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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

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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)

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

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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

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