1813_塑料肥皂盒注塑模的CADCAM

1813_塑料肥皂盒注塑模的CADCAM,塑料,肥皂盒,注塑,cadcam 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(zhuān) 業(yè) 名 稱(chēng) 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 designAbstract: 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 variationINTRODUCTIONBy 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 3l ?the value of the coefficient of travel speed variation is chosen in the permitted Krange. Meanwhile, the coordinate of the fixed hinge of crank possibly realized is Acalculated corresponding to value .K1.1 Length of bars of crank and rocker mechanismAs shown in Fig.1, left arc is the permitted field of point . The GC2coordinates of point are chosen by small step from point to point .A2CGThe coordinates of point are(1)02hycA??(2)ARxwhere , the step, is increased by small increment within range(0, ). If the 0h Hsmaller the chosen step is, the higher the computational precision will be. is the Rradius of the design circle. is the distance from to .d2CG(3)cos)2cos(cs33 ????????????lRlCalculating the length of arc and , the length of the bars of the 1A2mechanism corresponding to point is obtained[1,2].1.2 Minimum transmission angle min?Minimum transmission angle (see Fig.2) is determined by the equations[3]i (4)32214min)(cosll????(5)ax(6)axin180??where ——Length of crank(mm)l——Length of connecting bar(mm)2——Length of rocker(mm)3l——Length of machine frame(mm)4Firstly, we choose minimum comparing with . And then we record all min?in?values of greater than or equal to and choose the maximum of them.min??40Secondly, we find the maximum of corresponding to any oscillating angle inwhich is chosen by small step in the permitted range (maximum of is different ? min?oscillating angle and the coefficient of travel speed variation ).?KFinally, we change the length of rocker by small step similarly. Thus we may 3lobtain the maximum of corresponding to the different length of bars, different min?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 3llocation that the maximum of arises is only related to the ratio of the length of min?rocker and the length of machine frame / , while independent of .3l4 3l2 DESIGN METHOD2.1 Realizing the optimal transmission design given the coefficient of travel speed variation and the maximum oscillating angle of the rockerThe design procedure is as follows.(1) According to given and , taken account to the formula the extreme K?included angle is found. The corresponding ratio of the length of bars / is ? 3l4obtained consulting Fig.3.(7)?????180(2) Choose the length of rocker according to the work requirement, the length 3lof the machine frame is obtained from the ratio / .3l4(3) Choose the centre of fixed hinge as the vertex arbitrarily, and plot an Disosceles triangle, the side of which is equal to the length of rocker (see Fig.4), and 3l. Then plot , draw , and make angle ???21DC212CM?N1. Thus the point of intersection of and is gained. ???90NM2NC1Finally, draw the circumcircle of triangle .1P?(4) Plot an arc with point as the centre of the circle, as the radius. The arc D4lintersections arc at point . Point is just the centre of the fixed hinge of the GC2Acrank.Therefore, from the length of the crank(8)2/)(11Cl??and the length of the connecting bar(9)12lAlwe will obtain the crank and rocker mechanism consisted of , , , and .Thus 1l234lthe 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 variationWe take the following steps.(1) The appropriate ratio of the bars / can be chosen according to given . 3l4 KFurthermore, we find the length of machine frame (the length of rocker ).l 3l(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 2.13 DESIGN EXAMPLEThe known conditions are that the coefficient of travel speed variation and maximum oscillating angle . The crankandrocker mechanism 18.?K??40?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 ??15?find consulting Fig.3 according to the values of and .93.0/43?l ??If evaluate mm, then we will obtain mm.5l 76539.0/4?lNext, draw sketch(omitted).As result, the length of bars is mm, mm, mm, 16?l2l3l.4lmm.The minimum transmission angle is????698.42)(arcos3214minll?The results obtained by computer are mm, mm, 7.61 503.2lmm, mm.0.53?l 896.534?lProvided 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(lái)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(guò)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(guò)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(guò)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(kāi)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3l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)。 ?K同時(shí)，曲柄固定的鉸接座標(biāo) 可能被認(rèn)為是任意值 。A1.1 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)GC2 A2C。G點(diǎn) 的座標(biāo)是A(1)02hycA??(2)ARx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。0hH是設(shè)計(jì)圓的半徑。 是從 到 的距離。Rd2CG(3)2cos)cos(cs33 ????????????lRl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 ] 。1A2 A1.2 極小的傳動(dòng)角度 min?極小的傳動(dòng)角度 (參見(jiàn) Fig.2) 由 equations[3]確定i(4)32214min)(cosll????(5)32214ax)(ll(6)maxmin180??????由于 ——曲柄的長(zhǎng)度(毫米)1l ——連桿的長(zhǎng)度（毫米）2l——搖臂的長(zhǎng)度（毫米）3——機(jī)器的長(zhǎng)度（毫米）4l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min?i? min?，然后選擇他們之間的最大值。?40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in(最大值 是不同于擺動(dòng)的角度和滑移速度變化系數(shù) ) 。?min? K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)3l min?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3lmin?機(jī)械的長(zhǎng)度的比率 / 有關(guān) , 當(dāng)確定 時(shí)。3l42 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K??對(duì)應(yīng)的比率 / 是從圖 Fig.3 獲得的 。3l4(7)?????180?(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。3l 3l4(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D一條邊與搖臂的長(zhǎng)度 相等 (參見(jiàn) Fig.4)，令 。 然后做3l ???21C， 連接 ，并且做角度 。 因而增加了交點(diǎn)212CM?N1 ???9012N和 。 最后， 畫(huà)三角形 。1CP?(4)以點(diǎn) 作為圓的中心， 為半徑畫(huà)圓弧。 弧 交點(diǎn)在 點(diǎn)。 點(diǎn) 是D4l GC2A曲柄的固定鉸接的中心。所以, 從曲柄的長(zhǎng)度(8)/)(211Al??并且連桿的長(zhǎng)度(9)12lCl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134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K3l4 4l長(zhǎng)度 ) 。3l(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18.?K??40?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(guò)公式(7) ，我們能計(jì)算出極限角度 。 然后，我們通過(guò)表?5?格 Fig.3 查出 以及 和 的值。93.0/43?l??假設(shè) mm, 然后我們將得出 mm。5l 76.39.0/54?l然后, 做 sketch(omitted) 。最后, 算出桿的長(zhǎng)度分別是 mm, mm, mm, 16?l2l503l76.34?lmm.極小傳動(dòng)角度是 ?????3698.42)(arcos3214minll?結(jié)果由計(jì)算可得 mm， mm， mm， 7.16l 50.2l 0.53?lmm。896.534?l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。