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Unformatted text preview: pect to frequency as follows. TODO: Show example F(w), show resulting X(w). (FRF continuous, F&X discrete) Express F(w) = CONST …. Show units of FRF = 1/k, compliance. Work example. Show MDOF FRF for comparison. Lecture Notes 3906/16/06 12:25 PM * Mechanical Vibrations I d d ( H (ω ) ) = dω dω 1 ( k − mω ) + (ω c )
2 2 2 1 2 2 d d 2 ( k − mω 2 ) + (ω c ) H (ω ) ) = ( dω dω ( ) − 2 2 2 2 d ( H (ω ) ) = − 1 ( k − mω 2 ) + (ω c )2 2 dω ( ) − 3 2 d dω (( k − mω ) + (ωc ) ) 2 d ( H (ω ) ) = − 1 ( k − mω 2 ) + (ω c )2 dω 2 ( ) − 3 2 d 2 ( k − mω 2 ) + 2 (ω c ) dd (ω c ) 2 ( k − mω ) dω ω 2 d ( H (ω ) ) = − 1 ( k − mω 2 ) + (ω c )2 2 dω ( ) ( 2 ( k − mω ) ( −2m) + 2 (ωc ) c )
− 3 2 2 In order to satisfy the maximum (or minimum), the derivative of the magnitude of the d frequency response function, H (ω p ) = 0 , must equal zero. Since only the second dω term in the above relation can be equal to zero (except at ω = ∞ ), the solution can be reduced to solving the following equation. ( ) ( 2 ( k − mω ) ( −2mω ) + 2 (ω c ) c ) = 0
2 p p p ( −2 ( km − m ω ) + c )ω
2 2 p 2 p =0 By dividing by the mass, m 2 , the expression can be simplified by recognizing that k c Ω 2 = , 2ζΩ = and ω p = 0 cannot be the peak response for all possible MCK m m combinations.
k c 2 −2 − ω p + = 0 m m
2 −2 ( Ω 2 − ω 2 ) + ( 2ζΩ ) = 0 p
2 2 Ω 2 − ω p − 2ζ 2Ω 2 = 0 ω 2 = (1 − 2ζ 2 ) Ω 2 p Lecture Notes 40 06/16/06 12:25 PM Mechanical Vibrations I Therefore, the frequency of maximum (peak) response is as follows. ω p = 1 − 2ζ 2 Ω
For comparison, recognize that the damped natural frequency is ω r = 1 − ζ 2 Ω r . Resonant Response
Starting with the definition of resonance, ω = Ω , (i.e. the response of the system to an excitation at the undamped natural frequency of the system.), the amplitude of the system response can be calculated. Starting with the single degreeoffreedom frequency response function, H (ω ) = 1 −mω + jω c + k
2 from the definition of resonance, ω = Ω , (and k = mΩ 2 ) the resonant response, H (Ω) , equals
H (Ω) = 1 1 1 1 1 = = = = jΩc jΩζ cc jΩζ 2 km jΩ 2ζ k j 2ζ k Ω If the damping is “small” (i.e. ζ < 0.1 ), then the peak response can be approximated as H (ω p ) ≈ H (Ω) = 1 j 2ζ k Half Power Method
After defining the resonant frequency, it is possible to estimate the damping, ζ , from the shape of the frequency response function in the vicinity of the resonance. Again, assuming that the damping is “small” (i.e. ζ < 0.1 ), the halfpower bandwidth points can be calculated and then damping can be estimated from the bandwidth and the resonance location. Starting with the definition of the halfpower bandwidth points,
H ωp ( ) H ωp
2 ( ) Lecture Notes 41ωL ω pωH 06/16/06 12:25 PM Mechanical Vibrations I 1 2 H (ω p ) = H (ωL ) = H (ωH ) and recognizing that for a light...
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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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