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1、 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。基于電動液壓的比例閥門的流體運動方程，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。0MPa。然后假設(shè)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)合起來建立了液壓缸以等質(zhì)量等為參數(shù)的受力平衡參

2、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(yīng).根據(jù)實驗曲線，閥門的流量增益系數(shù)被確定為2.825×10—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(yīng)用于礦山,道路建設(shè)，民事和軍事建設(shè),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作器。 每種使用方法，研究員必須知道操作

3、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(xué)模型有利于控制器的設(shè)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.NGUYEN利用模糊的滑動方式和阻抗來控制挖掘機動臂的運動，SHAHRAM等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(fù)雜和昂貴的設(shè)計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 挖掘機機械臂概述 液壓挖掘機的挖掘研究結(jié)

4、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分別是動臂，斗桿和鏟斗的重心；m1,m2，m3是它們各自的質(zhì)量且能通過實驗給定（m1=868。136kg，m2=357.115kg and m3=210.736kg）; Ol，O2 和O3是鉸接點；G1′，G2′和 G3′分別是G1 , G2和G3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(三個測斜儀分別裝在動臂，

5、斗桿和鏟斗上）。在跟蹤控制實驗中，其目標(biāo)軌跡是根據(jù)挖掘機機械手運動學(xué)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(yīng)自動控制，普通液壓控制挖掘機應(yīng)改造電動液壓控制挖掘機。 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(dǎo)電液控制系統(tǒng)中，先導(dǎo)電液比例閥是在原始的SX-l4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是可移動的活塞的位移；Q1

6、 和 Q2分別代表流進和流出液壓缸的流量;pl，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;A1 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動力學(xué)特性 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）=KI／(1+bs）

7、 (1) 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 . 3。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： （2） (3) = 其中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；ρ是油密度，單位為 k

8、g/m3；和分別為二個管口壓力，單位為 MPa;當(dāng)挖掘機流程沒有飽和時，是一幾乎恒定。在本文中，其值由實驗測試得到。 在圖4中,ps，p1s，和分別表示系統(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p1s隨著荷載而改變,但是他們的區(qū)別不會隨著荷載而改變，其值接近對2.0MPa.因此，對橫跨閥門的流量的作用可以被忽略。假設(shè)，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： Q1=Kqxv(t），I(t）≥0

9、 (4) 其中Kq是閥門流量系數(shù)，單位為m2/s；并且 壓力 時間 圖4 動臂移動壓力曲線圖 3.3 液壓缸的連續(xù)性方程 一般來說，工程機械不允許外泄.當(dāng)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(dāng)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(dāng)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: (5) 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m3；是有效體積模量(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

10、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： (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(4)—(6）在拉伯拉斯變換以后，簡化的模型可以表達為： （7) 其中Y是y拉伯拉斯變換得到的；;bf=V1V2；a0=V1V2m； a1=BcV1V2；。 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

11、些參量是時變。因此要得到這些參量的準確值和數(shù)學(xué)等式是相當(dāng)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鉸接點旋轉(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。 考慮到機械手的坐標(biāo)軸心O1，機械手的轉(zhuǎn)矩和角加速度可考慮如下： (8) 其中的M 和 分別是工作裝置對O1的轉(zhuǎn)矩和角加速度.是點O1到點B的長度；由轉(zhuǎn)

12、動定律M=J可得：，即： (9） 其中的J是工作裝置指向O1的等效轉(zhuǎn)動慣量，單位是kg·m2;并且寫成如下式子： (10） J1， 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J1=450.9N·m，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：

13、 （11) 4。2 液壓缸負載的估算 工作裝置對于O1等效力矩等式為： （12) 其中和分別表示O1點到 G1′ ，G2′和 G3′三點的距離；那么反力負荷為： (13） 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(yīng)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KqKl

14、的范圍。那么根據(jù)圖4中的數(shù)據(jù)我們可得出：KqKl=2。825×10—4m3/(s·A) . 流量（L/min） 時間 圖5 動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 5 結(jié)論 （1）電液控制系統(tǒng)的數(shù)學(xué)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,并忽略液壓系統(tǒng)的內(nèi)部和外部泄漏影響。簡化模型可以得到： ，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，承載力，流量增益系數(shù)的值KqKl=2.825×10-4m3/（s·A）,其中KI 是電液比例閥的增益系數(shù). 出自:中南大學(xué)學(xué)報(英文版）200

15、8年第15卷第3期382-386頁 Modeling and parameter estimation for hydraulic system of excavator’s arm HE Qing—hua，HAO Peng,ZHANG Da-qing Abstract A retrofitted electro—hydraulic proportional system for hydraulic excavator was introduced firstly。 According to the principle and characteristic of load indepe

16、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 hydraulic cylinder were set up。 Based on the flow equation of electro-hydraulic propo

17、rtional valve, the pressure passing through the valve and the difference pressure were tested 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 displacem

18、ent and is not influenced by load， a simplified model of electro—hydraulic system was put 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 suc

19、h parameters as equivalent mass and bearing force of hydraulic— cylinder were set up. Finally, the step response of flow of boom cylinder was tested when the electro—hydraulic proportional valve was controlled by the step current。 Based on the experiment curve, the flow gain coefficient of valve uni

20、dentified as 2。825×10-4m3/（s·A) and the mode is verified. Key words： Excavator, Hydraulic-cylinder proportional system, Load independent flow distribution (LUDV） system， Modeling， Parameter estimation 1 Introduction For its high efficiency and multifunction， hydraulic excavator is widely used in

21、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 mobilization have been the latest trend for the construction machines．So，the automatic excavator gradually beco

22、mes popular in many countries and is considered a focus．Many control methods can be used to automatically control the manipulator of excavator．Whichever method is used， the researchers must know the structure of manipulator and the dynamic and static characteristics of hydraulic system．That is, the

23、exact mathematical models are helpful to design controller． However, it is difficult to model on time-variable parameters in mechanical structures and various nonlinearities in hydraulic actuators， and disturbance from outside．Researches on time delay control for excavator were carried out in Refs．N

24、GUYE used fuzzy sliding mode control and impedance control to automate the motion of excavator’s manipulator． SHAHRAM et al adopted impedance control to the teleported excavator．Nonlinear models of hydraulic system were developed by some researchers. However, it is complicated and expensive to desig

25、n controller, which 1imits its application．In this paper, based on the proposed model，the model of boom hydraulic system of excavator was simplified according to engineering and by considering the force equilibrium， continuous equation of hydraulic cylinder and flow equation of electro—hydraulic pro

26、portional valve;at the same time，the estimation methods and equations for the parameters of model were developed． 2 Overview of robotic excavator The backhoe hydraulic excavator studied is shown in Fig.1．In Fig。1,Fc presents the resultant force of hydraulic cylinder， gravity of boom，dipper, bucket

27、 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 direction is same to that of Fc，and ac can be decomposed into acl an d ac2 too；G1，G2 and G3 are th

28、e gravity centers of boom，dipper and bucket，respectively；ml，m2 and m3 are the masses of them，and their values can be given by experiment( m1=868。136kg，m2=357.115kg and m3=210。736kg）；Ol，O2 and O3 are the hinged points;G1′，G2′and G3′are projections of Gl,G2 and G3 on x axis，respectively． The arm of

29、excavator was considered a manipulator 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 manipulator．Then，the motion of boom,dipp

30、er 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． Based on original hydraulic system of SW E—85．The hydraulic pilot control system was replaced by an electro—hydraulic pilot c

31、ontrol system．The retrofitted hydraulic system is shown in Fig.2．In this work，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 derive

32、d from adding proportional relief valves on the original SX-l4 main valve，and hydraulic pilot handle was substituted by electrical one．The retrofitted system of excavator was still the LUDV system (Fig。3）of Rexroth with good controllability．In Fig.3，y is the displacement of piston；Q1 and Q2 are the

33、flows in and out to the cylinder respectively；pl，p2，ps and pr are the pressures of head and rod sides of cylinder, system and return oil，respectively；A1 and A2 are the areas of piston in the head and rod sides of cylinder， respectively; xv is the displacement of spool；m is the equivalent mass of loa

34、d。 Flg。1 Schematic diagtam of excavator's arm Flg。2 Schematic diagram of retrofitted electro-hydraulic system of excavator Flg.3 Schematic diagram of LUDV hydraulic system after retrofitting 3 Model of electro-hydraulic proportional system 3。1 Dynamics of electro—hydraulic proportional valve

35、In this work， the electro-hydraulic proportional valve consists of proportional relief valves and SX-14 main valve．A transfer function from input current to the displacement of spool can be obtained as follows： Xv（s）／Iv（s）=KI／（1+bs） (1） where Xv

36、is the Laplace transform of xv，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(t）and Id are respectively the control current of proportional valve and the current to overcome dead band，A． 3.2 Flow equation of ele

37、ctro—hydraulic proportional valve In this work，LUDV system was adopted in the experimental robotic excavator．According to the theory of LUDV system，the flow equation can be gotten: (2) (3) = where is the spring—setting pressure of load sense valve，MPa；cd is the flow coefficient

38、 m5／（N·s）；w is the area gradient of orifice，m2／m；ρ is the oil density， kg／m3;and are the two orifices pressure，respectively， M Pa．When the flow of excavator is not saturated，is a nearly constant．In this work,the value was tested and gotten by experiment．In Fig。4，ps,p1s，andrepresent the system press

39、ure，the load sense valve pressure and the diference of pressure, respectively. The pressure experiment curves of the system show the variation of three kinds of pressures．Although Ps and pls change with load,their difference does not change with load,the value approximates to 2。0MPa。So,the effect of

40、 on the flow across the valve can be neglected．It is assumed 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 as Q1=Kqxv（t),I(t)≥0 （4) wher

41、e is the flow gain coefficient of valve， m2／s, and Flg.4 Curves of pressure experiment under boom moving condition 3。3 Continuity equation of hydraulic cylinder Generally speaking，construction machine does not permit external leakage．At present，the external leakage can be controlled by seal

42、ing 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 system can be ignored．When the oil flows into head side of cylinder and discharges from rod side， the continuity e

43、quation can be written as (5) where V1 and V2 are the volumes of fluid flowing into and out the hydraulic cylinder, m3 ； is the effective bulk modulus（including liquid，air in oil and so on)，N/m2． 3.4 Force equilibrium equation of hydraulic cylinder It is assumed that the mass of oil in hy

44、draulic 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: （6） where Bc is the viscous damping coefficient，N·s／m． 3.5 Simplified model of electro-hydraulic

45、 proportional system After the Laplace transform of Eqns．(4）—（6)，the simplified model can be expressed as （7） where Y（s） is the Laplace transform of y； ;b1=V1V2；a0=V1V2m;a1=BcV1V2;. 4 Parameters estimation From the process of modeling and Eqn．(7)，it is clear that all

46、parameters in the simplified model are related to the structure。the motional situation and 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 p

47、arameters of model were estimated approximately by the estimation equation and method proposed in this work． 4.1 Equivalent mass estimation for load on hydraulic cylinder The load of boom hydraulic cylinder（it is assumed there is no external load）consists of boom，dipper and bucket．In Fig。1，boom，di

48、pper and bucket rotate around points O1，O2 and O3,respectively．So their motions are not straight 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 boom，dipper and bucket． Considering

49、O1 at an axis of manipulator， the torque and angular acceleration can begiven as follows: (8) where M and are the torque and angular acceleration of manipulator to O1，respectively; is the length from point O1 to point B．According to the rotating law： M=J,we get that is

50、 （9） where J is the equivalent moment inertia of manipulator to point O1,kg·m2,and it can be written as follows: (10） J1， J2 and J3 are the moment inertia of boom，dipper and bucket to their own bary center respectively．The values of them can

51、 be obtained by dynamic simulation based on the dynamic mode, J1=450.9N·m, J2=240.2N·m, J3=94。9N·m。 Comparing Eqn．(9）with Fc=mac，the equivalent mass at point B can be given： （11) 4.2 Estimation for load on hydraulic cylinder The equivalent

52、moment equation of manipulator to O1 is （12) where and are the length from pointO1 to point G1′ ，G2′and G3′；，respectively．Then，the counter force of load is （13） 4。3 Estimation for flow gain coefficient of valve The flow of pump can be me

53、asured by flow transducer． The instrument used in this work was Multi—system 5050．The step response of flow of boom cylinder 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 r

54、ange of KqKl can be identified according to Eqns。（1)and（4)．And then，according to data in Fig.4,we can get:KqKl=2。825×10—4m3/(s·A）． Flg.4 Flow of boom cylinder under electro-hydeaulic proportional value controlled by step current 5 Conclusions （1)The mathematic model of electro—hydraulic system is

55、 developed according to the characteristics of excavator．It is assumed that the flow across the valve is directly 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： where represent the displ

56、acement of piston and the displacement of spool． （2）From the model of electro—hydraulic system,we can obtain the equivalent mass ,bearing force , flow gain coefficient of 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