2022年毕业设计方案翻译zhongwen .pdf

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1、1 / 19 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

2、些零件有时会有非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

3、，一个具有复杂形状的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 这是按照这种技术生产最终设计完成前的模型Conceptland Ltd., Raanana, 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。精选学习资料 - - - - - - - - - 名师归纳总结 - - - -

4、 - - -第 1 页，共 19 页2 / 19 根据这种概念，在创造半层体中一个可能的方案如图2.20所示。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

5、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2.19b 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 可以被表示为：【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【3.3 】其中： a 为重力加

6、速度r 为滑轮半径为滑轮的角加速度精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页，共 19 页4 / 19 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，由于滑轮的影响可以忽略不计），公式3.6 可以写成下面的形式【3.7 】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来表示

7、。 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到。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 4 页，共 19 页5 / 19 【3.8】方程式还可以写为：【3.9】从方程 3.9 中我们得到：【3.10 】运动 L 距离所需时间【3.11】注：当或者时，刚体将没有运动，时间趋向于无穷长。在这里，我们分析弹性运动。这种运动的原理图如图3.4a 所示。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没有外力作用；b）有外力F 作用并且【3.13 】作

8、用力 P 总是与 X 方向相反。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 5 页，共 19 页6 / 19 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【3.14 】这个微分方程有一个简单的计算方法【3.15】一定要确定未知参数A、B 和。把公式3.15 带入公式3.14，我们可以得到【3.16 】并且参数为系统的固有频率。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,并且。然后我们把这些时刻代入公式3.15就可以直接得到，。因此公式3.14 的完全解为【3.17

9、 】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精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 6 页，共 19 页7 / 19 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【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，这个力可以被描述为解读力的形势。【3.19】方程式 3.19 的图解如图3.7 所示。图 3.7 刚体快速运动中由干摩擦产生的力质量块

10、 m 的运动可以被描述为【3.20 】可以被一系列的方程来代替【3.21】在公式 3.21 中用 K 来代替,可以得到【3.22 】方程式两边同时平方，可以得到【3.23 】R 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。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 7 页，共 19 页8 / 19 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，弹簧使质量块从点运动L 的距离到点，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【3.

11、24 】并且【3.25 】对 K=0 的情况，根据方程3.25 得到【3.26 】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 cutting of a variety of 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 thre

12、e-dimensional processing of parts. This concept is rich in content and industrial potential, and it is therefore worthwhile discussing it in brief. It is based on a principle that has been possible to formulate largely as a consequence of the power of thecomputer. The productivity of the first group

13、 of manufacturing processes mentioned aboveis vastly improved by the application of computers, although, at least in principle, theseprocesses 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 larg

14、e number of molded parts made of plastics and metals. These parts sometimes have very complicated shapes and ornate surfaces. Such shapes 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

15、use of a mold to produce individual patterns, which may require changesafter they are examined, is even more expensive (this is the case in whichnonconventionaltools areused and the process is expensive and time and labor consuming.In recent years, a new concept for providing the solution to this pr

16、oblem has been proposed.It is known as rapid prototyping, stereolithography, quick prototype tooling, orrapid modeling, and isdescribed in the book Solid Freeform Manufacturing, by H. D.Kochan (Technical University Dresden, Germany, Elsevier Scientific Publishers. To explain the idea underlying this

17、 manufacturing process, we use the model shown in Figure 2.19a. The model represents a helical wheel provided with specially formedteeth, 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 an

18、umber of thin, planar, and simply shaped layers. 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 10 页，共 19 页11 / 19 FIGURE 2.19 a Illustration of a rapidly modeled subject. Pay attention to the clearly visible layers of the material comprising the wheel. Each layer is displaced by a certain angle, th

19、us creating the image of a helical gear (here, for purposes of illustration, the thickness of the layers is exaggerated, b Examples of patterns made by this technique before the final design (production of Conceptland Ltd., Raanana, Israel. There are a number of different techniques that exploit thi

20、s idea for the computeraided processing of spatially cumbersome parts. We will describe here, in brief, theessence of the concept. The memory of the computer is loaded with geometric information about the partto be processed so that the configuration of each thin (say, 0.3-0.5 mm slice of thepart ca

21、n be numerically defined. A possible layout for a processbased on this concept for creating lamellar bodiesfor an intricate three-dimensional shape is shown in Figure 2.20. This layout consistsof a vessel 1 filled with a special liquid 2, which polymerizes to a solid under ultravioletirradiation. Th

22、e surface of the liquid covers a plate 3, the vertical location of whichis controlled by the systems computer 4. The ultraviolet beam generated by means of laser 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 acco

23、rding to a given program. As 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 11 页，共 19 页12 / 19 a result of this operation, a thin plane layer is created with a predetermined shape. In the next step, the plate 3 moves down for a distance corresponding to the thickness of one layer, and the procedure

24、is repeated. At this point in the process, the trajectory of the beam may be changed according to the configuration of the new layer. Thus, the body grows, layer by layer, to form a model of the desired shape. Figure 2.19b shows examples of possible units produced in this way Exercises Try to design

25、 the kinematic layout of a: 1. Sewing machine. 2. Machine for producing the chain shown in Figure 2.1 in accordance with theproduction 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

26、. Machine gun. 8. Automatic record player. 9. Photocopying machine.3 Dynamic Analysis of Drives In this chapter we shall discuss examples illustrating the operation time computation techniques for drives of different physical natures. We begin with the simplest a purely mechanical drive. 3.1 Mechani

27、cally Driven BodiesThe first case we shall consider in this section may be classified as a free-fall phenomenon. This is the situation 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,

28、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 write 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

29、 during processing. The weight M acts on the slider 2 via a cable Iwhich 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 12 页，共 19 页13 / 19 passes over a roller with moment of inertia /. The slider 2 pushes the rod mwhich is supported by frictional guide 3. Thus, the acting force F=Mg must overcometh

30、e 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 roller with moment of inertia /. Therefore, theequilibrium equation of forces takes the form 【3.3 】where a = the linear acceleration of t

31、he weight (or rod, r = the radius of the roller, and a = the angular acceleration of the roller. Since a=r 【3.4 】from Equation (3.3 we can derive an expression for a in the form 【3.5 】The time t needed to displace the rod through distance L can be calculated from the Formula 【3.6 】Obviously, for I /

32、 ? (M+ m (i.e., the influence of the roller is negligible in comparison with that of the moving masses, Equation (3.6 can be rewritten in the form 【 3.7】In this case we analyze movement along an inclined plane. This is the case thatoccurs when, for instance, parts slide along a tray from a feeder, a

33、s is shown in Figure3.3. Here is the inclination angle of the tray. The friction between the parts and thetray is described by the force Fl =frng cos 0 (here again, /= the dry friction coefficientwhich 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 13 页，共 19 页14 / 19 resists the movement along the t

34、ray. The driving force F in this case can befound from the known formula 【 3.8】The equilibrium equation thus has the form 【3.9 】From Equation (3.9 we obtain 【3.10 】The time t required to displace a part through a distance L equals 【3.11】Note: When sin =/cos 0 or/= tan 0, no movement will occur. The

35、time tends to infinitely long values. Here we analyze the movement of a mass driven by a previously 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 t

36、he force P developed by the spring on the values of the deformationjc (in both the stretched 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 proporti

37、onality coefficient tyingthe deformation of the spring to the force P it develops. It also defines the value of theslope of the characteristic and can be described as 【3.12 】And 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 14 页，共 19 页15 / 19 【3.13 】The force P always acts in the direction opposite

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

39、we obtain 【 3.16 】And 【3.17 】The parameter co is known as the natural frequency of the system. 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 express

40、ion (3.15 and obtain directly A = x0 and B = 0. Thus the complete solution of Equation (3.14 is 【3.18 】精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 15 页，共 19 页16 / 19 Expression (3.17 is interpreted graphically in Figure 3.6. To find the time needed to move the mass from the point JCQ to any other

41、 point xllocated at a distance L from JCG, we rewrite Expression (3.17 in the following way: 【3.18 】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.

42、 For example, if it is caused by dry friction, the force maybe described analytically in the form 【3.19】This graphic interpretation of Equation (3.19 is given in Figure 3.7. The movement ofmass m can be described by 【3.20 】which can be replaced by a system of equations in the form 【3.21】Substituting

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

44、s the behavior of the 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 = x In accordance with the diagramgiven in Figu

45、re 3.8, the value R equals cox0 - cok. This enables us to rewrite the 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 16 页，共 19 页17 / 19 first ofthe two Equations (3.23 in the following way：【 3.24 】And 【3.25 】For the case k = 0, it follows from Equation (3.25 that 【3.26 】Thus, Equation (3.26 coincide

46、s 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 force F shown in Figure 3.4b is constant. Such a case can arise when, for instance, the force of gravity acts on the massdriven by the spring. This is

47、 the situation which occurs in Figure 2.16 where the spring9 lifts the lever 6, the cutter 7, and the armature of the magnet 5. The following equation describes this situation: 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 17 页，共 19 页18 / 19 【3.27 】The solution of this equation consists of two comp

48、onents: 【 3.28】where, x is the solution of the homogeneous equation and therefore has the form given in Equation (3.15, while the partial solution x must be some constant value jc = D = const. Substituting x = D into Equation (3.27, we obtain 【3.29 】Thus, Equation (3.28 can be rewritten in the form

49、【3.30】For initial conditions t= 0, x = x0, and x = 0, Solution (3.30 gives 【3.31 】Naturally, the initial deformation of the spring at the beginning of the motion must include the deformation caused by the force F which entails the appearance of the initial coordinate x0 of the mass location. The abo

50、ve-considered spring-driven mechanisms can also be rotating in nature,as in Figure 3.9. Equation (3.27, rewritten in terms of angular motion, takes the form 【3.32 】where I- the moment of inertia of the rotating body, CQ = stiffness of the spring lumped to the angular displacement, T= resisting torqu