Performance analysis of the door of the hottest hi

Performance analysis of high-speed train doors

I. Introduction

with the rapid development of computers, the design and production mode of products have changed. The work that used to be done only by hand can now be done with the help of computers, while the work that could not be done or was difficult to be done before can also be done with the help of the high-speed computing ability of computers. Every enterprise hopes that its products can be favored by customers and occupy the market to the greatest extent. However, due to the fierce and cruel market competition, it is very difficult to achieve these. Therefore, how to respond to the rapid changes of the market with the fastest speed and the lowest cost is not only a problem that enterprise managers need to constantly think about, but also a problem related to the development of enterprises

II. Raising problems

for transportation, railway transportation is the transportation artery of every country, and the development of high-speed trains is the mainstream of China's railway transportation development. When the train door is closed, the tightness of the door during operation causes a pressure difference between the inside and outside of the door. The pressure difference acting on the door will cause certain deformation of the door. The faster the speed is, the greater the deformation. In addition, when two high-speed trains run opposite, the pressure difference will increase sharply, and the deformation of the door will reach the maximum. Therefore, the manufacturer must understand the actual working conditions of the door (such as the stress and deformation of the door) to ensure that their products are in good working condition in the actual use process

considering the weight and strength, the train doors generally adopt composite structure. On high-speed trains (the train running speed in this paper is 270km/h), the train door is required to have sound insulation and heat insulation functions to ensure the comfort inside the carriage. However, when the train runs at high speed, the pressure difference inside and outside the carriage will deform the door. Once the deformation is greater than the compensation value of the door itself, the door will not have the effect of sound insulation and heat insulation. Therefore, the specific structure of the door can be designed more reasonably by simulating the deformation of the door according to the operation of the train in advance

the door in this paper is composed of five layers of materials, with a door frame around it and a glass window for observation above it. Among the five layers of materials, the outermost and innermost are aluminum alloy, the honeycomb structure layer is connected with the aluminum alloy, the middle layer is polyethylene, and a special glue is used between layers to ensure the continuity of deformation between layers. The internal structure and properties of the above materials are different, including both metallic and non-metallic materials, both isotropic and orthotropic

III. establishment of calculation model

calculation model of door, as shown in Figure 1

Figure 1 calculation model

according to the actual structure of the door, this paper adopts three element types:

(1) the door frame adopts beam189 (shown in B in Figure 1)

(2) the door panel adopts shell99, with 3 layers (as shown in a in Figure 1)

(3) Shell91 is selected for the glass window (as shown in C in Figure 1)

among them, the number of nodes of beam189 is 3, and the degrees of freedom are 6 to 7; Both shell99 and Shell91 have 8 nodes and 6 degrees of freedom

IV. force and constraint conditions

the solid model of the door is shown in Figure 2

Figure 2 solid model of door

because the door needs to be connected with the body, this paper assumes that the body is fixed. According to the requirements of the drawing, pull the door with hooks on both sides of the door (places a, B, C and D in Figure 2) to avoid falling off during train operation. Above the door, there is a beam (E in Figure 2), on which there is a cylinder. The opening and closing of the door are completed by the cylinder

according to the above constraints of the door, set the coordinate axis of the door, that is, the X axis along the longitudinal direction of the door, the Z axis along the transverse direction of the door, and the Y axis along the normal direction of the door

according to the above settings, there are Y-direction constraints at door a, B, C and D and full constraints at door E in this paper

during the operation of the train, the whole train is in a fully enclosed state, and there is a pressure difference inside and outside the carriage. The pressure difference acts on the whole carriage, so it also acts on the door. This force is characterized by uniform distribution. From the perspective of force action, it is a uniformly distributed load, and the direction is along the Y axis. In addition, before the train runs, in order to make the door close to the body, there is a preload at a, B, C, D and E in Figure 2, and the direction is also along the Y axis

according to the actual stress condition of the door, its load distribution is as follows:

(1) there is a force along the Y axis at a, B, C, D and E in Figure 2

(2) there is a uniform force along the y-axis on the whole door

(3) the prediction shows that the action direction of the self weight of the door is along the x-axis direction

v. solution ideas and steps

from the perspective of force and deformation, the train is divided into three working states: static state, normal operation state and train intersection state. The door must work well in each state to achieve the expected purpose of the design. Therefore, it is necessary to calculate the deformation of the above three states respectively

1. The actual structure of the door material involved in this paper is very complex, and it will be very difficult to use the actual section for calculation, and even because of the adjacent size changes. The biggest feature of the ladder reciprocating single screw extruder is that the high filling processing degree of different materials is too large to be divided into grids. Therefore, this paper simplifies the section and puts forward the concept of equivalent section

the main basis for adopting the equivalent section is that the mechanical properties (moment of inertia and product of inertia) of the generated equivalent section under stress are the same as those of the original section, so as to simplify the complex section

by solving the following nonlinear equations, the relevant parameters of the equivalent section are obtained, as shown in formula (1)

in the above equations, s, Ixx, Iyy, IXY, XC and YC are the area, moment of inertia, product of inertia and section centroid coordinates of the section respectively, and T1, T2, T3, T4, W1 and W2 are the parameters of the equivalent section to be calculated respectively

from the above equations, it can be seen that the four equations have six variables, which requires setting the values of two variables in advance according to the actual situation, and then calculating the values of the other four variables until the values of the six variables are reasonable. Therefore, experience plays a certain role in the specific calculation process

2. Set the load step

because the pre tightening force has been applied to the door before the train starts, the deformation distribution of the door under the pre tightening force must be calculated first. In this paper, there are three working states, namely three load steps. The constraints and stress conditions of each state are different. The three load steps are:

(1) the train speed is zero

(2) the train speed is 75m/s, that is, the normal running state

(3) the two trains run in opposite directions at the speed of 75m/s respectively, that is, the two trains intersect

3. Calculation

in this paper, considering that the force and deformation of each carriage are the same during the running of the train, two carriages are taken as the research object in the calculation. At the same time, the air around the train is regarded as a large flow field. With the operation of the train, the air flow field is different at different times. In order to ensure the reliability of the calculation results, the air flow field is required to be large enough. In this paper, the size of the air flow field in front of the train is taken as 100 times the size of the carriage, 35 times the size of the rear, and 60 times the size of the left and right sides. The number of nodes after the grid of this model is very large. In order to save machine time, the grid density is larger in the area close to the carriage, and smaller in the area away from the carriage

since the speed of the train is 270km/h, which is far from the sound speed, and the relative speed is 540km/h when the two vehicles intersect, which is still far from the sound speed, it is assumed that the research object is an incompressible flow field and is treated as a flow around a non lifting object. According to the theory of fluid mechanics, for two-dimensional incompressible inviscid fluids, their potential functions Φ Sum stream function ψ Both should satisfy Laplace equation, as shown in formula (2)

potential function and stream function meet the following boundary conditions respectively, as shown in formula (3)

the sum in formula (3) is respectively Γ 2 tangential and normal velocities on the boundary

in the specific calculation, in order to simplify the calculation, the steady-state analysis method is adopted, the coupling algorithm of ANSYS software is used, and the corresponding boundary conditions are substituted according to the above requirements

VI. results and result analysis

after the above model is qualified, the total number of nodes is 42978, and the total number of elements is 14925. The calculation results of three load steps are shown in figures 3 to 5

Figure 3 Y-direction displacement distribution diagram of the first load step

Figure 4 Y-direction displacement distribution diagram of the second load step

Figure 5 Y-direction displacement distribution diagram of the third load step

through the above calculation, the results obtained are consistent with the actual test results, but there is still a certain gap in specific values. The author believes that the reasons for the gap are as follows

(1) the error between the model and the actual door

from the perspective of stress and deformation, although the nodes of the shell element and the beam element have been coupled together, there is still a gap between the working state of the model and the actual working state of the door. On the other hand, in the model calculation, especially in the calculation of shell99 composite shell, there are certain differences between the interface between layers and the working condition of the actual door, as well as the uniformity of the actual material, which will cause the differences between the two

(2) error of material properties

in the calculation process, most of the performance constants of various materials involved in the shell element are classical coefficients extracted from the material manual, and some are empirical data. These values are different from the actual material values. In particular, some non-metallic materials and orthotropic materials are used in this paper. The minimum and maximum properties of these materials are very different, and some properties are difficult to determine, so it is difficult to determine the characteristic coefficient of these materials

in addition, the shell (1) in this question is treated as a composite material by turning off the oil pump electromechanical unit. The mechanical properties of the composite material are very complex, such as anisotropy, material heterogeneity, low interlaminar shear modulus/shear tensile strength, and a certain degree of nonlinearity. When the interlaminar stress is not considered σ z、 τ ZL and τ For the interaction between ZT, the strength criterion can be used to consider the interaction between interlaminar stresses and determine whether spalling occurs, as shown in formula (4)

in formula (4), ZT and ZC are the tensile and compressive strength between layers respectively, and s' and s "correspond to τ ZL、 τ Interlaminar shear strength of ZT. In fact, due to various reasons, the interlaminar strength of composite materials is low, and local cracks often appear under the action of force, which is also one of the reasons for errors

for polymer non-metallic materials, due to the fiber particle size of the material, the uniformity of various material components and the instability of various industrial parameters in the manufacturing process, the non-uniformity of the material itself and the discreteness of material constants also affect the calculation accuracy to a certain extent. The relationship between the elastic modulus of rigid foam and the fiber content is shown in Figure 6

Figure 6 Relationship between elastic modulus of polyurethane foam and continuous fiber content