Unformatted text preview: sponse functions is easy to show. Simply recall the definition of the frequency response function, . For the given figure, there H (ω ) ≡ X (ω ) F (ω ) are force (input) locations and two response (output) locations. It is then clear that there are four possible frequency response functions that can be measured (or calculated.) Therefore, the expression for the frequency response must be extended to express which force/response pair is being presented. The frequency response function for multi degreeoffreedom system is X (ω ) written as H pq (ω ) ≡ p , where p is Fq (ω ) the input degree of freedom and q is the output degree of freedom. Using this definition, the multi degreeoffreedom frequency response
Lecture Notes 64 k1 m1 c1 k2 m2 c2 x1 (t ) f1 (t ) x2 (t ) f 2 (t ) 06/16/06 12:25 PM Mechanical Vibrations I functions plotted above make more sense. The plots represent a three degreeoffreedom system. If all the FRFs were plotted, there would be nine FRFs displayed, however only the three FRFs associated with the first input degreeoffreedom were actually shown, hence the notation used in the legend ( H11 , H 21 , H 31 ). Expanding the region around the third peak allows a more clear observation that the multi degreeoffreedom frequency response shown looks much like a single degreeoffreedom in the vicinity of the resonance.
H 180
11 .. H 21 .. H 31 H 0 .0 3 0 .0 2 11 .. H 21 .. H 31 90 Phas e [deg] 0 Im ag [m /N] 0 .0 1 0 0 .0 1 0 .0 2 9 0 1 8 0 8 8 .2 8.4 8.6 8 .8 9 9.2 9 .4 9.6 9.8 10 0 .0 3 8 8.2 8.4 8.6 8 .8 9 9.2 9.4 9.6 9.8 10 10 1 0 .0 3
H
11 21 M agnitude [m /N] 10 2 H H 0 .0 2 Real [m /N] 0 .0 1 0 0 .0 1 0 .0 2 H H H 11 21 31 31 10 3 10 4 10 5 8 8 .2 8.4 8.6 8 .8 9 9.2 F req u e n c y [H z ] 9 .4 9.6 9.8 10 0 .0 3 8 8.2 8.4 8.6 8 .8 9 9.2 F re q u e n c y [ H z ] 9.4 9.6 9.8 10 To estimate the modal parameters: • Natural Frequency – To identify the natural frequency ( Ω r ), it is first necessary to identify one of the following: o The ±90° phase change location (frequency) o The peak imaginary response location (frequency) o The zero real response location (frequency) • Damping – To estimate the damping, take the peak FRF magnitude, multiply by the square root of two ( 12 H (ω ) ), find the locations (frequencies), one above and one below, that have that FRF magnitude, the locations are called the halfpower points. Divide the difference of the two frequencies by twice the peak frequency (natural frequency). ζ r = ∆f 2 fc Mode Shape – To estimate the mode shape, either o Take the value of the peak imaginary part (at the natural frequency) for each output (response) location keeping the input (force) location fixed. – OR – o Take the value of the peak imaginary part (at the natural frequency) for each input (force) location keeping the output (response) location fixed. • For the third mode shown above: The natural frequency is estimated, using the peak imaginary response, as 9.0...
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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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