Ch5 - Response of MDOF Systems

7/28/2019 Ch5 - Response of MDOF Systems

1/28

Response of MDOF systems

Degree of freedom (DOF): The minimum number of

independent coordinates required to determine completelythe positions of all parts of a system at any instant of time.

Two DOF systems Three DOF systems


2/28

The normal mode analysis (EOM-1)

Example: Response of 2 DOF system

m 2mk k k

x1 x2

FBD m 2mkx1 k(x1-x2) kx2

EOM 1211 )( xmxxkkx

2221 2)( xmkxxxk

In matrix form, EOM is

0

0

2

2

20

0

2

1

2

1

x

x

kk

kk

x

x

m

m


3/28

EOM -2 (example)

0

0

2

2

20

0

2

1

2

1

x

x

kk

kk

x

x

m

m

EOM

M K Fx x

)()()()( tttt FKxxCxM In general form

M is the inertia of mass matrix (n x n)

C is the damping matrix (n x n)K is the stiffness matrix (n x n)

F is the external force vector (n x 1)

x is the position vector (n x 1)


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Synchronous motion

From observations, free vibration of undamped MDOF system is a

synchronous motion.

All coordinates pass the

equilibrium points at the

same time

All coordinates reach

extreme positions at the

same time

Relative shape does notchange with time

21 xx constant

time

x1 x2 x1x2

No phase diff. between x1 and x2

)sin(11 tAx

)sin(22 tAx

)(

1

tjeA)(

2

tj

eA

or

or


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Response of 2DOF system (example-1)

0

0

2

2

20

0

2

1

2

1

x

x

kk

kk

x

x

m

m

EOM

Synchronous motion

Sub. into EOM

0

0

2

2

20

0

2

1

2

1

2

2

x

x

kk

kk

x

x

m

m

0

0

22

2

2

1

2

2

A

A

mkk

kmk

0KxMx )()(2 tt

0xMK )()(2

t

022

22

2

mkk

kmk0)det( 2 MK Characteristic

equation (CHE)

)sin(11 tAx

)sin(22 tAx

)(

1

tjeA)(

2

tjeA

or

or


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022

2

2

2

mkk

kmk

02

3

3

2

24

m

k

m

k

m

k634.01

m

k366.22

CHE

Solve the CHENatural frequenciesof the system;

0

0

22

2

2

1

2

2

A

A

mkk

kmk

k

mk

mk

k

A

A 2

2

2

1 22

2

1

731.0

)634.0(2

)1(

2

1

m

m

kk

k

A

A

2

73.2

)366.2(2

)2(

2

1

m

m

kk

k

A

A

From


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1

731.0

)1(

2

1

A

A73.2

)2(

2

1

A

AAmp. ratio Amp. ratio

1

731.0)(1 x

1

73.2)(2 x

The first mode shape The second mode shape

0.731 1

-2.73

1

2

same directionOpposite

direction


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In general, the free vibration contains both modes

simultaneously (vibrate at both frequencies simultaneously)

)sin(1

73.2)sin(

1

732.0222111

2

1

tctcx

x

2121 ,,, cc are constants (depended on initial conditions)


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Initial conditions (1)

)sin(1

73.2

)sin(1

732.0

2221112

1

tctcx

x

4

2

)0(

)0(

2

1

x

x

0

0

)0(

)0(

2

1

x

x

Initial conditions and

)cos(1

73.2)cos(

1

732.022221111

2

1

tctcx

x

Velocity response

42

)0()0(

2

1

xx 2211 sin

173.2sin

1732.0

42

cc

0

0

)0(

)0(

2

1

x

x

222111 cos1

73.2cos

1

732.0

0

0

cc


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2211

sin1

73.2sin

1

732.0

4

2

cc

222111 cos1

73.2cos

1

732.0

0

0

cc

4 Eqs.,

4 unknowns

Solve for four unknowns

,732.31 c ,268.02 c 2/21

)2

sin(173.2268.0)

2sin(

1732.0732.3 21

2

1

ttxx

The response is

ttx

x21

2

1cos

268.0

732.0cos

732.3

732.2


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)sin(1

73.2)sin(

1

732.0222111

2

1

tctcx

x

2

464.1

)0(

)0(

2

1

x

x

0

0

)0(

)0(

2

1

x

x

(a) Initial conditions and

1

73.2

)0(

)0(

2

1

x

x

0

0

)0(

)0(

2

1

x

x

(b) Initial conditions and

Try to do


12/28

Summary (Free-undamped) (1)

0KxxM )()( tt

The motion is synchronous:

constant and

0KxMx )()(2 tt

0xMK )()( 2 t

Eigen value problem

0)det( 2 MKCharacteristics equation

2

n Eigen value

nNnn ,,, 21 Nnatural freq.

0xMK ini )(2

ix Eigen vector

)sin( tAx )( tjeAor

N mode shapesNxxx ,,, 21

EOM1

2

3

4 5

Direct Method


13/28

Summary (Free-undamped) (2)

Free-undamped response

)sin()sin()sin()( 22221111 NNNN tAtAtAt xxxx

N

i

iiii tAt

1

)sin()( xx

6

where A and are from initial condition x(0) and v(0)

Direct Method


14/28

Example

1k

2k

x

2l

1l

Determine the normal modes of vibration of an automobile simulated by

simplified 2-dof system with the following numerical values

lb3220W

ft5.41 l lb/ft24001 k

ft5.52 l lb/ft26002 k

ft4r2rg

WJC


15/28

Rigid body mode

Rigid body mode is the mode that the system moves as a

rigid body. The system moves as a whole without any relative motion

among masses. There is no oscillation. 0n


16/28

Forced harmonic vibration (1)

Example

EOMt

F

x

x

kk

kk

x

x

m

m

sin

00

0 1

2

1

2221

1211

2

1

2

1

System is undamped, the solution can be assumed as

tX

X

x

x

sin

2

1

2

1

Sub. into EOM

0

1

2

1

222221

12

2

111F

X

X

mkk

kmk

0)(

1

2

1 F

X

XZSimpler notation,

0

)(11

2

1 FZ

X

X


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0)(

)(adj

0)(

111

2

1 F

Z

ZF

ZX

X

))(()( 22222

121 mmZWhere

1 and 2 are natural frequencies

0)(

1 12

11121

12

2

222

2

1 F

mkk

kmk

ZX

X

The amplitudes are))((

)( 222

22

121

1

2

2221 mm

FmkX

))(( 22222

121

1212

mm

FkX


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))((

)2(22

2

22

1

2

1

2

1

m

FmkX

))((

22

2

22

1

2

12

m

kFX

m mk k k

x1 x2

F1sint

tF

x

x

kk

kk

x

x

m

m

sin

02

2

0

0 1

2

1

2

1

m

k

m

k 3, 21

EOM

Force response of

a 2 DOF system

0 1 2 3

0

1

2

3

4

5

-1

-2

-3

-4

-5

F

Xk

1

1

2

F

kX

1

1

F

kX

1 2

Same

direction

Opposite

direction


19/28

Solving methods


20/28

Modal analysis

is a method for solving for both transient and steady state

responses of free and forced MDOF systems through

analytical approaches.

Uses the orthogonality property of the modes to

decouple the EOM breaking EOM into independent

SDOF equations, which can be solved for response

separately.

Introduction


21/28

Coordinate coupling

1k

2k

2l

1l

mg

x Ref.

)( 11 lxk)( 22 lxk

1k

2k1

l mg

1x Ref.

11xk

)( 12 lxk

0

0

0

02

22

2

111122

112221 x

lklklklk

lklkkkx

J

m

0

012

22

2211

11

1 x

lklk

lkkkx

Jml

mlm


22/28

Concept of modal analysis

0

)(

0

02

22

2

111122

112221 tFx

lklklklk

lklkkkx

J

m

)()()( ttt FKxxM

)(

)(

0

0

10

01

2

1

2

1

2

2

2

1

2

1

tN

tN

r

r

r

r

n

n

)()()( ttt Nrr

EOM in physical coordinate (Coordinates are coupled)

EOM in modal coordinate (Independent SDOF equations)

Solve for )(tr

Transform r(t) back to x(t)


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Orthogonality

x = eigen vector (vector of mode shape)

iii

T

i MMxx

jiiTj ,0Mxx jiiTj ,0Kxx

iii

T

i KKxx

IfM and K are symmetric and then xi and xj are

said to be orthogonal to each other.njni


24/28

Normalization

u = normalized eigen vector (respect to mass matrix)

1iT

i Muu

jiiT

j ,0Muu

0xMK )()( 2 tFrom eigen value problem

constantis, CC ii xu

0uMK )()( 2 t

or iii MuKu2

22

ii

T

iii

T

i MuuKuu


25/28

Modal matrix

Modal matrix is the matrix that its columns are the mode

shape of the system

nuuuU 21

Then

100

010

001

IMUUT

2

2

2

2

1

00

00

00

nN

n

n

T

KUU

(Spectral matrix)


26/28

Modal analysis (undamped systems)-1

1. Draw FBD, apply Newtons law to obtain EOM

2. Solve for natural frequencies through CHE

3. Determine mode shapes through EVP

4. Construct modal matrix (normalized)

Procedures

)()()( ttt FKxxM

0xMK )()( 2 t0)det( 2 MK

nuuuU 21IMUU T

KUU T

5. Perform a coordinate transformation )()( tt Urx

)()()( ttt FKxxM )()()( ttt FKUrrMU

)()()( ttt TTT FUKUrUrMUU

)()()( ttt TFUrr


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)()()( ttt TFUrr

)(

)(

)(

)(

)(

)(

)(

)(

)(

)(

)(

00

00

00

)(

)(

)(

4

3

2

1

4

3

2

1

21

22221

11211

2

1

2

2

2

2

1

2

1

tN

tN

tN

tN

tF

tF

tF

tF

uuu

uuu

uuu

tr

tr

tr

tr

tr

trT

NNNN

N

N

NnN

n

n

N

Independent SDOF equations, can be solve forr(t)

6. Transform the initial conditions to modal coordinates

)()( ttUrx

IMUU T

)0()0(Urx

)0()0( MUrUMxU TT

)0()0( MxUr T

From

and )0()0( xMUr T


28/28


7. Find the response in modal coordinates

8. Transform the response in modal coordinateback to that in original coordinate )()( tt Urx

)(tr

)(tr)(tx

Ch5 - Response of MDOF Systems

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