文档介绍：目录
第一章设计任务········································3
第二章机构工作原理···································3第三章凸轮的设计
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第四章其他零件的设计
齿轮的设计·································
10
心轴的设计·································11
滚子轴的设计·······························11
圆柱凸轮轴的设计···························11
电机的选型·································12
减速器的选型·······························12
第五章强度校核
齿轮的校核··································12
轴的校核····································13
键的校核
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总装配图·········································14
总结·············································15
参考文献·········································16
工程图···········································16
致谢·············································19
第一章
设计任务
图1
结构组成——往复运动空间凸轮-齿条复合机构由1机架、2转盘、3齿条、4导轮、5托轮、6输出齿轮、7导轮、8托轮、9空间凸轮组成。
已知:
外扭矩
40kN·m
大转盘直径d
冲程
6m
冲次
4次/min
第二章
工作原理
外扭矩作用在转盘2上,转盘2与输出齿轮6同轴,并一起相对固定的机架1转动;空间凸轮9通过联轴器、减速器与驱动电机相连,进行动力输入;空间凸轮9转动的同时,推动其槽内与齿条3固结的导销沿空间凸轮9上的导槽运动;空间凸轮9上的导槽是封闭曲线,因此齿条3带着输出齿轮6往复运动,即实现了转盘1的往复回转运动。
第三章
凸轮设计
原因:柔性冲击,满足速度和力的需求,应用于中、低速重载。
余弦加速度规律,位移规律:
S=h[1-cos(πσ/σ0)]/2
代入升程角160°
S=h/2[1-cos(πσ/(160°×π/180o))]
S=h/2[1-cos()]
空间圆柱凸轮外表面展开曲线
S1=h/2[1-cos()]
S11=h/2sin()×
= sin()
S11=(×π/4)=tan38o=
得 h=,.
其中:最大压力角38o (满足力传动要求)。
远、近休止角各20o
由此得齿轮直径
h/d=6/ d=
得齿条受力
=40kN·m
F条=308kN
受力分析得:
Fx/308=tan380 Fx=
M凸=×
=·m
P=×=
圆柱凸轮的外圆柱半径应大于满足压力角条件下(α﹤[α]的许用最小外圆半径[r])
[r]=f·H /Φ
H——从动件最大升程(mm)
Φ——对应于最大升程时凸轮的转角(弧度)
f——凸轮的尺寸系数。