带式输送机传动装置设计!!!感激不尽

2025-02-07 14:56:46
推荐回答(1个)
回答1:

题目:传动装置,减速机设计及相关零件加工。
一.总体布置简图
1—电动机;2—带式运输机;3—齿轮减速器;4—联轴器;5—滚筒
二.工作情况:
载荷平稳、单向工作
三.原始数据
滚筒的扭矩T(Nm):520
滚筒的直径D(mm):260
运输带速度V(m/s):1.2
带速允许偏差(%):5
使用年限(年):8
工作制度(班/日):2
四.设计内容
传动方案的拟定及说明
一个好的传动方案,除了满足机器的功能要求外,还应当工作可靠,结构简单,尺寸紧凑,传动效率高,成本低廉以及使用维护方便。对比材料中2-1所示带式输送机的四种方案,再经由题目所知传动机由于工作载荷平稳,工作环境有轻尘,布局尺寸没有严格限制,将带传动放在高速级,即可缓冲吸振又能减小传动的尺寸。具体方案见图1-1。
电动机的选择
1.电动机类型和结构的选择
因为本传动的工作状况是:载荷平稳、单向旋转。所以选用常用的封闭式Y系列的三相异步电动机。
2.电动机容量的选择
1) 工作机所需功率Pw
Pw=2.4kW
2) 电动机的输出功率
Pd=Pw/η
η= =0.895
Pd=2.682kW
3.电动机转速的选择
nd=(i1’ i2’…in’)nw
则V带传动η1=0.96
初选为同步转速为1000r/min的电动机
4.电动机型号的确定
由表20-2查出电动机型号为Y132S-6,其额定功率为3kW,满载转速960r/min。基本符合题目所需的要求
计算传动装置的运动和动力参数
传动装置的总传动比及其分配
1.计算总传动比
由电动机的满载转速nm和工作机主动轴转速nw可确定传动装置应有的总传动比为:
i=nm/nw
nw=88.19
i=10.89
2.合理分配各级传动比
取V带传动的传动比i1,取i1=2.7,则单级圆柱齿轮减速器的传动比为i2=i/i1= =4.03,
取i2=4
各轴转速、输入功率、输入转矩
项 目 电动机轴 高速轴I 低速轴II 滚筒W轴
转速(r/min) 960 384 88.19 88.19
功率(kW) 2.682 2.575 2.471 2.422
转矩(N?m) 26.68 64.04 245.813 262.276
传动比 4 4.356 4.356 1
效率 0.96 0.97 0.99 0.99

传动件设计计算
(1)材料选择以斜齿圆柱齿轮传动方式
小齿轮:材料45钢,调质处理HBS1=230
大齿轮:材料45钢(或者ZG310~570)正火处理,HBS2=190
(2)参数选择
1)齿数,取Z1=22,则Z2=I Z=4.356×22=95.832 取Z2=96
2)齿宽系数, 查表6-6 取�0�1d=0.7
3)载荷系数, 由于载荷为平稳,且采用硬齿面,齿轮应取大值故取K=1.2
4)初选螺旋角, 取β=10°
(3)确定许用应力
小齿轮查表6-3 [σH1]=515 MPa
[σbb1]=310 MPa
大齿轮查表6-3, [σH2]=500 MPa
[σbb2]=250 MPa
(4)计算小齿轮转距 T1=9.55×106 P1/n1=9.55×106×2.575/384N�6�1mm
=6.40×104 N�6�1mm
(5)按齿轮弯曲强度计算 齿形系数,查表6-7
按当量齿数ZV = ,得YFS1=4.24 YFS2=3.97
齿形系数与许用弯曲力的比值为
因 较大,故将此值代入

查表6-1, 取M=3

(6)计算齿轮几何尺寸
齿轮传动中心距为
将中心距取为 180 ,以便加工,则螺旋角调整为

取 。由于去的标准模数比计算值大,故 向小圆整后依然能满足要求。

(7)校核接触强度

根据数据知道强度足够。

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