CN106545609B - The simulation calculation method for the offset frequencys progressive rate rigidity of plate spring characteristics such as two-stage auxiliary spring formula is non- - Google Patents

The simulation calculation method for the offset frequencys progressive rate rigidity of plate spring characteristics such as two-stage auxiliary spring formula is non- Download PDF

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CN106545609B
CN106545609B CN201710023252.XA CN201710023252A CN106545609B CN 106545609 B CN106545609 B CN 106545609B CN 201710023252 A CN201710023252 A CN 201710023252A CN 106545609 B CN106545609 B CN 106545609B
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CN106545609A (en
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周长城
杨腾飞
刘灿昌
赵雷雷
汪晓
邵明磊
王凤娟
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Shandong University of Technology
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    • F—MECHANICAL ENGINEERING; LIGHTING; HEATING; WEAPONS; BLASTING
    • F16—ENGINEERING ELEMENTS AND UNITS; GENERAL MEASURES FOR PRODUCING AND MAINTAINING EFFECTIVE FUNCTIONING OF MACHINES OR INSTALLATIONS; THERMAL INSULATION IN GENERAL
    • F16F—SPRINGS; SHOCK-ABSORBERS; MEANS FOR DAMPING VIBRATION
    • F16F1/00—Springs
    • F16F1/02—Springs made of steel or other material having low internal friction; Wound, torsion, leaf, cup, ring or the like springs, the material of the spring not being relevant
    • F16F1/18—Leaf springs
    • F16F1/185—Leaf springs characterised by shape or design of individual leaves
    • G—PHYSICS
    • G06—COMPUTING; CALCULATING OR COUNTING
    • G06F—ELECTRIC DIGITAL DATA PROCESSING
    • G06F30/00—Computer-aided design [CAD]
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    • G06F30/17—Mechanical parametric or variational design
    • G—PHYSICS
    • G06—COMPUTING; CALCULATING OR COUNTING
    • G06F—ELECTRIC DIGITAL DATA PROCESSING
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Abstract

本发明涉及两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，属于悬架钢板弹簧技术领域。本发明可根据各片主簧和副簧的结构参数，骑马螺栓夹紧距，弹性模量，主簧和各级副簧的初始切线弧高设计值，在接触载荷仿真计算的基础上，对两级副簧式非等偏频型渐变刚度板簧在不同载荷下的夹紧刚度特性进行仿真计算。通过样机试验可知，本发明所提供的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法是正确的，为两级副簧式非等偏频型渐变刚度板簧的夹紧刚度特性仿真计算提供了可靠的技术方法。利用该方法可得到可靠的夹紧刚度特性仿真计算值，提高产品的设计水平和性能及车辆行驶平顺性；同时，降低设计和试验费用，加快产品开发速度。

The invention relates to a simulation calculation method for the stiffness characteristics of a two-stage auxiliary spring type non-equal bias frequency gradually changing stiffness leaf spring, and belongs to the technical field of suspension leaf springs. According to the structural parameters of each main spring and auxiliary spring, the clamping distance of the saddle bolt, the elastic modulus, the design value of the initial tangent arc height of the main spring and each level of auxiliary spring, on the basis of contact load simulation calculation, the The clamping stiffness characteristics of the two-stage auxiliary spring non-equal bias frequency type gradient stiffness leaf spring under different loads are simulated and calculated. Through the prototype test, it can be seen that the simulation calculation method of the stiffness characteristics of the two-stage auxiliary spring type non-equal deviation frequency gradient stiffness leaf spring provided by the present invention is correct, and it is the clip of the two-stage auxiliary spring type non-equal deviation frequency gradient stiffness leaf spring. The simulation calculation of tight stiffness characteristics provides a reliable technical method. The method can be used to obtain reliable simulation calculation values of clamping stiffness characteristics, improve the design level and performance of products and the ride comfort of vehicles; at the same time, reduce design and test costs and speed up product development.

Description

两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法Simulation Calculation Method of Stiffness Characteristics of Non-equal Deviation Frequency Gradual Stiffness Leaf Springs with Two Stages of Secondary Springs

技术领域technical field

本发明涉及车辆悬架钢板弹簧，特别是两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法。The invention relates to a vehicle suspension leaf spring, in particular to a simulation calculation method for the stiffness characteristics of a two-stage secondary spring type non-equal bias frequency gradient stiffness leaf spring.

背景技术Background technique

为了进一步提高车辆在额定载荷下的行驶平顺性的设计要求，可将原一级渐变刚度板簧的副簧拆分设计为两级副簧，即采用两级副簧式渐变刚度板簧；同时，由于受主簧强度的制约，通常通过主簧初始切线弧高、第一级副簧和第二级副簧初始切线弧高及两级渐变间隙，使副簧适当提前承担载荷，从而降低主簧应力，在接触载荷下的悬架偏频不相等，即两级副簧式非等偏频型渐变刚度板簧，其中，两级副簧式渐变刚度板簧在不同载荷下的夹紧刚度特性，影响悬架偏频及车辆行驶平顺性和安全性，并且渐变刚度板簧的夹紧刚度，不仅与各片主簧和副簧的结构参数有关，而且还与接触载荷有关。然而，由于受两级副簧式非等偏频型渐变刚度板簧的根部重叠部分等效厚度和渐变刚度计算、及接触载荷仿真关键问题的制约，先前一直未能给出两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，因此，不能满足车辆行业快速发展和悬架弹簧悬架现代化CAD设计及软件开发的要求。随着车辆行驶速度及对车辆行驶平顺性和安全性要求的不断提高，对渐变刚度板簧悬架提出了更高要求，因此，必须建立一种精确、可靠的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，为两级副簧式非等偏频型渐变刚度板簧的夹紧刚度特性仿真计算提供可靠的技术方法，满足车辆行业快速发展、车辆行驶平顺性及对渐变刚度板簧的设计要求，提高两级副簧式非等偏频型渐变刚度板簧的设计水平、产品质量及车辆行驶平顺性；同时，降低设计和试验测试费用，加快产品开发速度。In order to further improve the design requirements of the ride comfort of the vehicle under the rated load, the auxiliary spring of the original one-stage gradual stiffness leaf spring can be split and designed into two-stage auxiliary springs, that is, two-stage auxiliary spring type gradient stiffness leaf springs are used; at the same time , due to the restriction of the strength of the main spring, usually through the initial tangent arc height of the main spring, the initial tangent arc height of the first-stage auxiliary spring and the second-stage auxiliary spring, and the two-stage gradual change gap, the auxiliary spring can properly bear the load in advance, thereby reducing the main load. Spring stress, the suspension bias frequency under contact load is not equal, that is, the two-stage secondary spring type non-equal bias frequency type gradual stiffness leaf spring, in which, the clamping stiffness of the two-stage secondary spring type gradual stiffness leaf spring under different loads The characteristics affect the suspension bias frequency and the ride comfort and safety of the vehicle, and the clamping stiffness of the gradient stiffness leaf spring is not only related to the structural parameters of each main spring and auxiliary spring, but also related to the contact load. However, due to the constraints of the calculation of the equivalent thickness and gradient stiffness of the root overlapping part of the two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf spring, and the key issues of contact load simulation, the two-stage secondary spring type has not been given before. The simulation calculation method of the stiffness characteristics of non-equal offset frequency gradient stiffness leaf springs cannot meet the requirements of the rapid development of the vehicle industry and the modern CAD design and software development of suspension springs and suspensions. With the continuous improvement of vehicle speed and the requirements for vehicle ride comfort and safety, higher requirements are put forward for the gradient stiffness leaf spring suspension. Therefore, it is necessary to establish an accurate and reliable two-stage secondary spring non-equal bias The simulation calculation method of the stiffness characteristics of the frequency gradient stiffness leaf spring provides a reliable technical method for the simulation calculation of the clamping stiffness characteristics of the two-stage auxiliary spring type non-equal frequency gradient gradient stiffness leaf spring, which meets the rapid development of the vehicle industry and the smoothness of vehicle driving And the design requirements for the gradient stiffness leaf spring, improve the design level, product quality and vehicle ride comfort of the two-stage secondary spring type non-equal bias frequency gradient gradient stiffness leaf spring; at the same time, reduce the design and test costs and speed up product development .

发明内容Contents of the invention

针对上述现有技术中存在的缺陷，本发明所要解决的技术问题是提供一种简便、可靠的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，仿真计算流程如图1所示。两级副簧式非等偏频型渐变刚度板簧的一半对称结构如图2所示，是由主簧1、第一级副簧2和第二级副簧3组成。采用两级副簧，主簧与第一级副簧之间和第一级副簧与第二级副簧之间设有两级渐变间隙δMA1和δA12，以提高额定载荷下的车辆行驶平顺性；为了确保满足主簧应力强度设计要求，第一级副簧和第二级副簧适当提前承担载荷，悬架渐变载荷偏频不相等，即将板簧设计为非等偏频型渐变刚度板簧。板簧的一半总跨度等于首片主簧的一半作用长度L1T，骑马螺栓夹紧距的一半为L0，宽度为b，弹性模量为E。主簧1的片数为n，主簧各片的厚度为hi，一半作用长度为LiT，一半夹紧长度Li＝LiT-L0/2，i＝1,2,…,n。第一级副簧片数为m1，第一级副簧各片的厚度为hA1j，一半作用长度为LA1jT，一半夹紧长度LA1j＝Ln+j＝LA1jT-L0/2，j＝1,2,…,m1，主簧与第一副簧的片数之和N1＝n+m1。第二级副簧片数为m2，第二级副簧各片的厚度为hA2k，一半作用长度为LA2kT，一半夹紧长度LA2k＝LN1+k＝LA2kT-L0/2，k＝1,2,…,m2。主副簧的总片数N＝n+m1+m2。根据各片主簧和副簧的结构参数、主簧和各级副簧的初始切弧高设计值、骑马螺栓夹紧距，在接触载荷仿真计算的基础上，对两级副簧式非等偏频型渐变刚度板簧在不同载荷下的夹紧刚度特性进行仿真计算。In view of the above-mentioned defects in the prior art, the technical problem to be solved by the present invention is to provide a simple and reliable simulation calculation method for the stiffness characteristics of the two-stage auxiliary spring type non-equal bias frequency gradient stiffness leaf spring. The simulation calculation flow is shown in the figure 1. The semi-symmetrical structure of the two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf spring is shown in Figure 2, which is composed of the main spring 1, the first secondary spring 2 and the second secondary spring 3. Two-stage auxiliary springs are adopted, and two-stage gradual gaps δ MA1 and δ A12 are set between the main spring and the first-stage auxiliary spring and between the first-stage auxiliary spring and the second-stage auxiliary spring to improve the driving performance of the vehicle under the rated load Ride comfort; in order to meet the stress strength design requirements of the main spring, the first-stage auxiliary spring and the second-stage auxiliary spring should bear the load in advance, and the suspension gradient load deviation frequency is not equal, that is, the leaf spring is designed as a non-equal deviation frequency type gradient stiffness leaf spring. Half of the total span of the leaf spring is equal to half of the working length L 1T of the first main spring, half of the saddle bolt clamping distance is L 0 , the width is b, and the elastic modulus is E. The number of pieces of the main spring 1 is n, the thickness of each piece of the main spring is h i , half of the working length is L iT , and half of the clamping length is L i =L iT -L 0 /2, i=1,2,…,n . The number of first-stage auxiliary reeds is m 1 , the thickness of each leaf of the first-stage auxiliary reed is h A1j , half of the working length is L A1jT , and half of the clamping length is L A1j ＝L n+j ＝L A1jT -L 0 /2 , j=1,2,...,m 1 , the sum of the number of pieces of the main spring and the first auxiliary spring is N 1 =n+m 1 . The number of second-stage auxiliary reeds is m 2 , the thickness of each second-stage auxiliary reed is h A2k , half of the working length is L A2kT , and half of the clamping length is L A2k = L N1+k = L A2kT -L 0 /2 , k=1,2,...,m 2 . The total number of primary and secondary springs N=n+m 1 +m 2 . According to the structural parameters of the main spring and auxiliary spring of each piece, the design value of the initial cutting arc height of the main spring and the auxiliary spring at each level, and the clamping distance of the saddle bolt, on the basis of the simulation calculation of the contact load, the two-stage auxiliary spring non-equal The clamping stiffness characteristics of the bias-frequency gradient-stiffness leaf spring under different loads were simulated and calculated.

为解决上述技术问题，本发明所提供的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，其特征在于采用以下仿真计算步骤：In order to solve the above-mentioned technical problems, the simulation calculation method of the stiffness characteristics of the two-stage secondary spring type non-equal bias frequency gradient stiffness plate spring provided by the present invention is characterized in that the following simulation calculation steps are adopted:

(1)两级副簧式非等偏频型渐变刚度板簧的主簧及其与各级副簧夹紧刚度的计算：(1) Calculation of the main spring of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring and its clamping stiffness with the auxiliary springs at all levels:

i步骤：不同片数重叠段的等效厚度hle计算Step i: Calculation of equivalent thickness h le of overlapping sections with different numbers of sheets

根据主簧片数n,主簧各片的厚度hi，i＝1,2,…,n；第一级副簧片数m1,第一级副簧各片的厚度hA1j，j＝1,2,…,m1；第二级副簧片数m2,第二级副簧各片的厚度hA2k，k＝1,2,…,m2；主簧与第一级副簧的片数之和N1＝n+m1，主副簧的总片数N＝n+m1+m2；对两级副簧式非等偏频型渐变刚度板簧的不同片数l重叠段的等效厚度hle进行计算，l＝1,2,…,N，即：According to the number of main reeds n, the thickness h i of each piece of the main reed, i=1,2,...,n; the number of first-level secondary reeds m 1 , the thickness of each piece of the first-level secondary reed h A1j , j= 1,2,…,m 1 ; the number of second-stage secondary reeds m 2 , the thickness of each second-stage secondary reed h A2k , k=1,2,…,m 2 ; the main spring and the first-stage secondary spring The sum of the number of sheets N 1 =n+m 1 , the total number of sheets of the main and auxiliary springs N=n+m 1 +m 2 ; for the different numbers of sheets of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring l The equivalent thickness h le of the overlapping section is calculated, l=1,2,...,N, namely:

其中，主簧根部重叠部分的等效厚度hMe、主簧与第一级副簧的根部重叠部分的等效厚度hMA1e、及主副簧总的根部重叠部分等效厚度hMA2e，分别为Among them, the equivalent thickness h Me of the root overlapping portion of the main spring, the equivalent thickness h MA1e of the root overlapping portion of the main spring and the first-stage auxiliary spring, and the equivalent thickness h MA2e of the root overlapping portion of the main and auxiliary springs are respectively

ii步骤：主簧夹紧刚度KM计算Step ii: Calculation of main spring clamping stiffness K M

根据两级副簧式非等偏频型渐变刚度板簧的宽度b，弹性模量E；主簧片数n，主簧各片的一半夹紧长度Li，i＝1,2,…,n；及i步骤中计算得到的hle，l＝i＝1,2,...,n，对主簧夹紧刚度进行计算，即According to the width b and elastic modulus E of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring; the number of main reeds n, half the clamping length L i of each piece of the main spring, i=1,2,..., n; and h le calculated in step i, l=i=1,2,...,n, to calculate the clamping stiffness of the main spring, namely

iii步骤：主簧与第一级副簧的夹紧复合刚度KMA1计算Step iii: Calculation of the clamping composite stiffness K MA1 of the main spring and the first-stage auxiliary spring

根据两级副簧式非等偏频型渐变刚度板簧的宽度b，弹性模量E；主簧片数n,主簧各片的一半夹紧长度Li，i＝1,2,…,n；第一级副簧片数m1，第一级副簧各片的一半夹紧长度为LA1j＝Ln+j，j＝1,2,…,m1；主簧与第一级副簧的片数之和N1＝n+m1，及i步骤中计算得到的hle，l＝1,2,...,N1，对主簧与第一级副簧的夹紧复合刚度KMA1进行计算，即According to the width b and elastic modulus E of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring; the number of main reeds n, half the clamping length L i of each piece of the main spring, i=1,2,..., n; the number of first-stage auxiliary reeds m 1 , the clamping length of half of each leaf of the first-stage auxiliary spring is L A1j ＝L n+j , j=1,2,...,m 1 ; the main spring and the first-stage The sum of the number of secondary springs N 1 =n+m 1 , and h le calculated in step i, l=1,2,...,N 1 , the clamping of the main spring and the first secondary spring Composite stiffness K MA1 is calculated as

iv步骤：主副簧的总复合夹紧刚度KMA2计算：Step iv: Calculation of the total composite clamping stiffness K MA2 of the primary and secondary springs:

根据两级副簧式非等偏频型渐变刚度板簧的宽度b，弹性模量E；主簧片数n，各片主簧的一半夹紧长度Li，i＝1,2,…,n；第一级副簧片数m1，第一级副簧各片的一半夹紧长度为LA1j＝Ln+j，j＝1,2,…,m1；第二级副簧片数m2，第二级副簧各片的一半夹紧长度LA2k＝LN1+k，k＝1,2,…,m2；主副簧的总片数N＝n+m1+m2，及i步骤中计算得到的hle，l＝1,2,...,N，对主副簧的总复合夹紧刚度KMA2进行计算，即According to the width b and elastic modulus E of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring; the number of main reeds n, half the clamping length L i of each main spring, i=1,2,..., n; the number of first-stage auxiliary reeds m 1 , the clamping length of half of each leaf of the first-stage auxiliary reed is L A1j = L n+j , j=1,2,...,m 1 ; the second-stage auxiliary reeds Number m 2 , half clamping length L A2k ＝L N1+k of each leaf of the second secondary spring, k=1,2,...,m 2 ; total number of primary and secondary springs N=n+m 1 +m 2 , and h le calculated in step i, l=1,2,...,N, calculate the total composite clamping stiffness K MA2 of the main and auxiliary springs, namely

(2)两级副簧式非等偏频型渐变刚度板簧的主簧及各级副簧的曲率半径的计算：(2) Calculation of the radius of curvature of the main spring of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring and the curvature radii of the auxiliary springs at all levels:

I步骤：主簧末片下表面初始曲率半径RM0b的计算Step I: Calculation of the initial radius of curvature R M0b of the lower surface of the main reed

根据主簧初始切线弧高HgM0，主簧首片的一半夹紧长度L1，主簧片数n，主簧各片的厚度hi，i＝1,2,…,n；对主簧末片下表面初始曲率半径RM0b进行计算，即According to the initial tangent arc height H gM0 of the main spring, half the clamping length L 1 of the first leaf of the main spring, the number of main spring leaves n, and the thickness h i of each leaf of the main spring, i =1,2,…,n; for the main spring The initial radius of curvature R M0b of the lower surface of the final piece is calculated, that is

II步骤：第一级副簧首片上表面初始曲率半径RA10a的计算Step II: Calculation of the initial radius of curvature R A10a on the upper surface of the first secondary reed

根据第一级副簧首片的一半夹紧长度LA11，第一级副簧的初始切线弧高设计值HgA10，对第一级副簧首片上表面初始曲率半径RA10a进行计算，即According to half the clamping length L A11 of the first leaf of the first auxiliary spring and the design value of the initial tangent arc height H gA10 of the first auxiliary spring, the initial curvature radius R A10a of the upper surface of the first auxiliary spring is calculated, namely

III步骤：第一级副簧末片下表面初始曲率半径RA10b的计算Step III: Calculation of the initial radius of curvature R A10b of the lower surface of the first secondary secondary reed

根据第一级副簧片数m1，第一级副簧各片的厚度hA1j，j＝1,2,…,m1；及II步骤中计算得到的RA10a，对第一级副簧末片下表面初始曲率半径RA10b进行计算，即According to the number m 1 of the first-stage secondary reeds, the thickness h A1j of each piece of the first-stage secondary reed, j=1,2,...,m 1 ; and the R A10a calculated in step II, for the first-stage secondary reed The initial radius of curvature R A10b of the lower surface of the final piece is calculated, that is

IV步骤：第二级副簧首片上表面初始曲率半径RA20a的计算Step IV: Calculation of the initial radius of curvature R A20a on the upper surface of the first secondary secondary reed

根据第二级副簧首片的一半夹紧长度LA21，第二级副簧的初始切线弧高设计值HgA20，对第二级副簧首片上表面初始曲率半径RA20a进行计算，即According to half the clamping length L A21 of the first leaf of the second-stage auxiliary spring and the design value of the initial tangent arc height H gA20 of the second-stage auxiliary spring, the initial curvature radius R A20a of the upper surface of the first leaf of the second-stage auxiliary spring is calculated, namely

(3)两级副簧式非等偏频型渐变刚度板簧的各次接触载荷的仿真计算：(3) The simulation calculation of each contact load of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring:

A步骤：第1次开始接触载荷Pk1的仿真计算Step A: start the simulation calculation of the contact load P k1 for the first time

根据两级副簧式非等偏频型渐变刚度板簧的宽度b，弹性模量E；主簧首片的一半夹紧长度L1，步骤(1)中计算得到的hMe，步骤(2)中计算得到的RM0b和RA10a，对第1次开始接触载荷Pk1进行仿真计算，即According to the width b and elastic modulus E of the two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf spring; half the clamping length L 1 of the first leaf of the main spring, h Me calculated in step (1), step (2 R M0b and R A10a calculated in ) are simulated for the initial contact load P k1 at the first time, namely

B步骤：第2次开始接触载荷Pk2的仿真计算Step B: start the simulation calculation of the contact load P k2 for the second time

根据两级副簧式非等偏频型渐变刚度板簧的宽度b，弹性模量E；步骤(1)中计算得到的hMA1e，步骤(2)中计算得到的RA10b和RA20a，及A步骤中仿真计算得到的Pk1，对第2次开始接触载荷Pk2进行仿真计算，即According to the width b of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring, the elastic modulus E; h MA1e calculated in step (1), RA10b and RA20a calculated in step (2), and The P k1 obtained by the simulation calculation in step A is simulated and calculated for the second initial contact load P k2 , namely

C步骤：第2次完全接触载荷Pw2的仿真计算Step C: Simulation calculation of the second full contact load P w2

根据A步骤中仿真计算得到的Pk1，B步骤中仿真计算得到的Pk2，对第2次完全接触载荷Pw2进行仿真计算，即According to the P k1 obtained from the simulation calculation in step A and the P k2 obtained from the simulation calculation in step B, the second full contact load P w2 is simulated and calculated, namely

(4)两级副簧式非等偏频型渐变刚度板簧的夹紧刚度特性的仿真计算：(4) Simulation calculation of the clamping stiffness characteristics of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring:

根据步骤(1)中计算得到的KM、KMA1和KMA2，步骤(3)中仿真计算所得到的Pk1、Pk2和Pw2，对两级副簧式非等偏频型渐变刚度板簧在不同载荷P下的夹紧刚度特性进行仿真计算，即According to K M , K MA1 and K MA2 calculated in step (1), and P k1 , P k2 and P w2 obtained in step (3), the gradient stiffness of the two-stage secondary spring non-equal deviation frequency type The clamping stiffness characteristics of the leaf spring under different loads P are simulated and calculated, namely

本发明比现有技术具有的优点Advantages of the present invention over prior art

由于受两级副簧式非等偏频型渐变刚度板簧的根部重叠部分等效厚度和渐变刚度计算及接触载荷仿真关键问题的制约，先前一直未能给出两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，因此，不能满足车辆行业快速发展和悬架弹簧悬架现代化CAD设计及软件开发的要求。本发明可根据各片主簧和副簧的结构参数，弹性模量，主簧和各级副簧的初始切线弧高设计值，在接触载荷仿真计算的基础上，对两级副簧式非等偏频型渐变刚度板簧在不同载荷下的夹紧刚度特性进行仿真计算。通过样机加载刚度特性试验可知，本发明所提供的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法是正确的，为两级副簧式非等偏频型渐变刚度板簧的夹紧刚度特性仿真计算提供了可靠的技术方法。利用该方法可得到可靠的夹紧刚度特性仿真计算值，提高产品的设计水平和性能及车辆行驶平顺性；同时，降低设计和试验费用，加快产品开发速度。Due to the constraints of the calculation of the equivalent thickness and gradient stiffness of the root overlapping part of the two-stage non-equal deflection frequency leaf spring and the key issues of contact load simulation, the two-stage non-equal deflection spring has not been given before. Therefore, it cannot meet the requirements of the rapid development of the vehicle industry and the modern CAD design and software development of suspension springs. According to the structural parameters of each main spring and auxiliary spring, the elastic modulus, the design value of the initial tangent arc height of the main spring and the auxiliary springs at all levels, and on the basis of contact load simulation calculation, the two-stage auxiliary spring non- The clamping stiffness characteristics of equal deviation frequency type gradient stiffness leaf springs under different loads were simulated and calculated. Through the load stiffness characteristic test of the prototype, it can be seen that the simulation calculation method of the stiffness characteristics of the two-stage auxiliary spring type non-equal deviation frequency gradient stiffness plate spring provided by the present invention is correct, and it is a two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness plate The simulation calculation of the clamping stiffness characteristics of the spring provides a reliable technical method. The method can be used to obtain reliable simulation calculation values of clamping stiffness characteristics, improve the design level and performance of products and the ride comfort of vehicles; at the same time, reduce design and test costs and speed up product development.

附图说明Description of drawings

为了更好地理解本发明，下面结合附图做进一步的说明。In order to better understand the present invention, further description will be made below in conjunction with the accompanying drawings.

图1是两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算流程图；Fig. 1 is the simulation calculation flow chart of the stiffness characteristics of the two-stage secondary spring type non-equal bias frequency gradient stiffness leaf spring;

图2是两级副簧式非等偏频型渐变刚度板簧的一半对称结构示意图；Fig. 2 is a schematic diagram of a semi-symmetrical structure of a two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring;

图3是实施例的两级副簧式非等偏频型渐变刚度板簧的夹紧刚度K随载荷P的变化特性曲线。Fig. 3 is a characteristic curve of the clamping stiffness K varying with the load P of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring of the embodiment.

具体实施方案specific implementation plan

下面通过实施例对本发明作进一步详细说明。The present invention will be described in further detail below by way of examples.

实施例：某两级副簧式非等偏频型渐变刚度板簧的宽度b＝63mm，骑马螺栓夹紧距的一半L0＝50mm，弹性模量E＝200GPa。主簧片数n＝3片，主簧各片的厚度h1＝h2＝h3＝8mm，一半作用长度分别为L1T＝525mm，L2T＝450mm，L3T＝350mm；主簧各片的一半夹紧长度分别为L1＝L1T-L0/2＝500mm，L2＝L2T-L0/2＝425mm，L3＝L3T-L0/2＝325mm。第一级副簧的片数m1＝1片，厚度hA11＝13mm，一半作用长度为LA11T＝250mm，一半夹紧长度为LA11＝L4＝LA11T-L0/2＝225mm。主簧与第一级副簧的片数之和N1＝n+m1＝4。第二级副簧的片数m2＝1，厚度hA21＝13mm，一半作用长度为LA21T＝150mm，一半夹紧长度为LA21＝L5＝LA21T-L0/2＝125mm。空载载荷P0＝1715N，额定载荷PN＝7227N。主副簧的总片数N＝5。主簧的初始切线弧高HgM0＝85.3mm，第一级副簧的初始切线弧高HgA10＝9.1mm，第二级副簧的初始切线弧高HgA20＝2.4mm。根据该渐变刚度钢板弹簧的各片主簧与第一级和第二级副簧的结构参数，弹性模量，空载载荷和额定载荷，主簧及各级副簧的初始切线弧高设计值，对该两级副簧式非等偏频型渐变刚度板簧在不同载荷下的夹紧刚度特性进行仿真计算。Example: The width of a two-stage auxiliary spring type unequal frequency gradient leaf spring with variable stiffness is b = 63mm, the half of the saddle bolt clamping distance L 0 = 50mm, and the modulus of elasticity E = 200GPa. The number of main reed pieces n=3 pieces, the thickness of each piece of main spring h 1 =h 2 =h 3 =8mm, half of the effective length is L 1T =525mm, L 2T =450mm, L 3T =350mm; each piece of main spring Half of the clamping lengths are L 1 =L 1T -L 0 /2 = 500 mm, L 2 = L 2T - L 0 /2 = 425 mm, L 3 = L 3T - L 0 /2 = 325 mm. The number of pieces of the first secondary spring is m 1 =1 piece, the thickness h A11 =13mm, half the working length is L A11T =250mm, and half the clamping length is L A11 =L 4 =L A11T -L 0 /2=225mm. The sum N 1 =n+m 1 =4 of the pieces of the main spring and the first secondary spring. The number of sheets of the secondary secondary spring is m 2 =1, the thickness h A21 =13mm, half the working length is L A21T =150mm, and half the clamping length is L A21 =L 5 =L A21T -L 0 /2=125mm. No-load load P 0 =1715N, rated load P N =7227N. The total number of sheets of primary and secondary springs N=5. The initial tangent arc height H gM0 of the main spring = 85.3 mm, the initial tangent arc height H gA10 of the first secondary spring = 9.1 mm, and the initial tangential arc height H gA20 of the second secondary spring = 2.4 mm. According to the structural parameters, elastic modulus, no-load load and rated load of each main spring and the first and second auxiliary springs of the gradual stiffness leaf spring, the design value of the initial tangent arc height of the main spring and each auxiliary spring , the clamping stiffness characteristics of the two-stage auxiliary spring non-equal deviation frequency type gradient stiffness leaf spring under different loads were simulated and calculated.

本发明实例所提供的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法，其仿真计算流程，如图1所示，具体仿真计算步骤如下：The simulation calculation method of the two-stage auxiliary spring type non-equal bias frequency gradient stiffness plate spring stiffness characteristics provided by the example of the present invention, its simulation calculation process, as shown in Figure 1, the specific simulation calculation steps are as follows:

(1)两级副簧式非等偏频型渐变刚度板簧的主簧及其与各级副簧夹紧刚度的计算：(1) Calculation of the main spring of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring and its clamping stiffness with the auxiliary springs at all levels:

i步骤：不同片数重叠段的等效厚度hle的计算Step i: Calculation of the equivalent thickness h le of overlapping segments with different numbers of sheets

根据主簧片数n＝3,主簧各片的厚度h1＝h2＝h3＝8mm；第一级副簧片数m1＝1,厚度hA11＝13mm；第二级副簧片数m2＝1,厚度hA21＝13mm；主副簧的总片数N＝5，对两级副簧式非等偏频型渐变刚度板簧的不同片数l重叠段的等效厚度hle进行计算，l＝1,2,…,N，即：According to the number of main reeds n=3, the thickness of each main reed h 1 =h 2 =h 3 =8mm; the number of first-level secondary reeds m 1 =1, the thickness h A11 =13mm; the second-level secondary reeds Number m 2 = 1, thickness h A21 = 13mm; the total number of primary and secondary springs N = 5, and the equivalent thickness h of overlapping sections with different numbers l of two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf springs le for calculation, l=1,2,…,N, that is:

h1e＝h1＝8.0mm；h 1e =h 1 =8.0 mm;

其中，主簧根部重叠部分的等效厚度hMe、主簧与第一级副簧的根部重叠部分的等效厚度hMA1e、及主副簧总的根部重叠部分等效厚度hMA2e，分别为Among them, the equivalent thickness h Me of the root overlapping portion of the main spring, the equivalent thickness h MA1e of the root overlapping portion of the main spring and the first-stage auxiliary spring, and the equivalent thickness h MA2e of the root overlapping portion of the main and auxiliary springs are respectively

ii步骤：主簧夹紧刚度KM的计算Step ii: Calculation of main spring clamping stiffness K M

根据两级副簧式非等偏频型渐变刚度板簧的宽度b＝63mm，弹性模量E＝200GPa；主簧片数n＝3,其中，各片主簧的一半夹紧长度L1＝500mm，L2＝425mm，L3＝325mm，及i步骤中计算得到的h1e＝8.0mm，h2e＝10.1mm，h3e＝11.5mm，l＝i＝1,2,...,n，对主簧夹紧刚度进行计算，即According to the width of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring b=63mm, the elastic modulus E=200GPa; the number of main reeds n=3, wherein, half of the clamping length L 1 of each main spring = 500mm, L 2 =425mm, L 3 =325mm, h 1e =8.0mm, h 2e =10.1mm, h 3e =11.5mm, l=i=1,2,...,n calculated in step i , to calculate the clamping stiffness of the main spring, namely

iii步骤：主簧与第一级副簧的复合夹紧刚度KMA1计算Step iii: Calculation of the composite clamping stiffness K MA1 of the main spring and the first secondary spring

根据两级副簧式非等偏频型渐变刚度板簧的宽度b＝63mm，弹性模量E＝200GPa；主簧片数n＝3,主簧各片的一半夹紧长度L1＝500mm，L2＝425mm，L3＝325mm；第一级副簧片数m1＝1，第一级副簧的一半夹紧长度为LA11＝L4＝225mm；主簧与第一级副簧的总片数N1＝n+m1＝4，及i步骤中计算得到的h1e＝8.0mm，h2e＝10.1mm，h3e＝11.5mm，h4e＝15.5mm，l＝1,2,...,N1，对主簧与第一级副簧的复合夹紧刚度KMA1进行计算，即According to the width of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring b=63mm, the elastic modulus E=200GPa; the number of main reeds n=3, half the clamping length of each piece of the main spring L 1 =500mm, L 2 ＝425mm, L 3 ＝325mm; the number of first-stage secondary springs m 1 =1, the clamping length of half of the first-stage secondary spring is L A11 ＝L 4 ＝225mm; the distance between the main spring and the first-stage secondary spring The total number of pieces N 1 =n+m 1 =4, and h 1e =8.0mm, h 2e =10.1mm, h 3e =11.5mm, h 4e =15.5mm, l=1,2, calculated in step i ..., N 1 , to calculate the composite clamping stiffness K MA1 of the main spring and the first-stage auxiliary spring, namely

iv步骤：主副簧的总复合夹紧刚度KMA2计算：Step iv: Calculation of the total composite clamping stiffness K MA2 of the primary and secondary springs:

根据两级副簧式非等偏频型渐变刚度板簧的宽度b＝63mm，弹性模量E＝200Gpa；主簧片数n＝3，主簧各片的一半夹紧长度L1＝500mm，L2＝425mm，L3＝325mm；第一级副簧片数m1＝1，第一级副簧的一半夹紧长度为LA11＝L4＝225mm；第二级副簧片数m2＝1，第二级副簧的一半夹紧长度LA21＝L5＝125mm；主副簧的总片数N＝5，及i步骤中计算得到的h1e＝8.0mm，h2e＝10.1mm，h3e＝11.5mm，h4e＝15.5mm，h5e＝18.1mm，l＝1,2,...,N，对主副簧的总复合夹紧刚度KMA2进行计算，即According to the width b=63mm of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring, the elastic modulus E=200Gpa; the number of main reeds n=3, half the clamping length L 1 of each piece of the main spring =500mm, L 2 ＝425mm, L 3 ＝325mm; the number of first-stage secondary reed m 1 =1, the clamping length of half of the first-stage secondary spring is L A11 ＝L 4 ＝225mm; the number of second-stage secondary reed m 2 =1, half the clamping length of the secondary secondary spring L A21 =L 5 =125mm; the total number of primary and secondary springs N=5, and h 1e =8.0mm, h 2e =10.1mm calculated in step i , h 3e ＝11.5mm, h 4e ＝15.5mm, h 5e ＝18.1mm, l＝1,2,...,N, calculate the total composite clamping stiffness K MA2 of the primary and secondary springs, namely

(2)两级副簧式非等偏频型渐变刚度板簧的主簧及各级副簧的曲率半径的计算：(2) Calculation of the radius of curvature of the main spring of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring and the curvature radii of the auxiliary springs at all levels:

I步骤：主簧末片下表面初始曲率半径RM0b的计算Step I: Calculation of the initial radius of curvature R M0b of the lower surface of the main reed

根据主簧的初始切线弧高HgM0＝85.3mm，主簧首片的一半夹紧长度L1＝500mm，主簧片数n＝3，主簧各片的厚度hi＝8mm，i＝1,2,…,n；对主簧末片下表面初始曲率半径RM0b进行计算，即According to the initial tangent arc height of the main spring H gM0 = 85.3mm, half the clamping length of the first leaf of the main spring L 1 = 500mm, the number of main spring leaves n = 3, the thickness of each leaf of the main spring h i = 8mm, i = 1 ,2,…,n; Calculate the initial radius of curvature R M0b of the lower surface of the main reed, that is

II步骤：第一级副簧首片上表面初始曲率半径RA10a的计算Step II: Calculation of the initial radius of curvature R A10a on the upper surface of the first secondary reed

根据第一级副簧首片的一半夹紧长度LA11＝225mm，第一级副簧的初始切线弧高HgA10＝9.1mm，对第一级副簧首片上表面初始曲率半径RA10a进行计算，即According to half the clamping length L A11 of the first secondary spring first piece = 225mm, and the initial tangent arc height H gA10 of the first secondary secondary spring = 9.1mm, the initial curvature radius R A10a of the upper surface of the first secondary secondary spring is calculated ,Right now

III步骤：第一级副簧末片下表面初始曲率半径RA10b的计算Step III: Calculation of the initial radius of curvature R A10b of the lower surface of the first secondary secondary reed

根据第一级副簧片数m1＝1，厚度hA11＝13mm，及II步骤中计算得到的RA10a＝2786.1mm，对第一级副簧末片下表面初始曲率半径RA10b进行计算，即According to the number of first-stage secondary reeds m 1 =1, thickness h A11 =13mm, and R A10a calculated in step II =2786.1mm, the initial radius of curvature R A10b of the lower surface of the first-stage secondary reed is calculated, which is

IV步骤：第二级副簧首片上表面初始曲率半径RA20a的计算Step IV: Calculation of the initial radius of curvature R A20a on the upper surface of the first secondary secondary reed

根据第二级副簧首片的一半夹紧长度LA21＝125mm，第二级副簧的初始切线弧高HgA20＝2.4mm，对第二级副簧末片上表面初始曲率半径RA20a进行计算，即According to half the clamping length L A21 of the first leaf of the second-stage auxiliary spring = 125mm, and the initial tangent arc height H gA20 of the second-stage auxiliary spring = 2.4mm, the initial radius of curvature R A20a of the upper surface of the second-stage auxiliary spring is calculated ,Right now

(3)两级副簧式非等偏频型渐变刚度板簧的各次接触载荷的仿真计算：(3) The simulation calculation of each contact load of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring:

A步骤：第1次开始接触载荷Pk1的仿真计算Step A: start the simulation calculation of the contact load P k1 for the first time

根据两级副簧式非等偏频型渐变刚度板簧的宽度b＝63mm，弹性模量E＝200GPa；主簧首片的一半夹紧长度L1＝500mm，步骤(1)中计算得到的hMe＝11.5mm，步骤(2)中计算得到的RM0b＝1532.1mm和RA10a＝2786.1mm，对第1次开始接触载荷Pk1进行仿真计算，即According to the width b=63mm of the two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf spring, the modulus of elasticity E=200GPa; half the clamping length L 1 of the first leaf of the main spring=500mm, calculated in step (1) h Me ＝11.5mm, R M0b ＝1532.1mm and RA10a ＝2786.1mm calculated in the step (2), carry out the simulation calculation on the initial contact load P k1 for the first time, that is

B步骤：第2次开始接触载荷Pk2的仿真计算Step B: start the simulation calculation of the contact load P k2 for the second time

根据两级副簧式非等偏频型渐变刚度板簧的宽度b＝63mm，弹性模量E＝200GPa；主簧首片的一半夹紧长度L1＝500mm，步骤(1)中计算得到的hMA1e＝15.5mm，步骤(2)中计算得到的RA10b＝2799.1mm和RA20a＝3256.4mm，及A步骤中仿真计算得到的Pk1＝1895N，对第2次开始接触载荷Pk2进行仿真计算，即According to the width b=63mm of the two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf spring, the modulus of elasticity E=200GPa; half the clamping length L 1 of the first leaf of the main spring=500mm, calculated in step (1) h MA1e ＝15.5mm, RA10b ＝2799.1mm and RA20a ＝3256.4mm calculated in step (2), and P k1 ＝1895N calculated by simulation in step A, simulate the second initial contact load P k2 calculation, ie

C步骤：第2次完全接触载荷Pw2的仿真计算Step C: Simulation calculation of the second full contact load P w2

根据A步骤中仿真计算得到的Pk1＝1895N，B步骤中仿真计算得到的Pk2＝2677N，对第2次完全接触载荷Pw2进行仿真计算，即According to P k1 = 1895N obtained by simulation calculation in step A and P k2 = 2677N obtained by simulation calculation in step B, the simulation calculation of the second full contact load P w2 is carried out, namely

(4)两级副簧式非等偏频型渐变刚度板簧的夹紧刚度特性的仿真计算：(4) Simulation calculation of the clamping stiffness characteristics of the two-stage auxiliary spring type non-equal deviation frequency type gradient stiffness leaf spring:

根据空载载荷P0＝1715N，额定载荷PN＝7227N，步骤(1)中计算得到的KM＝75.4N/mm、KMA1＝144.5N/mm和KMA2＝172.9N/mm，步骤(3)中仿真计算所得到的Pk1＝1895N、Pk2＝2677N和Pw2＝3781N，对两级副簧式非等偏频型渐变刚度板簧在不同载荷P下的夹紧刚度特性进行仿真计算，即According to no-load load P 0 =1715N, rated load P N =7227N, K M =75.4N/mm, K MA1 =144.5N/mm and K MA2 =172.9N/mm calculated in step (1), step ( 3) P k1 = 1895N, P k2 = 2677N and P w2 = 3781N obtained from the simulation calculation in 3), simulate the clamping stiffness characteristics of the two-stage auxiliary spring type non-equal bias frequency type gradient stiffness leaf spring under different loads P calculation, ie

利用Matlab计算程序，仿真计算所得到的该两级副簧式非等偏频型渐变刚度板簧的夹紧刚度K随载荷P变化特性曲线，如图3所示，其中，在第1次开始接触载荷Pk1、第2次开始接触载荷Pk2、第2次完全接触载荷Pw2和额定载荷PN情况下的夹紧刚度分别为Kk1＝KM＝75.4N/mm，Kk2＝KMA1＝144.5N/mm，Kw2＝KN＝KMA2＝172.9N/mm。Using the Matlab calculation program, the characteristic curve of the clamping stiffness K changing with the load P of the two-stage secondary spring type non-equal bias frequency type gradient stiffness leaf spring obtained by simulation calculation is shown in Fig. 3, in which, at the first time The clamping stiffness under the contact load P k1 , the second initial contact load P k2 , the second full contact load P w2 , and the rated load P N are K k1 =K M =75.4N/mm, K k2 =K MA1 =144.5 N/mm, K w2 =K N =K MA2 =172.9 N/mm.

通过样机试验可知，本发明所提供的两级副簧式非等偏频渐变刚度板簧刚度特性的仿真计算法是正确的，为两级副簧式非等偏频型渐变刚度板簧的刚度特性仿真计算提供了可靠的技术方法。利用该方法可得到可靠的夹紧刚度特性仿真计算值，提高产品的设计水平和性能及车辆行驶平顺性；同时，降低设计及试验费用，加快产品开发速度。It can be seen from the prototype test that the simulation calculation method of the stiffness characteristics of the two-stage auxiliary spring type non-equal deviation frequency gradient stiffness leaf spring provided by the present invention is correct, which is the stiffness of the two-stage auxiliary spring type non-equal deviation frequency gradient stiffness leaf spring The characteristic simulation calculation provides a reliable technical method. The method can be used to obtain reliable simulation calculation values of clamping stiffness characteristics, improve the design level and performance of products and the ride comfort of vehicles; at the same time, reduce design and test costs and speed up product development.

Claims (1)

1. the simulation calculation method for the offset frequencys progressive rate rigidity of plate spring characteristics such as two-stage auxiliary spring formula is non-, wherein, each leaf spring is in Heart mounting hole symmetrical structure, installation clamp away from half for U-bolts clamp away from half；Auxiliary spring is designed as two-stage pair Spring by main spring and the initial tangential camber of two-stage auxiliary spring and two-stage gradual change gap, improves traveling of the vehicle under rated load Ride comfort；In order to ensure meeting main spring stress intensity design requirement, first order auxiliary spring and second level auxiliary spring is made suitably to undertake in advance The offset frequencys type progressive rate leaf springs such as load, the offset frequency being suspended under gradual change load is unequal, i.e., two-stage auxiliary spring formula is non-；According to each The structural parameters of leaf spring, U-bolts clamp the initial tangential camber design load away from, elasticity modulus, main spring and two-stage auxiliary spring, On the basis of contact load simulation calculation, clamping of the offset frequencys type progressive rate leaf spring such as non-to two-stage auxiliary spring formula under different loads Stiffness characteristics carry out simulation calculation, and specific simulation calculation step is as follows：
(1) the main springs of offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-and its calculating that rigidity is clamped with auxiliary springs at different levels：
I steps：The equivalent thickness h of different the piece number overlay segmentsleIt calculates
According to main reed number n, the thickness h of main each of springi, i=1,2 ..., n；First order auxiliary spring the piece number m1, each of first order auxiliary spring Thickness hA1j, j=1,2 ..., m1；Second level auxiliary spring the piece number m2, thickness h that second level auxiliary spring is eachA2k, k=1,2 ..., m2； The sum of the piece number of main spring and first order auxiliary spring N1=n+m1, total the piece number N=n+m of major-minor spring1+m2；It is inclined to the non-grade of two-stage auxiliary spring formula The equivalent thickness h of the different the piece number l overlay segments of frequency type progressive rate leaf springleIt is calculated, l=1,2 ..., N, i.e.,：
<mrow> <msub> <mi>h</mi> <mrow> <mi>l</mi> <mi>e</mi> </mrow> </msub> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mroot> <mrow> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>l</mi> </munderover> <msubsup> <mi>h</mi> <mi>i</mi> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mroot> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>1</mn> <mo>&amp;le;</mo> <mi>l</mi> <mo>&amp;le;</mo> <mi>n</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mroot> <mrow> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msubsup> <mi>h</mi> <mi>i</mi> <mn>3</mn> </msubsup> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>l</mi> <mo>-</mo> <mi>n</mi> </mrow> </munderover> <msubsup> <mi>h</mi> <mrow> <mi>A</mi> <mn>1</mn> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mroot> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo>&amp;le;</mo> <mi>l</mi> <mo>&amp;le;</mo> <msub> <mi>N</mi> <mn>1</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mroot> <mrow> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msubsup> <mi>h</mi> <mi>i</mi> <mn>3</mn> </msubsup> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mn>1</mn> </msub> </munderover> <msubsup> <mi>h</mi> <mrow> <mi>A</mi> <mn>1</mn> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>l</mi> <mo>-</mo> <msub> <mi>N</mi> <mn>1</mn> </msub> </mrow> </munderover> <msubsup> <mi>h</mi> <mrow> <mi>A</mi> <mn>2</mn> <mi>k</mi> </mrow> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mroot> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <msub> <mi>N</mi> <mn>1</mn> </msub> <mo>+</mo> <mn>1</mn> <mo>&amp;le;</mo> <mi>l</mi> <mo>&amp;le;</mo> <mi>N</mi> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>;</mo> </mrow>
Wherein, the equivalent thickness h of main spring root lapMe, main spring and the root lap of first order auxiliary spring equivalent thickness Spend hMA1eAnd the root lap equivalent thickness h that major-minor spring is totalMA2e, it is respectively
<mrow> <msub> <mi>h</mi> <mrow> <mi>M</mi> <mi>e</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>h</mi> <mrow> <mi>n</mi> <mi>e</mi> </mrow> </msub> <mo>=</mo> <mroot> <mrow> <munderover> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msubsup> <mi>h</mi> <mi>i</mi> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mroot> <mo>;</mo> <msub> <mi>h</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>1</mn> <mi>e</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>h</mi> <mrow> <msub> <mi>N</mi> <mn>1</mn> </msub> <mi>e</mi> </mrow> </msub> <mo>=</mo> <mroot> <mrow> <munderover> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msubsup> <mi>h</mi> <mi>i</mi> <mn>3</mn> </msubsup> <mo>+</mo> <munderover> <mi>&amp;Sigma;</mi> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mn>1</mn> </msub> </munderover> <msubsup> <mi>h</mi> <mrow> <mi>A</mi> <mn>1</mn> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mroot> <mo>;</mo> <msub> <mi>h</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>2</mn> <mi>e</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>h</mi> <mrow> <mi>N</mi> <mi>e</mi> </mrow> </msub> <mo>=</mo> <mroot> <mrow> <munderover> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msubsup> <mi>h</mi> <mi>i</mi> <mn>3</mn> </msubsup> <mo>+</mo> <munderover> <mi>&amp;Sigma;</mi> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mn>1</mn> </msub> </munderover> <msubsup> <mi>h</mi> <mrow> <mi>A</mi> <mn>1</mn> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <mo>+</mo> <munderover> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mn>2</mn> </msub> </munderover> <msubsup> <mi>h</mi> <mrow> <mi>A</mi> <mn>2</mn> <mi>k</mi> </mrow> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mroot> <mo>;</mo> </mrow>
Ii steps：Main spring clamps stiffness KMIt calculates
According to the width b for the offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-, elastic modulus E；Main reed number n, main each of spring Half clamping length Li, i=1,2 ..., n；And the h being calculated in i stepsle, l=i=1,2 ..., n clamp main spring Rigidity is calculated, i.e.,
<mrow> <msub> <mi>K</mi> <mi>M</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mi>b</mi> <mi>E</mi> </mrow> <mrow> <mn>2</mn> <mo>&amp;lsqb;</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <msubsup> <mi>h</mi> <mrow> <mn>1</mn> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>l</mi> <mo>=</mo> <mn>2</mn> </mrow> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>l</mi> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msubsup> <mi>h</mi> <mrow> <mi>l</mi> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>+</mo> <mfrac> <mrow> <msubsup> <mi>L</mi> <mn>1</mn> <mn>3</mn> </msubsup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msubsup> <mi>h</mi> <mrow> <mi>n</mi> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mo>;</mo> </mrow>
Iii steps：The clamping complex stiffness K of main spring and first order auxiliary springMA1It calculates
According to the width b for the offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-, elastic modulus E；Main reed number n, main each of spring Half clamping length Li, i=1,2 ..., n；First order auxiliary spring the piece number m1, the half clamping length of each of first order auxiliary spring is LA1j=Ln+j, j=1,2 ..., m1；The sum of the piece number of main spring and first order auxiliary spring N1=n+m1And be calculated in i steps hle, l=1,2 ..., N1, to the clamping complex stiffness K of main spring and first order auxiliary springMA1It is calculated, i.e.,
<mrow> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>b</mi> <mi>E</mi> </mrow> <mrow> <mn>2</mn> <mo>&amp;lsqb;</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <msubsup> <mi>h</mi> <mrow> <mn>1</mn> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>l</mi> <mo>=</mo> <mn>2</mn> </mrow> <mrow> <msub> <mi>N</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>l</mi> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msubsup> <mi>h</mi> <mrow> <mi>l</mi> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>+</mo> <mfrac> <mrow> <msubsup> <mi>L</mi> <mn>1</mn> <mn>3</mn> </msubsup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <msub> <mi>N</mi> <mn>1</mn> </msub> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msubsup> <mi>h</mi> <mrow> <msub> <mi>N</mi> <mn>1</mn> </msub> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mo>;</mo> </mrow>
Iv steps：Total compound clamping stiffness K of major-minor springMA2It calculates：
According to the width b for the offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-, elastic modulus E；Main reed number n, each main spring Half clamping length Li, i=1,2 ..., n；First order auxiliary spring the piece number m1, the half clamping length of each of first order auxiliary spring is LA1j=Ln+j, j=1,2 ..., m1；Second level auxiliary spring the piece number m2, the half clamping length L of each of second level auxiliary springA2k=LN1+k, k =1,2 ..., m2；Total the piece number N=n+m of major-minor spring1+m2And the h being calculated in i stepsle, l=1,2 ..., N, to major-minor Total compound clamping stiffness K of springMA2It is calculated, i.e.,
<mrow> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>b</mi> <mi>E</mi> </mrow> <mrow> <mn>2</mn> <mo>&amp;lsqb;</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <msubsup> <mi>h</mi> <mrow> <mn>1</mn> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>l</mi> <mo>=</mo> <mn>2</mn> </mrow> <mrow> <mi>N</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mfrac> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mrow> <mi>l</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>l</mi> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msubsup> <mi>h</mi> <mrow> <mi>l</mi> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>+</mo> <mfrac> <mrow> <msubsup> <mi>L</mi> <mn>1</mn> <mn>3</mn> </msubsup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>N</mi> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msubsup> <mi>h</mi> <mrow> <mi>N</mi> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> </mfrac> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mo>;</mo> </mrow>
(2) calculating of the main spring for the offset frequencys type progressive rate leaf springs such as two-stage auxiliary spring formula is non-and the radius of curvature of auxiliary spring at different levels：
I steps：Main spring tailpiece lower surface initial curvature radius RM0bCalculating
According to main spring initial tangential camber HgM0, the half clamping length L of main first of spring1, main reed number n, the thickness of main each of spring hi, i=1,2 ..., n；To main spring tailpiece lower surface initial curvature radius RM0bIt is calculated, i.e.,
<mrow> <msub> <mi>R</mi> <mrow> <mi>M</mi> <mn>0</mn> <mi>b</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msup> <msub> <mi>L</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>+</mo> <msubsup> <mi>H</mi> <mrow> <mi>g</mi> <mi>M</mi> <mn>0</mn> </mrow> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>H</mi> <mrow> <mi>g</mi> <mi>M</mi> <mn>0</mn> </mrow> </msub> </mrow> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msub> <mi>h</mi> <mi>i</mi> </msub> <mo>;</mo> </mrow>
II steps：First upper surface initial curvature radius R of first order auxiliary springA10aCalculating
According to the first order auxiliary spring half clamping length L of firstA11, the initial tangential camber design load H of first order auxiliary springgA10, it is right First upper surface initial curvature radius R of first order auxiliary springA10aIt is calculated, i.e.,
<mrow> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>a</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msubsup> <mi>L</mi> <mrow> <mi>A</mi> <mn>11</mn> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>H</mi> <mrow> <mi>g</mi> <mi>A</mi> <mn>10</mn> </mrow> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>H</mi> <mrow> <mi>g</mi> <mi>A</mi> <mn>10</mn> </mrow> </msub> </mrow> </mfrac> <mo>;</mo> </mrow>
III steps：First order auxiliary spring tailpiece lower surface initial curvature radius RA10bCalculating
According to first order auxiliary spring the piece number m1, thickness h that first order auxiliary spring is eachA1j, j=1,2 ..., m1；And it is calculated in II steps The R arrivedA10a, to first order auxiliary spring tailpiece lower surface initial curvature radius RA10bIt is calculated, i.e.,
<mrow> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>b</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>a</mi> </mrow> </msub> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mn>1</mn> </msub> </munderover> <msub> <mi>h</mi> <mrow> <mi>A</mi> <mn>1</mn> <mi>j</mi> </mrow> </msub> <mo>;</mo> </mrow>
IV steps：First upper surface initial curvature radius R of second level auxiliary springA20aCalculating
According to the second level auxiliary spring half clamping length L of firstA21, the initial tangential camber design load H of second level auxiliary springgA20, it is right First upper surface initial curvature radius R of second level auxiliary springA20aIt is calculated, i.e.,
<mrow> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>20</mn> <mi>a</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msubsup> <mi>L</mi> <mrow> <mi>A</mi> <mn>21</mn> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>H</mi> <mrow> <mi>g</mi> <mi>A</mi> <mn>20</mn> </mrow> <mn>2</mn> </msubsup> </mrow> <mrow> <mn>2</mn> <msub> <mi>H</mi> <mrow> <mi>g</mi> <mi>A</mi> <mn>20</mn> </mrow> </msub> </mrow> </mfrac> <mo>;</mo> </mrow>
(3) simulation calculation for each secondary contact loads of offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-：
Step A：1st beginning contact load Pk1Simulation calculation
According to the width b for the offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-, elastic modulus E；The half of main first of spring clamps Length L1, the h that is calculated in step (1)Me, the R that is calculated in step (2)M0bAnd RA10a, to the 1st beginning contact load Pk1Simulation calculation is carried out, i.e.,
<mrow> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msubsup> <mi>Ebh</mi> <mrow> <mi>M</mi> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>a</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>R</mi> <mrow> <mi>M</mi> <mn>0</mn> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mn>6</mn> <msub> <mi>L</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mrow> <mi>M</mi> <mn>0</mn> <mi>b</mi> </mrow> </msub> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>a</mi> </mrow> </msub> </mrow> </mfrac> <mo>;</mo> </mrow>
Step B：2nd beginning contact load Pk2Simulation calculation
According to the width b for the offset frequencys type progressive rate leaf spring such as two-stage auxiliary spring formula is non-, elastic modulus E；It is calculated in step (1) hMA1e, the R that is calculated in step (2)A10bAnd RA20aAnd the P that simulation calculation obtains in step Ak1, start contact to the 2nd time and carry Lotus Pk2Simulation calculation is carried out, i.e.,
<mrow> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <mfrac> <mrow> <msubsup> <mi>Ebh</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>1</mn> <mi>e</mi> </mrow> <mn>3</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>20</mn> <mi>a</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mn>6</mn> <msub> <mi>L</mi> <mn>1</mn> </msub> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>10</mn> <mi>b</mi> </mrow> </msub> <msub> <mi>R</mi> <mrow> <mi>A</mi> <mn>20</mn> <mi>a</mi> </mrow> </msub> </mrow> </mfrac> </mrow>
Step C：2nd full contact load pw2Simulation calculation
The P obtained according to simulation calculation in step Ak1, simulation calculation obtains in step B Pk2, to the 2nd full contact load Pw2Simulation calculation is carried out, i.e.,
<mrow> <msub> <mi>P</mi> <mrow> <mi>w</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <msubsup> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> <mn>2</mn> </msubsup> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> </mfrac> <mo>;</mo> </mrow>
(4) simulation calculation of the clamping stiffness characteristics for the offset frequencys type progressive rate leaf springs such as two-stage auxiliary spring formula is non-：
According to the K being calculated in step (1)M、KMA1And KMA2, the obtained P of simulation calculation in step (3)k1、Pk2And Pw2, it is right Clamping stiffness characteristics of the offset frequencys type progressive rate leaf springs such as two-stage auxiliary spring formula is non-under different loads P carry out simulation calculation, i.e.,
<mrow> <mi>K</mi> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>K</mi> <mi>M</mi> </msub> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>0</mn> <mo>&amp;le;</mo> <mi>P</mi> <mo>&lt;</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mi>P</mi> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> </mfrac> <msub> <mi>K</mi> <mi>M</mi> </msub> <mo>+</mo> <mfrac> <mrow> <mi>P</mi> <mo>-</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mrow> <mo>(</mo> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mfrac> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> </mfrac> <msub> <mi>K</mi> <mi>M</mi> </msub> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>1</mn> </mrow> </msub> <mo>&amp;le;</mo> <mi>P</mi> <mo>&lt;</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mi>P</mi> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> </mfrac> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <mfrac> <mrow> <mi>P</mi> <mo>-</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mi>P</mi> <mrow> <mi>w</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mrow> <mo>(</mo> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <mfrac> <msub> <mi>P</mi> <mrow> <mi>w</mi> <mn>2</mn> </mrow> </msub> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> </mfrac> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>k</mi> <mn>2</mn> </mrow> </msub> <mo>&amp;le;</mo> <mi>P</mi> <mo>&lt;</mo> <msub> <mi>P</mi> <mrow> <mi>w</mi> <mn>2</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>K</mi> <mrow> <mi>M</mi> <mi>A</mi> <mn>2</mn> </mrow> </msub> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>w</mi> <mn>2</mn> </mrow> </msub> <mo>&amp;le;</mo> <mi>P</mi> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>.</mo> </mrow>
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