Download - Vibration Chapter02
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Mechanical Vibrations
Fifth Edition in SI UnitsSingiresu S. Rao
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2011 Mechanical Vibrations Fifth Edition in SI Units3
Chapter 2Free Vibration of Single-Degree-of-Freedom Systems
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Chapter Outline
2.1 Introduction
2.2 Free Vibration of an Undamped Translational System
2.3 Free Vibration of an Undamped Torsional System
2.4 Response of First-Order Systems and Time Constant
2.5 Rayleighs Energy Method
2.6 Free Vibration with Viscous Damping
2.7 Graphical Representation of Characteristic Roots and Corresponding Solutions
2.8 Parameter Variations and Root Locus Representations
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Chapter Outline
2.9 Free Vibration with Coulomb Damping
2.10 Free Vibration with Hysteretic Damping
2.11 Stability of Systems
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2.1Introduction
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2.1 Introduction
Free Vibration occurs when a system oscillates only under an initial disturbance with no external forces acting after the initial disturbance
Undamped vibrations result when amplitude of motion remains constant with time (e.g. in a vacuum)
Damped vibrations occur when the amplitude of free vibration diminishes gradually overtime, due to resistance offered by the surrounding medium (e.g. air)
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2.1 Introduction
Several mechanical and structural systems can be idealized as single degree of freedom systems, for example, the mass and stiffness of a system
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2.2Free Vibration of an Undamped Translational System
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion Using Newtons Second Law of Motion:
If mass m is displaced a distance when acted upon by a resultant force in the same direction,
If mass m is constant, this equation reduces to
where is the acceleration of the mass
)(tx)(tF
dttxdm
dtdtF )()(
(2.1))()( 22
xmdt
txdmtF
2
2 )(dt
txdx
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2.2 Free Vibration of an Undamped Translational System
For a rigid body undergoing rotational motion, Newtons Law gives
where is the resultant moment acting on the body and and are the resulting angular displacement and angular
acceleration, respectively.
For undamped single degree of freedom system, the application of Eq. (2.1) to mass m yields the equation of motion:
)2.2()( JtM M
22 /)( dttd
)3.2(0or )( kxxmxmkxtF
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2011 Mechanical Vibrations Fifth Edition in SI Units12
2.2 Free Vibration of an Undamped Translational System
Equation of Motion Using Other Methods:
1. DAlemberts PrincipleThe equations of motion, Eqs. (2.1) & (2.2) can be rewritten as
The application of DAlemberts principle to the system shown in Fig.(c) yields the equation of motion:
(2.4b) 0)(
)2.4a( 0)(
JtM
xmtF
)3.2(0or 0 kxxmxmkx
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2011 Mechanical Vibrations Fifth Edition in SI Units13
2.2 Free Vibration of an Undamped Translational System
Equation of Motion Using Other Methods:
2. Principle of Virtual DisplacementsIf a system that is in equilibrium under the action of a set of forces is subjected to a virtual displacement, then the total virtual work done by the forces will be zero.
Consider spring-mass system as shown, the virtual work done by each force can be computed as:
xxmWxkxW
i
S
)( force inertia by the done work Virtual)( force spring by the done work Virtual
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion Using Other Methods:
2. Principle of Virtual Displacements (Cont)When the total virtual work done by all the forces is set equal to zero, we obtain
Since the virtual displacement can have an arbitrary value, , Eq.(2.5) gives the equation of motion of the spring-mass system as
)5.2(0 xkxxxm 0x
)3.2(0 kxxm
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion Using Other Methods:
3. Principle of Conservation of EnergyA system is said to be conservative if no energy is lost due to friction or energy-dissipating nonelastic members.
If no work is done on the conservative system by external forces, the total energy of the system remains constant. Thus the principle of conservation of energy can be expressed as:
)6.2(0)(or constant UTdtdUT
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion Using Other Methods:
3. Principle of Conservation of Energy (Cont)The kinetic and potential energies are given by:
Substitution of Eqs. (2.7) & (2.8) into Eq. (2.6) yields the desired equation
)8.2( 21
)7.2( 21
2
2
kxU
xmT
)3.2(0 kxxm
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion of a Spring-Mass System in Vertical Position:
Consider the configuration of the spring-mass system shown in the figure.
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion of a Spring-Mass System in Vertical Position:
For static equilibrium,
The application of Newtons second law of motion to mass m gives
and since , we obtain
)9.2(stkmgW
Wxkxm st )(
)10.2(0 kxxm Wk st
where w = weight of mass m,= static deflection
g = acceleration due to gravityst
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion of a Spring-Mass System in Vertical Position:
Notice that Eqs. (2.3) and (2.10) are identical. This indicates that when a mass moves in a vertical direction, we can ignore its weight, provided we measure x from its static equilibrium position.
Hence, Eq. (2.3) can be expressed as
By using the identities
)15.2()( 21titi nn eCeCtx
)16.2( sincos)( 21 tAtAtx nn
where C1 and C2 are constants
where A1 and A2 are new constants
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2.2 Free Vibration of an Undamped Translational System
Equation of Motion of a Spring-Mass System in Vertical Position:
From Eq (2.16), we have
Hence,
Solution of Eq. (2.3) is subjected to the initial conditions of Eq. (2.17) which is given by
)17.2()0()0(
02
01
xAtxxAtx
n
)18.2( sincos)( 00 txtxtx nn
n
nxAxA / and 0201
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2011 Mechanical Vibrations Fifth Edition in SI Units21
2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Eqs.(2.15), (2.16) & (2.18) are harmonic functions of time. Eq. (2.16) can also be expressed as:
where A0 and are new constants, amplitude and phase angle respectively:
)23.2()sin()( 00 tAtx n0
)24.2(
2/12
0200
n
xxAA
)25.2(tan0
010
xx n
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2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
The nature of harmonic oscillation can be represented graphically as shown in the figure.
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2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Note the following aspects of spring-mass systems:
1. When the spring-mass system is in a vertical position
Circular natural frequency:
Spring constant, k:
Hence,
)26.2( 2/1
mk
n
)27.2(stst
mgWk
)28.2(2/1
stn
g
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2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Note the following aspects of spring-mass systems:
1. When the spring-mass system is in a vertical position (Cont)
Natural frequency in cycles per second:
Natural period:
)29.2(21
2/1
stn
gf
)30.2( 212/1
gfst
nn
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2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Note the following aspects of spring-mass systems:
2. Velocity and the acceleration of the mass m at time tcan be obtained as:
)(tx )(tx
)31.2()cos()cos()()(
)2
cos()sin()()(
222
2
tAtAtdtxdtx
tAtAtdtdxtx
nnn
nnnn
n
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2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Note the following aspects of spring-mass systems:
3. If initial displacement is zero,
If initial velocity is zero,
0x)32.2(sin
2cos)( 00 txtxtx n
nn
n
0x)33.2(cos)( 0 txtx n
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2011 Mechanical Vibrations Fifth Edition in SI Units27
2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Note the following aspects of spring-mass systems:
4. The response of a single degree of freedom system can be represented by:
By squaring and adding Eqs. (2.34) & (2.35)
)35.2()sin(
)34.2()sin()(
Ay
Axt
tAtx
nn
nn
)36.2(1
1)(sin)(cos
2
2
2
2
22
Ay
Ax
tt nn
nxy /where
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2.2 Free Vibration of an Undamped Translational System
Harmonic Motion
Note the following aspects of spring-mass systems:
4. Phase plane representation of an undamped system
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2.2 Free Vibration of an Undamped Translational System
Example 2.2Free Vibration Response Due to Impact
A cantilever beam carries a mass M at the free end as shown in the figure. A mass m falls from a height h on to the mass M and adheres to it without rebounding. Determine the resulting transverse vibration of the beam.
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2.2 Free Vibration of an Undamped Translational System
Example 2.2Free Vibration Response Due to ImpactSolution
Using the principle of conservation of momentum:
The initial conditions of the problem can be stated:
(E.1)2
)(
0
0
ghmM
mvmM
mx
xmMmv
m
m
(E.2) 2, 00 ghmMmx
kmgx
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2.2 Free Vibration of an Undamped Translational System
Example 2.2Free Vibration Response Due to ImpactSolution (Cont)
Thus the resulting free transverse vibration of the beam can be expressed as
where
)cos()( tAtx n
)(3 , tan , 3
0
01
2/12
020 mMl
EImM
kxxxxA n
nn
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2.2 Free Vibration of an Undamped Translational System
Example 2.5Natural Frequency of Pulley System
Determine the natural frequency of the system.Assume the pulleys to be frictionless and of negligible mass.
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2.2 Free Vibration of an Undamped Translational System
Example 2.5Natural Frequency of Pulley SystemSolution
The total movement of the mass m (point O) is
The equivalent spring constant of the system is
21
222kW
kW
(E.1))(4
)(4114
mass theofnt displacemeNet constant spring Equivalentmass theofWeight
21
21
21
21
21
kkkkk
kkkkW
kkW
kW
eq
eq
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2.2 Free Vibration of an Undamped Translational System
Example 2.5Natural Frequency of Pulley SystemSolution
By displacing mass m from the static equilibrium position by x, the equation of motion of the mass can be written as
Natural frequency is given by
(E.2) 0 xkxm eq
(E.3)rad/sec)(
2/1
21
21
2/1
kkmkk
mkeq
n
(E.4)cycles/sec)(4
12
2/1
21
21
kkmkkf nn
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2.3Free Vibration of an Undamped Torsional System
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2.3 Free Vibration of an Undamped Torsional System
From the theory of torsion of circular shafts, we have the relation:
)37.2(0l
GIMt
where
Mt = torque that produces the twist ,
G = shear modulus,
l = is the length of shaft,
I0 = polar moment of inertia of cross section of shaft
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2011 Mechanical Vibrations Fifth Edition in SI Units37
2.3 Free Vibration of an Undamped Torsional System
Polar Moment of Inertia:
Torsional Spring Constant:
)38.2(32
4
0dI
)39.2(32
40
lGd
lGIMk tt
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2.3 Free Vibration of an Undamped Torsional System
Equation of Motion:
Applying Newtons Second Law of Motion,
The natural circular frequency is
The period and frequency of vibration in cycles per second are:
)40.2(00 tkJ
)41.2(2/1
0
Jkt
n
)43.2(21 , )42.2(2
2/1
0
2/1
0
Jkf
kJ t
nt
n
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2.3 Free Vibration of an Undamped Torsional System
Note the following aspects of this system:
1) If the cross section of the shaft supporting the disc is not circular, an appropriate torsional spring constant is to be used.
2) The polar mass moment of inertia of a disc is given by
3) An important application of a torsional pendulum is in a mechanical clock
gWDDhJ832
44
0 where = mass density
h = thicknessD = diameterW = weight of the disc
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2.3 Free Vibration of an Undamped Torsional System
Example 2.6Natural Frequency of Compound Pendulum
Any rigid body pivoted at a point other than its center of mass will oscillate about the pivot point under its own gravitational force. Such a system is known as a compound pendulum as shown. Find the natural frequency of such a system.
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2.3 Free Vibration of an Undamped Torsional System
Example 2.6Natural Frequency of Compound PendulumSolution
For a displacement , the restoring torque (due to the weight of the body W) is (Wd sin ) and the equation of motion is
Hence, approximated by linear equation is
The natural frequency of the compound pendulum is
E.1)(0sin0 WdJ
E.2)(00 WdJ
(E.3)2/1
0
2/1
0
Jmgd
JWd
n
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2.3 Free Vibration of an Undamped Torsional System
Example 2.6Natural Frequency of Compound PendulumSolution
Comparing with natural frequency, the length of equivalent simple pendulum is
If J0 is replaced by mk02, where k0 is the radius of gyration of the body about O,
E.4)(0mdJl
(E.6) , (E.5) 22/1
20
0
dk
lkgd
n
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2.3 Free Vibration of an Undamped Torsional System
Example 2.6Natural Frequency of Compound PendulumSolution
If kG denotes the radius of gyration of the body about G, we have:
If the line OG is extended to point A such that
Eq.(E.8) becomes
(E.8) and E.7)( 2
2220
ddkldkk GG
(E.9)2
dkGA G
(E.10)OAdGAl
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2.3 Free Vibration of an Undamped Torsional System
Example 2.6Natural Frequency of Compound PendulumSolution
Hence, from Eq.(E.5), n is given by
This equation shows that, no matter whether the body is pivoted from O or A, its natural frequency is the same. The point A is called the center of percussion.
E.11)( /2/12/12/1
20
OAg
lg
dkg
n
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2.3 Free Vibration of an Undamped Torsional System
Example 2.6Natural Frequency of Compound PendulumSolution
Applications of centre of percussion
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2.4Response of First-Order Systems and Time Constant
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2011 Mechanical Vibrations Fifth Edition in SI Units47
2.4 Response of First-Order Systems and Time Constant
Consider a turbine rotor mounted in bearings as shown
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2.4 Response of First-Order Systems and Time Constant
The application of Newtons second law of motion yields the equation of motion of the rotor as
Assuming the trial solution as
Using the initial condition, , Eq. (2.48) can be written as
dtdww
wcwJ t
where
2.47 0
constantsunknown are s andA where
2.48 stAetw
00 wtw 2.49 0 stewtw
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2011 Mechanical Vibrations Fifth Edition in SI Units49
2.4 Response of First-Order Systems and Time Constant
By substituting Eq. (2.49) into Eq. (2.47), we obtain
Since leads to no motion of the rotor, we assume and Eq. (2.50) can be satisfied only if
Equation (2.51) is known as the characteristic equation which yields
. Thus the solution, Eq. (2.49), becomes
Because the exponent of Eq. (2.52) is known to be , the time constant will be equal to
2.50 00 tst cJsew00 w 00 w
2.51 0 tcJs
tJtcewtw 0Jcs t
Jct
2.53 tcJ
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2011 Mechanical Vibrations Fifth Edition in SI Units50
2.4 Response of First-Order Systems and Time Constant
For
Thus the response reduces to 0.368 times its initial value at a time equal to the time constant of the system.
t 2.54 368.0 0100 wewewtw Jtc
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2.5Rayleighs Energy Method
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2.5 Rayleighs Energy Method
The principle of conservation of energy, in the context of an undamped vibrating system, can be restated as
where subscripts 1 and 2 denote two different instants of time
If the system is undergoing harmonic motion, then
)55.2(2211 UTUT
)57.2(maxmax UT
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2011 Mechanical Vibrations Fifth Edition in SI Units53
2.5 Rayleighs Energy Method
Example 2.8Effect of Mass on wn of a Spring
Determine the effect of the mass of the spring on the natural frequency of the spring-mass system shown in the figure below.
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2011 Mechanical Vibrations Fifth Edition in SI Units54
2.5 Rayleighs Energy Method
Example 2.8Effect of Mass on wn of a SpringSolutionThe kinetic energy of the spring element of length dy is
The total kinetic energy of the system can be expressed as
(E.1)21 2
lxydy
lmdT ss
(E.2)32
121
21
21
)( spring ofenergy kinetic )( mass ofenergy kinetic
222
22
02 xmxm
lxydy
lmxm
TTT
sly
s
sm
where ms is the mass of the spring
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2011 Mechanical Vibrations Fifth Edition in SI Units55
2.5 Rayleighs Energy Method
Example 2.8Effect of Mass on wn of a SpringSolutionThe total potential energy of the system is given by
By assuming a harmonic motion
The maximum kinetic and potential energies can be expressed as
(E.3)21 2kxU
(E.4)cos)( tXtx n
(E.6)21 and (E.5)
321 2
max22
max kXUXmmT ns
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2011 Mechanical Vibrations Fifth Edition in SI Units56
2.5 Rayleighs Energy Method
Example 2.8Effect of Mass on wn of a SpringSolutionBy equating Tmax and Umax, we obtain the expression for the natural frequency:
Thus the effect of the mass of spring can be accounted for by adding one-third of its mass to the main mass.
(E.7)
3
2/1
sn mm
k
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2.6Free Vibration with Viscous Damping
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2.6 Free Vibration with Viscous Damping
Equation of Motion:
where c = damping
From the figure, Newtons law yields that the equation of motion is
)58.2(xcF
)59.2(0kxxcxmkxxcxm
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2011 Mechanical Vibrations Fifth Edition in SI Units59
2.6 Free Vibration with Viscous Damping
We assume a solution in the form
The characteristic equation is
The roots and solutions are
)60.2()( stCetx
)61.2(02 kcsms
)62.2(222
4 222,1 m
kmc
mc
mmkccs
)63.2()( and )( 21 2211tsts eCtxeCtx
where C and s are undetermined constants
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2011 Mechanical Vibrations Fifth Edition in SI Units60
2.6 Free Vibration with Viscous Damping
Thus the general solution is
where C1 and C2 are arbitrary constants to be determined from the initial conditions of the system
)64.2(
)(22
21
22
2
22
1
21
tmk
mc
mct
mk
mc
mc
tsts
eCeC
eCeCtx
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2011 Mechanical Vibrations Fifth Edition in SI Units61
2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
The critical damping cc is defined as the value of the damping constant c for which the radical in Eq.(2.62) becomes zero:
The damping ratio is defined as:
)65.2(22202
2
ncc mkm
mkmc
mk
mc
)66.2(/ ccc
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2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Thus the general solution for Eq.(2.64) is
Assuming that 0, consider the following 3 cases:
Case 1: Underdamped system
For this condition, (2-1) is negative and the roots are
)/2or or 1( mkmc/cc c
)69.2()(1
2
1
1
22 tt nn eCeCtx
nnisis
22
21
1
1
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2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 1: Underdamped system
The solution can be written in different forms:
)/2or or 1( mkmc/cc c
)70.2( 1cos 1sin1sin1cos
)(
02
0
2
22
21
12
11
1
2
1
122
22
teX
tXe
tCtCe
eCeCe
eCeCtx
nt
nt
nnt
titit
titi
n
n
n
nnn
nn
where (C1,C2), (X,),and (X0, 0) are arbitrary constants
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2011 Mechanical Vibrations Fifth Edition in SI Units64
2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 1: Underdamped system
For the initial conditions at t = 0,
and hence the solution becomes
)/2or or 1( mkmc/cc c
)71.2(1
and 2
00201
n
nxxCxC
)72.2(1sin1
1cos)( 22
0020
txxtxetx nn
nn
tn
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2011 Mechanical Vibrations Fifth Edition in SI Units65
2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 1: Underdamped system
Eq.(2.72) describes a damped harmonic motion. Its amplitude decreases exponentially with time, as shown in the figure below.
The frequency of damped vibration is:
)/2or or 1( mkmc/cc c
)76.2(1 2 nd
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2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 2: Critically damped system
In this case, the two roots are:
Due to repeated roots, the solution of Eq.(2.59) is given by
)77.2(221 n
c
mcss
)/2or or 1( mkmc/cc c
)78.2()()( 21tnetCCtx
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2011 Mechanical Vibrations Fifth Edition in SI Units67
2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 2: Critically damped system
Application of initial conditions gives:
Thus the solution becomes:
)/2or or 1( mkmc/cc c
)79.2( and 00201 xxCxC n
)80.2( )( 000 tn netxxxtx
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2011 Mechanical Vibrations Fifth Edition in SI Units68
2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 2: Critically damped system
It can be seen that the motion represented by Eq.(2.80) is a periodic (i.e., non-periodic).
Since , the motion will eventually diminish to zero, as indicated in the figure below.
)/2or or 1( mkmc/cc c
te tn as 0
Comparison of motions with different types of damping
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2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 3: Overdamped system
The roots are real and distinct and are given by:
In this case, the solution Eq.(2.69) is given by:
01 01222
1
n
n
s
s
)/2or or 1( mkmc/cc c
)81.2()(1
2
1
1
22 tt nn eCeCtx
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2011 Mechanical Vibrations Fifth Edition in SI Units70
2.6 Free Vibration with Viscous Damping
Critical Damping Constant and Damping Ratio:
Case 3: Overdamped system
For the initial conditions at t = 0,
)/2or or 1( mkmc/cc c
)82.2(121
121
20
20
2
20
20
1
n
n
n
n
xxC
xxC
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2011 Mechanical Vibrations Fifth Edition in SI Units71
2.6 Free Vibration with Viscous Damping
Logarithmic Decrement:
Using Eq.(2.70),
The logarithmic decrement can be obtained from Eq.(2.84):
)84.2(
)83.2()cos()cos(
1
1
2
1
020
010
2
1
dn
dn
n
n
n
eee
teXteX
xx
t
td
td
t
)85.2(2
212ln
22
1
mc
xx
dndn
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2011 Mechanical Vibrations Fifth Edition in SI Units72
2.6 Free Vibration with Viscous Damping
Logarithmic Decrement:
For small damping,
Hence,
or
Thus
)86.2( 1 if 2
)92.2( ln1
1
1
mxx
m
)87.2(2 22
)88.2( 2
where m is an integer
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2011 Mechanical Vibrations Fifth Edition in SI Units73
2.6 Free Vibration with Viscous Damping
Energy dissipated in Viscous Damping:
In a viscously damped system, the rate of change of energy with time is given by:
The energy dissipated in a complete cycle is:
)93.2( velocity force 2
2
dtdxccvFv
dtdW
)94.2()(cos 22022
2)/2(
0 XctdtcXdtdtdxcW ddddt d
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2011 Mechanical Vibrations Fifth Edition in SI Units74
2.6 Free Vibration with Viscous Damping
Energy dissipated in Viscous Damping:
Consider the system shown in the figure.
The total force resisting the motion is
If we assume simple harmonic motion is
Eq.(2.95) becomes
)95.2(xckxcvkxF
)96.2( sin)( tXtx d)97.2(cossin tXctkXF ddd
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2011 Mechanical Vibrations Fifth Edition in SI Units75
2.6 Free Vibration with Viscous Damping
Energy dissipated in Viscous Damping:
The energy dissipated in a complete cycle will be
)98.2( )(cos
)(cossin
2/2
0
22
/2
0
2
/2
0
XctdtXc
tdttkX
FvdtW
dt ddd
t dddd
t
d
d
d
-
2011 Mechanical Vibrations Fifth Edition in SI Units76
2.6 Free Vibration with Viscous Damping
Energy dissipated in Viscous Damping:
Computing the fraction of the total energy of the vibrating system that is dissipated in each cycle of motion,
The loss coefficient is defined as
)99.2(constant422
22
21 22
2
mc
Xm
XcWW
d
d
d
)100.2(2
)2/(tcoefficien lossWW
WW
where W is either the max potential energy or the max kinetic energy
-
2011 Mechanical Vibrations Fifth Edition in SI Units77
2.6 Free Vibration with Viscous Damping
Torsional systems with Viscous Damping:
Consider a single degree of freedom torsional system with a viscous damper as shown in figure.
-
2011 Mechanical Vibrations Fifth Edition in SI Units78
2.6 Free Vibration with Viscous Damping
Torsional systems with Viscous Damping:
The viscous damping torque is given by
The equation of motion can be derived as:
)101.2( tcT
)102.2(00 tt kcJ where J0 = mass moment of inertia of disc
kt = spring constant of system = angular displacement of disc
-
2011 Mechanical Vibrations Fifth Edition in SI Units79
2.6 Free Vibration with Viscous Damping
Torsional systems with Viscous Damping:
In the underdamped case, the frequency of damped vibration is given by
where
and
)103.2(1 2 nd
)104.2(0Jkt
n
)105.2(22 00 Jk
cJc
cc
t
t
n
t
tc
t ctc = critical torsional damping constant
-
2011 Mechanical Vibrations Fifth Edition in SI Units80
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a Motorcycle
An underdamped shock absorber is to be designed for a motorcycle of mass 200kg (shown in Fig.(a)). When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated in Fig.(b). Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 2 s and the amplitude x1 is to be reduced to one-fourth in one half cycle (i.e., x1.5 = x1/4). Also find the minimum initial velocity that leads to a maximum displacement of 250 mm.
-
2011 Mechanical Vibrations Fifth Edition in SI Units81
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a Motorcycle
-
2011 Mechanical Vibrations Fifth Edition in SI Units82
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a MotorcycleSolution
Since , the logarithmic decrement becomes
16/4/ ,4/ 15.1215.1 xxxxx
(E.1)127726.216lnln
22
1
xx
-
2011 Mechanical Vibrations Fifth Edition in SI Units83
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a MotorcycleSolution
From which can be found as 0.4037 and the damped period of vibration is given by 2 s. Hence,
rad/s 4338.3)4037.0(12
21222
2
2
n
ndd
-
2011 Mechanical Vibrations Fifth Edition in SI Units84
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a MotorcycleSolution
The critical damping constant can be obtained as
Thus the damping constant is
The stiffness is
s/m-N 54.373.1)4338.3)(200(22 nc mc
s/m-N 4981.554)54.1373)(4037.0( ccc
N/m 2652.2358)4338.3)(200( 22 nmk
-
2011 Mechanical Vibrations Fifth Edition in SI Units85
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a MotorcycleSolution
The displacement of the mass will attain its max value at time t1 is
sec 3678.0)9149.0(sin
9149.0)4037.0(1sinsin
1sin
1
1
211
21
t
tt
t
d
d
-
2011 Mechanical Vibrations Fifth Edition in SI Units86
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a MotorcycleSolution
The envelope passing through the max points is
Since x = 250mm,
The velocity of mass can be obtained by
(E.2)1 2 tnXex
m 4550.0)4037.0(125.0 )3678.0)(4338.3)(4037.0(2 XXe
(E.3))cossin()(
sin)(
ttXetx
tXetx
dddnt
dt
n
n
-
2011 Mechanical Vibrations Fifth Edition in SI Units87
2.6 Free Vibration with Viscous Damping
Example 2.11Shock Absorber for a MotorcycleSolution
When t = 0,
m/s4294.1 )4037.0(1)4338.3)(4550.0(
1)0(2
20
nd XXxtx
-
2.7Graphical Representation of Characteristic Roots and Corresponding Solutions
2011 Mechanical Vibrations Fifth Edition in SI Units88
-
2011 Mechanical Vibrations Fifth Edition in SI Units89
2.7 Graphical Representation of Characteristic Roots and Corresponding Solutions
Roots of the Characteristic Equation
The free vibration of a single-degree-of-freedom spring-mass-viscous-damper system is governed by Eq. (2.59):
whose characteristic equation can be expressed as (Eq. (2.61)):
2.106 0 kxxcxm
2.108 02022
2
nn wswskcsms
-
2011 Mechanical Vibrations Fifth Edition in SI Units90
2.7 Graphical Representation of Characteristic Roots and Corresponding Solutions
Roots of the Characteristic Equation
The roots of Eq. (2.107) or (2.108) are given by (see Eqs. (2.62) and (2.68)):
2.110 1,2
4,
221
2
21
nn iwwssm
mkccss
-
2011 Mechanical Vibrations Fifth Edition in SI Units91
2.7 Graphical Representation of Characteristic Roots and Corresponding Solutions
Graphical Representation of Roots and Corresponding Solutions
The response of the system is given by
Following observations can be made by examining Eqs. (2.110) and (2.111):1. The roots lying farther to the left in the s-plane indicate that the
corresponding responses decay faster than those associated with roots closer to the imaginary axis.
2. If the roots have positive real values of sthat is, the roots lie in the right half of the s-planethe corresponding response grows exponentially and hence will be unstable.
2.111 21 21 tsts eCeCtx
-
2011 Mechanical Vibrations Fifth Edition in SI Units92
2.7 Graphical Representation of Characteristic Roots and Corresponding Solutions
Graphical Representation of Roots and Corresponding Solutions
3. If the roots lie on the imaginary axis (with zero real value), the corresponding response will be naturally stable.
4. If the roots have a zero imaginary part, the corresponding response will not oscillate.
5. The response of the system will exhibit an oscillatory behavior only when the roots have nonzero imaginary parts.
6. The farther the roots lie to the left of the s-plane, the faster the corresponding response decreases.
7. The larger the imaginary part of the roots, the higher the frequency of oscillation of the corresponding response of the system.
-
2011 Mechanical Vibrations Fifth Edition in SI Units93
2.7 Graphical Representation of Characteristic Roots and Corresponding Solutions
Graphical Representation of Roots and Corresponding Solutions
-
2.8Parameter Variations and Root Locus Representations
2011 Mechanical Vibrations Fifth Edition in SI Units94
-
2011 Mechanical Vibrations Fifth Edition in SI Units95
2.8 Parameter Variations and Root Locus Representations
Interpretations of in the s-plane
The angle made by the line OA with the imaginary axis is given by
The radial lines pass through the origin correspond to different damping ratios
The time constant of the system is defined as
and ,, dn ww
2.113 sinsin
1
n
n
ww
nw1
-
2011 Mechanical Vibrations Fifth Edition in SI Units96
2.8 Parameter Variations and Root Locus Representations
Interpretations of in the s-plane and ,, dn ww
-
2011 Mechanical Vibrations Fifth Edition in SI Units97
2.8 Parameter Variations and Root Locus Representations
Interpretations of in the s-plane and ,, dn ww
-
2011 Mechanical Vibrations Fifth Edition in SI Units98
2.8 Parameter Variations and Root Locus Representations
Interpretations of in the s-plane
Different lines parallel to the imaginary axis denote reciprocals of different time constants
and ,, dn ww
-
2011 Mechanical Vibrations Fifth Edition in SI Units99
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
A plot or graph that shows how changes in one of the parameters of the system will modify the roots of the characteristic equation of the system is known as the root locus plot.
Variation of the damping ratio:We vary the damping constant from zero to infinity and study the migration of the characteristic roots in the s-plane.
From Eq. (2.109) when c = 0,
2.115 24
2,1 niwmk
mmks
-
2011 Mechanical Vibrations Fifth Edition in SI Units100
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the damping ratio:Noting that the real and imaginary parts of the roots in Eq. (2.109) can be expressed as
For , we have
2.116 12
4 and 2
22
dnn wwmcmkw
mc
10 2.117 222 nd ww
-
2011 Mechanical Vibrations Fifth Edition in SI Units101
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the damping ratio:
The radius vector will make an angle with the positive imaginary axis with
The two roots trace loci or paths in the form of circular arcs as the damping ratio is increased from zero to unity as shown
21with
cos , sin
n
n
nn
d
ww
www
-
2011 Mechanical Vibrations Fifth Edition in SI Units102
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the damping ratio:
-
2011 Mechanical Vibrations Fifth Edition in SI Units103
2.8 Parameter Variations and Root Locus Representations
Example 2.13Study of Roots with Variation of c
Plot the root locus diagram of the system governed by the equation by varying the value of c >0
0273 2 cs
-
2011 Mechanical Vibrations Fifth Edition in SI Units104
2.8 Parameter Variations and Root Locus Representations
Example 2.13Study of Roots with Variation of cSolution
The roots of equation are given by
We start with a value of C = 0 and the roots is as shown in the figure.
Eq. (E.2) gives the roots as indicated in the Table.
E.2 6
32422,1
ccs
-
2011 Mechanical Vibrations Fifth Edition in SI Units105
2.8 Parameter Variations and Root Locus Representations
Example 2.13Study of Roots with Variation of cSolution
-
2011 Mechanical Vibrations Fifth Edition in SI Units106
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the spring constant:
Since the spring constant does not appear explicitly in Eq. (2.108), we consider a specific form of the characteristic equation (2.107) as:
The roots of Eq. (2.121) are given by
2.121 0162 kss
2.122 6482
4256162,1 k
ks
-
2011 Mechanical Vibrations Fifth Edition in SI Units107
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the mass:
To find the migration of the roots with a variation of the mass m,we consider a specific form of the characteristic equation, Eq. (2.107), as
whose roots are given by
2.123 020142 sms
2.124 2
80196142,1
ms
-
2011 Mechanical Vibrations Fifth Edition in SI Units108
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the mass:
Some values of m and the corresponding roots given by Eq. (2.124) are shown in Table.
-
2011 Mechanical Vibrations Fifth Edition in SI Units109
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the mass:
-
2011 Mechanical Vibrations Fifth Edition in SI Units110
2.8 Parameter Variations and Root Locus Representations
Root Locus and Parameter Variations
Variation of the mass:
-
2.9Free Vibration with Coulomb Damping
2011 Mechanical Vibrations Fifth Edition in SI Units111
-
2011 Mechanical Vibrations Fifth Edition in SI Units112
2.9 Free Vibration with Coulomb Damping
Coulombs law of dry friction states that, when two bodies are in contact, the force required to produce sliding is proportional to the normal force acting in the plane of contact. Thus, the friction force F is given by:
Coulomb damping is sometimes called constant damping
)125.2( mgWNF where N is normal force,
is the coefficient of sliding or kinetic friction is 0.1 for lubricated metal, 0.3 for non-lubricated metal on metal, 1.0 for rubber on metal
-
2011 Mechanical Vibrations Fifth Edition in SI Units113
2.9 Free Vibration with Coulomb Damping
Equation of Motion:
Consider a single degree of freedom system with dry friction as shown in Fig.(a) below.
Since friction force varies with the direction of velocity, we need to consider two cases as indicated in Fig.(b) and (c).
-
2011 Mechanical Vibrations Fifth Edition in SI Units114
2.9 Free Vibration with Coulomb Damping
Equation of Motion:
Case 1. When x is positive and dx/dt is positive or when x is negative and dx/dt is positive (i.e., for the half cycle during which the mass moves from left to right) the equation of motion can be obtained using Newtons second law (Fig.b):
Hence
)126.2( or NkxxmNkxxm
)127.2( sincos)( 21 kNtAtAtx nn
where n = k/m is the frequency of vibration
A1 & A2 are constants
-
2011 Mechanical Vibrations Fifth Edition in SI Units115
2.9 Free Vibration with Coulomb Damping
Equation of Motion:
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 moves from right to left) the equation of motion can be derived from Fig. (c):
The solution of the equation is given by:
)128.2( or NkxxmxmNkx
)129.2(sincos)( 43 kNtAtAtx nn
where A3 & A4 are constants
-
2011 Mechanical Vibrations Fifth Edition in SI Units116
2.9 Free Vibration with Coulomb Damping
Equation of Motion:
Motion of the mass with Coulomb damping
-
2011 Mechanical Vibrations Fifth Edition in SI Units117
2.9 Free Vibration with Coulomb Damping
Solution:
Eqs.(2.107) & (2.109) can be expressed as a single equation using N = mg:
where sgn(y) is called the sigum function, whose value is defined as 1 for y > 0, -1 for y< 0, and 0 for y = 0.
Assuming initial conditions as
)130.2(0)sgn( kxxmgxm
)131.2(0)0()0( 0
txxtx
-
2011 Mechanical Vibrations Fifth Edition in SI Units118
2.9 Free Vibration with Coulomb Damping
Solution:
The solution is valid for half the cycle only, i.e., for 0 t /n. Hence, the solution becomes the initial conditions for the next half cycle. The procedure continued until the motion stops, i.e., when xn N/k. Thus the number of half cycles (r) that elapse before the motion ceases is:
)134.2(2
2
0
0
kNkNx
r
kN
kNrx
-
2011 Mechanical Vibrations Fifth Edition in SI Units119
2.9 Free Vibration with Coulomb Damping
Solution:
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 damping
2. The natural frequency of the system is unaltered with the addition of Coulomb damping, while it is reduced with the addition of viscous damping.
-
2011 Mechanical Vibrations Fifth Edition in SI Units120
2.9 Free Vibration with Coulomb Damping
Solution:
Note the following characteristics of a system with Coulomb damping:
3. The motion is periodic with Coulomb damping, while it can be nonperiodic in a viscously damped (overdamped) system.
4. The system comes to rest after some time with Coulomb damping, whereas the motion theoretically continues forever (perhaps with an infinitesimally small amplitude) with viscous damping.
-
2011 Mechanical Vibrations Fifth Edition in SI Units121
2.9 Free Vibration with Coulomb Damping
Solution:
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 4N/k, so the amplitudes at the end of any two consecutive cycles are related:
)135.2( 41 kNXX mm
-
2011 Mechanical Vibrations Fifth Edition in SI Units122
2.9 Free Vibration with Coulomb Damping
Torsional Systems with Coulomb Damping:
The equation governing the angular oscillations of the system is
The frequency of vibration is given by
)137.2(
)136.2(
0
0
TkJ
TkJ
t
t
)138.2(0Jkt
n
-
2011 Mechanical Vibrations Fifth Edition in SI Units123
2.9 Free Vibration with Coulomb Damping
Torsional Systems with Coulomb Damping:
The amplitude of motion at the end of the rth half cycle (r) is given by:
The motion ceases when
)140.2( 2
0
t
t
kTkT
r
)139.2(20t
r kTr
-
2011 Mechanical Vibrations Fifth Edition in SI Units124
2.9 Free Vibration with Viscous Damping
Example 2.15Pulley Subjected to Coulomb Damping
A steel shaft of length 1 m and diameter 50 mm is fixed at one end and carries a pulley of mass moment of inertia 25 kg-m2 at the other end. A band brake exerts a constant frictional torque of 400 N-m around the circumference of the pulley. If the pulley is displaced by 6and released, determine (1) the number of cycles before the pulley comes to rest and (2) the final settling position of the pulley.
-
2011 Mechanical Vibrations Fifth Edition in SI Units125
2.9 Free Vibration with Viscous Damping
Example 2.15Pulley Subjected to Coulomb DampingSolution
(1) The number of half cycles that elapse before the angular motion of the pullet ceases is:
The torsional spring constant of the shaft given by
)1.E(2
0
t
t
kTkT
r
m/rad-N 5.087,491
)05.0(32
)108( 410
lGJkt
where 0 = 6 = 0.10472 rad,
-
2011 Mechanical Vibrations Fifth Edition in SI Units126
2.9 Free Vibration with Viscous Damping
Example 2.15Pulley Subjected to Coulomb DampingSolution
With constant friction torque applied to the pulley = 400 N-m., Eq.(E.1) gives
Thus the motion ceases after six half cycles.
926.5
5.087,49800
5.087,4940010472.0
r
-
2011 Mechanical Vibrations Fifth Edition in SI Units127
2.9 Free Vibration with Viscous Damping
Example 2.15Pulley Subjected to Coulomb DampingSolution
(2) The angular displacement after six half cycles:
from the equilibrium position on the same side of the initial displacement.
39734.0rad 006935.05.087,49
4002610472.0
-
2.10Free Vibration with Hysteretic Damping
2011 Mechanical Vibrations Fifth Edition in SI Units128
-
2011 Mechanical Vibrations Fifth Edition in SI Units129
2.10 Free Vibration with Hysteretic Damping
Consider the spring-viscous damper arrangement shown in the figure below. The force needed to cause a displacement:
For a harmonic motion of frequency and amplitude X,
)141.2(xckxF
)143.2(
)sin(
cossin)(
22
22
xXckx
tXXckx
tcXtkXtF
-
2011 Mechanical Vibrations Fifth Edition in SI Units130
2.10 Free Vibration with Hysteretic Damping
Spring-viscous-damper system
-
2011 Mechanical Vibrations Fifth Edition in SI Units131
2.10 Free Vibration with Hysteretic Damping
When F versus x is plotted, Eq.(2.143) represents a closed loop, as shown in Fig(b). The area of the loop denotes the energy dissipated by the damper in a cycle of motion and is given by:
Hence, the damping coefficient:
Eqs.(2.144) and (2.145) gives
)144.2(coscossin 2/20 cXdttXtcXtkXFdxW
)145.2(hc where h = hysteresis damping constant
)146.2(2hXW
-
2011 Mechanical Vibrations Fifth Edition in SI Units132
2.10 Free Vibration with Hysteretic Damping
Hysteresis loop
-
2011 Mechanical Vibrations Fifth Edition in SI Units133
2.10 Free Vibration with Hysteretic Damping
Complex Stiffness
For general harmonic motion, , the force is given by
Thus, the force-displacement relation:
tiXex )147.2()( xcikiXeckXeF titi
)149.2()1(1 where
)148.2()(
ikkhikihk
xihkF
-
2011 Mechanical Vibrations Fifth Edition in SI Units134
2.10 Free Vibration with Hysteretic Damping
Response of the system
The energy loss per cycle can be expressed as
The hysteresis logarithmic decrement can be defined as
Corresponding frequency
)150.2(2XkW
)154.2()1ln(ln1
j
j
XX
)155.2(mk
-
2011 Mechanical Vibrations Fifth Edition in SI Units135
2.10 Free Vibration with Hysteretic Damping
Response of the system
Response of a hysteretically damped system
-
2011 Mechanical Vibrations Fifth Edition in SI Units136
2.10 Free Vibration with Hysteretic Damping
Response of the system
The equivalent viscous damping ratio
Thus the equivalent damping constant is
)156.2( 22
2kh
kh
eqeq
)157.2( 2
2 hkmkmkcc eqceq
-
2011 Mechanical Vibrations Fifth Edition in SI Units137
2.10 Free Vibration with Viscous Damping
Example 2.17Response of a Hysteretically Damped Bridge Structure
A bridge structure is modeled as a single degree of freedom system with an equivalent mass of 5 X 105 kg and an equivalent stiffness of 25 X106 N/m. During a free vibration test, the ratio of successive amplitudes was found to be 1.04. Estimate the structural damping constant () and the approximate free vibration response of the bridge.
-
2011 Mechanical Vibrations Fifth Edition in SI Units138
2.10 Free Vibration with Viscous Damping
Example 2.17Response of a Hysteretically Damped Bridge StructureSolution
Using the ratio of successive amplitudes,
The equivalent viscous damping coefficient is
0127.004.004.11
)1ln()04.1ln(ln1
or
XX
j
j
(E.1)km
mkkkceq
-
2011 Mechanical Vibrations Fifth Edition in SI Units139
2.10 Free Vibration with Viscous Damping
Example 2.17Response of a Hysteretically Damped Bridge StructureSolution
Using the known values of the equivalent stiffness and equivalent mass,
Since ceq < cc, the bridge is underdamped. Hence, its free vibration response is
s/m-N 109013.44)105)(1025()0127.0( 356 eqc
0063.0100678.7071
109013.40
1sin1
1cos)(
3
3
2
2002
0
c
eq
n
n
nn
t
cc
txxtxetx n
-
2.11Stability of Systems
2011 Mechanical Vibrations Fifth Edition in SI Units140
-
2011 Mechanical Vibrations Fifth Edition in SI Units141
2.11 Stability of Systems
Stability is one of the most important characteristics for any vibrating system
A asymptotically stable (called stable in controls literature) is when its free-vibration response approaches zero as time approaches infinity.
A system is considered to be unstable if its free-vibration response grows without bound (approaches infinity) as time approaches infinity.
A system is stable (called marginally stable in controls literature) if its free-vibration response neither decays nor grows, but remains constant or oscillates as time approaches infinity.
-
2011 Mechanical Vibrations Fifth Edition in SI Units142
2.11 Stability of Systems
-
2011 Mechanical Vibrations Fifth Edition in SI Units143
2.11 Stability of Systems
Example 2.18Stability of a System
Consider a uniform rigid bar, of mass m and length l, pivoted at one end and connected symmetrically by two springs at the other end, as shown in the figure. Assuming that the springs are unstretchedwhen the bar is vertical, derive the equation of motion of the system for small angular displacements of the bar about the pivot point, and investigate the stability behavior of the system.
-
2011 Mechanical Vibrations Fifth Edition in SI Units144
2.11 Stability of Systems
Example 2.18Stability of a System
-
2011 Mechanical Vibrations Fifth Edition in SI Units145
2.11 Stability of Systems
Example 2.18Stability of a System
The equation of motion of the bar, for rotation about the point O, is
For small oscillations, Eq. (E.1) reduces to
E.1 0sin2
cossin23
2
lWlklml
E.3 0
E.2 02
23
2
22
Wlklml
-
2011 Mechanical Vibrations Fifth Edition in SI Units146
2.11 Stability of Systems
Example 2.18Stability of a System
Where
The characteristic equation is given byThe solution of Eq. (E.2) depends on the sign of as indicated below.
Case 1. When
E.4 2
3122
22
mlWlkl
E.5 022 s2
02/312 22 mlWlkl
E.7 2
312 where
E.6 sincos2/12
21
mlWlklw
twAtwAt
n
nn
-
2011 Mechanical Vibrations Fifth Edition in SI Units147
2.11 Stability of Systems
Example 2.18Stability of a System
Case 2. When
For the initial conditions
Equation (E.9) shows that the system is unstable with the angular displacement increasing linearly at a constant velocity
E.8 21 CtCt 00 0 and 0 tt
02/312 22 mlWlkl
E.9 0 tt
-
2011 Mechanical Vibrations Fifth Edition in SI Units148
2.11 Stability of Systems
Example 2.18Stability of a System
Case 3. When
For the initial conditions
Equation (E.11) shows that increases exponentially with time; hence the motion is unstable.
E.10 e21 tt BeBt 00 0 and 0 tt
02/312 22 mlWlkl
E.11 21
0000tt eet