Top Banner

of 20

Vibration Calculation All

Jun 03, 2018

Download

Documents

Kha Phuc
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
  • 8/12/2019 Vibration Calculation All

    1/20

    99

    0 5 10 15 20 25 30 35 40-3

    -2

    -1

    0

    1

    2

    3

    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

  • 8/12/2019 Vibration Calculation All

    2/20

    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 =++

    100

  • 8/12/2019 Vibration Calculation All

    3/20

    (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)( .

    101

  • 8/12/2019 Vibration Calculation All

    4/20

    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.

    102

  • 8/12/2019 Vibration Calculation All

    5/20

    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

    103

  • 8/12/2019 Vibration Calculation All

    6/20

  • 8/12/2019 Vibration Calculation All

    7/20

    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)

    105

  • 8/12/2019 Vibration Calculation All

    8/20

    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):

    106

  • 8/12/2019 Vibration Calculation All

    9/20

    ( ) ( ) 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)

    107

  • 8/12/2019 Vibration Calculation All

    10/20

    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.

    108

  • 8/12/2019 Vibration Calculation All

    11/20

  • 8/12/2019 Vibration Calculation All

    12/20

    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.

    110

  • 8/12/2019 Vibration Calculation All

    13/20

    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.

    111

  • 8/12/2019 Vibration Calculation All

    14/20

    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=

  • 8/12/2019 Vibration Calculation All

    15/20

    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

    113

  • 8/12/2019 Vibration Calculation All

    16/20

    ( ) ( 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 .

    114

  • 8/12/2019 Vibration Calculation All

    17/20

    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)

    115

  • 8/12/2019 Vibration Calculation All

    18/20

    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:

    116

  • 8/12/2019 Vibration Calculation All

    19/20

    ==

    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

    117

  • 8/12/2019 Vibration Calculation All

    20/20

    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.