液压支架外文翻译

液压支架

学生姓名：

专业班级：机制07级09班

指导教师：

机电学院

2011年3月12日

液压支架的最优设计

摘要：这篇文章描述了两种在矿业行业中设计液压支架参数的最佳方法。这种方法是基于数学规程的设计方式。第一步，确保了支架四连杆机构在横向空间内预期移动最主要的参数。第二步，计算出了四连杆机构最大偏差的最主要的参数，如此便可设计出

的液压支架。

关键字：四连杆机构，最优设计，数学方程，近似方式，偏差。

1.绪论：

设计者打算为被考虑的机械系统找到最好的设计。这努力的一部份是挑选系统的最佳叁数。如果能使用一个合适的系统数学模型，可以使用数学上规画的方法。当然，它取决于系统的类型。基于这个公式，好的计算机软件能保证找到最佳的系统参数。此处被作者描述的液压支架是关于矿业产业中煤矿设备的一种。

Fig. 1 Hydraulic support

Velenje-斯洛文尼亚,为保护矿坑中的工作环境，他考虑使用如图2所示的双四连杆机构。机器AEDB决定了C点联结器运动的路径并且机器FEDG通过液压执行机构控制支架。这样就要求支架的运动更加的精确图2中C点的运动在垂直位移上有最小的横向变位。如果不是这种情况，支架将不会正常的工作，因为它的运动是基于以上因素决定的。在一个实验

室中测试了一种原始型号的液压支架。这种支架展现了大的 ransversal 变位,减少了它的可用范围。因此必须重新设计一种新型的支架。新的方案将尽可能的花费少的资金来改善原先的支架。设计人员决定使用数学上规画的方法找出引导四连杆机构AEDB运动最关键最复杂的叁数a1,a2,a4。否则必须改变这个计划，至少要改变四连杆机构AEDB。解决了以上问题我们将得到理想的液压支架系统。因为系统的各种叁数的偏差实际的情况和理想的运动轨迹将有一定的偏差，这就是为什么叁数a1,a2,a4被允许的最大误差将会由数学上规画的方法辅助计算。

Two four-bar mechanisms

2.液压支架形式的确定

首先必须发展一个合适的液压支架机械形式。这将基于以下的假设：

——刚性连杆

——各个连杆的动作相对地慢

液压支架是一个有一个自由度的机械装置。它的运动形式可以用两个四连杆机构FEDG和AEDB来进行同步模拟。四连杆机构AEDB对液压支架的运动有决定性的影响。机构2通过液压执行机构来控制液压支架。液压支架的运动轨道由连接点C精确的决定。因此，

Trajectory L of the point C

任务是根据近似与K 的运动轨道的点C 来确定连杆机构中杆1的最佳长度。四连杆机械装置

1 的各种形式可以由Ron 和 Dukkipati (1989)给的动作的运动学方程式来确定，点C 的

轨道L 的一般的情形在图3中被描述。图中C 点的轨迹L 的方程式中的坐标参数X 和Y 被用来

表示四连杆的主要参数a1a2…..a6。B 点和D 点的坐标是

B X =X -5?cos θ, (1)

B Y =Y -5?sin θ (2)

D X =X -6?cos(θ+λ) (3)

D Y =Y -sin(θ+λ) (4)

A1a2…..a6的参数之间的联系由下面的式子来确定。

2X +2Y =2? (5)

(D X -1?)2+2

Y D =2? (6)

把（1）——（4）代入（5）——（6）中得到下面的式子。

(X-5?cos θ)2+(Y-5?sin θ)2-2

?2=0 (7)

[X-6?cos(θ+λ)-1?]2+[Y-```6?sin(θ+λ)]2-?2

4=0 (8)

上面的参数描述了计算a1a2a4的最优参数进行数学建模的前提条件

数学建摸

Haug 和Arora 在1979年叙述了系统的数学模型。

Minf(u,v) （9）

限制条件：

（u,v ）≤0,I=1,2……..,m. （10）

和反映方程式

H j (u,v)=0,j=1,2,….,m. （11）

向量u=[u1…..un]t 称作不定向量，（9）中v=[v 1……v `m ]T 的作用是反映向量的不

定性。为了得到四连杆机构AEDB 的最优设计，不定向量被定义为： U=[1?2?4?]T (12)

下式反映了不定向量：

v=[xy]T (13)

相应杆件3?,5?,6?的尺寸保护不变，目标函数的功能被定义为测量类别，在轨迹L

和轨迹K

f(u,v)=max[g O (y)-f O (y)]2 (14)

x=go(y)是曲线K 的方程式而x=f o （y ）是曲线L 的方程式。选择合适的系统条件，这

个系统必须满足通常的Grass 条件

（3?+4?）-(1?+2?)0≤, (15)

（2?+3?）-(1? +4?)≤0. (16)

不等式（15）和（16）表达了一个四连杆机构2?4?的成立条件，连杆只能摆动，条

件：

-

u u ≤-≤u （17） 反映了较短的连杆和上连杆的设计不定限度，问题是（9）-（11）不能直接由优化方法

来确定。这个可以通过输入一个不定设计条件UN+1来解决，Hsieh 和Arora 用下面的式子

描述了一个更加方便的方法。

min 1+n u (18)

提出

g i (u,v)≤0,i= (19)

f(u,v)-n u 1+≤0 ， （20）

并且得到方程式

h j (u,v)=0,…..,m, （21）

当u=[u]…n u 1+n u ]T 和v=[v 1…v m ]T 一个导致四连杆机构AEDB 非线形的计算机程序

问题。由此被定义为

min ?7 (22)

提出限制条件：

(3?+4?)-(1?+2?)0≤, (23)

(2?+3?)-(1?+4?)0≤, (24)

-??1--

-

-??≤??≤????≤??≤????≤??≤.,,44422211 (25)

[g 0(y)-f 0(y)]2-7?,0≤ (26)

并且反映了

（x-5?cos θ）2+(y-5?sin θ)2-22?=0 (27)

[x-.0)]sin([])cos(2426216=?-+-+?-+?γθαγθy (28)

这个公式可以使C 点和K 点的轨迹横向误差缩减到最小，得到的结果是421,,???的最优参数。

3.液压支架的随机模型

公式（22）——（28）可以被用来计算参数421,,???,L 和K 的轨迹之间的误差很小。然而，计算出的结果可能与点C 的实际轨道L 之间有一定的差异，合适的数学模型可以补偿a1a2a4的参数误差公式（27）—（28）表示可以通过设计不定项U 来计算不定向量

v=~h(u),~h 的作用是基于数学模型（22）—（28）的，因此它表示了设计我们机械系统中不定向量U 和V 之间的联系，~h 的另一个作用是用来计算参数421,,???的最大偏差

421,,???,随机模型中不定向量,`]......[1T n U U U =

=V ~)(U h . (29)

在这个式子中猜测出不定项U ….UM 是独立的，在图中可能的点中，并且这个式子还表明

,,,.........2,1,n k U U k k k k K =?+≤≤?-μμμ (30)

是有规则分布的。主要参数 可以通过技术方法被限制，例如通过名义测量，通过规定偏差e,guk=36k

这些结果将出现概率分布，不定向量V 的概率分布函数要通过依靠不定向量的概率函数U 来确定，实际上不可能和实际计算的一模一样。因此不定向量V 将通过数学特征被描述，通过近似函数~H 或者通过min te carlo 方式

数学模型

液压支架偏差的数学模型可以通过独立的非线形程序来描述。

,][421T w ??????= （31）

并且目标函数：

421/1/1/1)(??+??+??=w f （32）

条件：

,0≤-E Y σ （33）

----

--??≤??≤????≤??≤????≤??≤??.,,444222111 （34）

在（33）中E 是最大的允许偏差，EY 通过点C 的坐标和1

611

1

2,_((_

g A J y ???∑=αμμσ }4,2,1{=A （35）

因此计算液压支架的最优偏差和非线形公式被描述为：

1/1min ??-

+1/42/1??+??， （36）

限制条件：

,0≤-E Y σ （37）

-??≤??≤??111，--??≤??≤??222，-

??≤??≤??444 （38）

4．数学例子

液压支架工作阻力是1600KN 双四连杆AEDG 和FEDG 必须满足下面要求：

——C 点的横向位移必须最小

——必须提供足够的侧向支护力

表格1中给出了液压支架的参数。下面的向量确定了行走装置FEDG

[),(]1310,1251),1325(,400[],4321mm d b b b b T T += （39）

和装置AEDB

).(]1310,382,1360,674[],,,[4321mm T T =???? （40）

在（39）中d 的参数是可变的，最大变动量是925mm ，连杆AEDB 的参数在表格2中给出了。

表1

液压支架参数

标号 长度（mm ）

M 110

N 510

O 640

P 430

Q 200

S 1415

T 380

表2

连杆装置AEDB 的参数

标号

5?

6?

α

β

γ

最佳连杆AEDB

式（22）—（28）确定了四连杆机构AEDB 的数学模型。一条长度为65mm 的直线确定了点C 的轨迹，这是限定条件（26）为什么是下式的原因

.0)65(7≤?--x （41）

连杆AB 和AE 之间的角度为度~94。8度。在图中式子（41）可以确定点1921....x x x 。

这些点2192221........???符合区间在[0~0]之间的角度，它们之间规则的相隔10，上限和下限的设计变动是

T u ]0,1280,1330,640[=（mm ） (42)

T u ]30,1340,1390,700=(mm) (43)

式子（22）—（28）表现了非线形的程序问题。Kegl et al.(1991)基于近似方程式解

决了这个问题，使用直线直接分化的办法，这个设计引出的问题的数学计算被完成了。开始、的设计变量是：

)(]30,1310,1360,674[],,,[70402010mm T T =???? （44）

在25以后重复的最佳设计参数是：

)(]65.3,88.1309,74.1360,42.676[*mm U T = （45）

图3中点C 的相关坐标参数X 和Y 被分别记录。点C 的坐标X 和Y 是

标号 x 初始 初始y 最终x 最终y

）（0

2? (mm) (mm) (mm) (mm)

图4表明了点C 的轨迹L 的最优设计和起始设计和直线K 相似。

Trajectories of the point C

4．2.机构AEDB 的最优偏差

在（30）~（38）中的非线形程序问题中，独立变量?1?，2??，4??的区间上下限 是：

T W ]001.0,001.0,001.0[=（)m （46）

)(]30,0.3,0.3[m W T = （47）

独立变量的起始标准是：

)(]1.0,1.0,1.0[0m W T = （48）

允许的轨迹偏离被限制在=E 和=E 两种状况中。在第一次情况下，表格4的最优偏差为=E

标号 大小（mm ）

1??

2??

4??

表格5中最优偏差为E =

标号 大小（mm ）

1??

2??

4??

设计变量421,,???的最优偏差被成服计算9次，E =时最优的数据在重复7次后获得，表4和5给出了结果，在图5中和图6中通过Monta 方式和近似方式计算了偏差数。

图5中 偏差标准为E =

图6中 偏差标准为E =

Standard deviations for E=

5．结论

依靠合适的系统数学模型和使用数学程序，液压支架的设计被改进了，并且得到了比较完美的结果。不过，由于最优偏差的结果，可以适当的考虑一个新的结构形式，对于机构AEDB 是特别有用的，因此设计的结果只有非常小的偏差。

Optimal design of hydraulic support

Abstract: this paper describes a procedure for optimal determination of two groups of parameters of a hydraulic support employed in the mining industry.

The procedure is based on mathematical programming methods. In the first step, the optimal vales of some parameters of the leading four-bar mechanism are found in order to ensure the desired motion of the support with minimal transversal displacements. In the second step, maximal tolerances of the optimal values of the leading four-bar mechanism are calculated, so the response of hydraulic support will be satisfying.

Key words: four-bar mechanism, optimal design, mathematical programming, approximation method, tolerance

The designer aims to find the best design for the mechanical system considered. Part of this effort is the optimal choice of some selected

Fig. 1 Hydraulic support

parameters of a system. Methods of mathematical programming can be used,

if a suitable mathematical model of the system is made. Of course, it depends on the type of the system. With this formulation, good computer support is assured to look for optimal parameters of the system. The hydraulic support described by Harl(1981) is a part of the mining

industry equipment in the mine Velenje-Slovenia, used for protection of working environment in the gallery. It consists of two four-bar mechanisms FEDG and AEDB as show in . The mechanism AEDB defines the part of

coupler point C and the mechanism FEDG is used to drive the support by

a hydraulic actuator.

It is required that the motion the support ,more precisely, the motion of point C in , is vertical with minimal transversal displacements.

If this is not the case, the hydraulic support will not work properly because it is stranded on removal of the earth machine. A prototype of

the hydraulic support was tested in a laborarory (Grm 1992). The support exhibited large ransversal displacements, which would reduce its employability. Therefore, a redesign was necessary. The project should

be improved with minimal cost if possible. It was decided to find the

best values for the most problematic parameters a1,a2,a4 of the leading

four-bar mechanism AEDB with methods of mathematical programming. Otherwise it would be necessary to change the project, at least mechanism AEDB. The solution of above problem will be different because of tolerances of varoious parameters of the system, whichis why the maximal allowed tolerances of parameters a1,a2,a4 will be calculate, with help of methods of mathematical programming.

Two four-bar mechanisms

deterministic model of the hydraulic support

At first it is necessary to develop an appropriate mechanical model of the hydraulic support. It could be based on the following assumptions: -the links are rigid bodies,

-the motion of individual links is relatively show.

The hydraulic support is a mechanism with one degree of freedom. Its kinematics can be modeled with synchronous motion of two four-bar mechanisms FEDG and AEDB (Oblak et . The leading four-bar mechanism AEDB has a decisive influence on the motion of the hydraulic support. Mechanism 2 is used to drive the support by a hydraulic actuator. The motion of the support is well described by the trajectory L of the coupler point C. Therefore, the task is to find the optimal values of link lengths of mechanism 1 by requiring that the trajectory of the point C is as possible to the desired trajectory K. The synthesis of the four-bar mechanism 1 has been performed with help of kinematics equations of motion given by Rao and Dukkipati(1989). The general situation is depicted in .

Trajectory L of the point C

Equation of trajectory L of the point C will be written in the coordinate frame considered. Coordinate x and y of the point C will be written with the typical parameters of a four-bar mechanism a1a2…..a6. The coordinates of points B and

D are

B X =X -5?cos θ, (1)

B Y =Y -5?sin θ (2)

D X =X -6?cos(θ+γ) (3)

D Y =Y -sin(θ+γ) (4)

The parameters A1a2…..a6 are related to each other by

2X +2Y =2? (5)

(D X -1?)2+2

Y D =2? (6)

By substituting （1）—（4）into （5）—（6）the response of the support

are obtained as

(X-5?cos θ)2+(Y-5?sin θ)2-2

?2=0 (7)

[X-6?cos(θ+λ)-1?]2+[Y-```6?sin(θ+λ)]2-?2

4=0 (8)

This equation represents the base of the mathematical model fore calculating the optimal values of parameters a1,a2,a4.

model

The mathematical model of the system will be in the from proposed by Haug and Arora (1979):

minf(u,v) （9）

subject to constraints ：

（u,v ）≤0,I=1,2……..,m. （10）

and response equations:

j h (u,v)=0,j=1,2,….,m. （11）

The vector u=T n u u ].....[1 is called the vector of design variables ，

v=[v 1……v `m ]T is the vector of response variables and f in (9) is the objective function. To perform the optimal design of the leading four-bar mechanism AEDB, the vector of design variables is defined as:

U=[1?2?4?]T (12)

And the vector of response variables as:

v=[xy]T (13)

The dimensions 3?,5?,6? of the corresponding links are kept objective function is defind as some “measure of difference ” between the trajectory L and the desired trajectory K as

f(u,v)=max[g O (y)-f O (y)]2 (14)

where x=go(y) is epuation of the curve K and x=f o （y ）is the equation of the curve L. Suitable limitations for our system will be chosen. The system must satisfy the well-known Grassh of conditions

（3?+4?）-(1?+2?)0≤, (15)

（2?+3?）-(1? +4?)≤0. (16)

Inequalities （15）and （16）express the property of a four bar mechanism, where the links 2?4?may only oscillate. The condition

-u u ≤-

≤u （17） Prescribes the lower and upper bounds of the design variables. The problem

(9)-(11) is not directly solvable with the usual gradient-based optimization methods. This could be circumvented by introducing an artificial design variable un+1as proposed by Hsieh and Arora (1984). The new formulation exhibiting a more convenient form may be written as

min 1+n u (18)

subject to

g i (u,v)≤0,i= (19)

f(u,v)-n u 1+≤0 ， （20） and response equations

h j (u,v)=0,…..,m, （21）

where u=[u]…n u 1+n u ]T and v=[v 1…v m ]T . A nonlinear programming problem of the leading four bar mechanism AEDB can be defined as

min ?7 (22)

subject to constraints

(3?+4?)-(1?+2?)0≤, (23)

(2?+3?)-(1?+4?)0≤, (24)

-??1--

-

-??≤??≤????≤??≤????≤??≤.,,44422211 (25)

[g 0(y)-f 0(y)]2-7?,0≤ (26)

And response equations

（x-5?cos θ）2+(y-5?sin θ)2-22?=0 (27)

[x-.0)]sin([])cos(2426216=?-+-+?-+?γθαγθy (28)

This formulation enables the minimization of the difference between the

transversal displacement of the point C and the prescribed trajectory K. the result is the optimal values of the parameters 421,,???.

stochastic model of the hydraulic support

The mathematical model (22)-(28) may be used to calculate such values of the parameters 421,,???,that the “difference between trajectories L and K ” is minimal. However, the real trajectory L of the point C could deviate from the calculated values because of different influences. The suitable mathematical model deviation could be treated dependently on tolerances of parameters 421,,???. The response

(27)-(28) allow us to calculate the vector of response variables v in dependence on the vector of design variables u. this implies v=~h(u). the function ~h is the base of the mathematical model (22)-(28), because it represents the relationship between the vector of the design variables u and response v of our mechanical system. The same function ~h can be used to calculate the maximal allowed values of the tolerances 421,,??????of parameters 421,,???. In the stochastic model the vector T n U U U ]......[1= of design variables is treated as a

random vector v=[v 1…v m ]T ,

=V ~)(U h . (29)

It is supposed that the design variables u1…..un are independent from the probability point of view and that they exhibit normal distribution, ),....,2,1)(,(*n k N U k k k =δμ. The main parameter k μ and ),......,2,1(n k k =δ could be bound with technological notions such as nominal measures, k k u =μ and tolerances, .k k u δ3=? ,so events

,,,.........2,1,n k U U k k k k K =?+≤≤?-μμμ (30)

Will occur with the chosen probability distribution function of the random vector V, that is searched for depends on the probability distribution function of the random vector U and it is practically impossible to calculate. Therefore, the random vector V will be described with help of “numbers of characteristics ”, that can be estimated by Taylor approximation of the function ~h in the point T n U U U ]......[1= or with help of the Monte Carlo method in the papers by Oblak (1982) and Harl(1998).

mathematical mode

l

The mathematical model for calculating optimal tolerances of the hydraulic support will be formulated as a nonlinear programming problem with independent variables

,][421T w ??????= （31）

And objective function

421/1/1/1)(??+??+??=w f （32）

with conditions

,0≤-E Y σ （33）

---

--

-??≤??≤????≤??≤????≤??≤??.,,444222111 （34）

In （33） is the maximal allowed standard deviation y ∑ of coordinate x of the point C and

1

6111

2,_((_

g A J y ???∑=αμμσ }4,2,1{=A （35）

The nonlinear programming problem for calculating the optimal tolerances could be therefore defined as

1/1min ??-

+1/42/1??+??， （36）

Subject to constraints

,0≤-E Y σ （37）

-??≤??≤??111，

--??≤??≤??222，-

??≤??≤??444 （38）

example

The carrying capability of the hydraulic support is 1600kN. Both four-bar mechanisms AEDB and FEDG must fulfill the following demand:

-they must allow minimal transversal displacements of the point C, and -they must provide sufficient side demand:

The parameters of the hydraulic support are given in table 1. The drive mechanism FEDG is specified by the vector

[),(]1310,1251),1325(,400[],4321mm d b b b b T T += （39）

And the mechanism AEDB by

).(]1310,382,1360,674[],,,[4321mm T T =???? （40）

In （39）, the parameter d is a walk of the support with maximal value of 952mm. parameters for the shaft of the mechanism AEDB are given in Table2

Table 1 Parameters of hydraulic support

Sign Length （mm ）

M 110

N 510

O 640

P 430

Q 200

S 1415

T 380

Table 2 parameters of the shaft for mechanism AEDB

Sign

5?

6?

α

β

γ

links of mechanism AEDB

With this data the mathematical of the four-bar mechanisms AEDB could be written in the form of (22)-(28). A straight line is defined by x=65(mm) for the desired trajectory of the point C. that is why condition (26) is

.0)65(7≤?--x （41）

The angle between links AB and AE may vary between ?8.76and ?8.94. The condition

(41) will be discretized by taking in to account only the points 1921....x x x in table

3. these points correspond to the angles 2192221........??? of the interval

[?8.76,?8.94] at regular intervals of ?1. The lower and upper bounds of design variables are

T

u ]0,1280,1330,640[=（mm ） (42)

T u ]30,1340,1390,700=(mm) (43) The nonlinear programming problem is formulated in the form of (22)-(28). The problem is solved by the optimizer described by Kegl et al.(1991) based on approximation method. the design derivatives are calculated numerically by using the direct differentiation method. The starting values of design variables are

)(]30,1310,1360,674[],,,[70402010mm T T =???? （44）

The optimal design parameters after 25 iterations are

)(]65.3,88.1309,74.1360,42.676[*mm U T = （45）

In Table 3 the coordinates x and y of the coupler point C are lested for the starting and optimal designs, respectively. Table 3 Coordinates x and y of the point C

Angle x start ystart xend yend

(mm) (mm) (mm) (mm)

）

（0

2

Figure 4 illustrates the trajectories L of the point C for the starting (hatched) and optimal (full) design as well as the straight line K.