Method for automatically detecting free vibration response of high-speed railway bridge for modal identification
11221273 · 2022-01-11
Assignee
Inventors
Cpc classification
G01H17/00
PHYSICS
International classification
Abstract
A method for automatically detecting the free vibration response segment of the high-speed railway bridges after trains passing. First, pre-select the test response sequence to be decomposed based on the maximum of the time instants corresponding to the absolute maximums of the response vectors at various measuring point. Then, Extract the single-frequency modal response from the test response by the iterative variational mode decomposition and fit the envelope amplitude of the modal response by Hilbert transform. Finally, the vibration features at each time instants are marked as decay vibration or non-decay vibration. The longest structural response segment that meets the decay vibration features is determined as the detected free vibration response segment for modal identification. This invention can effectively detect the free vibration data segment without human participation, which is of great significance for the real-time accurate modal analysis of high-speed railway bridges.
Claims
1. A method for automatically detecting free vibration response of high-speed railway bridge for modal identification, wherein steps are as follows: step 1: pre-selection of a test response to be decomposed an acceleration responses at different measuring points are given as y(t)=[y.sub.1(t),y.sub.2(t), . . . ,y.sub.s(t)].sup.T,t=Δt,2Δt, . . . ,NΔy, where Δt is the sampling time interval; N is the number of samples; s is the number of measuring points and the superscript indicates transposition; calculate a time instant corresponding to an absolute maximum value of a response vector at each measuring point i, i=1,2, . . . ,s, as
Description
DESCRIPTION OF DRAWINGS
(1)
(2)
(3)
DETAILED DESCRIPTION
(4) The present invention is further described below in combination with the technical solution.
(5) The numerical example of a simply supported beam model is employed, as shown in
(6) The procedures are described as follows:
(7) The acceleration responses collected at each measuring point is given as y(t)=[y.sub.1(t),y.sub.2(t), . . . ,y.sub.s(t)].sup.T, where the sampling time interval is Δt=1/f.sub.s=0.05; the number of samples is N=3501; the number of measuring points is s=5.
(8) Calculate the time instant corresponding to the absolute maximum of the response vector at each measuring point i (i=1,2, . . . ,s) as
(9)
Then the measuring point corresponding to the maximum of the time instants t.sub.i, i=1,2, . . . ,5, is obtained by
(10)
as h=1 with t.sub.h=t.sub.1=1022 Δt=51.1. Then the response sequence y.sub.1(t),t=t.sub.1, . . . ,NΔt is taken as the test response to be decomposed.
(11) The test response y.sub.1(t) is decomposed by the variational mode decomposition in Eq. (1) where the number of components is fixed as 2. After the first decomposition, two components y.sub.1,1.sup.[1](t) and y.sub.1,2.sup.[1](t) with the central angular frequencies ω.sub.1.sup.[1] and ω.sub.2.sup.[1] are obtained, as shown in
(12)
Since e.sub.1.sup.[1]>e.sub.2.sup.[1], the component y.sub.1,1.sup.[1](t) is considered as the new test signal to be decomposed by the variational mode decomposition. Repeat the above process until the angular frequency difference Δω.sup.[4]=|ω.sub.1.sup.[4]−ω.sub.2.sup.[4]|<2πf.sub.min/100. Since e.sub.1.sup.[4]>e.sub.2.sup.[4], the component y.sub.1,1.sup.[4](t) is selected as the modal response, which will be labelled as x(t)≡y.sub.1,1.sup.[4](t).
(13) Extend the modal response x(t) to obtain the continuation signal x.sub.e(t) to avoid the edge effect in the subsequent transform. Then the Hilbert transform is used to obtain the envelope amplitude of the continuation signal x.sub.e(t) as a.sub.e(t), as Eq. (2).
(14) The envelope a(t) corresponding to the modal response x(t) is intercepted from the envelope a.sub.e(t) of the continuation modal response x.sub.e(t), and the instantaneous amplitude difference is calculated as Δa(t)=a(t+1)−a(t). According to the characteristic that the envelope amplitude of the free vibration modal response is decreasing, the vibration feature corresponding to the time instants of Δa(t)≤0 is marked as “decay vibration” and expressed in 1. Otherwise, if Δa(t)>0, the vibration feature is marked as “non-decay vibration” and expressed in 0. Thus, the vibration feature from the selected time instants t=[t.sub.h,t.sub.h+Δt, . . . ,NΔt] will be labelled as a set of 0 or 1. Choose the time instants [t.sub.h+wΔt,t.sub.h+(w+1)Δt, . . . ,t.sub.h+κΔt] corresponding to the longest vibration feature sequence which are continuously marked as 1. And extract the structural vibration responses y(t), t=t.sub.h+wΔt, . . . ,t.sub.h+κΔt as the free vibration response segment {tilde over (y)}(k), k=1,2, . . . κ−w+1, as shown in
(15) The eigensystem realization algorithm with data correlation is performed on the detected free vibration response segment. First, the obtained free vibration responses are utilized to construct the Hankel matrix H(k) as Eq. (3). Set k=l=30, g=N.sub.b−2 l+1, and build the correlation function matrix as S=H(l)H(0).sup.T. Implement eigensystem realization algorithm for the correlation function matrix to solve the first 5 modal parameters of the structure, which are consistent with the numerical solution. The identified frequencies are f.sub.1=0.6727 Hz, f.sub.2=1.2995 Hz, f.sub.3=1.8378 Hz, f.sub.4=2.2508 Hz, f.sub.5=2.5104 Hz and the identified damping ratios are d.sub.1=0.8000%, d.sub.2=0.8000%, d.sub.3=0.9384%, d.sub.4=1.0706%, d.sub.5=1.1596%. As a comparison, the vibration responses corresponding to the time instants t=t.sub.h, . . . ,NΔt are implemented into the eigensystem realization algorithm to solve the first 5 modal parameters, which are a little dissimilarity with the numerical solution. The estimated frequencies are f′.sub.1=0.6712 Hz, f′2=1.2963 Hz, f′.sub.3=1.8413 Hz, f═.sub.4=2.2513 Hz, f′.sub.5=2.5132 Hz. The estimated damping ratios are d′.sub.1=0.7621%, d′.sub.2=2.1911%, d′.sub.3=0.6523%, d′.sub.4=0.8205%, d′.sub.5=1.3894%.