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Mechanical

VibrationsSingiresu S. Rao

SI Edition

Captulo 2Vibracin libre de

sistemas de ungrado de libertad


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2005 Pearson Education South Asia Pte Ltd. 2

2.1 Introduccin

Vibracin libre se produce cuando un sistema oscilasolamente bajo una perturbacin inicial con ningunafuerza externa

Vibraciones no amortiguadas resultan cuando la

amplitud de movimiento permanece constante conel tiempo(e.g. en el vaco) Podran producirse vibraciones amortiguadas

cuando la amplitud de la vibracin libre disminuye

gradualmente las horas extras, debido a laresistencia ofrecida por el medio ambiente(e.g. aire)


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2.1 Introduccin

Varios sistemas mecnicos y estructuralespueden ser idealizados como sistemas nicogrado de libertad, por ejemplo, la masa y larigidez de un sistema


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Si la masa m es desplazada una distanciacuando acte sobre l una fuerza resultanteen la misma direccin

2.2 Vibracin libre de un sistematraslacional no amortiguado

Ecuacin de movimiento usando la segunda leyde Newton del movimiento:)(t x

)(t F

dt t xd

mdt d

t F )(

)(

Si la masa m es constante, esta ecuacin sereduce a:

(2.1))(

)( 22

xmdt

t xd mt F


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donde 2

2

)(dt

t xd x Es la aceleracin de la masa

Para un cuerpo rgido sometidos a movimiento

de rotacin, le da la ley de Newton

)2.2()( J t M

donde es el momento resultante que acta enel cuerpo, y son losdesplazamiento angular resultante y laaceleracin angular resultante respectivamente.

M 22 /)( dt t d


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Para sistemas no amortiguados de un grado delibertad, la aplicacin, Eq. (2.1) a la masa m produce la ecuacin de movimiento :

)3.2(0

o

)(

kx xm

xmkxt F


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1)Principio de DAlemberts Las ecuaciones de movimiento, Eqs. (2.1) &(2.2) pueden volver a escribirse


Ecuacin de movimiento usando otros mtodos:

(2.4b) 0)(

)2.4a(0)(

J t M

xmt F

Aplicando el principio de DAlemberts al sistemamostrado de la Fig.(c) se obtiene la ecuacin demovimiento:

)3.2(0 o 0 kx xm xmkx


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Como el desplazamiento virtual puede tener un

valor arbitrario, Eq.(2.5) da la ecuacin delmovimiento del sistema masa resorte como

Cuando el trabajo virtual total de todas las fuerzases igual a cero, obtenemos

2.2 Vibracin libre de un sistema noamortiguado Traslacional

x xmW xkxW

i

S

)(inerciadefuerzala porrealizadovirtualTrabajo)(resortedelfuerzala porrealizadovirtualTrabajo

)5.2(0 xkx x xm

0 x

)3.2(0kx xm


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3) Principio de conservacin de la energa. Unsistema se dice que es conservadora si ningunaenerga se pierde debido a la friccin o disipacinde energa miembros.


Si no se es trabajar sobre el sistema conservador porfuerzas externas, la energa total del sistemapermanece constante. Por lo tanto el principio deconservacin de la energa puede ser expresado

como:

)6.2(0)(

constante

U T

dt

d

U T o


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2.2 Vibracin libre de un sistema noamortiguado TraslacionalLas energas cinticas y potenciales estn dada

por:)7.2(

21 2 xmT

o)8.2(

21 2kxU

La sustitucin de Eqs. (2.7) y (2.8) en la Eq.(2.6) da como resultado la ecuacin

)3.2(0kx xm


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2.2 Vibracin libre de un sistema noamortiguado Traslacional Ecuacin de movimiento de un sistema masa

resorte en posicin Vertical:

Considerar laconfiguracin delsistema masaresorte que semuestra en lafigura.


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2.2 Free Vibration of an UndampedTranslational System

Para equilibrio esttico,

)9.2( st k mg W

donde W = el peso de la masa m,= deflexin esttica

g = aceleracin de la gravedad st

La aplicacin de la segunda ley de Newton delmovimiento de masa m da

W xk xm st

)(

Y como , obtenemosW k st

)10.2(0kx xm


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Esto indica que cuando una masa se mueve en direccinvertical, podemos ignorar su peso, siempre y cuandomedimos x desde su posicin de equilibrio esttico.

2.2 Vibracin libre de un sistema noamortiguado TraslacionalNote que la Eqs. (2.3) y (2.10) son identicas.

)15.2()( 21t it i nn eC eC t x

Por lo tanto, la solucin general de la ecuacin (2.3)puede ser expresada como

donde C 1 y C

2 son constantes. Utilizando

identidades)16.2(sincos)( 21 t At At x nn

donde A1 y A2 son nuevas constantes .


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Por consiguiente, . Por lotanto la solucin de la ecuacin (2.3) sujeto a lascondiciones iniciales de EQ (2.17) est dada por

)17.2()0()0(

02

01

x At x x At x

n

n x A x A /y 0201

)18.2( sincos)(0

0 t x

t xt x nn

n


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La naturaleza dela oscilacinarmnica puedeser representadagrficamentecomo se muestraen la figura.


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)29.2(21

2/1

st n

g f

Por lo tanto, la frecuencia natural en ciclos porsegundo:

y, el perodo natural:

)30.2(21

2/1

g f

st

nn


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2) La velocidad y la aceleracin de lamasa m en el tiempo t pueden obtenerse como:


)(t x )(t x

)31.2()cos()cos()()(

)2

cos()sin()()(

222

2

t At At dt xd

t x

t At At dt dx

t x

nnn

nnnn

n

3) Si el desplazamiento inicial es cero, 0 x

)32.2(sin2cos)(00

t x

t x

t x nnn

n

Si la velocidad inicial es cero, 0 x)33.2(cos)( 0 t xt x n


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4) La respuesta del sistema de un solo grado delibertad puede ser representada en el plano deespacio o fase de estado:


)35.2()sin(

)34.2()sin()(

A y

A xt

t At x

nn

nn

n x y /

o

donde

Elevando al cuadrado y sumando las ecuacionesEqs. (2.34) & (2.35)

)36.2(1

1)(sin)(cos

2

2

2

2

22

A y

A x

t t nn


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Representacin del plano de un sistema no amortiguado


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Example 2.5Natural Frequency of Pulley SystemDetermine the natural frequency of the systemshown in the figure. Assume the pulleys to befrictionless and of negligible mass.


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Example 2.5 Solution

21

222

k W

k W

The total movement of the mass m (point O) is:

(E.1))(4

)(4114

masstheof ntdisplaceme NetconstantspringEquivalentmasstheof Weight

21

21

21

21

21

k k k k

k

k k k k W

k k W k W

eq

eq

The equivalent spring constant of the system:


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(E.2)0 xk xm eq

(E.3)rad/sec)(

2/1

21

21

2/1

k k m

k k m

k eqn

By displacing mass m from the static equilibriumposition by x, the equation of motion of the masscan be written as

Hence, the natural frequency is given by:

(E.4)cycles/sec)(4

12

2/1

21

21

k k mk k

f nn

2 3 F Vib i f U d d


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2.3 Free Vibration of an UndampedTorsinal System

)37.2(0l

GI M t

Por la teora de la torsin de ejes (flechas)circulares, tenemos la relacin:

dondeM t es el par de torsin queproduce la torsin , G esel modulo de cortante , es

la longitud del eje,I 0 es el momneto polar deinerciade la seccintransversal del eje

l

2 3 F Vib i f U d d


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)38.2(32

4

0d

I

Momento de inercia polar:

Constante de resorte torsional:

)39.2(32

40

l Gd

l GI M

k t t

2 3 F Vib i f U d d


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Ecuacin de movimiento:

)40.2(00 t k J Aplicando la segunda del de Newton,

Por lo tanto, la frecuencia natural circularnatural:

)41.2(2/1

0

J k t

n

El perodo y la frecuencia de la vibracin en ciclos

por segundo son:

)43.2(2

1

)42.2(2

2/1

0

2/1

0

J

k f

k J

t n

t n


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E l 2 6


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Example 2.6Natural Frequency of Compound Pendulum Any rigid body pivoted at a point other than itscenter of mass will oscillate about the pivot pointunder its own gravitational force. Such a system isknown as a compound pendulum (shown inFigure). Find the natural frequency of such asystem.


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For a displacement , the restoring torque (due tothe weight of the body W ) is ( Wd sin ) and theequation of motion is


E.1)(0sin0 Wd J

E.2)(00 Wd J Hence, approximated by linear equation:

The natural frequency of the compound pendulum:

(E.3)2/1

0

2/1

0

J

mgd J Wd

n


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Comparing with natural frequency, the length ofequivalent simple pendulum:


E.4)(0md J

l

If J 0 is replaced by m k 0 2

, where k 0 is the radius ofgyration of the body about O,

(E.6)

(E.5)

2

2/1

2

0

0

d

k l

k

gd n


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If k G denotes the radius of gyration of the bodyabout G, we have:


(E.8)

E.7)( 2

2220

d d

k

l

d k k

G

G

(E.9)2

d k GA G

If the line OG is extended to point A such that

and

Eq.(E.8) becomes(E.10)OAd GAl


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2.4 Condiciones de estabilidad

Consideremos una barra rgida uniforme que es girada en

un extremo y simtricamente conectada por dos muellesen el otro extremo, como se muestra en la figura. Asumirque la masa de la barra es m y que los resortes son sinestirar cuando la barra est vertical.


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2.4 Condiciones de estabilidad

)47.2(0sin2

cossin23

2

l

W l kl ml

)48.2(02312

02

23

2

2

22

ml Wl kl

l W kl

ml

Por lo tanto la ecuacin del movimiento de la barra, para la

rotacin sobre el punto O, puede ser escrita como:

Para pequeas oscilaciones, Eq.(2.47) reduce a:

o

La solucin para Eq.(2.48) depende de la seal,es como se explica a continuacin:22 2/)312( ml Wl kl


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Caso 3. Cuando , definimos

2.4 Stability Conditions

00 )0(y)0( t t

)52.2()( 00 t t

02/)312( 22 ml Wl kl

Para condiciones iniciales lasolucin ser

2/1

2

2

2123

ml kl Wl

y expresamos la solucin Eq. (2.48) como

)53.2()( 21t t e Be Bt

donde B1 y B2 son constantes. Para condiciones iniciales, Eq.(2.53) sera00 )0( and )0(

t t

)54.2(2

1)( 0000

t t eet


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El principio de conservacin de la energa, en elcontexto de un sistema vibratorio no amortiguado,puede ser replanteado como

donde los subndices 1 y 2 denotan dos instantesdistintos de tiempo. Si el sistema es sometido amovimiento armnico, entonces

2.5 Mtodo de E nerga de Rayleighs

)55.2(2211 U T U T

)57.2(maxmax U T


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(E.1)21

2

l x y

dyl

mdT s s

The kinetic energy of the spring element of lengthdy is

where ms is the mass of the spring. The total

kinetic energy of the system can be expressed as

(E.2)32

121

21

21

)(springof energykinetic)(massof energykinetic

22

2

22

0

2

xm

xm

l x y

dyl m

xm

T T T

s

l

y

s

sm


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where X is the maximum displacement of the massand n is the natural frequency, the maximumkinetic and potential energies can be expressed as


(E.3)21 2kxU

(E.4)cos)( t X t x n

The total potential energy of the system is given by

By assuming a harmonic motion

(E.6)2

1

(E.5)32

1

2max

22max

kX U

X m

mT n s


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donde c = constante de amortiguamiento

2.6 Free Vibration with Viscous Damping

Ecuacin de movimiento:)58.2( xc F

En la figura, la ley de Newton da la ecuacin delmovimiento:

)59.2(0kx xc xm

kx xc xm

or


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Por lo tanto la solucin general es:

)64.2(

)(22

21

22

2

22

1

21

t mk

mc

mc

t mk

mc

mc

t st s

eC eC

eC eC t x

donde C 1 y C 2 son constantes arbirtrarias que sondeterminadas por las condiciones iniciales delsistema


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Case1 . Underdamped system


)/2oror1( mk mc/ cc c

)69.2()(1

2

1

1

22 t t nneC eC t x

Thus the general solution for Eq.(2.64) is:

Assuming that 0 , consider the following 3 cases:

For this condition, ( 2 -1) is negative and the rootsare:

nn

i s

i s

2

2

21

1

1


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where (C 1,C 2 ), (X, ), and (X 0 , 0 ) are arbitraryconstants to be determined from initial conditions.

)70.2(1cos1sin

1sin1cos

)(

02

0

2

22

21

12

11

1

2

1

1

22

22

t e X

t Xe

t C t C e

eC eC e

eC eC t x

nt

nt

nnt

t it it

t it i

n

n

n

nnn

nn

and the solution can be written in different forms:


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Underdamped Solution

The frequency of damped vibration is:

)76.2(1 2 nd


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Due to repeated roots, the solution of Eq.(2.59) isgiven by


)77.2(221 n

c

mc

s s

)78.2()()( 21t net C C t x

)79.2( and 00201 x xC xC n

In this case, the two roots are:

Case2 . Critically damped system )/2oror1( mk mc/ cc c

Application of initial conditions gives:

Thus the solution becomes: )80.2()( 000 t n net x x xt x


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In this case, the solution Eq.(2.69) is given by:

The roots are real and distinct and are given by:


0101

22

21

n

n

s

s

)81.2()(1

2

1

1

22 t t nneC eC t x

For the initial conditions at t = 0 ,

Case3 . Overdamped system )/2oror1( mk mc/ cc c

)82.2(12

1

12

1

2

02

01

2

02

01

n

n

n

n

x xC

x xC


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)84.2(

)83.2()cos()cos(

1

1

2

1

020

010

2

1

d n

d n

n

n

n

ee

e

t e X t e X

x x

t

t

d t

d t

Logarithmic Decrement:Using Eq.(2.70),

The logarithmic decrement can be obtained fromEq.(2.84):

)85.2(2

2

1

2ln

22

1

mc

x x

d nd n


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b h


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The energy dissipated in a complete cycle will be

Thus, Eq.(2.95) becomes

Consider the system shown in the figure below.

The total force resisting the motion is:


)95.2( xckxcvkx F

)97.2(cossin t X ct kX F d d d

)96.2( sin)( t X t x d If we assume simple harmonic motion:

)98.2( )(cos

)(cossin

2/2

0

22

/2

0

2

/2

0

X ct d t X c

t d t t kX

Fvdt W

d t

d d d

t d d d d

t

d

d

d

2 6 F Vib i i h Vi D i


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where W is either the max potential energy or the maxkinetic energy.

Computing the fraction of the total energy of thevibrating system that is dissipated in each cycle ofmotion,


)99.2(constant422

22

2

1 22

2

mc

X m

X cW W

d

d

d

The loss coefficient, defined as the ratio of theenergy dissipated per radian and the total strainenergy:

)100.2(2

)2/(tcoefficienloss

W W

W W



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The equation of motion can bederived as:

Torsional systems with Viscous Damping:


)101.2( t cT

Consider a single degree of freedom torsionalsystem with a viscous damper, as shown in figure(a). The viscous damping torque is given by:

)102.2(00 t t k c J

where J 0 = mass moment of inertia of disckt = spring constant of system = angular displacement of disc



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In the underdamped case, the frequency ofdamped vibration is given by:


)103.2(1 2 nd

)104.2(0 J

k t n

where

and)105.2(

22 00 J k

c

J

c

c

c

t

t

n

t

tc

t

where c tc is the critical torsional damping constant

Example 2.11


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pShock Absorber for a Motorcycle An underdamped shock absorber is to bedesigned for a motorcycle of mass 200kg (shownin Fig.(a)). When the shock absorber is subjectedto an initial vertical velocity due to a road bump,the resulting displacement-time curve is to be asindicated in Fig.(b). Find the necessary stiffnessand dampeing constants of the shock absorber ifthe damped period of vibration is to be 2 s and the

amplitude x 1 is to be reduced to one-fourth in onehalf cycle (i.e., x 1.5 = x 1/4). Also find the minimuminitial velocity that leads to a maximumdisplacement of 250 mm.


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E l 2 11 S l ti


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From which can be found as 0.4037. Thedamped period of vibration given by 2 s. Hence,

Hence the logarithmic decrement becomes


16/4/,4/ 15.1215.1 x x x x x

(E.1)1

27726.216lnln

22

1

x x

Since ,

rad/s4338.3)4037.0(12

2

1

222

2

2

n

nd d

E l 2 11 S l ti


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The displacement of the mass will attain its maxvalue at time t

1, given by

Thus the damping constant is given by:

and the stiffness by:


s/m- N54.373.1)4338.3)(200(22 nc mc

s/m- N4981.554)54.1373)(4037.0( ccc

N/m2652.2358)4338.3)(200( 22 nmk

The critical damping constant can be obtained:

sec3678.0)9149.0(sin

9149.0)4037.0(1sinsin

1sin

1

1

211

21

t

t t

t

d

d

or

This gives:


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2 7 Free Vibration ith Co lomb Damping


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Coulombs law of dry friction states that, whentwo bodies are in contact, the force required toproduce sliding is proportional to the normalforce acting in the plane of contact. Thus, thefriction force F is given by:

2.7 Free Vibration with Coulomb Damping

)106.2(mg W N F where N is normal force,

is the coefficient of sliding or kinetic friction

is usu 0.1 for lubricated metal, 0.3 for nonlubricatedmetal on metal, 1.0 for rubber on metal

Coulomb damping is sometimes called constantdamping


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2 7 Free Vibration with Coulomb Damping


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2005 Pearson Education South Asia Pte Ltd.73


)107.2(or N kx xm N kx xm

Case 1 . When x is positive and dx/dt is positive or

when x is negative and dx/dt is positive (i.e., forthe half cycle during which the mass moves fromleft to right) the equation of motion can beobtained using Newtons second law (Fig.b):

Hence,

)108.2( sincos)( 21 k N

t At At x nn

where n = k/m is the frequency of vibration

A1 & A2 are constants



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Case 2 . When x is positive and dx/dt is negative

or when x is negative and dx/dt is negative (i.e.,for the half cycle during which the mass movesfrom right to left) the equation of motion can bederived from Fig. (c):


)109.2(or N kx xm xm N kx

The solution of the equation is given by:

)110.2(sincos)( 43 k N

t At At x nn

where A 3 & A4 are constants



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Fig.2.34 Motion of the mass with Coulomb damping



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Eqs.(2.107) & (2.109) can be expressed as asingle equation using N = mg:


Solution:

)111.2(0)sgn( kx xmg xm

where sgn(y) is called the sigum function, whosevalue is defined as 1 for y > 0, -1 for y< 0, and 0for y = 0.

Assuming initial conditions as

)112.2(0)0(

)0( 0t x

xt x


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Note the following characteristics of a system with

Coulomb damping:1. The equation of motion is nonlinear with Coulomb

damping, while it is linear with viscous damping2. The natural frequency of the system is unaltered with the

addition of Coulomb damping, while it is reduced with theaddition of viscous damping.

3. The motion is periodic with Coulomb damping, while itcan be nonperiodic in a viscously damped (overdamped)

system.4. The system comes to rest after some time with Coulomb

damping, whereas the motion theoretically continuesforever (perhaps with an infinitesimally small amplitude)with viscous damping.



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Note the following characteristics of a system with

Coulomb damping:5. The amplitude reduces linearly with Coulomb damping,

whereas it reduces exponentially with viscous damping.6. In each successive cycle, the amplitude of motion is

reduced by the amount 4 N/k, so the amplitudes at theend of any two consecutive cycles are related:

)116.2(4

1 k N

X X mm

As amplitude is reduced by an amount 4 N/k inone cycle, the slope of the enveloping straightlines (shown dotted) in Fig 2.34.



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)118.2(

)117.2(

0

0

T k J

T k J

t

t

)119.2(0 J

k t n

Torsional Systems with Coulomb Damping:The equation governing the angular oscillations ofthe system is

and

The frequency of vibration is given by



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and the amplitude of motion at the end of the r thhalf cycle ( r ) is given by:


The motion ceases when

)121.2(2

0

t

t

k T

k T

r

)120.2(2

0t

r k T

r

Example 2.14


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Pulley Subjected to Coulomb Damping A steel shaft of length 1 m and diameter 50 mm isfixed at one end and carries a pulley of massmoment of inertia 25 kg-m 2 at the other end. Aband brake exerts a constant frictional torque of400 N-m around the circumference of the pulley.

If the pulley is displaced by 6 and released,determine (1) the number of cycles before thepulley comes to rest and (2) the final settlingposition of the pulley.


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Example 2 14 Solution


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926.5

5.087,49

8005.087,49

40010472.0

r

and T = constant friction torque applied to the

pulley = 400 N-m. Eq.(E.1) gives

Thus the motion ceases after six half cycles.

(2) The angular displacement after six half cycles:

39734.0rad006935.0

5.087,494002610472.0

from the equilibrium posotion on the same side ofthe initial displacement.

2 8 Free Vibration with Hysteretic Damping


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Consider the spring-viscous damper arrangement

shown in the figure below. The force needed tocause a displacement:

For a harmonic motionof frequency andamplitude X,

2.8 Free Vibration with Hysteretic Damping

)122.2( xckx F

)123.2( sin)( t X t x

)124.2(

)sin(

cossin)(

22

22

x X ckx

t X X ckx

t cX t kX t F

Fig.2.35 Spring-Viscous damper system

2 8 Free Vibration with Hysteretic Damping


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)125.2(

coscossin

2

/2

0

cX

dt t X t cX t kX FdxW

When F versus x is plotted, Eq.(2.124) represents

a closed loop, as shown in Fig(b). The area of theloop denotes the energy dissipated by thedamper in a cycle of motion and is given by:

Hence, the dampingcoefficient:

)126.2( h

c

where h is called the hysteresisdamping constant.

Fig.2.36 Hysteresis loop



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Complex Stiffness .For general harmonic motion, , the force

is given by


)127.2(2hX W

t i Xe x

)128.2()( xcik iXeckXe F t it i

Eqs.(2.125) and (2.126) gives

Thus, the force-displacement relation:

)130.2()1(1

)129.2()(

ik k h

ik ihk

xihk F

where



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The hysteresis logarithmic decrement can bedefined as


)131.2(2 X k W

)135.2()1ln(ln1

j

j

X

X

Response of the system .

The energy loss per cycle can be expressed as

Corresponding frequency)136.2(

mk

Response of a hysteretically damped system



89/92


And thus the equivalent damping constant is


)137.2(22

2

k h

k h

eq

eq

The equivalent viscous damping ratio

)138.2( 2

2

hk mk mk cc eqceq


90/92



91/92



Using the ratio of successive amplitudes,

Eq.(2.135) yields the hysteresis logarithmicdecrement as

0127.004.0

04.11

)1ln()04.1ln(ln1

or

X

X

j

j

The equivalent viscous damping coefficient is

(E.1)km

mk

k k ceq



92/92

p .

Using the known values of the equivalent stiffnessand equivalent mass,

s/m- N109013.44)105)(1025()0127.0( 356eqc

Since c eq < c c, the bridge is underdamped.Hence, its free vibration response is

0063.0100678.7071

109013.40

1sin

1

1cos)(

3

3

2

2

0020

c

eq

n

n

nn

t

c

c

t x x

t xet x n

where

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