可移式臥式液壓千斤頂設(shè)計(jì)-手動(dòng)臥式千斤頂-抬升重量3噸【三維SW建模】【含8張CAD圖紙】,三維SW建模,含8張CAD圖紙,可移式,臥式,液壓,千斤頂,設(shè)計(jì),手動(dòng),抬升,重量,三維,SW,建模,CAD,圖紙 SJ002-1 2013屆畢業(yè)設(shè)計(jì)（論文）任務(wù)書 二級(jí)學(xué)院： 班 級(jí)： 學(xué) 生： 學(xué) 號(hào)： 指導(dǎo)教師： 職 稱： 課題名稱 3t車用手動(dòng)臥式千斤頂設(shè)計(jì) 課題類型 ? 畢業(yè)設(shè)計(jì)　 　　　　□ 畢業(yè)論文 課題內(nèi)容及其目標(biāo)(指標(biāo))要求 本課題要求設(shè)計(jì)的千斤頂為臥式液壓可移動(dòng)式千斤頂，主要用于重量在3噸左右的轎車的抬升。本課題要求綜合應(yīng)用機(jī)械設(shè)計(jì)和液壓理論知識(shí)完成3t車用手動(dòng)臥式千斤頂設(shè)計(jì)，主要包括輪系部分、支架部分、搖臂部分和液壓組件四個(gè)部分，并利用三維工程軟件Pro/E實(shí)現(xiàn)零件造型和虛擬裝配，同時(shí)撰寫設(shè)計(jì)說明書。 課題具體工作內(nèi)容包括： 1．課題調(diào)研，收集相關(guān)工程設(shè)計(jì)資料，擬訂總體設(shè)計(jì)方案，撰寫開題報(bào)告； 2．完成裝配總圖； 3．完成主要零件工程圖； 4．零部件Pro/E造型及裝配； 5．完成畢業(yè)設(shè)計(jì)說明書； 6．查閱翻譯課題相關(guān)英文資料，約15000英文字符。 設(shè)計(jì)指標(biāo)如下： 1．頂起重量3噸； 2．最小頂起高度為126mm，最大頂起高度為503mm； 3．操作力為950N。額定內(nèi)壓為89Mpa，負(fù)荷為3175/7000 Kg/LB，油壓揚(yáng)程為140mm； 4．要求整機(jī)重量較輕。 進(jìn)程安排 1．準(zhǔn)備工作（熟悉課題、調(diào)查研究、收集資料等）及撰寫開題報(bào)告：2周； 2．確定整體設(shè)計(jì)方案：2周； 3．方案細(xì)化、繪制工程圖及設(shè)計(jì)計(jì)算：6周； 4．三維建模及裝配：2周； 5．審核、修改課題相關(guān)資料及圖紙，整理編寫畢業(yè)設(shè)計(jì)說明書并審核：3周； 6．答辯準(zhǔn)備及答辯：1周。 指導(dǎo)教師： 年 月 日 系 主 任： 年 月 日 翻譯 中文翻譯 大型液壓頂支架的最優(yōu)化設(shè)計(jì) 摘要：本文介紹了從兩組不同參數(shù)的采礦工程所使用的液壓支架（如圖1）中選優(yōu)的流程。這種流程建立在一定的數(shù)學(xué)模型之上。第一步，尋找四連桿機(jī)構(gòu)的最理想的結(jié)構(gòu)參數(shù)以便確保支架的理想的運(yùn)動(dòng)軌跡有最小的橫向位移。第二步，計(jì)算出四連桿有最理想的參數(shù)時(shí)的最大誤差，以便得出最理想的、最滿意的液壓支架。 圖1 液壓支架 關(guān)鍵詞：四連桿機(jī)構(gòu)； 優(yōu)化設(shè)計(jì)； 精確設(shè)計(jì)； 模糊設(shè)計(jì)； 誤差 1 前言 設(shè)計(jì)者的目的時(shí)尋找機(jī)械系統(tǒng)的 最優(yōu)設(shè)計(jì)。導(dǎo)致的結(jié)果是一個(gè)系統(tǒng)所選擇的參數(shù)是最優(yōu)的。一個(gè)數(shù)學(xué)函數(shù)伴隨著一個(gè)合適的系統(tǒng)的數(shù)學(xué)模型的出現(xiàn)而出現(xiàn)。當(dāng)然這數(shù)學(xué)函數(shù)建立在這種類型的系統(tǒng)上。有了這種數(shù)學(xué)函數(shù)模型，加上一臺(tái)好的計(jì)算機(jī)的支持，一定能找出系統(tǒng)最優(yōu)的參數(shù)。 Harl描述的液壓支架是斯洛文尼亞的Velenje礦場(chǎng)的采煤設(shè)備的一個(gè)組成部分，它用來支護(hù)采煤工作面的巷道。它由兩組四連桿機(jī)構(gòu)組成，如圖2所示.四連桿機(jī)構(gòu)AEDB控制絞結(jié)點(diǎn)C的運(yùn)動(dòng)軌跡，四連桿機(jī)構(gòu)FEDG通過液壓泵來驅(qū)動(dòng)液壓支架。 圖2中，支架的運(yùn)動(dòng)，確切的說，支架上絞結(jié)點(diǎn)C點(diǎn)豎向的雙紐線的運(yùn)動(dòng)軌跡要求橫向位移最小。如果不是這種情況，液壓支架將不能很好的工作，因?yàn)橹Ъ芄ぷ髟谶\(yùn)動(dòng)的地層上。 實(shí)驗(yàn)室測(cè)試了一液壓支架的原型。支架表現(xiàn)出大的雙紐線位移，這種雙紐線位移的方式回見少支架的承受能力。因此，重新設(shè)計(jì)很有必要。如果允許的話，這會(huì)減少支架的承受能力。因此，重新設(shè)計(jì)很有必要。如果允許的話，這種設(shè)計(jì)還可以在最少的成本上下文章。它能決定去怎樣尋找最主要的 圖2 兩四連桿機(jī)構(gòu) 四連桿機(jī)構(gòu)數(shù)學(xué)模型AEDB的最有問題的參數(shù)。否則的話這將有必要在最小的機(jī)構(gòu)AEDB改變這種設(shè)計(jì)方案。 上面所羅列出的所有問題的解決方案將告訴我們關(guān)于最理想的液壓支架的答案。真正的答案將是不同的，因?yàn)橄到y(tǒng)有各種不同的參數(shù)的誤差，那就是為什么在數(shù)學(xué)模型的幫助下，參數(shù)允許的最大的誤差將被計(jì)算出來。 2 液壓支架的確定性模型 首先，有必要進(jìn)一步研究適當(dāng)?shù)囊簤褐Ъ艿臋C(jī)械模型。它有可能建立在下面所列假設(shè)之上： （1）連接體是剛性的， （2）單個(gè)獨(dú)立的連接體的運(yùn)動(dòng)是相對(duì)緩慢的. 液壓支架是只有一個(gè)方向自由度的機(jī)械裝置。它的運(yùn)動(dòng)學(xué)規(guī)律可以通過同步的兩個(gè)四連桿機(jī)構(gòu)FEDG和AEDB的運(yùn)動(dòng)來模擬。最主要的四連桿機(jī)構(gòu)對(duì)液壓支架的運(yùn)動(dòng)規(guī)律有決定性的影響。機(jī)構(gòu)2只是被用來通過液壓泵來驅(qū)動(dòng)液壓支架。絞結(jié)點(diǎn)C的運(yùn)動(dòng)軌跡L可以很好地來描述液壓支架的運(yùn)動(dòng)規(guī)律。因此，設(shè)計(jì)任務(wù)就是通過使點(diǎn)C的軌跡盡可能地接近軌跡K來找到機(jī)構(gòu)1的最理想的連接長(zhǎng)度值。四連桿機(jī)構(gòu)1的綜合可以通過 Rao 和 Dukkipati給出運(yùn)動(dòng)的運(yùn)動(dòng)學(xué)方程式的幫助來完成。 圖3 點(diǎn)C軌跡L 圖3描述了一般的情況。 點(diǎn)C的軌跡L的方程式將在同一框架下被打印出來。點(diǎn)C的相對(duì)應(yīng)的坐標(biāo)x和y隨著四連桿機(jī)構(gòu)的獨(dú)有的參數(shù)…一起被打印出來。 點(diǎn)B和D的坐標(biāo)分別是 xB=x -cos (1) yB=y -sin (2) xD=x -cos() (3) yD=y -sin() (4) 參數(shù)…也彼此相關(guān) xB2 +yB2= (5) (xD-α1)2+ yD2= (6) 把(1) － (4)代入（5）－（6）即可獲得支架的最終方程式 (x-cos)2+ (y- sin)2- =0 (7) [x- cos()-]2+[ y - sin()]2- =0 (8) 此方程式描述了計(jì)算參數(shù)的理想值的最基本的數(shù)學(xué)模型。 2.1 數(shù)學(xué)模型 Haug和Arora提議，系統(tǒng)的數(shù)學(xué)模型可以用下面形式的公式表示 min f(u,v), (9) 約束于 gi(u,v)0, i=1,2,…,l, (10) 和響應(yīng)函數(shù) hi(u,v)=0, j=1,2,…,m. (11) 向量 u=[u1,u2,…,un]T 響應(yīng)設(shè)計(jì)時(shí)的變量, v=[v1,v2,…,vm]T是可變響應(yīng)向量，(9)式中的f是目標(biāo)函數(shù)。 為了使設(shè)計(jì)的主導(dǎo)四連桿機(jī)構(gòu)AEDB達(dá)到最佳，設(shè)計(jì)時(shí)的變量可被定義為 u=[ ]T, (12) 可變響應(yīng)向量可被定義為 v=[x y]T. (13) 相應(yīng)復(fù)數(shù)α3，α5，α6的尺寸是確定的。 目標(biāo)函數(shù)被定義為理想軌跡K和實(shí)際軌跡L之間的一些“有差異的尺寸” f(u,v) =max[g0(y)-f0(y)]2, (14) 式中x= g0(y) 是曲線K的函數(shù)，x= f0(y)是曲線L的函數(shù)。 我們將為系統(tǒng)挑選一定局限性。這種系統(tǒng)必須滿足眾所周知的最一般的情況。 (15) (16) 不等式表達(dá)了四連桿機(jī)構(gòu)這樣的特性：復(fù)數(shù)只可能只振蕩的。 這種情況： (17) 給出了設(shè)計(jì)變量的上下約束條件。 用基于梯度的最優(yōu)化式方法不能直接的解決(9)–(11)的問題。 min un+1 (18) 從屬于 gi(u,v) 0, i=1,2,…,l, (19) f(u,v)- un+10, (20) 并響應(yīng)函數(shù) hj(u,v)=0, j=1,2,…,m, (21) 式中： u=[u1 … un un+1]T v=[v1 … vn vn+1]T 因此，主導(dǎo)四連桿機(jī)構(gòu)AEDB的一個(gè)非線性設(shè)計(jì)問題可以被描述為： minα7, (22) 從屬于約束 (23) (24) , ， (25) (26) 并響應(yīng)函數(shù)： (27) (28) 有了上面的公式，使得點(diǎn)C的橫向位移和軌跡K之間的有最微小的差別變得可能。結(jié)果是參數(shù)有最理想的值。 3 液壓支架的隨機(jī)模型 數(shù)學(xué)模型可以用來計(jì)算比如參數(shù)確保軌跡 L 和 K 之間的距離保持最小。然而端點(diǎn)C的計(jì)算軌跡L可能有些偏離，因?yàn)樵谶\(yùn)動(dòng)中存在一些干擾因數(shù)。看這些偏離到底合時(shí)與否關(guān)鍵在于這個(gè)偏差是否在參數(shù) 容許的公差范圍內(nèi)。 響應(yīng)函數(shù)（27）－（28）允許我們考慮響應(yīng)變量v的矢量，這個(gè)矢量依賴設(shè)計(jì)變量v的矢量。這就意味著v＝h (v)，函數(shù)h是數(shù)學(xué)模型（22）－（28）的基礎(chǔ)，因?yàn)樗枋龀隽隧憫?yīng)變量v的矢量和設(shè)計(jì)變量v的矢量以及和數(shù)學(xué)模型中v的關(guān)系。同樣，函數(shù)h用來考慮參數(shù)的誤差值 的最大允許值。 在隨機(jī)模型中，設(shè)計(jì)變量的矢量u=[u1,…,un]T可以被看作U=[U1,…,Un]T的隨機(jī)矢量，也就是意味著響應(yīng)變量的矢量v=[v1,…,vn]T也是一個(gè)隨機(jī)矢量V=[V1,V2,…,Vn]T v=h(u) (29) 假設(shè)設(shè)計(jì)變量 U1,…,Un 從概率論的觀點(diǎn)以及正常的分類函數(shù)Uk～ (k=1,2,…,n)中獨(dú)立出來。主要參數(shù)和 (k=1,2,…,n)可以與如測(cè)量這類科學(xué)概念和公差聯(lián)系起來，比如=,。所以只要選擇合適的存在概率 ， k=1,2,…,n (30) 式（30）就計(jì)算出結(jié)果。 隨機(jī)矢量 V 的概率分布函數(shù)被探求依賴隨機(jī)矢量 U 概率分布函數(shù)及它實(shí)際不可計(jì)算性。因此，隨意矢量 V 被描述借助于數(shù)學(xué)特性，而這個(gè)特性被確定是利用Taylor的有關(guān)點(diǎn) u=[u1,…,un]T 的函數(shù)h逼近描述，或者借助被Oblak和Harl在論文提出的Monte Carlo 的方法。 3.1 數(shù)學(xué)模型 用來計(jì)算液壓支架最優(yōu)化的容許誤差的數(shù)學(xué)模型將會(huì)以非線性問題的獨(dú)立的變量 w=[ ] (31) 和目標(biāo)函數(shù) (32) 的型式描述出來。 約束條件 (33) , ， (34) 在式（33）中，E是是坐標(biāo)C點(diǎn)的x 值的最大允許偏差，其中 A={1,2,4} (35) 非線性工程問題的計(jì)算公差定義式如下： (36) 它服從以下條件： (37) , (38) (39) 4 有數(shù)字的實(shí)列 液壓支架的工作阻力為1600kN。 以及四連桿機(jī)構(gòu)AEDB及FEDG 必須符合以下要求： －它們必須確保鉸接點(diǎn)C 的橫向位移控制在最小的范圍內(nèi)， －它們必須提供充分的運(yùn)動(dòng)穩(wěn)定性 圖2中的液壓支架的有關(guān)參數(shù)列在表1 中。 支撐四桿機(jī)構(gòu) FEDG 可以由矢量 (mm) (40) 來確定。 四連桿AEDB 可以通過下面矢量關(guān)系來確定。 (mm) 在方程(39)中，參數(shù)d是液壓支架的移動(dòng)步距，為925mm .四連桿AEDA的桿系的有關(guān)參數(shù)列于表2中。 表1 液壓支架的參數(shù) 表2 四連桿AEDA的參數(shù) 4.1 四連桿AEDA的優(yōu)化 四連桿的數(shù)學(xué)模型AEDA的相關(guān)數(shù)據(jù)在方程(22)-(28)中都有表述。(圖3)鉸接點(diǎn)C雙紐線的橫向最大偏距為65mm。那就是為什么式(26)為 (41) 桿AA與桿AE之間的角度范圍在76.8o和94.8o之間，將數(shù)…依次導(dǎo)入公式(41)中所得結(jié)果列于表3中。 這些點(diǎn)所對(duì)應(yīng)的角…都在角度范圍[76.8o,94.8o]內(nèi)而且它們每個(gè)角度之差為1o 設(shè)計(jì)變量的最小和最大范圍是 (mm) (42) (mm) (43) 非線性設(shè)計(jì)問題以方程(22)與(28)的形式表述出來。這個(gè)問題通過 Kegl et al(1991)提出的基于近似值逼近的優(yōu)化方法來解決。通過用直接的區(qū)分方法來計(jì)算出設(shè)計(jì)派生數(shù)據(jù)。 設(shè)計(jì)變量的初始值為 (mm)(44) 優(yōu)化設(shè)計(jì)的參數(shù)經(jīng)過25次反復(fù)計(jì)算后是 表3 絞結(jié)點(diǎn)C對(duì)應(yīng)的x與y 的值 角度 x初值(mm) y初值(mm) x終值(mm) y終值(mm) 76.8 66.78 1784.87 69.47 1787.50 77.8 65.91 1817.67 68.74 1820.40 78.8 64.95 1850.09 67.93 1852.92 79.8 63.92 1882.15 67.04 1885.07 80.8 62.84 1913.85 66.12 1916.87 81.8 61.75 1945.20 65.20 1948.32 82.8 60.67 1976.22 64.29 1979.44 83.8 59.65 2006.91 63.46 2010.43 84.8 58.72 2037.28 62.72 2040.70 85.8 57.92 2067.35 62.13 2070.87 86.8 57.30 2097.11 61.73 2100.74 87.8 56.91 2126.59 61.57 2130.32 88.8 56.81 2155.80 61.72 2159.63 89.8 57.06 2184.74 62.24 2188.67 90.8 57.73 2213.42 63.21 2217.46 91.8 58.91 2241.87 64.71 2246.01 92.8 60.71 2270.08 66.85 2274.33 93.8 63.21 2298.09 69.73 2302.44 94.8 66.56 2325.89 70.50 2330.36 (mm) (45) 在表3中C點(diǎn)x值與y 值分別對(duì)應(yīng)開始設(shè)計(jì)變量和優(yōu)化設(shè)計(jì)變量。 圖 4 用圖表示了端點(diǎn) C開始的雙紐線軌跡 L(虛線)和垂直的理想軌跡K(實(shí)線)。 圖4 絞結(jié)點(diǎn)C 的軌跡 4.2 四連桿機(jī)構(gòu)AEDA的最優(yōu)誤差 在非線性問題(36)-(38),選擇的獨(dú)立變量的最小值和最大值為 (mm) (46) (mm) (47) 獨(dú)立變量的初始值為 (mm) (48) 軌跡偏離選擇了兩種情況E=0.01和E=0.05。在第一種情況，設(shè)計(jì)變量的理想公差經(jīng)過9次反復(fù)的計(jì)算，已初結(jié)果。第二種情況也在7次的反復(fù)計(jì)算后得到了理想值。這些結(jié)果列在表 4和表5 中。 圖 5和圖 6的標(biāo)準(zhǔn)偏差已經(jīng)由Monte Carlo方法計(jì)算出來并表示在圖中(圖中雙點(diǎn)劃線示)同時(shí)比較泰勒近似法的曲線(實(shí)線)。 圖5 E=0.01時(shí)的標(biāo)準(zhǔn)誤差 圖6 E=0.05時(shí)的標(biāo)準(zhǔn)誤差 5 結(jié)論 通過選用系統(tǒng)的合適的數(shù)學(xué)模型以及采用數(shù)學(xué)函數(shù)，讓液壓支架的設(shè)計(jì)得到改良，而且產(chǎn)品的性能更加可靠。然而，由于理想誤差的結(jié)果的出現(xiàn)，將有理由再考慮一個(gè)新的問題。這個(gè)問題在四連桿的問題上表現(xiàn)的尤為突出，因?yàn)橐粋€(gè)公差變化稍微都能導(dǎo)致產(chǎn)品成本的升高。 12 Struct 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, the81Fig. 4 Trajectories of the point CTable 4 Optimal tolerances for E = 0.01SignValue (mm)a10.01917a20.00868a40.00933Table 5 Optimal tolerances for E = 0.05SignValue (mm)a10.09855a20.04339a40.04667Fig. 5 Standard deviations for E = 0.01optimal tolerances for the design variables a1, a2, a4werecalculated after 9 iterations. For E = 0.05 the optimumwas obtained after 7 iterations. The results are given inTables 4 and 5.In Figs. 5 and 6 the standard deviations are calculatedby the Monte Carlo method and with Taylor approxima-tion (full line represented Taylor approximation), respec-tively.Fig. 6 Standard deviations for E = 0.055ConclusionsWith a suitable mathematical model ofthe systemand byemploying mathematical programming, the design of the82hydraulic support was improved, and better performancewas achieved. However, due to the results of optimal tol-erances, it might be reasonable to take into considerationa new construction. This is especially true for the mech-anism AEDB, since very small tolerances raise the costsof production.ReferencesGrm, V. 1992: Optimal synthesis of four-bar mechanism. MSc.Thesis. Faculty of Mechanical Engineering MariborHarl, B. 1998: Stochastic analyses of hydraulic support 2S.MSc. Thesis. Faculty of Mechanical Engineering MariborHaug, E.J.; Arora, J.S. 1979: Applied optimal design. NewYork: WileyHsieh, C.; Arora, J. 1984: Design sensitivity analysis and op-timisation of dynamic response. Comp. Meth. Appl. Mech.Engrg. 43, 195219Kegl, M.; Butinar, B.; Oblak, M. 1991: Optimization of me-chanical systems: On strategy of non-linear first-order approx-imation. Int. J. Numer. Meth. Eng. 33, 223234Oblak, M. 1982: Numerical analyses of structures part II. Fac-ulty of Mechanical Engineering MariborOblak, M.; Ciglari c, I.; Harl, B. 1998: The optimal synthesis ofhydraulic support. ZAMM 3, 10271028Rao, S.S.; Dukkipati, R.V. 1989: Mechanism and machine the-ory. New Delhi: Wiley & Sons