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1、齿轮传动习题,10-17 10-21 10-23 10-26 10-27 10-28 10-34,10-17,试确定图a所示传动中蜗轮的转向、螺旋线的旋向,及图b所示传动中蜗杆和蜗轮的螺旋线的旋向。,解:a),b),右旋,右旋,10-21,在右图中,已知基圆半径rb=50mm,现需求: 1)当rk=65mm时,渐开线的展角q k 、压力角ak、和曲率半径rk 2)当qk=50时,渐开线的压力角ak、和向径rk的值。,解:,1) ak=arccos(50/65)=39.720,qk=tg(ak)- ak=tg39.720- 39.72p/180 =0.137rad=7.870,rk=rbtg a
2、k =50tg 39.720=41.54mm,2) qk=50=0.087266rad,查渐开线函数表或解方程ak=34045,rk=rb/cos(34045)=50/cos(34045)=60.85mm,10-23.,设有一渐开线标准齿轮,z=26,m=3mm, ha*=1 , a=200,求其齿廓曲线在分度圆和齿顶圆上的曲率半径 及齿顶圆压力角。,解:,r=mz/2=326/2=39mm,rb= rcosa=39cos200=36.648mm,r= rsina=39sin200=13.339mm,分度圆半径:,齿顶圆半径:,分度圆渐开线曲率半径:,ra= rasinaa=42sin 29.
3、2410=20.516mm,齿顶圆渐开线曲率半径:,齿顶圆压力角:,基圆半径:,ra= r+mha*=39+31=42mm,aa=arccos(rb /ra )= arccos(36.648 /42 )= 29.2410,10-26,已知一对渐开线标准外啮合圆柱齿轮传动的模数m=5mm、压力角a=20、中心距a=350mm、传动比i12=95,试求两轮的齿数、分度圆直径、齿顶圆直径、基圆直径以及分度圆上的齿厚和齿槽宽。,解:,z2/z1=9/5,m(z2+z1)/2=350,z2/z1=9/5,5(z2+z1)/2=350,即,z1=50 z2=90,d1=mz1=550=250mm d2=m
4、z2=590=450mm,da1= d1+2mha*=250+251=260mm da2= d2 +2mha* =450+251=460mm,db1= d1cos200=250cos200=234.923mm db2= d2cos200 =450cos200=422.862mm,s=e=pm/2=3.14165/2=7.85mm,10-27,当渐开线标准直齿圆柱齿轮的齿根圆与基圆重合时,齿数为多少?当齿数大于该值时,基圆与齿根圆哪个大?,解:,齿根圆大于基圆时,mz/2-ha*m-c*m mzcosa/2,rf rb,当齿数大于41时,齿根圆大于基圆.,渐开线标准直齿圆柱齿轮的齿根圆与基圆重合
5、不可能重合,10-28,已知一对标准外啮合直齿圆柱齿轮传动的。a=200、m=5mm、z1=19、z2=42,试求其重合度ea。问当有一对轮齿在节点P处啮合时,是否还有其他轮齿也处于啮合状态;又当一对轮齿在Bl点处啮合时,情况又 如何?,解:,d1=mz1= 519=95mm d2=mz2=542=210mm,da1=d1+2mha*=95+251=105mm da2= d2+2mha*=210+251=220mm,db1=d1cos200=95cos200=89.27mm db2= d2cos200=215cos200=197.34mm,aa1=arc cos(db1/da1 )= arc
6、cos(89.27/105 ) =31.770 aa2=arc cos(db2/da2 )= arc cos(197.34/220 ) =26.230,ea=z1(tgaa1-tga)+ z2(tgaa2-tga)/(2p),=19 (tg 31.770 tg200 )+42(tg26.230 -tg200)/(2p) =1.633,重合度几何意义,重合度几何意义,C1,C2,C1C2为单齿啮合区,B1C2、B2C1为双齿啮合区,当1ea2时,10-34,设已知一对斜齿轮传动的 z1=20, z2=40, mn=8mm,an=200 ,b=150(初选),B=30mm,ha*=1,试求a(应圆
7、整,并精确计算b)、eg及zv1、zv2。,解:,a=mn(z1+z2)/2cosb=8(20+40)/2cos150=248.5mm,圆整a=250mm,b=arc cosmn(z1+z2)/2a=arc cos 8(20+40)/(2250)= 16.2600,mt=mn/cosb=8/cos 16.2600=8.33mm,at=arc tg(tgan/cosb)=arc tg (tg200/cos16.2600)=20.7630,d1=mtz1=8.3320=166.6mm,d2=mtz2=8.3340=333.2mm,da2=333.2+281=349.2mm,da1= d1 +2mn
8、han*=166.6+281=184.6mm,aa1= arc cos(d1 cosat/da1)=arc cos(166.6cos20.7630/184.6)=32.440,aa2= arc cos(d2 cosat/da2)=arc cos(333.2cos20.7630/ 349.2)=26.850,ea=z1(tgaat1-tgat)+ z2(tgaat2-tgat)/(2p),=20(tg 32.440 tg20.7630 )+40(tg26.850 -tg20.7630)/(2p)=1.626,eb=Bsinb/pmn=30sin 16.2600/(3.148)=0.334,eg= ea +eb= 1.626 +0.334=1.960,zv1= z1/cos3b=20/ cos3 16.2600 =22.6 zv1=40/ cos3 16.2600 =45.2,