毕业设计方案翻译zhongwen

1、2.5快速成型一种新的生产产品的概念最近被引进制造过程。在这些观念中，一些是对已经存在的观念的修饰，但是另一些完全是革命性的。在前一类的实例中，我们可以运用数控机床来切割各种各样的材料，使用激光和喷气机可以切割从木头到陶瓷的各种材料。对于后一类，我们可以描述零件的建模和三维快速加工过程。这个概念在内容和工业上很有潜力，因此值得简易讨论一下。它在很大程度上是基于计算机能力发展的结果。对于第一组，工作效率的大幅度提高主要是由于计算机的应用，虽然，至少在原则上，这些过程可以以手动方式实现。对于第二组，没有计算机的帮助是很难执行这些过程的。现在制造依赖于大量的由塑料和金属做成的模具。这些零件有时会有非

2、常复杂的形状和华丽的表面。这些图形是不能在传统机床上处理的，因为仅完成单一部分的加工是非常浪费时间和金钱的。同样，使用模具去生产可能检验后需要改变的个人模式的工件也是非常昂贵的这种情况发生在非传统工具的生产中，并且这个过程非常昂贵，消耗时间和劳力）。在最近几年中，一个新的解决这种情况的概念被提出。它被称为快速成型，在H. D.Kochan编写的自由制造一书中描述了立体雕刻，模具快速原型和快速成型德累斯顿科技大学，德国，科学技术出版社）。我们用图2.19a中的模型来解释这种观念下的制造过程。这个模 型代表了具有特殊齿形的螺旋齿轮，齿是由相互之间具有角度的平 面层组成。换句话说，一个具有复杂形状的

3、三维模型是由简单形状 的薄平面层组成的。概念和布局图2.19 a ）快速原型演示工件。注意看齿轮上清晰可见的层。每一个层都按一定的角度旋转，从而形成了螺旋齿轮的形状在这里为了形象的说明，每个层的厚度变化是夸张的）b这是按照这种技术生产最终设计完成前的模型Conceptland Ltd., Ra'anana, Israel 的产品）。有好几种不同的技术，利用这种原理并应用计算机辅助加工具有复杂空间的零件。我们将在这里简单的描述这个概念的本质。电脑的内存用来存储要加工零件的几何图形信息，以至于零件 的每个几何薄层 通常0.3-0.5mm）都可以用数值定义。根据这种概念，在创造半层体中一个可

4、能的方案如图2.20所图2.20 快速建模布局图1 ）容器；2）聚合液体；3）金属板；4） 电脑；5）激光器；6）旋转镜这样的布局由容器1充满了一种特殊的液体2,这种液体在紫外线的 照射下会变成固体。液体的表面覆盖着板 3,板3的垂直位置由电脑 系统控制。紫外线由激光器5产生并且在镜子6的辅助作用下聚焦， 这也是由电脑系统控制的，因此，激光可以按照一定的程序在液体 表面移动。这种操作的结果就是，产生了一个事先确定了的薄层。下一步，板3向下移动一个薄层厚度的距离，并且重复上一步的程 序。在这个过程中，激光的运行轨迹可以根据新的薄层的形状需要而改变。因此，零件一层一层的被构成所期望的形状。图 2.

5、19b列举了一个用这种方法生产的一个产品的例子。练习题试着设计下面情况的运动布局：1. 缝纫机2. 按照图 2.2 中给出的生产布局来设计如图1 所示的设备生产链条3. 内燃发动机4. 家用面团搅拌器揉面机）5. 打字机6. 弹簧驱动或电动的机械玩具7. 机关枪8. 唱片机9. 影印机3刚体的动态分析在这一章 , 我们将举例子说明计算自然物理运动所需要的时间。我们从最简单的开始一个纯粹的机械传动。3.1 做机械运动的刚体我们所讨论的第一个例子可以作为一个自然现象。这种状况发生在例如当一大堆部件在料斗或分配器里垂直向下 移动的情况下。在图3.1中提出了最简单的例子，刚体落下L的高 度，假设下落过

6、程没有任何的阻力，那么我们就可以写出下落所需 时间的数学表达式：凶3.1 I图3.1自由落体刚体运动模型图3.2显示了一个用于自动机器（车床的进料装置。重物块M通过缠绕在滑轮上的线1作用在滑块2上。滑块2推动由摩擦面3支撑的杆 m因此，推力F= 重物必须克服的摩擦阻力F1, F1可以被表不为：日 【3.2】其中f=摩擦系数，m为棒的质量。另外，力F使滑轮有了转动惯量I因此，方程式以下面的方式达到平衡其中：a为重力加速度r为滑轮半径因为滑轮的角加速度因此a=r【3.4】从方程（3.3>,我们就可以推导出a的一种表达形式【3.5】杆运动L距离所需时间t可以用下面的方程计算出【3.6】非常明显

7、，由于I<滑轮的影响可以忽略不计），公式3.6可以写成下面的形式图3.2由重力驱动的运动布局原理在下面的例子中，我们分析刚体沿着斜面运动的情况这种情况发生在例如，零件像图 3.3所示那样随着送料机运动。其中。是送料机的斜度。零件和传送带之间的摩擦力可以由方程W!三来表示。在这里，f为抵抗零件在传送带上滑动的摩擦系数）图3.3重物在斜坡上的运动力F可以从已知公式里得至上I 'I【3.8】方程式还可以写为：3.9 从方程3.9中我们得到：一 .3.10运动L距离所需时间x I3.11 注：当 X 或者 时，刚体将没有运动,时间趋向3.4a 所于无穷长。在这里，我们分析弹性运动。这种运

8、动的原理图如图示。作为驱动源的弹簧的特性如图 3.5所示。这个特征表明力 P的大小由弹簧的变形决定 无论是拉伸还是压 缩）。当这个关系是线性的，像图3.5所示那样，弹簧的弹性系数c就是一个常量。换句话说，弹簧的弹性系数是表示弹簧的压力和 变形之间关系的比例常数。它也定义了斜坡角度的值并且可以被描 述为日 【3.12图3.4 由弹簧驱动的刚体运动 a没有外力作用；b）有外力F作用并且n 【3.13作用力P总是与X方向相反。图3.5弹簧的力与变形线性曲线因此，质量块m的运动可以由Dalamber方程式来描述 日【3.14这个微分方程有一个简单的计算方法一 3.151一定要确定未知参数A、B和凶。把

9、公式 3.15带入公式3.14,我们可以得到=一3.16并且一1参数凹为系统的固有频率。我们必须使用系统的原始条件去解决参数A和B。假设，在t=0时刻弹簧的变形回，并且回。然后我们把这些时刻代入公式3.15就可以直接得到 M m。因此公式3.14的完全解为汉【3.17公式3.17解释表达图见图3.6图3.6 弹簧驱动刚体运动的位移与时间关系为了找到把质量块从X0点移动到任何一个距离L的店X1所需 时间，我们把公式3.17写成一下形式H【3.18在实际情况中，我们必须要考虑质量块m在运动中所受到的阻力，如图3.4b所示。因为自然力可以是多种多样的，例如，如果它 是由干摩擦引起的，这个力可以被描述

10、为解读力的形势。H【3.19图3.7刚体快速运动中由干摩擦产生的力质量块m的运动可以被描述为凶【3.20可以被一系列的方程来代替 x| 团【3.21在公式3.21中用K来代替 百，可以得到国3.22方程式两边同时平方，可以得到3.23R是一个积分常数我们可以用图3.8来演示质量块在相位面上的运动图3.8 刚体运动中干摩擦引起的震动对速度和位移的影响当日时，质量块的震荡运动停止在我们的例子中，弹簧使质量块从 句 点运动L的距离到 二J 点，根据给定的图3.8, R的值等于 。这使得我们能够以以 下方式重写公式3.23中的第一个等式13.24并且3.25对K=0的情况，根据方程3.25得到|3.2

11、6因此，在K=0的相同情况下，公式3.26和公式3.18是相等的在n=0的情况下）现在我们考虑例子中力 F不变的情况 如图3.4b）。这种情况出现 在质量块的重力由弹簧驱动的情况下。例如在图2.16中，弹簧9拉动升降机6、刀具7和电枢的磁铁5.下面的方程描述了这样的状况【3.27该方程的解包括两个组成部分3.28x'是齐次方程的解并且方程 3.15已经给出了具体形式，剩下的 解x"必须是一个常量x"=D=常量。把x" = D代入方程3.27 ,我们可以得到国【3.29因此，我们就可以把方程式3.28写成一下的形式【3.30因为初始条件t=0,NI,那么结果

12、是3.31 通常，在运功的初始阶段弹簧 A的变形包括引起质量块改变初始 位置的力F产生的变形方程3.27写成角运动的形式-I 【3.32其中：I为旋转刚体的力矩CQ为弹簧的角位移T为抗力矩T= Fr,R力F的作用半径O为旋转角度我们不要忘记，这里的尺寸和公式 3.27中的不同。公式3.32的解有一个类似方程3.30的形式，如下所示：【3.33这里0是由初始条件£ = 0,0 = 00, and 0 = 0 计算出来的，并 且可以写成一下的形式I X 1【3.34图3.9弹簧驱动的旋转运动2.5 Rapid PrototypingNew production concepts of a

13、 different nature have recently been introduced intomanufacturing processes. Among these concepts, some are modifications of alreadyexisting ideas, but others are completely revolutionary. As examples of the formergroup, we may cite computerized numerically controlled(CNC> cutting of a variety of

14、materials, from wood to ceramics, with a laser beam and a water-plus-abrasive jet.With regard to the latter group, we may describe the process of rapid modeling or three-dimensional processing of parts. This concept is rich in content and industrialpotential, and it is therefore worthwhile discussin

15、g it in brief. It is based on a principlethat has been possible to formulate largely as a consequence of the power of thecomputer.The productivity of the first group of manufacturing processes mentioned aboveis vastly improved by the application of computers, although, at least in principle, thesepr

16、ocesses may be carried out in a manual mode. For the second group, it is impossibleto execute the processes without a computer.Modern manufacturing relies on a large number of molded parts made of plasticsand metals. These parts sometimes have very complicated shapes and ornate surfaces.Such shapes

17、cannot be processed on conventional machines, which makes any attemptto produce a single part of this kind very time and money consuming. For the samereason, the use of a mold to produce individual patterns, which may require changesafter they are examined, is even more expensive (this is the case i

18、n whichnonconventionaltools areused and the process is expensive and time and labor consuming>.In recent years, a new concept for providing the solution to this problem has been proposed.It is known as rapid prototyping, stereolithography, quick prototype tooling, orrapid modeling, and isdescribe

19、d in the book Solid Freeform Manufacturing, by H. D.Kochan (Technical University Dresden, Germany, Elsevier Scientific Publishers>.To explain the idea underlying this manufacturing process, we use the model shownin Figure 2.19a. The model represents a helical wheel provided with specially formedt

20、eeth, consisting of plane layers that are angularly shifted relative to one another. Inother words, a three-dimensional model with a complicated shape is composed of anumber of thin, planar, and simply shaped layers.FIGURE 2.19 a> Illustration of a rapidly modeled subject.Pay attention to the cle

21、arly visiblelayers of the material comprising the wheel. Each layer is displaced by a certain angle, thus creating the image of a helical gear (here, for purposes of illustration, the thicknessof the layers is exaggerated，b> Examples of patterns made by this technique before thefinal design (prod

22、uction of Conceptland Ltd., Ra'anana, Israel>.There are a number of different techniques that exploitthis idea for the computeraidedprocessing of spatially cumbersome parts. We will describe here, in brief, theessence of the concept.The memory of the computer is loaded with geometric informat

23、ion about the partto be processed so that the configuration of each thin (say, 0.3-0.5 mm> slice of thepart can be numerically defined.A possible layout for a processbased on this conceptfor creating lamellar bodies 一for an intricate three-dimensional shape is shown in Figure2.20. This layout con

24、sistsof a vessel 1 filled with a special liquid 2, which polymerizes to a solid under ultravioletirradiation. The surface of the liquid covers aplate 3, the vertical location of whichisfigure 2.20 Layout of the rapid modeling process.1) Vessel: 2 Polymerizing liquid: 3) Plate;4) Computer; 5) Laser;

25、6) Rotating mirror.controlled by the system's computer 4. The ultraviolet beam generated by means oflaser 5 is focussed with the aid of a mirror 6, which is also controlled by a computer, so that the beam moves on the surface of the liquid according to a given program. Asa result of this operati

26、on, a thin plane layer is createdwith a predetermined shape. Inthe next step, the plate 3 moves down for a distancecorresponding to the thickness ofone layer, and the procedure is repeated. At this point inthe process, the trajectory ofthe beam may be changed according to the configuration ofthe new

27、 layer. Thus, thebody grows, layer by layer, to form a model of the desiredshape. Figure 2.19b> showsexamples of possible units produced in this wayExercisesTry to design the kinematic layout of a:1. Sewing machine.2. Machine for producing the chain shown in Figure 2.1 in accordance with theprodu

28、ction layout given in Figure 2.2.3. Internal combustion engine.4. Domestic dough mixer (dough kneader>.5. Typewriter.6. Mechanical toy, spring or electrically driven.7. Machine gun.8. Automatic record player.9. Photocopying machine.3Dynamic Analysis of DrivesIn this chapter we shall discuss examp

29、les illustrating the operation time computation techniques for drives of different physical natures. We begin with the simplest a purely mechanical drive.3.1 Mechanically Driven BodiesThe first case we shall consider in this section may be classified as a free-fall phenomenon. This is the situation

30、which occurs, for instance, when a stack of parts movesvertically downwards in a magazine-type hopper (or dispenser>. The simplest exampleis presented in Figure 3.1, which shows a body falling from level I to level II through adistance L Assuming that there is no resistance of any kind, we can wr

31、ite the following expression for the time t required for this process:3.1 Figure 3.2 shows a mechanism used in automatic machines (lathes> for feedingrod-like material during processing. The weight M acts on the slider 2 via a cableII 、a figure 3.1 Model of a free-falling body.passes over a rolle

32、r with moment of inertia /. The slider 2pushes the rod mwhich is supported by frictional guide 3.Thus, the acting force F=Mg must overcomethe friction F1 in the guides 。Flmay be expressed as:【3.2】where/= the dry friction coefficient and m is the mass of the rod.In addition, the force F rotates the r

33、oller with moment of inertia /. Therefore, theequilibrium equation of forces takes the form【3.3】where a = the linear acceleration of the weight (or rod>, r = the radius of the roller, anda = the angular acceleration of the roller.Sincea=【3.4】from Equation (3.3> we can derive an expression for

34、a in the form【3.5】The time t needed to displace the rod through distance Lcan be calculated from theFormulaObviously, for I /【3.6】(M+ m> (i.e., the influenceof the roller is negligible in comparison with that of the moving massesA，Equation (3.6> can be rewritten in the form【3.7】In this case we

35、 analyze movement along an inclinedplane. This is the case thatoccurs when, for instance,parts slide along a tray from a feeder, as is shown inFigure3.3. Here <j> is the inclination angle of the tray.The friction between the parts and thetray is described bycoefficientwhichthe force Fl =frng c

36、os 0 (here again, /= the dry frictionfigure 3.2 Layout of a rodfeeding mechanism driven by the force of gravity.resists the movement along the tray>. The driving force Fin this case can befound from the known formulafigure 3.3 Model of gravitation drive on an inclined tray.【3.8】The equilibrium eq

37、uation thus has the form13.9From Equation (3.9> we obtain【3.10The time t required to displace a part through a distance L equalsI3.11Note: When sin <f> =/cos 0 or/= tan 0, no movement will occur. The time tends to infinitely long values.Here we analyze the movement of a mass driven by a pre

38、viously deformed spring.The layout of such a mechanism is shown in Figure 3.4a>. A spring as a driving sourceis described by its characteristic shown in Figure 3.5. This characteristic shows thedependence of the force P developed by the spring on the values of the deformationjc (in both the stret

39、ched and compressed modes>. When this dependence is linear, asshown in Figure 3.5, parameter c, which is the stiffness of the spring, is constant forthis case. In other words, stiffness of the spring is a proportionality coefficient tyingthe deformation of the spring to the force P it develops. I

40、t also defines the value of theslope of the characteristic and can be described as riAnd【3.12【3.13(3.14>, we obtainfigure 工4 Spring-driven body： a) Without and b) With resisting force Efigure 3.5 Characteristic of the spring: force versus deformation.The force P always acts in the direction oppos

41、ite to x.Thus, the movement of the mass m is described by the following equation basedon the Dalamber principle:【3.14This differential homogeneous equation has a simple solution:【335】where the unknown parameters A, B, and co must be determined. Substituting Expression (3.15> into Equation【3.16And

42、【3.17The parameter co is known as the naturalfrequency of thesystem. To find the unknownparameters A and B, we have to use the initial conditions of the system. Say, at themoment t = 0 the deformation of the spring x = x0 and x = 0. We then substitute thesedata into expression (3.15> and obtain d

43、irectly A = x0 and B =0. Thus the complete solution of【3.18Equation (3.14> isfigure 3.6 Displacement versus time for a body driven by a spring.Expression (3.17> is interpreted graphically in Figure 3.6.To find the time needed to move the mass from thepoint JCQ to any other point xllocated at a

44、 distance L fromJCG, we rewrite Expression (3.17> in the following way:【338】In a more realistic approach, we must consider a resisting force acting on the massm during its motion, as shown in Figure 3.4b>. Since the nature of the force can vary,so can its analytic description. For example, if

45、it is caused by dry friction, the force maybe described analytically in the form【3.19This graphic interpretation of Equation (3.19> is given infigure 3.7 The movement ofmass m can be described by0【3.20which can be replaced by a system of equations in the form x 区3.21 Substituting k =： , in Equati

46、ons (3.21, we obtain【3.22It is convenient to transform these equations multiplying them by 2 x and integratingthem into the following form:【3.23The value R is an integration constant which must be defined for every change of sgnx.This form of interpretation permits us to express the behavior of the

47、mass in theterms of the phase plane which is shown in Figure 3.8. The oscillating movement ofthe mass ceases at the moment when Rn = £ 2kco. In our case, the spring moves the massfrom a point x = x0 through a distance L to a point x = xA In accordance with the diagramgiven in Figure 3.8, the va

48、lue R equals cox0 - cok. This enables us to rewrite the first ofthe two Equations (3.23> in the following way【3.24iFFIGURE 3.7 Force developed by dry friction versus speed of the body.Andfigure 3.8 Speed versus displacement for an oscillating body influenced by dry friction.【3.25For the case k =

49、0, it follows from Equation (3.25> that【3.26Thus, Equation (3.26> coincides with the Formula (3.18> which describes the same situation k = 0 (as a result of n =0 in this case>.Let us consider the case in which the resisting forceF shown in Figure 3.4b> is constant. Such a case can arise whe