求解高速列車引起連續(xù)梁橋墩頂反力的半解析方法

1、induced by high-speed train over continuous girder bridgeCao Yanmei510152025303540(School of Civil Engineering, Beijing Jiaotong University, Beijing 100044)Abstract: In order to calculate the force transmitted to the top of pier induced by high-speed trainrunning through continuous girder bridge, th

2、e semi-analytical dynamic interaction model is establishedfor train-track-continuous girder bridge system, in which the dynamic character-istics of train, track,and bridge are taken into account. In the method, all the derivations are carried in the frequency domainand the vibrations in the time dom

3、ain are obtained by inverse Fourier transformation, which makes itvery convenient to observe dynamic characteristics of train-track-bridge structure from the point offrequency. Compared to the FEM model, the presented semi-analytical method is easier to calculateand the solution is more accurate.Key

4、words: High-speed train; continuous girder bridge; semi-analytical method; reaction on pier top0 IntroductionOver the last decades, as a rapid transportation, the high-speed railway has developed rapidlyin China and abroad. Since the elevated bridge has the advantages of less land use, small pier an

5、dabutment settlement, and convenient transfer with other forms of transports, high-speed railwaylines built on elevated bridges are being widely used. For example, in Japan, for the Shinkansen,the structure of various bridges takes up nearly half of the total railway length; in China, for the1038 km

6、 Beijing-Shanghai High-speed Railway, the length of bridges is 1016 km, occupying 80.5per cent of the total railway line. However, the environmental vibrations induced by the elevatedrailway system may cause human annoyance, damage to ancient and historical buildings, andinterruption of sensitive in

7、struments. Therefore, it is necessary to develop the prediction andassessment methodology for ground vibration induced by passing trains on bridge structures, inwhich the calculation of force transmitted to pier top is very important since the force, as anexcitation, is the primary step of predictin

8、g ground vibration.Currently, the train-track-bridge FEM model is commonly used in obtaining the force on piertop. Lei et al.1 established two FEM models including the train-track-bridge model and thebridge-ground-building model to study the dynamic responses of the Bangkok new airport raillinks. Zh

9、ai et al.2 studied high-speed train induced ground vibrations by means oftrain-track-ground numerical model. In the dynamic analysis of train-track interaction, Zakeri3proposed a combination of finite elements and infinite beam elements (two end elements) forrailway track modeling, which is an impro

10、vement to boundary condition effects in the finiteelement method. Xia et al.4, 5 and Cao6 studied the vibration effects of lightrail train-viaductsystem on surrounding environment in urban area, and experimentally studied the train-inducedstructural and environmental vibrations of rail transit over

11、elevated bridge with ladder tracks.In this paper, the semi-analytical dynamic interaction model is established fortrain-track-continuous girder bridge system, in which the dynamic characteristics of train, track,and bridge are taken into account. Compared to the FEM model, the presented semi-analyti

12、calmethod is easier to calculate and the solution is more accurate.Foundations: National Natural Science Foundation of China (51108023), Ph.D. Programs Foundation of Ministryof Education of China (20090009120021), and the Fundamental Research Fund of Beijing Jiaotong University(2011JBM088).Brief aut

13、hor introduction:Cao Yanmei, (1979-), Female, PhD.,Instructor,and the main research is about railwaytraffic induced environmental vibrations. E-mail: ymcao-1-1 Dynamic interaction model of train-track-continuous girderbridge system45The dynamic interaction model of train-track-continuous girder brid

14、ge system is shown inFig. 1, in which the x-axis is the train moving direction, the z-axis is the vertical direction to theground, and the origin of coordinate is located on the center of the first support on pier topconsidered in the calculation.VMci JcizciciCar bodyktictiktictikwiMti JticwiztijBog

15、ieRailMwikHzwilWheelwrkrbokbcrbxkbbridgekbwbkbzPierL1PierL2PierL3Pier5055Fig. 1 Dynamic interaction model of train-track-continuous girder bridge system1.1 Train modelingThe train model is composed of several vehicles, with each one modeled as a multiple-DOFs(degrees-of-freedom) vibration system con

16、sisting of one car-body, two bogies, four wheel-sets,which are connected with two groups of spring-damper suspension devices. As shown in Figure 1,two DOFs (floating and pitching) are considered for each car-body and each bogie, while only oneDOF for each wheel-set, respectively.On the basis of vehi

17、cle dynamics, the dynamic equilibrium equation of the ith vehicle can bewritten as:60Mi Zi + Ci Zi + KiZi = Pi (t)(1)where: Zi , Zi and Zi represent the displacement, velocity and acceleration vectors, andM i , Ki and Ci represent the mass, stiffness and damping matrices of the ith vehicle, respecti

18、vely;Pi (t ) denotes the vector formed by exciting forces acting on the ith vehicle.Eq. (1) can be transformed from the time domain to the frequency domain by means of the65iùt(Miù 2 + iCiù + K i )Zi (ù ) = P (ù )For a train vehicle, the load vector Pi (ù) can be expres

19、sed as:Pi (ù ) = DPiw/r (ù )(2)(3)06×4 I 4×4 70matrix; andw/r-2-PiNw/r (ù )Ô(4)Fourier transform F (ù ) = F (t )e dt as follows:iwhere D = is the factor matrix of the load vector for a four-axle vehicle; I is a unitP (ù) = P w/r (ù ) P w/r (ù ) Lwher

20、e Pilw/r (ù ) denotes wheel-rail interaction force corresponding to the lth wheel-set on theith vehicle.Based on Eq. (2), the displacement vector of vehicle vibration can be expressed as:75Zi (ù) = (K i Miù 2 + iCiù)1 DPw/r (ù)(5)Let Zwi (ù) represent the vector formed

21、by all wheel-set displacements of the ith vehicle,and then there exists the following relationship:Zwi (ù ) = BZi (ù )in which the factor matrix B = 04×6 I4×4 for a four-axis vehicle.(6)80Substituting Eq. (5) into Eq. (6) and utilizing B = D , the wheel-set displacement vector of

22、 theith vehicle can be written as:Zwi (ù) = DÔ (K i M iù 2 + iCiù )1 DPiw/r (ù )Eq. (7) can be further simplified as:Zwi (ù ) = A wiPiw/r (ù)(7)(8)85wÔ 2 1(m=1, 2, , Nwi; n=1, 2, , Nwi) represents the displacement of the nth wheel-set when a unitharmonic load

23、with frequency ù acts on the location of the mth wheel-set; Nwi is the number ofwheel-sets on the ith vehicle. The minus symbol “-” in the right hand side of Eq. (8) denotes thatthe direction of wheel-set displacement is opposite to that of wheel-rail interaction force.90Assuming the vibrations

24、 of different car-bodies self-independent, Eq. (8) can be applied tothe displacements of all wheel-sets and furthermore be expanded as the form of matrixmultiplication: A w1Zw = A w200O PA wN (9)95where Zw = Zw1 (ù ) Z w2 (ù ) Lis the number of vehicles of a train.w/rPNw/r (ù)Ô ,

25、 and Nv1.2 Track-bridge modelingIn this paper a ballastless slab track-bridge structure is considered. This kind of track-bridgesystem is composed by rail, fastening, slab of track, CA layer (cement-emulsified asphalt mortarfilling layer, for adjusting the slab elevation), concrete base and the gird

26、er of bridge, as shown in100Fig. 2.Fig. 2 Ballastless slab track-bridge structure (Unit: mm)In the simulation of track-bridge system, since the long welded rail is connected with the slab-3-iwhere A wi = D (K i Miù + iCiù) D is a complex matrix with Nwi × N wi order, and its elementam

27、ni w/rZwN v (ù )Ô , Pw/r = P1 (ù ) P w/r (ù ) L105110115of track by fasteners with certain space, the rail is considered as an infinite Euler-Bernoulli beamon the discrete vertical supports, as shown in Fig. 1. The supports representing the fasteners aremarked by the spring-dampe

28、r supporting system, with equivalent stiffness and dampingrepresented by k and c , respectively. According to the “Code for Design of High Speed Railway(TB 10621-2009)” published by Ministry of Railways of China7, the slab of track, CA layer,concrete base and girder are connected with each other so

29、closely that there are almost no relativevertical displacements between them, so they are simulated by an integrated part in the model oftrack-bridge structure.According to the theory of track irregularity, the sample function of track irregularity seriescan be regarded as the superposition of a ser

30、ies of harmonic components with differentfrequencies and phases. Let the spatial angular frequency be denoted by , and then thewavelength and the circular frequency of track irregularity can respectively be expressed asë = 2 / and ù = ÙV = 2V / ë , in which V is the train speed.

31、When considering a singlespatial angular frequency only, the track irregularity can be expressed as:k(10)120In actual analysis, the track irregularity at the position x = x + Vt is commonly describedas the function of time t:z (t) = z (ù )eù(11)where x is the initial position of the lth wh

32、eel-set of the ith vehicle, and z (ù ) is theamplitude of track irregularity corresponding to frequency ù , which can be written as:125z (ù ) = %( ) eù(12)For the whole train, the track irregularities corresponding to all positions of wheel-sets can beexpressed as the following v

33、ector:where z (ù ) e (ù ) = %( ) (13)130e (ù ) = eùeùLeù(14)z (ù ) = z (ù ) z (ù ) Lz (ù )(15)1.3 Coupling of vehicle and trackThe coupling of vehicle and track is accomplished by the wheel-rail contact relationshipassumed by the contact theory of He

34、rtz. Therefore, the wheel-rail interaction force for the lth135wheel-set of the ith vehicle can be expressed as follows:P (t ) = k z (t ) w ( x , t ) z ( x , t )(16)where kH is the stiffness coefficient of wheel-rail contact, in which the contact betweenwheelset and rail is considered to be linear,

35、so kH is actually the stiffness coefficient of the linearspring in Fig. 1; z (t) is the vertical displacement of the lth wheel-set of the ith vehicle;140w (x , t) and z (x , t ) are the vertical displacement of rail and the variation profile of trackirregularity at the location of the lth wheel-set

36、of the ith vehicle, respectively.Using the Fourier transform for the time t, Eq. (16) can be transformed as:-4-zr (x) = % k ) eik xr0zr zr 2 (ùk ) e 2 (ùk ) M M z rNv (ùk ) e N v (ùk ) zril(17)A set of vector equations in the frequency domain for all wheel-rail contact forces can

37、 be145obtained when Equation (17) is applied to all wheel-rail contact points:P = k (Z W Z )(18)in which Z = Z (ù ) Z (ù) Lis the displacement vector of all wheel-sets;Z is the profile vector of track irregularity; and W = W (ù) W (ù) LW (ù) is thevector formed by the displa

38、cements of rail at the locations of all wheel-sets, which will be derived150in the next section.1.4 Displacement of railWhen a single wheel-rail force P (t) is moving along the rail in the speed of V, thedifferential equation of rail beam in the vertical direction is:EIw (x, t)+ mw (x, t)(19)155160w

39、here: w (x, t ) represents the displacement of rail induced by a single wheel-railforce; EI is the bending stiffness of the rails; m is the mass of unit length of rail; andf (x, t) denotes the interaction forces between rail and bridge in the case of a single wheel-railforce P (t) ; N is the number

40、of discrete spring-damper supports considered; and x is the locationof the mth spring-damper support.The displacements of rail under several kinds of loads can be analytically derived as follows:(1) When the load is the unit impulse load, the displacement of rail is 19:G (x,ù) =14EIáeá

41、; ieá (20)in which á = mù / EI .(2) When the load is the single moving axle load P , the displacement of rail is:165w (x,ù ) =P % ùV Vù(21)rail is:w (x,ù) =V Vùù(22)(4) According to the generalized Duhamel integral, the displacement of rail induced by the

42、170rail-bridge interaction forces can be expressed as:Fig. 3 Sketch of arrangements of all wheels in the train(23)175For the whole train, all wheel-sets are orderly arranged showed in Fig. 3, in which 2a, 2b, and-5-P (ùk ) = kH zwil (ùk ) wril (xpil ,ùk ) zril (xpil ,ùk )Z wNv (&

43、#249;)xx= Pilw/r (t)(x xp0il Vt) f rbsingle (x, t)(x xm )Grail ( ,ù)e(3) When the load is the single wheel-rail interaction force P (ùk )eiù t , the displacement ofPilw/r (ùk ) % ù ùkGrail ( ,ù)ewrsingle-r/b (x,ù) = f rbsingle (xm ,ù)Grail (x xm ,ù )

44、l are respectively the distance between adjacent wheels, the distance between adjacent bogies, andthe length of a passage of the train; L1, L2, and L3 are three neighboring calculated span ofcontinuous girder.Therefore, the displacement of rail under the impact of all wheel-sets can be obtained by18

45、0means of the principle of superposition.(1) When the irregularity of track is not considered, the displacement of rail can be expressedas:w (x,ù) =V V(24)ùindicates the contribution to rail displacement of all185quasi-static axle loads; and f (x ,ù ) represents the interaction forces

46、 between rail and bridgeinduced by all wheel-rail forces.(2) When the irregularity of rail is considered, the displacement of rail can be expressed as:w (x,ù) =V Vùù(25)190all wheel-rail interaction forces.ùùindicates the contribution to the rail displacement ofWhen x = x ,

47、the rail displacement at the location of each wheel-set can be expressed. Thus,the rail displacement vector W can be written as: W(ù) G fG fM MG f LG f LM O M (26)where:195A(k ,ù ) =1 %V(27) e eMeMM M(28)G = G (x x ) G (x x ) LG ( x x )(29)f = f ( x,ù)f ( x ,ù) Lf ( x ,ù)(30

48、)and, when the contributions of axle loads are exclusively considered, k = ù / V and200C (ù) = C (ù ) .1.5 Displacement of the bridgeC (ù) = C (ù) ; when those of wheel-rail interaction forces are considered, k = (ù ù ) / V and1.5.1Greens function for continuous gi

49、rder bridgeIn order to obtain the displacement of bridge under the loading of arbitrary excitation, it is-6-1 % ùGrail 1 (ù ) f rb (xm ,ù)Grail (x xm ,ù)( , ù)e V Cin which: C1(ù) = P0 il e1 % ù ùkGrail ( ,ù)eC2 (ù) f rb (xm ,ù )Grail (x xm ,

50、49;)in which C2 (ù) = Pilw/r (ùk )e Wr2 (ù) 21Wr = = A(k x ,ù)Es WrN (ù) G rail frbG railv 2 frb LG1railN w1 frb fG 2rail rb frb G rail111222Grail (k x ,ù)C (ù) ix kEs = ixpN v 1k xeeeLee p2N W2 x LixpN v N WN kx L e 205necessary to firstly derive the Greens functi

51、on for displacement of bridge by solving the Eq. (31).G (x, t )x+ mG (x, t)t= (t)(x x )(31)in which G (x, t) is the displacement of bridge under unit impulse load, i.e. Greensfunction; E = E(1+ 2îi) is the complex elastic modulus considering the influence of internaldamping ( î denotes the

52、 damping ratio of bridge structure); I is the moment of inertia of the210bridge section; m is the mass of unit length of bridge; x is the acting position of (t) .Based on the superposition principle of vibration modes, the bridge displacement can beexpressed as:(32)where q (t) is the generalized coo

53、rdinate, and (x) is the ith vibration mode of the bridge, i.e.215 (x) = A sin(ë x) + A cos(ë x) + A sinh(ë x) + A cosh(ë x)(33)Let X = x x Lx represent the column vector comprised by the horizontalcoordinates of all joints in FEM bridge model, and Y = yy Ly represents the columnv

54、ector formed by the data sequence of the ith mode shape, and then the coordinates of all joints are220substituted into Eq. (33) to form the following matrix: sin ë x Mxcos ë xcos ë xMcos ë xsinh ë xsinh ë xMsinh ë xcosh ë x cosh ë x M xThen, the ith mode

55、shape function of the bridge can be fitted by means of the internalfunction linefit() of MATLAB, i.e. (x) = linefit(X, Y , N ) .Substitute Eq. (32) into Eq. (31), utilizing the mutual orthogonality of mode shapes, the225following system of equations can be derived: g qt tt tt tM= (t ) ( x )= (t ) (

56、x )= (t ) ( x )(34)in which g (i = 1, 2,L , n; j = 1, 2,L , n) is the expression about the known parameters.By using the Fourier integral transform to Eq. (34), the q (ù ) (i = 1, 2,L , n) can be got, andthen the Greens function of the continuous girder bridge G ( x, t ) can be obtained.2301.5.2Displacement of bridgeOn the basis of the generalized Duhamel integral, when a train is running through acontinuous girder bridge, the displacement of the