購(gòu)買設(shè)計(jì)請(qǐng)充值后下載，，資源目錄下的文件所見即所得，都可以點(diǎn)開預(yù)覽，，資料完整，充值下載可得到資源目錄里的所有文件。。?！咀ⅰ浚篸wg后綴為CAD圖紙，doc,docx為WORD文檔，原稿無(wú)水印，可編輯。。。具體請(qǐng)見文件預(yù)覽，有不明白之處，可咨詢QQ:12401814
畢業(yè)設(shè)計(jì)（論文）外文翻譯
題目 塑料肥皂盒注塑模的CAD/CAM
專 業(yè) 名 稱 機(jī)械設(shè)計(jì)制造及其自動(dòng)化
班 級(jí) 學(xué) 號(hào) 078105233
學(xué) 生 姓 名 徐 丁 昌
指 導(dǎo) 教 師 姚 坤 弟
填 表 日 期 2011 年 3 月 17 日
Novel Method of Realizing the optimal Mransmtssion of the crank-rocker Mechangism design
Abstract: A novel method of realizing the optimal transmission of the crank-and-rocker mechanism is presented. The optimal combination design is made by finding the related optimal transmission parameters. The diagram of the optimal transmission is drawn. In the diagram, the relation among minimum transmission angle, the coefficient of travel speed variation, the oscillating angle of the rocker and the length of the bars is shown, concisely, conveniently and directly. The method possesses the main characteristic. That it is to achieve the optimal transmission parameters under the transmission angle by directly choosing in the diagram, according to the given requirements. The characteristics of the mechanical transmission can be improved to gain the optimal transmission effect by the method. Especially, the method is simple and convenient in practical use.
Keywords：Crank-and-rocker mechanism, Optimal transmission angle, Coefficient of travel speed variation
INTRODUCTION
By conventional method of the crank-and-rocker design, it is very difficult to realize the optimal combination between the various parameters for optimal transmission. The figure-table design method introduced in this paper can help achieve this goal. With given conditions, we can, by only consulting the designing figures and tables, get the relations between every parameter and another of the designed crank-and-rocker mechanism. Thus the optimal transmission can be realized.
The concerned designing theory and method, as well as the real cases of its application will be introduced later respectively.
1 ESTABLISHMENT OF DIAGRAM FOR OPTIMAL TRANSMISSION DESIGN
It is always one of the most important indexes that designers pursue to improve the efficiency and property of the transmission. The crank-and-rocker mechanism is widely used in the mechanical transmission. How to improve work ability and reduce unnecessary power losses is directly related to the coefficient of travel speed variation, the oscillating angle of the rocker and the ratio of the crank and rocker. The reasonable combination of these parameters takes an important effect on the efficiency and property of the mechanism, which mainly indicates in the evaluation of the minimum transmission angle.
The aim realizing the optimal transmission of the mechanism is how to find the maximum of the minimum transmission angle. The design parameters are reasonably combined by the method of lessening constraints gradually and optimizing separately. Consequently, the complete constraint field realizing the optimal transmission is established.
The following steps are taken in the usual design method. Firstly, the initial values of the length of rocker and the oscillating angle of rocker are given. Then the value of the coefficient of travel speed variation is chosen in the permitted range. Meanwhile, the coordinate of the fixed hinge of crank possibly realized is calculated corresponding to value .
1.1 Length of bars of crank and rocker mechanism
As shown in Fig.1, left arc is the permitted field of point . The coordinates of point are chosen by small step from point to point .
The coordinates of point are
(1)
(2)
where , the step, is increased by small increment within range(0,). If the smaller the chosen step is, the higher the computational precision will be. is the radius of the design circle. is the distance from to .
(3)
Calculating the length of arc and , the length of the bars of the mechanism corresponding to point is obtained[1,2].
1.2 Minimum transmission angle
Minimum transmission angle (see Fig.2) is determined by the equations[3]
(4)
(5)
(6)
where ——Length of crank(mm)
——Length of connecting bar(mm)
——Length of rocker(mm)
——Length of machine frame(mm)
Firstly, we choose minimum comparing with . And then we record all values of greater than or equal to and choose the maximum of them.
Secondly, we find the maximum of corresponding to any oscillating angle which is chosen by small step in the permitted range (maximum of is different oscillating angle and the coefficient of travel speed variation ).
Finally, we change the length of rocker by small step similarly. Thus we may obtain the maximum of corresponding to the different length of bars, different oscillating angle and the coefficient of travel speed variation .
Fig.3 is accomplished from Table for the purpose of diagram design.
It is worth pointing out that whatever the length of rocker is evaluated, the location that the maximum of arises is only related to the ratio of the length of rocker and the length of machine frame /, while independent of .
2 DESIGN METHOD
2.1 Realizing the optimal transmission design given the coefficient of travel speed variation and the maximum oscillating angle of the rocker
The design procedure is as follows.
(1) According to given and , taken account to the formula the extreme included angle is found. The corresponding ratio of the length of bars / is obtained consulting Fig.3.
(7)
(2) Choose the length of rocker according to the work requirement, the length of the machine frame is obtained from the ratio /.
(3) Choose the centre of fixed hinge as the vertex arbitrarily, and plot an isosceles triangle, the side of which is equal to the length of rocker (see Fig.4), and . Then plot , draw , and make angle . Thus the point of intersection of and is gained. Finally, draw the circumcircle of triangle .
(4) Plot an arc with point as the centre of the circle, as the radius. The arc intersections arc at point . Point is just the centre of the fixed hinge of the crank.
Therefore, from the length of the crank
(8)
and the length of the connecting bar
(9)
we will obtain the crank and rocker mechanism consisted of , , , and .Thus the optimal transmission property is realized under given conditions.
2.2 Realizing the optimal transmission design given the length of the rocker (or the length of the machine frame) and the coefficient of travel speed variation
We take the following steps.
(1) The appropriate ratio of the bars / can be chosen according to given . Furthermore, we find the length of machine frame (the length of rocker ).
(2) The corresponding oscillating angle of the rocker can be obtained consulting Fig.3. And we calculate the extreme included angle .
Then repeat (3) and (4) in section 2.1
3 DESIGN EXAMPLE
The known conditions are that the coefficient of travel speed variation and maximum oscillating angle . The crankandrocker mechanism realizing the optimal transmission is designed by the diagram solution method presented above.
First, with Eq.(7), we can calculate the extreme included angle . Then, we find consulting Fig.3 according to the values of and .
If evaluate mm, then we will obtain mm.
Next, draw sketch(omitted).
As result, the length of bars is mm, mm, mm, mm.
The minimum transmission angle is
The results obtained by computer are mm, mm, mm, mm.
Provided that the figure design is carried under the condition of the Auto CAD circumstances, very precise design results can be achieved.
4 CONCLUSIONS
A novel approach of diagram solution can realize the optimal transmission of the crank-and-rocker mechanism. The method is simple and convenient in the practical use. In conventional design of mechanism, taking 0.1 mm as the value of effective the precision of the component sizes will be enough.
Signature of Supervisor:
譯文：
認(rèn)識(shí)曲柄搖臂機(jī)構(gòu)設(shè)計(jì)的最優(yōu)傳動(dòng)方法
摘要：一種曲柄搖臂機(jī)構(gòu)設(shè)計(jì)的最優(yōu)傳動(dòng)的方法被提出。這種優(yōu)化組合設(shè)計(jì)被用來找出最優(yōu)的傳遞參數(shù)。得出最優(yōu)傳遞圖。在圖中，在極小的傳動(dòng)角度之間, 滑移速度變化系數(shù)，搖臂的擺動(dòng)角度和桿的長(zhǎng)度被直觀地顯示。 這是這種方法擁有的主要特征。根據(jù)指定的要求，它將傳動(dòng)角度之下的最優(yōu)傳動(dòng)參數(shù)直接地表達(dá)在圖上。通過這種方法，機(jī)械傳動(dòng)的特性能用以獲取最優(yōu)傳動(dòng)效果。特別是， 這種方法是簡(jiǎn)單和實(shí)用的。
關(guān)鍵字：曲柄搖臂機(jī)構(gòu) 最優(yōu)傳動(dòng)角度 滑移速度變化系數(shù)
0 介紹
由曲柄搖臂機(jī)構(gòu)設(shè)計(jì)的常規(guī)方法, 在各種各樣的參量之間很難找出優(yōu)化組合的最優(yōu)傳動(dòng)。通過本文介紹的圖面設(shè)計(jì)方法可以幫助達(dá)到這個(gè)目的。在指定的情況下，通過觀查設(shè)計(jì)圖面, 我們就能得到每個(gè)參量和另外一個(gè)曲柄搖臂機(jī)構(gòu)設(shè)計(jì)之間的聯(lián)系。由因認(rèn)識(shí)最優(yōu)傳動(dòng)。
具體的設(shè)計(jì)的理論和方法， 以及它們各自的應(yīng)用事例將在以下介紹。
1 優(yōu)化傳動(dòng)設(shè)計(jì)的建立
優(yōu)化傳動(dòng)的設(shè)計(jì)一直是設(shè)計(jì)師改進(jìn)傳輸效率和追求產(chǎn)量的最重要的索引的當(dāng)中一個(gè)。曲柄搖臂機(jī)構(gòu)被廣泛應(yīng)用在機(jī)械傳動(dòng)中。如何改進(jìn)工作效率和減少多余的功率損失直接地與滑移速度變化系數(shù)，搖臂的擺動(dòng)角度和曲柄搖臂的比率有關(guān)系。這些參數(shù)的合理組合采用對(duì)機(jī)械效率和產(chǎn)量有重要作用, 這些主要體現(xiàn)在極小的傳輸角度上。
認(rèn)識(shí)機(jī)械優(yōu)化傳動(dòng)目的是找到極小的傳輸角度的最大值。設(shè)計(jì)參數(shù)是適度地減少限制而且分開的合理優(yōu)化方法的結(jié)合。因此，完全限制領(lǐng)域的優(yōu)化傳動(dòng)建立了。
以下步驟被采用在通常的設(shè)計(jì)方法。 首先，測(cè)量出搖臂的長(zhǎng)度和搖臂的擺動(dòng)角度的初始值。 然后滑移速度變化系數(shù)的值被定在允許的范圍內(nèi)。 同時(shí)，曲柄固定的鉸接座標(biāo)可能被認(rèn)為是任意值。
1.1 曲柄搖臂機(jī)構(gòu)桿的長(zhǎng)度
由圖Fig.1，左弧是點(diǎn)被允許的領(lǐng)域。點(diǎn)的座標(biāo)的選擇從點(diǎn)到點(diǎn)。
點(diǎn)的座標(biāo)是
(1)
(2)
當(dāng)，高度，在range(0 ，) 被逐漸增加。如果選的越小,計(jì)算精度將越高。 是設(shè)計(jì)圓的半徑。是從到的距離。
(3)
計(jì)算弧和的長(zhǎng)度，機(jī)械桿對(duì)應(yīng)于點(diǎn)的長(zhǎng)度是obtained[1,2 ] 。
1.2 極小的傳動(dòng)角度
極小的傳動(dòng)角度 (參見Fig.2) 由equations[3]確定
(4)
(5)
(6)
由于——曲柄的長(zhǎng)度(毫米)
——連桿的長(zhǎng)度（毫米）
——搖臂的長(zhǎng)度（毫米）
——機(jī)器的長(zhǎng)度（毫米）
首先, 我們比較極小值和。 并且我們記錄所有的值大于或等于，然后選擇他們之間的最大值。
第二, 我們發(fā)現(xiàn)最大值對(duì)應(yīng)于一個(gè)逐漸變小的范圍的任一個(gè)擺動(dòng)的角度 (最大值是不同于擺動(dòng)的角度和滑移速度變化系數(shù)) 。
最后, 我們相似地慢慢縮小搖臂的長(zhǎng)度。 因而我們能獲得最大值對(duì)應(yīng)于桿的不同長(zhǎng)度, 另外擺動(dòng)的角度和滑移速度變化系數(shù)。
Fig.3成功的表達(dá)設(shè)計(jì)的目的。
它確定了無(wú)論是搖臂的長(zhǎng)度,最大值出現(xiàn)的地點(diǎn)，只與搖臂的長(zhǎng)度和機(jī)械的長(zhǎng)度的比率/有關(guān), 當(dāng)確定時(shí)。
2 設(shè)計(jì)方法
2.1 認(rèn)識(shí)最優(yōu)傳動(dòng)設(shè)計(jì)下滑移速度變化系數(shù)和搖臂的最大擺動(dòng)的角度
設(shè)計(jì)步驟如下。
(1) 根據(jù)所給的和， 通常采取對(duì)發(fā)現(xiàn)極限角度的解釋。 桿的長(zhǎng)度的對(duì)應(yīng)的比率/是從圖Fig.3獲得的 。
(7)
(2) 根據(jù)工作要求選擇搖臂的長(zhǎng)度， 機(jī)械的長(zhǎng)度是從比率/獲得的。
(3) 任意地選擇固定的鉸接的中心作為端點(diǎn)，并且做一個(gè)等腰三角形,令一條邊與搖臂的長(zhǎng)度相等 (參見Fig.4)，令。 然后做， 連接，并且做角度。 因而增加了交點(diǎn)和。 最后， 畫三角形。
(4)以點(diǎn)作為圓的中心，為半徑畫圓弧。 弧交點(diǎn)在點(diǎn)。 點(diǎn)是曲柄的固定鉸接的中心。
所以, 從曲柄的長(zhǎng)度
(8)
并且連桿的長(zhǎng)度
(9)
我們將獲得曲柄搖臂機(jī)構(gòu)包括，，和。因而優(yōu)化傳動(dòng)加工會(huì)在指定的情況下進(jìn)行。
2.2 認(rèn)識(shí)優(yōu)化傳動(dòng)設(shè)計(jì)下?lián)u臂的長(zhǎng)度(或機(jī)械的長(zhǎng)度) 和滑移速度變化系數(shù)
我們采取以下步驟。
(1)根據(jù)選擇的確定桿的適當(dāng)比率/。 此外，我們得出機(jī)械 (搖臂的長(zhǎng)度) 。
(2) 搖臂對(duì)應(yīng)的擺動(dòng)的角度可以從圖Fig.3 獲得。 并且我們計(jì)算出極限角度。
然后根據(jù)2.1重覆(3) 和(4)
3 設(shè)計(jì)例子
已知的條件是, 滑移速度變化系數(shù)和最大擺動(dòng)角度。 提出曲柄搖臂機(jī)械優(yōu)化傳動(dòng)圖方法設(shè)計(jì)方案。
首先， 通過公式(7)，我們能計(jì)算出極限角度。 然后，我們通過表格Fig.3 查出以及和的值。
假設(shè)mm, 然后我們將得出mm。
然后, 做sketch(omitted) 。
最后, 算出桿的長(zhǎng)度分別是 mm, mm, mm, mm.
極小傳動(dòng)角度是
結(jié)果由計(jì)算可得 mm， mm， mm， mm。
在運(yùn)用Auto CAD 制圖設(shè)計(jì)的情況, 可達(dá)到非常精確設(shè)計(jì)結(jié)果。
4結(jié)論
認(rèn)識(shí)圖解法解答曲柄搖臂機(jī)構(gòu)的最優(yōu)傳動(dòng)。這種方法是簡(jiǎn)單和實(shí)用的。通常在機(jī)械設(shè)計(jì)中, 將0.1 毫米作為最小有效精度是足夠的。