Unformatted text preview: calculated by either of two approaches, either directly from the system pole or from the system’s mechanical properties (MCK). Lecture Notes -23- 06/16/06 12:25 PM Mechanical Vibrations I −2 rad s = 0.1414 -or- 14.14% 14.14 rad s Ωr c c 20 N ⋅s m ζr = = = = 0.1414 -or- 14.14% cc 2 mk 2 5kg ⋅1000 N m ζr = − σr =− Recognize one important point in the solution method. The direct (MCK) solutions are only valid for a single degree-of-freedom (SDOF) system while the solution methods based upon the system pole are valid for a system with any number of degrees-offreedom (MDOF). Lecture Notes -24- 06/16/06 12:25 PM Mechanical Vibrations I #5 – Estimating Modal Parameters – Time Domain
There are many techniques, both simple and advanced, for estimating modal parameters (frequency, damping, mode shape, and scaling) from measured experimental data. This section will focus upon the single degree-of-freedom time domain technique known as log decrement. Log Decrement
The log decrement method is based entirely upon the characteristics of a single degreeof-freedom free decay response.
4 3 2 1 A m plitude 0 -1 -2 -3 -4 0 1 2 3 4 5 6 7 8 Tim e (s ec ) 9 10 11
12 13 14 15 Recalling the form of solution, x(t ) = Xeλrt , it is possible to estimate the system pole ( λr = σ r + jω r ) directly from the observed response. Estimating the damped natural frequency ( ω r ) is easy. First estimate the period of one oscillation ( τ ), the reciprocal of the period, 1 , is the damped natural frequency ( ω r ) in τ Hz. ω r = 2π τ rad s -or- 1τ Hz
Estimating the damping ( σ r ) requires a little more effort. The development proceeds as follows. Begin by evaluating the position, x(t ) , for two time instants one cycle apart. Lecture Notes -25- 06/16/06 12:25 PM Mechanical Vibrations I x(t1 ) = Xeλrt1 = Xeσ rt1 e jω rt1 x(t2 ) = Xeλrt2 = Xeσ r t2 e jω rt2 Next evaluate the log decrement ( δ ) as: δ = ln x(t1 ) Xeσ rt1 e jωrt1 = ln σ rt2 jωrt2 Xe e x(t2 ) However, since the points are defined one cycle apart, t2 = t1 + τ . Substituting into the above expression yields, Xeσ rt1 e jωrt1 δ = ln σ r (t1+τ ) jωr (t1+τ ) e Xe Next, recall that e jθ = cosθ + j sin θ . Hence,
jω r ( t1 +τ ) =e jω r t1 + 2π ( ωr ) = e jω t +2π = e jω t e j 2π = e jω (t )
r1 r1 r 1 So the expression for the log decrement can be reduced to δ = ln eσ rt1 −σ t +τ = ln eσ rt1 e r ( 1 ) = ln ( e −σ rτ ) = −σ rτ σ r ( t1 +τ ) e ( ) Recalling that σ r = −ζ r Ω r , ω r = 1 − ζ r2 Ω r and ω r = 2π expression yields, τ . Substituting into the above
2πζ r 1 − ζ r2 δ = −σ rτ = ζ r Ω rτ = ζ r Ω r 2π ω =
r 2πζ r Ω r 1 − ζ Ωr
2 r = Therefore the log decrement equals, δ= 2πζ r 1 − ζ r2 Multiple Cycle Solution
If time points multiple (n) periods apart are used, the expression for the log decrement is modified as follows: 1 x(t ) 2πζ r δ = ln 0 = n x(tn ) 1 − ζ r2 where: n is the number of c...
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- Fall '11
- homogeneous solution