二级直齿圆柱齿轮减速器1

1、目录1. 前言·················································

2、;····················12. 设计任务书····························&#

3、183;··································13. 传动方案的分析和拟定（富传动方案简图）···········

4、83;························14. 电动机的选择························

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9、数计算···············································3··&#

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16、83;································77. 轴的设计计算················

17、;·············································8高速轴最小轴径计算···&

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24、;········108. 滚动轴承的选择和计算·······································&#

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27、3;109. 联轴器的选择················································

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29、83;··················11·······························&

30、#183;·······················1110. 键连接的选择和计算·······················

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33、······················1211. 润滑方式、润滑剂牌号及密封装置的选择························

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38、·····1212. 其他技术说明（如对装配、拆卸安装和维护的注意事项及需要采取的重要措施等）···1213. 结束语·································

39、83;·································12参考资料（“资料编号，作者。书名。出版社单位所在城市：名出版单位，出版年月”）附图一：二级直齿圆柱齿轮减速器总装配图附图二：减速器低速轴零件图附图三：减速器低速级大齿轮零件图一、前言机械设计基础

40、这门课是机械类专业很重要的一门基础课程，我们平时上课是按部就班地按照学，学的比较散，没有真正地联系贯通也没有实际用过，而课程设计将我们学的知识整个联系起来，让我们用自身学到的知识去处理实际问题，也真正贯彻我校 “学以致用” 的校训二、设计任务书（附纸）三传动方案的分析和拟定 1. 组成：传动装置由电动机、减速器、工作机组成。2. 特点：齿轮相对于轴承不对称分布，故沿轴向载荷分布不均匀，要求轴有较大的刚度。方案图如下：四电动机的选择Ped的选择电动机功率Ped，设：工作机（卷筒）所需功率PW，卷筒效率W，电机至卷筒轴的传动总效率a（减速器效率）电机需要的功率Pd 计算如下：=0.99 查手册取，

41、对于Ped=4kW的电动机型号有四种：型号Y112M-2Y112M-4Y132M1-6Y160M1-8同步转速3000 r/min1500 r/min1000 r/min750 r/min满载转速2890 r/min1440 r/min960 r/min720 r/min已知卷筒转速为60r/min，二级减速器的总传动比合理范围是ia=825。所以：电动机转速为nd=ia·n=4801500 r/min，该范围内的转速有 750 r/min, 1000 r/min, 1500 r/min。方案电动机额定功率Ped（kW）电动机转速r/min电动机重量（kg）参考价格减速器传动比ia型

42、号同步转速满载转速 1Y112M-441500144043242Y132M1-64100096073163Y160M1-8475072011812通过比较和计算，知道如果选用方案1，则在后面分配传动比时将出现i1=5.58,不符合教材P249页直齿传动比要求（i<3,最大可达5）。2号方案相对而言，其重量轻，价格便宜，传动比适中，故选2号方案。由上表读出Ia=16，按浸油润滑条件考虑，取高速级传动比i1·i2则 ia=1.3 i2·i2=·i22所以i2=3.51，i1 =ia/i2=4.56.五、传动装置运动和动力参数计算n1=nd=960 r/minn2

43、= n1/ i1 = r/minn3= n2/ i2= r/min功率的计算电动机：Pd高速轴×中间轴低速轴将各轴的运动和动力参数列于下表上：电动机轴轴轴轴滚筒轴转速n/(r/min)960960210.53 59.98 59.98 功率P/(kW)3.73 3.58 3.44 3.37 扭矩T/(N·m)37.50 37.13 162.58 548.02 537.11 传动比i11六.传动零件设计计算按教材表11.8选择齿轮的材料为：小齿轮选用45钢调质，硬度为220250HBS；大齿轮选用45钢正火，硬度为170210HBS。因为是普通减速器，由教材表11.20选8级精

44、度，要求齿面粗糙度Raum。2.确定设计准则由于该减速器为闭式齿轮传动，而且两齿轮均为齿面硬度HBS小于等于350的软齿面，齿面点蚀是主要的失效形式。应现按齿面接触疲劳强度进行设计计算，确定齿轮的主要参数和尺寸，然后再按弯曲强度校核齿根的弯曲强度。因两齿轮均为钢质齿轮，可应用教材P228页（11.23）求出d1值。确定有关参数与系数：（1）转矩T1由上表可以读出T1 N·m=37130 N·mm(2) 载荷系数K直齿，圆周速度高，精度相对较低，齿宽系数大，齿轮在两轴承间非对称（3）齿数z1和齿宽系数根据教材P249，在闭式软齿面齿轮传动中，齿轮的承受能力主要决定于齿面接触疲

45、劳强度，齿轮的弯曲强度总是足够的，因此齿数可多些，推荐z1=2440小齿轮的齿数z1取为24。此处小齿轮的齿数z1取24，则大齿轮的齿数z2= i1·z1=×24=109.44，因z1和z2最好互质，所以z2取109。实际齿数比为齿数传动比误差=11.2。（4）许用接触应力，由教材P224页=1。=60×960×1×（15×250××N1=1, ZN2=1根据教材P222页式（11.15）计算齿面接触疲劳许用应力故=则m1=2mm4.主要尺寸计算由此处取b2=50mmb1= b2+（510）取b1为55mm根据教材

46、P230页，若能得出,则校核合格。确定有关系数与参数：（1）齿形系数（2）应力修正系数YS(3)许用弯曲应力F根据教材P222页式（11.16）可得由教材230页式（11.25）可得齿根弯曲强度较和合格。由表11.21可知，选8级精度是合适的。小齿轮：da1=d1+2ha=48+2×1×2=52mm，直径较小做成齿轮轴。df1=d1-2hf=48-2×（1+0.25）=43mm大齿轮：da2=d2+2ha=218+2×1×2=222mm，由于200mm<da2<500mm,所以采用腹板式结构。df2=d2-2hf=218-5=213

47、mm整理求得的参数a12=133 m12=2 z1=24 b1=55 d1=48 da1=52 df1=43z2=109 b2=50 d2=218 da2= 222 df2=2136.2低速级齿轮的传动设计计算（1）齿数z3和齿宽系数齿数z3取为29。则大齿轮的齿数z2= i1·z1=×29=101.79，因z3和z4最好互质，所以z4取102。实际齿数比为齿数传动比误差=11.2。（2）许用接触应力由教材图11.23查得，=1。=60××1×（15×250××N3=1.07, ZN4根据教材P222页式（11.1

48、5）计算齿面接触疲劳许用应力=则m34由此处取b2=75mmb3= b4+（510）取b1为80mm整理求得的参数:a34=163.75 m34=2.5 z3=29 b3=80 d3=72.5 da3=77.5 df3z4=102 b4=75 d4=255 da4=260 df4由于200mm<da4<500mm,所以采用腹板式结构。七轴的设计计算计算（1）选择轴的材料，确定需用应力由前部分计算得知，减速器传递的功率属于中小功率，对材料无特殊要求，故选用45钢并经调质处理。（1） 按扭转强度估算轴径（最小直径）根据教材P341页表16.2得C=118107。又由式（16.2）得mm

49、考虑到轴的最小直径处要安装联轴器，会有键槽存在，故需将估算直径加大3%5%，取为17.319.5，再根据联轴器轴径要求，由设计手册取标准直径d1min=20mm。最小轴径计算mm考虑到轴的最小直径处要安装联轴器，会有键槽存在，故需将估算直径加大3%5%，取为42.547.8，再根据联轴器轴径要求，再初算转矩要求，直径需要取大一点，则由设计手册取标准直径d1min=50mm轴的结构如图所示。(1)各轴段直径的确定d31：最小直径，和联轴器相连d31=d1min=50mmd32：密封处轴段，d32= d31 +2a，a为轴肩高度，通常a（0.070.1）d，得d32在60左右，此处取d32=59m

50、m,以便于下一步轴承的选取。d33：d33= d32+15 mm，此处安装轴承，由设计手则选取d33=60 mmd34：此段起过渡作用d34= d33+15 mm,此处取65mmd35：d35 =d34+15 mm应与齿轮相配取d35=70mmd36：此处安装轴承，所以d36=d33=60 mm(2)各轴段长度确定l 31：由Y联轴器长度可以选得l 31 =110mml32：由箱体结构、轴承端盖、装配关系等确定l32=30mml33：由滚动轴承、挡油盘及装配关系等确定l33=50mml34：由装配关系，箱体结构等确定l34=50mml35：由齿轮位置关系等确定l35=12mml36：由齿轮宽度

51、确定l36=73mml37：由箱体结构、装配关系滚动轴承、挡油盘等关系确定l37=50mm作用在齿轮上的圆周力： 径向力： 则求垂直面的支反力：NN计算垂直弯矩：N·M N·M求水平面的支承力计算、绘制水平面弯矩图： N·MN·M求合成弯矩图： N·M求危险截面当量弯矩：由图可知，C-C处截面最危险，因减速器单项运转，故可以认为转矩为脉动循环变化，修正系数为0.6。其当量弯矩为： N·MMpa，许用弯曲应力，则则可以得出，所以轴是安全的。（略）最小轴径计算最小轴径处安装滚动轴承，根据设计手则滚动轴承的标准可以得出d21=30mm，其他

52、各段尺寸根据两齿轮宽度及箱体尺寸和安装要求可分别得出d22=48mm，d23=10mm，d24=77mm，d25=45mm八、滚动轴承的选择和计算因为设计的齿轮都是直齿，所以轴向不受力，再结合速度情况，选用深沟球轴承。高速轴安装轴承的轴段直径为25mm，由设计手则查得应选用深沟球轴承6205 GB/T276-94，中间轴安装轴承的轴段直径为30mm，由设计手则查得应选用深沟球轴承6206 GB/T276-94。选择：高速轴安装轴承的轴段直径为60mm，由设计手则查得应选用深沟球轴承6212 GB/T276-94校核：因为所以N应选1.1。受轴向载荷验算轴承寿命：h所选轴承具有足够的寿命。九、联轴器的选择(1)选择类型电动机与减速器高速轴联结用的联轴器，一般选用弹性可移式联轴器，所以此处与高速轴联接的联轴器选弹性柱销联轴器。（2）求计算转矩前面已经求出电动机的扭矩Td= N·M，由教材P435表19.1查得，工作情况系数K