted that is the flexibility matrix)

Pre-multiply both

t t

n n n

n n n

M I K

K M

I K M K

More on orthogonality relat

N

ions

o

2 1

1

2 1

1

1 2 1 1

1 1

sides by

0 for

is orthogonal to .

Consider

Pre-multiply both sides by

0 foris orthogonal to

thi

t

k

t t

k n n k n

n n n

t

k

t t

k n n k n

M

M MK M n k

MK M

I K M

MK M

MK M MK MK M n k MK MK M

s leads to an infinite family of orthogonality relations.


35/52

35

2

1

2

1

2

1

1

2

1

1

Consider

1


1 0 for

is orthogonal to

1Consider


n n n

n n

n

t

k

t t

k n k n

n

n n

n

t

k

t

k

M K

I M K

K

K KM K n k

KM K

I M K

KM K

KM

nd2 family

1 1

2

1 1

1 0 for

is orthogonal to

this leads to an infinite family of orthogonality relations.

t

n k n

n

K KM KM K n k

KM KM K


36/52

36

matrix satisfies two infinite families of orthogonality relations

with & being two special cases of these

family of orthgonality relations.

t t M I K

1 1 1

1 2 3

1 1 1

1 2 3

Do these additional orthogonality relations help is modeling

damping better?

: Rayleigh's damping model

C=

By virtue of d

C M K

M MK M MK MK M

K KM K KM KM K

Yes.

RecallGeneralization

iscussions in the preceding slides,

would be diagonal.

t C


37/52

37

The generalization admits as many free parameters as is neededto model damping. This number can match the number of modes for

which damping ratio is known.

In fact, damping can be specified i

Remarks

ndependently for all the modes

without worrying about the underlying matrix. Thus, for example,we can say that damping is 5% for all modes. A C matirx can be derived,

if needed, consistent with this

C

information, using generalized Rayleigh

damping model.

N


38/52

38

1

1

1

1 1

1

1 1

1

Diagonal 2

Naive approach: C=

More helpful approach:

Similarly

N

i i

t

i i

t

t t

t

t t t

t

t

C

M I M

M

M

M

C M M

Determination of a classical matrix given

This can be used even when all modes are

not included in the analysis (which invariably is the case).


39/52

39

Damping depends upon material, details of joints and supports.

Energy disspated in a cycle when a structure is oscillating harmonical

Modal damping when the sturcture is

made up of different materials

ly

is proportional to KE (or PE) in a cycle

(Excercise: show this for a sdof system).

Consider a structure that is oscillating in its -th mode. Let the structure

be made up of s number of mat

n

erials (e.g., RCC, steel, soil,...).

Let damping ratio for -th material in the -th mode.

(e.g., RCC~5%, steel ~2%, soil~10%).

Let equivalent damping ratio for the structure in the -th mode

n

r

n

eq

r n

n

.

Let total KE (or PE) stored in structral parts made up of r-th material

in the -th mode.

n

r E

n


40/52

40

1

1

can be computed based on undamped normal mode shapes.

sn n

r r n r eq s

nr

r

n

r

E

E

E


41/52

41

What happens if is not diagonal?t C

0 0; 0 ; 0

The matrix does not uncouple this set of equations.

We augment the above equation by the identity - 0

and write

- 0

0 - 0

0

MX CX KX F t X X X X

MX MX

MX CX KX F t

MX MX

M M X

M C K X

2 2 2 2 2 12 1

0

Introduce

00 - 0; ; ;

0 N N N N N N

X

F t X

M M X y A B p t

F t M C K X

A


42/52

42

is the new 2 1 response vector (state vector)

and B are the new structural matrices of size 2 2

and are symmetric ;

and are not positive defninite

and B are in gene

t t

Ay By p t

y t N

A N N

A B A A B B

A B

A

ral non-diagonal equations are coupled.

: Transform

2 2 transformation matrix to be selected such that

the transformation would uncouple the above equations of motion.

y Tz

T N N

y Tz ATz BTz p t

T

Strategy

(equations in the transformed coordinate system)

Aspiration: to select such that and are both diagonal so that

represents a set of 2 uncoupled equations wh

t t t ATz T BTz T p t q t

Az Bz q t

T A B

Az Bz q t N

ich can

be solved easily.

Ay By p t


43/52

43

: Determine T by determining the

eigensolutions associated with and matrices.

Free vibration analysis

0

Seek solutions of the form exp( )

exp( ) exp( ) 0

The

Ay By p t

A B

Ay By

y t R t

AR t BR t

BR AR

Strategy

required transofrmation matrix would be obtained by solving

this eigenvalue problem.

0

For nontrivial soutions the inverse of must not exist

characteristic equation 0

B A R

B A

B A

If is an eigenpair we haveR


44/52

44

* * * * *

* *

1

If , is an eigenpair we have

Taking conjugation on both sides we get

Note: and are real valued and &

If , is an eigenpair, , is also an eigenpair

Eigenvalues: ,

R

BR AR

BR AR A B A A B B

R R

* * *

2 1 2

* * *

1 2 1 2

, , , , , ,

Eigenvectors: , , , , , , ,

Recall

if exp exp

exp

N N

N N R R R R R R

X y X t t X t t X

y t R



45/52

45

Consider the th and th eigenpairs , & ,

Transpose both sides of the first of the above equations

S

m m n n

m m m

n n n

t t

n m m n m

t t

m n n m n

t t t t

m n m m n

m n R R

BR AR

BR AR

R BR R AR

R BR R AR

R B R R A R


ince &

Subtracting the above equations, we get 0

0 & 0

t t t t

m n m m n

t t

m n m m n

t t

m n n m n

t

m n m n

t t

m n m n

A A B B R BR R AR

R BR R AR

R BR R AR

R AR

R AR m n R BR m n


46/52

46

* * *1 2 1 2

* * * * * *

1 1 2 2 1 1 2 2

* * *

1 2 1 2

1

2

1 2

* *

Define the modal matrix

=

Let

0 0

0 0

= &

0

k k

k

k

N N

N N N N

N N

N

N

R R

R R R R R R


47/52

47

* * *

1 2 1 2

1 1 2

We have

0 & 0

Select the normalization constant such that

1 so that .

We also have

Define the modal matrix

=

t t

m n m n

t t

n n n n n

k k

k

k

N N

R AR m n R BR m n

R AR R BR

R R

R R R R R R

* * * * * *

2 1 1 2 2

* * *

1 2 1 2

N N N N

N N


48/52

48

1

2

1 2

* *

*

*

0 0

0 0Let = &

0

The orthogonality relations can now be written as

0

0

Use the transformation matirx .

N

N

t

t

A I

B

T


49/52

49

0

*

*

; 0

Transform

By virtue of orthogonality relations we have

0&

0

0

0

This represents a set of

t t t

t t

Ay By p t y y

y z

A z B z p t

A z B z p t q t

A I B

I z z q t

Forced response analysis

2 uncoupled first order equations.

Missions accomplished!

N


50/52

50

* *

* *

* * * *

* *

00

0

Let

00

0

( )

( )Initial conditions

0 0 0 0

t

t t

t t

t

t

t

I z z F t

u z

v

u u

F t v v

u u F t

v v F t

y z z Ay

Response when -th dof is driven by a unit harmonic excitationr


51/52

51

* *

1

* * *

1

0 0 1 -th dof 1 exp

( )

( )

( ) exp ; 1,2, ,

( ) exp ; 1,2, ,

li

t

t

t

N

j j j nj jr

n

N

j j j nj jr

n

F t r i t

u u F t

v v F t

u u F t i t j N

v v F t i t j N

*

*

* **

*

* *

*

*

1 1

exp expm & lim

lim exp

jr jr

j jt t

j j

N N kj jr kj jr

k kj j kj j k t

j j j j

i t i t u t v t

i i

x u x t u v

x v

x t u t v t x t i t

i i


52/52

52

* *

*1

* *

*1

lim exp exp

Frequency response function

N kj jr kj jr

k kr t

j j j

N kj jr kj jr

kr

j j j

x t i t i t i i

i i

Documents

Dynamics Ppt