Unformatted text preview: ly damped system, ω p ≈ Ω , then expanding the single degreeoffreedom frequency response yields, 1 1 1 = k − mω 2 + jω c 2 j 2ζ k Manipulating the expression yields a form from which the halfpower bandwidth points may be calculated. 1 1 = 2 2ζ 1 1− ω
Ω 2 2 + jω c k
2 2 = 1−
+ j 2ζ 1 ω
Ω 2 2 + j 2ζ ω
Ω 2 2ζ = 1 − ω
Ω ω
Ω Squaring both sides yields ω2 ω2 ω2 ω4 ω2 8ζ = 1 − 2 + 4ζ 2 2 = 1 − 2 2 + 4 + 4ζ 2 2 Ω Ω Ω Ω Ω 2 2 ω ω 2 2 + ( 4ζ − 2 ) + (1 − 8ζ ) = 0 Ω Ω ω Solving for yields, Ω
2 − ( 4ζ 2 − 2 ) ± ω = Ω 2 2 4 2 ( 4ζ 2 − 2 ) − 4 (1 − 8ζ 2 )
2 2 ω 2 2 = 1 − 2ζ ± 2ζ 1 + ζ Ω Assuming ζ is small (i.e. ζ approximated by, 1 ), then ζ 2 ≈ 0 , and the halfpower bandwidth points are ω ≈ 1 ± 2ζ Ω 2 Lecture Notes 42 06/16/06 12:25 PM Mechanical Vibrations I Taking the difference between the points yields,
2 2 ωH − ωL Ω 2 ω + ω L ω H − ω L ≈ H ≈ 4ζ Ω Ω Using the approximation, ω H + ω L ≈ 2Ω , allows reducing the expression to, ωH − ωL ≈ 2ζ Ω Therefore the damping factor can be approximated as, ζ ≈ ωH − ωL
2Ω ∆f BW or ζ ≈ . 2 fc 2 fc This approximation is also often written as ζ ≈ While this method appears particularly simple, in practice, there are several difficulties utilizing it effectively. Specifically, when using real sampled data, the ability to determine the actual magnitude and location of H (ω p ) is limited by the measured data. The actual sampled data will not in general include the exact frequency ( ω p ) and so both the location and magnitude will be in error. Since the magnitude of H (ω p ) is used to estimate the halfpower bandwidth points, the location of ω L and ω H will be in error. In general, the errors tend to be the following: under estimating the magnitude of H (ω p ) , under estimating the magnitude of the halfpower points, estimating ω L too low, and estimating ω H too high. The compound effect of these errors is that generally the damping factor ( ζ ) is estimated too high. Lecture Notes 43 06/16/06 12:25 PM Mechanical Vibrations I FRF Example A
Given a single degreeoffreedom (SDOF) system with m = 5kg , c = 20 N ⋅s m & k = 1000 N m , assuming a harmonic excitation of magnitude 20 N @ 10 Hz , determine the response of the system to the given force. Solution
First, identify the actual steadystate operating frequency. ω = 10 Hz or 20π rad s = 62.832 rad s
Then, evaluate the SDOF frequency response function. H (ω ) = H (20π rad s ) = 1 −mω + jω c + k 1
2 2 −5kg ⋅ ( 20π rad s ) + j 20π rad s ⋅ 20 N ⋅s m + 1000 N m 1 H (20π rad s ) = ( −18,739 + j1, 257 ) N m H (20π rad s ) = ( −53.125 − j 3.563) × 10−6 m N Finally, determine the response in the frequency domain. X (ω ) = H (ω ) ⋅ F (ω ) −6 m s ) = ( −53.125 − j 3.563 ) × 10 N ⋅ 20 N X (20π rad X (20π rad s ) = ( −1.063 − j 0.071) × 10−3 m or...
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This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani  Hyderabad.
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