挖掘机臂液压系统的模型化参量估计外文翻译


    挖掘机臂液压系统模型化参量估计

    首先介绍液压挖掘机改装电动液压例系统根负载独立流量分配( LUDV )系统原特点动臂液压系统例忽略液压缸中油量泄漏建立力衡方程液压缸连续性方程基电动液压例阀门流体运动方程测试分析穿阀门压力结果显示压力差异会改变负载时负载接20MPa然假设穿阀门液压油阀芯位移成正受负载影响提出电液控制系统简化模型时通分析结构承重动臂装置机械臂力矩等效方程旋转法参数估计估计法结合起建立液压缸等质量等参数受力衡参数方程阶跃电流控制电液例阀测试动臂液压缸中液压油阶跃响应根实验曲线阀门流量增益系数确定2825×104m3(s·A)验证该模型

    关键词:挖掘机电液例系统负载独立流量分配( LUDV )系统建模参数估计









    1 引言
    液压挖掘机具高效率功优点广泛应矿山道路建设民事军事建设危险废物清理领域液压挖掘机施工机械领域中发挥重作目前机电体化动化已成施工机械发展新趋势动挖掘机许国家逐渐变普遍认重点挖掘机许控制方法动控制操作器 种方法研究员必须知道操作器结构液压机构动态静态特征确切数学模型利控制器设计然外部干扰机械结构模型种非线性液压制动器时变参数难确定关挖掘机时滞控制研究已研究NGUYEN利模糊滑动方式阻抗控制挖掘机动臂运动SHAHRAM等采取阻抗挖掘机远距传物控制液压机构非线性模型已研究员开发出 然复杂昂贵设计控制器限制应文根提出模型根工程学受力衡挖掘机臂液压机构模型简化连续均衡液压缸流动均衡电液例阀时确定模型参量估计方法等式
    2 挖掘机机械臂概述
    液压挖掘机挖掘研究结果图1图中Fc表示液压缸动臂重力斗杆铲斗重力等B点合力方着液压缸AB方 Fc分解成Fc1Fc2 方分垂直行O1B 加速度ac方Fc相ac分解成ac1ac2G1 G2G3分动臂斗杆铲斗重心m1m2m3质量通实验定(m1868136kgm2357115kg and m3210736kg) OlO2 O3铰接点G1´G2´ G3´分G1 G2G3X轴投影
    挖掘机臂认三度机械手(三测斜仪分装动臂斗杆铲斗)踪控制实验中目标轨迹根挖掘机机械手运动学方程确定然动臂斗杆铲斗动作操作员控制适应动控制普通液压控制挖掘机应改造电动液压控制挖掘机
    基SW E85型原液压系统先导液压控制系统更换先导电液控制系统新改进液压系统图2示系统中动臂斗杆铲斗具相特点动臂液压系统作例子先导电液控制系统中先导电液例阀原始SXl4阀门基础增加例泄压阀衍生出电子手柄代液压手柄
    挖掘机改装系统具良控性LUDV系统(图3 )图3中 y移动活塞位移Q1 Q2分代表流进流出液压缸流量plp2ps pr分表示汽缸杆腔杆腔系统回油路压力A1 A2分表示汽缸杆腔杆腔面积xv代表阀芯位移m代表加载负载


    图1 挖掘机工作装示意图


    图2 挖掘机液压系统示意图

    图3 改造LUDV液压系统示意图
    3 模型电液例系统
    31 电动液压例阀门动力学特性
    文中电液例阀包括例减压阀SX14阀传递功输入液流阀芯位移:
    Xv(s)/Iv(s)KI/(1+bs) (1)
    中Xvxv拉普拉斯变换值单位mKI电液例阀获液流单位mA
    b阶系统时间常数单位sIvI(t)IdI(t) Id 分表示例阀门控制潮流克服静带潮流单位A
    32 电动液压例阀门流体运动方程
    文中实验性机器挖掘机采取LUDV系统根LUDV系统理流体运动方程:






    (2)

    (3)



    中负荷传感阀门压力差单位 MPacd径流系数单位m5/(N·s)w口面积梯度单位 m2/mρ油密度单位 kgm3分二口压力单位 MPa挖掘机流程没饱时恒定文中值实验测试
    图4中psp1s分表示系统压力负荷传感阀门压力压力差压力系统实验曲线显示三种压力值然psp1s着荷载改变区会着荷载改变值接20MPa横跨阀门流量作忽略假设流阀门流量口阀门成例荷载影响流量方程(2)简化:
    Q1Kqxv(t)I(t)≥0 (4)
    中Kq阀门流量系数单位m2s

    压力
    时间


    图4 动臂移动压力曲线图

    33 液压缸连续性方程
    般说工程机械允许外泄前外泄漏通密封技术控制方面实验证明挖掘机部泄漏相液压机构部外泄漏影响忽略油流进汽缸杆腔进入杆腔时连续性方程写成:
    (5)


    中 V1 V2 分表示流入流出液压缸液体体积单位m3效体积模量(包括液体油中空气等)单位Nm2
    34 液压缸力衡方程
    推测液压缸中油质量忽略负载刚性根牛顿法律液压缸力量衡等式:
    (6)
    中Bc黏阻止系数单位 N·sm
    35 电动液压例系统简化模型
    方程(4)—(6)拉伯拉斯变换简化模型表达:
    (7)
    中Yy拉伯拉斯变换bfV1V2a0V1V2m
    a1BcV1V2
    4 参量估计
    塑造程方程(7)中确切简化模型中结构运动情况挖掘机动臂体位关参量参量时变参量准确值数学等式相难解决问题文提出估计方程方法估算模型中重参数
    41 估算液压缸负载
    液压缸臂负载(假定没外部负载)动臂斗杆铲斗负载组成图1中动臂斗杆铲斗分绕着铰接点旋转运动着汽缸直线运动说运动方方程(5)中y方方程(6)中m简单认动臂斗杆铲斗质量总
    考虑机械手坐标轴心O1机械手转矩角加速度考虑:
    (8)


    中M 分工作装置O1转矩角加速度点O1点B长度转动定律MJ::
    (9)
    中J工作装置指O1等效转动惯量单位kg·m2写成式子:
    (10)
    J1 J2 J3分动臂斗杆铲斗中心惯性力矩值通模拟动态模型出J14509N·mJ22402N·mJ3949N·m
    较方程(9)Fcmac出点B等效质量:
    (11)
    42 液压缸负载估算
    工作装置O1等效力矩等式:
    (12)
    中分表示O1点 G1´ G2´ G3´三点距离反力负荷:
    (13)
    43增益系数阀流量估计
    流量传感器测量泵流量项工作仪器系统5050型动臂液压缸流量阶跃响应电液例阀控制结果图5示时该曲线验证等式(11) 根实验曲线等式(1)(4)确定KqKl范围根图4中数出:KqKl2825×104m3(s·A)


    流量(Lmin)
    时间

    图5 动臂液压缸流量阶跃响应电液例阀控制曲线图

    5 结
    (1)电液控制系统数学模型根挖掘机特点发展起假定流阀流量阀口成正忽略液压系统部外部泄漏影响简化模型: 中Y(s)Xv(s)分活塞阀芯位移
    (2)电液控制系统模型中等效质量承载力流量增益系数值KqKl2825×104m3(s·A)中KI 电液例阀增益系数

    出:中南学学报(英文版)2008年第15卷第3期382—386页










    Modeling and parameter estimation for hydraulic system of excavator’s arm
    HE QinghuaHAO PengZHANG Daqing
    Abstract
    A retrofitted electrohydraulic 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 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 electrohydraulic proportional 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 20MPa And then assume the flow across the valve id directly proportional to spool displacement and is not influenced by load a simplified model of electrohydraulic system was put forward At the same time by analyzing the structure and loadbearing 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 Finally the step response of flow of boom cylinder was tested when the electrohydraulic proportional valve was controlled by the step current Based on the experiment curve the flow gain coefficient of valve unidentified as 2825×104m3(s·A) and the mode is verified

    Key words Excavator Hydrauliccylinder 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 minesroad building civil and military constructionand 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.Sothe automatic excavator gradually becomes 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 exact mathematical models are helpful to design controller. However it is difficult to model on timevariable 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.NGUYE 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 design controller which 1imits its application.In this paper based on the proposed modelthe 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 electrohydraulic proportional valveat the same timethe estimation methods and equations for the parameters of model were developed.
    2 Overview of robotic excavator
    The backhoe hydraulic excavator studied is shown in Fig1.In Fig1Fc presents the resultant force of hydraulic cylinder gravity of boomdipper bucket and so on at point Bwhose direction is along cylinder AB Fc can be decomposed into Fcl and Fc2and their directions are vertical and parallel to that of O1Brespectivelyac is the acceleration whose direction is same to that of Fcand ac can be decomposed into acl an d ac2 tooG1G2 and G3 are the gravity centers of boomdipper and bucketrespectivelymlm2 and m3 are the masses of themand their values can be given by experiment( m1868136kgm2357115kg and m3210736kg)OlO2 and O3 are the hinged pointsG1´G2´and G3´are projections of GlG2 and G3 on x axisrespectively.
    The arm of excavator was considered a manipulator with three degrees of freedom (three inclinometers were set on the boomdipper and bucketrespectively).In tracking control experimentthe objective trajectories were planed based on the kinematic equation of excavator’s manipulator.Thenthe motion of boomdipper an d bucket was set by the controller.In order to suit for automatic contro1.the normal hydraulic control excavator should be retrofitted to electrohydraulic controller.
    Based on original hydraulic system of SW E85.The hydraulic pilot control system was replaced by an electrohydraulic pilot control system.The retrofitted hydraulic system is shown in Fig2.In this workbecause boomdipper an d bucket are of the same characteristicsthe hydraulic system of boom was taken as an example.In the electrohydraulic pilot control systemthe pilot electro—hydraulic proportional valves were derived from adding proportional relief valves on the original SXl4 main valveand hydraulic pilot handle was substituted by electrical one.The retrofitted system of excavator was still the LUDV system (Fig3)of Rexroth with good controllability.In Fig3y is the displacement of pistonQ1 and Q2 are the flows in and out to the cylinder respectivelyplp2ps and pr are the pressures of head and rod sides of cylinder system and return oilrespectivelyA1 and A2 are the areas of piston in the head and rod sides of cylinder respectively xv is the displacement of spoolm is the equivalent mass of load


    Flg1 Schematic diagtam of excavator’s arm

    Flg2 Schematic diagram of retrofitted electrohydraulic system of excavator

    Flg3 Schematic diagram of LUDV hydraulic system after retrofitting




    3 Model of electrohydraulic proportional system
    31 Dynamics of electro—hydraulic proportional valve
    In this work the electrohydraulic proportional valve consists of proportional relief valves and SX14 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 is the Laplace transform of xvmKI is the current gain of electrohydraulic proportional valvesm/Ab is the time constant of the first order systems:IvI(t)IdI(t)and Id are respectively the control current of proportional valve and the current to overcome dead bandA.
    32 Flow equation of electrohydraulic proportional valve
    In this workLUDV system was adopted in the experimental robotic excavator.According to the theory of LUDV systemthe flow equation can be gotten:





    (2)


    (3)



    where is the springsetting pressure of load sense valveMPacd is the flow coefficient m5/(N·s)w is the area gradient of orificem2/mρ is the oil density kg/m3and are the two orifices pressurerespectively M Pa.When the flow of excavator is not saturatedis a nearly constant.In this workthe value was tested and gotten by experiment.In Fig4psp1sandrepresent the system pressurethe 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 loadtheir difference does not change with loadthe value approximates to 20MPaSothe effect of 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 valveand the flow is not influenced by load.ThenEqn.(2) can be simplified as
    Q1Kqxv(t)I(t)≥0 (4)
    where is the flow gain coefficient of valve m2/s and

    Flg4 Curves of pressure experiment under boom moving condition
    33 Continuity equation of hydraulic cylinder
    Generally speakingconstruction machine does not permit external leakage.At presentthe external leakage can be controlled by sealing technology.On the other handit 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 equation 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 liquidair in oil and so on)Nm2.
    34 Force equilibrium equation of hydraulic cylinder
    It is assumed that the mass of oil in hydraulic cylinder is negligibleand 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 coefficientN·s/m.
    35 Simplified model of electro—hydraulic 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
    b1V1V2a0V1V2ma1BcV1V2
    4 Parameters estimation
    From the process of modeling and Eqn.(7)it is clear that all parameters in the simplified model are related to the structurethe motional situation and the posture of excavator’s arm.Moreoverthese parameters are time variable So it is quite difficult to get accurate values and mathematic equations of these parameters To solve this problemthose important parameters of model were estimated approximately by the estimation equation and method proposed in this work.
    41 Equivalent mass estimation for load on hydraulic cylinder
    The load of boom hydraulic cylinder(it is assumed there is no external load)consists of boomdipper and bucket.In Fig1boomdipper and bucket rotate around points O1O2 and O3respectively.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).Som in Eqn(6)cannot be simply regarded as the sum mass of boomdipper and bucket.
    Considering 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 O1respectively is the length from point O1 to point B.According to the rotating law:
    MJwe get

    that is (9)
    where J is the equivalent moment inertia of manipulator to point O1kg·m2and it can be written as follows:
    (10)
    J1 J2 and J3 are the moment inertia of boomdipper and bucket to their own bary center respectively.The values of them can be obtained by dynamic simulation based on the dynamic mode J14509N·m J22402N·m J3949N·m
    Comparing Eqn.(9)with Fcmacthe equivalent mass at point B can be given:
    (11)
    42 Estimation for load on hydraulic cylinder
    The equivalent moment equation of manipulator to O1 is
    (12)
    where and are the length from pointO1 to point G1´ G2´and G3´respectively.Thenthe counter force of load is
    (13)
    43 Estimation for flow gain coefficient of valve
    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 cylinder under the electro—hydraulic proportional valve controlled by the step curent is shown in Fig5.At the same timethe curve verifies Eqn.[11].Based on the experiment curvethe range of KqKl can be identified according to Eqns(1)and(4).And thenaccording to data in Fig4we can get:KqKl2825×104m3(s·A).


    Flg4 Flow of boom cylinder under electrohydeaulic proportional value controlled by step current
    5 Conclusions
    (1)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 directly proportional to the size of valve orificeand the influence of intemal and extemal leakage of hydraulic system is ignored.The simplified model can be obtained:

    where represent the displacement of piston and the displacement of spool.
    (2)From the model of electro—hydraulic systemwe can obtain the equivalent mass bearing force flow gain coefficient of value KqKl2825×104m3(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 382386
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