1、第1章 固定凸輪連桿機構(gòu)參數(shù)選取1. 確定驅(qū)動方案 圖1如上圖所示，設(shè)：與從動桿升程運動相對應(yīng)的曲柄轉(zhuǎn)角為，即;而與降程運動相對應(yīng)的曲柄轉(zhuǎn)角為，即,則：（1） 當(dāng)時，選用曲柄AB拉著BC桿運動的方案。（2） 當(dāng)時，選用曲柄AB推著BC桿運動的方案。（3） 當(dāng)時，任選其中一種驅(qū)動方案。已知數(shù)據(jù)，很明顯，所以選用方案2。2.確定e直動從動桿，取，取3. 確定h從結(jié)構(gòu)緊湊和減小凸輪壓力角考慮，應(yīng)將h值取小些。但h值愈小，對從動桿驅(qū)動力的壓力角也愈大。通常取，去4. 確定a若a值過小，會使凸輪壓力角明顯增大，甚至不能實現(xiàn)預(yù)期動動。可取a=0.60.9Sm或a=1.21.8lsin。取a=70mm6、確

2、定其值對凸輪的壓力角影響極大，過小，尤其是過大，會使壓力角急劇增加。在前述參數(shù)確定后，最好將優(yōu)化，目標(biāo)函數(shù)為 a1m() （a1m）min式中a1m為凸輪的最大壓力角。暫時取7. 求算b1、b2須先求算bmax、bmin。依據(jù)鉸銷B、D的坐標(biāo)，可建立它們之間距離的公式。B的坐標(biāo)為 D的坐標(biāo)為 式中 曲柄轉(zhuǎn)角，取升程起始時的 =0°；S 與相對應(yīng)的從動桿位移，即鉸銷D至其最低位置的距離。S值分為升程（=01）、最高位置停留（=11+2）、降程（=1+21+2+3）、最低位置停留（=1+2+3360°）四個階段求算。b值為 b= （1）用matlab編程畫出b與曲線圖，并算出：

3、clearsm=100;h=120;e=0;a=70;d=8*pi/180;fa1=110*pi/180;fa2=0*pi/180;fa3=150*pi/180;fa4=100*pi/180;fa01=0:0.001:fa1;s=sm/2*(1-cos(pi*fa01/fa1);xb=a*sin(d+fa01);yb=-a*cos(d+fa01);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);plot(fa01,b);max(b)min(b)hold on;fa02=fa1;s=sm;xb=a*sin(d+fa02);yb=-a*cos(d+fa02);xd=e

4、;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);plot(fa02,b,'r-d');max(b)min(b)hold on;fa03=fa1+fa2:0.001:fa1+fa2+fa3;s=sm*(1-(fa03-fa1-fa2)/fa3+1/(2*pi)*sin(2*pi*(fa03-fa1-fa2)/fa3);xb=a*sin(d+fa03);yb=-a*cos(d+fa03);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);plot(fa03,b,'g-');max(b)min(b)hold on;fa

5、04=fa1+fa2+fa3:0.001:fa1+fa2+fa3+fa4;s=0;xb=a*sin(d+fa04);yb=-a*cos(d+fa04);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);plot(fa04,b,'r-');max(b)min(b)xlabel('fa');ylabel('b');title('fa-b');運行結(jié)果：ans = 217.0095ans = 189.3564b = 197.0794b = 197.0794ans = 197.0794ans = 94.1923

6、ans = 190.0000ans = 136.7980由以上結(jié)果可以看出 并且b取最大值時，fa=1.21.4；b取最小值時，fa=3.53.7（2） 根據(jù)計算 解得： b1 =61.4086 b2 =155.60098、設(shè)計凸輪廊線固定凸輪的理論廊線就是滾子中心C的運動軌跡線，根據(jù)鉸銷B、D的位置及b1、b2值可確定C的位置。參閱1，令鉸銷B、D的連線BD與DOD1線（或y軸）的夾角為，BD與CD的夾角為，則 顯然，XBXD時為正值，反之則為負值，而始終為正值。這樣，鉸銷C的坐標(biāo)為 該式對直動和擺動兩種從動桿類型都適用，運算符號“+”和“”的確定原則是：令B=bmax 時的為 m， b=b

7、時的為m，則對于AB推動BC的驅(qū)動方案（如圖4所示），在 = mm區(qū)間，取“”號；在 =0 m和 =m360°區(qū)間，取“+”對于AB拉動BC的驅(qū)動方案，則剛好相反。（1） 用matlab求famax、faminclearsm=100;h=120;e=0;a=70;d=8*pi/180;fa1=110*pi/180;fa2=0*pi/180;fa3=150*pi/180;fa4=100*pi/180;fa01=1.2:0.01:1.4;s=sm/2*(1-cos(pi*fa01/fa1);xb=a*sin(d+fa01);yb=-a*cos(d+fa01);xd=e;yd=h+s;b1

8、=sqrt(xb-xd).2+(yb-yd).2);f=polyval(b1,fa01);fa03=3.5:0.01:3.7;s=sm*(1-(fa03-fa1-fa2)/fa3+1/(2*pi)*sin(2*pi*(fa03-fa1-fa2)/fa3);xb=a*sin(d+fa03);yb=-a*cos(d+fa03);xd=e;yd=h+s;b3=sqrt(xb-xd).2+(yb-yd).2);f=polyval(b3,fa03);運行結(jié)果：>> b1b1 = Columns 1 through 13 216.1970 216.3140 216.4223 216.5220

9、216.6128 216.6945 216.7672 216.8306 216.8846 216.9291 216.9639 216.9890 217.0042 Columns 14 through 21 217.0095 217.0046 216.9896 216.9642 216.9284 216.8821 216.8252 216.7575>> fa01fa01 = Columns 1 through 13 1.2000 1.2100 1.2200 1.2300 1.2400 1.2500 1.2600 1.2700 1.2800 1.2900 1.3000 1.3100 1

10、.3200 Columns 14 through 21 1.3300 1.3400 1.3500 1.3600 1.3700 1.3800 1.3900 1.4000>> b3b3 = Columns 1 through 13 94.6116 94.5076 94.4185 94.3443 94.2848 94.2401 94.2101 94.1946 94.1936 94.2070 94.2347 94.2766 94.3325 Columns 14 through 21 94.4023 94.4859 94.5832 94.6940 94.8182 94.9555 95.106

11、0 95.2693>> fa03fa03 = Columns 1 through 13 3.5000 3.5100 3.5200 3.5300 3.5400 3.5500 3.5600 3.5700 3.5800 3.5900 3.6000 3.6100 3.6200 Columns 14 through 213.6300 3.6400 3.6500 3.6600 3.6700 3.6800 3.6900 3.7000由以上數(shù)據(jù)可以看出： famax=1.33famin=3.58(2) 凸輪的設(shè)計clearsm=100;h=120;e=0;a=70;d=8*pi/180;fa1=1

12、10*pi/180;fa2=0*pi/180;fa3=150*pi/180;fa4=100*pi/180;famax=1.33;famin=3.58;b1=61.4086;b2=155.6009;fa01=0:0.002:famax;s=sm/2*(1-cos(pi*fa01/fa1);xb=a*sin(d+fa01);yb=-a*cos(d+fa01);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1+bet

13、a1);yc=yd-b2*cos(theta1+beta1);plot(xc,yc);hold on;fa02=famax:0.002:fa1;s=sm/2*(1-cos(pi*fa02/fa1);xb=a*sin(d+fa02);yb=-a*cos(d+fa02);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1-beta1);yc=yd-b2*cos(theta1-beta1);plot(xc,yc,

14、9;r');hold on;fa03=fa1+fa2:0.002:famin;s=sm*(1-(fa03-fa1-fa2)/fa3+1/(2*pi)*sin(2*pi*(fa03-fa1-fa2)/fa3);xb=a*sin(d+fa03);yb=-a*cos(d+fa03);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1-beta1);yc=yd-b2*cos(theta1-beta1);plot(

15、xc,yc);hold on;fa04=famin:0.002:fa1+fa2+fa3;s=sm*(1-(fa04-fa1-fa2)/fa3+1/(2*pi)*sin(2*pi*(fa04-fa1-fa2)/fa3);xb=a*sin(d+fa04);yb=-a*cos(d+fa04);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1+beta1);yc=yd-b2*cos(theta1+beta1);plot

16、(xc,yc,'r');hold on;fa05=fa1+fa2+fa3:0.002:fa1+fa2+fa3+fa4;s=0;xb=a*sin(d+fa05);yb=-a*cos(d+fa05);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1+beta1);yc=yd-b2*cos(theta1+beta1);plot(xc,yc);title('凸輪輪廓曲線'); 9.檢驗壓

17、力角（1）凸輪的壓力角1 參閱圖1，1為PC和Vc的夾角。PC為驅(qū)動鉸銷運動的力(不考慮摩擦力)，BC重合，vc為鉸鏈C的運動方向,，與C點的凸輪廊線切線重合。用K1、K2分別表示PC、vc的斜率，則K1 = 應(yīng)保證的最大值不超過許用值，即。可取clearsm=100;h=120;e=0;a=70;d=8*pi/180;fa1=110*pi/180;fa2=0*pi/180;fa3=150*pi/180;fa4=100*pi/180;b1=61.4086;b2=155.6009;famax=1.33;famin=3.58;fa01=0:0.001:famax;s=sm/2*(1-cos(pi*

18、fa01/fa1);xb=a*sin(d+fa01);yb=-a*cos(d+fa01);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1+beta1);yc=yd-b2*cos(theta1+beta1);k11=(yc-yb)./(xc-xb);for i=2:length(fa01)-1 k21(i)=(yc(i-1)-yc(i+1)./(xc(i-1)-xc(i+1); endk21(1)=k21(2)

19、;for i=2:length(fa01)-1 a1(i)=abs(atan(k21(i)-k11(i)./(1+k11(i)*k21(i)*180/pi;enda1(1)=a1(2);dfa01=fa01(2:length(fa01);plot(dfa01,a1);hold on;fa02=famax:0.002:fa1;s=sm/2*(1-cos(pi*fa02/fa1);xb=a*sin(d+fa02);yb=-a*cos(d+fa02);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.

20、2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1-beta1);yc=yd-b2*cos(theta1-beta1);k11=(yc-yb)./(xc-xb);for i=2:length(fa02)-1 k21(i)=(yc(i-1)-yc(i+1)./(xc(i-1)-xc(i+1); endk21(1)=k21(2);for i=2:length(fa02)-1 a2(i)=abs(atan(k21(i)-k11(i)./(1+k11(i)*k21(i)*180/pi;enda2(1)=a2(2);dfa02=fa02(2:length(fa02);p

21、lot(dfa02,a2);hold on;fa03=fa1+fa2:0.002:famin;s=sm*(1-(fa03-fa1-fa2)/fa3+1/(2*pi)*sin(2*pi*(fa03-fa1-fa2)/fa3);xb=a*sin(d+fa03);yb=-a*cos(d+fa03);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1-beta1);yc=yd-b2*cos(theta1-beta1);k

22、11=(yc-yb)./(xc-xb);for i=2:length(fa03)-1 k21(i)=(yc(i-1)-yc(i+1)./(xc(i-1)-xc(i+1); endk21(1)=k21(2);for i=2:length(fa03)-1 a3(i)=abs(atan(k21(i)-k11(i)./(1+k11(i)*k21(i)*180/pi;enda3(1)=a3(2);dfa03=fa03(2:length(fa03);plot(dfa03,a3);hold on;fa04=famin:0.002:fa1+fa2+fa3;s=sm*(1-(fa04-fa1-fa2)/fa3+

23、1/(2*pi)*sin(2*pi*(fa04-fa1-fa2)/fa3);xb=a*sin(d+fa04);yb=-a*cos(d+fa04);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1+beta1);yc=yd-b2*cos(theta1+beta1);k11=(yc-yb)./(xc-xb);for i=2:length(fa04)-1 k21(i)=(yc(i-1)-yc(i+1)./(xc(i-

24、1)-xc(i+1); endk21(1)=k21(2);for i=2:length(fa04)-1 a4(i)=abs(atan(k21(i)-k11(i)./(1+k11(i)*k21(i)*180/pi;enda4(1)=a4(2);dfa04=fa04(2:length(fa04);plot(dfa04,a4);hold on;fa05=fa1+fa2+fa3:0.002:fa1+fa2+fa3+fa4;s=0;xb=a*sin(d+fa05);yb=-a*cos(d+fa05);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(x

25、b-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);xc=xd+b2*sin(theta1+beta1);yc=yd-b2*cos(theta1+beta1);k11=(yc-yb)./(xc-xb);for i=2:length(fa05)-1 k21(i)=(yc(i-1)-yc(i+1)./(xc(i-1)-xc(i+1); endk21(1)=k21(2);for i=2:length(fa05)-1 a5(i)=abs(atan(k21(i)-k11(i)./(1+k11(i)*k21(i)*180/pi;enda5(1)=a5(2);dfa0

26、5=fa05(2:length(fa05);plot(dfa05,a5);hold on;xlabel('fa');ylabel('a');title('凸輪壓力角'); 由上圖可知，滿足許用值。（2） 從動桿的壓力角 為CD桿對鉸銷D的驅(qū)動力（不考慮摩擦力）與D的運動方向的夾角。對于直動從動桿，顯然 對于AB推動BC的驅(qū)動方案，在區(qū)間取“”號，其它區(qū)間取“+”號；而對于AB拉動BC的驅(qū)動方案，剛好相反。clearsm=100;h=120;e=0;a=70;d=8*pi/180;fa1=110*pi/180;fa2=0*pi/180;fa3=15

27、0*pi/180;fa4=100*pi/180;b1=61.4086;b2=155.6009;famax=1.33;famin=3.58;fa01=0:0.001:famax;s=sm/2*(1-cos(pi*fa01/fa1);xb=a*sin(d+fa01);yb=-a*cos(d+fa01);xd=e;yd=h+s;b=sqrt(xb-xd).2+(yb-yd).2);theta1=asin(xb-xd)./b);beta1=acos(b.2+b2.2-b1.2)./(2*b*b2);afa=abs(theta1+beta1);plot(fa01,afa);hold on; fa02=famax:0.002:fa1;s=sm/2*(1-cos(pi*fa02/fa1);xb=a*sin(d+fa02);yb=-a*cos(d+fa02);xd=e;yd=h+s;b