Rayleigh’s Classical Damping Revisited S. Adhikari and A. Srikantha Phani Department of Aerospace Engineering, University of Bristol, Bristol, U.K. Email: S.Adhikari@bristol.ac.uk URL: http://www.aer.bris.ac.uk/contact/academic/adhikari/home.html Classical Damping Revisited – p.1/26 B E College, India, January 2007
Bristol Aerospace Classical Damping Revisited – p.2/26 B E College, India, January 2007
Outline of the presentation Introduction Background of proportionally damped systems Generalized proportional damping Damping identification method Examples Summary and conclusions Classical Damping Revisited – p.3/26 B E College, India, January 2007
Introduction Equation of motion of viscously damped systems: y ( t ) + C˙ y ( t ) + Ky ( t ) = f ( t ) M¨ Proportional damping (Rayleigh 1877) C = α 1 M + α 2 K Classical normal modes Simplifies analysis methods Identification of damping becomes easier Classical Damping Revisited – p.4/26 B E College, India, January 2007
Limitations of proportional damping The modal damping factors: � α 1 � ζ j = 1 + α 2 ω j 2 ω j Not all forms of variation can be captured Classical Damping Revisited – p.5/26 B E College, India, January 2007
Damped Beam Example Damped free-free beam: L = 1 m, width = 39 . 0 mm thickness = 5 . 93 mm Classical Damping Revisited – p.6/26 B E College, India, January 2007
Damping factors −1 10 experiment fitted Pproportional damping Modal damping factor −2 10 −3 10 0 200 400 600 800 1000 1200 1400 1600 1800 Frequency (Hz) Classical Damping Revisited – p.7/26 B E College, India, January 2007
Our Objective Can we improve the Classical Damping proposed by Lord Rayleigh in 1877 so that we can take account of the frequency variation of the damping factors? Classical Damping Revisited – p.8/26 B E College, India, January 2007
Conditions for proportional damping Theorem 1 A viscously damped linear system can possess classical normal modes if and only if at least one of the following conditions is satisfied: (a) KM − 1 C = CM − 1 K , (b) MK − 1 C = CK − 1 M , (c) MC − 1 K = KC − 1 M . This can be easily proved by following Caughey and O’Kelly’s (1965) approach and interchanging M , K and C successively. Classical Damping Revisited – p.9/26 B E College, India, January 2007
Caughey series Caughey series: N − 1 � � � j M − 1 K C = M α j j =0 The modal damping factors: � α 1 � ζ j = 1 + α 2 ω j + α 3 ω 3 j + · · · 2 ω j More general than Rayleigh’s version of proportional damping Classical Damping Revisited – p.10/26 B E College, India, January 2007
Generalized proportional damping Premultiply condition (a) of the theorem by M − 1 : � � � � � � � � M − 1 K M − 1 C M − 1 C M − 1 K = Since M − 1 K and M − 1 C are commutative matrices M − 1 C = f 1 ( M − 1 K ) Therefore, we can express the damping matrix as C = M f 1 ( M − 1 K ) Classical Damping Revisited – p.11/26 B E College, India, January 2007
Generalized proportional damping Premultiply condition (b) of the theorem by K − 1 : � � � � � � � � K − 1 M K − 1 C K − 1 C K − 1 M = Since K − 1 M and K − 1 C are commutative matrices K − 1 C = f 2 ( K − 1 M ) Therefore, we can express the damping matrix as C = K f 1 ( K − 1 M ) Classical Damping Revisited – p.12/26 B E College, India, January 2007
Generalized proportional damping Combining the previous two cases � � � � M − 1 K K − 1 M C = M β 1 + K β 2 Similarly, postmultiplying condition (a) of Theorem 1 by M − 1 and (b) by K − 1 we have � KM − 1 � � MK − 1 � C = β 3 M + β 4 K Special case: β i ( • ) = α i I → Rayleigh damping. Classical Damping Revisited – p.13/26 B E College, India, January 2007
Generalized proportional damping Theorem 2 A viscously damped positive definite linear system possesses classical normal modes if and only if C can be represented by � � � � M − 1 K K − 1 M (a) C = M β 1 + K β 2 , or � KM − 1 � � MK − 1 � (b) C = β 3 M + β 4 K for any β i ( • ) , i = 1 , · · · , 4 . Classical Damping Revisited – p.14/26 B E College, India, January 2007
� M � 2 Example 1 Equation of motion: � − 1 K / 2 sinh( K − 1 M ln( M − 1 K ) 2 / 3 ) M ¨ q + − M e � √ � M − 1 K 4 + K cos 2 ( K − 1 M ) K − 1 M tan − 1 ˙ q + Kq = 0 π It can be shown that the system has real modes and � 1 � 1 � � ln 4 1 tan − 1 ω j j / 2 sinh 2 ξ j ω j = e − ω 4 + ω 2 j cos 2 3 ω j π . √ ω j ω 2 ω 2 j j Classical Damping Revisited – p.15/26 B E College, India, January 2007
Damping identification method To simplify the identification procedure, express the damping matrix by � � M − 1 K C = M f Using this simplified expression, the modal damping factors can be obtained as � � ω 2 2 ζ j ω j = f j � � 1 ω 2 = � or ζ j = f ( ω j ) (say) f j 2 ω j Classical Damping Revisited – p.16/26 B E College, India, January 2007
Damping identification method The function � f ( • ) can be obtained by fitting a continuous function representing the variation of the measured modal damping factors with respect to the frequency With the fitted function � f ( • ) , the damping matrix can be identified as 2 ζ j ω j = 2 ω j � f ( ω j ) �� � � � M − 1 K � M − 1 K or C = 2 M f Classical Damping Revisited – p.17/26 B E College, India, January 2007
Example 2 Consider a 3DOF system with mass and stiffness matrices 1 . 0 1 . 0 1 . 0 2 − 1 0 . 5 , M = K = 1 . 0 2 . 0 2 . 0 − 1 1 . 2 0 . 4 1 . 0 2 . 0 3 . 0 0 . 5 0 . 4 1 . 8 Classical Damping Revisited – p.18/26 B E College, India, January 2007
Example 2 0.02 0.018 0.016 0.014 Modal damping factor 0.012 0.01 0.008 0.006 0.004 0.002 0 0 1 2 3 4 5 Frequency ( ω ), rad/sec Damping factors Classical Damping Revisited – p.19/26 B E College, India, January 2007
Example 2 Here this (continuous) curve was simulated using the equation e − 2 . 0 ω − e − 3 . 5 ω � � � � � 1 + 0 . 75 ω 3 � f ( ω ) = 1 1 + 1 . 25 sin ω � 15 7 π From the above equation, the modal damping factors in terms of the discrete natural frequencies, can be obtained by e − 2 . 0 ω j − e − 3 . 5 ω j � � � � � � 2 ξ j ω j = 2 ω j 1 + 1 . 25 sin ω j 1 + 0 . 75 ω 3 . j 15 7 π Classical Damping Revisited – p.20/26 B E College, India, January 2007
Example 2 To obtain the damping matrix, consider the preceding equation as a function of ω 2 j and replace j by M − 1 K and any constant terms by that ω 2 constant times I . Therefore: � � √ √ � C = M 2 − 1 K − e − 3 . 5 − 1 K M M M − 1 K e − 2 . 0 15 � 1 � �� � I + 0 . 75( M − 1 K ) 3 / 2 � � M − 1 K I + 1 . 25 sin × 7 π Classical Damping Revisited – p.21/26 B E College, India, January 2007
Experimental Example 1 Natural frequencies, Hz Damping factors Natural frequencies, Hz (experimental) (in % of critical damping) (from FE) 33.00 0.6250 30.81 (-6.64 %) 85.00 0.2000 85.24 (0.29 %) 166.00 0.0833 167.61 (0.97 %) � � 276.00 0.0313 277.73 (0.63 %) b 409.00 0.0625 415.67 (1.63 %) 569.00 0.1250 581.42 (2.18 %) 758.00 0.1163 774.94 (2.24 %) 976.00 0.1786 996.20 (2.07 %) 1217.00 0.8621 1245.15 (2.31 %) 1498.00 0.7143 1521.77 (1.59 %) 1750.00 0.3571 1826.06 (4.35 %) Measured data for the beam example √ p 1 I + p 2 T + p 3 T 2 M − 1 K . C d = 2 MT = 2 p 2 K + 2( p 1 M + p 3 K ) Classical Damping Revisited – p.22/26 B E College, India, January 2007
Experimental Example 1 0.02 original inverse modal transformation 0.018 Rayleigh ′ s proportional damping polymonial fit 0.016 generalized proportional damping 0.014 Modal damping factors ( ζ j ) 0.012 0.01 0.008 0.006 0.004 0.002 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Natural frequencies ( ω j ), rad/sec x 10 4 Fitted and measured damping factors Classical Damping Revisited – p.23/26 B E College, India, January 2007
Summary 1. Measure a suitable transfer function H ij ( ω ) 2. Obtain the undamped natural frequencies ω j and modal damping factors ζ j 3. Fit a function ζ = � f ( ω ) which represents the variation of ζ j with respect to ω j for the range of frequency considered in the study √ M − 1 K 4. Calculate the matrix T = 5. Obtain the damping matrix using C = 2 M T � � f ( T ) Classical Damping Revisited – p.24/26 B E College, India, January 2007
Conclusions(1) Rayleigh s proportional damping is generalized. The generalized proportional damping expresses the damping matrix in terms of any non-linear function involving specially arranged mass and stiffness matrices so that the system still posses classical normal modes. This enables one to model practically any type of variations in the modal damping factors with respect to the frequency. Classical Damping Revisited – p.25/26 B E College, India, January 2007
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