按行程速比系數(shù)設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的解析法

1、 胡愛萍按行程速比系數(shù)設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的解析法胡愛萍陳權(quán)鄒日文祝海林(江蘇石油化工學(xué)院常州213016摘要利用平面曲柄滑塊機(jī)構(gòu)中各有關(guān)點(diǎn)的坐標(biāo)之間的關(guān)系,提出了一種按行程速比系數(shù)設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的解析法。與現(xiàn)有解析法相比,該方法解題原理簡(jiǎn)單明了,計(jì)算簡(jiǎn)便,易于被工程設(shè)計(jì)人員所接受,具有較好的實(shí)用價(jià)值。關(guān)鍵詞行程速比系數(shù)機(jī)構(gòu)設(shè)計(jì)解析法中圖號(hào)T H112引言眾所周知,按行程速比系數(shù)K設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的問(wèn)題一般歸納為:已知滑塊行程H、行程速比系數(shù)K,通常還帶有輔助條件,如給定偏距e(或給定曲柄長(zhǎng)度l1,或給定連桿長(zhǎng)度l2,來(lái)設(shè)計(jì)曲柄滑塊機(jī)構(gòu)(即確定未知長(zhǎng)度尺寸,最后校驗(yàn)最大壓力角 max

2、。對(duì)該問(wèn)題的求解傳統(tǒng)上采用圖解法。圖解法具有簡(jiǎn)單、直觀的特點(diǎn),長(zhǎng)期以來(lái)一直為大家普遍接受,沿用至今,但其設(shè)計(jì)精度較低。隨著計(jì)算機(jī)在工程中的廣泛應(yīng)用,解析法越來(lái)越顯示出它的優(yōu)越性。解析法可利用計(jì)算機(jī)迅速精確地設(shè)計(jì)曲柄滑塊機(jī)構(gòu)。近年來(lái)在一些教材和文獻(xiàn)中也都介紹了按行程速比系數(shù)K設(shè)計(jì)平面曲柄滑塊的解析法,但有的不夠完善,有的解析式子的得來(lái)比較繁復(fù),不易被設(shè)計(jì)人員所接受。本文利用平面曲柄滑塊機(jī)構(gòu)中各有關(guān)點(diǎn)的坐標(biāo)之間的關(guān)系,提出了一種按行程速比系數(shù)設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的解析法。與現(xiàn)有解析法相比,該方法解題原理簡(jiǎn)單明了,計(jì)算簡(jiǎn)便,易于被設(shè)計(jì)人員所接受,具有較好的實(shí)用價(jià)值。圖1按K設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的圖解

3、法的基本原理圖1建立坐標(biāo)系以l1、l2表示曲柄、連桿的長(zhǎng)度,e表示偏距,H表示滑塊行程。如圖1所示為大家熟知的按行程速比系數(shù)K設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的圖解法的基本原理圖。其中C1C2=H, 為極位夾角( =180°K-1K+1,過(guò)C1與C2點(diǎn)的圓周為滿足給定行程速比系數(shù)K的曲柄固定鉸鏈中心A所落居的圓周。由圖解法原理可知,l1、l2與A C1、A C2的關(guān)系為:l2-l1=A C1,l1+l2=A C2,且A點(diǎn)應(yīng)位于圓的弧C1E或C2F上。以圓的圓心為坐標(biāo)原點(diǎn)建立直角坐標(biāo)系ox y,如圖2所示,用(x c1,y c1、(x c2,y c2和(x A,y A分別表示圖中C1、C2和A(或

4、A 點(diǎn)的坐標(biāo)。由于H、K已知,所以點(diǎn)C1(x c1,y c1、C2(x c2,y c2的坐標(biāo)值已知,為x c1=-H/2y c1=(H/2ct gx c2=H/2y c2=(H/2ct g建立直角坐標(biāo)系后,可寫出:圓的方程x2+y2=r2(r為已知,r=H2sinA(或A 點(diǎn)的縱坐標(biāo)的計(jì)算式y(tǒng) A=H2 ctg -e2解題原理根據(jù)給定的輔助條件不同,解題原理分別敘述如下。2.1給定偏距e由于A(或A 點(diǎn)位于圓上,故該點(diǎn)應(yīng)滿足方程:x2A+y2A=r2(1對(duì)于給定偏距e,y A為已知,故由式(1可求出x A(x A有正負(fù)兩個(gè)根,負(fù)根為A點(diǎn)的橫坐標(biāo),正根為A 點(diǎn)的橫坐第18卷第4期1999年7月機(jī)

5、械科學(xué)與技術(shù)M ECHA N ICA L SCI ENCE A N D T ECHN OL O GYJuly1999 收稿日期:1998-08-04標(biāo)。再根據(jù)l 1、l 2與A C 1和A C 2的關(guān)系,有式:l 2-l 1=(x A -x c 12+(y A -y c 121/2l 2+l 1=(x A -x c 22+(y A -y c 2 21/2(2由式(2可求出曲柄長(zhǎng)度l 1和連桿長(zhǎng)度l 2。2.2給定曲柄長(zhǎng)度l 1由于2-1=2l 1,即點(diǎn)A 到C 1與C 2兩點(diǎn)的距離之差為常數(shù)。若取C 1與C 2點(diǎn)為雙曲線的兩定點(diǎn)(焦點(diǎn),則點(diǎn)A (或A 恰好為雙曲線上的一個(gè)點(diǎn),作雙曲線!如圖3所

6、示。圖3給定l 1時(shí)得到的雙曲線!與圓 的關(guān)系圖圖3中雙曲線!的方程為x 2a 2-(y -(H /2ct g 2b 2=1式中,a 、b 分別為為雙曲線的實(shí)半軸、虛半軸。由雙曲線實(shí)半軸和虛半軸的定義可知:a =l 1,b =(H /22-l 211/2根據(jù)點(diǎn)A (或A 既在圓 上又在雙曲線!上,以及y A 的計(jì)算式,有x 2A +y 2A =r2x 2Aa 2-(y A -(H /2ctg 2b 2=1y A =(H /2ctg -e(3由于r 、a 、b 、H 、 均為已知,故由式(3可求解出偏距e和A (或A 點(diǎn)的坐標(biāo)(x A ,y A 。再根據(jù)l 1、l 2與A C 1(或A C 2的

7、關(guān)系,有式l 2-l 1=(x A -x c 12+(y A -y c 121/2l 2+l 1=(x A -x c 22+(y A -y c 221/2由此可求出連桿長(zhǎng)度l 2。2.3給定連桿長(zhǎng)度l 2由于A C 2+A C 1=2l 2,即點(diǎn)A 到C 1與C 2兩點(diǎn)的距離之和為常數(shù)。若取C 1與C 2 點(diǎn)為橢圓的兩定點(diǎn)(焦點(diǎn),則點(diǎn)A (或A 恰好為橢圓上的一個(gè)點(diǎn),作橢圓!如圖4所示。圖4中橢圓!的方程為x 2a 2-(y -(H /2ct g 2b 2=1式中,a 、b 分別為橢圓的長(zhǎng)半軸、短半軸。由橢圓長(zhǎng)半軸和短半軸的定義可知:a =l 2,b =(l 22-(H /221/2圖4給定l

8、 2時(shí)得到的橢圓!與圓 的關(guān)系圖根據(jù)點(diǎn)A 既在圓 上又在橢圓!上,以及y A 的計(jì)算式,有式:x 2A +y 2A =r2x 2Aa 2-(y A -(H /2ctg 2b 2=1y A =(H /2ct g -e(4同前述一樣,由于r 、a 、b 、H 、 均為已知,故由式(4可求解出偏距e 和A (或A 點(diǎn)的坐標(biāo)(x A ,y A 。再根據(jù)l 1、l 2與A C 1(或A C 2的關(guān)系,可求出曲柄長(zhǎng)度l 1。4設(shè)計(jì)舉例設(shè)計(jì)一平面曲柄滑塊機(jī)構(gòu)。已知滑塊行程H =160mm,行程速比系數(shù)K =1.25,現(xiàn)分別給定:(1偏距e =90mm(2曲柄長(zhǎng)度l 1=75mm (3連桿長(zhǎng)度l 2=160m

9、 m試設(shè)計(jì)該機(jī)構(gòu),并校驗(yàn)最大壓力角 max 。計(jì)算結(jié)果如下:(1l 1=71.6mm ,l 2=217.4mm, max =sin-1(l 1+e l 2=48°(2e =55mm,l 2=177mm, ma x =47.3°(3e =42.3mm,l 1=76.2mm, max =48° max <=50°,設(shè)計(jì)結(jié)果滿足要求。參考文獻(xiàn)1劉遠(yuǎn)偉等.按行程速比系數(shù)K 設(shè)計(jì)平面曲柄滑塊機(jī)構(gòu)的解析法.機(jī)械設(shè)計(jì),1996(10科學(xué)與技術(shù),1992(1設(shè)計(jì)與制造,1997(34華大年主編.機(jī)械原理.北京:高等教育出版社,19945鄭文偉,吳克堅(jiān)主編.機(jī)械原理

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15、ngsha Comm unications Univer sity,Changsh a410076AbstractA finit e element m odel used t o simulate the de-for mation pr ocess of safety-cov er fo r gr inder w ith bro-ken gr inding-wheel is pr esent e.T he4-node qua dr ilat-eral shell element w ith a n one-point integ rat ion is em-ploy ed in the m

16、o del of safet y-co ver o f g r inder.T he mater ial is assumed to satisfy the elast o-plastic y ield conditio n w ith isotr opic har dening.T he8-node so lid element with an o ne-point quadr atur e is used for the model of gr inding-w heel,flange,spindle and bear ing.T he master-slav e algo rithm i

17、s used t o deal with the contact int erfaces and t he co ntact for ce is ev aluated by penalty method.F rictio n is based on the Coulomb for-mulation.A pr ior i stabilizatio n pr ocedur es is employ ed in t he numer ical contr ol of the hour glass instability.T he validity of this model is v erified

18、 by an indirect test.A numer ical ex ample using this model is given. KeywordsGr inderF EMSafety-cov er(上接第581頁(yè)Analytic Method to Design Plane Crank Block Mechanism According to theTravel-Velocity-RatioCoefficientHu AipingChen QuanZou M inZhu Hailin(Jiangsu Institute o f Petr ochemical T echnolog y,Chang zho u213016AbstractT his ar ticle g iv es an analyt ic met ho d to design plane crank-block mechanism accor ding to the tr avel-velocity-rat io co efficient and using the r alationship a-mong the coo rdinates of po ints co ncer ned.Compar ing w ith the ex isted methods,t