快速成型外文翻譯

1、2.5 快速成型一種新的生產(chǎn)產(chǎn)品的概念最近被引進(jìn)制造過程。在這些觀念中，一些是對已經(jīng)存在的觀念的修飾，但是另一些完全是革命性的。在前一類的實例中，我們可以運用數(shù)控機(jī)床來切割各種各樣的材料，使用激光和噴氣機(jī)可以切割從木頭到陶瓷的各種材料。對于后一類，我們可以描述零件的建模和三維快速加工過程。這個概念在內(nèi)容和工業(yè)上很有潛力，因此值得簡易討論一下。它在很大程度上是基于計算機(jī)能力發(fā)展的結(jié)果。對于第一組，工作效率的大幅度提高主要是由于計算機(jī)的應(yīng)用，雖然，至少在原則上，這些過程可以以手動方式實現(xiàn)。對于第二組，沒有計算機(jī)的幫助是很難執(zhí)行這些過程的。現(xiàn)在制造依賴于大量的由塑料和金屬做成的模具。這些零件有時會有

2、非常復(fù)雜的形狀和華麗的表面。這些圖形是不能在傳統(tǒng)機(jī)床上處理的，因為僅完成單一部分的加工是非常浪費時間和金錢的。同樣，使用模具去生產(chǎn)可能檢驗后需要改變的個人模式的工件也是非常昂貴的（這種情況發(fā)生在非傳統(tǒng)工具的生產(chǎn)中，并且這個過程非常昂貴，消耗時間和勞力）。在最近幾年中，一個新的解決這種情況的概念被提出。它被稱為快速成型，在h. d.kochan編寫的自由制造一書中描述了立體雕刻，模具快速原型和快速成型（德累斯頓科技大學(xué)，德國，科學(xué)技術(shù)出版社）。我們用圖2.19a中的模型來解釋這種觀念下的制造過程。這個模型代表了具有特殊齒形的螺旋齒輪，齒是由相互之間具有角度的平面層組成。換句話說，一個具有復(fù)雜形狀

3、的三維模型是由簡單形狀的薄平面層組成的。 概念和布局圖2.19 a）快速原型演示工件。注意看齒輪上清晰可見的層。每一個層都按一定的角度旋轉(zhuǎn)，從而形成了螺旋齒輪的形狀（在這里為了形象的說明，每個層的厚度變化是夸張的） b)這是按照這種技術(shù)生產(chǎn)最終設(shè)計完成前的模型（conceptland ltd., raanana, israel的產(chǎn)品）。有好幾種不同的技術(shù)，利用這種原理并應(yīng)用計算機(jī)輔助加工具有復(fù)雜空間的零件。我們將在這里簡單的描述這個概念的本質(zhì)。電腦的內(nèi)存用來存儲要加工零件的幾何圖形信息，以至于零件的每個幾何薄層（通常0.3-0.5mm）都可以用數(shù)值定義。 根據(jù)這種概念，在創(chuàng)造半層體中一個可能的

4、方案如圖2.20所示。 圖2.20 快速建模布局圖 1）容器；2）聚合液體；3）金屬板；4）電腦；5）激光器；6）旋轉(zhuǎn)鏡 這樣的布局由容器1充滿了一種特殊的液體2，這種液體在紫外線的照射下會變成固體。液體的表面覆蓋著板3，板3的垂直位置由電腦系統(tǒng)控制。紫外線由激光器5產(chǎn)生并且在鏡子6的輔助作用下聚焦，這也是由電腦系統(tǒng)控制的，因此，激光可以按照一定的程序在液體表面移動。這種操作的結(jié)果就是，產(chǎn)生了一個事先確定了的薄層。下一步，板3向下移動一個薄層厚度的距離，并且重復(fù)上一步的程序。在這個過程中，激光的運行軌跡可以根據(jù)新的薄層的形狀需要而改變。因此，零件一層一層的被構(gòu)成所期望的形狀。圖2.19b列舉了

5、一個用這種方法生產(chǎn)的一個產(chǎn)品的例子。練習(xí)題 試著設(shè)計下面情況的運動布局：1. 縫紉機(jī)2. 按照圖2.2中給出的生產(chǎn)布局來設(shè)計如圖1所示的設(shè)備生產(chǎn)鏈條3. 內(nèi)燃發(fā)動機(jī)4. 家用面團(tuán)攪拌器（揉面機(jī)）5. 打字機(jī)6. 彈簧驅(qū)動或電動的機(jī)械玩具7. 機(jī)關(guān)槍8. 唱片機(jī)9. 影印機(jī)3剛體的動態(tài)分析在這一章,我們將舉例子說明計算自然物理運動所需要的時間。我們從最簡單的開始一個純粹的機(jī)械傳動。3.1做機(jī)械運動的剛體我們所討論的第一個例子可以作為一個自然現(xiàn)象。這種狀況發(fā)生在例如當(dāng)一大堆部件在料斗或分配器里垂直向下移動的情況下。在圖3.1中提出了最簡單的例子,剛體落下l的高度，假設(shè)下落過程沒有任何的阻力，那么我

6、們就可以寫出下落所需時間的數(shù)學(xué)表達(dá)式： 【3.1】圖3.1 自由落體剛體運動模型圖3.2顯示了一個用于自動機(jī)器(車床)的進(jìn)料裝置。重物塊m通過纏繞在滑輪上的線1作用在滑塊2上?；瑝K2推動由摩擦面3支撐的桿m。因此，推力f=重物必須克服的摩擦阻力f1，f1可以被表示為： 【3.2】其中f=摩擦系數(shù)，m為棒的質(zhì)量。另外，力f使滑輪有了轉(zhuǎn)動慣量i因此，方程式以下面的方式達(dá)到平衡 【3.3】其中：a為重力加速度 r為滑輪半徑 為滑輪的角加速度因此 a=r 【3.4】 從方程(3.3),我們就可以推導(dǎo)出a的一種表達(dá)形式 【3.5】桿運動l距離所需時間t可以用下面的方程計算出 【3.6】非常明顯，由于（滑

7、輪的影響可以忽略不計），公式3.6可以寫成下面的形式 【3.7】圖3.2 由重力驅(qū)動的運動布局原理在下面的例子中,我們分析剛體沿著斜面運動的情況。這種情況發(fā)生在例如，零件像圖3.3所示那樣隨著送料機(jī)運動。其中是送料機(jī)的斜度。零件和傳送帶之間的摩擦力可以由方程來表示。（在這里，f為抵抗零件在傳送帶上滑動的摩擦系數(shù)）。圖3.3 重物在斜坡上的運動力f可以從已知公式里得到。 【3.8】方程式還可以寫為： 【3.9】從方程3.9中我們得到： 【3.10】運動l距離所需時間 【3.11】注：當(dāng)或者時，剛體將沒有運動，時間趨向于無窮長。在這里，我們分析彈性運動。這種運動的原理圖如圖3.4a所示。作為驅(qū)動源

8、的彈簧的特性如圖3.5所示。這個特征表明力p的大小由彈簧的變形決定（無論是拉伸還是壓縮）。當(dāng)這個關(guān)系是線性的，像圖3.5所示那樣，彈簧的彈性系數(shù)c就是一個常量。換句話說，彈簧的彈性系數(shù)是表示彈簧的壓力和變形之間關(guān)系的比例常數(shù)。它也定義了斜坡角度的值并且可以被描述為 【3.12】圖3.4 由彈簧驅(qū)動的剛體運動 a)沒有外力作用；b）有外力f作用并且 【3.13】作用力p總是與x方向相反。圖3.5 彈簧的力與變形線性曲線因此，質(zhì)量塊m的運動可以由dalamber方程式來描述 【3.14】這個微分方程有一個簡單的計算方法 【3.15】一定要確定未知參數(shù)a、b和。把公式3.15帶入公式3.14，我們可

9、以得到 【3.16】并且參數(shù)為系統(tǒng)的固有頻率。我們必須使用系統(tǒng)的原始條件去解決參數(shù)a和b。假設(shè)，在t=0時刻彈簧的變形,并且。然后我們把這些時刻代入公式3.15就可以直接得到，。因此公式3.14的完全解為 【3.17】公式3.17解釋表達(dá)圖見圖3.6圖3.6 彈簧驅(qū)動剛體運動的位移與時間關(guān)系為了找到把質(zhì)量塊從x0點移動到任何一個距離l的店x1所需時間，我們把公式3.17寫成一下形式 【3.18】在實際情況中，我們必須要考慮質(zhì)量塊m在運動中所受到的阻力，如圖3.4b所示。因為自然力可以是多種多樣的，例如，如果它是由干摩擦引起的，這個力可以被描述為解析力的形勢。 【3.19】方程式3.19的圖解如

10、圖3.7所示。圖3.7 剛體快速運動中由干摩擦產(chǎn)生的力質(zhì)量塊m的運動可以被描述為 【3.20】可以被一系列的方程來代替 【3.21】在公式3.21中用k來代替,可以得到 【3.22】方程式兩邊同時平方，可以得到 【3.23】r是一個積分常數(shù)我們可以用圖3.8來演示質(zhì)量塊在相位面上的運動。圖3.8 剛體運動中干摩擦引起的震動對速度和位移的影響當(dāng)時，質(zhì)量塊的震蕩運動停止在我們的例子中，彈簧使質(zhì)量塊從點運動l的距離到點，根據(jù)給定的圖3.8，r的值等于。這使得我們能夠以以下方式重寫公式3.23中的第一個等式 【3.24】并且 【3.25】對k=0的情況，根據(jù)方程3.25得到 【3.26】因此，在k=0

11、的相同情況下，公式3.26和公式3.18是相等的（ 在n=0的情況下）現(xiàn)在我們考慮例子中力f不變的情況（如圖3.4b）。這種情況出現(xiàn)在質(zhì)量塊的重力由彈簧驅(qū)動的情況下。例如在圖2.16中，彈簧9拉動升降機(jī)6、刀具7和電樞的磁鐵5.下面的方程描述了這樣的狀況 【3.27】該方程的解包括兩個組成部分 【3.28】x是齊次方程的解并且方程3.15已經(jīng)給出了具體形式，剩下的解x必須是一個常量x=d=常量。把x = d代入方程3.27，我們可以得到 【3.29】因此，我們就可以把方程式3.28寫成一下的形式 【3.30】因為初始條件t=0，那么結(jié)果是 【3.31】通常，在運功的初始階段彈簧a的變形包括引起

12、質(zhì)量塊改變初始位置的力f產(chǎn)生的變形方程3.27寫成角運動的形式 【3.32】其中：i為旋轉(zhuǎn)剛體的力矩cq為彈簧的角位移t為抗力矩t= fr,r力f的作用半徑o為旋轉(zhuǎn)角度我們不要忘記，這里的尺寸和公式3.27中的不同。公式3.32的解有一個類似方程3.30的形式，如下所示： 【3.33】這里0是由初始條件 = 0,0 = 00, and 0 = 0計算出來的，并且可以寫成一下的形式 【3.34】圖3.9 彈簧驅(qū)動的旋轉(zhuǎn)運動2.5 rapid prototypingnew production concepts of a different nature have recently been in

13、troduced 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 ofmaterials, from wood to ceramics, with a

14、laser beam and a water-plus-abrasive jet. with regard to the latter group, we may describe the process of rapid modeling orthree-dimensional processing of parts. this concept is rich in content and industrialpotential, and it is therefore worthwhile discussing it in brief. it is based on a principle

15、that has been possible to formulate largely as a consequence of the power of the computer.the productivity of the first group of manufacturing processes mentioned above is vastly improved by the application of computers, although, at least in principle, these processes may be carried out in a manual

16、 mode. for the second group, it is impossible to 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 cannot be processed on conventional m

17、achines, which makes any attempt to produce a single part of this kind very time and money consuming. for the same reason, the use of a mold to produce individual patterns, which may require changes after they are examined, is even more expensive (this is the case in which nonconventional tools are

18、used 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, or rapid modeling, and is described in the book solid freeform m

19、anufacturing, 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 formed teeth, consisting of plane layer

20、s that are angularly shifted relative to one another. in other words, a three-dimensional model with a complicated shape is composed of a number of thin, planar, and simply shaped layers.figure 2.19 a) illustration of a rapidly modeled subject. pay attention to the clearly visiblelayers of the mater

21、ial 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 (production of conceptland ltd., raana

22、na, israel). there are a number of different techniques that exploit this idea for the computeraidedprocessing of spatially cumbersome parts. we will describe here, in brief, the essence of the concept.the memory of the computer is loaded with geometric information about the part to be processed so

23、that the configuration of each thin (say, 0.3-0.5 mm) slice of the part can 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 consists of a vessel 1 filled with a speci

24、al liquid 2, which polymerizes to a solid under ultraviolet irradiation. the surface of the liquid covers a plate 3, the vertical location of which is controlled by the systems computer 4. the ultraviolet beam generated by means oflaser 5 is focussed with the aid of a mirror 6, which is also control

25、led by a computer,so that the beam moves on the surface of the liquid according to a given program. asa result of this operation, a thin plane layer is created with a predetermined shape. inthe next step, the plate 3 moves down for a distance corresponding to the thickness ofone layer, and the proce

26、dure is repeated. at this point in the process, the trajectory ofthe beam may be changed according to the configuration of the new layer. thus, thebody grows, layer by layer, to form a model of the desired shape. figure 2.19b) showsexamples of possible units produced in this wayexercisestry to desig

27、n the kinematic layout of a:1. sewing machine.2. machine for producing the chain shown in figure 2.1 in accordance with the production 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. ma

28、chine gun.8. automatic record player.9. photocopying machine.3dynamic analysis of drivesin this chapter we shall discuss examples illustrating the operation time computation techniques for drives of different physical natures. we begin with the simplesta purely mechanical drive.3.1 mechanically driv

29、en 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 moves vertically downwards in a magazine-type hopper (or dispenser). the simplest example is presented in figure 3.1, which s

30、hows a body falling from level i to level ii through a distance 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 feeding rod-like material duri

31、ng processing. the weight m acts on the slider 2 via a cable i which passes over a roller with moment of inertia /. the slider 2 pushes the rod m which is supported by frictional guide 3. thus, the acting force f=mg must overcome the friction f1 in the guides; fl may be expressed as: 【3.2】where/= th

32、e dry friction coefficient and m is the mass of the rod.in addition, the force f rotates the roller with moment of inertia /. therefore, the equilibrium 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 ac

33、celeration 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 theformula 【3.6】obviously, for i / (m+ m) (i.e., the influence of the roller is negligible in comparison with

34、 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 that occurs when, for instance, parts slide along a tray from a feeder, as is shown in figure 3.3. here j) is the inclination angle of the tray. th

35、e friction between the parts and the tray is described by the force fl =frng cos 0 (here again, /= the dry friction coefficientwhich .resists the movement along the tray). the driving force f in this case can be found from the known formula 【3.8】the equilibrium equation thus has the form 【3.9】from e

36、quation (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 time tends to infinitely long values.here we analyze the movement of a mass driven by a previously deformed spring. the layout of such

37、 a mechanism is shown in figure 3.4a). a spring as a driving source is described by its characteristic shown in figure 3.5. this characteristic shows the dependence of the force p developed by the spring on the values of the deformation jc (in both the stretched and compressed modes). when this depe

38、ndence is linear, as shown in figure 3.5, parameter c, which is the stiffness of the spring, is constant for this case. in other words, stiffness of the spring is a proportionality coefficient tying the deformation of the spring to the force p it develops. it also defines the value of the slope of t

39、he characteristic and can be described as 【3.12】and 【3.13】the force p always acts in the direction opposite to x.thus, the movement of the mass m is described by the following equation based on the dalamber principle: 【3.14】this differential homogeneous equation has a simple solution: 【3.15】where th

40、e unknown parameters a, b, and co must be determined. substituting expression (3.15) into equation (3.14), we obtain 【3.16】and 【3.17】the parameter co is known as the natural frequency of the system. to find the unknown parameters a and b, we have to use the initial conditions of the system. say, at

41、the moment t = 0 the deformation of the spring x = x0 and x = 0. we then substitute these data into expression (3.15) and obtain directly a = x0 and b = 0. thus the complete solution of equation (3.14) is 【3.18】expression (3.17) is interpreted graphically in figure 3.6.to find the time needed to mov

42、e the mass from the point jcq to any other point xl located 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 mass m during its motion, as shown in figure 3.4b). since the nature of the fo

43、rce can vary, so can its analytic description. for example, if it is caused by dry friction, the force may be described analytically in the form 【3.19】this graphic interpretation of equation (3.19) is given in figure 3.7. the movement of mass m can be described by 【3.20】which can be replaced by a sy

44、stem of equations in the form 【3.21】substituting k = , in equations (3.21), we obtain 【3.22】it is convenient to transform these equations multiplying them by 2 x and integrating them into the following form: 【3.23】the value r is an integration constant which must be defined for every change of sgnx.

45、this form of interpretation permits us to express the behavior of the mass in the terms of the phase plane which is shown in figure 3.8. the oscillating movement of the mass ceases at the moment when rn = 2kco. in our case, the spring moves the mass from a point x = x0 through a distance l to a poin

46、t x = x in accordance with the diagram given in figure 3.8, the value r equals cox0 - cok. this enables us to rewrite the first of the 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) 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 force f shown in figure 3.4b) is constant. such a case can arise when, for instance