Vibration Fundamental -Basic

7/27/2019 Vibration Fundamental -Basic

1/109

Mechanical

VibrationsSingiresu S. Rao

SI Edition

Chapter 1

Fundamentals of

Vibration


2/109


3/109

2005 Pearson Education South Asia Pte Ltd.3

8. Continuous Systems

9. Vibration Control

10. Vibration Measurement and Applications

11. Numerical Integration Methods in VibrationAnalysis

12. Finite Element Method

13. Nonlinear Vibration

14. Random Vibration

Course Outline


4/109

2005 Pearson Education South Asia Pte Ltd. 4

1.1 Preliminary Remarks

1.2 Brief History of Vibration

1.3 Importance of the Study of Vibration

1.4 Basic Concepts of Vibration

1.5 Classification of Vibration

1.6 Vibration Analysis Procedure

1.7 Spring Elements

Chapter Outline


5/109


1.8 Mass or Inertia Elements

1.9 Damping Elements

1.10 Harmonic Motion

1.11 Harmonic Analysis

Chapter Outline


6/109


1.1 Preliminary Remarks

Brief History of vibration

Examination of vibrations important role

Vibration analysis of an engineering system

Definitions and concepts of vibration

Concept of harmonic analysis for generalperiodic motions


7/109



Origins of vibration:

582-507 B.C.

Pythagoras, the Greek philosopher

and mathematician, is the first to

investigate musical sounds on ascientific basis. He conducted

experiments on a vibrating string by

using a simple apparatus called a

monochord. He further developed

the concept of pitch.


8/109



Around 350 B.C.

Aristotle wrote treatises on music and sound

In 320 B.C.

Aristoxenus wrote a three-volume work entitled

Elements of Harmony

In 300 B.C.

Euclid wrote a treatises Introduction to Harmonics

A.D. 132Zhang Heng invented the worldsfirst seismograph to measure

earthquakes


9/109



Galileo to Rayleigh:Galileo Galilei (1564 1642)

- founder of modern experimental science

- started experimenting on simple pendulum

- published a book, Discourses ConcerningTwo New Sciences, in 1638, describingresonance, frequency, length, tension anddensity of a vibrating stretched string

Robert Hooke (1635 1703)- found relation between pitch and frequency ofvibration of a string


10/109



Joseph Sauveur (1653 1716)

- coined the word acoustics for the science ofsound

- founded nodes, loops, harmonics and the

fundamental frequency

- calculated the frequency of a stretched string

from the measured sag of its middle point

Sir Isaac Newton (1642-1727)- published his monumental work, Philosophiae

Naturalis Principia Mathematica, in 1686,

discovering three laws of motion


11/109



Joseph Lagrange (1736 1813)- found the analytical solution of the vibratingstring and the wave equation

Simeon Poisson (1781 1840)

- solved the problem of vibration of arectangular flexible membrane

R.F.A. Clebsch (1833 1872)- studied the vibration of a circular membrane

Lord Baron Rayleigh

- founded Rayleigh-Ritz method, used to findfrequency of vibration of a conservativesystem and multiple natural frequencies


12/109



Recent contributions:1902 Frahm investigated the importance of

torsional vibration study in the design ofpropeller shafts of steamships

Aurel Stodola (1859 1943)- contributed to the study of vibration of beams,plates, and membranes.

- developed a method for analyzing vibrating

beams which is applicable to turbine bladesC.G.P. De Laval (1845 1913)

- presented a practical solution to the problemof vibration of an unbalanced rotating disk


13/109



1892 Lyapunov laid the foundations of modernstability theory which is applicable to all

types of dynamical systems

1920 Duffling and Van der Pol brought the first

definite solutions into the theory ofnonlinear vibrations and drew attention to

its importance in engineering

Introduction of the correlation function byTaylor

1950 advent of high-speed digital computers

generate approximate solutions


14/109



1950s developed finite element method enabledengineers to conduct numerically detailed

vibration analysis of complex mechanical,

vehicular, and structural systems

displaying thousands of degrees offreedom with the aid of computers

Turner, Clough, Martin and Topp presented thefinite element method as known today


15/109


1.3 Importance of the Study of Vibration

Why study vibration?Vibrations can lead to excessive deflections

and failure on the machines and structures

To reduce vibration through proper design of

machines and their mountingsTo utilize profitably in several consumer and

industrial applications

To improve the efficiency of certain machining,

casting, forging & welding processesTo stimulate earthquakes for geological

research and conduct studies in design ofnuclear reactors


16/109



Vibration = any motion that repeats itself after

an interval of time

Vibratory System consists of:

1) spring or elasticity

2) mass or inertia

3) damper

Involves transferof potential energy to kinetic

energy and vice versa


17/109



Degree of Freedom (d.o.f.) =

min. no. of independent coordinates required

to determine completely the positions of all

parts of a system at any instant of time

Examples ofsingle degree-of-freedomsystems:


18/109



Examples ofsingle degree-of-freedom

systems:


19/109



Examples ofTwo degree-of-freedom systems:


20/109



Examples ofThree degree of freedom systems:


21/109



Example ofInfinite-number-of-degrees-of-

freedom system:

Infinite number of degrees of freedom system

are termed continuous ordistributedsystems

Finite number of degrees of freedom are

termed discrete orlumpedparameter systems

More accurate results obtained by increasing

number of degrees of freedom


22/109



Free Vibration:

A system is left to vibrate on its own after an

initial disturbance and no external force acts on

the system. E.g. simple pendulum

Forced Vibration:A system that is subjected to a repeating

external force. E.g. oscillation arises from diesel

engines

Resonance occurs when the frequency of the

external force coincides with one of the

natural frequencies of the system


23/109



Undamped Vibration:

When no energy is lost or dissipated in friction

or other resistance during oscillations

Damped Vibration:

When any energy is lost or dissipated infriction or other resistance during oscillations

LinearVibration:

When all basic components of a vibratorysystem, i.e. the spring, the mass and the

damper behave linearly


24/109



NonlinearVibration:

Ifany of the components behave nonlinearly

Deterministic Vibration:

If the value or magnitude of the excitation (force

or motion) acting on a vibratory system isknown at any given time

Nondeterministic orrandom Vibration:

When the value of the excitation at a giventime cannot be predicted


25/109



Examples of deterministic and random

excitation:


26/109


27/109


1.6 Vibration Analysis Procedure

Example of the modeling of a forging hammer:

E l 1 1


28/109


Example 1.1

Mathematical Model of a Motorcycle

Figure 1.18(a) shows a motorcycle with a rider.

Develop a sequence of three mathematical

models of the system for investigating vibration in

the vertical direction. Consider the elasticity of the

tires, elasticity and damping of the struts (in thevertical direction), masses of the wheels, and

elasticity, damping, and mass of the rider.


29/109


Example 1.1 Solution

We start with the simplest model and refine it

gradually. When the equivalent values of the

mass, stiffness, and damping of the system are

used, we obtain a single-degree of freedom model

of the motorcycle with a rider as indicated in Fig.1.18(b). In this model, the equivalent stiffness (keq)

includes the stiffness of the tires, struts, and rider.

The equivalent damping constant (ceq) includes

the damping of the struts and the rider. Theequivalent mass includes the mass of the wheels,

vehicle body and the rider.


30/109




31/109



This model can be refined by representing the

masses of wheels, elasticity of tires, and elasticity

and damping of the struts separately, as shown in

Fig. 1.18(c). In this model, the mass of the vehicle

body (mv) and the mass of the rider (mr) areshown as a single mass, mv + mr. When the

elasticity (as spring constant kr) and damping (as

damping constant cr) of the rider are considered,

the refined model shown in Fig. 1.18(d) can beobtained.


32/109




33/109



Note that the models shown in Figs. 1.18(b) to (d)

are not unique. For example, by combining the

spring constants of both tires, the masses of both

wheels, and the spring and damping constants of

both struts as single quantities, the model shownin Fig. 1.18(e) can be obtained instead of Fig.

1.18(c).


34/109


1.1kxF

F= spring force,

k= spring stiffness or spring constant, and

x = deformation (displacement of one end

with respect to the other)

1.7 Spring Elements

Linearspring is a type of mechanical link that is

generally assumed to have negligible mass and

damping

Spring force is given by:


35/109


Work done(U) in deforming a spring or the

strain (potential) energy is given by:

When an incremental force Fis added to F:

2.12

1 2kxU

3.1...)(!2

1

)()(

)(

22

2

*

*

*

*

x

dx

Fd

xdx

dFxF

xxFFF

x

x

1.7 Spring Elements


36/109


1.7 Spring Elements


37/109


1.7 Spring Elements

Static deflection of a beam at the free end is

given by:

Spring Constantis given by:

6.13

3

EI

Wlst

W= mg is the weight of the mass m,

E = Youngs Modulus, andI = moment of inertia of cross-section of beam

7.13

3l

EIWk

st


38/109


1.7 Spring Elements

Combination of Springs:

1) Springs inparallel if we have n springconstants k1, k2, , kn inparallel, then theequivalent spring constant keq is:

11.121 ... neq kkkk


39/109


1.7 Spring Elements

Combination of Springs:

2) Springs in series if wehave n spring constants k1,

k2, , kn in series, then the

equivalent spring constantkeq is:

17.11

...

111

21 neqkkkk

Example 1 3


40/109


Example 1.3

Torsional Spring Constant of a Propeller Shaft

Determine the torsional spring constant of the

speed propeller shaft shown in Fig. 1.25.


41/109



We need to consider the segments 12 and 23 of

the shaft as springs in combination. From Fig.

1.25, the torque induced at any cross section of

the shaft (such asAA orBB) can be seen to be

equal to the torque applied at the propeller, T.Hence, the elasticities (springs) corresponding to

the two segments 12 and 23 are to be considered

as series springs. The spring constants of

segments 12 and 23 of the shaft (kt12 and kt23) aregiven by


42/109



m/rad-N109012.8

)3(32

)15.025.0()1080(

32

)(

6

449

23

4

23

4

23

23

23

23

l

dDG

l

GJk

t

m/rad-N105255.25

)2(32

)2.03.0()1080(

32

)(

6

449

12

4

12

4

12

12

12

12

l

dDG

l

GJk

t


43/109



Since the springs are in series, Eq. (1.16) gives

m/rad-N105997.6

)109012.8105255.25(

)109012.8)(105255.25(

6

66

66

2312

2312

tt

tt

t

kk

kkk

eq

Example 1 5


44/109


Example 1.5

Equivalent kof a Crane

The boomAB of crane is a uniform steel bar of

length 10 m and x-section area of 2,500 mm2.A weight Wis suspended while the crane is

stationary. Steel cable CDEBFhas x-sectional

area of 100 mm2. Neglect effect of cable CDEB,

find equivalent spring constant of system in the

vertical direction.


45/109


m3055.12

426.151135cos)10)(3(2103

1

2221

l

l


A vertical displacementx of pt B will cause the

spring k2 (boom) to deform byx2 =x cos 45 andthe spring k1 (cable) to deform by an amount

x1 =x cos (90). Length of cable FB, l1 is asshown.


46/109



The angle satisfies the relation:

The total potential energy (U):

0736.35,8184.0cos

10cos)3)((23 2122

1

ll

1.E)]90cos([2

1)45cos(

2

1 22

21 xkxkU

N/m106822.10355.12

)10207)(10100( 696

1

111

l

EAk

N/m101750.510

)10207)(102500( 796

2

222

l

EAk


47/109



Potential Energy of the equivalent spring is:

By setting U = Ueq, hence:

2.E2

1 2xkU eqeq

N/m104304.26 6eq k


48/109


Using mathematical model to represent the

actual vibrating system

E.g. In figure below, the mass and damping

of the beam can be disregarded; the system

can thus be modeled as a spring-masssystem as shown.



49/109



Combination of Masses

E.g. Assume that the

mass of the frame is

negligible compared to

the masses of the floors.The masses of various

floor levels represent the

mass elements, and the

elasticities of the verticalmembers denote the

spring elements.


50/109


Case 1: Translational Masses Connected by a

Rigid Bar

Velocities of masses can be expressed as:

18.111

331

1

22 x

l

lxx

l

lx



51/109


Case 1: Translational Masses Connected by a

Rigid Bar

By equating the kinetic energy of the system:

19.11xxeq

20.12

1

2

1

2

1

2

1 2eqeq

233

222

211 xmxmxmxm

21.13

2

1

32

2

1

21eq m

l

lm

l

lmm


1 8


52/109


meq = single equivalent translational mass

= translational velocity= rotational velocity

J0 = mass moment of inertia

Jeq = single equivalent rotational mass

x


Case 2: Translational and Rotational Masses

Coupled Together


53/109



Coupled Together

1. Equivalent translational mass:

Kinetic energy of the two masses is given

by:

Kinetic energy of the equivalent mass is

given by:

22.12

1

2

1 20

2 JxmT

23.12

1 2eqeqeq xmT


1 8 M I ti El t


54/109



Coupled Together

2. Equivalent rotational mass:

Here, and , equating Teq

and Tgives

eq Rx

25.1

2

1

2

1

2

1

20eq

20

22eq

mRJJor

JRmJ

24.120

eqR

Jmm

Since and , equating Teq & T

gives R

x xx eq


Example 1.6


55/109


Example 1.6

Equivalent Mass of a System

Find the equivalent mass of the system shown in

Fig. 1.31, where the rigid link 1 is attached to thepulley and rotates with it.

E l 1 6 S l i


56/109



Assuming small displacements, the equivalent

mass (meq) can be determined using theequivalence of the kinetic energies of the two

systems. When the mass m is displaced by a

distance , the pulley and the rigid link 1 rotate by

an angle . This causes the rigid link 2

and the cylinder to be displaced by a distance

. Since the cylinder rolls without

slippage, it rotates by an angle .The kinetic energy of the system (T) can be

expressed (for small displacements) as:

x

pprx /

1

pprxllx /

112

cpcc rrxlrx // 12

E l 1 6 S l ti


57/109



)E.1(2

1

2

1

2

1

2

1

2

1

2

12

2

22

22

2

11

22

xmJxmJJxmTcccpp

where Jp, J1, and Jc denote the mass moments of

inertia of the pulley, link 1 (about O), and cylinder,

respectively, indicate the angularvelocities of the pulley, link 1 (about O), and

cylinder, respectively, and represent the

linear velocities of the mass m and link 2,

respectively. Noting that ,Equation (E.1) can be rewritten as

cp and, 1

2and xx

2/2cccrmJ 3/and

2

111lmJ

E l 1 6 S l ti


58/109



)E.2(2

1

22

1

21

321

21

21

2

1

2

1

2

2

1

2

2

2

11

2

2

p

c

cp

cc

ppp

p

r

lxm

rr

lxrm

rlxm

rxlm

rxJxmT

By equating Equation (E.2) to the kinetic energy of

the equivalent system

E.3)(21 2xmT

eq

E l 1 6 S l ti


59/109



)E.4(2

1

3

12

2

1

2

2

1

2

2

12

2

2

11

2

p

c

p

c

ppp

p

eq

r

lm

r

lm

r

lm

r

lm

r

Jmm

we obtain the equivalent mass of the system as

Example 1.7


60/109


p

Cam-Follower Mechanism

A cam-follower mechanism is

used to convert the rotary motionof a shaft into the oscillating orreciprocating motion of a valve.

The follower system consists of a

pushrod of mass mp, a rocker armof mass mr, and mass moment ofinertia Jrabout its C.G., a valve ofmass mv, and a valve spring of

negligible mass.Find the equivalent mass (meq) of this cam-follower system by assuming the location ofmeqas (i) ptA and (ii) pt C.

E l 1 7 S l ti


61/109


The kinetic energy of the system (T) is:

Ifmeq denotes equivalent mass placed at ptA,

with , the kinetic energy equivalent masssystem Teq is:

1.E2

1

2

1

2

1

2

1 2222rrrrvvpp xmJxmxmT

xx eq

2.E21 2

eqeqeq xmT


Example 1 7 Solution


62/109


By equating Tand Teq, and note that

Similarly, if equivalent mass is located at point C,

hence,

11

3

1

2 and,,,l

x

l

lxx

l

lxxxx rrvp

3.E21

2

321

2

221

eql

lm

l

lm

l

Jmm rvrp

,eq v

xx

4.E2

1

2

1 2eq

2eqeqeq vxmxmT




63/109


Equating (E.4) and (E.1) gives

5.E

2

21

23

2

2

122

eq

l

lm

l

lm

l

Jmm rp

rv


1 9 Damping Elements


64/109



Viscous Damping:

Damping force is proportional to the velocity ofthe vibrating body in a fluid medium such as air,water, gas, and oil.

Coulomb orDry Friction Damping:

Damping force is constant in magnitude butopposite in direction to that of the motion of thevibrating body between dry surfaces

Material orSolid orHysteretic Damping:Energy is absorbed or dissipated by materialduring deformation due to friction betweeninternal planes



65/109


Hysteresis loop for elastic materials:




66/109


Shear Stress () developed in the fluid layer at

a distance y from the fixed plate is:

where du/dy = v/h is the velocity gradient. ShearorResisting Force (F) developed at the

bottom surface of the moving plate is:

where A is the surface area of the moving plate.

26.1dy

du

27.1cvhAv

AF

Where A is the surface area of the moving

plate and is the damping constanth

A

c




67/109



and

is called the damping constant.

If a damper is nonlinear, a linearization process

is used about the operating velocity (v*) and theequivalent damping constant is:

28.1

h

Ac

29.1*vdv

dFc

Example 1.9


68/109


Piston-Cylinder Dashpot

Develop an expression for the damping constant

of the dashpot as shown in Fig. 1.36(a).



69/109



The damping constant of the dashpot can be

determined using the shear stress equation forviscous fluid flow and the rate of fluid flow

equation. As shown in Fig. 1.36(a), the dashpotconsists of a piston diameterD and length l,

moving with velocity v0 in a cylinder filled with a

liquid of viscosity . Let the clearance between the

piston and the cylinder wall be d. At a distance y

from the moving surface, let the velocity and shearstress be vand , and at a distance (y + dy) let the

velocity and shear stress be (v dv) and (+ d),

respectively (see Fig. 1.36b).



70/109



The negative sign fordvshows that the velocity

decreases as we move toward the cylinder wall.The viscous force on this annular ring is equal to

E.1)(dy

dy

dDlDldF

But the shear stress is given by

(E.2)dy

dv

where the negative sign is consistent with a

decreasing velocity gradient. Using Eq. (E.2) in

Eq. (E.1), we obtain



71/109



E.3)(2

2

dy

vd

DldyF

E.4)(4

4

22 D

P

D

Pp

The force on the piston will cause a pressure

difference on the ends of the element, given by

Thus the pressure force on the end of the elementis

E.5)(4

dyD

PDdyp



72/109


where denotes the annular area between y

and (y+ dy). If we assume uniform mean velocityin the direction of motion of the fluid, the forces

given in Eqs. (E.3) and (E.5) must be equal. Thus

we get


2

2

4dyvdDldydy

DP

Ddy

or

E.6)(4

22

2

lD

P

dy

vd


73/109



74/109



The volume of the liquid flowing through the

clearance space per second must be equal to thevolume per second displaced by the piston. Hence

the velocity of the piston will be equal to this rate of

flow divided by the piston area. This gives

E.9)(

4

2

0

D

Qv

Equations (E.8) and (E.9) lead to



75/109



E.10)(4

213

03

3

vd

D

dlD

P

By writing the force as P = cv0, the damping

constant c can be found as

E.11)(2143

3

3

Dd

dlDc

Example 1.10 Equivalent Spring and Damping


76/109


Constants of a Machine Tool Support

A precision milling machine is supported on four

shock mounts, as shown in Fig. 1.37(a). The

elasticity and damping of each shock mount can

be modeled as a spring and a viscous damper, as

shown in Fig. 1.37(b). Find the equivalent springconstant, keq, and the equivalent damping

constant, ceq, of the machine tool support in terms

of the spring constants (ki) and damping constants

(ci) of the mounts.

Example 1.10 Equivalent Spring and Damping

C f S


77/109


Constants of a Machine Tool Support



78/109


The free-body diagrams of the four springs and

four dampers are shown in Fig. 1.37(c). Assumingthat the center of mass, G, is located

symmetrically with respect to the four springs and

dampers, we notice that all the springs will be

subjected to the same displacement, , and all thedampers will be subject to the same relative

velocity , where and denote the

displacement and velocity, respectively, of the

center of mass, G. Hence the forces acting on the

springs (Fsi) and the dampers (Fdi) can be

expressed as


x

x x x



79/109





80/109


Let the total forces acting on all the springs and all

the dampers be Fsand Fd, respectively (see Fig.

1.37d). The force equilibrium equations can thus

be expressed as


E.1)(4,3,2,1;4,3,2,1;

ixcFixkF

idi

isi

E.2)(4321

4321

ddddd

sssss

FFFFF

FFFFF



81/109



E.3)(xcF

xkF

eqd

eqs

where Fs + Fd= W, with Wdenoting the total

vertical force (including the inertia force) acting onthe milling machine. From Fig. 1.37(d), we have

Equation (E.2) along with Eqs. (E.1) and (E.3),

yield

E.4)(4

4

4321

4321

cccccc

kkkkkk

eq

eq



82/109


where ki= kand ci= cfori= 1, 2, 3, 4.Note: If the center of mass, G, is not located

symmetrically with respect to the four springs and

dampers, the ith spring experiences a

displacement of and the ith damper experiences

a velocity of where and can be related to

the displacement and velocity of the center of

mass of the milling machine, G. In such a case,Eqs. (E.1) and (E.4) need to be modified suitably.


x x

ix

ix

ix

ix



83/109


30.1sinsin tAAx

31.1cos tAdt

dx

32.1sin 222

2

xtAdt

xd


Periodic Motion: motion repeated after equal

intervals of time

Harmonic Motion: simplest type of periodic

motion

Displacement (x): (on horizontal axis)Velocity:

Acceleration:

1 10 Harmonic Motion


84/109



Scotch yokemechanism:

The similarity

between cyclic

(harmonic) andsinusoidal

motion.



85/109


Complex numberrepresentation of harmonic

motion:

where i = (1) and a and bdenote the real and

imaginaryx andy components ofX,respectively.

35.1ibaX



86/109


Also, Eqn. (1.36) can be expressed as

Thus,

43.1sincos

36.1sincos

iAeiAX

iAAX

47.12,1;)( 22 jbaA jjj

48.12,1;tan 1

ja

b

j

jj



87/109

2005 Pearson Education South Asia Pte Ltd.

87

Operations on Harmonic Functions:

Rotating Vector,

where Re denotes the real part

51.1tiAeX

54.1cos]Re[ntDisplaceme tAAe ti

55.190cossin]Re[Velocity

tAtAAei

ti

56.1180cos

cos

]Re[onAccelerati

2

2

2

tA

tA

Ae ti



88/109


88

Displacement, velocity, and accelerations as

rotating vectors

Vectorial addition ofharmonic functions

Example 1.11

Additi f H i M ti


89/109


89

Addition of Harmonic Motions

E.1)()()()cos()(21txtxtAtx

Find the sum of the two harmonic motions

).2cos(15)(andcos10)(21 ttxttx

Solution:

Method 1: By using trigonometric relations: Since

the circular frequency is the same for bothx1(t)

andx2(t), we express the sum as



90/109


90

p

E.2)()2sinsin2cos(cos15cos10

)2cos(15cos10sinsincoscosttt

ttttA

E.3)()2sin15(sin

)2cos1510(cos)sin(sin)cos(cos

t

tAtAt

That is,

That is,

By equating the corresponding coefficients of

costand sinton both sides, we obtain

E.4)(1477.14

)2sin15(2cos1510

2sin15sin2cos1510cos

22

A

AA



91/109


91

p

E.5)(5963.74

2cos1510

2sin15tan 1

and

Method 2: By using vectors: For an arbitrary value

oft, the harmonic motionsx1(t) and x2(t) can be

denoted graphically as shown in Fig. 1.43. By

adding them vectorially, the resultant vectorx(t)can be found to be

E.6)()5963.74cos(1477.14)( ttx



92/109


92

p

E.7)(15ReRe)(

10ReRe)()2()2(

22

11

titi

titi

eeAtx

eeAtx

Method 3: By using complex number

representation: the two harmonic motions can bedenoted in terms of complex numbers:

The sum ofx1(t) andx2(t) can be expressed as

E.8)(Re)(

)(

ti

Aetx

whereA and can be determined using Eqs. (1.47)

and (1.48) asA = 14.1477 and = 74.5963



93/109


93

Definitions of Terminology:

Amplitude (A) is the maximum displacement

of a vibrating body from its equilibrium

position

Period of oscillation (T) is time taken to

complete one cycle of motion

Frequency of oscillation (f) is the no. of

cycles per unit time

59.1

2

T

60.12

1

Tf



94/109


94


Natural frequency is the frequency which a

system oscillates without external forces

Phase angle () is the angular differencebetween two synchronous harmonic motions

62.1sin

61.1sin

22

11

tAx

tAx



95/109


95


Beats are formed when two harmonic

motions, with frequencies close to one

another, are added



96/109


96


Decibel is originally defined as a ratio ofelectric powers. It is now often used as a

notation of various quantities such as

displacement, velocity, acceleration,

pressure, and power

1.69)(log20dB

1.68)(log10dB

0

0

XX

P

P

where P0 is some reference value of power

andX0 is specified reference voltage.



97/109


97

y

Fourier Series Expansion:Ifx(t) is a periodic function with period , itsFourier Series representation is given by

1

0

21

21

0

)70.1()sincos(2

...2sinsin

...2coscos

2

)(

nnn

tnbtnaa

tbtb

tataa

tx



98/109


98

y

Gibbs Phenomenon:An anomalous behavior observed from aperiodic function that is being represented by

Fourier series.As n increases, the

approximation can be seen

to improve everywhere

except in the vicinity of the

discontinuity, P. The errorin amplitude remains at

approximately 9 percent,

even when .k



99/109


99

Complex Fourier Series:

The Fourier series can also be represented interms of complex numbers.

)79.1(sincos

)78.1(sincos

tite

tite

ti

ti

Also,

)81.1(2

sin

)80.1(2

cos

i

eet

eet

titi

titi

and



100/109


100

Frequency Spectrum:Harmonics plotted as vertical lines on a diagram

of amplitude (an and bn ordn and n) versus

frequency (n)



101/109


101

A periodic function:

Representation of a function in time andfrequency domain:



102/109


102

Even and odd functions:

1

0

)88.1(cos2

)(

)87.1()()(

nn tnaatx

txtx

Even function & its Fourierseries expansion

Odd function & its Fourier

series expansion

1

)90.1(sin)(

)89.1()()(

nn

tnbtx

txtx



103/109


103

Half-Range Expansions:

The function is extended to include

the interval as shown in the

figure. The Fourier series

expansions ofx1(t) andx2(t) areknown as half-range expansions.

0to



104/109


104

Numerical Computation

of CoefficientsIfx(t) is not in a simple

form, experimental

determination of the

amplitude of vibrationand numerical

integration procedure

like the trapezoidal or

Simpsons rule is usedto determine the

coefficients an and bn. )99.1(2

sin2

)98.1(2cos2

)97.1(2

1

1

10

iN

iin

iN

iin

N

ii

tnx

Nb

tnxN

a

xN

a

Example 1.12

Fourier Series Expansion


105/109


105

Fourier Series Expansion

Determine the Fourier series expansion of the

motion of the valve in the cam-follower systemshown in the Figure.



106/109


106

E.1)()()(

)()(tan

1

2

21

tyl

ltx

l

tx

l

ty

E.2)(0;)(

tt

Yty

Ify(t) denotes the vertical motion of the pushrod,

the motion of the valve,x(t), can be determinedfrom the relation:

or

where

and the period is given by .

2



107/109


107

1

2

l

Yl

A

E.3)(0;)( t

t

Atx

By defining

x(t) can be expressed as

Equation (E.3) is shown in the Figure.

To compute the Fourier coefficients an and bn, we

use Eqs. (1.71) to (1.73):

E.4)(2

)(

/2

0

2

/2

0

/2

00A

tAdt

tAdttxa



108/109


108

E.5)(..2,,1,0

sincos

2cos

coscos)(/2

0

22

/2

0

/2

0

/2

0

n

n

tnt

n

tnAdttnt

A

dttn

t

Adttntxan

E.6)(..2,,1,

cossin

2sin

sinsin)(

/2

0

22

/2

0

/2

0

/2

0

nn

A

n

tnt

n

tnAdttnt

A

dttnt

Adttntxbn



109/109

Therefore the Fourier series expansion ofx(t) is

E.7)(...3sin

3

12sin

2

1sin

2

...2sin2

2sin2

)(

tttA

tA

tAA

tx

The first three terms of the

series are shown plotted in the

figure. It can be seen that theapproximation reaches the

sawtooth shape even with a

Vibration Fundamental -Basic

Documents