transient vibration

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  • 7/29/2019 Transient Vibration

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    ME 535 - Vibrations

    Transient Vibration

    Impulse: time integral of force: $ ( )F F t dt =

    If and unit impulse or function$F = 1 0

    is a unit impulse at and has the following( ) t t = properties:

    ( )

    ( )

    t tt

    t dt

    = > =

    = < <

    0

    1 00

    for allanyassumed value for

    for

    If is multiplied by any time function , as shown in the following figure,( ) t f t( )the product will be zero everywhere except where , and its time integral will be:t =

    ( ) ( )f t t dt f( ) = < <

    0 0

    Since the impulse acting on a mass will result in a sudden change in itsF dt m dv= $Fvelocity (without an appreciable change in its displacement) given by:

    $F

    m

    dv

    dt=

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    ME 535 - Vibrations

    Transient Vibration

    Page 2

    Since, under free vibration, an undamped spring-mass system will have the following response:

    ,xx

    t x tn

    n n= +&( )

    sin ( ) cos0

    0

    the response of an undamped spring-mass system initially at rest, and excited by an impulse

    is:$F

    xF

    mt Fh t

    n

    n= =$

    sin $ ( )

    where: is the response to a unit impulse.h tm

    tn

    n( ) sin=1

    For a damped spring-mass system, the response to an impulse is:

    xF

    me t Fh t

    h tm

    e t

    n

    t

    n

    n

    t

    n

    n

    n

    =

    =

    =

    $

    sin $ ( )

    ( ) sin

    11

    1

    11

    2

    2

    2

    2

    Arbitrary Excitation

    We can deal with any arbitrary excitation as a series of

    impulses:

    Strength of impulse: $ ( )F f=

    Contribution to response at t is dependent on :( )t

    ( ) ( )f h t

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    ME 535 - Vibrations

    Transient Vibration

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    where: is the response to an impulse.( )h t

    If the system is linear, then superposition holds and:

    convolution integral( ) ( ) ( )x t f h t dt

    = 0

    Example (4.2.1 pg. 92):

    Determine the response of s single DOF system to

    the step excitation shown.

    h tm

    tn

    n( ) sin=1

    ( ) ( )

    ( )

    x tF

    mt d

    F

    mt d

    x t Fk

    t

    n

    t

    n

    n

    t

    n

    n

    ( ) sin sin

    ( ) cos

    = =

    =

    000

    0

    01

    or, using Laplace transforms:

    ( )

    mx kx F u t sk

    mX s

    F

    m s

    x s

    F

    m

    s s

    k

    m

    x tF

    ktn

    && ( ) ( )

    ( )( ) cos

    + = +

    =

    =

    +

    =

    0

    2 0

    0 2

    0

    1

    1