《机械设计基础》课程教学资源（PPT课件）第六章 齿轮传动 6.23齿轮设计例题

机械设计基础CFoundationof Machine Design武汉理工大学齿轮设计例题主讲：罗齐汉武汉理工大学物流工程学院机械设计与制造系

齿轮设计例题 主讲：罗齐汉 武汉理工大学 物流工程学院 机械设计与制造系 机械设计基础 Foundation of Machine Design

7例：已知单级闭式斜齿圆柱齿轮传动。P=10Kw，nj=1210r/min，i=4.3，电动机驱动，中等冲击载荷，双向传动，单班制工作，要求工作寿命为5年，试设计此单级斜齿轮传动。解：1.确定许用应力取小齿轮用40Cr调质大齿轮用45号钢调质表6-5：40Cr调质，HBW,=241~286，取27045号钢调质，HBWz=217~255，取230图6-31c:Hliml=720MPaC Hlim2 = 580MPa图6-33cC Fliml = 300MPaO Flim2 = 220MPa

例：已知单级闭式斜齿圆柱齿轮传动。P1=10Kw， n1=1210r/min，i=4.3 ，电动机驱动，中等冲 击载荷，双向传动，单班制工作，要求工作寿 命为5年，试设计此单级斜齿轮传动。 表6-5： 40Cr调质，HBW1=241~286，取270 45号钢调质，HBW2=217~255，取230 解： 1.确定许用应力 图6-31c： 图6-33c  Flim1 = 300MPa  Flim2 = 220MPa  H lim1 = 720MPa 取小齿轮用 40Cr调质 大齿轮用45号钢调质  H lim2 = 580MPa

循环次数：Nz=60nrt，=60×1210×1×5×300×8=8.712×NL2 = NLi/i= 2.03×108图6-32(曲线1)ZNTi =1.15ZNT2 =1.24YST = 2YNT1 = YNT2 = 1图6-34SFmin =1.25安全系数（一般可靠度）：SHmin=1.1720Hliml[oμ] = 21.15 = 752.7MPaVTSHmin1.1580Hlim2 ZNT2[oμ :<1.24=653.8MPa1.1SHmin300x2CFlimIYsTLY[o ].NT1X0.7x1×0.7=336MPaSFmin1.25220x2O Flim2YsT2 YN[o, VT2 ×0.7:x1x0.7= 246.4MPaSFmin1.25

图6-34 ZNT1 =1.15 YST = 2 SFmin =1.25 图6-32 (曲线1) YNT1 = YNT 2 =1 SH min =1.1   Y MPa S Y NT F F S T F 1 0.7 336 1.25 300 2 1 0.7 min lim1 1 1   =  =  =     Z MPa S NT H H H 1.15 752.7 1.1 720 1 min lim1 1 = =  =     Z MPa S NT H H H 1.24 653.8 1.1 580 2 min lim2 2 = =  =   循环次数： 8 NL1 = 60nrth = 601210153008 = 8.71210 8 NL2 = NL1 /i = 2.0310 ZNT 2 =1.24 安全系数（一般可靠度）：   Y MPa S Y NT F F S T F 1 0.7 246.4 1.25 220 2 2 0.7 min lim2 2 2   =  =  =  

C软齿面,2.3按接触强度计算1）计算名义转矩T10T =9550P= 9550x78.93N.m=78930N.mm1210n2）取载荷系数K表6-6，电动机驱动，中等冲击，K=1.2~1.6，取1.43）初选参数Z = 21, Z2 =iZ, = 4.3×21 = 90.3 = 90实际传动比：i。=90/21=4.29传动比误差：i-io—4.3-4.29 0.2% <2.5%4.3i可用

N m N m m n P T = =  = 78.93  = 78930  1210 10 9550 9550 1 1 1 2）取载荷系数K 表6-6，电动机驱动，中等冲击，K=1.2~1.6，取1.4 3）初选参数 Z1 = 21，Z2 = iZ1 = 4.321= 90.3 = 90 实际传动比： i 0 = 90/ 21= 4.29 传动比误差： 0.2% 2.5% 4.3 0 4.3 4.29 =  − = − i i i 可用 2．软齿面，按接触强度计算 1）计算名义转矩T1

C齿宽系数（对称布置，软齿面）：Va =0.8~1.4 ， 取1.1。β=100u= z2 / z, =90/21= 4.29按齿面接触疲劳强度计算公式：3.17ZKTu+ld, ≥[oH hydu3.17×189.81.4 x 789304.29+11.14.29653.8= 47.16mm

齿宽系数（对称布置，软齿面）：  d = 0.8 ~ 1.4 ， 取1.1。 0  = 10 u = z2 /z1 = 90/ 21= 4.29 按齿面接触疲劳强度计算公式：   m m u Z KT u d H d E 47.16 4.29 4.29 1 1.1 1.4 78930 653.8 3.17 189.8 3.17 1 3 2 3 1 2 2 1 =        +        =       +           

Cd, = ud, = 4.29x×47.16 = 202.32mma = (d, +d)/2 =(47.16+202.32)/2 = 124.74mm取（取尾数为0或5的整数）a =125mm2acosβ2×125×c0s100=2.22mn = 2.5mmmn21+90Z, + Z2mn(Z + Z2) 2.5(21+90)cosB==1.11>12a2x125 a =145mm重取，取2.5(21+ 90)cos β= m,(Z +Z)=0.9572a2×145β =16.884° =16°53'01"若 β>20°Z,→Z,+1β<8° Z, →Zz-1

d2 = ud1 = 4.2947.16 = 202.32mm a = (d1 +d2 )/ 2 = (47.16+202.32)/ 2 =124.74mm 取 a =125mm （取尾数为0或5的整数） m m m Z Z a mn 2.22 n 2.5 21 90 2 cos 2 125 cos10 1 2 = = +    = + =  1.11 1 2 125 2.5(21 90) 2 ( ) cos 1 2 =   + = + = a mn Z Z  重取，取 a =145mm 0.957 2 145 2.5(21 90) 2 ( ) cos 1 2 =  + = + = a mn Z Z  0 0 ' ''  =16.884 =16 5301 20 1 8 2 2 1 0 2 2 0 若   Z → Z +   Z → Z −

64）计算几何尺寸2.5×21d, =m,Z：54.859mmcos β0.9572.5×90m.Z2d,=235.110mmcos β0.957dal = dl +2mn =54.859+2×2.5 = 59.859mmda2 = d2 +2mn = 235.110+2×2.5= 240.110mmdl =dl -2.5mn =54.859-2.5×2.5= 49.859mmdf2 = d2 -2.5mn = 235.110- 2.5×2.5= 230.110mmbz = yad, =1.1×54.859 = 60.345 =62mmb = bz +(5 ~ 10) = 70mm

4）计算几何尺寸 54.859 0.957 2.5 21 cos 1 1 mm m Z d n =  = =  b2 = d d1 =1.154.859 = 60.345 = 62mm b1 = b2 +(5 ~10) = 70mm 235.110 0.957 2.5 90 cos 2 2 mm m Z d n =  = =  da1 = d1 + 2mn = 54.859+ 22.5 = 59.859mm da2 = d2 + 2mn = 235.110+ 22.5 = 240.110mm d f 1 = d1 − 2.5mn = 54.859 − 2.52.5 = 49.859m m d f 2 = d2 − 2.5mn = 235.110 − 2.52.5 = 230.110m m

1.6KTcos BYFaYsa03.校核弯曲强度bm,Z,Zvi = Zj /cos3 β= 21/ cos3 β = 22.32Zv2 = Z2 /cos3 β = 90/cos3 β= 95.74由表6-8:YFal=2.76 YFa2=2.20Ysal = 1.56 Ysa2 =1.781.6KTcoS βYFalYsalOF1bm,Z,1.6×1.4× 789300.957×2.76x1.5662 ×2.52 ×21=89.525MPa89.525OF1YV×2.2×1.78=81.424MPaOFa21Sa2YFalY sal2.76×1.56CF1<[Fl ,CF2<[o,l弯曲强度足够

3．校核弯曲强度 Fa S a n F Y Y bm Z KT  cos  1.6 1 2 1 = / cos 21/ cos 22.32 3 3 ZV1 = Z1  =  = / cos 90/ cos 95.74 3 3 ZV 2 = Z2  =  = 由表6-8： YFa1 = 2.76 YSa1 =1.56 YSa2 =1.78 MPa Y Y bm Z KT F a s a n F 89.525 0.957 2.76 1.56 62 2.5 21 1.6 1.4 78930 cos 1.6 2 1 1 1 2 1 1 =       =  =  Y Y MPa Y Y F a S a F a S a F F 2.2 1.78 81.424 2.76 1.56 89.525 2 2 1 1 1 2   =  = =   ∵  F1   F 1 ，  F2   F 2 ∴弯曲强度足够 YFa2 = 2.20

4.齿轮的圆周速度校核Td,n,3.14×54.859×1210=3.474m/s60×100060×1000由表6-11，精度合格注意事项：1.中心距a取尾数为0或5的整数2.模数mn>1.5~2mm，且取标准值3.啮合型参数：分度圆、齿根圆、齿顶圆直径，取到小数点后3位，螺旋角要算至秒4．结构型参数：齿宽b，取整数

4．齿轮的圆周速度校核 m s d n v 3.474 / 60 1000 3.14 54.859 1210 60 1000 1 1 =    =  =  由表6-11，精度合格 注意事项： 1．中心距a取尾数为0或5的整数 2．模数 ，且取标准值 3．啮合型参数：分度圆、齿根圆、齿顶圆直径， 取到小数点后3位，螺旋角要算至秒 4．结构型参数：齿宽b，取整数 mn 1.5 ~ 2mm