英文文献_及翻译 液压顶升支架的最优化设计.doc
外文翻译--大型液压顶支架的最优化设计【中英文文献译文】
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翻译 中文翻译大型液压顶支架的最优化设计摘要：本文介绍了从两组不同参数的采矿工程所使用的液压支架（如图1）中选优的流程。这种流程建立在一定的数学模型之上。第一步，寻找四连杆机构的最理想的结构参数以便确保支架的理想的运动轨迹有最小的横向位移。第二步，计算出四连杆有最理想的参数时的最大误差，以便得出最理想的、最满意的液压支架。图1 液压支架关键词：四连杆机构； 优化设计； 精确设计； 模糊设计； 误差 1 前言 设计者的目的时寻找机械系统的 最优设计。导致的结果是一个系统所选择的参数是最优的。一个数学函数伴随着一个合适的系统的数学模型的出现而出现。当然这数学函数建立在这种类型的系统上。有了这种数学函数模型，加上一台好的计算机的支持，一定能找出系统最优的参数。Harl描述的液压支架是斯洛文尼亚的Velenje矿场的采煤设备的一个组成部分，它用来支护采煤工作面的巷道。它由两组四连杆机构组成，如图2所示.四连杆机构AEDB控制绞结点C的运动轨迹，四连杆机构FEDG通过液压泵来驱动液压支架。图2中，支架的运动，确切的说，支架上绞结点C点竖向的双纽线的运动轨迹要求横向位移最小。如果不是这种情况，液压支架将不能很好的工作，因为支架工作在运动的地层上。实验室测试了一液压支架的原型。支架表现出大的双纽线位移，这种双纽线位移的方式回见少支架的承受能力。因此，重新设计很有必要。如果允许的话，这会减少支架的承受能力。因此，重新设计很有必要。如果允许的话，这种设计还可以在最少的成本上下文章。它能决定去怎样寻找最主要的图2 两四连杆机构四连杆机构数学模型AEDB的最有问题的参数。否则的话这将有必要在最小的机构AEDB改变这种设计方案。 上面所罗列出的所有问题的解决方案将告诉我们关于最理想的液压支架的答案。真正的答案将是不同的，因为系统有各种不同的参数的误差，那就是为什么在数学模型的帮助下，参数允许的最大的误差将被计算出来。2 液压支架的确定性模型首先，有必要进一步研究适当的液压支架的机械模型。它有可能建立在下面所列假设之上：（1）连接体是刚性的，（2）单个独立的连接体的运动是相对缓慢的.液压支架是只有一个方向自由度的机械装置。它的运动学规律可以通过同步的两个四连杆机构FEDG和AEDB的运动来模拟。最主要的四连杆机构对液压支架的运动规律有决定性的影响。机构2只是被用来通过液压泵来驱动液压支架。绞结点C的运动轨迹L可以很好地来描述液压支架的运动规律。因此，设计任务就是通过使点C的轨迹尽可能地接近轨迹K来找到机构1的最理想的连接长度值。四连杆机构1的综合可以通过 Rao 和 Dukkipati给出运动的运动学方程式的帮助来完成。图3 点C轨迹L图3描述了一般的情况。点C的轨迹L的方程式将在同一框架下被打印出来。点C的相对应的坐标x和y随着四连杆机构的独有的参数一起被打印出来。点B和D的坐标分别是xB=x -cos (1)yB=y -sin (2)xD=x -cos() (3)yD=y -sin() (4) 参数也彼此相关xB2 +yB2= (5)(xD-1)2+ yD2= (6)把(1) (4)代入（5）（6）即可获得支架的最终方程式(x-cos)2+ (y- sin)2- =0 (7)x- cos()-2+ y- sin()2- =0 (8)此方程式描述了计算参数的理想值的最基本的数学模型。2.1 数学模型Haug和Arora提议，系统的数学模型可以用下面形式的公式表示min f(u,v), (9)约束于gi(u,v)0, i=1,2,l, (10)和响应函数hi(u,v)=0, j=1,2,m. (11) 向量 u=u1,u2,unT 响应设计时的变量, v=v1,v2,vmT是可变响应向量，(9)式中的f是目标函数。为了使设计的主导四连杆机构AEDB达到最佳，设计时的变量可被定义为u= T, (12)可变响应向量可被定义为v=x yT. (13)相应复数3，5，6的尺寸是确定的。目标函数被定义为理想轨迹K和实际轨迹L之间的一些“有差异的尺寸”f(u,v) =maxg0(y)-f0(y)2, (14)式中x= g0(y) 是曲线K的函数，x= f0(y)是曲线L的函数。我们将为系统挑选一定局限性。这种系统必须满足众所周知的最一般的情况。 (15) (16)不等式表达了四连杆机构这样的特性：复数只可能只振荡的。这种情况： (17)给出了设计变量的上下约束条件。用基于梯度的最优化式方法不能直接的解决(9)(11)的问题。min un+1 (18)从属于gi(u,v) 0, i=1,2,l, (19) f(u,v)- un+10, (20)并响应函数hj(u,v)=0, j=1,2,m, (21)式中： u=u1 un un+1T v=v1 vn vn+1T因此，主导四连杆机构AEDB的一个非线性设计问题可以被描述为：min7, (22)从属于约束 (23) (24) , ， (25) (26)并响应函数： (27) (28) 有了上面的公式，使得点C的横向位移和轨迹K之间的有最微小的差别变得可能。结果是参数有最理想的值。3 液压支架的随机模型数学模型可以用来计算比如参数确保轨迹 L 和 K 之间的距离保持最小。然而端点C的计算轨迹L可能有些偏离，因为在运动中存在一些干扰因数。看这些偏离到底合时与否关键在于这个偏差是否在参数 容许的公差范围内。响应函数（27）（28）允许我们考虑响应变量v的矢量，这个矢量依赖设计变量v的矢量。这就意味着vh (v)，函数h是数学模型（22）（28）的基础，因为它描述出了响应变量v的矢量和设计变量v的矢量以及和数学模型中v的关系。同样，函数h用来考虑参数的误差值 的最大允许值。 在随机模型中，设计变量的矢量u=u1,unT可以被看作U=U1,UnT的随机矢量，也就是意味着响应变量的矢量v=v1,vnT也是一个随机矢量V=V1,V2,VnT v=h(u) (29)假设设计变量 U1,Un 从概率论的观点以及正常的分类函数Uk (k=1,2,n)中独立出来。主要参数和 (k=1,2,n)可以与如测量这类科学概念和公差联系起来，比如=,。所以只要选择合适的存在概率， k=1,2,n (30)式（30）就计算出结果。随机矢量 V 的概率分布函数被探求依赖随机矢量 U 概率分布函数及它实际不可计算性。因此，随意矢量 V 被描述借助于数学特性，而这个特性被确定是利用Taylor的有关点 u=u1,unT 的函数h逼近描述，或者借助被Oblak和Harl在论文提出的Monte Carlo 的方法。3.1 数学模型用来计算液压支架最优化的容许误差的数学模型将会以非线性问题的独立的变量 w= (31)和目标函数 (32)的型式描述出来。约束条件 (33) , ， (34)在式（33）中，E是是坐标C点的x 值的最大允许偏差，其中 A=1,2,4 (35)非线性工程问题的计算公差定义式如下： (36)它服从以下条件： (37) , (38) (39)4 有数字的实列液压支架的工作阻力为1600kN。以及四连杆机构AEDB及FEDG 必须符合以下要求：它们必须确保铰接点C 的横向位移控制在最小的范围内，它们必须提供充分的运动稳定性图2中的液压支架的有关参数列在表1 中。支撑四杆机构 FEDG 可以由矢量 (mm) (40)来确定。四连杆AEDB 可以通过下面矢量关系来确定。 (mm) 在方程(39)中，参数d是液压支架的移动步距，为925mm .四连杆AEDA的杆系的有关参数列于表2中。表1 液压支架的参数 表2 四连杆AEDA的参数4.1 四连杆AEDA的优化四连杆的数学模型AEDA的相关数据在方程(22)-(28)中都有表述。(图3)铰接点C双纽线的横向最大偏距为65mm。那就是为什么式(26)为 (41)杆AA与杆AE之间的角度范围在76.8o和94.8o之间，将数依次导入公式(41)中所得结果列于表3中。这些点所对应的角都在角度范围76.8o,94.8o内而且它们每个角度之差为1o设计变量的最小和最大范围是 (mm) (42) (mm) (43)非线性设计问题以方程(22)与(28)的形式表述出来。这个问题通过Kegl et al(1991)提出的基于近似值逼近的优化方法来解决。通过用直接的区分方法来计算出设计派生数据。设计变量的初始值为 (mm)(44)优化设计的参数经过25次反复计算后是表3 绞结点C对应的x与y 的值角度x初值(mm)y初值(mm)x终值(mm)y终值(mm)76.866.781784.8769.471787.5077.865.911817.6768.741820.4078.864.951850.0967.931852.9279.863.921882.1567.041885.0780.862.841913.8566.121916.8781.861.751945.2065.201948.3282.860.671976.2264.291979.4483.859.652006.9163.462010.4384.858.722037.2862.722040.7085.857.922067.3562.132070.8786.857.302097.1161.732100.7487.856.912126.5961.572130.3288.856.812155.8061.722159.6389.857.062184.7462.242188.6790.857.732213.4263.212217.4691.858.912241.8764.712246.0192.860.712270.0866.852274.3393.863.212298.0969.732302.4494.866.562325.8970.502330.36 (mm) (45)在表3中C点x值与y 值分别对应开始设计变量和优化设计变量。图 4 用图表示了端点 C开始的双纽线轨迹 L(虚线)和垂直的理想轨迹K(实线)。图4 绞结点C 的轨迹4.2 四连杆机构AEDA的最优误差在非线性问题(36)-(38),选择的独立变量的最小值和最大值为 (mm) (46) (mm) (47)独立变量的初始值为 (mm) (48)轨迹偏离选择了两种情况E=0.01和E=0.05。在第一种情况，设计变量的理想公差经过9次反复的计算，已初结果。第二种情况也在7次的反复计算后得到了理想值。这些结果列在表 4和表5 中。图 5和图 6的标准偏差已经由Monte Carlo方法计算出来并表示在图中(图中双点划线示)同时比较泰勒近似法的曲线(实线)。图5 E=0.01时的标准误差图6 E=0.05时的标准误差5 结论通过选用系统的合适的数学模型以及采用数学函数，让液压支架的设计得到改良，而且产品的性能更加可靠。然而，由于理想误差的结果的出现，将有理由再考虑一个新的问题。这个问题在四连杆的问题上表现的尤为突出，因为一个公差变化稍微都能导致产品成本的升高。12Struct Multidisc Optim 20, 7682 Springer-Verlag 2000Optimal design of hydraulic supportM. Oblak, B. Harl and B. ButinarAbstract This paper describes a procedure for optimaldetermination of two groups of parameters of a hydraulicsupport employed in the mining industry. The procedureis based on mathematical programming methods. In thefirst step, the optimal values of some parameters of theleading four-bar mechanism are found in order to ensurethe desired motion of the support with minimal transver-sal displacements. In the second step, maximal tolerancesof the optimal values of the leading four-bar mechanismare calculated, so the response of hydraulic support willbe satisfying.Key words four-bar mechanism, optimal design, math-ematical programming,approximationmethod, tolerance1IntroductionThe designer aims to find the best design for the mechan-ical system considered. Part of this effort is the optimalchoice of some selected parameters of a system. Methodsof mathematical programming can be used, if a suitablemathematical model of the system is made. Of course, itdepends on the type of the system. With this formulation,good computer support is assured to look for optimal pa-rameters of the system.The hydraulic support (Fig. 1) described by Harl(1998) is a part of the mining industry equipment inthe mine Velenje-Slovenia, used for protection of work-ing environment in the gallery. It consists of two four-barReceived April 13, 1999M. Oblak1, B. Harl2and B. Butinar31Faculty of Mechanical Engineering, Smetanova 17, 2000Maribor, Sloveniae-mail: maks.oblakuni-mb.si2M.P.P. Razvoj d.o.o., Ptujska 184, 2000 Maribor, Sloveniae-mail: bostjan.harluni-mb.si3Faculty of Chemistry and Chemical Engineering, Smetanova17, 2000 Maribor, Sloveniae-mail: branko.butinaruni-mb.simechanisms FEDG and AEDB as shown in Fig. 2. Themechanism AEDB defines the path of coupler point Cand the mechanism FEDG is used to drive the support bya hydraulic actuator.Fig. 1 Hydraulic supportIt is required that the motion of the support, moreprecisely, the motion of point C in Fig. 2, is vertical withminimal transversal displacements. If this is not the case,the hydraulic support will not work properly because it isstranded on removal of the earth machine.A prototype of the hydraulic support was tested ina laboratory (Grm 1992). The support exhibited largetransversal displacements, which would reduce its em-ployability. Therefore, a redesign was necessary. Theproject should be improved with minimal cost if pos-77Fig. 2 Two four-bar mechanismssible. It was decided to find the best values for the mostproblematic parameters a1,a2,a4of the leading four-barmechanism AEDB with methods of mathematical pro-gramming. Otherwise it would be necessary to change theproject, at least mechanism AEDB.The solution of above problem will give us the re-sponse of hydraulic support for the ideal system. Realresponse will be different because of tolerances of vari-ous parameters of the system, which is why the maximalallowed tolerances of parameters a1,a2,a4will be calcu-lated, with help of methods of mathematical program-ming.2The deterministic model of the hydraulic supportAt first it is necessary to develop an appropriate mechan-ical model of the hydraulic support. It could be based onthe following assumptions: the links are rigid bodies, the motion of individual links is relatively slow.The hydraulic support is a mechanism with one de-gree of freedom. Its kinematics can be modelled with syn-chronous motion of two four-bar mechanisms FEDG andAEDB (Oblak et al. 1998). The leading four-bar mech-anism AEDB has a decisive influence on the motion ofthe hydraulic support. Mechanism 2 is used to drive thesupport by a hydraulic actuator. The motion of the sup-port is well described by the trajectory L of the couplerpoint C. Therefore, the task is to find the optimal valuesof link lengths of mechanism 1 by requiring that the tra-jectory of the point C is as near as possible to the desiredtrajectory K.The synthesis of the four-bar mechanism 1 has beenperformed with help of kinematics equations of motiongivenby Rao and Dukkipati (1989).The generalsituationis depicted in Fig. 3.Fig. 3 Trajectory L of the point CEquations of trajectory L of the point C will be writ-ten in the coordinate frame considered. Coordinates xand y of the point C will be written with the typicalparameters of a four-bar mechanism a1,a2, ., a6. Thecoordinates of points B and D arexB= xa5cos,(1)yB= ya5sin,(2)xD= xa6cos(+),(3)yD= ya6sin(+).(4)The parameters a1,a2, ., a6are related to eachother byx2B+y2B= a22,(5)(xDa1)2+y2D= a24.(6)By substituting (1)(4) into (5)(6) the responseequations of the support are obtained as(xa5cos)2+(ya5sin)2a22= 0,(7)xa6cos(+)a12+ya6sin(+)2a24= 0.(8)This equation representsthe base of the mathematicalmodel forcalculatingthe optimalvalues ofparametersa1,a2, a4.782.1Mathematical modelThe mathematical model ofthe systemwill be formulatedin the form proposed by Haug and Arora (1979):min f(u,v),(9)subject to constraintsgi(u,v) 0,i = 1,2,. ,?,(10)and response equationshj(u,v) = 0,j = 1,2,. ,m.(11)The vector u = u1.unTis called the vector of designvariables, v = v1.vmTis the vector of response vari-ables and f in (9) is the objective function.To perform the optimal design of the leading four-barmechanism AEDB, the vector of design variables is de-fined asu = a1a2a4T,(12)and the vector of response variables asv = x yT.(13)The dimensions a3, a5, a6of the corresponding links arekept fixed.The objective function is defined as some “measure ofdifference” between the trajectory L and the desired tra-jectory K asf(u,v) = maxg0(y)f0(y)2,(14)where x = g0(y) is the equation of the curve K and x =f0(y) is the equation of the curve L.Suitable limitations for our system will be chosen. Thesystem must satisfy the well-known Grasshoffconditions(a3+a4)(a1+a2) 0,(15)(a2+a3)(a1+a4) 0.(16)Inequalities (15) and (16) express the property of a four-bar mechanism, where the links a2,a4may only oscillate.The conditionu u u(17)prescribes the lower and upper bounds of the design vari-ables.The problem (9)(11) is not directly solvable with theusual gradient-based optimization methods. This couldbe circumvented by introducing an artificial design vari-able un+1as proposed by Hsieh and Arora (1984). Thenew formulation exhibiting a more convenient form maybe written asmin un+1,(18)subject togi(u,v) 0,i = 1,2,. ,?,(19)f(u,v)un+1 0,(20)and response equationshj(u,v) = 0,j = 1,2,. ,m,(21)where u = u1.unun+1Tand v = v1.vmT.Anonlinearprogrammingproblemofthe leading four-bar mechanism AEDB can therefore be defined asmin a7,(22)subject to constraints(a3+a4)(a1+a2) 0,(23)(a2+a3)(a1+a4) 0,(24)a1 a1 a1,a2 a2 a2,a4 a4 a4,(25)g0(y)f0(y)2a7 0,(y ?y,y?),(26)and response equations(xa5cos)2+(ya5sin)2a22= 0,(27)xa6cos(+)a12+ya6sin(+)2a24= 0.(28)This formulation enables the minimization of the differ-ence between the transversal displacement of the point Cand the prescribed trajectory K. The result is the optimalvalues of the parameters a1, a2, a4.793The stochastic model of the hydraulic supportThe mathematical model (22)(28) may be used to cal-culate such values of the parameters a1, a2, a4, thatthe “difference between trajectories L and K” is mini-mal. However, the real trajectory L of the point C coulddeviate from the calculated values because of differentinfluences. The suitable mathematical model deviationcould be treated dependently on tolerances of parametersa1,a2,a4.The response equations (27)(28) allow us to calcu-late the vector of response variables v in dependence onthe vector of design variables u. This implies v =h(u).The functionh is the base of the mathematical model(22)(28), because it represents the relationship betweenthe vector of design variables u and response v of ourmechanical system. The same functionh can be used tocalculate the maximal allowed values of the tolerancesa1, a2, a4of parameters a1, a2, a4.In the stochastic model the vector u = u1.unTofdesign variables is treated as a random vector U = U1.UnT, meaning that the vector v = v1.vmTof re-sponse variablesis alsoa randomvectorV = V1.VmT,V =h(U).(29)It is supposed that the design variables U1, ., Unareindependent from the probability point of view and thatthey exhibit normal distribution, Uk N(k,k) (k =1,2,. ,n). The main parameters kand k(k = 1, 2,. , n) could be bound with technological notions suchas nominal measures, k= ukand tolerances, e.g. uk=3k, so eventskuk Uk k+uk,k = 1,2,. ,n,(30)will occur with the chosen probability.The probability distribution function of the randomvector V, that is searched for depends on the probabil-ity distribution function of the random vector U and itis practically impossible to calculate. Therefore, the ran-dom vector V will be described with help of “numberscharacteristics”, that can be estimated by Taylor approx-imation of the functionh in the point u = u1.unTorwith help of the Monte Carlo method in the papers byOblak (1982) and Harl (1998).3.1The mathematical modelThe mathematical model for calculating optimal toler-ances of the hydraulic support will be formulated asa nonlinear programmingproblemwith independent vari-ablesw = a1a2a4T,(31)and objective functionf(w) =1a1+1a2+1a4(32)with conditionsYE 0,(33)a1 a1 a1,a2 a2 a2,a4 a4 a4.(34)In (33) E is the maximal allowed standard deviation Yof coordinate x of the point C andY=16?jA?g1aj(1,2,4)?2aj,A = 1,2,4.(35)The nonlinear programming problem for calculatingthe optimal tolerances could be therefore defined asmin?1a1+1a2+1a4?,(36)subject to constraintsYE 0,(37)a1 a1 a1,a2 a2 a2,a4 a4 a4.(38)4Numerical exampleThe carrying capability of the hydraulic support is1600kN. Both four-bar mechanisms AEDB and FEDGmust fulfill the following demand: they must allow minimal transversal displacements ofthe point C, and they must provide sufficient side stability.The parameters of the hydraulic support (Fig. 2) aregiven in Table 1.The drive mechanism FEDG is specified by the vectorb1,b2,b3,b4T= 400,(1325+d),1251,1310T(mm),(39)and the mechanism AEDB bya1,a2,a3,a4T= 674,1360,382,1310T(mm).(40)In (39), the parameter d is a walk of the support withmaximal value of 925mm. Parameters for the shaft of themechanism AEDB are given in Table 2.80Table 1 Parameters of hydraulic supportSignLength (mm)M110N510O640P430Q200S1415T380Table 2 Parameters of the shaft for mechanism AEDBSigna51427.70 mma61809.68 mm179.340.520.144.1Optimal links of mechanism AEDBWith this data the mathematical model of the four-barmechanisms AEDB could be written in the form of (22)(28). A straight line is defined by x = 65 (mm) (Fig. 3) forthe desired trajectory of the point C. That is why condi-tion (26) is(x65)a7 0.(41)The angle between links AB and AE may vary be-tween 76.8and 94.8. The condition (41) will be dis-cretized by taking into accountonly the points x1, x2, .,x19in Table 3. These points correspond to the angles 21,22, ., 219ofthe interval 76.8, 94.8 at regularinter-vals of 1.The lower and upper bounds of design variables areu = 640,1330,1280,0T(mm),(42)u = 700,1390,1340,30T(mm).(43)The nonlinear programming problem is formulated inthe form of (22)(28). The problem is solved by the op-timizer described by Kegl et al. (1991) based on approx-imation method. The design derivatives are calculatednumerically by using the direct differentiation method.The starting values of design variables are?0a1,0a2,0a4,0a7?T= 674,1360,1310,30T(mm).(44)The optimal design parameters after 25 iterations areu= 676.42,1360.74,1309.88,3.65T(mm).(45)In Table 3 the coordinates x and y of the coupler pointC are listed for the starting and optimal designs, respec-tively.Table 3 Coordinates x and y of the point CAnglexstartystartxendyend2()(mm)(mm)(mm)(mm)76.866.781784.8769.471787.5077.865.911817.6768.741820.4078.864.951850.0967.931852.9279.863.921882.1567.041885.0780.862.841913.8566.121916.8781.861.751945.2065.201948.3282.860.671976.2264.291979.4483.859.652006.9163.462010.2384.858.722037.2862.722040.7085.857.922067.3562.132070.8786.857.302097.1161.732100.7487.856.912126.5961.572130.3288.856.812155.8061.722159.6389.857.062184.7462.242188.6790.857.732213.4263.212217.4691.858.912241.8764.712246.0192.860.712270.0866.852274.3393.863.212298.0969.732302.4494.866.562325.8970.502330.36Figure4illustrates the trajectoriesLofthe pointC forthe starting (hatched) and optimal (full) design as well asthe straight line K.4.2Optimal tolerances for mechanism AEDBIn the nonlinear programming problem (36)(38), thechosen lower and upper bounds of independent variablesa1, a2, a4arew = 0.001,0.001,0.001T(mm),(46)w = 3.0,3.0,3.0T(mm).(47)The starting values of the independent variables arew0= 0.1,0.1,0.1T(mm).(48)The allowed deviation of the trajectory was chosen fortwo cases as E = 0.01 and E = 0.05. In the first case, the8
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