Vibroacoustic Simulation. Alexander Peiffer

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Vibroacoustic Simulation - Alexander Peiffer

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2ζω0 ηω0 ω0/Q 1 12τ Decay time τ 1/ζω0 2/ω0η 2Q/ω0 2/Δω0 1

      In tools and software for vibroacoustic simulations many different quantities are used. The overview of all those different criteria shall help to avoid mistakes and confusion.

      1.3 Two Degrees of Freedom Systems (2DOF)

       StartLayout 1st Row m 1 ModifyingAbove u 1 With two-dots plus c Subscript v Baseline 1 Baseline ModifyingAbove u 1 With dot plus k Subscript s Baseline 1 Baseline u 1 plus k Subscript s c Baseline left-parenthesis u 1 minus u 2 right-parenthesis equals upper F Subscript x Baseline 1 EndLayout (1.69)

       StartLayout 1st Row m 2 ModifyingAbove u 2 With two-dots plus c Subscript v Baseline 2 Baseline ModifyingAbove u 1 With dot plus k Subscript s Baseline 2 Baseline u 2 plus k Subscript s c Baseline left-parenthesis u 2 minus u 1 right-parenthesis equals upper F Subscript x Baseline 2 EndLayout (1.70)

      Figure 1.11 Two degrees of freedom system. Source: Alexander Peiffer.

      By introducing harmonic motion for u1=u1ejωt and u2=u2ejωt we get

       StartLayout 1st Row 1st Column left-parenthesis minus omega squared m 1 plus j omega c Subscript v Baseline 1 Baseline plus k Subscript s Baseline 1 Baseline plus k Subscript s c Baseline right-parenthesis bold-italic u 1 minus k Subscript s c Baseline bold-italic u 2 2nd Column equals 3rd Column bold-italic upper F Subscript x Baseline 1 EndLayout (1.71)

       StartLayout 1st Row 1st Column minus k Subscript s c Baseline bold-italic u 1 plus left-parenthesis minus omega squared m 2 plus j omega c Subscript v Baseline 2 Baseline plus k Subscript s Baseline 2 Baseline plus k Subscript s c Baseline right-parenthesis bold-italic u 2 2nd Column equals 3rd Column bold-italic upper F Subscript x Baseline 2 EndLayout (1.72)

      neglecting the time dependence ejωt. It is practical to write this in matrix form:

      In the following, the square brackets and the curly brackets denote a coefficient matrix and vector, respectively.

      1.3.1 Natural Frequencies of the 2DOF System

      We start with a simplified system without damping and external forces in order to get the natural frequencies of the system.

       Start 2 By 2 Matrix 1st Row 1st Column minus omega squared m 1 plus k Subscript s Baseline 1 Baseline plus k Subscript s c Baseline 2nd Column minus k Subscript s c Baseline 2nd Row 1st Column minus k Subscript s c Baseline 2nd Column minus omega squared m 2 plus k Subscript s Baseline 2 Baseline plus k Subscript s c Baseline EndMatrix StartBinomialOrMatrix bold-italic u 1 Choose bold-italic u 2 EndBinomialOrMatrix equals StartBinomialOrMatrix 0 Choose 0 EndBinomialOrMatrix (1.74)

       StartLayout 1st Row 1st Column Start 2 By 2 Matrix 1st Row 1st Column k Subscript s Baseline 1 Baseline plus k Subscript s c 2nd Column minus k Subscript s c 2nd Row 1st Column minus k Subscript s c 2nd Column k Subscript s Baseline 2 Baseline plus k Subscript s c EndMatrix StartBinomialOrMatrix bold-italic u 1 Choose bold-italic u 2 EndBinomialOrMatrix minus omega squared Start 2 By 2 Matrix 1st Row 1st Column m 1 2nd Column 0 2nd Row 1st Column 0 2nd Column m 2 EndMatrix StartBinomialOrMatrix bold-italic u 1 Choose bold-italic u 2 EndBinomialOrMatrix 2nd Column equals ellipsis 2nd Row 1st Column left-bracket Start 1 By 1 Matrix 1st Row upper K EndMatrix minus omega squared Start 1 By 1 Matrix 1st Row upper M EndMatrix right-bracket Start 1 By 1 Matrix 1st Row bold-italic u EndMatrix 2nd Column equals StartBinomialOrMatrix 0 Choose 0 EndBinomialOrMatrix EndLayout (1.75)

      The non trivial solutions of this are given by:

       det left-brace Start 1 By 1 Matrix 1st Row upper K EndMatrix minus omega Subscript n Superscript 2 Baseline Start 1 By 1 Matrix 1st Row upper M EndMatrix right-brace equals 0 (1.76)

      This leads to the characteristic equation with λ=ω2

       m 1 m 2 lamda squared minus left-bracket k Subscript s Baseline 1 Baseline m 2 plus k Subscript s Baseline 2 Baseline m 1 plus k Subscript s c Baseline left-parenthesis m 1 plus m 2 right-parenthesis right-bracket lamda plus k Subscript s Baseline 1 Baseline k Subscript s Baseline 2 Baseline plus left-parenthesis k Subscript s Baseline 1 Baseline plus k Subscript s Baseline 2 Baseline right-parenthesis k Subscript s c Baseline equals 0 (1.77)

      With ω12=ks1/m1, ω12=ks1/m1 and ωc2=ksc(m1+m2)m1m2 the solutions are:

       omega Subscript n Baseline 1 slash n Baseline 2 Superscript 2 Baseline equals StartFraction omega 1 squared plus omega 2 squared plus omega Subscript c Superscript 2 Baseline Over 2 EndFraction plus-or-minus StartRoot StartFraction left-parenthesis omega 1 squared plus omega 2 squared plus omega Subscript c Superscript 2 Baseline right-parenthesis squared Over 4 EndFraction minus omega 1 squared omega 2 squared plus omega Subscript c Superscript 2 Baseline StartFraction omega 1 squared m 1 plus omega 2 squared m 2 Over m 1 plus m 2 EndFraction EndRoot (1.78)

      The eigenvalues shall be entered into the equations to solve for {Ψi}.

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