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8/12/2019 Vibration Calculation All

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4.4.1 Forced Vibrations From Harmonic Excitation

As discussed earlier, forced vibrations are one very important practical mechanism for theoccurrence of vibrations.

x

m

F(t)

c k

Fig. 4.10: Sdof Oscillator with Viscous Damping and External Force

The equation of motion of the damped linear sdof oscillator with an external force is:

( )t F kx xc xm =++ &&& (4.4.1)

The general solution of this differential equation is:

( ) ( ) ( )321321 forceexternal fromresults

part vibrations free

t xt xt x += hom (4.4.2)

which consists of the homogeneous part resulting from the free vibration and the particular part resulting from the external disturbance F (t ). The homogeneous solution has already beentreated in the last chapter.

Fig 4.11: Homogeneous and particular part of the solution and superposition


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While the homogeneous part of the solution will decay to zero with time we are especiallyinterested in the stationary solution.

4.4.2 Excitation with Constant Force Amplitude

4.4.1.1 Real ApproachThe excitation function is harmonic, is the frequency of excitation

t F t F = cos)( (4.4.3)

Eqn. 4.4.1 becomes

t F kx xc xm =++ cos&&& (4.4.4)

Dividing by the mass m

t m

F x

mk

xmc

x =++ cos

&&& (4.4.5)

Introducing again the dimension less damping and the natural circular frequency

02

mc

D = andmk =20

and the amplitude

m F

f

= (4.4.6)

This yields:

(4.4.7)t f x x D x =++ cos2 200 &&& To solve this differential equation, we make an approach with harmonic functions

t Bt At x += sincos)( (4.4.8)

This covers also a possible phase lag due to the damping in the system. Differentiating (4.4.8)to get the velocity and the acceleration and putting this into eqn. 4.4.7 leads to

t f

t Bt At Bt A Dt Bt A

=++++cos

)sincos()cossin(2sincos 20022

(4.4.9)After separating the coefficients of the sin- and cos-functions and comparing the coefficientswe get:

(4.4.10a) f A B D A 2 2002 =++

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(4.4.10b)02 2002 =+ B A D B

From the second equation we see that

A D B B =+ 020

2

2

which leads to

( ) A D

B22

0

02

=

and we put this result into eqn.(4.4.10a):

f A A D

D A )(

22 2022

0

00

2 =++

f A D

)(

4)(

220

220

222

0 =+

)(4)( 22022

02222

0 =+ f A D

This yields the solution for A and B:

[ ]2202222022

04)(

)(

+

=

D

f A (4.4.11a)

and

[ ]220222200

4)(

)2(

+=

D

D f B (4.4.11b)

Introducing the dimensionless ratio of frequencies

0

==

frequency Natural

frequency Excitation (4.4.12)

2222

220

4)1(

)1)(/(

D

f A

+= (4.4.13a)

and

2222

20

4)1(

)2)(/(

D

D f B

+

= (4.4.13b)

With A and B we have found the solution for t Bt At x += sincos)( .

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Another possibility is to present the solution with amplitude and phase angle:

)cos()( = t C t x (4.4.14)

The amplitude is

20

2222

22

4)1(

1

f

D B AC

+=+= (4.4.15)

Considering that and we getm F f / = mk /20 =

k F

DC

4)1(

12222 +

= (4.4.16)

Introducing the dimensionless magnification factor V 1 which only depends on the frequencyratio and the damping D :

22221 4)1(

1),(

D DV

+= (4.4.17)

we get the amplitude as

k F

V C

1= (4.4.18)

and the phase angle (using trigonometric functions similar as in chap.4.2.2) :

21

2)(tan

== D

A B

(4.4.19)

We can see that as approaches 1 the amplitude grows rapidly, and its value near or at theresonance is very sensitive to changes of the damping D.

The maximum of the magnification curve for a given D can be found at

0

221

resres D == (4.4.20)

If D is very small then 1res . The maximum amplitude for this D then is

21max 12

1),(

D Dk F

DV k

F C res

== (4.4.21)

For 0: V 11: the system behaves quasi-statically, for very large values of : V

1 0: the

vibrations are very small.

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0=

22221 D4)1( 1V +=

D=0 7071

D=0 5

D=0 3

D=0 2

D=0 1

D=0 05

54.543.532.52 1.5 1 0.5 0 0 1 2 3 4 5 6 7 8

9 10

V1

0=

2

1

D2arctan

=

D=0,7071D=0,5

D=0

D=0,05 D=0,2D=0,1

180

150

120

90

60

30

054.43.32.21.10.0

Fig.4.12: Magnification factor V 1 and phase angle to describe the vibration behavior of the

damped oscillator under constant force amplitude excitation

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As we can see, the solution of one part is the conjugate complex of the other:

21 x x = (4.4.29)

The solution for x(t ) is combined from the two partial solutions, which we just have found:

t it i e xe xt x += 21 )( (4.4.30)

This can be resolved:

t xit xt xit xt x +++= sincossincos)( 2211

and using the fact that 21 x x = , we finally get

{ } { } t xt xt x = sinIm2cosRe2)( 11 (4.4.31)

The factor of 2 compensates the factor associated with the force amplitude. All theinformation can be extracted from only so that only this part of the solution has to besolved.

1 x

t k

F

D

Dt

k F

Dt x

++

+= sin

4)1(

2cos

4)1(

)1()(

22222222

2

(4.4.32)

which is the same result as eqn. (4.4.8) with (4.4.13).Also the magnitude ( 1 x ) and phase can be obtained in the same way and yield the previous

results:

Magnitude:( )

( ) F k

DV F k D

x

V factor ionmagnificat

1,1

41

1 1

1

=+

=4 4 4 34 4 4 21

(4.4.33)

Phase: { }{ } 12Re Imtan 11

== D x x (4.4.34)

4.4.1.3 Complex Approach, Alternative

Instead of eqn. (4.4.24), we can write

=+== t it it i e F ee F t F t F 2

Re2)(

2

cos)(

or

{ t ie F t F t F == Recos)( } (4.4.35)

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According to this approach, we formulate the steady state response as

{ t ie X t x = Re)( } (4.4.36)

The complex amplitude X is determined from the equation of motion, solving

( ){ } { }t it i e F e X k cim =++ ReRe 2 (4.4.37)

The real parts are equal if the complex expression is equal:

(4.4.38)( ) t it i e F e X k cim =++ 2 Elimination of the time function yields:

(4.4.39)( F X k cim 2 =++ )

)

)

The expression in brackets is also called the dynamic stiffness

( cimk k dyn += 2)( (4.4.40)

Now we solve (4.4.39) to get the complex amplitude:

( cimk F

X +

=2

(4.4.41)

The expression

( ) Input Output

F

X

cimk H ==

+=

1)(

2 (4.4.42)

is the complex Frequency Response Function (FRF). Introducing the dimensionless frequency as before yields:

( ) k F

Di X

21

12 +

= (4.4.43)

Because( ) t it iit i e X ee xe xt x === ReReRe)cos( (4.4.44)

we take the magnitude and phase lag of this complex result x

(4.4.45) ie x X = )( which leads to the same result as before, see (4.4.33) and (4.4.34):

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( ) ( ) F

k DV F

k D x

V factor ionmagnificat

1,1

41

1 1

1

=+

=4 4 4 34 4 4 21

(4.4.33)

{ } 12

ReImtan

== D X X (4.4.34)

4.4.2 Harmonic Force from Imbalance Excitation

c

m k 2

m k 2

k 2

k 2

m M

x

Fig. 4.13: Sdof oscillator with unbalance excitation

The total mass of the system consists of the mass mM and the two rotating unbalance massesmu :

22 U M

mmm += (4.4.46)

The disturbance force from the unbalance is depending on the angular speed , is the

excentricity:

( ) t mt F U Unbalance = cos (4.4.47)

Now, following the same way as before (real or complex) leads to the solution:

)cos()( = t C t x

where

Amplitude: ( ) ( ) mm

DV mm

DC x U U

,41

3=+==

(4.4.48)

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Phase:1

2tan

= D (4.4.49)

Magnification factor ( )( ) 41

,3

D D

+=V (4.4.50)

The phase is the same expression as in the previous case, however, the magnification factor isdifferent, because the force amplitude is increasing with increasing angular speed.

0=

22222

3 D4)1(V

+=

D=0,7071

D=0,5

D=0,3

D=0,2

D=0,1

D=0,05

5 4.5 43.532.521.51 0.5 0 0 1 2 3 4 5V3 6

7 8 9

10

Fig. 4.14: Magnification factor V 3 for the case of imbalance excitation

As can be seen: for 0: V 10: there is no force if the system is not rotating or rotates onlyslowly, for very large values of : V 1 1: that means that the mass m is vibrating with anamplitude ( mu/m), but the common center of gravity of total system m and mu does notmove.

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4.4.3.2 Case 2

c k

m x

u(t)

Fig. 4.16: Excitation of the sdof oscillator by harmonic motion of the spring/dampercombination

The equation of motion now also contains the velocity u :&

kuuckx xc xm +=++ &&&& (4.4.55)

Amplitude of vibration and phase shift becomes

Amplitude:( )

( )u DV u D

D x

V function Magn

,41

41 2

. 2

=+

+=4 4 4 34 4 4 21

(4.4.56)

Phase: ( ) 412tan

D D

+= (4.4.57)

As can be seen the phase now is different due to the fact that the damper force depending onthe relative velocity between ground motion and motion of the mass plays a role. Theamplitude behaviour is described by the magnification factor V 2.

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1

0=

222222

2 D4)1(

D41V+

+=

2

D=0 7071

D=0 5

D=0 3

D=0 2

D=0 1

D=0 05

543210

1

8

V2 6 4

2

0

Fig. 4.17: Magnification factor V 2 for the case of ground excitation via spring anddamper

Notice that all curves have an intersection point at 2= which means that for 2> higher damping does not lead to smaller amplitudes but increases the amplitudes. This is dueto the fact that larger relative velocities (due to higher frequencies ) make the damper stifferand hence the damping forces.Further cases of ground motion excitation are possible.

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4.4 Excitation by Impacts

4.5.1 Impact of finite duration

T i

F(t)

t

F

F(t)

k c

x

m

Fig. 4.18: Sdof Oscillator under impact loading

We consider an impact of finite length T i and constant force level during the impact Theimpact duration T i is much smaller than the period of vibration T :

Di T

T 2=


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and000 == A x D

i v

D

v B

mT F

0

0

00

1

=

==v (4.5.3)

so that the system response to the impact is a decaying oscillation where we have assumed

that the damping D < 1:

( ) ( t evt x Dt D D

sin00 = ) (4.5.4)

4.5.2 DIRAC-Impact

F

t

Fig. 4.19: DIRAC-Impact

The DIRAC-Impact is defined by

( ) ( ) ( ) ( ) 1,000 ====

dt t but

t t t t F t F (4.5.5)

is the Kronecker symbol. The duration of this impact is infinitely short but the impact isinfinitely large. However, the integral is equal to 1 or , respectively. For the initialdisplacement and calculation of the initial velocity following the previous chapter, weget

F 00 = x

( ) ( t em

F t D

t D

D

sin

0= ) x (4.5.6)

For , the response x(t) is equal to the impulse response function (IRF) h(t )1 = F

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( ) ( t em

t Dt D

D

sin

1 0= )h (4.5.7)

The IRF is an important characteristic of a dynamic system in control theory.

4.5 Excitation by Forces with Arbitrary Time Functions

F( )

+

F

t

t

x

Fig. 4.20: Interpretation of an arbitrary time function as series of DIRAC-impulses

Using the results of the previous chapters we can solve the problem of an arbitrary timefunction F (t ) as subsequent series of Dirac-impacts, where the initial conditions follow fromthe time history of the system.

The solution is given by the Duhamel-Integral or convolution integral :

== t t

Dt D

Dd F t hd F t e

mt

00

)( )()()())(sin(1

)( 0

x (4.6.1)

As can be seen, the integral contains the response of the sdof oscillator with respect to aDIRAC-impact multiplied with the actual force F ( ), which is integrated from time 0 to t .

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4.7 Periodic Excitations

4.7.1 Fourier Series Representation of Signals

Periodic signals can be decomposed into an infinite series of trigonometric functions, calledFourier series.

Fig. 4.21 : Scheme of signal decomposition by trigonometric functions

F

t

T

Fig. 4.22 : Example of a periodic signal: periodic impacts

The period of the signal is T and the corresponding fundamental frequency is

T

2= (4.7.1)

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Now, the periodic signal x(t) can be represented as follows

)sin()cos(2

)(1

0

=++=

k k k t k bt k a

at x

(4.7.2)

The Fourier-coefficients a0, ak and bk must be determined. They describe how strong thecorresponding trigonometric function is present in the signal x(t ). The coefficient a 0 is thedouble mean value of the signal in the interval 0 T:

=T

dt t xT

a0

0 )(2

(4.7.3)

and represents the off-set of the signal. The other coefficients can be determined from

( )=T

k dt t k t xT a

0cos)(2 (4.7.4)

( )=T

k dt t k t xT b

0sin)(

2 (4.7.5)

The individual frequencies of this terms are

T

k k k

2== (4.7.6)

for k = 1 we call the frequency 1 fundamental frequency or basic harmonic and thefrequencies for k = 2,3, the second, third, harmonic (or generally higher harmonics).

4.7.1.1 Alternative real Representation

We can write the Fourier series as a sum of cosine functions with amplitude ck and a phaseshift k

)cos()(1

0 =

++=k

k k t k cct x (4.7.7)

22k k k bac += and )arctan(

k

k k a

b= (4.7.8)

4.7.1.2 Alternative complex Representation

The real trigonometric functions can also be transformed into complex exponentialexpression:

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==

k

t ik k e X t x

)( (4.7.9)

The X k are the complex Fourier coefficients which can be determined by solving the integral:

=T

t ik k dt et xT

X 0

)(1 (4.7.10a)

or

[ =T

k dt t k it k t xT X

0sincos)(

1 ] (4.7.10b)

which clearly shows the relation to the real Fourier coefficients series given by eqns.(4.7.4)and (4.7.5):

{ } { }2

Im;2

Re k k k

k b

X a

X ==

The connection to the other real representation (chap. 4.7.11) is

k k c X = { }{ }

)ReIm

(tank

k k X

X = (4.7.11)

The coefficients with negative index are the conjugate complex values of the corresponding

positive ones:

(4.7.12)*k k X X =

4.7.2 Forced Vibration Under General Periodic Excitation

x

m

F(t)

c k

Fig. 4.23: Sdof oscillator under periodic excitation

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Let us use once more the single dof oscillator but now the force is a periodic function whichcan be represented by a Fourier series

=++=

1

0 )sin()cos(2

)(k sk ck

t k F t k F F

t F (4.7.13)

The F ck and F sk are the Fourier coefficients which can be determined according to the lastchapter (eqns. 4.7.3.-4.7.5). The response due to such an excitation is

=++=

111

0 )sin(),()cos(),(2

)(k

k sk

k k ck

k t k k F

DV t k k

F DV

k F

t x (4.7.14)

with the frequency ratio

0

= k k = ,...2,1k (4.7.15)

Each individual frequency is considered with its special amplification factor V and individual phase shift, which in the present case can be calculated from

22221 4)1(1),(

k k

k D

DV

+

= (4.7.16)

21

2tan

k

k k

D

= (4.7.17)

For the other cases of mass unbalance excitation or ground excitation the procedure worksanalogously. The appropriate V -functions have to be used and the correct pre-factors (which isin the present case 1/ k ) have to be used.

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