Mee5206 Chapter 3

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MEE 5206:

VIBRATION

Chapter 3Harmonically Excited Vibration

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Mechanical Vibrations

Fourth EditionWilliam T. Thomson

REFERENCES

Third EditionDaniel J. INMAN

Fifth EditionSingiresu S. Rao

TEXT BOOK

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Chapter 3Harmonically Excited Vibration

Introduction

Equation of Motion

Response of an Undamped System Under Harmonic Force

Response of a Damped System Under Harmonic Force

Response of a Damped System Under

Response of a Damped System Under Rotating Unbalance

Transfer-Function ApproachSolutions Using Laplace Transforms

Frequency Transfer Functions

tieFtF 0)(

Chapter Outline

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4

Learning Objectives

After completing this chapter, the reader should be able

to do the following:

Find the responses of undamped and viscously damped single-degree-of-

freedom systems subjected to different types of harmonic force, including base

excitation and rotating unbalance.

Distinguish between transient, steady-state, and total solutions.

Understand the variations of magnification factor and phase angles with the

frequency of excitation and the phenomena of resonance and beats.

Find the response of systems involving Coulomb, hysteresis, and other types of

damping.

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Introduction

Forced vibrationsoccurs when external energy is supplied to thesystem during vibration

The external force can be supplied through either an applied force

or an imposed displacement excitation, which may be harmonic,

nonharmonic but periodic, nonperiodic, or random in nature.

Harmonic responseresults when the system responses to a

harmonic excitation

Transient responseis defined as the response of a dynamic system

to suddenly applied nonperiodic excitations

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2011 Mechanical Vibrations Fifth Edition in SI Units6

From the figure below, the equation of motion using NewtonsSecond Law of Motion states that

The homogeneous solution of the equation is

)1.3()(tFkxxcxm

)2.3(0 kxxcxm

A spring-mass-damper system

Equation of Motion

Since Eq. (3.1) is nonhomogeneous, its general solution x(t) is given by thesum of the homogeneous solution xh(t) and the particular solution, xp(t).

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Equation of Motion

The variations of homogeneous and general solutions with time for

a typical case are shown in the figure below.

Homogenous and general solutions of Eq. (3.1) for an underdamped case

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Response of an Undamped System UnderHarmonic Force

Consider an undamped system subjected to a harmonic force. If aforce acts on the mass mof the system,

The homogeneous solution is given by:

tFtF cos)( 0

)3.3(cos0 tFkxxm

)4.3(sincos)( 21 tCtCtx nnh 2/1)/( mkn where is the natural frequency

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Because the exciting force and particular solution is harmonic andhas the same frequency, we can assume a solution in the form:

Thus, the total solution (General solution) will be:

where X is the max amplitude ofxp(t)

)5.3(cos)( tXtxp

)6.3(

1

220

n

st

mkFX

kFst /0where denotes the static deflection

)7.3(cossincos)(2

021 t

mk

FtCtCtx

nn


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Using initial conditions

Hence

The max amplitude can be expressed as

)8.3(, 0220

01

n

xC

mk

FxC

)9.3(cos

sincos)(

2

0

0

2

00

tmk

F

t

x

tmk

F

xtx nn

n

)10.3(

1

12

n

st

X


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The quantity is the ratio of the dynamic to static amplitude ofmotion and called the magnification factor, amplification factor, or

amplitude ratio.

The variation of the amplitude ratio with

the frequency ratio is shown in the figure.

The response of the system can be

identified to be of three types.

stX /

Magnification factor of an undamped system


n

r

frequency ratio

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Case 1:

When 0 < < 1, the denominator in Eq.(3.10) is positive and

the response is given by Eq.(3.5) without change. The harmonic

response of the system is in phase with external force, shown in

figure.

n/

Harmonic response when 1/0 n


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Case 2:

When > 1, the denominator in Eq.(3.10) is negative and the

steady-state solution can be expressed as

where the amplitude is

The variations are shown in figure.

n/

)11.3(cos)( tXtxp

)12.3(

1

2

n

stX

Harmonic response when 1/ n


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Case 3:

When , the amplitude X given by Eq.(3.10) or (3.12)

becomes infinite.

The forcing frequency is equal to the natural frequency of thesystem is called resonance.

The total response if the

system at resonance is

Harmonic response when 1/ n

)15.3(sin2

sincos)( 00 tt

tx

txtx nnst

n

n

n

1/ n


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Total Response

The total response of the system, Eq.(3.7) or Eq.(3.9), can also be

expressed as

)16.3(1for;cos

1

)cos()(2

n

n

st

n ttAtx


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Total Response

And

)17.3(1for;cos

1

)cos()(2

n

n

st

n ttAtx


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Beating Phenomenon

If the forcing frequency is close to, but not exactly equal to, the

natural frequency of the system, beatingmay occur. The amplitude

builds up and diminishes in a regular pattern.

The phenomenon of beating can be expressed as:

)22.3(sinsin2

/)( 0 tt

mFtx

Phenomenon of beats


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Beating Phenomenon

The time between the points of zero amplitude or the points of

maximum amplitude is called theperiod of beatingand is given by

The frequency of beating defined as

nb 2

)23.3(222

nb


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Example 3.1Plate Supporting a Pump

A reciprocating pump, having a mass of 68kg, is mounted at themiddle of a steel plate of thickness 1cm, width 50cm, and length250cm, clamped along two edges as shown in Figure. During operation

of the pump, the plate is subjected to a harmonic force, F(t)= 220 cos62.832tN. Find the amplitude of vibration of the plate.


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Solution

The plate can be modeled as a fixed-fixed beam having Youngsmodulus (E) = 200GPa, length = 250cm, and area moment of inertia,

The bending stiffness of the beam is given by:

49322 m10667.41)10)(1050(12

1)( I

(E.1)N/m82.102400)10250(

)10667.41)(10200)(192(19232

99

3

lEIk


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Solution

Thus,

(E.2)mm32487.1m00132487.0

)832.62(6882.102400

2202

2

0

g

mk

FX


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Response of Damped System Under HarmonicForce

The equation of motion can be derived as

Using trigonometric relations, we obtain

)26.3(cossincos 02 tFtctmkX

)27.3(0cossin

sincos2

02

cmkX

FcmkX

)24.3(cos0 wtFkxxcxm The equation of motion for forcing function is:wtFtF cos)( 0

The particular equation of motion can be:

(3.25))cos()( wtXtx p

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The figure shows typical plots of the forcing function and steady-

state response.

(a) Graphical representation (b) Vectorial representation


)28.3(2/1

2222

0

cmk

FX

)29.3(tan2

1

mk

c

The solution gives

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Substituting the following,

We obtain

;m

kn ;2 n

m

c ;0

k

Fst

n

r

)30.3()2()1(

1

21

12222/1

22

2 rrX

nn

st

)31.3(1

2tan

1

2tan

2

1

2

1

r

r

n

n


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The following characteristics of the magnification factor ( ) can be

noted from the figure as follows:


st

XM

magnification factor

M

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1. For an undamped system , Eq.(3.30) reduces to Eq.(3.10),and as .

2. Any amount of damping reduces the magnification factor

(M) for all values of the forcing frequency.

3. For any specified value of r, a higher value of damping reduces the

value of M.

4. In the degenerate case of a constant force (when r= 0), the value

of M= 1.

)0( M 1r

)0(


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5. The reduction in M in the presence of damping is very significant ator near resonance.

6. The amplitude of forced vibration becomes smaller with increasing

values of the forcing frequency (that is, as ).

7. For , the maximum value of M occurs when

which can be seen to be lower than the undamped natural

frequency and the damped frequency

M 1r

2

10

)32.3(21or21 22 nr

21 nd


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8. The maximum value ofX(when ) is given by:

and the value of X at by

9. For when r= 0. For , the graph of M

monotonically decreases with increasing values of r.

2

21 r)33.3(

12

1

2max

st

X

)34.3(2

1

nst

X

0,2

1

dr

dM

2

1

n


f d d i

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The following characteristics of the phase angle can be observed

from the figure and Eq.(3.31) as follows:


R f D d S t U d H i

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1.For an undamped system , Eq.(3.31) shows that the phaseangle is 0 for 0 < r < 1 and 180for r > 1. This implies that the

excitation and response are in phase for 0 < r < 1 and out of phase

for r > 1 when .

2. For and 0 < r < 1, the phase angle is given by 0 < < 90,implying that the response lags the excitation.

3. For and r > 1, the phase angle is given by 90< < 180,

implying that the response leads the excitation.

)0(

0

0

0


R f D d S t U d H i

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4.For and r = 1, the phase angle is given by = 90

,implying that the phase difference between the excitation and the

response is 90.

5. For and large values of r, the phase angle approaches 180,implying that the response and the excitation are out of phase.

0

0


R f D d S t U d H i

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32

Total response

For an underdamped system,

For the initial conditions, Eq.(3.35) yields

)35.3()cos()cos()( 00 tXteXtx

d

tn

)36.3(1 2nd

)36.3(sinsincos

coscos

00000

000

XXXx

XXx

dn


(3.37))cos(

sincostan

)sincos(1)cos[(

0

0

.

00

20

.

022

00

Xx

XXxx

XXxxXxX

d

nn

nn

d

Response of Damped System Under Harmonic

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Example 3.3

Total Response of a System

Find the total response of a single degree of freedom system with m=10kg, c= 20 N-s/m, k= 4000 N/m,x0= 0.01 m, under thefollowing conditions:

a. An external force acts on the system withand .

b. Free vibration with F(t)= 0.

00 x

tFtF cos)( 0N1000 F

rad/s10



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Example 3.3


Solution

a. From the given data,

(E.1)m03326.0

5.05.0205.01

025.0

21

5.0

rad/s974984.192005.011

05.0

m025.0

rad/s20

222222

2010

22

1040002

20

2

4000100

104000

0

rr

X

r

st

nd

km

ccc

kF

st

mk

n

n

c



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Example 3.3


Solution

We have

Using initial conditions

(E.2)814075.35.01

5.005.02tan

1

2tan

2

1

2

1

r

r

(E.3)023186.0cos 00 X

(E.4))814075.3sin()10)(03326.0(

sin)974984.19(cos)20)(05.0(0 0000

XX

0and01.0 00 xx



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Example 3.3


Solution

Substituting Eq.(E.3) into (E.4),

Hence,

(E.5)002268.0sin 00 X

(E.7)586765.5

0978176.0cos

sintan

0

00

000

X

X

(E.6)023297.0)sin()cos( 2/12002000 XXX



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Example 3.3


Solution

b. For free vibration, the total response is

Using the initial conditions,

(E.9)010012.0974984.19

01.02005.001.0

2/12

2

2/12

02

00

d

nxxX

(E.8))cos()( 00

teXtx dt

n

(E.10)865984.2974984.19

2005.0tantan 1

0

001

0

x

xx

d

n


Response of Damped System Under tieFtF 0)(

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Response of Damped System Under

The equation of motion becomes

Assuming the particular solution

Substituting,

)47.3(0tieFkxxcxm

)48.3()( tip Xetx

)49.3()( 2

0

icmk

FX

)50.3()()( 22222222

2

0

cmkci

cmkmkFX

eFtF 0)(

Response of Damped System Under tieFtF 0)(

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p p y


Using the relation

Hence, the steady-state solution becomes

eFtF 0)(

)51.3(

)(2/12222

0

ie

cmk

FX

)52.3(tan2

1

mk

c

iAeiyx

)53.3(

)()()( )(

2/1222

0

tip ecmk

Ftx

tieFtF 0)( Response of Damped System Under

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Frequency Response

The complex frequency response is given by:

The absolute value becomes,

Thus, the steady-state solution becomes,

et 0)(

)54.3(21

1)(

2

0 rirF

kXiH

)56.3()()( ieiHiH )57.3(1

2tanwhere

2

1

r

r

)58.3()()( )(0 tip eiHk

Ftx

p p y


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Frequency Response

If , the corresponding steady-state solution is given

by the real part of Eq.(3.53)

0)(

)59.3()(Re)(Re

)cos()()(

)(

)(00

2/1222

0

titi

p

eiHk

FeiH

k

F

tcmk

Ftx

tFtF cos)( 0

p p y


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Frequency Response

If , the corresponding steady-state solution is given

by the imaginary part of Eq.(3.53)

0)(

)60.3()(Im

)sin()()(

)(

)(0

2/1222

0

ti

p

eiHk

F

tcmk

Ftx

tFtF sin)( 0

p p y


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Complex Vector Representation of Harmonic Motion

Differentiating Eq.(3.58) with respect to time,

The various terms of the equation of motion

can be represented in the figure.

0)(

)61.3()()()()(onAccelerati

)()()(Velocity

2)(02

)(0

txeiHk

Fitx

txieiHk

Fitx

p

ti

p

p

ti

p

Response of a Damped System Under Rotating

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p p y gUnBalance

The equation of motion can be derived by the usual procedure:

The solution can be expressed as

The amplitude and phase angle is given by

)78.3(sin2 tmekxxcxM

)79.3()(Im)sin()()(

2

ti

n

p eiHM

me

tXtx

)(

)()(

2

2/1222

2

iH

M

me

cMk

meX

n

)80.3(tan2

1

Mk

c


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By defining ncc Mccc 2and/

)(

)2()1(

2

2/1222

2

iHrrr

r

me

MX

)81.3(

1

2tan

2

1

r

r

p p y gUnBalance


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The following observations can be made from Eq.(3.81) and the

figure above:

1. All the curves begin at zero amplitude. The amplitude near

resonance is markedly affected by damping. Thus if the machine is

to be run near resonance, damping should be introduced

purposefully to avoid dangerous amplitudes.

2. At very high speeds (large), MX/meis almost unity, and the

effect of damping is negligible.

3. For 0 < < 1/2 , the maximum of MX/me

)82.3(0

me

MX

dr

d

UnBalance


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The solution gives:

With corresponding maximum value:

Thus the peaks occur to the right of the resonance value of r = 1

4. For , does not attain a maximum. Its value grows

from 0 at r = 0 to 1 at r .

5. The magnitude (or maximum value) of F can be derived as

1

21

1

2

r

2max 12

1

me

MX

me

MX,21

(3.84)4141

2/1

2222

222

rr

r

meF

UnBalance


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Example 3.7

Francis Water Turbine

The schematic diagram of a Francis water turbine is shown in thefigure in which water flows from A into the blades B and down into thetail race C. The rotor has a mass of 250 kg and an unbalance (me) of

5kg-mm. The radial clearance between the rotor and the stator is5mm. The turbine operates in the speed range 600 to 6000rpm. Thesteel shaft carrying the rotor can be assumed to be clamped at thebearings. Determine the diameter of the shaft so that the rotor isalways clear of the stator at all the operating speeds of the turbine.Assume damping to be negligible.

UnBalance


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Example 3.7


UnBalance

Response of a Damped System Under Rotatingl

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Example 3.7


Solution

The max amplitude can be obtained from Eq.(3.80) by setting c = 0 as

The value of ranges from:

While the natural frequency is given by

(E.1))1()( 2

2

2

2

rkme

MkmeX

rad/s20060

260006000rpmtorad/s20

60

2600rpm600

(E.2)rad/s625.0250

kk

M

kn

UnBalance

Response of a Damped System Under RotatingU B l

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Example 3.7


Solution

For = 20, Eq.(E.1) gives

For = 200, Eq.(E.1) gives

(E.3)N/m1004.10102

004.0

)20(1

)20()100.5(005.0 2425

2

2

23

kk

kk

(E.4)N/m1004.1010

200

004.0

)200(1

)200()100.5(005.0 2627

2

2

23

kk

kk

UnBalance

Response of a Damped System Under RotatingU B l

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Example 3.7


Solution

The stiffness of the cantilever beam is given by

Diameter of the beam is

(E.5)64

33 4

33

d

l

E

l

EIk

(E.6)mm127m1270.0

m106005.2)1007.2(3

)2)(1004.10)(64(

3

64 4411

32434

d

E

kl

d

UnBalance

Transfer-Function Approach

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The transfer-function approach is used for the formulation and

solution of dynamic problems in the controls literature.

Also used for solving forced-vibration problems

The transfer function of a linear, time-invariant differential equation

is defined as the ratio of the Laplace transform of the output or

response function to the Laplace transform of the input or forcing

function, assuming zero initial conditions.


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Example 3.13

Transfer Function Corresponding to a Differential Equation

Consider the following nth-order linear, time-invariant differential

equation that governs the behavior of a dynamic system:

E.101

1

1

01

1

1

tfb

dt

tfdb

dt

tfdb

txadt

txda

dt

txda

m

m

mm

m

m

n

n

nn

n

n


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Example 3.13


Solution

By taking Laplace transforms of both sides of Eq. (E.1), we obtain

If all initial conditions are assumed to be zero, Eq. (E.2) reduces to the

following form:

E.2involvingconsitionsinitial

involvingconsitionsinitial

0

1

1

0

1

1

tfsXasFsbsFsbtxsXasXsasXsa

m

m

m

m

n

n

n

n

E.301

10

1

1 sFbsbsbsXasasa m

m

m

m

n

n

n

n


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Example 3.13


Solution

Transfer function of the system evaluated at zero initial conditions is

By taking the inverse Laplace transform of Eq. (E.5), we can find the

output of the system in the time domain for any known input.

E.5sFsTsXsFsXsT

Solutions Using Laplace Transforms

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Example 3.15

Response of a Damped System Using Laplace Transforms

Derive an expression for the complete response of a damped single-

degree-of-freedom system subjected to a general force, f(t), as shown

using Laplace transforms.


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Example 3.15


Solution

The Laplace transform of Eq. (3.1) leads to the relation

For convenience,

E.10

210

22

2 222222 x

wswsx

wswsws

wswsmsFsX

nnnn

n

nn

E.32

122

nn

s

i

wswsmsF

sFsF


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Example 3.15


Solution

The inverse Laplace transform of will be equal to the known

forcing function

Inverse Laplace transform of the first term on the right-hand side of

Eq. (E.1) can be expressed as

nddtw

d

si wwtwemw

tfwtFtf n 20 1 wheresin1

andcos

E.7sin1

00

1

dtwefmw

tffsFsF dtw

t

rd

s

t

r

iSin

sFi


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Example 3.15


Solution

The inverse Laplace transform of the coefficient of x(0) in Eq. (E.1)

yields

E.91tantanwhere

E.8cos2

2

211-1

122

1

d

n

d

tw

d

n

nn

n

w

w

twew

w

wsws

wsn


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Example 3.15


Solution

The complete response of the system can also be expressed as

E.13sin1cossin1 10

twew

twewwdwetf

mwtx d

w

d

d

w

d

nd

w

t

d

nnn

Frequency Transfer Functions

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The steady-state response of a linear system subjected to a

sinusoidal (or harmonic) input will also be sinusoidal (or harmonic)

of the same frequency.

Mee5206 Chapter 3

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