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ID:9855427
大小:1.06 MB
页数:30页
时间:2018-05-12
《课程设计--二级圆锥-圆柱齿轮减速器》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、29机械设计基础课程设计计算说明书设计题目:二级圆锥-圆柱齿轮减速器设计者:学号:同组者:学院:专业班级:指导教师:29二○一四年○六月二十一日目录一、设计任务书················································································2二、总体设计计算·············································································41.电机型号选择2.各级传动比分配3.
2、各轴的运动参数和动力参数计算(转速、功率、转矩)三、传动机构设计计算·····································································61.直齿圆柱传动2.圆锥齿轮传动四、轴系零件设计计算·····································································101.输入轴的设计计算2.中间轴的设计计算3.输出轴的设计计算29五、滚动轴承的选择与寿命校核计算··················
3、···························20六、键连接的强度校核计算··························································23七、润滑和密封方式的选择····························································24八、箱体的设计···············································································25九、附件
4、的结构设计和选择····························································25十、设计总结···················································································26参考文献···························································································2729一、设计任务书
5、1、二级圆锥-圆柱设计方案(1)已知条件:输送带牵引力F=3500N带速V=1.7m/s卷筒直径D=200mm(2)整体方案如下:图1-1二级圆锥-圆柱齿轮减速器整体外观参考图29图1-2二级圆锥-圆柱齿轮减速器内部结构参考图图1-3二级圆锥-圆柱设计运动方案简图29二、总体设计计算1、电机型号选择(1)电动机类型选择:Y系列三相异步电动机(2)电动机功率计算:输出功率:P输出=F×V/1000=5.95KW按《机械设计基础课程设计》P7表2-4取η联轴器=0.99轴承的效率的确定:圆锥齿轮处选用圆锥滚子轴承直齿圆柱处选用圆锥滚
6、子轴承按《机械设计基础课程设计》P7表2-4取η轴承=0.98圆锥齿轮效率的确定:按《机械设计基础课程设计》P7表2-4取η锥齿=0.96直齿圆柱齿轮效率的确定:按《机械设计基础课程设计》P7表2-4取η圆柱=0.97传动装置的总效率:η总=η3轴承×η2联轴器×η锥齿×η圆柱=0.993×0.992×0.95×0.97=0.85电机所需的工作功率:P工作=P输出/η总=7.02KW(2)确定电动机转速卷筒转速:n筒=60×1000V/πD=60×1000×1.7/(π×200)=162.33r/min按《机械设计基础课程设计》
7、P4表2-1推荐的传动比合理范围,取圆柱齿轮传动一级减速器传动比范围i’a=3~6。取圆锥传动比i’1=2~3,则总传动比理时范围为i’a=6~18。故电动机转速的可选范围为n’d=i’a×n2=(6~18)×162=972~2916r/min,符合这一范围的同步转速有1000和1500r/min。根据容量和转速,由有关手册查出有两种适用的电动机型号:因此有两种传动比方案:1000和1500r/min。综合考虑电动机传动平稳和传动装置尺寸、重量、价格和带传动、减速器的传动比,可见第2方案比较适合,所以选同步转速为1000r/mi
8、n 。(3)确定电动机型号按手册P167表12-1,选用Y160M-6型三相异步电动机。Ped=7.5KW,n1=970r/min,额定转矩2质量119kg。2、计算传动装置总传动比和分配各级传动比(1)传动装置总传动比:i总=n1/n筒=6(2)分配各级传动比
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