Experimental modal analysis of a beam travelled by a moving mass - PowerPoint PPT Presentation

Experimental modal analysis of a beam travelled by a moving mass using Hilbert Vibration Decomposition Mathieu BERTHA Jean-Claude GOLINVAL University of Liège 30 June, 2014

The research is focused on the identification of time-varying systems M ( t ) ¨ x ( t ) + C ( t ) ˙ x ( t ) + K ( t ) x ( t ) = f ( t ) Dynamics of such systems is characterized by : ◮ Non-stationary time series ◮ Instantaneous modal properties ◮ Frequencies : ω r ( t ) ◮ Damping ratio’s : ξ r ( t ) ◮ Modal deformations : q r ( t ) Mathieu BERTHA (ULg) EURODYN 2014, June 2014 1

The Hilbert Transform The Hilbert transform H of a signal x ( t ) is the convolution product of 1 this signal with the impulse response h ( t ) = π t H ( x ( t )) = ( h ( t ) ∗ x ( t )) � + ∞ = p.v. x ( τ ) h ( t − τ ) d τ −∞ � + ∞ 1 x ( τ ) = π p.v. t − τ d τ −∞ It is a particular transform that remains in the time domain It corresponds to a phase shift of − π 2 of the signal Mathieu BERTHA (ULg) EURODYN 2014, June 2014 2

The Hilbert transform and the analytic signal The analytic signal z is built as z ( t ) = x ( t ) + i H ( x ( t )) A ( t ) e i φ ( t ) = The instantaneous properties of the signal can then be obtained  A ( t ) = | z ( t ) |   φ ( t ) = ∠ z ( t )  ω ( t ) = d φ    dt Mathieu BERTHA (ULg) EURODYN 2014, June 2014 3

The Hilbert Vibration Decomposition (HVD) method x ( t ) It is an iterative process The sifting of the signal extracts monocomponents Analytic signal z ( t ) = x ( t ) + i H ( x ( t )) from higher to lower instantaneous amplitude Frequency extraction ω ( t ) = d φ ( t ) = d ∠ z ( t ) dt dt It is applicable to single channel measurement Lowpass filtering ω ( t ) → ω k ( t ) Crossing monocomponents may be a problem Synchronous demodulation x k ( t ) Sifting process x ( t ) := x ( t ) − x k ( t ) Mathieu BERTHA (ULg) EURODYN 2014, June 2014 4

The experimental set-up 2.1 meter aluminum beam Steel block ( ≈ 3.5 kg, 38.6%) 1 shaker (random force) 7 accelerometers LMS SCADAS & LMS Test.Lab system Mathieu BERTHA (ULg) EURODYN 2014, June 2014 5

Time invariant modal identification of the beam subsystem 9.8 Hz 30.43 Hz 39.23 Hz 53.32 Hz 99.22 Hz CMIF : 1% ε f s s s s s 40 s s s s s 39 ε ζ : 1% s s s s s 38 s s s s s : 1% 37 ε V s s s s s 36 v s s s v 35 s s s s s 34 s s s s s 33 s s s s s 32 s s s s s 31 s s s s s 30 v s s s v 29 v s s s s 28 v s s s s 27 s s s s s 26 v s s s s 25 v s s s s 24 v s s s s 23 s s s s s 22 v s s s v 21 s s s s s 20 s s s s s 19 v s s s s 18 v s s s s 17 v s s s s 16 v s s s s 15 o s s s s 14 s v s s 13 s s s s 12 v s s s 11 v v s v 10 v v s s 9 o o v v 8 v v 7 v v 6 o s 5 0 10 20 30 40 50 60 70 80 90 100 110 120 130 Frequency [Hz] Mathieu BERTHA (ULg) EURODYN 2014, June 2014 6

Time-varying dynamics of the system Mathieu BERTHA (ULg) EURODYN 2014, June 2014 7

The sifting process and the benefit of the source separation x ( t ) Source separation x ( t ) → s ( t ) Analytic signal z ( t ) = s 1 ( t ) + i H ( s 1 ( t )) Phase extraction φ ( t ) = ∠ z ( t ) Trend extraction φ ( t ) → φ ( k ) ( t ) VKF x ( k ) ( t ) , V k ( t ) Sifting process x ( t ) := x ( t ) − x ( k ) ( t ) Mathieu BERTHA (ULg) EURODYN 2014, June 2014 8

Other modes are extracted after few iterations x ( t ) Source separation x ( t ) → s ( t ) Analytic signal z ( t ) = s 1 ( t ) + i H ( s 1 ( t )) Phase extraction φ ( t ) = ∠ z ( t ) Trend extraction φ ( t ) → φ ( k ) ( t ) VKF x ( k ) ( t ) , V k ( t ) Sifting process x ( t ) := x ( t ) − x ( k ) ( t ) Mathieu BERTHA (ULg) EURODYN 2014, June 2014 9

Monocomponents and complex amplitudes are extracted with a Vold-Kalman filter The Vold-Kalman model and the modal expansion are very similar. The extracted complex amplitudes are then considered as unscaled mode shapes e i φ k ( t ) Vold-Kalman filter: x ( t ) = � a k ( t ) k � � Modal expansion: x ( t ) = � V k ( t ) η k ( t ) k Mathieu BERTHA (ULg) EURODYN 2014, June 2014 10

The moving mass affects both frequencies and mode shapes 120 Frequency [Hz] 100 80 60 40 20 z 0 x 0 5 10 15 20 25 30 35 40 Time [s] t = 5 s t = 10 s t = 15 s t = 20 s t = 25 s t = 30 s t = 35 s Mathieu BERTHA (ULg) EURODYN 2014, June 2014 11

Thank you for your attention Mathieu BERTHA (ULg) EURODYN 2014, June 2014 12