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Vibration

Transcript of Vibration Chapter02

  • Mechanical Vibrations

    Fifth Edition in SI UnitsSingiresu S. Rao

  • 2011 Mechanical Vibrations Fifth Edition in SI Units3

    Chapter 2Free Vibration of Single-Degree-of-Freedom Systems

  • 2011 Mechanical Vibrations Fifth Edition in SI Units4

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units5

    Chapter Outline

    2.9 Free Vibration with Coulomb Damping

    2.10 Free Vibration with Hysteretic Damping

    2.11 Stability of Systems

  • 2011 Mechanical Vibrations Fifth Edition in SI Units6

    2.1Introduction

  • 2011 Mechanical Vibrations Fifth Edition in SI Units7

    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)

  • 2011 Mechanical Vibrations Fifth Edition in SI Units8

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units9

    2.2Free Vibration of an Undamped Translational System

  • 2011 Mechanical Vibrations Fifth Edition in SI Units10

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units11

    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

  • 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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units14

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units15

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units16

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units17

    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.

  • 2011 Mechanical Vibrations Fifth Edition in SI Units18

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units19

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units20

    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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units22

    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.

  • 2011 Mechanical Vibrations Fifth Edition in SI Units23

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units24

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units25

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units26

    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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units28

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units29

    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.

  • 2011 Mechanical Vibrations Fifth Edition in SI Units30

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units31

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units32

    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.

  • 2011 Mechanical Vibrations Fifth Edition in SI Units33

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units34

    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

  • 2.3Free Vibration of an Undamped Torsional System

    2011 Mechanical Vibrations Fifth Edition in SI Units35

  • 2011 Mechanical Vibrations Fifth Edition in SI Units36

    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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units38

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units39

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units40

    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.

  • 2011 Mechanical Vibrations Fifth Edition in SI Units41

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units42

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units43

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units44

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units45

    2.3 Free Vibration of an Undamped Torsional System

    Example 2.6Natural Frequency of Compound PendulumSolution

    Applications of centre of percussion

  • 2.4Response of First-Order Systems and Time Constant

    2011 Mechanical Vibrations Fifth Edition in SI Units46

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units48

    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

  • 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

  • 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

  • 2.5Rayleighs Energy Method

    2011 Mechanical Vibrations Fifth Edition in SI Units51

  • 2011 Mechanical Vibrations Fifth Edition in SI Units52

    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

  • 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.

  • 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

  • 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

  • 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

  • 2.6Free Vibration with Viscous Damping

    2011 Mechanical Vibrations Fifth Edition in SI Units57

  • 2011 Mechanical Vibrations Fifth Edition in SI Units58

    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

  • 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

  • 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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units62

    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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units63

    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

  • 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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units66

    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

  • 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

  • 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

  • 2011 Mechanical Vibrations Fifth Edition in SI Units69

    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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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