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5

Vibration AnalysisAll real engineering structures exhibit some amount of energy

dissipation. A helical spring in the suspension system of an

automobile will dissipate some energy due to the internal interaction

of grains of steel within the spring. A leaf spring adds friction

between the leaf elements to the internal energy dissipation. hile

both of these do aid the damping effect in a suspension system! they

are not significant because of the presence of the shoc" absorber.

At other times these internal effects are the only energy dissipating

means a#ailable. $ne example of this situation is the cable used in the

tether satellite experiments conducted %ointly by the &nited States and

'taly in the late 1(()*s. +hese experiments in#estigated the possibilityof using a single cable attached to a controlled mass as an alternati#e

to solar cells for generating electricity on future space flights.

,y dragging a conductor through the magnetic field of the earth! it is

possible to create an electric current which will pro#ide power to the

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-

Vibration Analysisspacecraft. $ne ma%or point of concern related to the potential problem

that drag on the approximately sixty miles of cable might create a s"ip

rope effect that would lead to instability andor brea"ing of the ).5 mm

diameter cable.

+his situation was compounded by the fact that the only a#ailable

energy dissipation was the mo#ement between the wire and the

insulation. /ortunately! this small amount of damping pro#ed to be

sufficient enough to pre#ent any problems.

As will be seen as the e0uations are de#eloped! most engineering

systems exhibit only a small amount of damping e#en thought they

may be extremely large and complex. $nce again! this amount of

damping is a builtin characteristic that comes about from the materials

and construction methods employed. 't is the result of effects such as

friction between connected elements! internal friction that occurs

during deformation! and windage.

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Vibration Analysis,ecause the amount of damping is generally small! it is acceptable to

ignore its presence when determining the #ibrational characteristics of

the system. owe#er! the need to "now the amplitude of #ibration

re0uires that the damping be included. +his is especially true when the

damping effect is large.

ue the complex nature of most engineering systems! it is generally

difficult to determine or estimate damping exactly especially due to the

interaction of these damping effects. +his situation can be dealt with by

studying each type damping indi#idually and then determining which

type of damping dominates in the system of interest. +his procedure

generally results in the ability to do an ade0uate analysis.

+he most common types of damping are #iscous! dry friction! and

hysteretic. ysteretic or structural6 damping arises in structural

elements due to hysteresis losses in the material. +he type and amount

of damping has a large effect on the dynamic response le#el.

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11

Vibration Analysis+he roots are

( )2

1!2

4

2 2

c kmc

s m m

= ence

1 2

1 2 6

s t s tx t X e X e= +

whereX1andX2are arbitrary constants which are found by using the

initial conditions. +he system response depends on both the algebraicsign of cand whether c2is greater than!less than! or e0ual to 4km.

'f cis positi#e!xt6 will decrease in

amplitude o#er time. 'f cis negati#e!xt6 will increase in amplitude o#er

time meaning the system is unstable. 'f

c2is less than 4km!xt6 will oscillate.

'f c2is e0ual to or greater than 4km!

xt6 will not oscillate.

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12

Vibration Analysis+o establish that these effects do occur! substitute the #alues fors1and

s2intoxt6 and manipulate the resulting expression as follows9

( ) ( )2 2

1 2

4 4

2 2 2 2

1 2 1 2 6

c km c kmc c

t tm m m m

s t s tx t X e X e X e X e

+ = + = +

+his e0uation must be e#aluated for each of the three possibilities. +o

accomplish this! the transitional #alue of c28 4kmis used to define the

critical damping coefficient! cc! which represents the #alue of damping

where oscillation ceases. +his yields2

2 2 2c

kmc km m

m = = =

+o eliminate the need to ha#e a different e0uation for e#ery possible

combination of m! c! and k! the damping ratio! ! is defined as

2

2 2c

c c c c m

c m m

= = = =

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13

Vibration Analysis

( ) ( )

2 2

2 2 2 2 2 2

1 2

1 1

1 2

6

6 : ;

t t

t t t

OR

x t X e X e

x t e X e X e

+

+

= +

= +

+his allows the expression forxt6 to be rewritten as follows9

1 are complex andxt6 willoscillate for at least a short time! depending on the #alue of .?uler*sidentity allows the complex exponential terms to be rewritten in the

following manner9

cos sini

e i =

Substitution of this e0ui#alence into the e0uation forxt6 yields

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14

Vibration Analysis2 2

1 1

1 2

2 2

1

2 2

2

2 2

1 2 1 26

6 : ;

6 : cos 1 6 sin 1 6

cos 1 6 sin 1 6 ;

6 cos 1 6 6sin 1 6

t i t i t

t

t

OR

OR

x t e X e X e

x t e X t i t

X t i t

x t e X X t i X X t

+

= +

= +

+

= + +

Sincext6 is a real 0uantity i.e. it can be felt and obser#ed6! the #alues

forX1

andX2

must be defined in such a way as to produce a real #alue

forxt6. +his means thatX1andX2must be complex con%ugates or

X18 a ib X28 a@ ib

Substituting this result into the e0uation forxt6gi#es

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15

Vibration Analysis2 2

6 2 cos 1 6 2 sin 1 6tx t e a t b t

= +

't is also con#enient to define the damped natural fre0uency as

21d

=

6 cos sint

d dx t e A t B t = +

Substituting this definition into the e0uation forxt6 yields

xt6 is a harmonic

function that decays

exponentially with time

as shown in the figure

to the right. +he #alues

for A and , depend on

the initial conditions

and the decay rate on .

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1-

Vibration Analysis

12 2where and tan

6 sin 6

6 cos 66 cos 62

t

d

t t

d d

OR

BX A BA

x t e X t

x t e X t e X t

= + =

= +

= =

A more compact form forxt6 can be had by including a phase angle

as shown below

/rom the system

response shown! it is

easily seen that both

the initial displacement

and #elocity were

positi#e and that wassmall enough to allow

complete oscillations

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1

Vibration Analysisto occur before reaching a le#el that can be considered effecti#ely

ero.

+here are three things about any transient response that are important.

+hese are 16 rise time which indicates how 0uic"ly the system

response reaches a specified percentage of its maximum #alue! 26

maximum #alue! 36 settling time which indicates how 0uic"ly thesystem response decays to within a specified percentage abo#e the

e0uilibrium or steadystate

#alue. +hese #alues ha#e

an impact on one anotherand it is important to "now

what range is acceptable

for each when designing a

system.

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17

Vibration Analysis

Vibrational Motion

-1

-0.5

0

0.5

1

0 2 4 6 8 10

Time

Amplitude

= 0 = 0 01 = 0 05 = 0 1 = 0 5 = 0 71

Shown below are plots of the damped transient response of the same

system with increasingly amounts of damping. Botice how the

maximum #alue and settling time both decrease as increases. ,ut theamount of energy needed to begin mo#ing the system also increases

with damping.

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1(

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2)

Vibration Analysis

1 are ero andxt6 will simply decay exponentiallybac" to the e0uilibrium or steadystate #alue.

( ) ( )

( ) ( )

2 2

1 2

2 2 2 2 2 2

4 4

2 2 2 2

1 2 1 2

1 2

1 2

6

6

6 : ;

c km c kmc c

t tm m m m

s t s t

t t

t

OR

OR

x t X e X e X e X e

x t X e X e

x t e X X t

+

+

= + = +

= +

= +

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21

Vibration Analysis

Vibrational Motion

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4

Time

Amplitude

=

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Vibration Analysis

1 are positi#e! real numbers andxt6 willsimply decay exponentially bac" to the e0uilibrium or steadystate

#alue but at a rate that is slower than the critical damping case.

( ) ( )

( ) ( )

( ) ( )

2 2

1 2

2 2 2 2 2 2

2 2

4 4

2 2 2 2

1 2 1 2

1 2

1 1

1 2

6

6

6 : ;

c km c kmc c

t tm m m m

s t s t

t t

t tt

OR

OR

x t X e X e X e X e

x t X e X e

x t e X e X e

+

+

+

= + = +

= +

= +

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Vibration Analysis+o understand what happens toxt6 as increases! each exponentneeds to be examined.

2

2

12

12

lim 1 1

lim 1 )

t

t

e

e

+

+ +

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24

Vibration Analysis

Vibrational Motion

-1

-0.5

0

0.5

1

0 0.5 1 1.5 2

Time

Amplitude

= 2 = 5 = 20 = 50 = 100 = 1000

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2-

Vibration Analysis+he transient response can be used to determine the amount of

damping present in a single degreeoffreedom model of a system by

imparting an initial displacement andor #elocity to the system andmonitoring the amplitude of the response.

As shown below! a single degreeoffreedom was disturbed from

e0uilibrium and the amplitude of the motion plotted #ersus time. +he

motion is gi#en by

6 sin 6t

dx t e X t = +

+he maximum #alue

will occur whensin 6 1

dt + =

2

1 2 2

1d

ddf

= = =

And the period! d! is

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2

Vibration Analysis+herefore!

1 ''' '

'' '''''' ''

6 sin 6

6 sin 6

t t

d

t t

d

A!

x t e X t X e X

x t e X t X e X

= + = =

= + = =,ut!

'' ' ' 2

2

1

dt t t

= + = +

Substituting yields!

2

'

'

2 6'

1

''

t

t

A!

X X e

X X e

+

=

=i#idingX'byX''yields

2

2

1'

''

6X e

X

=

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27

Vibration Analysis+a"ing the natural logarithm of each side gi#es the logarithmic

decrement! ! as

Sol#ing for yields!

'f is small! it is possibleto use

'

''

2ln21

XX

= =

22 2 2

2 2 2 2

2

22

1 4

4 6

4 6

OR

OR

=

= +

= +

2

=

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3)

Vibration Analysis?#en if is large enough that 1 > 2cannot be set e0ual to 1! goodaccuracy for the estimate of can be obtained using the approximatee0uation as shown below.As seen in the figure! the error in the estimate of is small up to a#alue of of about ).45 or ).5.

Damping Ratio vs. Logarithmic Decrement

0

0.2

0.4

0.6

0.8

1

0 4 8 12 16 20

Logarithmic Decrement

Damp

ingRatio

Exact

Approx.

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Vibration AnalysisGiven -+he machine shown below weighs 3-)) lb and rests on a set

of #ibration isolators. As the machine is lowered onto the #ibrationisolators! the four springs one on each corner6 each deflect 3 inches.

etermine the undamped natural fre0uency of #ibration! ! and thestiffness of each spring. hat should the damping coefficient for this

set of isolators be if a damper is located with each spring and the

damped natural fre0uency of #ibration! d! is to be 1)D less than E

Solution Since all four springsdeflect the same amount! the springs

are in parallel and kin the figurerepresents the sum of the indi#idual

stiffnesses. +herefore! the #alue of k

can be found using

3-)) 12)) .3 .

" lbk lb inx in

= = =

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ib i A l i

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34

Vibration Analysis

2

2 2 2

2 2

12)) .63-)) 62

32.14 12 . 6

1117(.1- . 1)5.7 .

c

c

c

OR

OR

k"

c km g

lb in lbc

in s

c lb s in lb s in

= =

=

= =

+he critical damping coefficient for the set is found using

+he damping ratio for the set is found using

2

22 2

1

1 1 ).( ).1(

).43-

d

d

OR

OR

=

= = =

=

Vib i A l i

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Vibration Analysis+he damping coefficient for the set is found using

?ach damper needs a damping coefficient of 11.53 lbsin.

).43- 1)5.7 .

4-.11 .

cc

OR

OR

c c cc

c lb s in

c lb s in

= =

=

=

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3-

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ib i A l i

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37

Vibration AnalysisSolution 'f the rod is rotated slightly in cloc"wise direction! the

spring and damper will exert a restoring force which is #erticallyupward and they will create a counter cloc"wise moment about O

e0ual to 2 2 2

2 2 2

)

O

OR

ka ca $ m#

m# ca ka

= =

+ + =

& &&

&& &

/or the critical damping case! the solution has the form of

)6

)

)6 )6

)6 )66

)6 ) )66

O

A!

OR

A B e A

A B e Be

B

= = + =

= = + +

=

&

+o determineAandBwe need

6 6 tt A Bt e = +

Vib ti A l i

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3(

Vibration Analysis+herefore!

)

6 1 6 tt t e = +

/or )8 2o! 8 3 rads! and t 8 1 s!

+he logarithmic decrement is gi#en by

'

2''

2ln1

XX

= =

o 3 o16 2 1 36 ).4e = + =

So if is ).7! is

2 2

2 2 ).76

1 1 ).76

5.)2 7.37).-

OR

= =

= =

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Vib ti A l i

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Vibration AnalysisGiven Shown below is the schematic representation of a drop

hammer forge. +his system is composed of an an#il which weighs

5!))) B and is mounted on a foundation that has a stiffness of 5 x 1)-Bm and a #iscous damping coefficient of 1)4Bsm6 and a drop

hammer called the tub which weighs 1)3B. uring a particular

operation the tub falls 2 m before stri"ing the an#il. 'f the an#il is at

rest prior to impact by the tub! determine the response of the an#il afterimpact. Assume the coefficient of

restitution between the an#il and

the tub is ).4.

Solution +he first tas" is todetermine the #elocity of the tub

%ust prior to impact and the

#elocities of both the tub and an#il

%ust after impact.

Vib ti A l i

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Vibration Analysis

+o determine these #elocities! we use the principle of conser#ation of

momentum with the tub #elocity before and after impact representedby vt1and vt2! respecti#ely. +he #elocity of the an#il before and after

impact are gi#en by va1and va2! respecti#ely. +he principle of

conser#ation of momentum states

%va2 va16 8 mvt2> vt16

,oth va1and vt1ha#e #alues which

are either "nown or can be easily

found. As the an#il is initially at

rest! va1is ero. +he #elocity of the

tub %ust prior to impact is found

using

G mv2t18 mg&

Vib ti A l i

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Vibration Analysis

Substituting these #alues for vt1and va1into the momentum e0uation

yields

1 2 2 (.7)--5 2 -.2-1t m sv g&= = =

$r the #alue for vt1is

( ) ( )2 25))) 1)))) -.2-1

(.7)--5 (.7)--5a tv v =

2 251).2)41 -37.7 1)2.)41

a tv v=

$H

+he definition of the coefficient of

restitution is

2 2 2 2

1 1

).4) -.2-1

a t a t

a t

v v v vr

v v

= =

Vib ti A l i

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Vibration Analysis

Sol#ing these two e0uations simultaneously yields

2 22.5)4

a tv v= +

$H

2 2 1.4-1 1.)435

a tm s m sv A! v= =

+hus the initial conditions for the an#il are

+he damping coefficient is e0ual to

)6 ) )6 1.4-1m sx A! x= =&

-

1)))

2 5)))2 5 1)(.7)--5

).)(7((5

c

k %

= =

=

Vib ti A l i

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45

Vibration Analysis

-

2

5 1) (7.((55)))(.7)--5

1 (7.)25

n

d n

rad s

A!

rad s

k%

= = =

= =

+he undamped and damped natural fre0uencies are found using

+he displacement of the an#il is gi#en

by

) ))

)

6 cos sin

6 sin

tn nd d

d

tn

d

d

OR

x xx t e x t t

xx t e t

+= +

=

&

&

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