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  • 8/15/2019 Vib-4wk1Open Access Computer Rules and Regulations V2016Open Access Computer Rules and Regulations V2016Open Access Computer Rules and Regulations V2016Open A…

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    MAE 351AE 351:: Wk 4

    CHAPTER 3CHAPTER 3

    Harmonically ExcitedHarmonically Excited Vibration Vibration

    m

    ck 

     xF(t)

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    MAE 351AE 351:: Wk 4

    3.1. General3.1. General

    • System subject to external dynamic forces

     – Forcing function, Exciting function

    • Harmonic excitation Harmonic response

    Nonperiodic excitation Transient response

    • Harmonic excitations

    • Governing equation

    (nonhomogeneous equation)

    General sol’n:  x (t ) = x h(t ) + x  p(t )

    where  x h(t ) = homogeneous sol’n of

     x  p(t ) = particular sol’n

    •  x h(t ) : exponentially decaying (initial transient) vibration

     x  p(t ) : steady-state vibration

    ( )

    ( )0

    0

    0

    ( )

    ( ) cos

    ( ) sin( )

    i t  F t F e

     F t F t 

     F t F t 

    ω φ 

    ω φ 

    ω φ 

    +=

    = +

    = +

    ( )mx cx kx F t  + + =

    0mx cx kx+ + =

    m

    ck 

     xF(t)

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    MAE 351AE 351:: Wk 4

    Homogeneous, Particular, General Solutions

    (Underdamped 1-DOF Vibration)

    S3 2 S iti P i i l

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    MAE 351AE 351:: Wk 4

    3.2. Superposition Principle3.2. Superposition Principle

    x(t) = output (response); F(t) = input (excitation)

    Introduce a linear differential operator 

    Then, G = ‘Black-box’ of the 2nd order system

    ( )mx cx kx F t  + + = 2

    2  ( )

    d x dxm c kx F t  

    dt dt  + + =

    2

    2

    d d G m c k  

    dt dt  

    ≡ + +

    [ ]( ) ( )G x t F t  =

    or 

    Consider two excitations and responses:

    F 1(t )=G [ x 1(t )], F 2 (t )=G [ x 2 (t )]

    Next, consider 

    F 3(t )=c 1F 1(t ) + c 2 F 2 (t ) : linear combination of F 1(t ) & F 2 (t )c 1,c 2 : known constants

    If x 3(t ) = c 1 x 1(t ) +c 2  x 2 (t ), the system is “linear ”: otherwise “nonlinea

    G [ x 3(t )] = G [c 1 x 1(t ) + c 2  x 2 (t )] = c 1G [ x 1(t )] + c 2 G [ x 2 (t )]= c 1F 1(t ) + c 2 F 2 (t) = F 3(t ) ; principle of superpositio

    F (t)

     x (t)

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    . . esponse o an. . esponse o an n ampen ampe ys emys emunder Harmonic Forceunder Harmonic Force

    For

    F (t ) : harmonic  x  p(t ): harmonic

    with ω with ω

    Assume  x  p

    (t ) = X cos ω

    t, then

    cosomx kx F t  ω + =

    1 2

    0,

    ( ) cos sinh n n

    n

    mx kx

     x t C t C t k 

    m

    ω ω 

    ω 

    + =

    = +

    =

    2

    0( cos cos ) cos X m t k t F t ω ω ω ω  − + =

    0

    2

     F  X 

    k mω 

    =

    m

     xF(t)

    ; Amplitude of particular 

    solution

    Q: What is the physical meaning of ‘particular solution’? 

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    x(t) = total solution

    =

    Then,

    01 2cos sin cosn n  F C t C t t  

    k mω ω ω 

    ω 2+ +

    1 2

    0 0

    0 01 0 22

    Two I.C.'s of (0)& (0) ,Let (0) & (0)

    ;n

     x x C C  x x x x

     F x

    C x C k mω ω 

    ⇒= =

    = − =−

    0 00   2

    0

    2

    ( ) cos sin

    cos

    n n

    n

     F x x t x t t k m

     F 

    t k m

    ω ω ω ω 

    ω ω 

    = − +   −  

    +

    Homogeneous

    solutionRecall: free vib.

    Recall:

    In free vib. case:

    01 0 2;

    n

     xC x C ω = = 

    General Solution

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    From  0

    02

    2   2

    0

    n

    and

    1 11

    1

    where static deflection,

    = frequency ratio ,

    Dynamic ampl.

    Static ampl.

    magnification factor 

    or amplification factor 

    or amplification ratio

     st 

     st 

    n

     st 

     st 

     F  X F k 

    k m

     X r 

     F 

     X 

    δ ω 

    δ  ω 

    ω 

    δ 

    ω ω 

    δ 

    = =−

    ⇒ = ≡−

    = =

    =

    =

    Look! X   infinity as r   1

    “RESONANCE”

    a ac e s csa ac e s cs

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    . arac er s cs. arac er s cs

     

    : 0 < ω /ωn < 1

     

    :ω /ωn > 1

    :ω /ωn = 1

     

    F (t ) & x  p(t ) are in phase

     

    out-of phase ( r > 1 ) x  p(t ) = -X cosωt 

    ( 0 < r < 1

    (r=1)

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    (r=1)

     X   ; Resonance   ω =ωn

    1 2   2

    00   2

    2

    2

    00

    ( ) cos sin cos1   ( )

    1cos sin (cos cos )

    1 ( )cos cos   ( sin ) 1

    lim lim sin (L'Hospital's rul1 ( ) 2

    2

    ( ) cos sin

    n n

     st 

    n nn

    n n st n

    n n

    nn n

    n

    n

     sn n

    n

     x t C t C t t 

     x x t t t t 

    t t    t t t t 

     x x t x t t 

    ω ω ω ω  

    δ 

    ω ω ω ω ω 

    ω ω δ ω ω  

    ω ω ω ω ω    ω 

    ω ω ω ω ω 

    ω δ 

    ω ω ω 

    → →

    = + +− /

    = + + −− /

    −   −= =

    − / −

    ⇒ = + +

    sin

    2t n

    n

    t t 

    ω ω 

    B T t l RB T t l R

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    B. Total ResponseB. Total Response

    2

    2 2 1   21 2

    1

    ( ) cos( ) cos1 ( )

    where ; tan

     st n

    n

     x t A t t 

    C  A C C 

    δ ω φ ω 

    ω ω 

    φ    −

    = − ±

    − /

    = + =

    + for ω/ωn < 1

    - for ω/ωn > 1

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    C. BeatingC. Beating

     

    Linear superposition of individual vibrations:

    1 1 1 2 2 2exp[ ( )] exp[ ( )] x A i t A i t ω φ ω φ  = + + +

    Nonperiodic (aperiodic) oscillations in general

    When ω2 = ω1 + ω, then1 2 1 1( ) ( )

    1 2[ ]i i t i t i t   x A e A e e Aeφ φ ω ω ω φ  +∆ += − ≡

    2 2 1/ 2

    1 2 1 2[ 2 cos( )] A A A A t φ φ ω 1 2= + + − − ∆

    1   1 1 2 2

    1 1 2 2

    sin sin( )

    tan s cos( )

     A t 

    co A t  

    φ φ ω 

    φ  φ φ ω 

    −   + + ∆

    = + + ∆

    where

    ; Approximately simple harmonic vibratio

    with slowly varying ampl. A and phase φ

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    Consider a special case of A1 = A2 and φ1 = φ2 = 0

    Then,

    1/ 21 1

    1

    1 2

    (2 2cos ) 0 2

    sintan

    1 cos

    Beat frequency b

     A t A A

     f f f f  

    ω 

    ω φ 

    ω 

    ω 

    = + ∆ ≤ ≤∆ = + ∆

    ≡ = ∆ = − = ∆

     

    (Forcing freq., ω) ≈ (Natural freq., ωn (system))

    0 0

    0

    2

    0

    2

    0,

    ( / )( ) (cos cos )

    ( / )2sin sin

    2

    n

    n

    n n

    n

    When x x

     F m x t t t 

     F mt t 

    ω ω ω ω 

    ω ω ω ω  

    ω ω 

    2

    2

    = =

    = −−

    + −= ⋅

    − 2

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    As ωn - ω = ω ≡ 2  (  : very small),

    ω +ωn ≈ 2ω & ωn2 - ω 2 = (ωn + ω)(ωn - ω ) ≈ 4 ω

    Then,0 /( ) sin sin

    2

     F m x t t t ε ω 

    εω 

    ≈ 

    variable amplitude2 2 2

     period of beating

    2

    b

    n

    b n

    π π π τ 

    ε ω ω ω  

    ω ε ω ω ω  

    ≡ = = =2 − ∆

    = = − = ∆