哈工大机械原理大作业――凸轮――2号

HarbinInstituteofTechnology机械原理大作业课程名称：机械原理设计题目：凸轮机构设计一、设计题目(1)凸轮机构运动简图：(2)凸轮机构的原始参数序号升程升程运动角升程运动规律升程许用压力角回程运动角回程运动规律回程许用压力角远休止角近休止角1490°120°余弦加速度35°90°3-4-5多项式65°80°70°(1)推杆升程、回程运动方程如下：A.推杆升程方程：设为1rads升程位移为：01cos451cos1.52hs203升程速度为：1100sin67.5sin1.52hv203升程加速度为：2221100cos101.25cos1.52ha203B.推杆回程方程：回程位移为：345111110156shTTT1029918回程速度为：22111103012hvTTT1029918回程加速度为：2211112060132haTTT1029918其中：010sT1029918(2)利用Matlab绘制推杆位移、速度、加速度线图A.推杆位移线图clcclearx1=linspace(0,2*pi/3,300);x2=linspace(2*pi/3,10*pi/9,300);x3=linspace(10*pi/9,29*pi/18,300);x4=linspace(29*pi/18,2*pi,300);T1=(x3-10*pi/9)/(pi/2);s1=45*(1-cos(1.5*x1))s2=90;s3=90*(1-(10*T1.^3-15*T1.^4+6*T1.^5));s4=0;plot(x1,s1,'r',x2,s2,'r',x3,s3,'r',x4,s4,'r')xlabel('角度ψ/rad');ylabel('位移s/mm')title('推杆位移线图')gridaxis([0,7,-10,100])得到推杆位移线图：B.推杆速度线图clcclearx1=linspace(0,2*pi/3,300);x2=linspace(2*pi/3,10*pi/9,300);x3=linspace(10*pi/9,29*pi/18,300);x4=linspace(29*pi/18,2*pi,300);T1=(x3-10*pi/9)/(pi/2);v1=67.5*1*sin(1.5*x1);v2=0;v3=-30*90*1*T1.^2/(pi/2).*(1-2*T1+T1.^2);v4=0;plot(x1,v1,'r',x2,v2,'r',x3,v3,'r',x4,v4,'r')xlabel('角度ψ/rad');ylabel('速度v/(mm/s)')title('推杆速度线图')Grid得到推杆速度线图：C.推杆加速度线图clcclearx1=linspace(0,2*pi/3,300);x2=linspace(2*pi/3,10*pi/9,300);x3=linspace(10*pi/9,29*pi/18,300);x4=linspace(29*pi/18,2*pi,300);T1=(x3-10*pi/9)/(pi/2);a1=101.25*1.^2.*cos(1.5*x1);a2=0;a3=-60.*90.*T1./((pi/2).^2).*(1-3*T1+2*T1.^2);a4=0;plot(x1,a1,'r',x2,a2,'r',x3,a3,'r',x4,a4,'r')xlabel('角度ψ/rad');ylabel('加速度a/')title('推杆加速度线图')Grid得到推杆加速度线图：三、凸轮机构的ds/dψ-s线图，并依次确定凸轮的基圆半径和偏距.1、凸轮机构的ds/dψ--s线图：x1=linspace(0,2*pi/3,300);x2=linspace(2*pi/3,10*pi/9,300);x3=linspace(10*pi/9,29*pi/18,300);x4=linspace(29*pi/18,2*pi,300);T1=(x3-10*pi/9)/(pi/2);s1=45*(1-cos(1.5*x1))s2=90;s3=90*(1-(10*T1.^3-15*T1.^4+6*T1.^5));s4=0;v1=67.5*1*sin(1.5*x1);v2=0;v3=-30*90*1*T1.^2/(pi/2).*(1-2*T1+T1.^2);v4=0;plot(v1,s1,'r',v2,s2,'r',v3,s3,'r',v4,s4,'r')xlabel('ds/dψ');ylabel('(位移s/mm)')title('ds/dψ—s曲线')gridaxis([-120,80,-10,100])得到ds/dψ—s曲线：2、确定凸轮的基圆半径和偏距：在dssd线图中，右侧曲线为升程阶段的类速度-位移图，作直线Dtdt与其相切，且与位移轴正方向呈夹角[1]=350,故该直线斜率：032sin2=tan5533cos2ok通过编程求其角度。%求升程切点位置转角f=sym('tan(55/180*pi)*3*cos(3*k/2)-2*sin(3*k/2)=0');k=solve(f);pretty(k);x=67.5*sin((3*k)/2);y=45*(1-cos(3/2*k));求的转角近似值k=0.7560（rad）进而求的切点坐标（x,y）=(61.1625，25.9633)左侧曲线为回程阶段的类速度-位移图，作直线D’td’t与其相切，它与位移轴正方向的夹角为[2]=65,故该直线斜率tan（-25)°令2209tm则23222264smmmkmmtan(25)。通过编程求其角度。%求回程切点位置转角f=sym('tan(25/180*pi)*2/pi*(2-6*(2*k/pi-20/9)+4*(2*k/pi-20/9)^2)+((2*k/pi-20/9)-2*(2*k/pi-20/9)^2+(2*k/pi-20/9)^3)=0');k=solve(f);pretty(k);x=-90*2/pi*(10*3*(2*k/pi-20/9).^2-15*4*(2*k/pi-20/9).^3+6*5*(2*k/pi-20/9).^4);y=90*(1-(10*(2*k/pi-20/9).^3-15*(2*k/pi-20/9).^4+6*(2*k/pi-20/9).^5));求得转角近似值k=4.5627,切点为（-80.7164,16.8313）因此：直线Dtdt：y-25.9633=tan(55/180*pi)*(x-61.1625);直线Dt’dt’:y-16.8313=-tan(25/180*pi)*(x+80.7164);又因为，在从动件推程起始点，s=0，且/0dsd时，有0tan/es，为保证此时的[]，作直线00Bd与纵坐标交角为[]，凸轮轴心只能在线上或在其左下方选取。易求直线00Bd：y=-tan(55/180*pi)*x;编程如下：x1=linspace(0,2*pi/3,300);x2=linspace(2*pi/3,10*pi/9,300);x3=linspace(10*pi/9,29*pi/18,300);x4=linspace(29*pi/18,2*pi,300);T1=(x3-10*pi/9)/(pi/2);s1=45*(1-cos(1.5*x1))s2=90;s3=90*(1-(10*T1.^3-15*T1.^4+6*T1.^5));s4=0;v1=67.5*1*sin(1.5*x1);v2=0;v3=-30*90*1*T1.^2/(pi/2).*(1-2*T1+T1.^2);v4=0;plot(v1,s1,'r',v2,s2,'r',v3,s3,'r',v4,s4,'r')xlabel('ds/dψ');ylabel('(位移s/mm)')title('ds/dψ—s曲线')gridaxis([-120,80,-10,100])holdonx=linspace(-120,80,300);x1=linspace(0,80,300);y1=tan(55/180*pi)*(x-61.1625)+25.9633;y2=-tan(25/180*pi)*(x+80.7164)+16.8313;y3=-tan(55/180*pi)*x1;plot(x,y1,'g',x,y2,'g',x1,y3,'g')axis([-120,80,-100,120])holdonplot(25,-75,'*r')gridaxisequal在轴心公共许用区内取轴心位置，能够满足压力角要求，由于三条直线近似交于一点，现取直线Dt’dt’与直线Dtdt的交点为轴心位置，通过解二元一次方组:55tan61.162525.963318025tan80.716416.8313180yxyx可以求得:21.419630.7955xy最小基圆对应的轴心坐标大致为（21.4196，-30.7955）为方便可取：偏距e=20mm,075smm，220077.6209resmm。下图红点所在位置即设为轴心位置。四、滚子半径的确定及凸轮理论轮廓和实际轮廓的绘制.小滚子半径的确定的MATLAB程序：%确定最小曲率半径clcclearv=[];symsx1x2x3x4x5s0=75;e=20;s1=45*(1-cos(1.5*x1));t1=(s1+s0).*cos(x1)-e*sin(x1);y1=(s0+s1).*sin(x1)-e*cos(x1);tx1=diff(t1,x1);txx1=diff(t1,x1,2);yx1=diff(y1,x1);yxx1=diff(y1,x1,2);forxx1=0:(pi/100):(2*pi/3);k1=subs(abs((tx1*yxx1-txx1*yx1)/(tx1^2+yx1^2)^1.5),{x1},{xx1});v=[v,1/k1];ends2=90;t2=(s2+s0).*cos(x2)-e*sin(x2);y2=(s0+s2).*sin(x2)-e*cos(x2);tx2=diff(t2,x2);txx2=diff(t2,x2,2);yx2=diff(y2,x2);yxx2=diff(y2,x2,2);forxx2=(2*pi/3):(pi/100):(10*pi/9);k2=subs(abs((tx2*yxx2-txx2*yx2)/(tx2^2+yx2^2)^1.5),{x2},{xx2});v=[v,1/k2];endT1=(x3-10*pi/9)/(pi/2);s3=90*(1-(10*T1.^3-15*T1.^4+6*T1.^5));t3=(s3+s0).*cos(x3)-e*sin(x3);y3=(s0+s3).*sin(x3)-e*cos(x3);tx3=diff(t3,x3);txx3=diff(t3,x3,2);yx3=diff(y3,x3);yxx3=diff(y3,x3,2);forxx3=(10*pi/9):(pi/100):(29*pi/18);k3=subs(abs((tx3*yxx3-txx3*yx3)/(tx3^2+yx3^2)^1.5),{x3},{xx3});v=[v,1/k3];ends4=0;t4=(s4+s0).*cos(x4)-e*sin(x4);y4=(s0+s4).*sin(x4)-e*cos(x4);tx4=diff(t4,x4);txx4=diff(t4,x4