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圆柱体相贯线焊接专机工作台设计计算说明书
一 圆柱齿轮设计
2.4.1蜗轮轴(即小锥齿轮轴)的设计
1轴的材料的选择,确定许用应力
2按扭转强度,初步估计轴的最小直径
3轴的结构设计
考虑到减速器为普通中用途中小功率减速传动装置,轴主要传递蜗轮的转矩。
选取轴的材料为45钢,调质处理。根据
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
15-3,取A0=100,于是得:
d≥
轴的最小直径为d1,与联轴器的孔径相适应,故需同时选取联轴器的型号
计算转矩
=
,查表14-1,选取
=1.3,则有
=KT=1.3×9.550×
×3.78/54.60
=859500N•mm
最小直径d1=48mm
根据d2=50mm,初步选取0基本游隙组,标准精度级的单列圆锥滚子轴承30212,其尺寸为:
d×D×T=50mm×110mm×23.75mm
故选d3=60mm L6=23.75mm
查GB/T294-94得:
圆锥滚子轴承da=69(30212)即轴肩为
h=
mm=4.5mm 取3 所以d5=69+3=72mm
又:轴环的亮度b=1.4h,即b≥1.4×6=8.4
b取12mm,即L5=12mm
(4)蜗轮的轴段直径
蜗轮轴段的直径的右端为定位轴肩。
故d4=d5-2h,求出d4=64mm
与传动零件相配合的轴段,略小于传动零件的轮毂宽。
蜗轮轮毂的宽度为:
B2=(1.2~1.5)d4=(1.2~1.5)×64
=76.8~96,取b=80mm,即L4=80mm
(5)轴承端盖的总宽度为20mm。取端盖的外端面与半联轴器右端端面的距离为l=35mm。故
L2=20+35=55mm
(6)取蜗轮与箱体内壁距离为a=16mm,滚动轴承应距箱体内壁一段距离s(5~8)。取s=8mm,已知滚动轴承宽度为T=23.75mm,蜗轮轮毂长为L=80mm,则:
L3=T+s+a+(80-78)=49.75mm
选用45号钢, [σb]=600MPa [σb-1]1=55MPa
=1.3
=859500N•mm
d1=48mm
d2=50mm
L1=82mm
d3=60mm
L6=23.75mm
d5=72mm
轴环L5=12mm
d4=64mm
L4=80mm
L2=55mm
L3=49.75mm
至此已初步确定了轴端各段直径和长度,
轴的总长为:L总=82+55+49.75+80+12+36=315mm
4轴的强度校核
(1) 轴向零件的同向定位
蜗轮,半联轴器与轴的同向定位均采
用平键链接。按d4由表6-1查得平键截面 b×h=20mm×12mm,键槽用键槽铣刀加工,长为63mm,同时为了保证蜗轮与轴配合有良好的对中性,故选择蜗轮轮毂与轴端配合为
;同样,半联轴器与轴的连接,选用平键14mm×9mm×70mm,半联轴器与轴的配合为
,滚动轴承与轴的同向定位是由过渡配合来保证的,此处选轴的直径尺寸公差为m6。
(2) 确定轴上的圆角和倒角尺寸
参考15-2,取的倒角2×45°,各轴
肩处的圆角半径为(见附图)。
(3.1)确定各向应力和反力
蜗轮分度圆直径d=340.2 mm
转矩T=892.9 N·m
蜗轮的切向力为:
Ft=2T/d=2×892.9×103/340.2=5249.9 N
蜗轮的径向力为:
Fr=Ft×
=5249.9×tan20°/cos11°18′35″
=1853.5 N
蜗轮的轴向力为:
Fa=Ft×
=5249.9×tan11°18′35″
=1050 N
T=892.9N·m
Ft=5249.9 N
Fr =4853.5 N
Fa=1050N
反力及弯矩、扭矩见10.3反力及弯局矩、扭矩图所示:
5轴的强度校核
(3.2)垂直平面上:
支撑反力:
=
=2182 N
其中132为两轴承中心的跨度,59为蜗轮
中心到右边轴承中心的距离。
N
水平平面:
N
N
(3) 确定弯距
=59
EMBED Equation.3 =59
2902.9=171271 N·mm
垂直弯矩:
N·mm
N·mm
合成弯矩:
= 233893N·mm
=172357 N·mm
扭矩T=892.9 N·mm
(4) 按弯矩合成应力校核该轴端强度
进行校核时,通常只校核轴上承受
最大弯矩和扭矩的截面(即危险截面)的强度。轴单向旋转扭转切应力为脉动循环变应力。取α=0.6 轴端计算应力:
=27MPa<[σ-1]=60MPa
故是安全的。
=2182 N
=
N
=2347 N
=2902.9N
=171271 N·mm
=233893 N·mm
=172357 N·mm
=27MPa
合格
三 圆锥齿轮设计
圆锥齿轮传动比为1:1
3.1确定齿数Z及校核
(1)选Z
。软齿面应尽量选大些。
(2)Z
= i
EMBED Equation.3 Z
。且Z
为整数。
(3)计算U=
(4)
=
EMBED Equation.3 5%
3.2按接触强度计算d1
1.计算公式
EMBED Equation.3
2.计算T1
T1=95500
Pd-------Kw
nd-----r/min
η1=0.99
3.计算K
K=KAKVK
(1)由表4-8选用系数KA
(2)选动载荷系数KV记为KVt
(3)取
值。一般取
=0.3
=
(4)由土4-45查出齿向载荷分布系数K
(5)计算
K=KAKVK
取KV=KVt故Kt=KAKVK
4.弹性系数ZE由表4-9查得
5.节点系数ZH由表4-48查得
6.许用应力[
]H=ZN
ZW
(1)由图4-58查得
(2)由已知条件计算
N1=60n1*r*tn
N2=N1/U
式中:n----啮和次数
n1-----r/min
tn----每天工作小时
N-----年
300天/年
小时/天
(3)由图4-59查得寿命系数
ZN1 ZN2
(4)由表4-11查得安全系数SH
(5)由图查得工作硬化系数Zw
(6)计算
[
]H1=ZN
ZW
[
]H2=ZN
ZW
(7)计算d1
d1
试选Kt=Kvt
3.3 校核d1
因为试选的Kv可能与实际不符合。
(1)模数m=
取标准值。可改变Z1而达到选用适当的m的目的,但 u有变则需重新计算d1。
(2)按几何关系计算d1
d1=m Z1 dm1= d1(1-0.5
)
(3)圆周速度Vm(平均直径dm)
Vm=
计算
由
查图4-43得Kv
(4)校核d1
d1=
d1与d1t相差太大,则需重新选Kvt,再计算d1t
3.4 校核齿根弯曲强度
(1)计算公式
(2)当量齿数计算 Zv=
a.
b.
c.由当量齿数Zv查图4-55得齿形系数YFa1,YFa2
查图4-56得齿根应力修正系数Ysa1,Ysa2.
d.确定[
]F=YH
Yx
查图4-61得
和
查图4-62得YN1, YN2
查图4-63得尺寸系数Yx
查图4-11得安全系数SF
计算
比较
,
的大小,取较大值
校核弯曲强度
3.5 几何尺寸计算
1.分度圆直径d
d1 =mZ1
d2=mZ2
2.节锥
=arctan
=90
-
3.节锥距R
R=
=
4.齿宽b=R
5.周节P=
m
6.齿顶高ha ha=m
7.齿根高hf hf=1.2m
8.齿顶间隙 c=0.2m
9.齿顶圆直径
=m(Z+2
)
=m(Z+2
)
10.齿根圆直径
= m(Z-2.4
)
= m(Z-2.4
)
3.6 受力分析
Ft1=-Ft2=
Fr1=-Fa2= Ft1*tan
EMBED Equation.3
Fa1=-Fr2= Ft1*tan
EMBED Equation.3
小齿轮为45钢,调质217HBS~255HBS。取240HBS。大齿轮为45钢正火163HBS~217HBS。取200HBS。8级精度
Z
选20
Z
=2.65
20=53
U=2.65
=0<5%
T1=9550000
5.5
0.99/720=72221.9 N*mm
KA=1.0
KVt=1.1
=0.3
=0.500
K
=1.03
Kt=1.133
ZE=189.8
ZH=2.5
=570MPa
=460MPa
N1=1.27
EMBED Equation.3
N2=4.76
EMBED Equation.3
tn =29200
ZN1=1
ZN2=1
SH=1
Zw=1
[
]H1=570MPa
[
]H2=460MPa
d1t
m=4.395
取m=4.5
d1=90mm
dm1=76.5mm
Vm=2.88m/s
=0.576
Kv=1.0
d1=85.14mm
故d1与d1t相差不大,符合要求。
=0.936
=20.67
=0.353
=69.33
=21.37
=150.14
YFa1=2.63 YFa2=2.16
Ysa1=1.56 Ysa2=1.89
=230MPa
=190MPa
YN1=YN2=1
Yx=1
SF=1
=230MPa
=190MPa
<
故取大齿轮计算
合格
d1=90mm
d2=238.5mm
=20.674
=69.326
R=127.46mm
b=38.238mm
取b=40mm
P=14.13mm
ha=4.5mm
hf=5.4mm
c=0.9mm
=91.9mm
=241.7mm
=79.9mm
=234.7mm
Ft1=Ft2= Ft
Ft=1888.15N
Fr1=-Fa2=643.25N
Fa1=-Fr2=242.59N
3.7 锥齿轮轴(即工作台转轴)的设计
1.齿轮轴的设计
(1)确定轴上零件的定位和固定方式 (如图)
(2)按扭转强度估算轴的直径
选用45#调质,硬度217~255HBS
轴的输入功率为PⅠ=5.445 Kw
转速为nⅠ=1.5r/min
根据课本P205(13-2)式,并查表13-2,取c=117
d≥
(3)确定轴各段直径和长度
(4)求齿轮上作用力的大小、方向
Ft=1888.15N
Fr=Ft·tanα=1888.15×tan200=643.25N
Ft,Fr的方向如下图所示
(5)轴长支反力
根据轴承支反力的作用点以及轴承和齿轮在轴上的安装位置,建立力学模型。
水平面的支反力:RA=
=944.08N RB=
=2832.23 N
垂直面的支反力: RA’=
=321.67N
RB’ =
=964.88 N
(6)画弯矩图
右起第四段剖面处的弯矩:
水平面的弯矩:M水平=RA×0.08=37.76 Nm
垂直面的弯矩:M垂直= RA’×0.08=12.87 Nm
合成弯矩:
(7)画转矩图: T= Ft×d1/2=84.59 Nm
(8)画当量弯矩图
因为是单向回转,转矩为脉动循环,α=0.6
可得右起第四段剖面C处的当量弯矩:
(9)判断危险截面并验算强度
已知M当=93.87Nm ,由课本表13-1有:
[σ-1]=60Mpa 则:
σe= M当/W= M当/(0.1·D43)
=93.87×1000/(0.1×453)= 10.30MPa<[σ-1]
σe= MD/W= MD/(0.1·D13)
=50.75×1000/(0.1×283)=33.12 Nm<[σ-1]
所以确定的尺寸是安全的 。
受力图如下:
D1=Φ28mm
L1=40mm
D2=Φ36mm
L2=40mm
D3=Φ45mm
L3=20mm
D4=Φ50mm
L4=80mm
D5=Φ45mm
L5=20mm
D6=Φ42mm
L6= 100mm
Ft=1888.15Nm
Fr=643.25Nm
RA=944.08N
RB=2832.23N
RA’=321.67N
RB’=964.88 N
M水平=37.76 Nm
M垂直= 12.87 Nm
M合=39.89Nm
T=84.59 Nm
α=0.6
M当=93.87Nm
[σ-1]=60Mpa
MD=50.75Nm
四 键联接设计
4.1.输入轴与小齿轮联接采用平键联接
此段轴径d1=28mm,L1=40mm
查手册得,选用A型平键,得:
另外本人甩卖全部机械毕业设计5元一份(全部甩卖)
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