購買設(shè)計(jì)請充值后下載，，資源目錄下的文件所見即所得，都可以點(diǎn)開預(yù)覽，，資料完整，充值下載可得到資源目錄里的所有文件。。?！咀ⅰ浚篸wg后綴為CAD圖紙，doc,docx為WORD文檔，原稿無水印，可編輯。。。具體請見文件預(yù)覽，有不明白之處，可咨詢QQ:12401814
南昌航空大學(xué)科技學(xué)院學(xué)士學(xué)位論文
中文譯文
液壓支架的最優(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)?？催@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。
10
南昌航空大學(xué)科技學(xué)院學(xué)士學(xué)位論文
Optimal design of hydraulic support
m. oblak. Harl and b. butinar
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 values 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 response of hydraulic support wil be satisfying.
Keywords: four-bar mechanism, optimal design, mathematical programming \, approximation method, tolerance
1 Introduction
The designer aims to find the best design for the mechanical system considered. Part of thie effort is the optimal choice of some selected parameters of a system. Methods of mathematical programming can be used, Of course, it depends on the type of the systemWith this foemulation, good computer support is assured to look for optimal parameters of the system.
The hydraulic support (Fig.1) described by Harl (1998) is a part of the mining industry equipmenr port in the mine Velenje-Slovenia, used for protection of working environment in the gallery. It consists of four-bar mechanisms FEDG and AEDB as shown in Fig.2. The mechanism AEDB defines the path of coupler point C and the mechanism FEDG is used to drive the support by a hydraulic actuator。
Fig. 1 Hydraulic support
It is required that the motion of the support,more precisely, the motion of the point C in Fig.2, 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 laboratory (Grm 1992). The support exhibited large transversal displacements, which reduce its employability. Thetefore, 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
Fig.2 Two four-bar mechanisms
Parameters of the leading four-bar me AEDB with methods of mathematical programming. Otherwise it chanisms would be necessary to change the project, at least mechanism AEDB.
The solution of above problem will give us the response of hydranlic support for the ideal system. Real response will be different because of tolerances of various parmeters of the system, which is why the maximal allowed tolerances of paramentsa1,a2,a3,a4 will be calculated support. ,with help of mathematical programming.
2 The deterministic model of the hydraulic support
At fist it is necessary to develop an appropriate metical model of the hydraulic support.It could be based on the following assumptions:
- the links are rigid bodies,
- the motion of individual is relatively slow.
The hydraulic support is a mechanism with one degree of freedom. Its kinematics can be model consists of four-bar mechanisms FEDG and AEDB (Oblak et al. 1998).The leading four-bar mechanisms AEDB with methods of mathematical programming. Otherwise it would be necessary to change the project, at least mechanism AEDB. It is required that the motion of the support,more precisely, the motion of the poit C. Therefore, the path of coupler point C is as near as possible to the desired trajectory k.
The synthesis of the four-bar mechanism 1 has been performed with help of motion given by Rao Dukkipati(1989). The general situation is depicted in Fig,3.
Fig.3 Trajectory L of the point C
Equations of trajectory L of the point C will be written in the coordinate frame considered. Coordinates x and y of the point C will be written with the typical parameters of a four-bar mechanism a1,a2,….a6.The coordinates of points B and D are
xBcos (1)
yB=sin (2)
xD=cos() (3)
yD=sin() (4)
The parameters …are related to each other by
xB2+ (5)
α1)2+ yD2= (6)
By substituting (1) － (4) into （5）－（6）the response equations of the support are obtained as
(xcos)2+ (y- sin)2- =0 (7)
[x- cos()-]2+[ y- sin()]2- =0 (8)
This equation represents the mathematical model for calculating the optimal values of paramerters a1,a2,a4.
2.1 Mathematical model
The mathemtial model of the system will be formulated in the from proposed by Haug and Arora (1979):
gi(u,v)0, i=1,2,…,l, (10)
and response equations
hi(u,v)=0, j=1,2,…,m. (11)
The vector u=[u1,u2,…,un]T is called the vector of design variables, v=[v1,v2,…,vm]Tis the vector of response variables and f in(9)is the objective function.
Tobperform the optimal design of the leading four-bar mechanism AEDB,the vector of design variables is defined as
u=[ ]T, (12)
and the vector of response variables as
v=[xy]T. (13)
The dimensions α3，α5，α6 of the corresponding links are kept fixed.
The
f(u,v) =max[g0(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.The system must satisfy the well-known Grasshoff conditions
(15)
(16)
Inequalities (15) and (16) express the property of a four-bar mechanism, where the links may only oscillate.
The condition：
(17)
Prescribes the lower upper bounds of the design variables.
The problem (9)–(11)is not dirrctly solvable with the usual gradient-based optimization methods. This could be cirumvented by int express the property of the objective function be written with the typical parameters be written as
minun+1 (18)
sobject to
gi(u,v) 0, i=1,2,…,l, (19)
f(u,v)- un+10, (20)
and response equations
hj(u,v)=0, j=1,2,…,m, (21)
where：
u=[u1 … un un+1]T
v=[v1 … vn vn+1]T
A nonlinear programming problem of the leading four-bar mechanism AEDB can therefore be difined as
mina7, (22)
sobject to constraints
(23)
(24)
,
(25)
(26)
And respose equationt
(27)
(28)
3.The stochastic model of the hydraulic support
The mathematical model can be used to calculate the parameters of L and K to ensure that the track such as to maintain the distance between the minimum. However the endpoint C calculation L may deviate from the track, because of the movements in the presence of some interference factor. Look at these deviations from what should or not lies in the deviation is in the parametric tolerance tolerance range.Response function (27) - (28) allows us to consider the response variable V vector, the vector of dependent variable V vector design. This means that v = H ( V, H ) function is a mathematical model (22) - (28) foundation, because it describes a response variable V vector and V vector as well as design variables and the mathematical model of the relationship between v. Similarly, the function H used to consider the parameter errors in the value of the maximum permissible value.In the stochastic model, design variable vector u=[u1, ... , un]T can be viewed in U=[U1, ... , Un]T random vector, which means the response variable vector v=[v1, ... , vn]T is a random vector V=[V1, V2, ... , Vn]TV=h ( U ) (29)Suppose design variable U1, ... , Un from probability theory and the classification of normal function of Uk ～( k=1,2, ... , n ) of independence. The main parameters and ( k=1,2, ... , n ) can be associated with such as the measurement of this kind of scientific concepts and tolerance to link, such as a =,. So as long as the choice of suitable existence probability, k=1,2, ... , n (30)Type (30) is calculated the results of.Random vector V probability distribution function is search for dependent random vector U probability distribution function and its actual computability. Therefore, random vector V is described by mathematical properties, and the properties were identified using Taylor on u=[u1, ... , un]T h approximation function description, or with the aid of Oblak and Harl in the Monte Carlo method.
3.1 The mathematical model
Used to calculate the allowable error of hydraulic support optimization mathematical model will be nonlinear problem of independent variable
w=[ ] (31)
and objective function
(32)
With conditions
(33)
,
(34)
In（33），E is the maximal allowed standard deviation of coordinate x of the point C and
A={1,2,4} (35)
The nonlinear programming problem for calculating the optmal tolerances could be therefore defined as ：
(36)
Subject to constraints
(37)
,
(38)
4.Numerical examply
The carrying of the hydraulic support is 1600kN. Both four-bar AEDB andFEDG must fulfill the following demand：
－they must allow minimal transversal displacements of the point C, and，
－they must provide sufficient side stability.
The parameters of the hydraulic support (Fig.2) are given in Table 1.
The drive mechanism FEDG is specified by the vector
(mm) (39)
And the mechanism AEDB by
(mm)
In(39),the parameter d is a walk of the support with maximal value of 925 mm. Parameters for the shaft of the mechanism AEDB are given in Table 2.
4.1Four connecting rod AEDA optimization
Four link model AEDA related data in equation (22) - (28) are expressed. ( Fig 3). C lemniscate of maximum horizontal offset for65mm. That is why type (26) for the
(41)
Rod and bar between AE AA angle in the range of 76.8o and 94.8o, will be a number… successively introduced formula (41) obtained results are listed in table 3.
These points corresponding to the angle of…in the range of [76.8o,94.8o] and they each angle difference of1
The design variables of the minimum and maximum range is
(mm) (42)
(mm) (43)
Nonlinear design problems in equation (22) and (28) in the form of. This problem by
Kegl et al (1991) based on the approximation of the optimal approximation solution. By using the direct method of distinguishing to calculate design derived data.
Design variables for the initial value
(mm) (44)
Optimization of design parameters through calculation is repeated 25 times
Table3 node C corresponding to the X and Y values
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)
In Table 3 of C x values and Y values respectively corresponding to the design variables and the variables of optimization design line )。
Fig 4diagram represents the endpoint C started the lemniscate locus L ( dotted line) and perpendicular to the ideal trajectory K ( solid line).
Lateral displacement and the trajectory of Figure 4graph represents the endpoint C started the lemniscate locus L ( dotted line) and perpendicular to the ideal trajectory K ( solid line).
Fig.4 Trajectories of the point C
4.2 Four connecting rod mechanism AEDA optimal error
In the nonlinear problem (36) - (38), selection of independent variables of the minimum and maximum value
(mm) (46)
(mm) (47)
Independent variable initial value
(mm) (48)
Trajectory deviation to chose two cases E=0.01and E=0.05. In the first case, the design variables of the ideal of tolerance after 9times of repeated calculations, early results. Second situations are7 times repeated calculation to obtain the ideal value. These results are listed in Table 4and table 5.
Figure 5 and Figure6of the standard deviation from Monte Carlo method is calculated and expressed in the diagram (shown in double dots line below) while Taylor approximation curve ( solid line).
Fig.5 Standard deviations for E=0.01
Optimal tolerances for the design variables a1,a2,a4 were calculated after 9 iterations. For E=0.05 the optimum was obtained after 7 iterations.The results are given in Table 4 and 5.
In Fig.5 and 6 the staylor ndard deviations are calculated by the Monte Carlo method and with Taylor approximation (full line represented Taylor approximation),respectively.
Fig.6 Standard deviations for E=0.05
5.Conclusins
With a suitable mathematical model of the system and by employing mathematical programming,the design of the hydraulic support was improved, and better performance was achieved.However, due to the results of optimal tolerances,it might be reasonable to take into consideration a new construction. This is especially true for the mechanism AEDB, since very small tolerances raise the costs of production.
12