mechanical vibrations chapter 5 - faculty server contact |...

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 5

Mechanical VibrationsChapter 5

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell


Multiple Degree of Freedom Systems

• Referred to as a Multiple Degree of Freedom• An NDOF system has ‘N’ independent degrees offreedom to describe the system

• There is one natural frequency for every DOF inthe system description

Systems with more than one DOF:



• Each natural frequency has a displacementconfiguration referred to as a ‘normal mode’

• Mathematical quantities referred to as‘eigenvalues’ and ‘eigenvectors’ are used todescribe the system characteristics

• While the resulting motion appears morecomplicated, the system set of equations canalways be decomposed into a set of equivalentSDOF systems for each mode of the system.

Properties of a MDOF system:



• lumped mass• stiffness proportionalto displacement

• damping proportional tovelocity

• linear time invariant• 2nd order differentialequations

Assumptions

m

m

k

k

c

c

1

1

2

2

1

2

f 1

f 2

x1

x2



Free body diagram

f 1

f 2

(k 2 x1x2 )- (c 2 x1x2 )-

x1

x2

k 1x1 c1x1

m2

m1



Newton’s Second Law

Rearrange terms

( ) ( ) ( )( ) ( ) ( )122122222

1221112211111

xxkxxctfxmxxkxkxxcxctfxm

−−−−=−+−−+−=

&&&&

&&&&&

( ) ( ) ( )( )tfxkxkxcxcxm

tfxkxkkxcxccxm

22212221222

1221212212111

=+−+−=−++−++

&&&&

&&&&

f 1

f 2

(k 2 x1x2 )- (c 2 x1x2 )-

x1

x2

k 1x1 c1x1

m2

m1



Matrix Formulation

( )

( )

=

−

−++

−

−++

)t(f)t(f

xx

kkkkk

xx

ccccc

xx

mm

2

1

2

1

22

221

2

1

22

221

2

1

2

1

&

&

&&

&&Matrices andLinear Algebraare important !!!


MDOF - Frequencies and Mode Shapes

Example 5.1.1

(5.1.1)( )

( ) 2122

1211

kxxxkxm2xxkkxxm−−−=−+−=

&&

&&



For normal mode type of oscillation, we can write

substituting into the differential equation yields

(5.1.2)ti

222

ti111

eAortsinAx

eAortsinAxω

ω

ω=

ω=

( )( ) 0Am2k2kA

0kAAmk2

22

1

212

=ω−+−

=−ω−(5.1.3)



In matrix form this is

and the determinant of the matrix is

whose solution yields the eigenvalues

(5.1.4)

(5.1.5)

( )( )

=

ω−−−ω−

00

AA

m2k2kkmk2

2

12

2

( )( ) 0

m2k2kkmk2

2

2=

ω−−−ω−

0mk

23

mk3

mk

23

mk3 2

22

24 =

+λ−λ=

+ω−ω (5.1.6)



The frequencies of the system are

and the general ratio of response is

(5.1.7)

(5.1.8)

mk366.2

mk366.2

mk634.0

mk634.0

22

11

=ω⇒=λ

=ω⇒=λ

km2k2

mk2k

AA 2

22

1 ω−=

ω−=



The ratio for the first frequency , ω1, is

The ratio for the second frequency, ω2, is

(5.1.8a)

(5.1.8b)

731.0mk2

kAA

2

)1(

2

1 =ω−

=

73.2mk2

kAA

2

)2(

2

1 −=ω−

=



The mode shape for the two different modes is

Each mode oscillates according to

(5.1.8)

(5.1.8b

−

=φ

=φ1

73.2)x(;

1731.0

)x( 21

( )111

)1(

2

1 tsin1731.0

Axx

ψ+ω

=

( )221

)2(

2

1 tsin1

73.2A

xx

ψ+ω

−

=


MDOF - Initial Conditions

The general description of the system(for the example considered) is

and the initial conditions are specified as

(5.2.1)

−

=φ=ω

=φ=ω

173.2

)x(;mk366.2

1731.0

)x(;mk634.0

22

11

( ) 2,1itsincxx

iiii

)i(

2

1 =ψ+ωφ=



The displacement is written as

The velocity is written as

(5.2.3)

( )

( )222

1112

1

tsin1732.2

c

tsin1731.0

cxx

ψ+ω

−

+

ψ+ω

=

( )

( )2222

11112

1

tcos1732.2

c

tcos1731.0

cxx

ψ+ω

−

ω+

ψ+ω

ω=

&

&

(5.2.2)



Example 5.2.1 - Initial conditions are:

which correspond to

(5.2.3a)

(5.2.2a)

=

=

0.00.0

)0(x)0(x

;0.40.2

)0(x)0(x

2

1

2

1

&

&

( ) ( )2211 sin1732.2

csin1731.0

c42

ψ

−

+ψ

=

( ) ( )222111 cos1732.2

ccos1731.0

c00

ψ

−

ω+ψ

ω=



Example 5.2.1 - upon solving these equations forthe response due to the specified initial conditionsyields:

( ) ( )tcos1732.2

268.0tcos1731.0

732.3xx

212

1 ω

−

+ω

=


MDOF - Coordinate Coupling

Coordinate coupling exists in many problems.Either static coupling, dynamic coupling or bothstatic and dynamic coupling can exist.The equations of motion are:

and can be cast in matrix form as:

(5.3.2)

(5.3.1)0xkxkxmxm

0xkxkxmxm

222121222121

212111212111

=+++=+++

&&&&

&&&&

=

+

00

xx

kkkk

xx

mmmm

2

1

2221

1211

2

1

2221

1211

&&

&&



Coordinate coupling can be eliminated through atransformation to a different coordinate systemwherein the independent variables are not couplingeither statically or dynamically.These coordinates are referred to as

‘principal coordinates’or

‘normal coordinates’



For systems with general damping, this is noteasily possible unless the damping is of a specialform or the system is first converted to the statespace formulation of the system equations

(5.3.3)

=

+

+

00

xx

k00k

xx

cccc

xx

m00m

2

1

22

11

2

1

2221

1211

2

1

22

11

&

&

&&

&&



Example 5.3.1 Static Coupling

( ) ( )( ) ( )

=

θ

+−−+

+

θ

00x

lklklklklklkkkx

J00m

222

2111122

112221&&

&&



Example 5.3.1 Dynamic Coupling

( )( )

=

θ

+

++

θ

00x

lklk00kkx

Jmemem c

242

231

21c

c &&&&



Example 5.3.1 Static & Dynamic Coupling

( )

=

θ

++

θ

00x

lklklkkkx

Jmlmlm 1

222

2211

1

1&&

&&


MDOF - Forced Harmonic Vibration

Consider a system excited by a harmonic force

which has a solution assumed to be

(5.4.1)tsin0F

xx

kkkk

xx

mm 1

2

1

2221

1211

2

1

22

11 ω

=

+

&&

&&

tsinXX

xx

2

1

2

1 ω

=



Substituting into the differential equation yields

which is generally written in terms of theimpedance matrix as

(5.4.2)( )

( )

=

ω−ω−

0F

XX

mkkkmk 1

2

12

222221

122

1111

[ ]

=

ω0F

XX

)(Z 1

2

1



Solving this yields

where the adjoint matrix and determinant are

(5.4.3)[ ] [ ]

ωω

=

ω=

−

0F

)(Zdet)(ZAdj

0F

)(ZXX 111

2

1

[ ] ( )( )

ω−−−ω−=ω 2

111121

122

2222

mkkkmk)(ZAdj

( )( )222

22121mm)(Zdet ω−ωω−ω=ω



The general equation becomes

and the amplitudes of response are

(5.4.6)

(5.4.3)

( )( )

( )( )

ω−ωω−ω

ω−−−ω−

=

0F

mmmkkkmk

XX 1

222

22121

2111121

122

2222

2

1

( )( )( )

( )( )222

22121

1212

222

22121

12

22221

mmFkX

mmFmkX

ω−ωω−ω−

=

ω−ωω−ωω−

=

Example 5.4.1 and 5.4.2 are good examples


MDOF - Vibration Absorber

A very common, practical application of a 2 DOFsystem is that of the ‘tuned absorber’. This iscommonly used to minimize objectionable resonanceRecall

The amplitude of response for X1 is

(5.3.2)

(5.6.1)

2

222

1

121 m

k;mk

=ω=ω

1

22

2

2

11

2

2

2

0

11

kk1

kk1

1

FkX

−

ωω

−

ωω

−+

ωω

−

=


MDOF - Tuned Absorber

1

22

2

2

11

2

2

2

0

11

kk1

kk1

1

FkX

−

ωω

−

ωω

−+

ωω

−

=


MATLAB Examples - VTB4_2

VIBRATION TOOLBOX EXAMPLE 4_2

>> clear>> m=[1 0;0 1];k=[2 -1;-1 1];x0=[1;0];v0=[0;0];tf=5;plotpar=1;>> [x,v,t]=VTB4_2(m,k,x0,v0,tf,plotpar);

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Dis

plac

emen

t of X

1

time (s ec)

P res s any key to continue

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Dis

plac

emen

t of X

2

time (s ec)

P res s any key to continue




>> clear>> m=1; c=.1; k=2;>> VTB5_1(m,c,k)>>

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

5

10

15Trans mis s ibility plot for ζ = 0.035355 ω = 1.4142 rad/s

Tran

smis

sibi

lity

Rat

io

Dimens ionles s Frequency




>> beta=1

beta =

1

>> VTB5_4(beta)>>

0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Mas s ratio vers us s ys tem natural frequency for β = 1

mas

s ra

tio - µ

normalized frequency ω/ωa

mechanical vibrations chapter 5 - faculty server contact |...

Documents