multidegree of freedom systems

7/29/2019 Multidegree of Freedom Systems

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Multi-Degree of Freedom Systems

1

1. General n-degree of freedom vibration system

1.1. Matrices in the equations of motion

[ ]{ } [ ]{ } [ ]{ } { }( )x C x K x F t+ + =

when [ ]

1 11 12 1

2 21 22 2

1 2

0 0

0 0

0

0

n

n

n n n nn

m m m m

m m m mM

m m m m

= =

" "

"

# % # # % #

" " "

[ ]

11 12 1

21 22 2

1 2

=

n

n

n n nn

k k k

k k kK

k k k

"

"

# # % #

"

1.2. Positive definite & Positive semi-definite.

Positive definiteness

An nn matrix

[ ]

11 12 1

21 22 2

1 2

n

n

n n nn

m m m

m m mM

m m m

=

"

"

# # % #

"

is positive definite if for any given vector

{ }

1

2

n

x

=

#, { } [ ]{ } 0

Tx M x and { } [ ]{ } 0

Tx M x = if and only if { } 0x = .

usually diagonal forlumped-mass &

stiffness systems

In general, it is an n by npositive definite matrix.

In general, its an nxn matrix.

It usually has a narrow diagonal bandwidth.In general, it is positive semi-definite.If there is no rigid body mode allowed, it

should be positive definite.


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2

Positive Definiteness of the mass matrix

The mass matrix [ ]

1

2

0 00 0

0

0 0 n

mm

M

m

=

""

# # %

"

is positive definite.

[proof]

Consider the kinetic energy of a vibration system

( ) ( ) ( ) { } [ ]{ }2 2 2

1 1 2 2

1 1 1 1= + + + =

2 2 2 2

T

k n nE m x m x m x x M x "

where { }

1

2=

n

x

xx

x

#

and [ ]

1

2

0 0

0 0=

0

0 0 n

m

mM

m

"

"

# # %

"

.

Since 0kE and =0kE if and only if { } { }= 0x , the mass matrix [ ] is positive definite.

Positive Semi-Definiteness

A nn matrix [ ]

11 12 1

21 22 2

1 2

=

n

n

n n nn

k k k

k k kK

k k k

"

"

# # % #

"

is positive semi-definite if for any given vector

{ }

1

2

=

n

x

# , { } [ ]{ } 0T

x K x .


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3

Positive semi-definiteness of the stiffness matrix

The stiffness matrix [ ]

11 12 1

21 22 2

1 2

=

n

n

n n nn

k k kk k k

K

k k k

""

# # % #

"

is positive semi-definite.

[Proof]

Consider the process which make the system change from { }0

0

0

0

x

=

#to { }

1

2

n

n

x

xx

x

=

#

1st process: { }

0

0

0

0

x

=

# { }

1

1

0

0

x

=

# i.e.x1 changes from 0 tox1 whilexj = 0 for all

j > 1. (xj is constrained such thatxj = 0.)

2nd process : { }

1

1

0

0

x

x

=

# { }

1

2

20

0

x

x

=

#

i.e.x2 changes from 0 tox2 whilex1= x1 and

xj = 0 for allj > 2.

i-th process : { }

1

1

1

1 0

0

0

i

i

xx

=

#

#

{ }

1

1

1

0

0

i

ii

x

x

x

=

#

#

i.e.xi changes from 0 toxi whilex1= x1,

x2= x2 , .. ,xi-1= xi-1 , andxj = 0 for allj > i.

i-th


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n-th process : { }

1

2

1

1

0

n

n

x

x

=

# { }

1

2

1

n

n

n

x

x

x

x

x

=

# i.e.xn changes from 0 toxn while and

xj =xj for allj < n so that the final of the system the whole process become the desired state {x}n.

Now calculate the potential energy. The increase of the potential energy through the each sub-process

can be calculated as follows:

1st process:1 1 2

1 1 1 11 1 1 11 10 0

1( ) ( )2

x x

pE F dx k x dx k x = = =

( Since 0 1jx j= , no contribution is made by the force exerted on 1jm j )

2nd process:2 2 2

2 2 2 21 1 22 2 2 21 1 2 22 20 0

1( ) ( ) ( )

2

x x

pE F dx k x k x dx k x x k x = = + = +

# # #

i-th process:

1 1 2 20 0

2

1 1 2 2 ( 1) 1

( )

1 ( )2

i ix x

p i i i i i ii i i

i i i i i i i i ii i

E F dx (k x k x k x )dx

k x x k x x k x x k x

= = + + +

= + + + +

"

"

# # #

n-th process:

1 1 2 20 0

2

1 1 2 2 ( 1) 1

( )

1( )

2

n nx x

p n n n n n nn n n

n n n n n n n n nn n

E F dx (k x k x k x )dx


= = + + +

= + + + +

"

"

Therefore, the potential energy of the final state will be

=

===

+

==n

i

i

j

jiji

n

i

iii

n

i

ipp xxkxkEE2

1

11

2

1

)(2

1)(

If [K] is symmetric, then

1 12

2 1 2 1 1 1 1

1 1( )

2 2

n i n i n n n

ji i j ij i j ij i j ii i

i j i j i j i


= = = = = = =

= =

11

21 22

31 32 33

41 42 43 44

1 2 3 4n n n n nn

k

k k

k k k

k k k k

k k k k k

# # # # %

"


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5

Therefore { } [ ]{ }1 1

1 1

2 2

n nT

p ij i j

i j

E k x x x K x= =

= =

SinceEp is always non-negative, [K] is positive semi-definite.Note that Ep= 0 when the system moves in a rigid body translation. Hence, Ep= 0 even when

{x}{0}. This implies that [K] is not positive definite sometimes. Therefore, we can guarantee that

[K] is positive semi-definite only.

The stiffness matrix is symmetric.[Proof] Now we will try to prove that [K] is symmetric.

Consider the two processes. Process and Process .

Process

-1st process: { }

0

0

0

0

Ix

=

# { }

1

0

0

0

0

00

0

0

i

I

x

x

=

#

#

#

-2nd process:{ }

1

0

0

0

0

0

0

0

0

i

Ix

=

#

#

#

{ }2

0

0

0

0

0

0

0

i

I

j

x

=

#

#

#

j-th

j-th

i-th i-th

i-th

j-th


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Process:

-1st process: { }

0

00

0

0

0

0

0

0

0

IIx

=

#

#

#

{ }1

00

0

0

0

0

0

0

II

j

x

x

=

#

#

#

-2nd

process :{ }1

0

0

0

0

0

0

0

0

II

j

x

=

#

#

#

{ }2

0

0

0

0

0

0

0

i

II

j

x

=

#

#

#

The potential energy obtained by the process, Process

2 21 1( ) ( ) ( )2 2

P I ii i jj j ji i jE k x k x k x x= + +

The potential energy obtained by process

2 21 1( ) ( ) ( )2 2

P II jj j ii i ii j iE k x k x k x x= + +

Since the final state is identical and the potential energy is determined by the state independently

the process through which the final state is achieved

ijjiPP kkEE == )()(

i-th

j-th

i-th

j-th

i-th

j-th

i-th

j-th


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Since i,j are arbitrarily chosen, ,ij jik k i j= [ K] is symmetric.

This completes the proof of the theorem [K] is positive semi-definite.

2. Responses ofn-degree of freedom systems

z It is time to go back to the equation of motion for the multi-degree of freedom vibration system.

z The main idea of the method is that we can solve the coupled n-differential equations by a proper

transformation and by solving the uncoupled n-differential equations.

z The problem or questions is how we can find an appropriate transformation that makes the coupled

n-differential equations into uncoupled n-differential equations

2.1. Solving Technique

Suppose we can find a linear transformation such that

{ } [ ]{ }z= i.e.

1 111 12 1

2 21 22 2 2

1 2

( ) ( )

( ) ( )

( ) ( )

n

n

n n nnn n

x t z t

x t z t

x t z t

=

"

"

# # % ## #"

and

[ ] [ ][ ] [ ]1

1 0 0

0 1 0

0 0 1

M I

= =

"

"

# # % #

"

and

[ ] [ ][ ]

1 1

1 2 2

2 0 00

0 2 0

20

0 0 2

n n

n n

C

= =

"%

"

# # % #%

"

and

[ ] [ ][ ]

2

1

21 2 2

2

0 00

0 0

00 0

n

n

K

= =

"%

"

# # % #%

"

.


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The equation of motion

[ ]{ } [ ]{ } [ ]{ } { }x C x K x F+ + =

Since { } { }[ ] z= , { } { }[ ]x z= , { } { }[ ]x z= , Eq. can be re-written as

[ ][ ]{ } [ ] [ ]{ } [ ][ ][ ] { }z C z K z F + + =

Now Multiply the both side of Eq by [ ]T

to yield

[ ] [ ][ ]{ } [ ] [ ][ ]{ } [ ] [ ][ ][ ] [ ] { }T T T T

z C z K z F + + =

By the assumption, becomes

{ } { } { } [ ] { }20 0

2

0 0

T

n n nz z z F

+ + =

% %

% %

or2

1

2 1,2, ,3n

i i i i ji j

j

z z z F i =

+ + = = "

since ji is the component of []T

at i-th row andj-th column.

Eq implies a set of independent n-differential equations whose solutions are known as

( ) ( ) ( )

( ) ( ) ( ) ( )

0 00

1

01

cos sin

1sin

i i

i i

t i i i ii i id id

id

nt t

j idij

jid

z zz t e z t t

F e t dt

=

+= +

+

Once all thezi(t)s are obtained, the response of the system {x(t)} can be obtained by using the

transformation {x(t)} = []{z(t)} i.e. ( ) ( )1

n

i ij j

j

t z t=

= .

Therefore, the response (, whether free or forced, ) of the system can be easily obtained. The

remaining question is Does there exist such a matrix [] ?

The answer is Yes. So, we will show how to find it!

2.2. Normalization with respect to Mass Matrix [M]

For many vibration systems, the mass matrix is a diagonal matrix, i.e.


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[ ]

1

2

3

0 0 0

0 0 0

0 0 0

0 0 0 n

m

m

M m

m

=

"

"

"

# # # % #

"

.

Note that [M] is symmetric. Now consider

1

2

1/ 2

3

1/ 0 0 0

0 1/ 0 0

0 0 1/ 0

0 0 0 1/ n

m

m

M m

m

=

"

"

"

# # # % #

"

.

Then, [ ] [ ]1/ 2 1/ 2

1 0 0

0 1 0

0 0 1

T

M M M I

= =

"

"

# # % #

"

.

Consider a new variables { }

1

2

( )

( )( )

( )n

y t

y ty t

y t

=

#such that { } { }1/ 2M y = .

Then, the equation of motion will be

[ ] { } [ ] { } [ ] { } { }1/ 2 1/ 2 1/ 2M y C M y K M y F + + = 4

Multiply the both sides of Eq. by1/ 2

T

M to yield

[ ] { } [ ] { }

[ ] { } { }

1/ 2 1/ 2 1/ 2 1/ 2

1/ 2 1/ 2 1/ 2

T T

T T

M M M y M C M y

K M y M F

+

+ =

{ } [ ] { } [ ] { } { }1/ 2 1/ 2 1/ 2 1/ 2 1/ 2T T T

y M C M y M K M y M F + + =

Let [ ]1/ 2 1/ 2T

K M K M = . Note that K is symmetric.


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2.3. Diagonalization of K

For simplicity, consider the case that the damping force is zero. i.e. [ ] 0C . Then, Eq can be

reduced to the following:

{ } { } { }1/ 2T

y K y M F + =

Does there exist a transformation matrix [] such that

[ ] [ ]

2

1

21 2 2

2

0 00

0 0

00 0

n

n

K

= =

"%

"

# # % #%"

.

Now we can use the important theorem in Linear Algebra.

SimilarityEvery real symmetric matrix is orthogonally similar to a real diagonal

matrix.

Orthogonal Matrix.For a nXn matrix [], [] is orthogonal if []T [] = [I]

This implies that []T= []-1 ! Similarity.Two nXn matrices [A] and [B] are similar if there exists a (transformation)

matrix[] such that[]-1[A][]=[B].

The theorem Every real symmetric matrix is orthogonally similar to a real diagonal matrix.

implies that: For a real nn symmetric matrix K , there exists a transformation matrix []

that is orthogonal such that


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[ ] [ ] [ ] [ ]

2

1

21 2 2

2

0 00

0 0

00 0

T

n

n

K K

= = =

"%

"

# # % #%

"

In order that i (I= 1, 2, , n ) are real, K should have real non-negative eigenvalues.

This is true because [K] is positive semi-definite.

The proof of this claim is simple. Suppose there exists a negative eigenvalue i2

of K ,

{ } { } { } [ ] [ ]{ } { } { } ( ) ( )2 22

1

0

0

nTT T T

n i i

i

y K y z K z z z z =

= = =

%

%

.

Consider the case that 0i

z and 0iz = i j .

Then, { } { } ( ) ( )2 2

0T

i iy K y w z =


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{ } { } [ ] { }2 1 20

0

TT

nz z M F

+ =

%

%

or ( )2

1

( )n

i i ji j

j

z z F t =

+ = (i = 1, 2, 3, , n)

where [] = [M-1/2][] and ij is the i-th row andj-th column component of [].

Note that { } { } [ ]{ } [ ]{ }1 2 1 2M y M z z = = = .

In other words, [ ]{ } [ ]{ } { }x K x F+ = [ ][ ]{ } [ ] [ ]{ } { }z K z F + =

[ ] [ ][ ]{ } [ ] [ ][ ]{ } [ ] { }T T T

z K z F + =

{ } { } [ ] { }20

0

T

nz z F

+ =

%

%

For the given initial conditions, { }

10

200

0

t

n

xx=

=

#, { }

10

200

0

t

n

x

x

x

=

=

#

and

a given forcing function { }

1

2

( )

( )

( )n

F t

F tF

F t

=

#,

since []-1 = ([M-1/2][])-1 = []-1[M-1/2] = []T[M1/2] = []T[M-1/2][M1/2][M1/2] = []T[M-1/2]T[M] =

([M-1/2][])T[M] = []T[M], we can find the corresponding initial condition for {z(t)} by the following

formula:


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{ } [ ] { }

1 0

1

10

2 02010 0

0

0

1

( ) [ ] ( )

n

j j j

j

n

T j j j

jt t

n n

jn j j

j

m x

zm xz

z t M x t

z

m x

=

== =

=

= = =

##

{ } [ ] { }

1 0

1

10

2 02010 0

0

0

1

( ) [ ] ( )

n

j j j

j

n

T j j j

jt t

n n

jn j j

j

m x

z

m xzz t M x t

z

m x

=

== =

=

= = =

##

and

{ } [ ] { }1

( )n

T

ji j

j

Q F F t =

= =

or

( )0 01

n

i ji j j

j

z m x=

= , ( )0 01

n

i ji j j

j

z m x=

= ,1

( ) ( )n

i ji j

j

Q t F t =

= .

2.4. Solution

Solve the n-ordinary differential equations:

( )2

( )i i i iz z Q t+ = with the I.C. 0(0)i iz z= , 0(0)i iz z= (i = 1, 2, 3, , n)

When ( )1( ) ( )n

i ji j

jQ t F t

== [ ] [ ] [ ]

00

0

1cos sin ( )sin ( )

ti

i i i i i i

i i

zz z t t Q t d = + +

.

Since { } [ ]{ }( ) ( )ix t z t= , [ ]1

( ) ( )n

i il l

l

t z t=

= .


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2.5. Modal Analysis

Let { }i

be the i-th column of the matrix [ ] . i.e. { }

1

2

i

i

i

ni

=

#.

Note that

{ } [ ]{ } { } { } { } { }

{ } { }

1 2

1 1

( ) ( ) , , , ( )

( ) ( )

i in

n n

i ii ii i

x t z t z t

z t z t

= =

= =

= =

"

For free vibrations, ( ) 0iF t i . [ ] [ ]0

0( ) cos sini

i i i i

i

zz t z t t

= + .

If the I.C. is given such that0

0i

z and 0 0jz = j i and 0 0jz = j .

Then, { } [ ]{ }0( ) cosi i ix t z t = .

This implies that the response of the system has only one frequency component and the

amplitudes of vibration at other masses are determined if the amplitude of vibration at onemass is given.

In other words, the free vibration of system can be represented as a superposition ofn-fixed-

pattern vibrations with frequencies 1, 2, , n.

The fixed pattern of vibration {}i corresponding to a natural frequency i is called the

corresponding mode shape to a natural frequency i.

The combination of a natural frequency, i, and its corresponding mode shape is called a

(vibration) mode of the given vibration system.

In the case of forced vibrations, (If {x}t=0 = {0} and 0{ } {0}tx = = )

0

1( ) ( ) sin[ ( )]

t

i i i

i

z t Q t d

=

01 1

1{ ( )} [ ]{ } ( ){ } { } ( ) sin[ ( )]

n nt

i i i i i

t t i

x t z z t Q t d = =

= = =

Therefore, in the case of forced vibrations, the response of the system can be seen as a


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superposition of modal responses.

i-th modal response would be zi(t){}i. Note that the mode shape is fixed in spite of any

given forcing function.

2.6. Damped System

Now it is time to consider a mechanical vibration system with a damper. The equation of motion will

be as follows;

[ ]{ } [ ]{ } [ ]{ } { }x C x K x F+ + =

There are various mechanisms for damping forces. Hence, it is impossible to show that [C] is

symmetric, or so. But the damping forces are usually very small compared to the spring forces in most

case on structural dynamics.

Now we propose a special case in which the equation of motion can be solved by the

similar method to that for undamped system response.

Suppose [ ] [ ] [ ]C M K = + , i.e. The damping coefficient matrix [C] can be

represented by a sum of a constant () times [M] and a constant () times [K]. Then,

2

1

22

2 2

3

2

[ ] [ ][ ] [ ] [ ][ ] [ ] [ ][ ] [ ] [ ][ ] [ ] [ ][ ]

0 0 0

0 0 0

[ ] [ ] 0 0 0

0 0 0

T T T T T

n

n

C M K M K

I

= + = +

+

+ = + = + +

%%

"

"

"

# # # % #

"

Another important case is that [C] is such a matrix that

1 1

2 22

2 0 00

0 2 0[ ] [ ][ ] 2

00 0 2

T

n n

n n

C

= =

"%

"

# # % #%"

Note that if [ ] [ ] [ ]C M K = + ,2 1

2 2 2

ii i

i i

+= = + . i.e. This is a

special case of the above case. ( i.e.

0

[ ] [ ][ ] 2

0

T

n nC

=

%

%

)

Then, The equations of motion will be:


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[ ] [ ][ ]{ } [ ] [ ][ ]{ } [ ] [ ][ ]{ } [ ] { }T T T T z C z K z F + + =

2{ } [ 2 ]{ } [ ]{ } { }n n nz z z Q + + =% %

% % or

22 ( ) ( )i i i i i i iz z z Q t + + = i = 1,2,3,,n.

The ordinary differential equations22 ( )i i i i i i iz z z Q t + + = can be solved. For

given initial conditions,

( )( )

( )0 00

0

1( ) { cos[ ] sin[ ]} ( ) sin[ ( )]

( ) ( )

ipih

i i i i

Z tZ t

tt ti i i i

i i id id i id

id id

ih ip

z zz t e z t t Q e t d

z t z t

+= + +

= +

Since1 1

free vibration terms force vibration termsinduced by the initial induced by the externalcondition forces

{ ( )} [ ]{ ( )} ( ){ } ( ){ }n n

ih i ip i

i i

x t z t z t z t = =

= = +

.

2.7. How to find []: modal matrix

Now the question is how we can find the modal matrix [] where [] = [M

1/2

][]. Note that[] = [M-1/2][], consider the i-th column of [], {}i. Since [] is orthogonal,

{ } { }Ti j ij = . Since2

0

[ ] [ ][ ]

0

T

nK

=

%

%

Note that since1 12 2[ ] [ ] [ ][ ]TK M K M

= , 2

0 if{ } [ ]{ }

ifT

i j

j

i jK

i j

=

=

Since [ ]{ }jK is an n-dimensional vector, [ ]{ }jK should be linearly dependent on

{}j, i.e.2[ ]{ } { }

j j jK = . Then

2 2 2{ } [ ]{ } { } { } { } { }T T Ti j i j j j i j j

K = = =

Therefore, if we find all the eigenvalues and eigenvectors of [ ]K , we can construct pre-

modal matrix such that1 2

[ ] [{ } ,{ } , ,{ } ]n = " .

The eigenvectors of [ ]K can be found by solving the eigenvalue problem


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2[ ]{ } { }K = .

Since1 12 2[ ] [ ] [ ][ ]TK M K M

= ,

12[ ]{ } { }M

= and

1 12 21[ ] [ ]M

= ,

Eq becomes1 12 22[ ][ ]{ } [ ]{ }M K M

= .

Multiply by12[ ]

1 12 22 2[ ]{ } [ ][ ]{ } [ ]{ }K M M M = =

Hence, we can find the modal matrix [] by solving the associated eigenvalue problem,2[ ]{ } [ ]{ }K M =

Now lets solve the associated eigenvalue problem

2.8. Solution Method for Eigenvalue Problem

[ ] [ ] [ ] [ ]2 2{ } { } { } 0K M K M = =

In order to { } {0} , [ ] [ ]2det 0K M = --- or

211 1 12 1

2

12 22 2 2

2

1 2

det 0

n

n

n n nn n

k m k k

k k m k

k k k m

=

"

"

# # % #

"

---

This gives n-th order polynomial equation of2. Therefore, we can find n-values for2.

Let 12, 2

2, . , n

2be the n-roots of the polynomial equation. Then, we can find the eigenvector

(or modal vector) by solving the following equation:

2111 1 12 1

2212 22 2 2

2

1 2

0

0

0

ii n

ii n

nin n nn i n

k m k k

k k m k

k k k m

=

"

"

# ## # % #

"

---


18/23


18

Equation

provide the direction ofi-th modal vector. Hence, first set { }

1

2

i

i

ii

ni

=

#

and determine i by the orthogonality of the modal matrix, i.e.{ } [ ]{ } 1T

i i = and

obviously { } [ ]{ } 0T

j i = if i j .

The modal matrix [] can be constructed by collection the eigenvectors {}i, i.e.

[ ] { } { } { }1 2, , , n = " .

3. Exercises

3.1.

1 1 22 0mx kx kx+ =

2 1 22 0mx kx kx + =

1 1

2 2

0 2 0

0 2 0

x xm k k

x xm k k

+ =

[Step 1] Solve the associated eigenvalue problem. [ ]{ } [ ]{ }2K x M x= .

1 12

2 2

2 1

1 2

x xk

x xm

=

2 1det 0

1 2

=

where2

2

0

= and 20

km

=

( )2 22 1 4 3 0 = + = , ( ) ( )3 1 0 = 1,3 = .

1 0 2 0, 3 3

k k

m m = = = =

[Step 2] Find eigenvectors

Case (i)0 =

m

m

k

2k

k


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19

{ }11 11 1112 12

2 1 1 1 1 10 0

1 2 1 1 1 1

= = =

Normalization1 1

1 0 11

1 0 1

Tm

m

=

.

2

1

1 11

1 1

T

m =

.2

12 1m = .

Since 11

2m = , ( )

1

1 21

1 2m

=

Case (ii)0

3 =

{ }2

1 10

1 1

=

{ } 221

1

=

Normalization:2 2

2 2

1 12 1

1 1

T

m m

= =

. Hence,2

1

2m =

{ }2

1 21

1 2m

=

[Step 3] Construct a Modal matrix [].

[ ] { } { }1 2

1 11,

1 12m

= =

[Step 4] Solve the uncoupled differential equations.

( ) ( ) ( )2 101 0 1 1 10 0 00

0 cos sinz

z z z t z t t

+ = = +

( ) ( ) ( )2 202 0 2 2 20 0 00

3 0 cos 3 sin 3z

z z z t z t t

+ = = +

[ ]110 10

20 20

z x

z x

=

, [ ]

110 10

20 20

z x

z x

=

or

( )( )

( ) ( ) { } ( ) ( ) { }

( ) ( ) ( ) ( )

1

1 0 1 0 2 0 2 01 22

1 0 1 0 2 0 2 0

cos sin cos 3 sin 3

1 11 1cos sin cos 3 sin 3

1 12 2

x tt t t t

x t

t t t t m m

= + + +

= + + +

Apply I.C.s

m1

m2

m1 m2

1st mode

2nd mode


20/23


20

101 2

20

100 01 2

20

1 1

1 12 2

1 13

1 12 2

x

xm m

x

xm m

+ =

+ =

Solve these equations to get 1,

2,

1,

2.

3.2. : Three-degree of freedom system

Equation of motion1 1 1 1 2 1 2 1

2 2 2 2 1 3 2 3 2

3 3 3 3 2 3

( ) ( )

( ) ( ) ( )

( ) ( )

m x k x k x x f t

m x k x x k x x f t

m x k x x f t

= +

= +

= +

Express the above equations in matrix forms.

1 1 1 2 2 1 1

2 2 2 2 3 3 2 2

3 3 3 3 3 3

0 0 0 ( )

0 0 ( )

0 0 0 ( )

m x k k k x f t

m x k k k k x f t

m x k k x f t

+ + + =

. (1)

Consider the case that m1 = m2 = m3 = m, k1 = k2 = k3 = k. Then, Eq. (1) becomes:

1 1 1

2 2 2

3 3 3

1 0 0 2 1 0 ( )

0 1 0 1 2 1 ( )

0 0 1 0 1 1 ( )

x x f t

m x k x f t

x f t

+ =

Step I : solve the associated eigen-value problem.

2 2 2

2 1 0 1 0 0

[ ] [ ] 1 2 1 0 1 0 det [ ] [ ] 0

0 1 1 0 0 1

K M k m K M

= =


21/23


21

2 1 0

det 1 2 1 0

0 1 1

=

2(2 ) (1 ) (1 ) (2 ) 0 = 3 25 6 1 0 3.24698, 1.55496, 0.198062 + =

1

2 1 1

3

1.801938 1 0 0 0.554958

1 1.801938 1 0 { } 1.000

0 1 0.801938 0 1.24698

x

x

x

= =

1

2 2 2

3

0.44504 1 0 0 1.0001 0.44504 1 0 { } 0.44504

0 1 0.55496 0 0.80193

xx

x

= =

1

2 3 3

3

1.24698 1 0 0 1.000

1 1.24698 1 0 { } 1.24698

0 1 2.24698 0 0.55496

x

x

x

= =

Step II : Construct a normal matrix

[ ]1 2 3

1 2 3 1 2 3

1 2 3

0.554958

[ ] { } { } { } 0.44504 1.24698

1.24698 0.80193 0.55496

= =

Since [ ] [ ][ ] [ ]T I = , 1 1{ } [ ]{ } 1T M = , hence,

2 2 2 2 2

1 1{ (0.554958 1.0000 1.24698 )} 2.862938m m + + = . 1

0.59101

m =

2 2 2

2 2 2{ } [ ]{ } 1 { (1 0.44504 0.80193 )} 1T M m = + + = 2

0.736979

m =

2 2 2

3 3 3{ } [ ]{ } 1 { (1 1.24698 0.55496 )} 1T M m = + + = 3

0.591009

m =

Therefore,

0.327986 0.736979 0.5910091

[ ] 0.59101 0.327985 0.736976

0.73698 0.591006 0.327986m

=

.


22/23


22

Step IV : Solve the uncoupled differential equations2 2

1 0 1 10.44504 ( )z z Q t+ = where { ( )} [ ] { }TQ t f= .

Therefore,1 11 1 21 2 31 3( ) ( ) ( ) ( )Q t f t f t f t = + + . i.e.,

1 1 1 2 3( ) { } { } [0.327986 ( ) 0.59101 ( ) 0.73698 ( )] /TQ t f f t f t f t m= = + + .

2 2

2 0 2 21.24698 ( )z z Q t+ = where

2 2 1 2 3( ) { } { } [0.736979 ( ) 0.327985 ( ) 0.591006 ( )] /TQ t f f t f t f t m= = +

2 2

3 0 3 31.80194 ( )z z Q t+ = where

3 3 1 2 3( ) { } { } [0.591009 ( ) 0.736976 ( ) 0.327986 ( )] /TQ t f f t f t f t m= = + .

For the given initial conditions,x1(0) =x2(0) =x3(0) = 0, 1 2 3(0) (0) (0) 0x x x= = = ,

1 1 2 3 00

0

1 1( ) [0.327986 ( ) 0.59101 ( ) 0.73698 ( )]sin[0.44504 ( )]0.44504

t

z t f f f t dm

= + +

2 1 2 3 00

0

1 1( ) [0.736979 ( ) 0.327985 ( ) 0.591006 ( )]sin[1.24698 ( )]

1.24698

t

z t f f f t dm

= +

3 1 2 3 00

0

1 1( ) [0.591009 ( ) 0.736976 ( ) 0.327986 ( )]sin[1.80194 ( )]

1.80194

t

z t f f f t dm

= +

Step V : Transform and obtain the solutionSince {x} = []{z} =z1(t){}1 +z2(t){}2 +z3(t){}3 ,


23/23


23

1

2 1 2 3 00

0

3

1 2 30

0

( ) 0.3279861

( ) [0.327986 ( ) 0.59101 ( ) 0.73698 ( )]sin[0.44504 ( )] 0.591010.44504

( ) 0.73698

1[0.736979 ( ) 0.327985 ( ) 0.591006 ( )]sin[1

1.24698

t

t

x t

x t f f f t dm

x t

f f fm

= + +

+ +

0

1 2 3 00

0

0.736979

.24698 ( )] 0.327985

0.591006

0.5910091

[0.591009 ( ) 0.736976 ( ) 0.327986 ( )]sin[1.80194 ( )] 0.7369761.80194

0.327986

t

t d

f f f t dm

+ +

11 12 13

21 22 23 0

31 32 33

1Let [ ] ,

i im

= =

. Then,

3

1 1 2 2 3 3

1

1 1( ) [ ( ) ( ) ( )] ( )i i i i ji j

j

Q t f t f t f t f t m m

=

= + + = . Hence,

3

00

10

1 1( ) [ ( )]sin[ ( )]

t

i ji j i

ji

z t f t dm

=

=

and

3 3

00

0 10

1 1{ ( )} ( ) sin[ ( )] { }

t

ji j i i

i ji

x t f t dm

= =

=

.

multidegree of freedom systems

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