资源描述
湖 南 科 技 大 学
课程设计报告
课程设计名称: 机械设计
学 生 姓 名: 瓮马杰
学 院: 机电工程学院
专业及班级: 材料成型及控制工程二班
学 号: 0803040227
指导教师: 廖先禄
2011年6月7日
内容摘要
本设计是两级圆锥-圆柱齿轮减速器的课程设计,根据设计任务书的相关要求,并结合自己的实习经验及课程上学习的理论知识来独立设计完成的。本文发扬了优秀课程设计的系统严密、数据精确、图标规范、文笔流畅、可读性好的优点。通过这一次的设计可以初步掌握一般简单机械的一套完整的设计方法,构成减速器的通用零件。
选用减速器时应根据工作机的选用条件,技术参数,动力机的性能,经济性等因素,比较不同类型、品种减速器的外廓尺寸,传动效率,承载能力,质量,价格等,选择最适合的减速器。与减速器联接的工作机载荷状态比较复杂,对减速器的影响很大,是减速器选用及计算的重要因素。
这次设计主要介绍了减速器的构成及设计参数,灵活并全面的运用了所学过的知识。并进一步培养了工程设计的独立能力,树立正确的设计思想掌握常用的机械零件,机械传动装置和简单机械设计的方法和步骤,要求综合的考虑使用经济、工艺等方面的要求。
设计中存在的不足请老师能给予意见和建议。
目录
一、设计基本参数··············································································1
1工况·························································································1
2.原始数据···················································································1
二、传动方案的拟定···········································································1
三、电动机的选择·················································································2
1.选择电动机的类型··········································································2
2.选择电动机功率············································································2
3.确定电动机转速············································································2
四、传动比的计算及分配··········································································3
1.总传动比····················································································3
2.分配传动比··················································································3
五、传动装置运动、动力参数的计算·······························································4
1.各轴转速·····················································································4
2.各轴功率·····················································································4
3.各轴转矩·····················································································4
4.结果参数·····················································································4
六、传动件的设计计算··············································································5
(一)高速级锥齿轮传动的设计计算·······························································5
1.选择材料·····················································································5
2.初步计算传动的主要尺寸·····································································5
3.确定传动尺寸·················································································6
4.校核齿根弯曲疲劳强度·······································································7
(二)、低速级斜齿圆柱齿轮的设计计算····························································8
1.选择材料······················································································8
2.按齿面接触疲劳强度进行设计································································8
3.按齿根弯曲强度进行设计····································································11
4.几何尺寸计算················································································13
七、轴的设计计算··················································································13
(一)输入轴(I轴)的设计······································································13
1.轴的设计计算················································································14
2.轴的结构设计················································································15
3.校核轴的强度·················································································16
(二)、中间轴(Ⅱ轴)的设计·····································································17
1.轴的设计计算·················································································17
2.轴的结构设计·················································································18
3.校核轴的强度·················································································20
(三)、输出轴(Ⅲ轴)的设计 ··································································21
1.轴的设计计算················································································· 21
2.轴的结构设计··················································································22
3.校核轴的强度··················································································24
八、轴承的校核·······················································································28
1输入轴滚动轴承计算············································································28
2中间轴滚动轴承计算············································································29
3输出轴轴滚动轴承计算···········································································30
九、键联接的选择及校核计算···········································································31
1.输入轴键计算····················································································31
2.中间轴键计算····················································································32
3.输出轴键计算····················································································32
十、联轴器的选择·······················································································33
十一、润滑与密封·······················································································33
十二、减速器附件的选择···············································································34
十二、设计体会·························································································34
十三、参考文献·························································································34
35
一、设计基本参数
1.工况:
设备工作环境有粉尘、常温连续工作,该传送设备的传动系统由电动机—减速器—运输带组成。工作寿命10年(设每年工作300天),连续工作每天24小时。
2.原始数据:
传送带拉力F(KN)
传送带速度V(m/s)
鼓轮直径D(mm)
使用年限(年)
2400
1.2
300
10
二、传动方案的拟定
运动简图 如图2-1:
图2-1 运动结构简图
由图2-1可知该设备原动机为电动机,传动装置为减速器,工作机为型砂运输设备。
减速器为展开式直齿圆锥—斜齿圆柱齿轮的两级传动,轴承初步选用圆锥滚子轴承,联轴器选用弹性柱销联轴器。
三、电动机的选择
计算项目
计算及说明
计算结果
1.选择电动机的类型
按电动机的工作要求和条件,选用Y系列鼠笼三相一步电机
Y系列鼠笼三相一步电机
2.选择电动机功率
工作机要求的电动机输出功率为:
P d=P w/ɳ
其中 P w=Fѵ/(1000 ɳ w)
则 P d= Fѵ/(1000 ɳ wɳ)
由电动机至运输带的传动总效率为:
ɳ wɳ= ɳ 1ɳ 3 2ɳ 3ɳ 2 4ɳ5ɳ 6
ɳ 1是圆锥齿轮传动(8级精度,油润滑)的效率;
ɳ 2是一对滚动轴承传动的效率;
ɳ 3是一对球轴承传动的效率;
ɳ 4是联轴器传动的效率;
ɳ 5是卷筒传递的效率;
ɳ 6是一般齿轮传动的效率(8级精度,油润滑)。
查《机械设计课程设计》附表1-1有:
ɳ 1=0.96,ɳ 2=0.98,ɳ 3=0.99,
ɳ 4=0.99,ɳ 5=0.96,ɳ 6=0.97
则ɳ wɳ= ɳ 1ɳ 3 2ɳ 3ɳ 2 4ɳ5ɳ 6=0.81
P w=Fѵ/(1000 ɳ w)=3.85kW
由《机械设计课程设计》附录九选取电动机额定功率P=4kW。
总=0.81
Pw=3.85Kw
P=4Kw
3.确定电动机转速
卷筒轴工作转速为:
ṉ w=60*1000ѵ/(∏D)=76.19r/min
计算输入转速
由于传动比 i=12.6
则输入转速n m= ṉ w*i=959 r/min
由《机械设计课程设计》附录九知,应选用的电动机转速为1000r/min,型号为Y132M1-6。
所选电动机的主要性能见 图3-1。
外观尺寸见 图3-2。
nw=76.19r/min
nm=959r/min
额定功率
Ped/kW
同步转速
n/(r.min-1 )
满载转速
n m/ r.min-1 )
电动机总重
/N
启动转矩/额定转矩
最大转矩/额定转矩
4
1000
960
730
2.0
2.0
图3-1电动机的主要参数
电动机(型号 Y132M1——6)的主要外形尺寸和安装尺寸
中心高
H
外形尺寸
L*(AC/2+AD)*HD
底脚安装尺寸
A*B
地脚螺栓直径
K
轴外伸尺寸
D*E
132
515*345*315
216*178
12
38*80
图3-2 电动机的主要外形尺寸和安装尺寸
四、传动比的计算及分配
计算项目
计算及说明
计算结果
1.总传动比
传动装置的总传动比为:i= ṉ m / ṉ w=12.6
i=12.6
2.分配传动比
一级传动比i 1的计算
因为圆锥-圆柱齿轮减速器,i 1≈0.25i,且i 1≤3,式中i 1为圆锥齿轮的传动比,i为减速器的总传动比。
则i 1=0.24i=3
二级传动比i2的计算
则i= i 1 *i 2
可算出i 2=4.2
i1=3
i2=4.2
五、传动装置运动、动力参数的计算
计算项目
计算及说明
计算结果
1.各轴转速
Ⅰ轴的输入转速 n 1= n m=960 r/min
Ⅱ轴的输入转速 n 2= n 1/ i 1=960 /3 =320 r/min
Ⅲ轴的输入转速 n 3= n 2/ i 2=320 /4.2=76 r/min
Ⅳ轴的输入转速 n 4= n 3=76 r/min
n1=n0=960r/min
n2=320r/min
nw=n3=76r/min
2.各轴功率
Ⅰ轴的输入功率 P 1= P dɳ 01= P dɳ 4=3.85kW*0.99=3.81kW()
Ⅱ轴的输入功率 P 2= P 1ɳ 12= P 1ɳ 1ɳ 2=3.81kW*0.96*0.98=3.58kW
Ⅲ轴的输入功率 P 3= P 2ɳ 23= P 2ɳ 2ɳ 6=3.58kW*0.98*0.97=3.40kW
Ⅳ轴的输入功率 P 4= P 3ɳ 34= P 3ɳ 3ɳ 4=3.40kW*0.99*0.99=3.3kW
p1=3.81kw
P2=3.58kw
P3=3.40kw
Pw=3.3kw
3.各轴转矩
电动机的输出转矩
T d=(9550* P d)/ n m=(9550*3.85)/960N·m=38.30 N·m
Ⅰ轴的输入转矩
T 1= T di 0ɳ 01= T dɳ 4=38.30 N·m*1*0.99=37.92 N·m
Ⅱ轴的输入转矩
T 2= T 1i 1ɳ 12= T 1i 1ɳ 1ɳ 2=37.92 N·m*3*0.96*0.98=107.02 N·m
Ⅲ轴的输入转矩
T 3= T 2i 2ɳ 23= T 1i 2ɳ 2ɳ 6=107.02 N·m*4.2*0.98*0.97=427.28 N·m
Ⅳ轴的输入转矩
T 4= T 3ɳ 34= T 3ɳ 3ɳ 4=427.28 N·m*0.99*0.99=418.78 N·m
Td=38.30N·mm
T1=37.92N·mm
T2=37.92N·mm
T3=427.28N·mm
T4=418.78N·mm
4.结果参数
运动和动力参数如下 表5-1
轴名
参数
电动机轴
Ⅰ轴
Ⅱ轴
Ⅲ轴
Ⅳ轴
转速n/(r/min)
960
960
320
76
76
功率P/kW
3.85
3.81
3.58
3.4
3.3
转矩T/(N•m)
38.3
37.92
107.02
427.28
418.78
传动比i
1
3
4.2
1
效率ɳ
0.99
0.94
0.95
0.98
表5-1 运动和动力参数
六、 传动件的设计计算
(一)、高速级锥齿轮传动的设计计算
计算项目
计算及说明
计算结果
1.选择材料、热处理方式和公差等级
考虑到带式运输机为一般机械,大、小锥齿轮均选用45钢,小齿轮调质处理,大齿轮正火处理,由《机械设计》第八版 表10-1得齿面硬度HBW1=217~255,HBW2=162~217.平均硬度HBW1=236,HBW2=190.HBW1-HBW2=46.在30~50HBW之间。选用8级精度。
45钢
小齿轮调质处理
大齿轮正火处理
8级精度
2.初步计算传动的主要尺寸
由以下公式求锥齿轮的最小直径
d1t≥
确定公式内各参数的值:
小齿轮传递转矩为T1=N·mm
因v值未知,Kv值不能确定,可初步选载荷系数Kt=1.3
由《机械设计》第八版表10-6,查得弹性系数
ZE=189.8
齿数比=i=3
取齿宽系数=0.3
选小齿轮齿数Z 1=25,大齿轮齿数Z 2= Z 1i 1=25*3=75
许用接触应力可用下式公式
由《机械设计》第八版 图10-21、c、d表查得接触疲劳极限应力为
小齿轮与大齿轮的应力循环次数分别为
N1=60n1jLh=60*960*1*24*300*10=4.147*109
N2=N1/i1=4.147*109/3=1.382*109
)由《机械设计》第八版图10-18查得寿命系数
KN1=1,
KN2=0.98;
由表8-20取安全系数SH=1,则有
取
初算小齿轮的分度圆直径d1t,有
d1t≥90.3mm
3.确定传动尺寸
计算载荷系数
由《机械设计》第八版 表10-2查得使用系数KA=1.0;
齿宽中点分度圆直径为
Dm1t=d1t(1-0.5)=90.3*(1-0.5*0.3)mm=76.76mm
故vm1=πdm1tn1/60*1000=π*76.76*960/60*1000m/s=3.86m/s
由《机械设计》第八版 图10-8降低1级精度,按9级精度查得动载荷系数Kv=1.26;
由《机械设计》第八版 表查得轴承系数
;
则
则载荷系数
K=KAKvKß=1.0*1.25*1.875=2.34
对d1t进行修正
因K与Kt有较大的差异,故需对Kt计算出的d1t进行修正 ,即
d1=≥90.3 mm =109.8mm
大端模数m
查《机械原理》第七版 表10-6,取标准模数m=4.5mm
大端分度圆直径为
d1=mZ1=4.5*25mm=112.5mm>109.8mm
d2=mZ2=4.5*75mm=337.5mm
锥齿距为
R=
齿宽为
b==0.3*177.88mm=53.364mm
取B2=55mm B1=60
K=2.34
m=4.5mm
d1=113mm
d2=339mm
Z1=29
Z2=87
R=178mm
B2=55mm
B1=60mm
4.校核齿根弯曲疲劳强度
齿根弯曲疲劳强度条件为
K、b、m和同前
圆周力为
Ft=
齿形系数YF和应力修正系数YS
即当量齿数为
由《机械设计》第八版 表10-5查得
YFa1=2.60, YFa2=2.06
YSa1=1.595,YSa2=1.97
许用弯曲应力
由图8-11查得寿命系数KN1=1,KN2=0.98;取安全系数SH=1.25
满足齿根强度要求
满足齿根弯曲强度
(二)、低速级斜齿圆柱齿轮的设计计算
计算项目
计算及说明
计算结果
1.选择材料、热处理方式和公差等
大、小锥齿轮均选用45钢,小齿轮调质处理,大齿轮正火处理,由《机械设计》表10-1得齿面硬度HBW1=217~255,HBW2=162~217.平均硬度HBW1=236,HBW2=190.HBW1-HBW2=46.在30~50HBW之间。选用8级精度。
45钢
小齿轮调质处理
大齿轮正火处理
8级精度
2. 按齿面接触疲劳强度进行设计
按齿面接触疲劳强度进行设计
因为是软齿面闭式传动,故按齿面接触疲劳强度进行设计其设计公式为
1)确定公式内各参数的值:
小齿轮传递转矩为T2=107020N·mm
因v值未知,Kv值不能确定,可初步选载荷系数Kt=1.6
由《机械设计》图10-30,查得弹性系数ZE=189.8
初选螺旋角,由图10-30选取区域系数ZH=2.433
齿数比=i=4.2
由《机械设计》第八版图10-21按齿面硬度查得小齿轮的接触疲劳强度极限及大齿轮的接触疲劳强度极限分别为
查《机械设计》第八版 表10-7,取齿宽系数=1.1
小齿轮与大齿轮的应力循环次数分别为
N1=60n1jLh=60*320*1*24*300*10=1.382*109
N2=N1/i2=1.382*109/4.2=3.291*108
由《机械设计》第八版 图10-18查得寿命系数KN1=1,KN2=0.98;由表8-20取安全系数SH=1,则有
许用接触应力
初选Z3=23,则Z4=uZ3=4.2*23=96.6,取Z4=97
由《机械设计》第八版 图10-26查得
2)设计计算
小齿轮分度圆直径
计算圆周速度
计算齿宽b及模数m nt
b/h=74.91/6.41=11.686
计算纵向重合度
计算载荷系数K
已知使用系数KA=1.0;
由《机械设计》第八版 图10-8降低1级精度,按9级精度查得动载荷系数Kv=1.08;
由《机械设计》第八版 表10-4查得;
由《机械设计》第八版 图10-13查得;
由《机械设计》第八版 表10-3查得
则载荷系数
按实际的载荷系数校正所算得的分度圆直径
d 3==68.1 mm =75.8mm
计算模数m n
T2=107020N.mm
Kt=1.6
ZE=189.8
ZH=2.433
=i=4.2
=1.1
Z3=23
Z4=97
=1.65
=68.1mm
=
b=74.91mm
=2.85mm
h=6.41mm
b/h=11.686
=2.307
K=2.21
d 3=75.8mm
=3.2mm
3.按齿根弯曲强度进行设计
按齿根弯曲强度进行设计
按齿根弯曲强度设计公式
1)确定公式内各参数的值:
根据重合度,从《机械设计》第八版 图10-28查得螺旋角影响系数
计算当量齿数
由《机械设计》第八版 表10-5查取齿形系数
YFa3=2.60,YFa4=2.18,
YSa3=1.595,YSa4=1.79
由图8-11查得寿命系数KN1=1,KN2=0.98;取安全系数SH=1.25
许用弯曲应力
计算大小齿轮的并比较
小齿轮的数值大
2)设计计算
计算模数
对比计算结果,由齿面接触疲劳强度计算的发面模数大于有齿根弯曲疲劳强度计算的发面模数,取m n=2.5mm,已满足弯曲强度。
但是为了同时满足接触疲劳强度,许按接触疲劳强度算得的分度圆直径来计算应有的齿数。于是有
取z 3=29 ,则z 4=uz 3=122
=0.0241
m n=2.5mm
z 3=29
z 4=122
4.几何尺寸计算
计算中心距
=196.4mm
取整,
螺旋角为
因值与初选值相差不大,故对与有关的参数不进行修正
计算大小齿轮分度圆直径
圆整后B 4=85mm B 3=90mm
318.331mm
B 3=90mm
B 4=85mm
七、轴的设计计算
(一)、输入轴(I轴)的设计
计算项目
计算及说明
计算结果
1.轴的设计计算
1)I轴I上力的计算
求输入轴上的功率、转速和转矩
=3.81 kw =960r/min =37920Nmm
求作用在齿轮上的力
已知高速级小圆锥齿轮的平均分度圆直径为
则
圆周力、径向力及轴向力的方向 见 图7-1
图7-1 输入轴载荷图
初步确定轴的最小直径
先初步估算轴的最小直径。选取轴的材料为45钢(调质),根据《机械设计(第八版)》表15-3,取,得
输入轴的最小直径为安装联轴器的直径,为了使所选的轴直径与联轴器的孔径相适应,故需同时选取联轴器型号。联轴器的计算转矩,查《机械设计(第八版)》表14-1,由于转矩变化很小,故取,则
=1.3*37920N.mm=49296N.mm
查GB/T 14653-93得知选用型号LX3型弹性柱销联轴器,其公称转矩1250000N.mm.而电动机轴的直径为38mm所以联轴器的孔径不能太小。取=30mm,半联轴器长度L=82mm,半联轴器与轴配合的毂孔长度为60mm。
=17.73
Ft=789.58N
Fr=272.50N
Fa=90.95N
2.轴的结构设计
2)轴的结构设计
拟定轴上零件的装配方案(见图7-2)
图7-2 输入轴轴上零件的装配
根据轴向定位的要求确定轴的各段直径和长度
为了满足半联轴器的轴向定位,12段轴右端需制出一轴肩,故取23段的直径。左端用轴端挡圈定位,12段长度应适当小于L所以取=58mm
初步选择滚动轴承。
因轴承同时受有径向力和轴向力,故选用单列圆锥滚子轴承,参照工作要求并根据,由《机械设计课程设计》表13-1中初步选取0基本游隙组,标准精度级的单列圆锥滚子轴承30308,其尺寸为40mm90mm25.25mm
所以而=25.25mm
这对轴承均采用轴肩进行轴向定位,由《机械设计课程设计》表13-1查得30308型轴承的定位轴肩高度
故取
取安装齿轮处的轴段67的直径;为使套筒可靠地压紧轴承,56段应略短于轴承宽度
故取=24mm,
轴承端盖的总宽度为20mm。
根据轴承端盖的装拆及便于对轴承添加润滑油的要求,求得端盖外端面与半联轴器右端面间的距离,取=50mm。
锥齿轮轮毂宽度为50mm,为使套筒端面可靠地压紧齿轮取由于
故取
轴上的周向定位
圆锥齿轮的周向定位采用平键连接,按由《机械设计(第八版)》表6-1查得平键截面,键槽用键槽铣刀加工,长为45mm,同时为保证齿轮与轴配合有良好的对中性,故选择齿轮轮毂与轴的配合为;
同样,半联轴器处平键截面为
与轴的配合为;
滚动轴承与轴的周向定位是由过渡配合来保证的,此处选轴的尺寸公差为k5。
确定轴上圆角和倒角尺寸
取轴端倒角为,轴肩处的倒角可按R1.6-R2适当选取。
=30mm
=58mm
=50mm
=25.25mm
=24mm
3.校核轴的强度
3)校核轴的强度
求轴上的载荷(30307型的a=15.7mm。所以俩轴承间支点距离为109.5mm 右轴承与齿轮间的距离为54.25mm。)(见图7-3)
载荷
水平面H
垂直面V
支反力F
弯矩M
总弯矩
=100970.1N.mm
扭矩T
=37.92N.M
图7-3 轴上的载荷
按弯扭合成应力校核轴的强度
根据图7-1 可知右端轴承支点截面为危险截面,由上表中的数据及轴的单向旋转,扭转切应力为脉动循环变应力,取,轴的计算应力为
= 15.78Mpa
前已选定轴的材料为45钢(调质),由《机械设计(第八版)》表15-1查得,故安全。
满足强度要求
(二)、中间轴(Ⅱ轴)的设计
计算项目
计算及说明
计算结果
1.轴的设计计算
1)I轴I上力的计算
求输入轴上的功率P、转速n和转矩T
=3.58kw =320r/min =107.02N.M
求作用在齿轮上的力
已知小斜齿轮的分度圆直径为
已知圆锥直齿轮的平均分度圆直径
dm2=d2(1-0.5)=337.5*(1-0.5*0.3)mm=286.875mm
圆周力、,径向力、及轴向力、的方向如图7-4所示
图7-4、中间轴受载荷图
初步确定轴的最小直径
先初步估算轴的最小直径。选取轴的材料为40Cr(调质),根据《机械设计(第八版)》表15-3,取,得,
中间轴最小直径显然是安装滚动轴承的直径和
=72.5
=2952N
=1120.6N
=874.4N=24.6mm
2.轴的结构设计
2)轴的结构设计
拟定轴上零件的装配方案(见图7-5)
图7-5中间轴上零件的装配
根据轴向定位的要求确定轴的各段直径和长度
初步选择滚动轴承。因轴承同时受有径向力和轴向力,故选用单列圆锥滚子轴承,参照工作要求并根据,由《机械设计课程设计》表13.1中初步选取0基本游隙组,标准精度级的单列圆锥滚子轴承30306,其尺寸为, 。
这对轴承均采用套筒进行轴向定位,由《机械设计课程设计》表13.1查得30306型轴承的定位轴肩高度37mm,因此取套筒直径37mm。
取安装齿轮的轴段,锥齿轮左端与左轴承之间采用套筒定位,已知锥齿轮轮毂长L=55mm,为了使套筒端面可靠地压紧端面,此轴段应略短于轮毂长,故取L2-3=50mm,齿轮的右端采用轴肩定位,轴肩高度,故取,则轴环处的直径为。
已知圆柱直齿轮齿宽B3=85,为了使套筒端面可靠地压紧端面,此轴段应略短于轮毂长,故取L4-5=80mm。
齿轮距箱体内比的距离为a=16mm,大锥齿轮于
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