2025年機械設計原理綜合訓練項目

CCHG指導老師：試設計偏置直動滾子推桿盤形凸輪機構凸輪的理論輪廓曲線和工作廓線，已知凸輪軸置于推桿軸線右側，偏距e=20mm,基圓半徑ro=50mm,滾子半徑r=10mm。凸輪以等角速度沿順時針方向回轉(zhuǎn)，在凸輪轉(zhuǎn)過角δ?=120°的過程中，推桿按正弦加速度運動規(guī)律上升h=50mm,凸輪繼續(xù)轉(zhuǎn)過δ?=30°時，推桿保持不動，其后凸輪再回轉(zhuǎn)角度δ3=60°時，推桿又按余弦加速度運動規(guī)律下降至起始位置；凸輪轉(zhuǎn)過一周的其他角度時，推桿又靜止不解：對于偏置直動滾子推桿盤形凸輪機構，凸輪理論輪廓上B點的直角坐推程階段：s?=h[(δ?/δ?1)-sin(2πδ?/δ?1)/(2π)]=h[(3δ?/遠休止階段近休止階段①推程階段dx/dδ=-{3h[1-cos(3δ?)]/(2π)-e}sinδdy/dδ={3h[1-cos(38?)]/(2π)-e}cosδ?-(dy/dδ=-ecos(2π/3+δ?)-(s?+s)sidx/dδ=[3hsin(3δ?)/2+e]sin(5π/6+δ?)-(S?+s)cdy/dδ=-[3hsin(3δ?)/2+e]cos(5π/6+δ?)-(s?+s)sdx/dδ=esin(7π/6+δ?)-(S?+s)co三、程序如下：%清除變量%定義基圓半徑%推桿上升高度%滾子半徑xl=-((sO+s1).*sin(al)+e*cosyl=(sO+s1).*cos(a1)-e*sinx2=-((sO+s2).*sin(a2+2*pi/s3=h*(1+cos(3*a3))/2;x3=-((sO+s3).*sin(a3+5*pi/6)+e*cos(a3+5*pi/6));x4=-((sO+s4).*sin(a4+7*pi/6)+e*cos(a4+7*pi/6));y4=(sO+s4).*cos(a4+7*pi/6)-e*sin(a4+7*pi/6);ml=-(h*3/2/pi*(1-cos(3*al))-e).*sin(a1)-(sO+sl).*cosn1=(h*3/2/pi*(1-cos(3*al))-e).*cos(al)-(sO+s1).*sinpl=-ml./sqrt(m1.^2+n1.^中間變量dy/d$m2=-(sO+s2).*cos(a2+2*pi/3)+e*sin(a2+2*n2=-(sO+s2).*sin(a2+2*pi/3)-e*cos(a2+2*pp2=-m2./sqrt(m2.^2+n2.^2y7=y2-r*p2;%y'函數(shù)m3=(h*3/2*sin(3*a3)+e).*sin(a3+5*pi/6)-(sO+s3).*cos(a3+5*n3=-(h*3/2*sin(3*a3)+e).*cos(a3+5*pi/6)-(sO+s3).*sin(a3+5*pm4=-(sO+s4).*cos(a4+7*pi/6)+e*sin(a4+7*n4=-(sO+s4).*sin(a4+7*pi/6)-e*cos(a4+7*pjl=yl(1)+r*sin(a0);j5=yl(75)+r*sin(a0);j6=y1(90)+r*sin(a0);j7=y2(1)+r*sin(a0);j8=y2(50)+r*sin(a0);j9=y3(1)+r*sin(a0);j10=y3(25)+r*sin(a0);jll=y3(40)+r*sin(a0);j12=y3(50)+r*sin(a0);j13=y3(75)+r*sin(a0);j14=y4(1)+r*sin(a0);j15=y4(50)+r*sin(a0);plot(xl,yl,'b-',x2,y2,'g-',x3,y3,'m-',x4,y4,'c-',.x6,y6,'b-',x7,y7,'g-',x8,y8,'m-',x9,y9,'c-',...plot(x5,y5,'r—',gl,jl,'k-',g2,j2,'k-',g3,j3,'k-',g4,j4,'k-',g5,j5,'k-',g6,j6,'k-',g7,j7,'k-',...g8,j8,'k-',g9,j9,'k-',g10,j10,'k-',gl1,j11,'k-',...g12,j12,'k-',g13,j13,'k-',g14,j14,'k-',g15,j15,'k-','Lin%標題%標題points=[x6',y6',zeros(100,1);x7,y7',zeros(100,1);:...由MATLAB分析后獲得