计算机辅助公差设计

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1、计算机辅助公差设计已知:h=30±0.0lmm,d=30±0.02mm,a=90±0.05°DrwpnSpec试编写matlab程序，并求:1)F的尺寸及其公差2)h,d,仅的标准差对刁的敏感度和贡献率，画出直方图。解:symsfhdthidtjtfdhdddjvfghgdgjtheta=sym(，theta')%所求尺寸F%尺对1【的敏感度%F对D的敏感度%F对theta的敏感度f-d/(2*sin(theta/2))+d/2~hdh=diff(f,h)dd=diff(f,d)dj=diff(f,theta)vf二sqrt((dh*th/3)，2+(dd*td/3)"2+(dj*tj*

2、pi/540厂2)tf=3*vf%F的偏差gh二(dh*th/3厂2/((tf/3厂2)*100%H对F的贡献度gd二(dd*td/3)"2/((tf/3)"2)*100%D对F的贡献度gj=(dj*tj*pi/540)"2/((tf/3)"2)*100%theta对F的贡献度%代入数据进行计算h二30*30theta=pi/2th二0.01td=0.02tj=0.000278*pif=eval(f)dh=eval(dh)dd=eval(dd)dj=eval(dj)vf=eval(vf)tf二eval(tf)gh=eval(gh)gd=eval(gd)gj=eval(gj)MATLAB计

3、算结果如下：theta=thetaf=d/2-h+d/(2*sin(theta/2))dh=-1dd=1/(2*sin(theta/2))+1/2dj=-(d*cos(theta/2))/(4*sin(theta/2)2)1/2厂2)/9vf=(th"2/9+(tcT2*(l/(2*sin(theta/2))+(pi"2*cT2*tj"2*cos(theta/2)"2)/(4665600*sin(theta/2)"4))"(1/2)tf=3*(th八2/9+(tcT2*(l/(2*sin(theta/2))+1/2)2)/9+(pi"2*d“2*tj"2*cos(theta/2)"2)/

4、(4665600*sin(theta/2厂4))"(1/2)gh=(100*th^2)/(9*(th^2/9+(td，2*(l/(2*sin(thcta/2))+1/2厂2)/9+(pi*2*d2*tj"2*cos(theta/2)2)/(4665600*sin(theta/2)4)))gd=(100*td"2*(l/(2*sin(theta/2))+1/2厂2)/(9*(th，2/9+(td"2*(1/(2*sin(theta/2))+1/2厂2)/9+(pi*2*d*2*tj*2*cos(theta/2)*2)/(4665600*sin(theta/2)*4)))(pi"2*d2*t

5、j2*cos(theta/2)2)/(46656*sin(theta/2)4*(th2/9+(t(T2*(l/(2*sin(theta/2))+1/2厂2)/9+(p「2*cT2*tj"2*cos(theta/2)"2)/(4665600*sin(theta/2)*4)))h=30d=30theta=1.5708th=0.0100td=0.0200tj=&7336e-004f=6.2132dh二-1dd=1.2071dj=-10.6066vf二0.00870.0261gh二14.6441gd85.35210.0038由计算可得如下结果:F对II的敏感度F对D的敏感度F对theta的敏感度

6、F的偏差H对卜'的负•献度D对F的贡献度theta对F的贡献度F二6・2132±0.0261dh=-ldd=l.2071dj=-10.6066tf=0.0261gh=14.6441gd二85.3521gj=0.0038直方图如下:1、h,d,a的标准差对07的敏感度程序：X=[dh,dd,dj]Bar(x)图:XSQ£s®回fgFileEditviewInsertToolsDesktopWindowHelpDE4fe+二空列”復□S_■QONotenewtoolbarbuttons:databrushina&linksplotsPlm”vidEO20・2・4・6■8-10-12123H的

7、标准差D的标准差theta的标准差2、h,d,a的标准差对＜yF的贡献度程序：y=[gh,gd,gj]bar(y)图:801K的标准差X注QaJ訓聖丽FileEditviewInsertToolsDesktopWindowHelp口曰£釦弐黔⑧退"・

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9、■旦ONotenewtoolbarbuttons:databrushing&link"plotsPlm”vidEO90706050403020102D的标准差3theta的标