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8/13/2019 Vibration Lectures Part 1
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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
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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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Free Vibration with Coulomb Damping
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
-
Free Vibration with Coulomb Damping
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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
-
Free Vibration with Coulomb Damping
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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
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2 6 4 Energy dissipated in Viscous Damping:
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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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95
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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96
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
==