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


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


Chapter Outline

2.9 Free Vibration with Coulomb Damping

2.10 Free Vibration with Hysteretic Damping

2.11 Stability of Systems


2.1Introduction


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)


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


2.2Free 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



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



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




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




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




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




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



Equation of Motion of a Spring-Mass System in Vertical Position:

Consider the configuration of the spring-mass system shown in the figure.




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




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




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



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



Harmonic Motion

The nature of harmonic oscillation can be represented graphically as shown in the figure.



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



Harmonic Motion


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



Harmonic Motion


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



Harmonic Motion


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



Harmonic Motion


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



Harmonic Motion


4. Phase plane representation of an undamped 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.



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



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



Example 2.5Natural Frequency of Pulley System

Determine the natural frequency of the system.Assume the pulleys to be frictionless and of negligible mass.



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



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




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



Polar Moment of Inertia:

Torsional Spring Constant:

)38.2(32

4

0dI

)39.2(32

40

lGd

lGIMk tt



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



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



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.



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




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




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




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




Applications of centre of percussion

2.4Response of First-Order Systems and Time Constant




Consider a turbine rotor mounted in bearings as shown



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



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



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




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



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.



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



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



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




Equation of Motion:

where c = damping

From the figure, Newtons law yields that the equation of motion is

)58.2(xcF

)59.2(0kxxcxmkxxcxm



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



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



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




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





The solution can be written in different forms:


)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





For the initial conditions at t = 0,

and hence the solution becomes


)71.2(1

and 2

00201

n

nxxCxC

)72.2(1sin1

1cos)( 22

0020

txxtxetx nn

nn

tn





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:


)76.2(1 2 nd




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


)78.2()()( 21tnetCCtx





Application of initial conditions gives:

Thus the solution becomes:


)79.2( and 00201 xxCxC n

)80.2( )( 000 tn netxxxtx





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.


te tn as 0

Comparison of motions with different types of damping




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


)81.2()(1

2

1

1

22 tt nn eCeCtx




Case 3: Overdamped system

For the initial conditions at t = 0,


)82.2(121

121

20

20

2

20

20

1

n

n

n

n

xxC

xxC



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



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



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




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




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




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



Torsional systems with Viscous Damping:

Consider a single degree of freedom torsional system with a viscous damper as shown in figure.




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




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



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.



Example 2.11Shock Absorber for a Motorcycle



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




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




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




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




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




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




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



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



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




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.

2.8Parameter 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



Interpretations of in the s-plane and ,, dn ww



Interpretations of in the s-plane

Different lines parallel to the imaginary axis denote reciprocals of different time constants

and ,, dn ww



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




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




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




Variation of the damping ratio:



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



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



Example 2.13Study of Roots with Variation of cSolution




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




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





Some values of m and the corresponding roots given by Eq. (2.124) are shown in Table.

2.9Free 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



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).



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



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



Equation of Motion:

Motion of the mass 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



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



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.



Solution:


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.



Solution:


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



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



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



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.



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,




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




(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




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



Spring-viscous-damper system



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



Hysteresis loop



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



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




Response of a hysteretically damped system




The equivalent viscous damping ratio

Thus the equivalent damping constant is

)156.2( 22

2kh

kh

eqeq

)157.2( 2

2 hkmkmkcc eqceq



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.



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



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




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.



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.




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




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




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




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

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