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单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,10-2 设有一渐开线标准齿轮 z=20，m=8mm，=20，h,a,*,=1，试求：1)其齿廓曲线在分度圆及齿顶圆上的曲率半径、,a,及齿顶圆压力角,a,；2)齿顶圆齿厚 s,a,及基圆齿厚 s,b,；3）若齿顶变尖( s,a,=0)时，齿顶圆半径 r,a,又应为多少？并完成图示各尺寸的标注。,解：,1）,求,、,a,、,a,d = m z =820=160(mm),d,a,= d +2 h,a,=(z+2h,a,*,)m=(20+21)8= 176(mm),d,b,= d cos,= 160cos20= 1600.9397= 150.35(mm),= r,b,tan,= 75.175 tan20= 75.1750.3640= 27.36(mm),a,=arccos( r,b,/r,a,) = arccos0.8543=31.32,a,=,r,b,tan,a,=75.175tan31.32= 75.1750.6085=45.74(mm),2）,求,s,a,及,s,b,s,a,=s(r,a,/r),-,2r,a,(inv,a,-,inv,)=,8/2,88/80,-,176(inv31.32,-,inv20),= 5.55(mm),inv31.32= tan31.32,-,31.32,inv20= tan20,-,20,/180=0.3640,s,b,=s(r,b,/r),-,2r,b,(0,-,inv,)= cos,(s+mz,inv,),=14.05(mm),3）,求当,s,a,=0,时的,r,a,令,s,a,=s(r,a,/r),-,2r,a,(inv,a,-,inv,)= 0, inv,a,= s/2r + inv,a,= 35.48,r,a,= r,b,/ cos,a,=75.175/ cos 35.48,=92.32(mm),10,-,3,试问渐开线标准齿轮的齿根圆与基圆重合时，其齿数,z,应为多少？又当齿数大于以上求得的齿数时，基圆与齿根圆哪个大？,解：,d,b,= d cos,=,m zcos,d,f,= d,-,2 h,f,=(z,-,2h,a,*,-,2 c,*,)m,由,d,f,d,b,，,有：,z2(h,a,*,+,2 c,*,) / (1,-,cos,) =41.45（,不能圆整,）,当齿根圆与基圆重合时,，z=41.45；,当,z 42,时，齿根圆大于基圆。,10,-,5,已知一对渐开线标准外啮合圆柱齿轮传动，其模数,m=10mm,， 压力角,=20，,中心距,a=350mm，,传动比,i,12,=9/5，,试计算这对齿轮传动的几何尺寸。,解：,1）,确定两轮的齿数,a = m ( z,1,+z,2,) /2 = 10 ( z,1,+z,2,) /2 =350,i,12,= z,2,/ z,1,= 9/5,联立解得：,z,1,=,25,z,2,=,45,2）,计算两轮的几何尺寸,分度圆直径：,d,1,= m z,1,= 250 d,2,= m z,2,= 450,齿顶圆直径：,d,a,1,= m(z,1,+2h,a,*,) = 270 d,a,2,= m(z,2,+2h,a,*,) = 470,齿根圆直径：,d,f,1,= m(z,1,-,2h,a,*,-,2 c,*,) = 225 d,f,2,= m(z,2,-,2h,a,*,-,2 c,*,) = 425,基 圆直径：,d,b,1,= m z,1,cos,=234.92,d,b,2,= m z,2,cos,齿 全 高：,h = h,a,+ h,f,=(2h,a,*,+ c,*,)m = 22,.5,齿顶、根高：,h,a,= h,a,*,m = 10 h,f,= (h,a,*,+ c,*,齿 距：,齿厚、槽宽：,s = e = p / 2=,基 圆 齿 距：,p,b,= p cos,=,m cos,节 圆直径：,d,1,= d,1,= 250 d,2,= d,2,= 450,10,-,6,已知一对标准外啮合直齿圆柱齿轮传动，=20，m=5mm， z,1,= 19，z,2,= 42，试求该传动的实际啮合线B,1,B,2,的长度及重合度,。如果将中心距a加大直到刚好能连续传动（,=19 ），试求此种情况下传动的啮合角、中心距a、两轮节圆半径r,1,及r,2,、顶隙c及周向侧隙c,n,。,解：,1）,求,B,1,B,2,及,d,b1,= m z,1,cos,=,5,d,b2,= m z,2,cos,=,5,d,a,1,= m(z,1,+2h,a,*,) = 5,(19+2,1) = 105,d,a,2,= m(z,2,+2h,a,*,) = 5,(42+2,1) = 220,a1,=arccos(d,b1,/d,a1,105) =31.77,a2,=arccos(d,b2,/d,a2,220) =26.23,= z,1,(tan,a1,-,tan,)+z,2,(tan,a2,-,tan,) /(2),=,= 19(tan31.77,-,tan20) +42,(tan26.23,-,B,1,B,2,=,p,b,= ,m cos,=,5,2）,当刚好能连续传动时,= z,1,(tan,a1,-,tan,)+z,2,(tan,a2,-,tan,) /(2),= 19(tan31.77,-,tan,) +42,(tan26.23,-,tan,) /(2) =1,解得：,=23.229,a= a cos,/,cos,= m ( z,1,+z,2,) cos,/(2cos,),=5,( 19+42) cos20/ (2cos23.229) =155.945(mm),r,1,= r,1,cos,/,cos,= m z,1,cos,/(2cos,),=5,19cos20/ (2cos23.229) =48.573(mm),r,2,= a,-,r,1,= 155.945,-,48.573 = 107.372(mm),c = a,-,a + c = 155.945,5 = 4.695(mm),c,n,= p,-,(s,1,+ s,2,) = 2a(inv,-,inv,),= 2,155.945 (inv23.229,-,inv20),= 2.767(mm),10,-,7 图示为以,l,= 1mm/mm绘制的一对渐开线标准齿轮传动，设轮1为原动件，轮2为从动件，两轮的转向如图所示，现要求：,1）据图上尺寸，确定两齿轮的基本参数(m、h,a,*,、z,1,及z,2,)；,2）标出两齿轮的齿顶圆、齿根圆、基圆、分度圆、节圆和分度圆压力角及啮合角；,3）标出理论啮合线N,1,N,2,、开始啮合点、终止啮合点及实际齿廓工作段（标在K点处齿廓上）；,4）标出实际啮合线B,1,B,2,及基圆齿距p,b,，并估算重合度,。,解：1）,作O,1,N,1,啮合线，量得= 20；,量得r,1,=72 ， r,2,=108，p,b,=23.5 ，,h,a,=8， h,f,=10,由p,b,= m cos得m =7.96，,取标准值m =8；,由r,1,= m z,1,/2，得z,1,=18；,由r,2,= m z,2,/2，得z,2,=27；,由h,a,= h,a,*,m，得h,a,*,= 1；,由h,f,= (h,a,*,+ c,*,) m ，得c,*,=0.25。,2）如图,3）如图,4）量得B,1,B,2,= 38， p,b,= B,1,B,2,/ p,b,10,-,9,已知一对外啮合变位齿轮传动，z,1,= z,2,= 12， m=10mm， = 20 ，m=5mm，h,a,*,= 1 ， a= 130mm，试设计这对齿轮传动，并验算重合度及齿顶厚（ s,a,应大于0.25m，取x,1,= x,2,）。,解：,1）,确定传动类型,a = m ( z,1,+z,2,) /2 = 10 ( 12+12,) /2 =120 a= 130mm,故此传动应为,正,传动。,2）,确定两轮变位系数, = arccos(a cos,/,a) = arccos(120 cos20,/,130) = 29.83,x,1,+ x,2,= (z,1,+z,2,) (inv,-,inv,) / (2tan,),= (12+12) (inv29.83,-,取,x = x,1,= x,2,=,0.6245，x,min,= h,a,*,(z,min,-,z) / z,min,=1,(17,-,x x,min,3）,计算几何尺寸,分度圆分离系数：,y =,(a,-,a) / m = 1,齿顶高变动系数：,= x,1,+ x,2,-,y,齿 顶 高：,h,a1,= h,a2,=,(,h,a,*,+ x,-,),m= 13.755(mm),齿 根 高：,h,f1,= h,f2,=,(,h,a,*,+ c,*,-,x),m,= 6.255(mm),分 度 圆 直 径：,d,1,= d,2,= m z,1,=,120 (mm),齿 顶 圆 直 径：,d,a,1,= d,a,2,= d,1,+2h,a1,= 147.51 (mm),齿 根 圆 直 径：,d,f,1,= d,f,2,= d,1,-,2h,f1,= 107.49 (mm),基 圆 直 径：,d,b1,= d,b2,= d,1,cos,= 112.763(mm),分 度 圆 齿 厚：,s,1,= s,2,= (,/2 + 2 x tan,) m = 20.254 (mm),4）,检验重合度及齿顶厚,a1,=,a2,= arccos(d,b1,/d,a1,147. 51) = 40.13,= z,1,(tan,a1,-,tan,)+z,2,(tan,a2,-,tan,) /(2) = 1.0297 1,s,a1,=,s,a2,= s(r,a1,/r,1,),-,2r,a1,(inv,a1,-,inv,故可用。,10,-,11,设已知一对斜齿轮传动，z,1,=20， z,2,= 40， m,n,= 8mm， ,n,= 20，h,a,n,*,= 1 ，c,n,*,= 0.25， B = 30mm，并初取=15，试求该传动的中心距a（a值应圆整为个位数为0或5，并相应重算螺旋角）、几何尺寸、当量齿数和重合度。,解：,1）,计算中心距,初取,=15，,则,a = m,n,( z,1,+z,2,) /(2cos) = 8 ( 20+40) / (2cos 15),取,a = 250mm，,则,= arccosm,n,( z,1,+z,2,) /(2a),= arccos8 ( 20+40) /(2,250) =16.26,2）,计算几何尺寸及当量齿数,分度圆直径：,d,1,= m,n,z,1,/ cos= 166.67(mm) d,2,= 333.33(mm),齿顶圆直径：,d,a,1,= d,1,+2h,a,n,*,m,n,= 182.67(mm) d,a,2,= 349.33(mm),齿根圆直径：,d,f,1,= d,1,2(h,a,n,*,+,c,n,*,)m,n,=146.67(mm) d,f,2,= 313.33(mm),基 圆直径：,d,b,1,= d,1,cos,t,(mm) d,b,2,= 311.69(mm),齿顶、根高：,h,a,= h,a,n,*,m,n,= 8 (mm) h,f,= (h,a,n,*,+ c,n,*,) m,n,= 10(mm),法、端面齿厚：,s,n,=,m,n,/2=12.57(mm) s,t,=,m,n,/(2cos) =13.09(mm),法、端面齿距：,p,n,=m,n,=25.14(mm) p,t,=m,n,/,cos,=26.19(mm),当 量 齿 数：,z,v1,= z,1,/ cos,3,=22.61 z,v2,= z,2,/ cos,3,3）,计算重合度,t,=arctan(tan,n,/ cos) =,arctan(tan20/ cos16.26) =20.764,at1,= arccos(d,b1,/d,a1,) =31.447,at2,= arccos(d,b2,/d,a2,) =26.843,= z,1,(tan,at1,-,tan,t,)+z,2,(tan,at2,tan,t,) /(2),=20(tan31.447,-,tan20.764)+40(tan26.843,tan20.764)/(2),=B sin/m,n,=,+,10,-,13,在图示的各蜗杆传动中，蜗杆均为主动，试确定图示蜗杆、蜗轮的转向或螺旋线的旋向。,10,-,14,已知一对等顶隙直齿圆锥齿轮传动，z,1,=15， z,2,= 30， m,= 5mm， h,a,*,= 1 ，c,*,= 0.2， = 90，试确定这对圆锥齿轮的几何尺寸。,解：,分度圆锥角：,1,= arctan(z,1,/ z,2,)=26.57 ,2,=,1,=63.43 ,分度圆直径：,d,1,= m z,1,= 75(mm) d,2,= m z,1,=150(mm),齿顶圆直径：,d,a,1,= d,1,+2h,a,cos,1,= 83.94(mm) d,a,2,= 154.47(mm),齿根圆直径：,d,f,1,= d,1,2h,f,cos,1,=64.27(mm) d,f,2,= 144.63(mm),齿 顶 高：,h,a1,= h,a,*,m = 5(mm) h,a2,= 5(mm),齿,根 高：,h,f1,= (h,a,*,+ c,*,) m = 6(mm) h,f2,= 6(mm),顶 隙：,c = c,*,m = 1(mm),齿厚、槽宽：,s =,m /2 = 7.85(mm) e =,m /2 = 7.85(mm),锥距、齿宽：,R= m /2= 83.85(mm) B=R/3 28(mm),齿顶、根角：,a,=,f,= arctan(h,f,/ R) = 4,.09,齿顶圆锥角：,a,1,=,1,+,f,= 30,.66, ,a,2,=,2,+,f,= 67,.52,齿根圆锥角：,f,1,=,1,f,= 22,.48, ,f,2,=,2,f,= 59,.34,当 量 齿 数：,z,v1,= z,1,/ cos,1,=16,.77,z,v2,= z,2,/ cos,2,= 67,.08,
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