挖掘機外文翻譯外文文獻中英翻譯

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1、挖掘機外文翻譯外文文獻中英翻譯挖掘機臂液壓系統(tǒng)的模型化參量估計摘要首先介紹了液壓挖掘機的一個改裝的電動液壓的比例系統(tǒng)。根據(jù)負載獨立流量分配( LUDV )系統(tǒng)的原則 和特點，以動臂液壓系統(tǒng)為例并忽略液壓缸中的油大量 泄漏，建立一個力平衡方程和一個液壓缸的連續(xù)性方 程?；陔妱右簤旱谋壤y門的流體運動方程，測試的 分析穿過閥門的壓力的不同。結(jié)果顯示壓力的差異并不 會改變負載，此時負載接近 2.0MP&然后假設穿過閥門 的液壓油與閥芯的位移成正比并且不受負載影響，提出 了一個電液控制系統(tǒng)的簡化模型。同時通過分析結(jié)構(gòu)和 承重的動臂裝置，并將機械臂的力矩等效方程與旋轉(zhuǎn) 法、參數(shù)估計估計法結(jié)合起來建立了

2、液壓缸以等質(zhì)量等 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 比例閥來測試動臂液壓缸中液壓油的階躍響應。根據(jù)實 驗曲線，閥門的流量增益系數(shù)被確定為2.825 X10- 4m3/(s A),并驗證了該模型。關(guān)鍵詞：挖掘機，電液比例系統(tǒng)，負載獨立流量分配(LUDV )系統(tǒng)，建模，參數(shù)估計1引言由于液壓挖掘機具有高效率、多功能的優(yōu)點，所以 被廣泛應用于礦山，道路建設，民事和軍事建設，危險 廢物清理領(lǐng)域。液壓挖掘機在施工機械領(lǐng)域中也發(fā)揮了 重要作用。目前，機電一體化和自動化已成為施工機械 發(fā)展的最新趨勢。因此，自動挖掘機在許多國家逐漸變 得普遍并被認為重點。挖掘機可以用許多控制方法自動 地控制操

3、作器。 每種使用方法，研究員必須知道操作器 結(jié)構(gòu)和液壓機構(gòu)的動態(tài)和靜態(tài)特征。即確切的數(shù)學模型 有利于控制器的設計。然而，來自外部的干擾使得機械 結(jié)構(gòu)模型和各種非線性液壓制動器的時變參數(shù)很難確 定。關(guān)于挖掘機時滯控制的研究已經(jīng)有人在研究了。NGUYE利用模糊的滑動方式和阻抗來控制挖掘機動臂的 運動，SHAHRAM采取了阻抗對挖掘機遠距傳物的控制液壓機構(gòu)非線性模型已經(jīng)由研究員開發(fā)出來了。然而, 復雜和昂貴的設計控制器限制了它的應用。在本文，根 據(jù)提出的模型，根據(jù)工程學和受力平衡，挖掘機臂液壓 機構(gòu)模型簡化為連續(xù)均衡的液壓缸和流動均衡的電液比 例閥；同時，確定了模型的參量的估計方法和等式。2挖掘機機

4、械臂概述液壓挖掘機的挖掘研究結(jié)果如圖1。在圖中，F(xiàn)c表 示液壓缸，動臂的重力，斗桿，鏟斗的重力等在B點合力,其方向是沿著液壓缸 AB方向；Fc可分解成Fci和 Fc2 ,他們的方向分別為垂直于和平行于 OB ,加速度ac 的方向與Fc是相同的，并且ac也可以分解成aci和ac2； G , G2和G分別是動臂，斗桿和鏟斗的重心；m, m,m3是它們各自的質(zhì)量且能通過實驗給定(m=868.136kg , m2=357.115kg and m 3=210.736kg) ; Oi, Q 和 Q是較接 點；G ,G和G3分別是G , G2和G在X軸上的投 影。挖掘機的臂被認為是一個三個自由度的的機械手(

5、三個分別裝在動臂，斗桿和鏟 斗上)0在跟蹤控制實驗中，其目標軌跡是根據(jù)挖掘機機械手運動學方程確定的。然 后，動臂，斗桿和鏟斗的動作有操作員控制。為了適應自動控制，普通液壓控制挖掘 機應改造電動液壓控制挖掘機?；赟W E-85型原有的液壓系統(tǒng)，把先導液壓控制系統(tǒng)更換為先導電液控制系 統(tǒng)。新改進的液壓系統(tǒng)如圖2所示。在這系統(tǒng)中，因為動臂，斗桿和鏟斗具有相同的 特點，將動臂的液壓系統(tǒng)作為一個例子。在先導電液控制系統(tǒng)中，先導電液比例閥是在原始的SX-14主要閥門基礎(chǔ)上增加比例泄壓閥衍生出的并且用電子手柄替代液壓手 柄。挖掘機的改裝系統(tǒng)仍是具有良好的可控性的 LUDV系統(tǒng)（圖3）。在圖3中，y 是可移

6、動的活塞的位移；Qi和Q2分別代表流進和流出液壓缸的流量；pi, p2, ps和pr 分別表示汽缸的有桿腔和無桿腔，系統(tǒng)和回油路的壓力；Ai和A2分別表示汽缸的有桿腔和無桿腔的面積；xv代表閥芯的位移；m代表加載的負載；圖1挖掘機工作裝示意圖圖2挖掘機液壓系統(tǒng)示意圖圖3改造后LUDV液壓系統(tǒng)示意圖3模型的電液比例系統(tǒng)3.1 電動液壓的比例閥門在本文中，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：Xv(s)/Iv(s尸Ki/(1+bs)(1)其中Xv是Xv的值，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)的時間常數(shù)，單位為 s； Iv=

7、I(t)-Id, I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 Ao3.2 電動液壓的比例閥門的流體運動方程在本文中，實驗性機器人挖掘機采取了LUDV系統(tǒng)。根據(jù)LUDV系統(tǒng)的理論，可以得到流體運動方程：QiCdWXv2piCdWXvj2 pj , I 0CdWXv J2 PiPr / , I (t) 0CdWXv J2 P2Pr / , I (t) 0Q2CdWXv - P2CdWXvj2 p/ , I(t) 0其中P是負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 s)； w是管口的面積梯度，單位為 m2/m； p是油密度，單位為kg/m3;

8、Pi和P2分別為二個管口壓力，單位為 MPa；當挖掘機流程沒有飽和時，p是一幾乎包定。在本文中，其值由實驗測試得到。在圖4中，Ps, Pis,和p分別表示系統(tǒng)壓力、負荷傳感閥門壓力和它們的壓力 差；壓力系統(tǒng)的實驗曲線顯示三種不同的壓力值。雖然ps和pis隨著荷載而改變，但是他們的區(qū)別不會隨著荷載而改變，具值接近對2.0MPa。因此，對橫跨閥門的流量的作用p可以被忽略。假設，流過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)能被簡化為：Qi = KqXv(t),I(t) 0(4)其中Kq是閥門流量系數(shù)，單位為 m2/s；并且Kqcdwj2 p/圖4動臂移動壓力曲線圖3.3

9、 液壓缸的連續(xù)性方程一般來說，工程機械不允許外泄。當前，外在泄漏可以通過密封技術(shù)控制。另一 方面，由實驗證明了挖掘機內(nèi)部泄漏是相當小的。因此，液壓機構(gòu)內(nèi)部和外在泄漏的 影響可以被忽略。當油流進汽缸無桿腔并且進入到有桿腔內(nèi)時，連續(xù)性方程可以寫 成：?-Qi Ai y Vi pl cT?LQ2 A2 y V 2 P2/ c其中Vi和V2分別表示流入及流出的液壓缸液體的體積，單位是m3； c是(包括液體，油中的空氣等)，單位是 N/m2。3.4 液壓缸力的平衡方程據(jù)推測，液壓缸中油的質(zhì)量可以忽略，而且負載是剛性的。那么可以根據(jù)牛頓的法律得到液壓缸的力量平衡等式：?PiAi p2A2 my Bc y

10、Fc(6)其中Bc是黏阻止的系數(shù)，單位是N s/m。3.5 電動液壓的比例系統(tǒng)簡化的模型方程一(6)在拉伯拉斯變換以后，簡化的模型可以表達為：Y sbiXv s bfsFc s sa0s2 a1s a22其中Y是y拉伯拉斯變換得到的；bicKq?AV2 4 /A ； bf=ViV2；ao=ViV2m；22ai=BcViV2； a2cV2A ViA,。4參量估計從塑造的過程和方程(7)中可以得到在確切的簡化的模型中與結(jié)構(gòu)，運動情況以及挖 掘機動臂的體位有關(guān)的所有參量。而且，這些參量是時變。因此要得到這些參量的準確值和數(shù)學等式是相當難的。要解決這個問題，本文提出了估計方程和方法來估算模 型中的這些

11、重要參數(shù)。4.1估算液壓缸負載液壓缸臂上的負載(假定沒有外部負載)由動臂，斗桿和鏟斗上的負載組成。在 圖1中，動臂，斗桿和鏟斗分別繞著各自的較接點旋轉(zhuǎn)。因此他們的運動不是沿著汽 缸的直線運動，也就是說他們的運動方向與方程(5)中的y的方向是不同的。因此 方程(6)中的m不能簡單的認為是動臂，斗桿和鏟斗質(zhì)量的總和?？紤]到機械手的坐標軸心Oi,機械手的轉(zhuǎn)矩和角加速度可考慮如下：Fclo1BSin ac sin &(8)其中的M和分別是工作裝置對Oi的轉(zhuǎn)矩和角加速度。Qb是點Oi到點B的長度；由轉(zhuǎn)動定律M=J可得：FclOiBsinJacsin /loiB ,即：Fc aJ/l2OiB(9)其中的J

12、是工作裝置指向Oi的等效轉(zhuǎn)動慣量，單位是kg m2;并且寫成如下式子：222J J imlOiGiJ 2rm21O2J 3rm31O3(i0)Ji, J2和J3分別是動臂，斗桿和鏟斗對各自的中心的慣性力矩；它們的值可以通過 模擬動態(tài)模型得出 Ji=450.9Nm, J2=240.2N m, J3=94.9N m。比較方程(9)和Fc=mac,可以得出點B的等效質(zhì)量：2m J/12OiB(ii)4.2 液壓缸負載的估算工作裝置對于Oi等效力矩等式為：FclOiB sinmiglOiGi 仁m3glOiG3(i2)其中I。，L 和he分別表示Oi點至1J Gi，G2和G3三點的距離；那么反力負 O

13、iGi O|G2OiG 3荷為：FcmiglOiGi m2glOiG2m3glO lOiB sin(i3)4.3 增益系數(shù)閥流量的估計流量傳感器可以測量泵的流量。用于這項工作的儀器為多系統(tǒng) 5050型。動臂液 壓缸流量的階躍響應在電液比例閥控制下的結(jié)果如圖 5所示。同時，該曲線驗證等式 (11)。根據(jù)實驗曲線和等式(1)和(4)可確定KqKi的范圍。那么根據(jù)圖4中的數(shù)據(jù)我們 可得出：KqKi=2.825 10-4m3/(s A)。圖5動臂液壓缸流量的階躍響應在電液比例閥控制下的曲線圖5結(jié)論(1)電液控制系統(tǒng)的數(shù)學模型是根據(jù)挖掘機的特點發(fā)展起來的。假定流過閥的 流量與閥口大小成正比，并忽略液壓系

14、統(tǒng)的內(nèi)部和外部泄漏影響。簡化模型可以得 到：Y(s) bXv(s) b1sFc(s)/s(%s2 as a2),其中 Y (s)和 Xv(s)分別是活塞和 閥芯的位移。(2)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/12of,承載力Fl(mMoGm2g16Gm31GJ ,流量增益系數(shù)的值 KqKi=2.825 10-4m3/(s A),其中O1G1O1G2QG3Ki是電液比例閥的增益系數(shù)出自：中南大學學報(英文版)2008年第15卷第3期382 386頁Modeling and parameter estimation for hydraulic system of excavat

15、or s armHE Qing-hua, HAO Peng, ZHANG Da-qingAbstractA retrofitted electro-hydraulic proportional system for hydraulic excavator was introduced firstly. According to the principle and characteristic of load independent flow distribution (LUDV) system, taking boom hydraulic system as an example and ig

16、noring the leakage of hydraulic cylinder and the mass of oil in it ,a force equilibrium equation and a continuous equation of hydraulic cylinder were set up. Based on the flow equation of electro-hydraulic proportional valve, the pressure passing through the valve and the difference pressure were te

17、sted and analyzed. The results show that the difference of pressure does not change with load and it approximates to 2.0MPa. And then, assume the flow across the valve id directly proportional to spool displacement and is not influenced by load, a simplified model of electro-hydraulic system was put

18、 forward. At the same time, by analyzing the structure and load-bearing of boom instrument, and combining moment equivalent equation of manipulator with rotating law, the estimation methods and equations for such parameters as equivalent mass and bearing force of hydraulic- cylinder were set up. Fin

19、ally, the step response of flow of boom cylinder was tested when the electro-hydraulicproportional valve was controlled by the step current. Based on the experiment curve, the flow gain coefficient of valve unidentified as 2.825 10-4m3/(s A) and I shutt】。valvtXad Sen sc I vaIvc .p Jshiltiiw) -IRcxro

20、lh SX1-1 valve，r=。Flg.3 Schematic diagram of LUDV hydraulic system after retrofitting3 Model of electro-hydraulic proportional system3.1 Dynamics of electro hydraulic proportional valveIn this work, the electro-hydraulic proportional valve consists of proportional relief valves and SX-14 main valve

21、A transfer function from input current to the displacement of spool can be obtained as follows Xv(s)/Iv(s尸KI/(1+bs)(1)where Xv is the Laplace transform of x/, m； Ki is the current gain of electro-hydraulic proportional valves, m/A ； b is the time constant of the first order system s： Iv=I(t)- Id, I(

22、t)and Id are respectively the control current of proportional valve and the current to overcome dead band A.3.2 Flow equation of electro-hydraulic proportional valveIn this work, LUDV system was adopted in the experimental robotic excavator. According to the theory of LUDV system , the flow equation

23、 can be gotten, cdwxv2 p/ , I(t) 0Qi cd wxvPi -J(2)Icdwxvj2 phpr / , I(t) 02cd WXv 72Pr / ，I (t) 0Q2cdwXvj p2 =YIcdw.j2 p, I(t) 0where p is the spring-setting pressure of load sense valveMPa； cd is the flow coefficient m5/(N s)； w is the area gradient of orifice, m2/m； p is the oil densitykg/ 3m ; p

24、1 and p2 are the two orifices pressure respectively, M Pa. When the flow of excavator is not saturated p is a nearly constant In this work , the value was tested and gotten by experiment In Fig.4 , ps, p1s,and p represent the system pressure the load sense valve pressure and the diference of pressur

25、e, respectively. The pressure experiment curves of the system show the variation of three kinds of pressure sAlthough Ps and pls change with load, their difference does not change with load the value approximates to 2.0MPa.So, the effect of on the flow across the valve can be neglected It is assumed

26、 that the flow across the valve is proportional to the size of orifice valve and the flow is not influenced by load. Then, Eqn. (2) can be simplified asQ=Kqxv(t),I(t) 0(4)where is the flow gain coefficient of valve, m2/s, and Kqow 2 p/04 H 1216Tims Flg.4 Curves of pressure experiment under boom movi

27、ng condition3.3 Continuity equation of hydraulic cylinderGenerally speaking construction machine does not permit external leakage At present, the external leakage can be controlled by sealing technology On the other hand, it has been proven that the internal leakage of excavator is quite little by e

28、xperiment tsSo, the influence of internal and external leakage of hydraulic system can be ignore d When the oil flows into head side of cylinder and discharges from rod side the continuity equation can be written as?-QiAi y Vi Pi/ cn?Q Q2 A2 y V 2 P2/ cwhere Vi and V2 are the volumes of fluid flowin

29、g into and out the hydraulic cylinder,m3 ;c is the effective bulk modulus(including liquid , air in oil and so on), N/m2.3.4 Force equilibrium equation of hydraulic cylinderIt is assumed that the mass of oil in hydraulic cylinder is negligible,and the load is rigid.Then the force equilibrium equatio

30、n of hydraulic cylinder can be calculated from theNewton s second la w ?P1A1P2 A2my Bcy Fc(6)where Bc is the viscous damping coefficient, N s/m.3.5 Simplified model of electro hydraulic proportional systemAfter the Laplace transform of Eqns. (4) (6), the simplified model can be expressed asY sb1Xv s

31、 bfs艮 s s a0s2 a1s a2(7)where Y(s) is the Laplace transform of y 222b(cKq?AV2A2/ A1;bi=ViV2；ao=ViV2m;ai=BcViV2； a2cv2AViA24 Parameters estimationFrom 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 t

32、he posture ofexcavators arrMfloreover, these parameters are time variable. So it is quite difficult toget accurate values and mathematic equations of these parameters. To solve this prob |em those important parameters of model were estimated approximately by the estimation equation and method propos

33、ed in this work4.1 Equivalent mass estimation for load on hydraulic cylinderThe load of boom hydraulic cylinder(it is assumed there is no external load)consists of boom, dipper and bucket In Fig.1, boom, dipper and bucket rotate around points O, O2 and O3, respectively. So their motions are not stra

34、ight line motions about the cylinders, that is to say, their motion directions are different from Y in Eqn.(5). So, m in Eqn.(6)cannot be simply regarded as the sum mass of boq mdipper and bucketConsidering Oi at an axis of manipulator, the torque and angular acceleration can begiven as follows：1 MF

35、ciloiBFcMsin1-11(8)I _aci lo1Bac sinlo1Bwhere M and are the torque and angular acceleration of manipulator to O respectively;lQB is the length from point Oi to point B. According to the rotating law: M=J ,we getFclOi B 1 nvJ c 1/ lO|B-2，一、that isFc acJ/l OiB(9)where J is the equivalent moment inerti

36、a of manipulator to point O,kg m2, and it can be written as follows：222J JimJ01GlJ2m2l01G2J3m3l01G3(i0)Ji, J2 and S are the moment inertia of boon dipper and bucket to their own bary center respectively The values of them can be obtained by dynamic simulation based on the dynamic mode, J=450.9N m, J

37、2=240.2N m, J3=94.9N m.Comparing Eqn. (9)with Fc=mac, the equivalent mass at point B can be given2m J/l2OiB(11)4.2 Estimation for load on hydraulic cylinderThe equivalent moment equation of manipulator to O isFclOiB sinmlglOiGim2glOiG2m3glOiG3(12)where ,l” and l” are the length from point。to point G

38、i, G2 and , 。1 Gi OG 2OG 3respectively. Then, the counter force of load isFCm1glQGm2gloGm3glQGlO1Bsin(13)GiO1G2G314.3 Estimation for flow gain coefficient of valveThe 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

39、boom cylinder under the electro hydraulic proportional valve controlled by the step curent is shown in Fig.5 At the same timq the curve verifies Eqn. 11 . Based on the experiment curve the range of KqKl can be identified according to Eqns.(1)and(4) And then, according to data in Fig.4, we can get Kq

40、Kl=2.825 10-4m3/(s A).Flg.4 Flow of boom cylinder under electro-hydeaulic proportional value controlled by stepcurrent5 Conclusions(1) The mathematic model of electro hydraulic system is developed according tothe characteristics of excavator It is assumed that the flow across the valve is directly p

41、roportional to the size of valve orifice, and the influence of intemal and extemal leakage ofhydraulic system is ignored. The simplified model can be obtainedY(s) hXv(s) hsFc(s)/s(a0s2 &s a?)where represent the displacement of piston and the displacement of spo ol(2) From the model of electro hydrau

42、lic system, we can obtain the equivalent2mass m J /l qb , bearing forceFl(m1g10Gm2g101G m3lOG3) , flow gain coefficientof value KqKl=2.825 10-4m3/(s A) , where Ki is the current gain of electro- hydraulic proportional valves.From : Journal of Central South University (English) 2008 Vol 15 No. 3 pages 382-386