# Analysis of Dropper Stress in a Catenary System for a High-Speed Railway

Authors: Liming Chen; Juan Sun; Like Pan; Fan He
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

When the pantograph slides over the contact wire at a high speed, there are cyclic stresses acting on droppers. The cyclic stresses are related with the fatigue life and fracture of the dropper. Thus, the stress calculation of the dropper is particularly important to ensure the safe operation of a train. In this paper, a simple model based on a span of five droppers is constructed. The contact wire is treated as a beam element and the vibration equation under a dynamic force is deduced. Then, the stresses are performed numerically using the finite difference method. The stress-time relationship under a dynamic force is obtained and compared with that under a constant force. The results show that there is a large shock effect during the deformation of dropper from straightening to compression and then bending. The dynamic force could cause much higher stresses than a constant force. It indicates that a dynamic force has a more significant impact on the stress value of dropper than a constant force.


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## Body

## 1. Introduction

With the rapid development of electrified railways, the safety and reliability of the catenary system need to be further improved. Dropper is loaded, and its stress is produced when pantograph passes through contact wire. Obviously, the fracture of dropper mainly depends on its stress level. Certainly, many factors have important impacts on ensuring the safety of high-speed railway. For example, the contact loss reduction of pantograph in high-speed operation is also crucial [1]. The pantograph-catenary contact force has to be kept as uniform as possible to reduce contact loss [2]. In addition, there are many factors affecting the reliability of catenary, such as pantograph arcing [3], arc resistance [4, 5], wind speed [6], and so on.Much work has studied the fatigue properties of catenary system. For example, Huh et al. observed the fracture surface of copper film by electron microscope and found that the fracture of copper film was transgranular under cyclic load [7]. Similar research works of copper have also been carried out by other authors [8, 9]. Liu et al. found that the stress concentration existed at the end of the dropper, which is the main reason for the fatigue damage of dropper [10]. Contact force and contact loss have been also investigated [11, 12]. Song et al. used an analytical model to study wave propagation property in the catenary system subjected to a moving pantograph [13]. Lee and Chung also analyzed wave propagation in a catenary system by using the equations of motion and developing a finite element computer program for the static and dynamic analyses [14]. Huang et al. developed a new detection method of the pantograph-catenary system based on image process [15]. Benet et al. constructed a mathematical model of the pantograph-catenary dynamic interaction [16]. Bucca et al. proposed experimental and numerical models to perform the pantograph-catenary dynamic interaction [17]. Zhu et al. established a set of arc plasma conservation equations to solve the temperature distribution in the arc [18]. Li et al. provided some key applications in vehicle dynamics control [19]. Chen et al. presented a novel dynamic event-triggered scheme for the load frequency regulation [20].Though many researchers have focused on mechanical characteristics of catenary system, including material, wave propagation, arc, and load, the detailed understandings of the mechanical analysis of dropper are still absent. It is very necessary to provide a theoretical basis for the fracture mechanism of dropper. In this paper, we theoretically and numerically analyze the stress distributions under a dynamic force based on a span of five droppers. This purpose is to find out how the stress distributions of different droppers change. Furthermore, we compare the difference between both conditions of dynamic and constant forces in order to investigate the effects of a dynamic force on the stress distribution of dropper.


## 2. Model and Methods

### 2.1. Response Equation

In this paper, a simple model based on a span of five droppers is constructed, as shown in Figure1. We assume that both ends of contact wire are free. In Figure 1, a denotes the distance between the origin and the location of a load.Figure 1
The model diagram of five droppers under a dynamic force.The contact wire is treated as a beam element and the free vibration equation of a beam is as follows.(1)EI∂4yx,t∂x4+ρA∂2yx,t∂t2=0,where ρ and EI are the density and the bending rigidity of contact wire and A is the cross-sectional area.When an elastic body is subjected to a certain order of main vibration, its mass points should move at the same frequency and phase, and each point should also pass through the static equilibrium position and reach the maximum deviation position at the same time. That is, the system has a certain time-independent mode of vibration.Therefore, the solution of the equation can be assumed to be(2)yx,t=YxTt=Yxbsinωt+ϕ.Substituting Equation (1), then(3)d4Yxdx4−k4Yx=0.where k4=ρAω2/EI.The basic solution of Equation (3) is Yx=eλx. After substituting Equation (3), the characteristic equation is obtained as follows:(4)λ4−k4=0.Its characteristic roots are listed as follows:(5)λ1=k,λ2=−k,λ3=ki,λ4=−ki.Owing to(6)e±kx=coshkx±sinhkx,e±ikx=coskx±isinkx.Hence, the basic solution of the equation issinkx,coskx,sinhkx,coshkx. Therefore, the general solution of Equation (3) can be expressed as a linear combination of these solutions.(7)Yx=A1sinkx+A2coskx+A3sinhkx+A4coshkx.The bending moment and cross section shear force are zero when both ends of the beam are free, thus(8)Y″0=0,Y‴0=0,Y″l=0,Y‴l=0.Substituting Equation (7), then(9)A1=A3,A2=A4,A1sinhkl−A1sinkl+A2coshkl−A2coskl=0a,A1coshkl−A1coskl+A2sinhkl+A2sinkl=0b.The following could be obtained by Equation (9) (a) and (b):(10)A1=A2coskl−coshklsinhkl−sinkl,1−coskl⋅coshkl=0.The solution is expressed as(11)knl≈n+12π,n=1,2,3,….Thus, the natural frequency is obtained as(12)ωn=EIρAkn2=n+1/22π2l2EIρA,n=1,2,3,….The corresponding mode function is written as(13)Yx=A1sinkx+sinhkx+A2coskx+coshkx=A2coskl−coshklsinhkl−sinklsinkx+sinhkx+A2coskx+coshkx=A2coshkx+coskx−sinhkx+sinkxcoshkl−cosklsinhkl−sinkl.Thus, the modal function is obtained as(14)Ynx=cosknx+coshknx−sinknx+sinhknxcoshknl−cosknlsinhknl−sinknl.If a dynamic force,Ft is loaded on the beam at a distance from the coordinate origin, the forced vibration equation of the beam is written as(15)EI∂4yx,t∂x4+ρA∂2yx,t∂t2=Ftδx−a.We express the solution as(16)yx,t=∑n=1∞YnxTnt.Substituting the above forced vibration equation, thus(17)∑n=1∞EITnYn4+ATn2Yn=Ftδx−a.MultiplyingYm at both sides and integrating along the beam length, thus(18)∑n=1∞Tn∫0lEIYn4Ymdx+Tn2∫0lρAYnYmdx=Ft∫0lδx−aYmdx.Using the orthogonality condition of modal function, then(19)Tn″+ωn2Tn=FtYna.Here, we setFt=Fsinωt, where F is a constant force.Thus,(20)Tn=FYnaωn21−ω/ωn2sinωt,n=1,2,3,….Hence(21)yx,t=∑n=1∞YnxTnt=Fsinωt∑n=1∞YnaYnxωn21−ω/ωn2,n=1,2,3…….We setF=100N and f=1Hz. Only the first five orders are calculated and higher orders are neglected. Therefore, the response of the contact wire is as follows.(22)yx,t=100sin2πt∑n=15YnaYnxωn21−2π/ωn2.


### 2.2. Theoretical Equations for Stress Calculation

In our previous work, the theoretical equations for stress calculation of dropper were deduced in detail, which could be found in reference [21]. Here, we directly give them as follows according to Figure 2.(23)∂x¯∂s¯2+∂y¯∂s¯2−1=fx¯cosθ+fy¯sinθ,∂θ∂s¯=λ2m¯,∂m¯∂s¯=fx¯sinθ−fy¯cosθ,∂fx¯∂s¯=μ∂2x¯∂t¯2,∂fy¯∂s¯=μ∂2y¯∂t¯2+μ,dy¯dx¯=tanθ,where x¯=x/L, y¯=y/L, s¯=s/L, vx¯=vx/gL, vy¯=vy/gL, fx¯=fx/EdAd, fy¯=fy/EdAd, m¯=m/EdAdL, μ=ρdgL/Ed, t¯=t/L/g, λ2=EdAdL2/EdId, x, y are the position coordinates and s is the arc coordinate, θ is the angle between the minor segment of dropper and the axis x, fx, fy are the forces acting on the string along the x and y directions, m is the bending moment, EdId is the bending rigidity, ρd is the dropper density, vx, vy are the velocities in the x and y directions, g is the gravity acceleration, Ed is the elastic modulus of the dropper, Ad is the cross-sectional area of the dropper, L is the length of the dropper, and t is time.Figure 2
Force analysis diagram of microelement segment of the dropper.Based on the above equations, the stress could be calculated as(24)σ=FNAd+mWZ,where FN is the axial force and FN=fxcosθ+fysinθ and WZ is dropper’s section modulus in bending.


### 2.3. Initial and Boundary Conditions of Each Dropper

The initial and boundary conditions of each dropper can be uniformly expressed as(25)x¯s¯,t¯=xpLpy¯s¯,t¯=s¯whent¯=0,x¯s¯,t¯=xpLpy¯s¯,t¯=ZpLpsin2πLpg⋅t¯whens¯=0,x¯s¯,t¯=xpLpy¯s¯,t¯≤1,whens¯=1.where the subscript p denotes the number of dropper, xp is the distance away from the coordinate origin, and Lp and Zp are, respectively, the length and amplitude of each dropper. Zp could be obtained by Equation (22).


### 2.4. Numerical Method

The finite difference method is employed and the set of Equation (23) could be transformed into a simple form using the following scheme:(26)∂u∂s¯=ui+1−uiΔs¯u=x¯,y¯,θ,m¯,fx¯,fy¯,(27)dy¯dx¯=y¯i+1−y¯ix¯i+1−x¯i,(28)∂2v∂t¯2=vj+2−2vj+1+vjΔt¯2v=x¯,y¯,where i and j stand for the variables with respect to the ith element and the jth timestep, respectively. Considering the accuracy and the time, we select the number of elements is 32 and the dimensionless timestep is Δt¯=7.5×10−10. Each dropper is discredited into 32 elements with the same length. Thus, the dimensionless length of each element is Δs¯=1/32. Equation (23) plus the associated initial and boundary conditions could be solved by substituting the expressions (26)–(28) and writing a MATLAB code.


### 2.5. Given Parameters

The parameters are given for the numerical calculation according to our previous work [21]. The material properties of five droppers are the same. The elastic modulus and the bending rigidity are Ed=83.29 GPa and EdId=2.53 Pa·m4, respectively. The section modulus in bending is WZ=6.7663×10−9 m3. The density is ρd=8.9×103 kg·m−3, and the cross-sectional area is Ad=1.29×10−5 m2. However, droppers I and V have L1 = L5 = 1.6 m, droppers II and IV have L2 = L4 = 1.295 m, and dropper III has L3 = 1.2547 m in length. The density and the bending rigidity of contact wire are ρ=8.9×103 kg·m−3 and EI=233.92 Pa·m4, and the cross-sectional area is A=1.58×10−4 m2.


## 2.1. Response Equation

In this paper, a simple model based on a span of five droppers is constructed, as shown in Figure1. We assume that both ends of contact wire are free. In Figure 1, a denotes the distance between the origin and the location of a load.Figure 1
The model diagram of five droppers under a dynamic force.The contact wire is treated as a beam element and the free vibration equation of a beam is as follows.(1)EI∂4yx,t∂x4+ρA∂2yx,t∂t2=0,where ρ and EI are the density and the bending rigidity of contact wire and A is the cross-sectional area.When an elastic body is subjected to a certain order of main vibration, its mass points should move at the same frequency and phase, and each point should also pass through the static equilibrium position and reach the maximum deviation position at the same time. That is, the system has a certain time-independent mode of vibration.Therefore, the solution of the equation can be assumed to be(2)yx,t=YxTt=Yxbsinωt+ϕ.Substituting Equation (1), then(3)d4Yxdx4−k4Yx=0.where k4=ρAω2/EI.The basic solution of Equation (3) is Yx=eλx. After substituting Equation (3), the characteristic equation is obtained as follows:(4)λ4−k4=0.Its characteristic roots are listed as follows:(5)λ1=k,λ2=−k,λ3=ki,λ4=−ki.Owing to(6)e±kx=coshkx±sinhkx,e±ikx=coskx±isinkx.Hence, the basic solution of the equation issinkx,coskx,sinhkx,coshkx. Therefore, the general solution of Equation (3) can be expressed as a linear combination of these solutions.(7)Yx=A1sinkx+A2coskx+A3sinhkx+A4coshkx.The bending moment and cross section shear force are zero when both ends of the beam are free, thus(8)Y″0=0,Y‴0=0,Y″l=0,Y‴l=0.Substituting Equation (7), then(9)A1=A3,A2=A4,A1sinhkl−A1sinkl+A2coshkl−A2coskl=0a,A1coshkl−A1coskl+A2sinhkl+A2sinkl=0b.The following could be obtained by Equation (9) (a) and (b):(10)A1=A2coskl−coshklsinhkl−sinkl,1−coskl⋅coshkl=0.The solution is expressed as(11)knl≈n+12π,n=1,2,3,….Thus, the natural frequency is obtained as(12)ωn=EIρAkn2=n+1/22π2l2EIρA,n=1,2,3,….The corresponding mode function is written as(13)Yx=A1sinkx+sinhkx+A2coskx+coshkx=A2coskl−coshklsinhkl−sinklsinkx+sinhkx+A2coskx+coshkx=A2coshkx+coskx−sinhkx+sinkxcoshkl−cosklsinhkl−sinkl.Thus, the modal function is obtained as(14)Ynx=cosknx+coshknx−sinknx+sinhknxcoshknl−cosknlsinhknl−sinknl.If a dynamic force,Ft is loaded on the beam at a distance from the coordinate origin, the forced vibration equation of the beam is written as(15)EI∂4yx,t∂x4+ρA∂2yx,t∂t2=Ftδx−a.We express the solution as(16)yx,t=∑n=1∞YnxTnt.Substituting the above forced vibration equation, thus(17)∑n=1∞EITnYn4+ATn2Yn=Ftδx−a.MultiplyingYm at both sides and integrating along the beam length, thus(18)∑n=1∞Tn∫0lEIYn4Ymdx+Tn2∫0lρAYnYmdx=Ft∫0lδx−aYmdx.Using the orthogonality condition of modal function, then(19)Tn″+ωn2Tn=FtYna.Here, we setFt=Fsinωt, where F is a constant force.Thus,(20)Tn=FYnaωn21−ω/ωn2sinωt,n=1,2,3,….Hence(21)yx,t=∑n=1∞YnxTnt=Fsinωt∑n=1∞YnaYnxωn21−ω/ωn2,n=1,2,3…….We setF=100N and f=1Hz. Only the first five orders are calculated and higher orders are neglected. Therefore, the response of the contact wire is as follows.(22)yx,t=100sin2πt∑n=15YnaYnxωn21−2π/ωn2.


## 2.2. Theoretical Equations for Stress Calculation

In our previous work, the theoretical equations for stress calculation of dropper were deduced in detail, which could be found in reference [21]. Here, we directly give them as follows according to Figure 2.(23)∂x¯∂s¯2+∂y¯∂s¯2−1=fx¯cosθ+fy¯sinθ,∂θ∂s¯=λ2m¯,∂m¯∂s¯=fx¯sinθ−fy¯cosθ,∂fx¯∂s¯=μ∂2x¯∂t¯2,∂fy¯∂s¯=μ∂2y¯∂t¯2+μ,dy¯dx¯=tanθ,where x¯=x/L, y¯=y/L, s¯=s/L, vx¯=vx/gL, vy¯=vy/gL, fx¯=fx/EdAd, fy¯=fy/EdAd, m¯=m/EdAdL, μ=ρdgL/Ed, t¯=t/L/g, λ2=EdAdL2/EdId, x, y are the position coordinates and s is the arc coordinate, θ is the angle between the minor segment of dropper and the axis x, fx, fy are the forces acting on the string along the x and y directions, m is the bending moment, EdId is the bending rigidity, ρd is the dropper density, vx, vy are the velocities in the x and y directions, g is the gravity acceleration, Ed is the elastic modulus of the dropper, Ad is the cross-sectional area of the dropper, L is the length of the dropper, and t is time.Figure 2
Force analysis diagram of microelement segment of the dropper.Based on the above equations, the stress could be calculated as(24)σ=FNAd+mWZ,where FN is the axial force and FN=fxcosθ+fysinθ and WZ is dropper’s section modulus in bending.


## 2.3. Initial and Boundary Conditions of Each Dropper

The initial and boundary conditions of each dropper can be uniformly expressed as(25)x¯s¯,t¯=xpLpy¯s¯,t¯=s¯whent¯=0,x¯s¯,t¯=xpLpy¯s¯,t¯=ZpLpsin2πLpg⋅t¯whens¯=0,x¯s¯,t¯=xpLpy¯s¯,t¯≤1,whens¯=1.where the subscript p denotes the number of dropper, xp is the distance away from the coordinate origin, and Lp and Zp are, respectively, the length and amplitude of each dropper. Zp could be obtained by Equation (22).


## 2.4. Numerical Method

The finite difference method is employed and the set of Equation (23) could be transformed into a simple form using the following scheme:(26)∂u∂s¯=ui+1−uiΔs¯u=x¯,y¯,θ,m¯,fx¯,fy¯,(27)dy¯dx¯=y¯i+1−y¯ix¯i+1−x¯i,(28)∂2v∂t¯2=vj+2−2vj+1+vjΔt¯2v=x¯,y¯,where i and j stand for the variables with respect to the ith element and the jth timestep, respectively. Considering the accuracy and the time, we select the number of elements is 32 and the dimensionless timestep is Δt¯=7.5×10−10. Each dropper is discredited into 32 elements with the same length. Thus, the dimensionless length of each element is Δs¯=1/32. Equation (23) plus the associated initial and boundary conditions could be solved by substituting the expressions (26)–(28) and writing a MATLAB code.


## 2.5. Given Parameters

The parameters are given for the numerical calculation according to our previous work [21]. The material properties of five droppers are the same. The elastic modulus and the bending rigidity are Ed=83.29 GPa and EdId=2.53 Pa·m4, respectively. The section modulus in bending is WZ=6.7663×10−9 m3. The density is ρd=8.9×103 kg·m−3, and the cross-sectional area is Ad=1.29×10−5 m2. However, droppers I and V have L1 = L5 = 1.6 m, droppers II and IV have L2 = L4 = 1.295 m, and dropper III has L3 = 1.2547 m in length. The density and the bending rigidity of contact wire are ρ=8.9×103 kg·m−3 and EI=233.92 Pa·m4, and the cross-sectional area is A=1.58×10−4 m2.


## 3. Results and Discussion

Leta=0, which means the dynamic force is loaded at dropper I. Under the condition, the stresses of the droppers are calculated. The results could be obtained through numerical calculations. The stresses of five different droppers are performed numerically when a dynamic force is directly loaded at dropper I. The stress-time relationships of five droppers under a dynamic force are identical. Figure 3 represents the stress distribution of dropper I. It is seen from Figure 3 that in a period, the dropper begins to flex and sag when the dropper is loaded, the stress maintains very low and then the dropper is straightened, the stress attains to the highest value from nearly zero in an instant time. The highest stress is about 460 MPa. Then, the inertia causes the dropper to be in a compression state. The dropper behaves the damped sinusoidal vibration like a spring oscillator. At the bending stage, the stress is almost zero due to the tiny bending stiffness of the dropper. The results show that there is a large shock effect during the deformation of each dropper from straightening to compression and then bending.Figure 3
The stress distribution of dropper I.Though the dynamic force is not directly loaded at other four droppers, the response could be produced due to the wave propagation. Hence, the vibration of other four droppers could be provoked and the similar deformation could be induced. Figure4 shows that there are also different stages of dropper II from rebound to vibration and then bending stage. The highest stress of dropper II is 240 MPa or so, which is lower 220 MPa than that of dropper I. It shows that they differ from the maximum tensile stresses. However, the lowest stresses of the two droppers are not a great difference, which makes a little distinction between the maximum compressive stresses. Figure 5 depicts the stress variation of dropper III. Compared with Figure 4, the highest stress decreases a little while the lowest stress is almost not changed.Figure 4
The stress distribution of dropper II.Figure 5
The stress distribution of dropper III.It is seen from Figure6 that the highest and lowest stresses of dropper IV decrease a half in contrast to dropper III. In Figure 7, the highest and lowest stresses of dropper V further decrease and the maximum tensile and compressive stresses are approximately 85 MPa and 50 MPa. According to the stress distributions of the five droppers, it is found that the maximum tensile stress gradually decreases due to the energy loss of vibration and the attenuation of wave propagation.Figure 6
The stress distribution of dropper IV.Figure 7
The stress distribution of dropper V.In our previous work, the stress of a dropper under a constant force was simulated and its maximum tensile stress was about 65 MPa [21]. The comparison of the stress of loaded dropper subjected to a dynamic force and a constant force is made, as shown in Figure 8. It is clearly seen that the stress distributions under the two conditions are identical. The unique difference is the stress level. The stress value under a dynamic force is extremely higher than that under a constant force. Actually, the constant force in our previous work is set to 200 N while the dynamic force in this work is set to 100sin2πt N. That is, the lower dynamic force provokes very high stress. The maximum tensile stress under the dynamic force is almost seven times as high as that under the constant force. Undoubtedly, the maximum tensile stress plays a key role in fracture. Therefore, it is more possible to cause a fracture of dropper under a dynamic force. Furthermore, it is found that the maximum stress of dropper V is also higher than that under the constant force according to Figures 7 and 8. As described in this paper, dropper V is not directly loaded, and it is far away from the loaded dropper I. Thus, it is deduced that a dynamic force has a more significant impact on the stress of dropper than a constant force, which is the motivations and the original contribution in this work.Figure 8
Comparison of stress under constant and dynamic forces.In the present study, we use a simple model and neglect many factors, such as sleeve ring and clamping pipe. In addition, a sinusoidal dynamic force is applied in order to make a contrast to a subsequent experiment. As well known, there are junction claws between dropper and contact wire and they could cause reflection of wave propagation. Wind perturbation also leads to the additional vibration. These complicated factors would be progressively considered in our future work.


## 4. Conclusion

In this work, a span of five droppers is constructed using a simple model and their stresses are performed numerically. The results suggest that a dynamic force could cause much higher stresses than a constant force. It indicates that a dynamic force has a more significant impact on stress value of dropper than a constant force. A large shock effect could be provoked by a dynamic force. Other nonloaded droppers are also excited to produce very high stresses due to wave propagation. This work would be improved to simulate more complicated conditions for corresponding to reality.



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*Source: 9663767-2022-09-01.xml*