四己知一对外啮合标准安装的渐开线标准齿轮传动,a=20°,h,=1,c=0.25,模数 m=5mm,Z1=20,Z2=40,试求: (1)传动比i12和中心距a。 (2)两轮的分度圆半径,齿顶圆半径,齿根圆半径,基圆半径,节圆半径。 (3)两轮的齿顶圆压力角和齿顶圆曲率半径 (4)若将两轮的中心距加大至a!=152mm,问此时的啮合角a为多大? (5)标准安装时的实际啮合线长度和基圆齿距。 解: 2 40 2 12 (x1+z2)=×(20+40)=150mm 2 r;=mz1/2=5×202=50mm r2=mz2/2=5×40/2=100mm ra1=m(z1/2+h12)=5×(202+1)=55mmra2=5×(402+1)=105mm rn=m(z1/2-hac)=5×(20/2-140,25)=4375mmrn=5×(402-1-0.25)93.75mm r=r, cosa=50cos200=46.98mm bn2=100c0s209397mm r1,=r1=50mm r2,=r2=100mm (3)a1=cos4(rb11)=cos1(46.98/55=31330a2=c0s1(93:97105)=26.50 p1=rn1×tg(3133)=2860mmp2=rb2×tg(26.50)=4685mm (4) acos=a'cosa’150c0s20-152csa,a’=2198 (5)Pb=mcos20=14.76mm Ea=(1(goxr1goa)+4tga2-g)(2n)=(20×1g3131g20(9265020)(2m=1.636 B1B2=pb6a=14.76×1636=24.15mm四 解: 1 2 12 z z i = (20 40) 150mm 2 5 2 =  + = 20 40 = = ( ) 2 1 2 z z m a = + r1=mz1 /2=520/2=50mm r2=mz2 /2=540/2=100mm ra1=m(z1 /2+ha * )=5(20/2+1)=55mm ra2=5(40/2+1)=105mm rf1= m(z1 /2-ha * -c * ) =5(20/2-1-0.25)=43.75mm rf2=5(40/2-1-0.25)=93.75mm rb1=r1 cosa=50cos200=46.98mm rb2=100cos200=93.97mm r1’= r1=50mm r2’= r2=100mm (1) (2) (3) acosa= a’cosa’ 150cos200=152cosa’ a’ =21.98 (4) 0 aa1=cos-1 (rb1/ra1)=cos-1 (46.98/55)=31.330 aa2=cos-1 (93.97/105)=26.500 pb=pmcos200 =14.76mm ea=(z1 (tgaa1-tga)+z2 (tgaa2-tga))/(2p)=(20(tg31.330 -tg200 )+20(tg26.500 -tg200 ))/(2p)=1.636 B1B2= pb ea=14. 76 1.636=24.15mm r1=rb1×tg(31.330 )=28.60mm r2=rb2×tg(26.500 )=46.85mm (5)