chapter 02chapter 02 review of vibration … 02chapter 02 review of vibration fundamentals me 7me...

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CHAPTER 02CHAPTER 02CHAPTER 02CHAPTER 02Review of Vibration FundamentalsReview of Vibration Fundamentals

ME 7ME 70808 Techniques for Vibration Control and IsolationTechniques for Vibration Control and Isolation

Prepared by Prepared by Asst.Prof.DrAsst.Prof.Dr. G. Gökhan O. Özgenökhan O. ÖzgenFall 2012Fall 2012--20120133

Monday 09:40-11:30 @ B202

ME 7ME 70808 Techniques for Vibration Control and IsolationTechniques for Vibration Control and Isolation1

Department of Mechanical EngineeringDepartment of Mechanical EngineeringMiddle East Technical UniversityMiddle East Technical University

06531 AnkaraTURKEY06531 AnkaraTURKEY

1. Introduction1. Introduction

Mechanical vibrations

Oscillatory displacements or deformations of a mechanical system around a defined equilibrium position.


Animations courtesy of Dr. Dan Russell, Kettering University

2


Vibrations can occur in the engineering systems where dynamic forces are existing.

In order for vibrations to occur, mechanical system must have an elastic behavior. Elasticity is the ability of a mechanical system which is deformed under the

action of external forces, to recover its original undeformed state when it is unloaded.

The simplest example for an elastic member is a simple elastic spring.


Inertia plays an important role in vibration behavior (recall basic dynamics)

3


Most real structures and systems behave as multiple degrees of freedom (DOF) systems.

Many resonances, many mode shapes.

0 200 400 600 800 1000 1200 140010

-2

100

102

|H1 1|


Animations courtesy of Dr. Dan Russell, Kettering University

3


Great deal of time is spent on single DOF (SDOF) systems before working on multi DOF (MDOF) systems.

WHY?

SDOF system is the building block for modeling MDOF systems.



SDOF SystemSDOF System

Single Equation of motion

Second order ordinary differential equation (ODE)

MDOF SystemMDOF System (with N DOF)

N equations of motion

N Second order ordinary differential equations (coupled)

)(tfkxxcxm


differential equations (coupled)

THROUGH MODAL ANALYSIS

N uncoupled ODES

N equivalent SDOF systems defined in MODAL COORDINATES

6

)(tfxKxCxM

4

1. Introduction1. IntroductionForced Response of a SDOF system (Particular Forced Response of a SDOF system (Particular

Solution)Solution) Excitation in form of a force persisting for an extended period of time.

RESULT: FORCED RESPONSE (PARTICULAR SOLUTION)

Response due to initial condiitons: TRANSIENT RESPONSE, FREE VIBRATION RESPONSE (HOMOGENIOUS SOLUTION)

FORCED RESPONSE+ TRANSIENT RESPONSE = TOTAL RESPONSE

W ill t t ith th i l f f it ti SIMPLE HARMONIC


We will start with the simples form of excitation: SIMPLE HARMONIC

Then comes PERIODIC excitation (Can be defined in terms of multiple harmonic functions using Fourier Series)

NON-PERIODIC excitation: CONVOLUTION INTEGRAL

LAPLACE TRANSFORM may also be used to calculate total response for any non-periodic excitations for given initial conditions.

7

Laplace TransformationLaplace Transformation

• Consider a sample system with an input r(t) and an output y(t)

The physical input-output relation is not explicit.

We have to apply a suitable ODE solution technique to find y(t) for a given r(t), which can be quite time consuming.

• Applying Laplace transformation to both sides of this equation we obtain an algebraic equation that relates the input and output

Input-output relationship is more explicit: Physical insight.

Inverse Laplace transformation gives us the output y(t) (solution of the DE) in the time domain.

)()()()()(

012

2

23

3

3 trtyadt

tdya

dt

tyda

dt

tyda =+++

)()()()()()( Solve

sRsGsYsRsMsYY(s)

=→=

3

Laplace Transformation IILaplace Transformation II

• Laplace transformation is an integral transformation:

[ ] ∫==∞ −

0)()()( dtetfsFtf stL

Function of time, t Function of Laplace variable, s

0

)( function real for the where ∫∞ −→ dttetff(t) σ

4

ExamplesExamples

• Laplace transform of a unit step function:

[ ] [ ]s

es

dtetf stst 11)1(1)(

00

=−===∞

−∞

−∫LL

Unit step function : f(t)=1 for t >0

f(t)=0 for t <0f(t)

time, t

5

Examples IIExamples II

• Exponential Function:

Some transforms must be evaluated by integration by parts, partial fractions, or other creative approaches.

• Unit ramp function:

• Cosine function:

There are transform tables available such as the one on the inside back cover of the book.

[ ]αα

αααα+

=+

−=∫=∫=∞

+−∞

+−∞

−−−s

es

dtedteee tstssttt 11

0

)(

0

)(

0

L

[ ]2

0

1

sdttet st =∫=

∞−L

[ ]22

0 ) cos( ) cos(

ωωω

+=∫=

∞−

s

sdtett stL

6

Properties of Laplace TransformProperties of Laplace Transform

• Laplace transforms take time-domain functions and transform them into s-domain functions, where s is a complex variable:

• Multiplication by a Constant:

• Sum and Difference:

For more properties See Table 2-1.

ωσ js +=

Real part Imaginary part

[ ] )()( skFtkf =L

[ ] ( )sFsFtftf 2121 )()()( ±=±L

7

DifferentiationDifferentiation

)0()()(

fssFdt

tdf−=⎥⎦

⎤⎢⎣⎡L

Subtract the initial condition • Differentiation:

• Example:

First Derivative Multiply by s

[ ]

[ ]

αα

αα

αα

αα

αααα

α

αα

αα

+−

=++

−+

=−+

=−

−=+−

=

+−

=∫−=∫ −=

−===

∞ +−∞ −−

−−

ss

s

s

s

s

sfssF

fssFs

tf

sdtedteetf

etffetf

tsstt

tt

1)0()(

)0()( )(

)( )(

)( and 1)0( with )(

?

0

)(

0

&

&

&

L

L

8

Differentiation in GeneralDifferentiation in General……

• Second Order Differentiation:

• General form for higher order derivatives is similar to the expression above. See Table 2-1

• Differentiation in the time domain is equivalent to multiplication of the Laplace transform of the original function by s.

In the time domain, we work with differential equations

In the s-domain, we work with algebraic equations

)0()0()()( 2

2

2

fsfsFsdt

tfd &−−=⎥⎦

⎤⎢⎣

⎡L

9

Example Example

• Consider the second order vibrating system (say, this is our plant):

y(t), output (disp.)

r(t), input (force)

m

ck

[ Assuming zero initial conditions ]

( ) )()( Transform Laplace

)()()()( 2 sRsYkcsmstrtkytyctym =++⇒=++ &&&

kcsms

sRsY

++=

2

)()(

Solution of the sytemresponse (output) in the s-Domain

Can we go back to the time domain?

10

Inverse Laplace TransformInverse Laplace Transform……

• …is, given the Laplace transform F(s), obtaining the time function f(t) through the following integral equation.

• Solving this integral is not a practical approach for a complicated function like Y(s).

• If we can break a function like Y(s) into simple components which exist in the Transform pair table, we could avoid using equation above.

• Partial-fraction expansion can be used for this purpose.

[ ] ∫∞+

∞−

==jc

jc

st- dsesFj

F(s)tf )(2

1)( 1

πL

11

PartialPartial--Fraction ExpansionFraction Expansion

• Consider the rational function G(s) (similar to Y(s)):

it is assumed that P(s) and Q(s) are polynomials with constant coefficients.

also, the order of P(s) is assumed to be greater than that of the Q(s).

)(

)()(

sP

sQsG =

011

1)( asasassP nn

n ++++= −− L

12

PartialPartial--Fraction Expansion IIFraction Expansion II

• CASE 1: Y(s) has n simple poles.

• Y(s) can be rewritten as follows:

• Question: How do we find unknown coefficients of the fractions?

Pole: values of s that make the denominator of Y(s) equal to zero.

)())((

)()(

21 nssssss

sQsG

+++=

L

)()()()(

2

2

1

1

n

n

ss

C

ss

C

ss

CsG

+++

++

+= L

13

PartialPartial--Fraction Expansion IIIFraction Expansion III

• To solve for C1:

Multiply both sides by (s+s1)

Set s equal to -s1

Find C1

Repeat the same thing for all C’s.

• Once all constants are found, Laplace transform table can be used to find the time domain counter parts of all fractional terms, i.e.

• Compare PFE approach withthis integral:

tnsn

tsts

n

n eCeCeCtgss

C

ss

C

ss

CsG −−− +++=→

+++

++

+= LL 2

21

1

Table Transform

2

2

1

1 )( )()()(

)(

∫ ⎥⎦

⎤⎢⎣

⎡+++

=∞+

∞−

jc

jc

st

ndse

ssssss

sQ

jtg

)())((

)(

2

1)(

21 Lπ

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Example: PFEExample: PFE

• Example: Inverse Laplace transform the following s-domain function:

• …Applying the PFE technique….

• From Table of transform pairs:

sss

ssG

127

126)(

23 ++

+=

4

3

3

21

)4)(3(

126

127

126)(

23 +−

++=

+++

=++

+=

ssssss

s

sss

ssG

0for 321)( 43 <−+= −− teetg tt

15

PartialPartial--Fraction Expansion IVFraction Expansion IV

• CASE 2: Y(s) has simple but complex conjugate poles.

• Apply the same procedure where you have simple real poles.

• Coefficients of the fractions with conjugate complex poles will probably turn out to be complex valued too.

)())((

)()(

21 nssssss

sQsG

+++=

L ωσ

ωσ

js

js

−−=

+−=

2

1

16

Example: PFE IIExample: PFE II




22

1)(

22 ++=

sssG

)1(2

1

)1(2

1

)1)(1(

1

22

1)(

2 jsjjsjjsjssssG

−++

++−=

−+++=

++=

0for sin(t))( >= − tetg t

17

PartialPartial--Fraction Expansion VFraction Expansion V

• CASE 3: Y(s) has multiple-order poles.

• Y(s) can be rewritten as follows:

• Apply the same procedure for the terms for single poles (for C’s).

rirn ssssssss

sQsG

))(())((

)()(

21 ++++=

−L3)2)(4)(3(

1)(

EXAMPLE

+++=

sssssG

ri

r

iirn

n

ss

A

ss

A

ss

A

ss

C

ss

C

ss

CsG

)()()()()()()(

221

2

2

1

1

+++

++

++

+++

++

+=

−LL

18

PartialPartial--Fraction Expansion VIFraction Expansion VI

• For the coefficients of the multiple-order pole terms use the following formulas:

ri

r

iirn

n

ss

A

ss

A

ss

A

ss

C

ss

C

ss

CsG

)()()()()()()(

221

2

2

1

1

+++

++

++

+++

++

+=

−LL

[ ]

[ ]

[ ]

[ ] 43]-2 [Eq. )()()!1(

1

42]-2 [Eq. )()(!2

1

41]-2 [Eq. )()(

40]-2 [Eq. )()(

)1(

)1(

1

2

2

1

1

i

i

i

i

ss

rir

r

ss

rir

ss

rir

ss

rir

sGssds

d

rA

sGssds

dA

sGssds

dA

sGssA

−=−

−

−=−

−=−

−=

+−

=

+=

+=

+=

M

19

Example: PFE IIIExample: PFE III




2)2)(1(

1)(

++=

sssG

22 )2(

1

)2(

1

)1(

1

)2)(1(

1)(

+−

+−

+=

++=

ssssssG

0for )( 22 >−−= −−− tteeetg ttt

20

Solution of Linear ODE via LaplaceSolution of Linear ODE via Laplace……

• Transform the differential equation to the s-domain.

• Manipulate the transformed algebraic equation to solve for the output Y(s).

The input R(s) is specified ( step function, ramp function, harmonic, etc. ).

Initial conditions should also be taken into account.

• Perform Partial Fraction Expansion (PFE) of the transformed algebraic equation.

Perform this step if needed (if the output expression is not easily inverse Laplace transformable).

PFE allows us to decompose the solution into its characteristic modes.

This is equivalent to identifying the homogeneous and the particular parts of the solution.

The characteristic modes can be clearly interpreted in the s-domain.

• Obtain the inverse Laplace transform of the response using the Laplace transform table.

2

Example I: 1Example I: 1stst order DE order DE

• Recall the 1st order differential equation for the spring-damper component.

• Define initial conditions as:

• Question: Find the solution of this DE for a unit step input.

rkydt

dyc =+

ent)(displacemoutput : )(

(force)input : )(

ty

tr

00

)0( and 0 y

dt

dyy)y( &==

(t)ur(t)?y(t) s== for

3

Example I : 1Example I : 1stst order DE order DE

• Laplace transforming both sides, we obtain:

• Finally the response in Laplace domain can be solved for as:

][][ rkydt

dyc LL =+

][][][ rykdt

dyc LLL =+

)()()( 0 sRskYcyscsY =+−

)()(

)(1)( 0

cks

y

cks

sR

csY

++

+=

4


• Recall that the input is a unit step function, i.e. R(s)=1/s. Also let’s assume that the initial conditions are zero (y0 = 0) .

• Rewrite the response in PFE form.

)/(

11)(

cksscsY

+=

5

)()/(

11)( 21

cks

C

s

C

cksscsY

++=

+=

)(1 tuC s⋅ teC )ck(2

−


• PFE coefficients are found as follows:

• Finally the response is obtained as:

[ ]kcksc

ssYCs

s

1

)(

11)(

0

01 =⎥⎦

⎤⎢⎣

⎡+

===

=

)(

1111)(

cksksksY

+−=

[ ]ksc

sYcksCcks

cks

111)( )(2 −=⎥⎦

⎤⎢⎣⎡=+=

−=−=

tcks e

ktu

kty )(1

)(1

)( −−=

6


• Same as the solution from the “classical approach” that we have discussed last lecture [ subsitute b0=1, a1=c and a0=k ].

• Plot the response for k=1 and c=1.

tcks e

ktu

kty )(1

)(1

)( −−= 0>t

0 2 4 6 8 100

0.5

1

1.5

Time (s)

y(t)

Unit step response of a sample 1st order system

t = 0.:0.05:10;k=1;c=1;y=1/k-1/k*exp(-(k/c)*t);figureplot(t,y);xlabel('Time (t)')ylabel('y(t)')title('Unit step response of a sample 1st order system')

7

Example II : 2Example II : 2ndnd order DE order DE

• Consider the following differential equation.

• Define initial conditions as:

• Question: Find the solution of this DE for a unit step input.

rdt

dry

dt

dy

dt

yd126127

2

2

+=++ output : )(

input : )(

ty

tr

00

)0( and 0 y

dt

dyy)y( &==

(t)ur(t)?y(t) s== for

8


• Laplace transforming both sides…

]126[]127[2

2

rdt

dry

dt

dy

dt

yd+=++ LL

0002 7)(12)(6)(12)(7)( yysysRssRsYssYsYs &++++=++

][12][6][12][7][2

2

rdt

dry

dt

dy

dt

yd LLLLL +=++

)(12)0(6)(6)(127)(7)( 0002 sRrssRsYyssYysysYs +−=+−+−− &

002 )7()()126()( )127( yyssRssYss &++++=++

0

9


• The response can be solved as follows.

• Recall that the input is a unit step function, i.e. R(s)=1/s.

• Factoring the denominator

)127(

)7(

)127(

)()126()(

200

2 ++++

+++

+=

ss

yys

ss

sRssY

&

)127(

)7(

)127(

)126()(

200

2 ++++

+++

+=

ss

yys

sss

ssY

&

)4)(3(

)7(

)4)(3(

)126()( 00

++++

+++

+=

ss

yys

sss

ssY

&

10


• For zero initial conditions, response reduces to:

• Using PFE, the response can be written as the summation of threefractions:

)4)(3(

)126()(

+++

=sss

ssY

)4()3( )4)(3(

)126()( 321

++

++=

+++

=s

C

s

C

s

C

sss

ssY

11


• The solution in the time domain would be

• Compute the PFE coefficients

tt- eCeCCsYty 43

321

1 )]([)( −− ++==L

[ ] 112

12

)4)(3(

)126()(

0

01 ==⎥⎦

⎤⎢⎣

⎡++

+==

==

s

s ss

sssYC

[ ] 23

6

)4(

)126()()3(

3

32 =−−

=⎥⎦

⎤⎢⎣

⎡++

=+=−=

−=s

s ss

ssYsC

[ ] 34

12

)3(

)126()()4(

4

43 −=−

=⎥⎦

⎤⎢⎣

⎡++

=+=−=

−=s

s ss

ssYsC

12


• Finally, the response is obtained as:

tts eetuty 43 32)(1)( −− −+⋅=

Transient parts of the response –homogeneous solution

Forced part of the response –particular solution

0>t

13


• What if the initial conditions were not zero?

• Find the response for

• Response with non-zero initial conditions

• Combining two terms…

1 and 1 00 == yy &

)4)(3(

8

)4)(3(

)126(

)4)(3(

)7(

)4)(3(

)126()( 00

+++

+++

+=

++++

+++

+=

ss

s

sss

s

ss

yys

sss

ssY

&

)4)(3(

)1214()(

2

++++

=sss

sssY

14


• Compared to the response expression for zero initial conditions, the numerator has changed.

• The denominator is the same, which means the functional form of the response in the time domain does not change.

• For the PFE, the coefficients are going to be different for this case.

)4(

~

)3(

~~

)4)(3(

)1214()( 321

2

++

++=

++++

=s

C

s

C

s

C

sss

sssY

15


• PFE coefficients are found as:

• The response becomes:

• Initial conditions changes the transient behavior of the response.

• The steady state value is not affected (as time goes to infinity)

7~

and 7~

and 1~

321 −=== CCC

tts eetuty 43 77)(1)( −− −+⋅=

tts eetuty 43 32)(1)( −− −+⋅=

Solution with zero initial conditionsSolution with non-zero

initial conditions

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