液壓外文翻譯---液壓系統(tǒng)的模型化參量估計(jì)

1、<p>　　本 科 畢 業(yè) 論文</p><p><b>　　外文翻譯</b></p><p>　　畢業(yè)論文題目：1000噸四柱液壓機(jī)臺(tái)面及頂出結(jié)構(gòu)設(shè)計(jì)</p><p><b>　　學(xué)生姓名： </b></p><p><b>　　學(xué) 號(hào)：</b></p&g

2、t;<p><b>　　系 別： </b></p><p>　　專業(yè)班級(jí)：機(jī)械設(shè)計(jì)制造及其自動(dòng)化</p><p>　　液壓系統(tǒng)的模型化參量估計(jì)</p><p>　　何清華，赫鵬，張大慶</p><p>　　摘 要 首先介紹了液壓挖掘機(jī)的一個(gè)改裝的電動(dòng)液壓的比例系統(tǒng)。根據(jù)負(fù)載獨(dú)立流量分配（ LUDV ）系

3、統(tǒng)的原則和特點(diǎn)，以動(dòng)臂液壓系統(tǒng)為例并忽略液壓缸中的油大量泄漏，建立一個(gè)力平衡方程和一個(gè)液壓缸的連續(xù)性方程?；陔妱?dòng)液壓的比例閥門的流體運(yùn)動(dòng)方程，測(cè)試的分析穿過(guò)閥門的壓力的不同。結(jié)果顯示壓力的差異并不會(huì)改變負(fù)載，此時(shí)負(fù)載接近2.0MPa。然后假設(shè)穿過(guò)閥門的液壓油與閥芯的位移成正比并且不受負(fù)載影響，提出了一個(gè)電液控制系統(tǒng)的簡(jiǎn)化模型。同時(shí)通過(guò)分析結(jié)構(gòu)和承重的動(dòng)臂裝置，并將機(jī)械臂的力矩等效方程與旋轉(zhuǎn)法、參數(shù)估計(jì)估計(jì)法結(jié)合起來(lái)建立了液壓缸以等質(zhì)量

4、等為參數(shù)的受力平衡參數(shù)方程。最后用階躍電流控制電液比例閥來(lái)測(cè)試動(dòng)臂液壓缸中液壓油的階躍響應(yīng)。根據(jù)實(shí)驗(yàn)曲線，閥門的流量增益系數(shù)被確定為2.825×10-4m3/(·A)，并驗(yàn)證了該模型。 </p><p>　　關(guān)鍵詞：挖掘機(jī)，電液比例系統(tǒng)，負(fù)載獨(dú)立流量分配（ LUDV ）系統(tǒng)，建模，參數(shù)估計(jì) </p><p><b>　　1 引言</b><

5、/p><p>　　由于液壓挖掘機(jī)具有高效率、多功能的優(yōu)點(diǎn)，所以被廣泛應(yīng)用于礦山，道路建設(shè)，民事和軍事建設(shè)，危險(xiǎn)廢物清理領(lǐng)域。液壓挖掘機(jī)在施工機(jī)械領(lǐng)域中也發(fā)揮了重要作用。目前，機(jī)電一體化和自動(dòng)化已成為施工機(jī)械發(fā)展的最新趨勢(shì)。因此，自動(dòng)挖掘機(jī)在許多國(guó)家逐漸變得普遍并被認(rèn)為重點(diǎn)。挖掘機(jī)可以用許多控制方法自動(dòng)地控制操作器。 每種使用方法，研究員必須知道操作器結(jié)構(gòu)和液壓機(jī)構(gòu)的動(dòng)態(tài)和靜態(tài)特征。即確切的數(shù)學(xué)模型有利于控制器的設(shè)計(jì)。

6、然而，來(lái)自外部的干擾使得機(jī)械結(jié)構(gòu)模型和各種非線性液壓制動(dòng)器的時(shí)變參數(shù)很難確定。關(guān)于挖掘機(jī)時(shí)滯控制的研究已經(jīng)有人在研究了。NGUYEN利用模糊的滑動(dòng)方式和阻抗來(lái)控制挖掘機(jī)動(dòng)臂的運(yùn)動(dòng)，SHAHRAM等采取了阻抗對(duì)挖掘機(jī)遠(yuǎn)距傳物的控制。液壓機(jī)構(gòu)非線性模型已經(jīng)由研究員開(kāi)發(fā)出來(lái)了。 然而，復(fù)雜和昂貴的設(shè)計(jì)控制器限制了它的應(yīng)用。在本文，根據(jù)提出的模型，根據(jù)工程學(xué)和受力平衡，挖掘機(jī)臂液壓機(jī)構(gòu)模型簡(jiǎn)化為連續(xù)均衡的液壓缸和流動(dòng)均衡的電液比例閥；同時(shí)，確定

7、了模型的參量的估計(jì)方法和等式。</p><p>　　2 挖掘機(jī)機(jī)械臂概述</p><p>　　液壓挖掘機(jī)的挖掘研究結(jié)果如圖1。在圖中，F(xiàn)c表示液壓缸，動(dòng)臂的重力，斗桿，鏟斗的重力等在B點(diǎn)合力，其方向是沿著液壓缸AB方向； Fc可分解成Fc1和Fc2 ，他們的方向分別為垂直于和平行于O1B ，加速度ac的方向與Fc是相同的，并且ac也可以分解成ac1和ac2；G1 ， G2和G3分別是動(dòng)臂，

8、斗桿和鏟斗的重心；m1，m2，m3是它們各自的質(zhì)量且能通過(guò)實(shí)驗(yàn)給定（m1=868.136kg，m2=357.115kg 和m3=210.736kg； Ol，O2 和O3是絞 接點(diǎn)；G1′，G2′和 G3′分別是G1 ， G2和G3在X軸上的投影。 挖掘機(jī)的臂被認(rèn)為是一個(gè)三個(gè)自由度的的機(jī)械手（三個(gè)測(cè)斜儀分別裝在動(dòng)臂，斗桿和鏟斗上）。在跟蹤控制實(shí)驗(yàn)中，其目標(biāo)軌跡是根據(jù)挖掘機(jī)機(jī)械手運(yùn)動(dòng)學(xué)方程確定的。然后，動(dòng)臂，斗桿

9、和鏟斗的動(dòng)作有操作員控制。為了適應(yīng)自動(dòng)控制，普通液壓控制挖掘機(jī)應(yīng)改造電動(dòng)液壓控制挖掘機(jī)。 基于SW E-85型原有的液壓系統(tǒng)，把先導(dǎo)液壓控制系統(tǒng)更換為先導(dǎo)電液控制系統(tǒng)。新改進(jìn)的液壓系統(tǒng)如圖2所示。在這系統(tǒng)中，因?yàn)閯?dòng)臂，斗桿和鏟斗具有相同的特點(diǎn)，將動(dòng)臂的液</p><p>　　圖2 挖掘機(jī)液壓系統(tǒng)示意圖</p><p>　　圖3 改造后LUDV液壓系統(tǒng)示意圖</p>

10、<p>　　3 模型的電液比例系統(tǒng)</p><p>　　3.1 電動(dòng)液壓的比例閥門動(dòng)力學(xué)特性 </p><p>　　在本文中，電液比例閥包括比例減壓閥和SX-14主要閥.傳遞功能從輸入液流的閥芯位移可如下：</p><p>　　Xv(s)／Iv(s)=KI／(1+bs) (1)<

11、;/p><p>　　其中Xv是xv的拉普拉斯變換值，單位為m；KI是電液比例閥獲得的液流，單位為m/A； b是一階系統(tǒng)的時(shí)間常數(shù)，單位為s；Iv=I(t)-Id，I(t)和 Id 分別表示比例閥門的控制潮流和克服靜帶的各自潮流，單位為A 。</p><p>　　3.2 電動(dòng)液壓的比例閥門的流體運(yùn)動(dòng)方程 </p><p>　　在本文中，實(shí)驗(yàn)性機(jī)器人挖掘機(jī)采取了LUDV系

12、統(tǒng)。根據(jù)LUDV系統(tǒng)的理論，可以得到流體運(yùn)動(dòng)方程：</p><p>　　= w = （2） </p><p>　　= w = （3） </p><p>　　其中Δp是負(fù)荷傳感閥門的壓力差，單位為 MPa；cd是徑流系數(shù)，單位為m5／(N

13、3;s)；w是管口的面積梯度，單位為 m2／m；ρ是油密度，單位為 kg/m3；Δp1和Δp2分別為二個(gè)管口壓力，單位為 MPa；當(dāng)挖掘機(jī)流程沒(méi)有飽和時(shí)，Δp是一幾乎恒定。在本文中，其值由實(shí)驗(yàn)測(cè)試得到。 在圖4中，ps，p1s,和Δp分別表示系統(tǒng)壓力、負(fù)荷傳感閥門壓力和它們的壓力差；壓力系統(tǒng)的實(shí)驗(yàn)曲線顯示

14、三種不同的壓力值。雖然ps和p1s隨著荷載而改變，但是他們的區(qū)別不會(huì)隨著荷載而改變，其值接近對(duì)2.0MPa。因此，對(duì)橫跨閥門的流量的作用Δp可以被忽略。假設(shè)，流過(guò)閥門的流量與管口閥門的大小成比例，并且荷載不影響流量。那么方程(2)能被簡(jiǎn)化為：</p><p>　　Q1=Kqxv(t),I(t≥0 (4) </p><p&

15、gt;　　其中Kq是閥門流量系數(shù)，單位為m2/s；并且Kq=CdW </p><p>　　圖4 動(dòng)臂移動(dòng)壓力曲線圖</p><p>　　3.3 液壓缸的連續(xù)性方程 </p><p>　　一般來(lái)說(shuō)，工程機(jī)械不允許外泄。當(dāng)前，外在泄漏可以通過(guò)密封技術(shù)控制。另一方面，由實(shí)驗(yàn)證明了挖掘機(jī)內(nèi)部泄露是相當(dāng)小的，因此，液壓機(jī)構(gòu)內(nèi)部和外在泄露的影響可以被忽略。當(dāng)油流進(jìn)氣缸無(wú)桿腔并

16、且進(jìn)入到有桿腔內(nèi)時(shí)，連續(xù)性方程可以寫成：</p><p>　　其中V1和V2分別表示流入及流出的液壓缸液體的體積，單位是 : 是有效體積模量（包括液體，油中的空氣等），單位是N/ 。</p><p>　　3.4 液壓缸力的平衡方程</p><p>　　據(jù)推測(cè)，液壓缸中油的質(zhì)量可以忽略，而且負(fù)載是剛性的。那么可以根據(jù)牛頓定律得到液壓缸的力量平衡等式：</p>

17、;<p>　　=m + + （6） </p><p>　　其中 是黏阻止的系數(shù)，單位是 </p><p>　　3.5電動(dòng)液壓的比例系數(shù)簡(jiǎn)化的模型</p><p>　　方程（4） （6）在拉布拉斯變換以后，簡(jiǎn)化的模型可以表達(dá)為：</p>&l

18、t;p>　　(7) </p><p>　　其中Y是y拉布拉斯變換后得到的； = .( ); = ; = m; = ;</p><p><b>　　= ( )。</b></p><p><b>　　 4 參量估計(jì)</b></p><p>　　從塑造的過(guò)程和方程（7）中可以得到在確

19、切的簡(jiǎn)化的模型中與結(jié)構(gòu)，運(yùn)動(dòng)情況以及挖掘機(jī)動(dòng)臂的體位有關(guān)的所有參量。而且，這些參量是時(shí)變。因此要得到這些參量的準(zhǔn)確值和數(shù)學(xué)等式是相當(dāng)難得。要解決這個(gè)問(wèn)題，本文提出了估計(jì)方程和方法來(lái)估算模型中的這些重要參數(shù)。</p><p>　　4.1估算液壓缸負(fù)載</p><p>　　液壓缸臂上的負(fù)載（假定沒(méi)有外部負(fù)載）由動(dòng)臂，斗桿和鏟斗上的負(fù)載組成。在圖1中，動(dòng)臂，斗桿和鏟斗分別繞著各自的鉸接點(diǎn)旋轉(zhuǎn)。因

20、此他們的運(yùn)動(dòng)不是沿著氣缸的直線運(yùn)動(dòng)，也就是說(shuō)他們的運(yùn)動(dòng)方向與方程（5）中的y的方向是不同的。因此方程（6）中的m不能簡(jiǎn)單的認(rèn)為是動(dòng)臂，斗桿和鏟斗質(zhì)量的總和。 考慮到機(jī)械手的坐標(biāo)軸心O1，機(jī)械手的轉(zhuǎn)矩和角加速度可考慮如下：</p><p>　　(8) </p><p>　　其中的M 和 ω分別是工作裝置對(duì)O1的轉(zhuǎn)矩和角加速度。 是點(diǎn)O1到點(diǎn)B的</p&g

21、t;<p>　　長(zhǎng)度；由轉(zhuǎn)動(dòng)定律M=Jω可得： ，即： </p><p>　　(9) </p><p>　　其中的J是工作裝置指向O1的等效轉(zhuǎn)動(dòng)慣量，單位是kg·m2;并且寫成如下式子：</p><p>　　J= + + + (10) </p>

22、<p>　　J1, J2 和 J3分別是動(dòng)臂，斗桿和鏟斗對(duì)各自的中心的慣性力矩；它們的值可以通過(guò)模擬動(dòng)態(tài)模型得出J1=450.9N·m，J2=240.2N·m，J3=94.9N·m。 </p><p>　　比較方程（9）和 =m ，可以得出點(diǎn)B的等效質(zhì)量：</p><p>　　M=J/

23、 (11) </p><p>　　4.2 液壓缸負(fù)載的估算 </p><p>　　工作裝置對(duì)于O1等效力矩等式為： </p><p>　　+ + (12) 其中 和 分別表示O1點(diǎn)到 G1′ ，G2′和 G3′三點(diǎn)的距離；</

24、p><p>　　那么反力負(fù)荷為： + + / (13) </p><p>　　4.3增益系數(shù)閥流量的估計(jì) 流量傳感器可以測(cè)量泵的流量。用于這項(xiàng)工作的儀器為多系統(tǒng)5050型。</p><p>　　動(dòng)臂液壓缸流量的階躍響應(yīng)在電液比例閥控制下的結(jié)果如圖5所示。同時(shí)，該線</p><p>

25、;　　驗(yàn)證等式(11) 。根據(jù)實(shí)驗(yàn)曲線和等式(1)和(4)可確定 的范圍。那么根據(jù)圖4</p><p>　　中的數(shù)據(jù)我們可得出： =2.825×10-4m3/(s·A) 。 </p><p>　　圖5 動(dòng)臂液壓缸流量的階躍響應(yīng)在電液比例閥控制下的曲線圖</p><p><b>　　5 結(jié)論</b></p>

26、<p>　　（1）電液控制系統(tǒng)的數(shù)學(xué)模型是根據(jù)挖掘機(jī)的特點(diǎn)發(fā)展起來(lái)的。假定流過(guò)閥的流量與閥口大小成正比，并忽略液壓系統(tǒng)的內(nèi)部和外部泄漏影響。簡(jiǎn)化模型可以得到： </p><p>　　，其中Y（s）和Xv(s)分別是活塞和閥芯的位移。 </p><p>　?。?）從電液控制系統(tǒng)的模型中，我們可以得到等效的質(zhì)量M=J/ ，承載力 + + ,流量增益系數(shù)的值 =2.825×

27、10-4m3/(s·A)，其中 是電液比例閥的增益系數(shù)。 </p><p>　　出自：中南大學(xué)學(xué)報(bào)（英文版）2008年第15卷第3期382—386頁(yè)</p><p>　　Modeling and parameter estimation for hydraulic system of excavator’s arm </p><p>　　HE Qing

28、-hua，HAO Peng，ZHANG Da-qing</p><p>　　Abstract A retrofitted electro-hydraulic proportional system for hydraulic excavator was introduced firstly. According to the principle and characteristic of load indepe

29、ndent flow distribution (LUDV) system, taking boom hydraulic system as an example and ignoring the leakage of hydraulic cylinder and the mass of oil in it ,a force equilibrium equation and a continuous equation of hydra

30、ulic cylinder were set up. Based on the flow equation of electro-hydraulic proportional valve, the pressure pa</p><p>　　1 Introduction</p><p>　　For its high efficiency and multifunction, hydraul

31、ic excavator is widely used in mines，road building, civil and military construction，and hazardous waste cleanup areas．The hydraulic excavator also plays an important role in construction machines．Nowadays, macaronis and

32、mobilization have been the latest trend for the construction machines．So，the automatic excavator gradually becomes popular in many countries and is considered a focus．Many control methods can be used to automatically con

33、trol the m</p><p>　　2 Overview of robotic excavator</p><p>　　The backhoe hydraulic excavator studied is shown in Fig.1．In Fig.1，F(xiàn)c presents the resultant force of hydraulic cylinder, gravity of

34、boom，dipper, bucket and so on at point B，whose direction is along cylinder AB； Fc can be decomposed into Fcl and Fc2，and their directions are vertical and parallel to that of O1B，respectively；ac is the acceleration whose

35、 direction is same to that of Fc，and ac can be decomposed into acl an d ac2 too；G1，G2 and G3 are the gravity centers of boom，dipper and bucket，resp</p><p>　　The arm of excavator was considered a manipulator

36、with three degrees of freedom (three inclinometers were set on the boom，dipper and bucket，respectively)．In tracking control experiment，the objective trajectories were planed based on the kinematic equation of excavator’s

37、 manipulator．Then，the motion of boom，dipper an d bucket was set by the controller．In order to suit for automatic contro1．the normal hydraulic control excavator should be retrofitted to electro-hydraulic controller． <

38、/p><p>　　Based on original hydraulic system of SW E-85．The hydraulic pilot control system was replaced by an electro-hydraulic pilot control system．The retrofitted hydraulic system is shown in Fig.2．In this wor

39、k，because boom，dipper an d bucket are of the same characteristics，the hydraulic system of boom was taken as an example．In the electro-hydraulic pilot control system，the pilot electro—hydraulic proportional valves were de

40、rived from adding proportional relief valves on the original SX-l4 main valve，a</p><p>　　Flg.1 Schematic diagtam of excavator’s arm</p><p>　　Flg.2 Schematic diagram of retrofitted electro-hydra

41、ulic system of excavator</p><p>　　Flg.3 Schematic diagram of LUDV hydraulic system after retrofitting</p><p>　　3 Model of electro-hydraulic proportional system</p><p>　　3.1 Dynam

42、ics of electro—hydraulic proportional valve </p><p>　　In this work， the electro-hydraulic proportional valve consists of proportional </p><p>　　reliefvalves and SX-14 main valve．A transfer funct

43、ion from input current to the displacement of spool can be obtained as follows： Xv(s)／Iv(s)=KI／(1+bs) (1) where Xv is the Laplace transform of xv，m；KI is the current gain of electro-h

44、ydraulic proportional valves，m／A；b is the time constant of the first order system，s：Iv=I(t)-Id，I(t)and Id are respectively the control current of proportional valve and the current to overcome dead band，A． </p>&

45、lt;p>　　3.2 Flow equation of electro-hydraulic proportional valve </p><p>　　In this work，LUDV system was adopted in the experimental robotic excavator．According to the theory of LUDV system，the flow equati

46、on can be gotten： = w = (2)</p><p>　　= w = （3）</p><p>　　where Δp is the spring-setting pressure of load sense valve，MP

47、a；Cd is the flow coefficient m5／(N·s)；w is the area gradient of orifice，m2／m；ρ is the oil density, kg／m3; Δp1 and Δp2 are the two orifices pressure，respectively, M Pa．When the flow of excavator is not saturated，Δp

48、 is a nearly constant．In this work，the value was tested and gotten by experiment．In Fig.4，ps，p1s,andΔp represent the system pressure，the load sense valve pressure and the difference of pressure, respectively. The pressur

49、e e</p><p>　　Q1= (t),I(t≥0 (4) where is the flow gain coefficient of valve, m2／s, and =CdW </p><p>　　Flg.4 Curves of pressure experiment und

50、er boom moving condition</p><p>　　3.3 Continuity equation of hydraulic cylinder </p><p>　　Generally speaking，construction machine does not permit external leakage．At present，the external le

51、akage can be controlled by sealing technology．On the other hand，it has been proven that the internal leakage of excavator is quite little by experiments．So, the influence of internal and external leakage of hydraulic sys

52、tem can be ignored．When the oil flows into head side of cylinder and discharges from rod side， the continuity equation can be written as </p><p>　　where V1 and V2 are the volumes of fluid flowing into and o

53、ut the hydraulic cylinder, m3 ； is the effective bulk modulus(including liquid，air in oil and so on)，N/m2． </p><p>　　3.4 Force equilibrium equation of hydraulic cylinder </p><p>　　It is assum

54、ed that the mass of oil in hydraulic cylinder is negligible,and the load is rigid. Then the force equilibrium equation of hydraulic cylinder can be calculated from the Newton’s second law： =m + +

55、 （6）</p><p>　　where Bc is the viscous damping coefficient，N·s／m． 3.5 Simplified model of electro—hydraulic proportional system After the Laplace transform of Eqns．(

56、4)—(6)，the simplified model can be expressed as </p><p>　　(7) </p><p>　　where Y(s) is the Laplace transform of y; = .( ); = ; = m; = ;</p><p><b>　　= ( )。</b>&

57、lt;/p><p>　　4 Parameters estimation</p><p>　　From the process of modeling and Eqn．(7)，it is clear that all parameters in the simplified model are related to the structure。the motional situation and

58、 the posture of excavator’s arm．Moreover，these parameters are time variable. So it is quite difficult to get accurate values and mathematic equations of these parameters. To solve this problem，those important parameters

59、of model were estimated approximately by the estimation equation and method proposed in this work． 4.1 Equivalent mass estimati</p><p>　　Considering O1 at an axis of manipulator, the torque and angular acce

60、leration can begiven as follows： </p><p><b>　　(8) </b></p><p>　　where M and ω are the torque and angular acceleration of manipulator to O1，respectively;BQl1 is the length from po

61、int O1 to point B．According to the rotating law： </p><p>　　M=Jω,we get ，</p><p>　　that is： (9) </p><p>　　where J is the

62、 equivalent moment inertia of manipulator to point O1,kg·m2，and it can be written as follows： </p><p>　　J= + + + (10) </p><p>　　J1, J2 and J3 are the moment

63、inertia of boom，dipper and bucket to their own bary center respectively．The values of them can be obtained by dynamic simulation based on the dynamic mode, J1=450.9N·m, J2=240.2N·m, J3=94.9N·m. </p>

64、<p>　　Comparing Eqn．(9)with Fc=m ，the equivalent mass at point B can be given： M=J/ (11) </p>&l

65、t;p>　　4.2 Estimation for load on hydraulic cylinder </p><p>　　The equivalent moment equation of manipulator to O1 is </p><p>　　+ + (12) </p><p>

66、　　where and are the length from pointO1 to point G1′ ，G2′and G3′；，respectively．Then，the counter force of load is </p><p>　　+ + / (13) </p><p>　　4.3

67、 Estimation for flow gain coefficient of valve </p><p>　　The flow of pump can be measured by flow transducer． The instrument used in this work was Multi—system 5050．The step response of flow of boom cylinde

68、r under the electro—hydraulic proportional valve controlled by the step curent is shown in Fig.5．At the same time，the curve verifies Eqn．[11]．Based on the experiment curve，the range of KqKl can be identified according to

69、 Eqns.(1)and(4)．And then，according to data in Fig.4，we can get： =2.825×10-4m3/(s·A)．</p><p>　　Flg.4 Flow of boom cylinder under electro-hydeaulic proportional value controlled by step current <

70、;/p><p>　　5 Conclusions</p><p>　?。?）The mathematic model of electro—hydraulic system is developed according to the characteristics of excavator．It is assumed that the flow across the valve is direc

71、tly proportional to the size of valve orifice，and the influence of intemal and extemal leakage of hydraulic system is ignored．The simplified model can be obtained： </p><p>　　where represent the displacement

72、of piston and the displacement of spool． </p><p>　?。?）From the model of electro—hydraulic system，we can obtain the equivalent mass M=J/ ，bearing force + + , flow gain coefficient of value =2.825×10-4m3