Vibroacoustic Simulation. Alexander Peiffer

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Vibroacoustic Simulation - Alexander Peiffer


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as would be the case with viscous damping. Examples for damping processes are:

       structural or hysteretic damping

       coulomb or dry-friction damping

       velocity-squared or aerodynamic drag damping.

normal upper Delta upper E Subscript cycle Baseline equals integral Subscript 0 Superscript upper T Baseline c Subscript v Baseline ModifyingAbove x With dot StartFraction d x Over d t EndFraction d t equals c Subscript v Baseline ModifyingAbove u With caret squared omega squared integral Subscript 0 Superscript 2 pi slash omega Baseline s i n squared left-parenthesis omega t plus phi right-parenthesis d t equals pi c Subscript v Baseline omega ModifyingAbove u With caret squared

      With the dissipated energy per cycle ΔEcycle for any particular damping type the equivalent viscous damping cveq can be determined from

      1.2.5.1 Hysteretic Damping

      In many cases damping is caused by structural damping. If materials like aluminium or steel are cyclically stressed they form a hysteresis loop. Experimental observations show, that the energy dissipated per cycle is proportional to the square of the strain (displacement):

      Comparing this with Equation (1.56) gives:

       c Subscript v normal e normal q Baseline equals StartFraction alpha Subscript x Baseline Over pi omega EndFraction (1.58)

      Entering this into the equation of motion in complex form (1.27)

      and rewriting (1.59) leads to

       minus m omega squared bold-italic u e Superscript j omega t Baseline plus k Subscript s Baseline left-parenthesis 1 plus j eta right-parenthesis bold-italic u e Superscript j omega t Baseline equals bold-italic upper F Subscript x Baseline e Superscript j omega t (1.60)

      with the structural loss factor

      and

       bold-italic k Subscript s Baseline equals k Subscript s Baseline left-parenthesis 1 plus j eta right-parenthesis (1.62)

      In this case the displacement response reads:

       StartFraction ModifyingAbove u With caret Over ModifyingAbove u With caret Subscript 0 Baseline EndFraction equals StartFraction 1 Over StartRoot left-bracket 1 minus left-parenthesis omega slash omega 0 right-parenthesis squared right-bracket squared plus eta squared EndRoot EndFraction (1.64)

      and

       phi 0 equals arc tangent StartFraction negative eta Over 1 minus left-parenthesis omega slash omega 0 right-parenthesis squared EndFraction (1.65)

      Amplitude and phase resonance occur at the same frequency ω0. At resonance the viscously damped system amplification is u^/u^0=1/2ζ. Hence

       eta equals 2 zeta equals StartFraction 1 Over upper Q EndFraction (1.66)

      There is a further interpretation of the loss factor. ΔEcycle is the energy dissipated per cycle. The dissipated power given by Πdiss=ΔEd/T=ΔEcycleω/2π. Using equations (1.57) and (1.61) we get:

       normal upper Pi Subscript diss Baseline equals eta omega one-half k Subscript s Baseline ModifyingAbove upper X With caret squared equals eta omega upper E (1.67)

      At the beginning of the cycle the total energy E is stored as potential energy in the spring. The dissipated power is a product of damping loss, frequency and the total energy of the system. This will be frequently used in the following sections, but particularly in Chapter 6 about statistical energy methods. The energy aspect leads to an equivalent definition of the loss factor:

       eta equals StartFraction 1 Over 2 pi EndFraction StartFraction normal upper Delta upper E Subscript cycle Baseline Over upper E EndFraction (1.68)

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