vibration lectures part 1

8/13/2019 Vibration Lectures Part 1

1/71

1

Dr. Mostafa Ranjbar

Fundamentals of Vibration


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-

Outline Why vibration is important?

Definition; mass, spring (or stiffness)dashpot

Newtons laws of motion, 2nd

order ODE

Alternative way to find eqn of motion:energy methods

Three types of vibration for SDOF sys.

Examples


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Vibrations can lead to excessive deflectionsand failure on the machines and structures

To reduce vibration through proper design ofmachines and their mountings To utilize profitably in several consumer and

industrial applications To improve the efficiency of certainmachining, casting, forging & weldingprocesses

To stimulate earthquakes for geologicalresearch and conduct studies in design ofnuclear reactors

Why to study vibration


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Why to study vibration

Imbalance in the gas or diesel engines

Blade and disk vibrations in turbines Noise and vibration of the hard-disks in

your computers

Vibration testing for electronic

packaging to conform Internatioalstandard for quality control (QC)

Safety eng.: machine vibration causes

parts loose from the body

Cooling fan in the power supply


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Stiffness From strength of materials (Solid Mech) recall:

20 mm0

10 3 N

f k

x

g

x 0x 1 x 2 x 3

Force

Displacement


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-

Free-body diagram and equations of motion

Newtons Law:

00 )0(,)0(

0)()(

)()(

v x x x

t kxt xm

t kxt xm

==

=+

=

&

&&

&&k

c

m

y

x

mg

N

x (t )

f k

f c Friction-freesurface

Friction-free

surface

k

c=0


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-

2nd Order Ordinary Differential Equationwith Constant Coefficients

)sin()(

rad/sinfrequencynatural:

0)()(: byDivide 2

+=

=

=+

t At xmk

t xt xm

n

n

n&&


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-

Periodic Motion

n 2

Amplitude, A

Time, t

period

Max velocity

Initial displacement

Phase =

x(t )


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-

Periodic Motion

2 4 6 8 10 12

-1.5

-1

-0.5

0

0.5

1

1.5

Time [s]

x(t) [mm]

2 n___

A

-A

x(t )

n 2

Time, t


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-

Frequency n is in rad/s is the natural frequency

f n = n rad/s2 rad/cycle

= n cycles2 s

= n2

Hz

T = 2 n

s is the period

We often speak of frequency in Hertz , but weneed rad/s in the arguments of the trigonometricfunctions ( sin and cos function).


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Amplitude & Phase from the initialconditions

x0 = A sin( n 0 + ) = Asin v0 = n A cos( n 0 + ) = n A cos

Solving yields

A = 1 n

n2 x0

2 +v02

Amplitude

1 24 44 34 4 4, = tan 1 n x0

v0

Phase

1 24 4 34 4


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Phase Relationship between x, v, a

t

A

A

O

v n A

v n A

O

v n 2A

v 2n A

O

t

t

Displacementx (t ) = A sin(v n t + f )

Velocityx (t ) = v n A cos (v n t + f )

Accelerationx (t ) = v n 2A sin(v n t + f )

)cos( += t A xn

)sin( += t A x nn&

)cos(2 += t A x nn&& Acceleration

Velocity

Displacement


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Example For m= 300 kg and n =10 rad/scompute the stiffness:

n = k

m k = m n2

= (300)10 2 kg/s 2

=3 104

N/m


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-

Other forms of the solution:

x(t ) = Asin( nt + ) x(t ) = A1 sin nt + A2 cos n t x(t ) = a 1e j nt + a 2e j nt

Phasor : representation of a complex number in terms of acomplex exponentialRef: 1) Sec 1.10.2, 1.10.3

2) http://mathworld.wolfram.com/Phasor.html


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Some useful quantities

A = peak value x = lim

T

1

T x(t )dt = average value

0

T

x 2 = limT

1

T

x 2 ( t )dt 0

T

= mean - square value

xrms = x 2 = root mean square value


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-

Peak Values

A x

A x

A x

2max

max

max

:onaccelerati

:velocity

:ntdisplaceme:of peak valueormax

=

==

&&

&


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Example Hardware store spring, bolt: m= 49.2x10-3

kg,k=857.8 N/m and x0 =10 mm. Compute n and maxamplitude of vibration.

n = k

m= 857.8 N/m

49.2 10 -3 kg=132 rad/s

f n = n2 =21 Hz

T = 2 n

= 1 f n

= 121 cyles sec

0.0476 s

x(t )max = A =1

n n

2 x02 + v02 = x 0 =10 mm


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Compute the solution and max velocity andacceleration

v(t )max = n A =1320 mm/s = 1.32 m/sa( t )max = n

2 A =174.24 10 3 mm/s 2

=174.24 m/s2

17.8 g ! =tan 1 n x0

0

=

2

rad

x(t ) =10sin(132 t + / 2) =10 cos(132 t ) mm


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Derivation of the solution

jt jt

jt jt

n

t t

t

nn

nn

eaeat xeat xeat x

j jmk

mk

k m

kaeaem

kx xmaet x

+= ==

===

=+=+

=+=

21

21

2

2

)()( and )(

0

0

0 into )( Substitute &&


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Viscous Damping:Damping force is proportional to the velocityof the vibrating body in a fluid medium suchas air, water, gas, and oil.

Coulomb or Dry Friction Damping:Damping force is constant in magnitude butopposite in direction to that of the motion ofthe vibrating body between dry surfaces

Material or Solid or Hysteretic Damping:Energy is absorbed or dissipated by materialduring deformation due to friction betweeninternal planes

Damping Elements


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Shear Stress ( ) developed in the fluid layer at adistance y from the fixed plate is:

where d u/ d y = v/h is the velocity gradient.

Shear or Resisting Force (F) developed at the bottomsurface of the moving plate is:

where A is the surface area of the moving plate.

( )26.1dydu =

( )27.1cvh Av

A F ===

is the damping constanth A

c =

Viscous Damping


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-

and

is called the damping constant.

If a damper is nonlinear , a linearization processis used about the operating velocity ( v*) and theequivalent damping constant is:

( )28.1h Ac =

( )29.1*vdv

dF c =

Viscous Damping


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Linear Viscous Damping

A mathematicalform

Called a dashpot orviscous damper

Somewhat like ashock absorber

The constant c hasunits: Ns/m or kg/s

)(t xc f c &=


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-

Spring-mass-damper systems

From Newtons law:

00 )0( ,)0(

0)()()(

)()(

)(

v x x x

t kxt xct xm

t kxt xc

f f t xm k c

==

=++=

=

&

&&&

&

&&k

c

m

y

x

mg

N

x (t )

f k

f c Friction-freesurface

Friction-free

surface

k

c

f k f c


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-

Derivation of the solution

t t

t t

n

t t t

t

eaeat x

eat xeat x

j jmk

mk

k cm

kaeaecaem

kx xc xmaet x

21

21

21

21

2

2

)(

)( and )(

0

0

0 into )( Substitute

+===

===

=++=++

=++= &&&

1 22,1 = nn


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Solution (dates to 1743 by Euler)

less)(dimensionratiodamping2=

and where

0)()(2)(

bymotionof equationDivide2

=

=

=++

km

cm

k

t xt xt x

m

n

nn

&&&


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-

Let x(t ) = ae t & subsitute into eq. of motion 2ae t + 2 n ae

t + n2ae t = 0

which is now an algebraic equation in :

1,2 = n n 2 1

from the roots of a quadratic equationHere the discriminant 2 1, determines

the nature of the roots

-


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-

Three possibilities:

00201

21

,

:conditionsinitialtheUsing

)(

221=

damped criticallycalled

repeated &equalareroots1)1

xva xa

teaeat x

mkmcc

n

t t ncr

nn

+==

+====

=

Sec. 2.6

-


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Critical damping continued

No oscillation occurs Useful in door mechanisms , analog gauges

0.8

0.6

0.4

0.2

0.0

0.2 0.5 1 .0 1.5 2.0 2 .5 3.0 3.5

Time (sec )

D i s p l a c e m

e n t ( m m )

x(t ) = [ x0 +(v

0 +

n x

0)t ]e nt

Di s

pl a c em en t

Time, t

-


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12

)1(

12

)1(

where

)()(

1

:rootsrealdistincttwo-goverdampincalled ,1)2

2

02

02

2

02

0

1

12

11

22,1

22

++=

++

=

+=

=>

n

n

n

n

t t t

nn

xva

xv

a

eaeaet x nnn

-


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The overdamped response

0.4

0.2

0.0

0.4

0.2

0 1

1

2

3

2 3 4 5 6

Time (s ec )

D i s p

l a c e

m e

n t ( m m ) 1. x 0 = 0.3,2. x 0 = 0,

3. x 0 = 0.3,v 0 = 0v 0 = 1v 0 = 0

1: x0 =0.3, v0 = 02: x0 =0.0, v0 = 13: x0 =-0.3, v0 = 0

Di s

pl a c em en t

Time, t

-


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3)


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Underdamping

x(t )=

e nt (a1e j nt 1

2

+ a

2e j nt 1

2

)

= Ae nt sin( d t + ) d = n 1 2 , damped natural frequency

A = 1 d

(v0 + n x0 )2 +( x0 d )2

= tan1 x

0

d

v0 + n x0

-


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Underdamped-oscillation

1.0

0.0

1.010 15 Time

(sec)

D i s p l a c e m e n t ( m m )

Gives an oscillatingresponse withexponential decay

Most natural systemsvibrate with andunderdamped response

See textbook for detailsand otherrepresentations

Di s

pl a c em en t

Time, t

-


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Example consider the spring in Ex., if c = 0.11 kg/s,determine the damping ratio of the spring-bolt system.

m = 49.2 103 kg, k =857.8 N/m

ccr = 2 km =2 49.2 103 857.8

=12.993 kg/s =

cccr

= 0.11 kg/s12.993 kg/s

=0.0085

the motion is underdamped

and the bolt will oscillate

-


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ExampleThe human leg has a measured natural frequency of around20 Hz when in its rigid (knee locked) position, in thelongitudinal direction (i.e., along the length of the bone) with adamping ratio of = 0.224 .

Calculate the response of the tip if the leg bone to v0(t=0)= 0.6 m/s and x 0(t=0)=0

This correspond to the vibration induced while landing on yourfeet, with your knees locked from a height of 18 mm) and plot

the response. What is the maximum accelerationexperienced by the leg assuming no damping?

-


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Solution:V0=0.6, X 0=0, = 0.224

n = 20

1

cycles

s

2 rad

cycles=125.66 rad/s

d =125.66 1 .224( )2 =122.467 rad/s

A =0.6 + 0.224( )125.66( )0( )( )2 + 0( )122.467( )2

122.467 = 0.005 m

= tan -10( ) d ( )

v0 +

n0( )

= 0

x t ( )= 0.005 e28.148 t sin 122.467 t ( )

20

200 )()(

1d n

d

x xv A

++=

+=

00

01tan xv

x

n

d

-


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Use undamped formula to get max acceleration:

( ) ( )( ) 2222

0

00

2

020

m/s396.75m/s66.1256.06.0

max

m6.0

m

0,6.0,66.125,

==

==

==

===

+=

nnn

nn

n

n

A x

v A

xvv

x A

&&

maximum acceleration = 75.396 m/s2

9.81 m/s 2 g =7.68g' s

-


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Plot of the response:

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14-5

-4

-3

-2

-1

0

1

2

3

4

5

Time (s)

Displacement (mm)Displacement

Time, t

)467.122sin(005.0)(148.28

t et x t

=

-


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Example Compute the form of the response of anunderdamped system using the Cartesian form

++=

+=

++= +

+=

=+==+=+=

=+

t xt xv

et x

xv A

x A x Avt At Ae

t At Ae x

x A A Ae x x

t At Aet Aet x

y x y x y x

d d d

nt

d

n

d n

d d

t

d

d d t

n

d d t

d t

n

n

n

nn

cossin)(

)0sin0cos()0cos0sin()sincos(

)cossin(

))0cos()0sin(()0(

)cossin()sin()(

sincoscossin)sin(

000

001

01010

21

21

02210

0

21

&

Eq. 2.72

-


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MODELING AND ENERGY

METHODS An alternative way to determine the

equation of motion and an alternative wayto calculate the natural frequency

-


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Modelling

Newtons Laws

&&

&&

00 I M

xm F

i

xi

=

=

-


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Energy Methods

0)(or

constant2

1=donework

EnergyKinetic

12

2

1

2

EnergyPotential

21

=+

=+

==

=

U T dt

d

U T

T T xmU U

dx xm Fdx

321&

48476

&&

Alternate method of getting the eq. of motion

-


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Rayleighs Method

T 1+ U 1= T 2+ U 2

Let t 1 be the time at which m moves throughits static equilibrium position , then U 1=0, reference point Let t 2 be the time at which m undergoes its

max displacement (v=0 so T 2=0), U 2 is max(T 1 must be max ),

Thus U max =T max

Ref: Section 2.5

-


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Example The effect of including the mass of thespring on the value of the frequency.

x(t )

m s, k

y y +dy

l

m

Ex. 2.8


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What about gravity?

m

m

+ x(t )

0

k

mg

k

+ x(t )

2

2

21

)(21

xmT

mgxU

xk U

grav

spring

&==

+=

mg k =0, from FBD, and static equilibrium

-


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0

0)()()(

0)(21

21

0)(use Now

equ.staticfrom0

22

=+

=++ ++

=

++

=+

kx xm

mg k xkx xm x x xk xmg x xm

xk mgx xmdt d

U T dt d

&&

43421&&&&&&&&&

&

-


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More on springs and stiffness

Longitudinal motion A is the cross sectional area(m 2)

E is the elastic modulus(Pa=N/m 2)

l is the length (m)

k is the stiffness (N/m) x(t )

m

k = EA

ll

-


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Torsional Stiffness

J p is the polar moment of

inertia of the rod J is the mass moment of

inertia of the disk

G is the shear modulus, l is the length

Jp

J (t)

0 k =

GJ pl

Sec. 2.3

-


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Example compute the frequency of a shaft/masssystem { J = 0.5 kg m 2}

rad/s2.2

)mm)(0.5kg2(]32/m)105.0()[ N/m108(

cm0.5of diametershaft,steelm2aFor32

,

0)()(

0)()(

2

42210

4

=

==

===

=+

=+=

J

GJ

d J J

GJ J k

t J k

t

t k t J J M

pn

p p

n

l

l

&&

&&&&

-


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Transverse beam stiffness

f

m

x

Strength of materials andexperiments yield:

k = 3 EI l 3

n = 3 EI ml 3

-


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Samples of Vibrating Systems

Deflection of continuum (beams, plates,bars, etc) such as airplane wings, truckchassis, disc drives, circuit boards

Shaft rotation Rolling ships

See text for more examples.

-

E l ff t f f l


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Example : effect of fuel onfrequency of an airplane wing

Model wing as transversebeam

Model fuel as tip mass

Ignore the mass of the wingand see how the frequencyof the system changes as

the fuel is used upx(t)

l

E, I m

-


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Mass of pod 10 kg empty 1000 kg full I = 5.2x10 -5 m 4, E =6.9x10 9 N/m, l = 2 m

Hence thenaturalfrequencychanges by anorder ofmagnitudewhile it emptiesout fuel.

full =3 EI ml 3

= 3(6.9 109 )(5.2 105 )

1000 23

=11.6 rad/s = 1.8 Hz

empty = 3 EI ml 3 = 3(6.9 10 9 )(5.2 10 5 )

10 2 3

=115 rad/s =18.5 Hz

Pod= a streamlined external housing that enclose engines or fuel

-


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Does gravity effect frequency?

kx

mg

0

+x

Static equilibrium:

!00

)(:equationDynamic

0

=+=++

++==

+==

kx xmmg k kx xm

mg xk xm F

mg k F

&&&&

&&

-

S i D fl i


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Static Deflection

k mg

m

s ===

=

it.tomassaattaching bygravity

of forceunder thecompressed or stretched isspringdistance,

Many symbols in use including xs and x0

-

C bi i S i


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Combining Springs

Equivalent Spring

series : k AC =1

1k 1 + 1k 2 parallel : k

ab = k

1 + k

2

-

Use these to design from


59/71

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Use these to design fromavailable parts

Discrete springs available in standardvalues Dynamic requirements require specific

frequencies Mass is often fixed or + small amount

Use spring combinations to adjust wn Check static deflection

-


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ExampleDesign of a spring mass system

using available springs: series vs parallel

Let m = 10 kg Compare a series and parallel

combination a) k 1 =1000 N/m, k 2 = 3000 N/m,

k 3 = k 4 =0

b) k 3 =1000 N/m, k 4 = 3000 N/m,k 1 = k 2 =0

k1 k2

k3k

4

m

-


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Case a) parallel connection :k 3 = k 4 =0, k eq = k 1 + k 2 =1000 +3000 = 4000 N/m

parallel = k eg

m= 4000

10=20 rad/s

Case b) series connection :

k 1 = k 2 =0, k eq =1

(1 k 3 ) + (1 k 4 )= 3000

3 +1= 750 N/m

series = k eg

m= 750

10=8.66 rad/s

Same physical components, very different frequency Allows some design flexibility in using off-the-shelf components

-

Free Vibration with Coulomb 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:

)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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Equation of Motion:Consider a single degree of freedom system withdry friction as shown in Fig.(a) below.

Since friction force varies with the direction ofvelocity , we need to consider two cases asindicated in Fig.(b) and (c).

p g

-



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)107.2(or N kx xm N kx xm =+=&&&&

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

p g


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Free Vibration with Hysteretic Damping


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Consider the spring-viscous damper arrangementshown in the figure below. The force needed tocause a displacement:

For a harmonic motion

of frequency andamplitude X ,

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

=

=+=

22 x X c X c

X c )(t F )(t x


67/71


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2 6 4 Energy dissipated in Viscous Damping:


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94

The energy dissipated in a complete cycle is:

)93.2(velocityforce2

2

====

dt dx

ccv Fvdt

dW

2.6.4 Energy dissipated in Viscous Damping:

In a viscously damped system, the rate of change

of energy with time is given by:

)94.2(

)(cos

2

2

0

222

)/2(

0

X c

t d t cX dt dt

dxcW

d

d d d t

d

=

=

=

=

-

Energy dissipation


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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

=+

=

=

=

=

=

-

Energy dissipation and Loss Coefficient


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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, Specific Damping Capacity

)99.2(constant422

22

21 22

2

==

==

mc

X m

X cW W

d

d

d

The loss coefficient , defined as the ratio of the

energy dissipated per radian and the total strainenergy:)100.2(

2)2/(

tcoefficienlossW W

W W

==

vibration lectures part 1

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