Vibroacoustic Simulation. Alexander Peiffer
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Название: Vibroacoustic Simulation

Автор: Alexander Peiffer

Издательство: John Wiley & Sons Limited

Жанр: Отраслевые издания

Серия:

isbn: 9781119849865

isbn:

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       Start 1 By 1 Matrix 1st Row Start 1 By 1 Matrix 1st Row upper K EndMatrix minus omega Subscript n i Superscript 2 Baseline Start 1 By 1 Matrix 1st Row upper M EndMatrix EndMatrix Start 1 By 1 Matrix 1st Row normal upper Psi Subscript i Baseline EndMatrix equals StartBinomialOrMatrix 0 Choose 0 EndBinomialOrMatrix with i equals 1 comma 2 (1.79)

      leading to the surprisingly simple eigenvalues after some painful math

       StartLayout 1st Row Start 1 By 1 Matrix 1st Row normal upper Psi Subscript i Baseline EndMatrix equals StartBinomialOrMatrix StartFraction k Subscript s Baseline 1 Baseline plus k Subscript s c Baseline minus omega Subscript n i Superscript 2 Baseline m 1 Over k Subscript s c Baseline EndFraction Choose 1 EndBinomialOrMatrix EndLayout (1.80)

      The results present the modes of the system or the shape of movement for this natural frequency. Let us simplify the above expression by additional conditions: ks1=ks2=ks and m1=m2=m.

      So the modal frequencies read with ω02=ks/m

       omega Subscript n Baseline 1 slash n Baseline 2 Superscript 2 Baseline equals omega 0 squared plus StartFraction omega Subscript c Superscript 2 Baseline Over 2 EndFraction minus-or-plus StartFraction omega Subscript c Superscript 2 Baseline Over 2 EndFraction (1.81)

      with the eigenvectors:

normal upper Psi 1 equals StartBinomialOrMatrix 1 Choose 1 EndBinomialOrMatrix normal upper Psi 2 equals StartBinomialOrMatrix negative 1 Choose 1 EndBinomialOrMatrix

      Figure 1.12 Mode shapes of the 2DOF example. Source: Alexander Peiffer.

      When we enter numerical figures with m = 0.1 kg, ks=10 N/m and ksc=2 N/m we get the modal frequencies ωn1=10.0 s−1(fn1=1.59 Hz) and ωn2=11.83 s−1(fn2=1.88 s−1).

      1.3.1.1 Forced Vibration of the 2DOF System

       Start 1 By 1 Matrix 1st Row bold-italic upper D left-parenthesis omega right-parenthesis EndMatrix Start 1 By 1 Matrix 1st Row bold-italic u left-parenthesis omega right-parenthesis EndMatrix equals Start 1 By 1 Matrix 1st Row bold-italic upper F Subscript x Baseline left-parenthesis omega right-parenthesis EndMatrix (1.82)

      The solution can be written using the inverse of the stiffness matrix

      Figure 1.13 Magnitude and phase of response to unit force at mass 1. Source: Alexander Peiffer.

      1.3.1.2 Dynamic Vibration Absorber

      Figure 1.14 DVA mounted on resonant master system. Source: Alexander Peiffer.

       Start 2 By 2 Matrix 1st Row 1st Column minus omega squared m plus k Subscript s b Baseline plus k Subscript s Baseline plus j omega c Subscript v Baseline 2nd Column minus k Subscript s Baseline minus j omega c Subscript v Baseline 2nd Row 1st Column minus k Subscript s Baseline j omega c Subscript v Baseline 2nd Column minus omega squared m Subscript s Baseline plus j omega c Subscript v Baseline plus k Subscript s Baseline EndMatrix StartBinomialOrMatrix bold-italic u 1 Choose bold-italic u 2 EndBinomialOrMatrix equals StartBinomialOrMatrix bold-italic upper F Subscript x Baseline 1 Baseline Choose 0 EndBinomialOrMatrix (1.84)

      with the following transfer function

       bold-italic u 1 equals minus StartFraction k left-parenthesis i c Subscript v Baseline omega plus k Subscript s Baseline minus m Subscript s Baseline omega squared right-parenthesis Over left-parenthesis i c Subscript v Baseline omega plus k Subscript s Baseline right-parenthesis squared minus left-parenthesis i c Subscript v Baseline omega plus k Subscript s Baseline minus m Subscript s Baseline omega squared right-parenthesis left-parenthesis i c Subscript v Baseline omega plus k Subscript s b Baseline plus k Subscript s Baseline minus m omega squared right-parenthesis EndFraction bold-italic upper F Subscript x Baseline 1 (1.85)

      Assuming zero damping gives the characteristic equation for the combined resonances

       left-parenthesis k Subscript s Baseline minus omega Subscript i Superscript 2 Baseline m Subscript s Baseline right-parenthesis left-parenthesis k Subscript s b Baseline plus k Subscript s Baseline minus omega Subscript i Superscript 2 Baseline m right-parenthesis minus k Subscript s Superscript 2 Baseline equals 0 (1.86)

      With the resonance frequencies СКАЧАТЬ