一级链传动设计怎样做呀,急求!!!

2025-03-01 22:14:16
推荐回答(1个)
回答1:

链传动设计计算
一、原始数据
传递的功率P,转速n1、n2(或n1、传动比i),原动机种类、载荷性质、传动用途等。
二、设计计算内容
链轮齿数、链节距、传动中心距、链节数、链轮毂孔直径、压轴力等
三、设计步骤和方法
设计类型 中、高速(v>=0.6m/s)链传动的设计
[步骤] [内容(按功率曲线设计)]
1 ◇假定链速,按表3选择小链轮齿Z1
◇确定从动轮链轮齿数Z2=Z1n1/n2 (Z2必须≤120)
2 ◇按表4取工作情况系数KA
◇确定计算功率:Pca=KAP
3 ◇按表5查取小链轮齿数系数KZ、链长系数KL;按表6查取多排链系数Kp(查Kz、KL要先估计工作点在功率曲线顶点的左侧还是右侧)
◇计算单根链条所需的额定功率P0
P0=Pca/(KZKLKp)
4 ◇按图1(功率曲线)查取链节距p(同时核实原工作点位置的估计是否合适)
◇按图2确定润滑方式
5 ◇初定中心距ao=(30-50)p
◇计算链条长度(链节数)Lp,圆整并尽量取偶数

6 ◇计算理论中心距

◇计算保持合适的安装垂度所需的中心距减小量△a=(0.002-0.04)a
◇确定实际安装中心距a'=a-△a
7 验算链速,核实原假定是否恰当

8 ◇按表7确定链轮各部分尺寸
◇按表8验算小链轮榖孔直径dkmax
9 ◇确定链传动有效圆周力:Fe=1000Pca/v
◇取压轴力系数:KFP=1.15(水平传动)或1.05(垂直传动)
◇计算压轴力:
Fp≈KFPFe
10 写出滚子链标记:链号-排数×整链节数 标准号

设计类型 低速(v<0.6m/s)链传动的设计
[步骤] [内容(按静强度设计)]
1.2 同中、高速链传动的设计步骤1.2
3 估取链节距p(无法估取时,可参考上述步骤3初定一个节距p)
4 计算链的有效圆周力:Fe=1000Pca/v
5 ◇按表1查取单位长度链条质量q
◇计算链的离心拉力:
6 ◇确定中心距a(方法同中、高速链传动的设计步骤5、6)
◇取两轮中心线与水平面的夹角α
◇按图3查取垂度系数Kf
◇计算链的悬垂拉力Ff,取以下两式中的大者:

7 计算链的紧边拉力F1=Fe+Fc+Ff
8 ◇选择静强度许用安全系数[S]=4-8,令:

◇计算单排链极限拉伸载荷Flim,按表1检验原估计的链号是否合适
9 按图2确定润滑方式

四、设计计算说明
1、小链轮齿数Z1
小链轮的齿数可根据链速按表3选择。Z1少可减小外廓尺寸,但齿数过少,将导致:
1)传动的不均匀性和动载荷增大;
2)链条进入和退出啮合时,链节间的相对转角增大,铰链磨损加剧;
3)链传动的圆周力增大,从而加速了链条和链轮的损坏。
增加小链轮齿数对传动有利,但如Z1选得太大时,大链轮齿数Z2将更大,除增大了传动的尺寸和质量外,还易发生跳齿和脱链,使链条寿命降低。链轮齿数的取值范围为17≤Z≤120。由于链节数通常是偶数,为考虑磨损均匀,小链轮齿数一般应取奇数。Z2=iZ1,通常限制链传动的传动比i≤6,推荐的传动比i=2~3.5。

2、工作情况系数
查表4,当工作情况特别恶劣时,值较表值要大得多。
3、链的节距
链的节距越大,承载能力就越高,但传动的多边形效应也要增大,振动冲击和噪声也越严重。所以设计时应尽量选取小节距的单排链或多排链。链条节距p可根据功率P0和小链轮转速n1由额定功率曲线选取。
4、修正系数
式P0=Pca/(KZKLKp)表明单排链的额定功率为P0KZKLKP,这是考虑到链传动的实际工作条件与标准实验条件的不同而引入系数KZKL和KP对P0进行修正。
5、链传动的中心距和链节数
中心距过小,链速不变时,单位时间内链条绕转次数增多,链条曲伸次数和应力循环次数增多,因而加剧了链节距的磨损和疲劳。同时,由于中心距小,链条在小链轮上的包角变小,在包角范围内,每个轮齿所受载荷增大,且容易出现跳齿和脱链现象;
中心距过大,会引起从动边垂度过大,传动时造成松边颤动。因此在设计时,若中心距不受其它条件限制,一般可初选a0=(30~50)p,最大取a0max=80p。
6、小链轮毂孔最大直径
根据小链轮的节距和齿数由链轮毂孔直径表确定链轮毂孔的最大直径dkmax,若dkmax小于安装链轮处的轴径,则应重新选择链传动的参数(增大Z1或p)。
7、设计计算类型
对于链速v<0.6m/s的低速链传动,因抗拉静力强度不够而破坏的几率很大,故常按下式进行抗拉静力强度计算。

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