2012年哈工大--单级圆柱齿轮减速器和一级带传动说明书.doc
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1、Manipulator is now used as a industrial robots in use, the control objectives often appear often in industrial automation. Industrial automation technology has gradually matured, as mature a technology line has been rapid development in industrial automation as a separate subject. Manipulator applic
2、ation began to filter into welding, logistics, mechanical processing, and other industries. Especially at high or very low temperatures, full of poisonous gases, high radiation case, robot in similar circumstances showed great use also brings great convenience to the staff. Precisely because of this
3、 robot to get peoples attention began to be a high degree of development. Labor rates, working conditions, labor intensive aspects of promoting development. Both at home and abroad to develop the PLC (programmable logic controller) is in various special circumstances and under special conditions set
4、 for mechanical devices. Now turned on the development of the microelectronics automatic control technology and the rapid development of the trains, the success of PLC hardware software and simulation control win big and successful development, now continues to develop as a factory automation standa
5、rds. Because robots are good development of the technology makes a good optimization of productive capital, and robot shows this unique advantages, such as: has good compatibility, wide availability, hardware is complete, and programming that can be mastered in a short time, so in the context of ind
6、ustrial PLC applications became ubiquitous. Manipulator in many developed country agriculture and industry has been applied, such as the use of mechanical harvesting large areas of farmland, repeated operations on the high-speed line that uses a robotic arm, and so on. Today, the high level of autom
7、ation combined with restrictions on the manipulator development level is slightly lower than the international. The design is mainly arm welding machine by PLC Automation control. This of design let designers on in school by learn of has a must of consolidation, understand has some usually didnt opp
8、ortunities awareness in world range within some leading level of knowledge has has must awareness, hope designers can in yihou of design in the can success of using in this design in the proceeds of experience 1.2 manipulator in both at home and abroad of research profile automation mechanical arm r
9、esearch began Yu 20th century medium-term, after years with with computer and automation technology of development, Makes mechanical arm on the Grand stage of industrial automation and shine, gradually became an industrial evaluation standards, and its importance can be seen. Now original robotic ar
10、m spent most of mass production and use on the production line, which is programmed robotic arm. As the first generation of manipulator position control systems main features, although not back several generations that can detect the external environment, but can still successfully complete like wel
11、ding, painting, delivery as well as for materials simple movements. Second generation mechanical arms are equipped with sensors and manipulators have the environment there is a certain amount of sense, when the mechanical arm is to use the program as a basis. Difference is that the robot begand目录第1章
12、 电动机的选择计算 -2第2章 传动装置运动、动力参数计 3第3章 齿轮及V带传动设计 4第4章 轴的设计及校核计算 7第5章 滚动轴承选择与寿命校核18第6章 键连接的选择及校核计算20第7章 减速器的附件及其说明-20第8章 箱体结构设计-20第9章 联轴器的选择 20第10章 润滑与密封22参考文献-22第1章 电动机的选择计算1.1 电动机类型和结构的选择因为本传动的工作状况是:中等冲击载荷、单向旋转。所以选用常用的封闭式Y系列三相异步电动机。1.2 电动机功率的选择1.2.1 传动装置的总功率由设计手册得,V型带传动的效率:0.96滚动轴承(一对)的效率:0.98圆柱齿轮(闭式)的效
13、率:0.97齿轮联轴器的效率:0.99绞车卷筒的效率为:0.96=v带2轴承齿轮联轴器卷筒=0.96*0.982*0.97*0.99*0.96=0.8671.2.2 电动机所需的工作功率: 1.3 确定电动机转速计算卷筒的工作转速:按手册推荐的传动比合理范围,取圆柱齿轮传动二级减速器传动比范围,作为总传动比范围,即I1 =840,故电动机转速的可选范围为:n筒=(840)76=6083040r/min符合这一范围的同步转速有750、1000、和1500r/min。根据容量和转速,由有关手册查出有三种适用的电动机型号:因此有三种传动比方案:通过综合考虑电动机和传动装置尺寸、重量、价格和带传动、减
14、速器的传动比,选择n=750r/min。1.4 电动机型号的确定根据以上选用的电动机类型,及所需的额定功率及同步转速,选定电动机型号为Y132M-8。主要性能:额定功率:3KW,满载转速710r/min,额定转矩2.0。主要性能如下表:电动机型号额定功率/kW满载转速/(r/min)起动转矩 额定转矩最大转矩 额定转矩Y132M-83.07102.02.0第2章 传动装置运动、动力参数计算2.1 总传动比的确定由已知数据可知:2.2 分配各级传动比查阅资料可知,单级减速器i=36合理,由i齿轮=1.4i带,计算得齿轮i齿轮=3.62则有 ,可得: 2.3 计算各轴转速电动机轴 : 小齿轮轴 :
15、 大齿轮轴 : 卷 筒 轴 : 2.4 计算各轴的功率电动机轴 : 小齿轮轴 : 大齿轮轴 : 卷 筒 轴 : 2.5 计算各轴转矩电动机轴 : 小齿轮轴 : 大齿轮轴 : 卷 筒 轴 : 2.6 传动装置运动、动力参数汇总表表3.1 传动装置运动、动力参数计算轴名参数电动机轴小齿轮轴大齿轮轴卷筒轴转速r/min710275.27676输入功率P/kw32.372.282.23输入转矩T/Nmm3350082100传动比i2.583.621.0 效率0.920.950.95第3章 齿轮及V带传动设计3.1 v带设计计算(一)选择普通V带截型每天工作8小时,查表得:kA=1.1PC=KAP=1.
16、13=3.3KW由机械设计基础课本查表可知:选用A型V带;(二)确定带轮基准直径,并验算带速查表: 取dd1=140mmdd2=i带dd1(1-0.02)=2.581400.98=354mm带速V:V=dd1n1/601000=140710/(601000) =5.2m/s在525m/s范围内,带速合适。(三)验算传动比i带=2.58,i计算=dd2/dd1(1-0.02)=2.57传动比误差=(2.58-2.57)/2.58=0.38%1200(适用)(六)确定带的根数由课本查表可得: P0=1.41KW P0=0.09KW K=0.93 KL=1.01Z=PC/P=PC/(P1+P1)KK
17、L=3.3/(1.41+0.09) 0.931.01=2.34取Z=3,即采用三根V带(七)计算轴上压力查得q=0.1kg/m,单根V带的初拉力:F0=500PC/ZV(2.5/K-1)+qV2=5003.3/35.2(2.5/0.93-1)+0.15.22N=181.26N则作用在轴承的压力FQ: FQ=2ZF0sin1/2=23181.26sin155.44/2=1060.6N3.2齿轮传动的设计计算 (1)选择齿轮材料及精度等级 考虑减速器传递功率不大,且为闭式齿轮传动,所以齿轮采用软齿面。小齿轮选用45号碳素钢调质,齿面硬度为236HBS。大齿轮选用45号钢正火,齿面硬度190HBS;
18、齿轮选择8级精度,齿面精糙度Ra1.63.2m (2)按齿面接触疲劳强度设计齿轮材料及其力学性能材料牌号热处理方式硬度接触疲劳极限 弯曲疲劳极限45调质197286HBS550620410480由d176.6(kT1(u+1)/duH2)1/3确定有关参数如下:传动比i齿=3.62 取小齿轮齿数Z1=25,则大齿轮齿数:Z2=iZ1=3.6225=90.5,取Z2=90 实际传动比i0=90/25=3.6传动比误差:(i-i0)/i=(3.62-3.6)/3.62=0.552%2.5% (符合)齿数比:u=3.6取d=1.0 (3)转矩T1T1=9.55106P/n1=9.551062.37/
19、275.2 Nmm =8.21104Nmm (4)载荷系数k:取k=1.2 (5)许用接触应力HH= Hlim/SHHlim1=570MPa Hlim2=390MPaFlim1=220MPa HFlim2=170MPa通用齿轮和一般工业齿轮,按一般可靠度要求选取安全系数SH=1.0, SF=1.25 H1=Hlim1/SH=570/1.0Mpa=570Mpa H2=Hlim2/SH=390/1.0Mpa=390MpaF1=220/1.25=176MPa H2=170/1.25=136MPaHmin=minH1 , H2=390MPa故得:d176.6kT1(u+1)/duH2)1/3=76.6
20、1.282100(3.6+1)/1.03.639021/3mm = 71.9mm取d1=75mm模数:m=d1/Z1=75/25=3mm 取标准模数:m=3mm中心距:a=0.5m(Z1+Z2)=172.5mm分度圆直径:d1=m Z1=75mm d2=m Z2=270mm齿宽:b=dd1=75mm 大齿轮齿宽为75mm,小齿轮齿宽应比大齿轮齿宽长5-10mm,取小齿轮齿宽为80mm(6)校核齿根弯曲疲劳强度由齿轮齿数,查表得:YF1=2.72 YF2=2.22YF1/F1=0.0155 YF2/F2=0.0163由于YF1/F1 YF2/F2,故取YF2=2.22代入公式计算F=(2kT1/
21、bm2d1)YF2HF=(2kT1/bm2d1)YF2=25.92F2=136MPa因此齿根弯曲疲劳强度满足使用条件 (7)其他尺寸小齿轮:da1= d1+2ha1=81mm df=d1-2hf=67.5mm大齿轮:da2=d2+2ha2=276mm df=d2-2hf2=262.5mm由于大齿轮分度圆半径较大采用锻造毛坯的腹板式结构大齿轮的有关尺寸计算如下(需要根据后面轴的设计来确定大齿轮的详细参数):齿轮结构图如下:第4章 轴的设计及校核计算4.1 高速轴的设计计算:(一)高速轴4.1.1 按扭矩初算轴径选用45#调质,硬度197286HBS,取c=118考虑有键槽,将直径增大7%,则因此
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- 2012 哈工大 圆柱齿轮 减速器 一级 传动 说明书
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