急求一份 一级链式直齿圆柱齿轮减速器课程设计说明书

2025-03-16 16:18:52
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目 录

设计任务书……………………………………………………
一、传动方案的拟定及电动机的选择……………………………2
二、V带选择 ………………………………………………………4
三.高速级齿轮传动设计……………………………………………6
四、轴的设计计算 …………………………………………………9
五、滚动轴承的选择及计算………………………………………13
六、键联接的选择及校核计算……………………………………14
七、联轴器的选择…………………………………………………14
八、减速器附件的选择……………………………………………14
九、润滑与密封……………………………………………………15
十、设计小结………………………………………………………16
十一、参考资料目录………………………………………………16

设计题号:3
数据如下:
已知带式输送滚筒直径320mm,转矩T=130 N•m,带速 V=1.6m/s,传动装置总效率为ŋ=82%。
一、拟定传动方案
由已知条件计算驱动滚筒的转速nω,即
r/min
一般选用同步转速为1000r/min或1500r/min 的电动机作为原动机,因此传动装置传动比约为10或15。根据总传动比数值,初步拟定出以二级传动为主的多种传动方案。
2.选择电动机
1)电动机类型和结构型式
按工作要求和工作条件,选用一般用途的Y(IP44)系列三相异步电动机。它为卧式封闭结构。
2)电动机容量
(1)滚筒输出功率Pw

(2)电动机输出功率P

根据传动装置总效率及查表2-4得:V带传动ŋ1=0.945;滚动轴承ŋ2 =0.98;圆柱齿轮传动 ŋ3 =0.97;弹性联轴器ŋ4 =0.99;滚筒轴滑动轴承ŋ5 =0.94。
(3)电动机额定功率Ped
由表20-1选取电动机额定功率Ped =2.2kw。
3)电动机的转速
为了便于选择电动机转速,先推算电动机转速的可选范围。由表2-1查得V带传动常用传动比范围i1 =2~4,单级圆柱齿轮传动比范围i2 =3~6,则电动机转速可选范围为nd= nω•i1•i2 =573~2292r/min

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