lect02 - 2dof spring mass systems [compatibility mode]

8/9/2019 LECT02 - 2DOF Spring Mass Systems [Compatibility Mode]

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NOISE AND VIBRATION

BDC 4013

Lecture 2 : Two Degree of FreedomLecture 2 : Two Degree of FreedomSystemsSystems


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1.Introduction


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2. 2-DOF System (a basic understanding

in multi-DOF)


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5

3.Deriving 2-DOF mathematical

model Equation of Motion

F1 F2

m1 m2

k1x1

1 1c x&

k3x2

3 2c x&

k2 (x2-x1)

( )2 2 1c x x& &

1x&&

2x&&

1F 2F


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m1k1x1

1 1c x&

k2 (x2-x1)( )2 2 1c x x& &

1x&&1F

m23 2

c x&

k2 (x2-x1)

( )2 2 1c x x& &

2x&&

2F

1 1 1 1 1 1 1 2 2 1 2 2 1

1 1 1 2 1 2 2 1 2 1 2 2 1

( ) ( )

( ) ( )

m x F k x c x k x x c x x

m x c c x c x k k x k x F

= + +

+ + + + =

&& & & &

&& & &

2 2 2 2 2 1 2 2 1 3 2 3 2

2 2 2 1 2 3 2 2 1 2 3 2 2

( ) ( )

( ) ( )

m x F k x x c x x k x c x

m x c x c c x k x k k x F

=

+ + + + =

&& & & &

&& & &

k3x2


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1 1 1 2 1 2 2 1 2 1 2 2 1( ) ( )m x c c x c x k k x k x F + + + + =&& & &

2 2 2 1 2 3 2 2 1 2 3 2 2( ) ( )m x c x c c x k x k k x F + + + + =&& & &

Rearrange in matrix notation

1 1 1 1

2 2 2 2

x x x F

x x x F

+ + =

&& &

&& &

( )( )

( )( )

1 2 2 1 2 21 1 1 1 1

2 2 3 2 2 32 2 2 2 2

00

c c c k k k m x x x F

c c c k k k m x x x F

+ + + + = + +

&& &

&& &


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General Multi-DOF Equation

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

[ ] [ ] [ ]

{ } { } { }

, and are symmetric matrices n x n

x , and are vectors n row

n is the number of the degrees of freedom

n is the number of natural frequencies

M C K

x x&& &


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4.Free Vibration of an Undamped 2-DOF

System

k2

k1

x1

x2

m1

m2

1 1 1 1 2 2 1 1

1 1 1 2 1 2 2

( ) 0

( ) 0

m x k x k x x F

m x k k x k x

+ = =

+ + =

&&

&&

2 2 2 2 1 2

2 2 2 1 2 2

( ) 0

0

m x k x x F

m x k x k x

+ = =

+ =

&&

&&

1 1 1 2 1 2 2

2 2 2 1 2 2

1 1 1 2 2 1

2 2 2 2 2

( ) 0

0

can be written in matrix

0 ( ) 0

0 0

m x k k x k x

m x k x k x

m x k k k x

m x k k x

+ + =

+ =

+ + =

&&

&&

&&

&&

1 1 2 2sin( ) sin( )x A t x A t = =

2 2

1 1 2 2sin( ) sin( )x A t x A t = = && &&


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10

1 1 1 2 2 1

2 2 2 2 2

0 ( ) 0

0 0

m x k k k x

m x k k x

+ + =

&&

&&

1 1

2 2

sin( )

sin( )

x A t

x A t

=

=

2

1 1

2

2 2

sin( )

sin( )

x A t

x A t

=

=

&&

&&2

1 1 2 2 11

2

2 2 2 22

( ) 00

00

A k k k Am

A k k Am

+ + =

2

1 2 1 2 1

222 2 2

( ) 0

0

k k m k A

Ak k m

+ =

Can be solvedonly if

2

1 2 1 2

2

2 2 2

( )0

k k m k

k k m

+ =


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[ ]

2

1 2 1 2

2

2 2 2

4 2 2

1 2 1 2 2 2 1 1 2 2 2

( )

0

( ) ( ) ( ) 0

k k m k

k k m

m m k k m k m k k k k

+ =

+ + + + =

[ ] [ ]2 2

1 2 2 2 1 1 2 2 2 1 1 2 1 2 2 22 2

1 2

1 2

( ) ( ) 4( ) ( ),

2( )

k k m k m k k m k m m m k k k k

m m

+ + + + + =

1 1

2 2

first natural frequency

first natural frequency

n

n

= =

= =


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2

1 2 1 1 2 1

22

2 2 2 1

( ) 0

0

n

n

k k m k A

Ak k m

+ =

1at first natural frequency = n

2 (1) (1)

1 2 1 1 1 2 2( ) 0nk k m A k A + =

2(1)

1 2 1 121 (1)

1 2

( )n

k k mAr

A k

+ = =

2 (2) (2)

1 2 1 2 1 2 2( ) 0nk k m A k A + =

2(2)1 2 1 22

2 (2)

1 2

( )n

k k mAr

A k

+ = =

Modal vector

1 1

1 1 1

1 1

2 1 1

A A

A r A

= =

( ) ( )

( )

( ) ( )A

(2) (2)

(2) 1 1

(2) (2)

2 2 1

A A

A r A

= =

A

First mode

Second mode


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k

k

x1

x2

m

m

Example 1;

Draw FBD, determine equation of motion and find thenatural frequencies of the system shown below

[ ]2 4 2 2( ) 3 0m km k + =

2 2 2 2 2 22

12 2

2

2

1

2

3 9 4 3 5

2 2

3 5 3 5

2 4 2 4

1.618

0.618

n

n

n

km k m m k km m k

m m

k k k k

m m m m

k

m

k

m

+ += =

= + = +

=

=

[ ]4 2 21 2 1 2 2 2 1 1 2 2 2( ) ( ) ( ) 0m m k k m k m k k k k + + + + =

m1 = m2 = m

k1 = k2 = k

Given;


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Free vibration with initial conditions

For a specific condition, the system can be made tovibrate in its ith normal mode (i = 1, 2) by subjecting it to

the specific initial conditions

.0)0(

,)0(

,0)0(

constant,some)0(

2

)(

12

1

)(

11

==

==

==

===

tx

Artx

tx

Atx

i

i

i

&

&


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However, for any other general initial conditions, bothmodes will be excited. The resulting motion can be

obtained by a linear superposition of the two normalmodes

)()()( 2211 txctxctx rrr

+=

)cos()cos()()()(

)cos()cos()()()(

22

)2(

1211

)1(

11

)2(

2

)1(

22

22

)2(

111

)1(

1

)2(

1

)1(

11

+++=+=

+++=+=

tArtArtxtxtx

tAtAtxtxtx

asexpressedbecanvectortheofcomponentstheThus

.generalityoflossnowithchoosecanweand

constantsunknowntheinvolvealreadyandSinceconstants.areandwhere

)(

1,

)()(

21

)2(

1

)1(

1

)2()1(

21

tx

ccAA

txtx

cc

r

rr

==


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:conditionsinitialthefrom

determinedbecanandconstantsunknownthewhere 21)2(

1

)1(

1 ,, AA

)0()0(),0()0(

)0()0(),0()0(

2222

1111

xtxxtx

xtxxtx

&&

&&

====

====

2

)2(

1221

)1(

1112

2

)2(

121

)1(

112

2

)2(

121

)1(

111

2

)2(

11

)1(

11

sinsin)0(

coscos)0(

sinsin)0(

coscos)0(

ArArx

ArArx

AAx

AAx

=

+=

=

+=

&

&

Four algebric equation

with unknowns..


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17

Example 2. Free Vibration Response of a Two

DOF System

).0()0()0(,1)0( 2211 xxxx && ===

Find the free vibration response of the system shown in

Fig.5.3(a) with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 andc1 = c2 = c3 = 0 for the initial conditions


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Solution: For the given data, the eigenvalue

problem, Eq.(5.8), becomes

(E.100

55-53510

00

2

1

2

2

2

1

32

2

22

221

2

1

=

+

+

=

++

++

X

X

X

X

kkmk

kkkm

or


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{ }

{ } )8.5(0)(0)(

2322

212

22121

2

1

=++

=++

XkkmXk

XkXkkm

{ }

{ } )9.5(0))((

)()()(

223221

132221

4

21

=+++

+++

kkkkk

mkkmkkmm

)15.5()cos()cos(

)()()(

)cos()cos()()()(

22

)2(

1211

)1(

11

)2(2

)1(22

22

)2(

111

)1(

1

)2(

1

)1(

11

+++=

+=

+++=+=

tXrtXr

txtxtx

tXtXtxtxtx

Related equations involved.


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Related equations involved.

)17.5(sinsin)0(

coscos)0(

sinsin)0(

coscos)0(

2

)2(

1221

)1(

1112

2

)2(

121

)1(

112

2

)2(

121

)1(

111

2

)2(

11

)1(

11

XrXrx

XrXrx

XXx

XXx

=

+=

=

+=

&

&

=

=

+=

=

)(

)0()0(sin,

)(

)0()0(sin

)0()0(cos,

)0()0(cos

122

2112

)2(

1

121

2121

)1(

1

12

2112

)2(

1

12

2121

)1(

1

rr

xxrX

rr

xxrX

rr

xxrX

rr

xxrX

&&&&


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from which the natural frequencies can be found as

By setting the determinant of the coefficient matrix in

Eq.(E.1) to zero, we obtain the frequency equation,

Example 1 Solution

(E.2)01508510

24=+

E.3)(4495.2,5811.1

0.6,5.2

21

2

2

2

1

==

==

The normal modes (or eigenvectors) are given by

E.5)(5

1

E.4)(2

1

)2(

1)2(

2

)2(

1)2(

)1(

1)1(

2

)1(

1)1(

X

X

XX

XX

X

X

=

=

=

=

r

r


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22

By using the given initial conditions in Eqs.(E.6) and (E.7),

we obtain

The free vibration responses of the masses m1 and m2 are

given by (see Eq.5.15):

Example 1.Solution

(E.7))4495.2cos(5)5811.1cos(2)((E.6))4495.2cos()5811.1cos()(

2

)2(

11

)1(

12

2

)2(

11

)1(

11

++=

+++=

tXtXtx

tXtXtx

(E.11)sin2475.121622.3)0(

(E.10)sin4495.2sin5811.10)0(

(E.9)cos5cos20)0(

(E.8)coscos1)0(

2

)2(

1

)1(

12

2

)2(

11

)1(

11

2

)2(

11

)1(

12

2

)2(

11

)1(

11

XXtx

XXtx

XXtx

XXtx

+==

===

===

+===

&

&


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23

while the solution of Eqs.(E.10) and (E.11) leads to

The solution of Eqs.(E.8) and (E.9) yields

Example 1. Solution

(E.12)

7

2cos;

7

5cos 2

)2(

11

)1(

1 == XX

(E.13)0sin,0sin 2)2(

11

)1(

1 == XXEquations (E.12) and (E.13) give

(E.14)0,0,7

2,7

521

)2(

1

)1(

1 ==== XX


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Example 1.Solution

Thus the free vibration responses of m1 and m2 are given

by

(E.16)4495.2cos7

105811.1cos7

10)(

(E.15)4495.2cos7

25811.1cos

7

5)(

2

1

tttx

tttx

=

+=


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lect02 - 2dof spring mass systems [compatibility mode]

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