1225 pm mechanical vibrations i the interesting

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Unformatted text preview: tio as: c ζ = cc We can now express the damping and damped natural frequency as functions of the undamped natural frequency and damping ratio, as follows: Lecture Notes -2106/16/06 12:25 PM Mechanical Vibrations I σr = − c = −ζ r Ω r 2m 2 k c 2 ωr = ± − = ± 1 − ζ r Ωr m 2m The single degree of freedom equation can now be written as: 2 x(t ) + 2ζ r Ω r x(t ) + Ω r x(t ) = f (t ) m While each of these parameters has been developed from a single degree of freedom (1 DOF) point of view, the concepts are equally applicable to the multi-degree of freedom case, as well. For this, we need to use a nomenclature that reflects the fact that there is more than one system pole (with its conjugate). We do this by using the subscript r (indicating resonance) with each of the modal parameters, as follows. λr = σ r + jω r is the system pole for mode r, where σ r is the damping and ω r is the damped natural frequency for mode r. From this, we define Ω r = σ r2 + ω r2 as the undamped natural frequency for mode r. Then, the damping ratio is simply defined as, ζr = − σr Ωr . The general homogeneous solution then takes on the following form. { x(t )} = ∑ { X }r e r λr t + { X *} e r λ* t r The outgrowth of this approach is that it is no longer necessary to know the actual mass, stiffness and damping of the system in order to identify its modal characteristics. This is a significant advantage when applied to experimental modal analysis because, as will be shown later, the system poles ( λr ’s) can be obtained directly from measured Frequency Response Functions. Lecture Notes -22- 06/16/06 12:25 PM Mechanical Vibrations I Homogeneous Example Given a single degree-of-freedom mass/spring/damper system (MCK) with 5kg mass, 20 N ⋅s m damping, and 1000 N m stiffness: identify the pole of the system ( λr ), the undamped natural frequency ( Ω r ), and the damping ratio ( ζ r ). To calculate the system pole, evaluate the characteristic equation, ms 2 + cs + k = 0 5s 2 + 20 s + 1000 = 0 which has the following roots. Note that the roots of the characteristic equation are the system poles. 1000 N m 20 N ⋅s m −20 N ⋅s m ±j − λr = 2 ⋅ 5kg 5kg 2 ⋅ 5kg 2 λr = −2 rad s ± j 200 rad 2 s2 − ( 2 rad s ) 2 λr = −2 ± j14 rad s The pole gives the damping ( σ r ) and the damped natural frequency ( ω r ) directly. Recall, that to express the answer in units of Hertz, it is necessary to divide by 2π . σ r = −2 rad s -or- − Hz = −0.318Hz π 7 ω r = ±14 rad s -or- ± Hz = ±2.228Hz π For this problem, the undamped natural frequency ( Ω r ) may be calculated from either the magnitude of the system pole, or by evaluating the characteristic equation for c = 0 . Both approaches yield the same solution. Ω r = ± λr = ± −2 ± j14 rad s = ± 200 rad Ωr = ± 2 1 s2 = ±14.14 rad s -or- 2.251Hz k 1000 N m 2 =± = 200 rad s2 = ±14.14 rad s -or- 2.251Hz m 5kg Again, for this problem, the damping ratio may be...
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