c9.distributed mass and elasticity2 compatibility mode

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  • 7/29/2019 C9.DIstributed Mass and Elasticity2 Compatibility Mode

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    DYNAMIC OF STRUCTURES

    CHAPTER 9

    SYSTEMS WITH DISTRIBUTED MASS

    AND ELASTICITY

    Department of civil engineering, University of North Sumatera

    Ir. DANIEL RUMBI TERUNA, MT;IP-U

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    We return to the partial diffential equation of simply beamsubjected to force

    The displacement is given by a linier combination of the modes:

    ( )( )

    ( )( )

    ( ) )1(,,,

    2

    2

    2

    2

    2

    2

    txpt

    txuxm

    x

    txuxEI

    x=

    +

    Subtituting Eq.(2) in Eq.(1) gives

    ( )txp ,

    ( ) ( ) ( ) )2(,1

    tqxtxu rr

    r

    =

    =

    ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) )3(,11

    txptqxxEItqxxm rr

    rr

    r

    r =+

    =

    =

    &&

    Multiply each term by , integrate it over the length of the beam gives( )xn

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) )4(,01 01 0

    dxxtxpdxxxEIxtqdxxxxmtqL

    nr

    r

    L

    nrr

    r

    L

    nr =+

    =

    =

    &&

    This equation can be written as

    nr =By use orthogonality properties of modes, all term left side vanish exceptthe one term for which . Thus, Eq.(4) becomes

    ( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ] ( ) ( ) )5(,00

    2

    0

    =+

    L

    nn

    L

    nn

    L

    nn dxxtxpdxxxEIxtqdxxxmtq &&

    ( ) ( ) ( ) )6(tptqKtqM nnnnn =+&&where

    ( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] dxxxEIxdxxxEIxKdxxxmM nL

    nn

    L

    nn

    L

    nn

    2

    00

    2

    0

    , ===

    ( ) ( ) ( ) )7(,0

    =L

    nn dxxtxptP

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    Equation (6) governs the modal coordinate , and the generalized

    properties , , and depend only on the mode .Thus we have infinite number of equation like Eq.(6), one for each mode. Thepartial diffrential equation equation (1) in the unknown fungtion hasbeen transformed to an infinite set of ordinary diffrential equation (6) inunknowns .

    Once the have been determined, the contribution of the mode to the

    displacement is given by:

    The generalized mass and generalized stiffness for modeare related

    ( )txu ,

    nM nKthn

    )8(2 nnn MK =

    ( )tqnthnnM nK tPn

    th

    n ( )xn

    tqntqn

    ( ) ( ) ( ) )9(, tqxtxu nnn =

    thn

    txu ,

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    ( ) ( ) ( ) ( ) )10(,,11

    tqxtxutxu nn

    n

    n

    n

    =

    =

    ==

    ( )xu

    ( ) ( ) ( ) ( ) ( ) ( )[ ] )11(== xuxEIxVxuxEIxM

    4. Specialize

    The bending moment and shear force at any section along the length of thebeam are related to displacement as follows:

    The static relationships apply at each instant of time with replaced by

    which is given by Eq.(10). Thus

    xu

    ( )txu ,

    ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) )12(,,11

    =

    =

    ==n

    nnn

    nn tqxxEItxVtqxxEItxM

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    EXAMPLE

    ( ) )(sin2

    22

    aL

    xnx

    m

    EI

    L

    nnn

    ==

    0p

    L

    EIm,

    x

    0p

    t

    SOLUTION

    1. Determined the natural frequencies and modes

    2. Determined the generalized mass, stiffness and force

    ( ) ( )[ ] )(2

    sin0

    2

    0

    bmL

    dxL

    xnmdxxxmM

    LL

    nn =

    ==

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    EXAMPLE

    ( ) ( ) ( )[ ] )(2

    sin3

    442

    0

    4

    44

    0

    cL

    EIndx

    L

    xn

    L

    nEIdxxxEIxK

    L

    n

    L

    nn

    =

    =

    =

    ( ) ( ) ( ) )(, 00

    epdxxtxpP n

    L

    nn ===

    3. Set up modal equation

    The n th modal equation is

    ( ) ( )( )

    n

    nnnn

    nnnnn

    M

    ptqtqor

    ftptqKtqM

    0

    2

    )(

    =+

    =+

    &&

    &&

    )(2

    3

    442 d

    L

    EInMKor nnn

    ==

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    EXAMPLE

    ( ) ( ) ( ) )(sincos 0 gK

    ptBtAtq

    n

    nnnnnn

    ++=

    ( ) 00 =nq

    4. Solved modal equation

    If at , and , we get0=t00

    =nq&

    )(00 hBK

    pA n

    n

    nn =

    =

    Subtituting Eq.(h) in Eq.(g) lead to

    ( ) ( ) ( ) ( ) )(2cos1 443

    00 inEI

    LptK

    ptq nnn

    nn

    ==

    Subtituting Eq.(i) in Eq.(10) , we obtained the displacement response txu ,

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    ( )( )

    ( ) )(sincos12/2

    ,1

    44

    3

    0 hL

    xnt

    n

    L

    EI

    Lptxu

    nn

    n

    =

    =

    4. Specialiaze for

    where

    )(

    ,...11,7,31

    ,...9,5,11

    ,...6,4,20

    2j

    n

    n

    nL

    n

    =

    =

    =

    =

    2/L=

    Subtituting Eq. (j) in Eq.(h) gives

    ( )

    )(...........7

    sin2401

    cos15sin

    625

    cos1

    3sin

    81

    cos1sin

    1

    cos12,

    75

    31

    4

    3

    0

    kL

    xt

    L

    xt

    L

    xt

    L

    xt

    EI

    Lptxu

    +

    +

    =

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    Displacement at mid span is

    The displacement for the first mode contribution only can be expressed as

    )(...........2401

    cos1

    625

    cos1

    3sin

    81

    cos1

    1

    cos12,

    2

    75

    31

    4

    3

    0

    ltt

    L

    xtt

    EI

    Lpt

    Lu

    +

    +

    +

    +

    =

    ( ) )(cos16.48

    ,2

    1

    3

    0 mtEI

    Lpt

    Lu =

    dynamic displacementamplication factor

    Staticdisplacement

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    ( )

    )(...........7sin49cos15sin

    25cos1

    3sin

    9

    cos1sin

    1

    cos12,

    75

    31

    2

    0

    kL

    xtL

    xt

    L

    xt

    L

    xtLptxM

    +

    +

    =

    The bending moment are obtained by subtituting Eq.(h) in Eq.(12)

    The bending moment at mid span is

    )(...........49

    cos1

    25

    cos1

    9

    cos1

    1

    cos12

    ,2

    75

    31

    2

    3

    0

    ott

    s

    ttLp

    t

    L

    M

    +

    +

    +

    +

    =

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    MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

    The bending moment for the first mode contribution only can be expressed as

    ( ) )(cos193.4

    ,2

    10 pt

    EI

    Lpt

    LM =

    Moment dynamicamplication factor

    Static bendingmoment

    Solution for by use Duhamel integral

    ( ) ( ) ( ) ( ) ( )

    dtpm

    dtpm

    tqn

    t

    nn

    n

    t

    nn

    ==

    sin

    1sin

    1

    00

    0

    tqn

    ( ) ( )[ ] ( )tn

    pt

    m

    ptq n

    nt

    n

    n

    nn

    cos1

    2cos

    44

    0

    02

    0 ==