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习题答案
带传动作业
4.1 解:
1000P1000,8F,,,533.3Nev15
12FFFFFF,,,,,e1211133由题意得: ,
33F,F,,533.3,800N1e22所以
4.3 解:
一、确定计算功率
P,K,P,1.1,5,5.5NcaA由
表
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4.6(P)取K=1.1 , 则 69A
二、选择带的型号
根据Pn由图4.11(P)选A型V带。 、ca 1 71
三、确定带轮基准直径d d
1. 确定小带轮基准直径dd1
由表4.4(P)及图4.11(P)取d= 90 mm 6171d1
2. 验算带速
nd,,1440,90,1d1v,,,6.79m/s,5m/s60,100060,1000 合适 3.计算大带轮基准直径d d2
n14401d,d,,90,381mmd2d1n3402 由表4.4(P)取d =400 mm 。 61d24. 验算实际传动比
dn4001440d21,,,4.44n',,,324.3rpmi122d90i4.44d112
n',n324.3,340220000,100,,100,4.6,50000n3402 合适 四、确定中心距及带的基准长度
1.初选中心距
a,(0.7~2)(d,d),(0.7~2)(90,400),343~980mm0d1d2
a,600mm0取
2. 确定带的基准长度
2,(d,d)21ddL,2a,(d,d),0012dd24a0
2(400,90),,2,600,(90,400),,2009.7mm24,600 由表4.2(P)取L = 2000 mm 60d
3. 计算实际中心距
L,L2000,2009.7d0a,a,,600,,595mm022 五、验算包角
d,d400,90d2d1,,180:,,60:,180:,,60:,151.5:,120:1a600 合适
六、确定带的根数
由表4.5(P)得P=0.95 , 由表4.7(P)得?P=0.165, 680 720 由表4.8(P)得K=0.92, 由表4.2(P)得K=1.07, α73 60L
P5.5caz,,,4.9根
PPKK(,,)(1,0.165),0.92,1.0100,L则
z,5根取
七、确定有效拉力
P2.55.52.522caF,500(,1),qv,500(,1),0.1,6.79,143.7N0vzK6.79,50.92,
八、求对轴的压力
,148.71F,2zFsin2,5,143.7sin,1383.7NQ022
链传动作业
5.1 解:
1. 选择链轮齿数
n9601i,,,2.9
n3302 1)根据查表5.5(P)取z = 23 ; 9212)则z = i z =2.9×23 = 66.7,取z = 68 211
2. 确定计算功率
由表5.6(P)得 K=1.0 ,则P= KP =1.0×7=7 kw 93Aca A
3. 初定中心距
由题意取 a= 40p 0
4. 计算链条的节数
a2ZZZZ,p,201221L,,,(),p,pa220
pp2,4023,6868,232,,(),126.78
pp,2402
取 L= 128 节 p
5. 选择链条的型号和确定链条的节距
由表5.7(P94)得K=1.23; Z
由表5.8 (P94) 得K=1.06; L
由表5.9 (P94) 得K=1 m
P7caP,,,5.37kw0KKK1.23,1.06,1ZLm则
根据所求出的功率P和小链轮转速n由图5.10(P91)查得链号为12A,再根据链号由表01
5.1(P85)得到链条节距为p=19.05 mm 。
6. 验算链速
Znp23,960,19.0511v,,,7.01m/s,15m/s
60,100060,1000 合适
7. 确定链传动的实际中心距
,,z,zz,zz,zp22121221a,(L,),(L,),8(),,PP,4222,,
,,19.0523,6823,6868,2322,(128,),(128,),8(),768.39mm,,,4222,,
8. 链传动作用在轴上的力即压轴力
1000P1000,7F,,,998.57Nev7.01
F= 1.2F=1.2×998.57=1198.29 N Qe
5.3 解:
1. 选择链轮齿数
1)根据i查表5.5(P)取z = 23 ; 921
2)则z = i z =3×23 = 69 21
2. 确定计算功率
由表5.6(P)得 K=1.4 ,则P= KP =1.4×22=30.8 kw 93Aca A
3. 初定中心距
由题意取 a= 40p 0
4. 计算链条的节数
a2ZZZZ,p,201221L,,,(),p,pa220
pp2,4023,6969,232,,(),127.33
pp,2402
取 L= 128 节 p
5. 选择链条的型号和确定链条的节距
由表5.7(P94)得K=1.23; Z
由表5.8 (P94) 得K=1.06; L
由表5.9 (P94) 得K=1 m
P30.8caP,,,23.62kw0KKK1.23,1.06,1ZLm则
根据所求出的功率P和小链轮转速n由图5.10(P91)查得链号为12A,再根据链号由表01
5.1(P85)得到链条节距为p=19.05 mm 。
6. 验算链速
Znp23,750,19.0511v,,,5.78m/s,15m/s
60,100060,1000 合适
7. 确定链传动的实际中心距
,,z,zz,zz,zp22121221a,(L,),(L,),8(),,PP,4222,,
,,19.0523,6923,6969,2322,(128,),(128,),8(),,4222,,,
,768.39mm,800mm
合适。
8. 链传动作用在轴上的力即压轴力
1000P1000,22F,,,3806.2Nev5.78 F= 1.2F=1.2×3806.2=4567.5 N Qe
5.4 解:
全为顺时针转。
齿轮传动作业 6.1 解:
一、确定大齿轮齿面硬度及齿轮许用应力
1. 由表6.1(P106)选:大齿轮45钢调质 HBS=210 。
222. 由图6.8b(P110)得:σ= 600 N/mm , σ= 560 N/mm。 Hlin1 Hlin2
83. N = 60nj L= 60×1460×1×10000 = 8.76×10 ; 11 h
i =z/z=108 / 32 =3.375 2 1
88 N = N / i = 8.76×10/ 3.375=2.6×10; 21
由图6.8(P108)得:K=1 , K=1.05 。 HN1HN2
24. 取S=1 , 则: [σ]= 600 N/mm HH1
2[σ]= 560× 1.05 = 588 N/mm H2
5. 计算K
1) 由表6.2(P112)得:K=1 A
2a2,210m,,,3
z,z32,10812 mm , d= zm = 32×3= 96 mm 2) 计算11
44v=πdn /6×10=π×96×1460/6×10=7.34m/s , 1 1
由图6.10(P114)得:K=1.23 v
3) 取K=1.0 α
b70,,,,0.73dd9614) , 由图6.13a(P115)得:K=1.02 β 5) K= K K KK=1×1.23×1.0×1.02=1.25 αβA v
二、按齿面接触疲劳强度校核计算
1.由表6.3(P117)得:Z=189.8 ,取Z=2.445 E H
KF2KTu,1u,12t1,,,,ZZ.,ZZ.,,N/mmHEHEHH2bdubdu112.由 ;
22bdu58870,963.375,,,22H21T,(),()1ZZ2Ku,1189.8,2.52,1.253.375,1EH
,305688.03N,mm得:
P61T,9.55,101n13. 由 ,得:
Tn305688.03,146011P,,,46.73kw1669.55,109.55,10 6.2 解:
一、选择齿轮材料、精度及许用应力
1.由表6.1(P106)选:小齿轮45钢调质 HBS=260 ,
大齿轮45钢调质 HBS=220 。
2. 取7级精度。
z542i,,,2.08
z2613. 22 3. 由图6.9b(P111)得:σ=230 N/mm , σ= 210 N/mm; Flin1 Flin2
22 由图6.8b(P110)得:σ= 600 N/mm , σ= 560 N/mm。 Hlin1 Hlin2
94. N = 60nj L= 60×1440×1×5×16×300 = 2.07×10 ; 11 h
98 N = N / i = 2.107×10/2.08 =9.97×10; 21
5. 由图6.7(P109)得:K= K=1 ;由图6.8(P108)得:K=1 , K=1 。 FN1FN2HN1HN2
6. 取S=1 ,S=1.4 ,则: HF
2 2 [σ]= 230×0.7/1.4=115 N/mm,[σ]= 210×0.7/1.4=105 N/mm F1F2
2 2[σ]=600 N/mm, [σ]= 560 N/mm H1H2
二、按齿面接触疲劳强度设计计算
1. 试选 K= 1.3 t
P7.56621T,9.55,10,9.55,10,49739.58N/mm1n144012. 3. 由表6.5(P121)取 φ=1.1 (不对称布置) d
4. 由表6.3(P117)得:Z=189.8 , 且Z=2.5 E H
2KTZZu,121tEHd,3.()1tu,,,,dH
2,1.3,49739.582.08,1189.8,2.523,(),50mm
1.12.085605. 计算: 6. 求K
1) 由表6.2(P112)得:K=1 A
442) v=πdn /6×10=π×50×1440/6×10=3.77 m/s , 1 1
由图6.10(P114)得:K=1.1 v
3) 取K=1.2 α
4) 由图6.13a(P115)得:K=1.1 β
5) K= K K KK=1×1.1×1.2×1.1=1.452 αβA v
7. 修正
K1.45233d,d,50,,51.88mm11tK1.3t 三、几何尺寸计算
d51.881m,,,1.995mmm,2mmz2611. , 取
d,mz,2,26,52mmd,mz,2,54,108mm11222.,
m2a,(z,z),(26,54),80mm12223. ,
b,,d,1.1,52,57.2mmd14.
取 b=65 mm , b=60 mm 1 2
四、按齿根弯曲疲劳强度校核
1. 由表6.4(P120)得:Y=2.60 , Y=1.595 , Fa1 Sa1
Y=2.3 , Y=1.71 Fa2 Sa2
2KT2,1.452,49739.581,,YY,2.6,1.595FFaSa111bdm60,52,21
22,,,95.99N/mm,,,115N/mmF12.
YY2.3,1.71FaSa22,,,,95.99FF21YY2.6,1.595FaSa11
22,,,91.04N/mm,,,105N/mmF2
所以强度足够。
五、结构设计(略)
6.4 解:其螺旋线方向和各受力方向如图所示。
6.5 解:一、求螺旋角
2b2,40,,,,0.3R250du,12120,5(),1
201.
z201,,arctan(),arctan(),21.8:1z5022.
2T1,,F,F,Ftancos,tan20:cos21.8:2111artd(1,0.5),1R
2T1,tan20:cos21.8:,2T,0.004120,5(1,0.5,0.3)3.
50T,iT,T,2.5T,T2111320 4.
2T2T32,,,,F,Ftan,tan,costana3t3dzmn33
2T,2.51,sin,,2T,0.018sin,123,6 5.
F,2T,0.004,F,2T,0.018sin,a21a31 6.
0.004,,arcsin(),12.84:
0.18 7.
二、其各受力方向如图所示。
蜗杆传动作业 7.1 解:各受力方向如图所示。
7.2 解:
一、确定许用应力及计算参数 2 查表7.10(P161)得[σ]=200 N/mm, 0H7N = 60nj L= 60×46×1×15000 = 4.14×10 2 h
77101088Z,,,0.84N7N4.14,10 2[σ]=Z[σ]=0.84×200=168 N/mm HN0H
查表7.9(P160)得Z=155,查表7.8(P160)得K=1.1 E
60n,in,,46,1380r/min122取z=2 ,则 1
二、求传递的扭矩和功率
dmz,,,505,60,16822H12T,(),(),454996.73N,mm2K3.25Z1.13.25,155E
mz5,21tg,,,,0.2,1d50,,tg(0.2),11.31:1,
dn,50,1380,,11v,,,3.68m/ss60,1000cos60000cos11.31:,
,,1.5:v由表7.5(P154)得 ,
,tgtg11.31:,,,,0.881,,tg(,)tg(11.31:,1.5:)v则
,,,,0.9923取
,,,,,,0.88,0.99,0.99,0.86123则
PTn454996.73,46222P,,,,2.55kw166,,9.55,10,9.55,10,0.86所以
滚动轴承作业
8.2 解:1.求轴承所受载荷
,,15:06'34"由轴承手册得, C= 97.8 KN ,C= 74.5 KN r 0r
Y,0.4cot,,0.4cot15:06'34",1.48由表8.10(P196)得
FFFrrrF,,,S221.482.96Y,由表8.11(P197)得
FF48002200r1r2F,,,1621.62NF,,,743.24NS1S22.962.962.962.96所以,
F+F= 650+1621.62=2271.62 N > F=742.24 N aes1s2
所以 F=F= 1621.62 N , F= F+F=2271.62 N a1s1a2aes1
2.求轴承的当量动载荷
e,1.5tan,,1.5tan15:06'34",0.4由表8.10(P196)得
FF1621.622271.62a1a2,,0.34,e,0.4,,1.03,e,0.36
F4800F2200r1r2而 ,
由表8.10(P196)得X=1 , Y= 0 11
X=0.4 , Y=1.48 22
由表8.7(P194)得 f=1.3 P
P= f(XF+YF) =1.3(1×4800+0×1621.62)=6240 N 1 P 1r11a1
P= f(XF+YF) =1.3(0.4×2200+1.48×2271.62)=5514.6 N 2 P 2r22a2
因为P>P ,所以取P =P=6240 N 121
3. 求轴承的额定动载荷
,60nL'P624060,960,15000h3.33,C',,,47535.77N,47.54KN66f11010t
而C′= 47.54 KN < C= 97.8 KN 故该对轴承合适。 r
8.3 解:1.求轴承所受的径向载荷
F,802800,80reF,,,933.3Nrc240240
F,F,F,2800,933.3,1866.7Nrdrerc 2.求轴承所受的派生轴向力
由表8.11(P197)得 F= 0.68F s r
F=0.68F=0.68×933.3=634.64 N scrc
F=0.68F=0.68×1866.7=1269.36 N sdrd
3.求轴承所受的轴向载荷
F+F= 750+1269.36=2019.36 N > F=634.64 N aesd sc
所以 F= F+F= 2019.36 N , F= F= 1269.36 N ac aesd ad sd
8.4 解:1.将作用在齿轮上的力转化到轴上
Fd960,212.5aem2M,,,102000N,mmae22 则 2. 求轴承所受的径向载荷
F,195,M2710,195,102000reaeF,,,2472.35N1r255255 F,F,F,2710,2472.35,237.65Nr2rer1
3.求轴承所受的派生轴向力
初选轴承型号为30207,
,,14:02'10"由轴承手册得, C= 51.5 KN ,C= 37.2 KN r 0r
Y,0.4cot,,0.4cot14:02'10",1.6由表8.10(P196)得
FFFrrrF,,,S221.63.2Y,由表8.11(P197)得
FF2472.35237.65r1r2F,,,772.61NF,,,74.27NS1S23.23.23.23.2所以, 4.求轴承所受的轴向载荷
F+F= 960+47.27=1034.27 N > F=772.61 N aes2s1
所以 F= F+F=1034.27 N , F= F= 74.27 N a1aes2a2s2
5.求轴承的当量动载荷
e,1.5tan,,1.5tan14:02'10",0.37(P196)得 由表8.10
FF1034.2774.27a1a2,,0.42,e,0.37,,0.31,e,0.37
F237.65F2472.35r1r2而 , 由表8.10(P196)得X=0.4 , Y=1.6 , X=1 , Y=0 1122
由表8.7(P194)得 f=1.2 P
P= f(XF+YF) =1.2(0.4×2472.35+1.6×1034.27)=3172.53 N 1 P 1r11a1
P= f(XF+YF) =1.2(1×237.65+0×74.27)=285.18 N 2 P 2r22a2
因为P>P ,所以取P =P=3172.53 N 121
3. 求轴承的额定动载荷
,60nL'P3172.5360,500,30000h3.33,C',,,24466.14N,24.47KN66f11010t
而C′=24.47 KN < C= 51.5 KN 故该对轴承合适。 r
所以轴承型号为30207。
滑动轴承作业
9.2 解:
1. 验算轴承的工作能力
,,,,,,p,8MPav,3m/spv,15MPam/s由表9.2(P221)得,,
F36000r,,p,,,10MPa,p,8MPa
dB60,60
Fn36000,150,,pv,,,4.71MPam/s,pv,15MPam/s
19100B19100,60
dn,60,150,,,,v,,,0.47m/s,v,3m/s
60,100060000 所以轴承的工作能力不够。
2. 计算轴的允许转速
dn,,,v,,v,3m/s
60,1000根据
60000v60000,3,,n,,,954.93r/min
,,d,60得
Fnr,,pv,,pv,15
19100B根据
,,19100Bpv19100,60,15n,,,477.5r/min
F36000r得 所以轴承允许的转速n为477.5 r/min 。
3. 计算轴承能承受的最大载荷
Fr,,p,,p,8MPa
dB根据
,,F,dBp,60,60,8,28800Nr得
Fnr,,pv,,pv,15MPam/s
19100B根据
19100Bpv19100,60,15,,F,,,19100Nrn900得 所以轴承能承受的最大载荷F为 19100 N 。 r
4. 计算轴所允许的最大转速
Fr,,p,,p,8MPa
dB根据
,,F,dBp,60,60,8,28800Nr得
dn,,,v,,v,3m/s
60,1000根据
60000v60000,3,,n,,,954.93r/min
,,d,60得
Fnr,,pv,,pv,15
19100B根据
,,19100Bpv19100,60,15n,,,596.88r/min
F28800r得 所以轴所允许的最大转速n为596.88 r/min 。 max
9.3 解:选择轴承材料为ZcuPb5Sn5Zn5 , 则由表9.8(P236)得:
,,,,p,7.5MPapv,2MPam/s,,
Fa,,p,,p,7.5,22z(d,d)214根据
2222,,z(d,d)p,1,(120,60),7.5,21F,,,63617.25Na44得
所以该轴承所能承受的最大轴向载荷F为63617.25 N。 a
轴作业
11.2 解:其轴系的正确结构如下图所示。
11.4 解:一、求轴上的载荷
1. 计算齿轮上的力
mz4,18n1d,,,74.5mm1,coscos15: 齿轮的分度圆:
mz4,82n2d,,,339.57mm2,coscos15:
662T2,9.55,10P2,9.55,10,221F,,,,3916.85N1tddn74.5,1440111 圆周力:
,Ftan3916.85,tan20:t1nF,,,1475.91Nr1,coscos15: 径向力:
F,Ftan,,3916.85,tan15:,1049.52Na1t1 轴向力:
轴向力对轴产生的弯矩:
Fd1049.52,339.57a12M,,,178192.75N.mma122
2. 求轴的支反力 (如上图所示)
F,903916.85t1F,F,,,1958.43NNH1NH21802轴左、右端水平的支反力: 轴左端垂直的支反力:
MF,901475.91178192.75a1r1F,,,,,1727.91NNV11801802180 轴右端垂直的支反力:
F,F,F,1475.91,1727.91,,252NNV2r1NV1
轴左端的支反力:
2222F,F,F,1958.43,1727.91,2611.73NNNHNV111 轴右端的支反力:
2222F,F,F,1958.43,(,252),1974.58NNNHNV222
二、绘制轴的弯矩图和扭矩图(如上图所示)
M,F,90,1958.43,90,176258.7N,mmHNH1 截面处C处水平弯矩:
截面处C处垂直弯矩:
M,F,90,1727.91,90,155511.9N,mmV1NV1
M,F,90,,252,90,,22680N,mmV2NV2
截面处C初合成弯矩:
2222M,M,M,176258.7,(155511.9),235055.48N,mmHV11
2222M,M,M,176258.7,(,22680),177711.88N,mmHV22
P2266T,9.55,10,9.55,10,145902.78N,mm1n14401轴的扭矩:
,,0.97,,0.9912 取联轴器效率为:,滚动轴承效率为:,
,,0.983齿轮效率为:,
,,,,,,0.97,0.99,0.98,0.94123则总效率为:
82T,i,T,,0.94,145902.78,624788.12N,mm2118所以 三、弯扭合成强度校核
,,0.6 考虑启动、停机影响,扭矩为脉动循环变应力,则取,
截面处C处计算弯矩:
2222M,M,(,T),235055.48,(0.6,624788.12),442471.18N,mmca112
2222M,M,(,T),177711.88,(0.6,624788.12),414862.85N,mmca221
2,,,,60N/mm,1由表11.2(P288)得:
MM2caca11,,,,,,,N/mmca,13W0.1d由
M442471.18ca33d,,,41.94mm
,,,0.10.1,60,1得截面处C处的直径为: 考虑键槽的影响则截面处C处的直径为:
d,105%d,1.05,41.91,44mmc
11.5 解:一、求轴上的载荷
1. 计算齿轮上的力
mz3,1122nd,,,341.98mm2,coscos10:44'2 齿轮的分度圆:
mz4,233nd,,,93.24mm3,coscos9:22'3
662T2,9.55,10P2,9.55,10,5.52F,,,,1706.57Nt2ddn341.98,180222 圆周力:
,Ftan1706.57,tan20:2tnF,,,632.2N2r,coscos10:44'2 径向力:
F,Ftan,,1706.57,tan10:44',323.49Na2t22 轴向力:
662T2,9.55,10P2,9.55,10,5.52F,,,,6259.24Nt3ddn93.24,180332圆周力:
,Ftan6259.24,tan20:3tnF,,,2345.85N3r,coscos9:22'3径向力:
F,Ftan,,6259.24,tan9:22',1032.47Na3t33轴向力: 轴向力对轴产生的弯矩:
Fd323.49,341.98a22M,,,55313.56N.mma222
Fd1032.47,93.24a33M,,,48133.75N.mma322
2. 求轴的支反力 (如上图所示)
轴左端水平的支反力:
F,80,F,2101706.57,80,6259.24,210t2t3F,,,4680.54NNH1310310 轴右端水平的支反力:
F,F,F,F,1706.57,6259.24,4680.54,3285.27NNH2t2t3H1 轴左端垂直的支反力:
F,80,M,M,F,210r2a2a3r3F,NV1310
632.2,80,55313.56,48133.75,2345.85,210,,1092.27N
310 轴右端垂直的支反力:
F,F,F,F,2345.85,632.2,1092.27,621.38NNV2r3r2NV1 轴左端的支反力:
2222F,F,F,4680.54,1092.27,4806.3NNNHNV111 轴右端的支反力:
2222F,F,F,3285.27,621.38,3343.52NNNHNV222 二、绘制轴的弯矩图和扭矩图(如下图所示)
截面处?处水平弯矩:
M,F,100,4680.54,100,468054N,mmH1NH1
截面处?处水平弯矩:
M,F,80,3285.27,80,262821.6N,mmH2NH2 截面处?处垂直弯矩:
M,F,100,1092.27,100,109227N,mmV1NV1
M,M,M,109227,48133.75,157360.75N,mmV2V1a3 截面处?处垂直弯矩:
M,F,80,621.38,80,49710.4N,mmV4NV2
M,M,M,55313.56,49710.4,5603.16N,mmV3a2v4
截面处?初合成弯矩:
2222M,M,M,468054,109227,480629.88N,mmHV111
2222M,M,M,468054,157360.75,493798.49N,mmHV212
2222M,M,M,468054,5603.16,468087.54N,mmHV323
2222M,M,M,468054,49710.4,480686.38N,mmHV424
P5.566T,9.55,10,9.55,10,291805.56N,mm2n1802轴的扭矩:
三、弯扭合成强度校核
,,0.6 考虑启动、停机影响,扭矩为脉动循环变应力,则取,
M,M,480629.88N,mmca11 各截面处处计算弯矩:
2222M,M,(,T),493798.49,(0.6,291805.56),523919N,mmca222
2222M,M,(,T),468087.54,(0.6,291805.56),499760.06N,mmca332 M,M,480686.38N,mmca44
2,,,,60N/mm,1由表11.2(P288)得:
截面?处的强度:
MM52391922ca2ca2,,,,,,,41.9N/mm,,,60N/mm,ca1133W0.1d0.1,50 截面?处的强度:
MM480686.3822ca4ca4,,,,,,,52.78N/mm,,,60N/mm,ca2133W0.1d0.1,45 所以截面?、?处强度足够。
轴毂联接
12.1 解:
1. 确定键的类型和尺寸
两处都采用平键联接,由手册得:
b×h×L =25×14×80, k=h,t =14,9=5 mm 11111
b×h×L =20×12×125 , k=h,t =12,7.5=4.5 mm 22222
l,L,b,80,25,55mml,L,b,125,20,105mm111222 ,
2. 确定许用应力
22,,,,60N/mm,,,,110N/mmp2p1 由表12.1(P325)得:,
3. 校核其连接强度
2T2,100000022,,,,,,80.81N/mm,,,110N/mmpp11dkl90,5,55111
2T2,100000022,,,,,,60.47N/mm,,,60N/mmpp22dkl70,4.5,105222
在半联轴器处采用C型键联接,故其联接强度为:
b202l,L,,125,,115mm2222
2T2,100000022,,,,,,55.21N/mm,,,60N/mmpp22dkl70,4.5,115222
所以两处平键连接的强度都足够。
螺纹连接
13.2 解:
1. 确定螺栓所需的预紧力
kTkFD/2kFD1.2,50,400ffQfQF,,,,,50kN,50000N0fzrfzD/2fzD0.12,8,50000
2. 确定螺栓的许用应力
2,,240N/mmSmin 由表13.5(P347)得:
由表13.7(P348)得:S = 2 (不控制预紧力,静载、碳钢、M16~M30)
,2402Smin,,,,,120N/mm,
S2
3. 确定螺栓直径
5.2F5.2,500000d,,,26.26mm1,,,,,,120
由机械设计手册得:取M30的螺栓,其d=27.835 mm,p =2 mm 。 1
13.3 解:一、计算螺栓受力
取螺栓组个数为20个 。
22Dp,500,2,,F,,,392699.08NQ44 1. 汽缸盖所受合力F : Q
2. 单个螺栓所受最大工作载荷Fmax
F392699.08QF,,,19634.95Nmaxz20
F',1.8F,1.8,19634.95,35342.92N 0max 3. 剩余预紧力F′: 0
4. 螺栓所受最大拉力F ?
F=F+ F′= 19634.95 + 35342.92 = 54977.87 N ? max 0
5. 相对刚度系数
C1,0.3
C,C12由表13.9(P351)得:
6. 预紧力F 0
C1F,F',F,35342.92,0.3,19634.95,41233.41N00maxC,C12
二、设计螺栓尺寸
1. 选择螺栓材料及等级
因螺栓受变载荷作用,故按静强度条件进行设计,按变载荷情况校核螺栓疲劳强度。螺
栓选用45钢,强度等级为5.6级。
2. 计算许用应力
2,,340N/mmSmin由表13.5(P347)得:
由表13.7(P348)得:S = 3 (不控制预紧力,变载、碳钢、M16~M30)
,3402Smin,,,,,113.3N/mm,
S3
3. 确定螺栓直径
5.2F5.2,54977.87,d,,,28.34mm1,,,,,,113.3 由机械设计手册得:取M30的螺栓,其d=28.376 mm, 1
d=29.026 mm , p =1.5 mm 。 2
三、校核螺栓的疲劳强度
1. 计算螺栓的应力幅
C2F2,19634.952max,,,0.3,,4.66N/mma22C,C,,d,28.376121 2. 计算许用应力
2,,220N/mm,T1由表13.4(P347)得:
,,0.64,k,3.45,并取S,3,, 由表13.7(P348)得:
,,0.64,2202,1T,,,,,13.6N/mm,aSk3,3.45,, 则
3. 疲劳强度校核
22,,,,4.66N/mm,,,13.6N/mmaa因 ,所以疲劳强度满足要求。
故该螺栓组选用直径为:M30的20个螺栓。