thiet ke he dan dong thung tron - deso5

8/3/2019 Thiet Ke He Dan Dong Thung Tron - Deso5

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THIET KE TRAM DAN ONG THUNG TRON.

1

T 0,8 T

P H N G A N :

.

1. ong c ien 3 pha khongong bo.

2. Bo truyen ai thang.

3. Hop giam toc banh rang tru 2cap tach oi cap nhanh.

4. Noi truc vong an hoi.

5. Thung tron.

0,6 tCK 0,4tCK

S o tai trong

So lieu:

Cong suat tren thung tron P(kW) :..........35 ..................So vong quay cua thung tron n(v/ph) :..........45 ..................

Thi gian phuc vu a(nam) : ..........8 ....................

Quay 1 chieu, lam viec 2 ca, tai va ap nhe.

( 1 nam lam viec 300 ngay, 1 ca lam viec 8 gi )

45

1

3


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2

Cong suat tren truc trung tron: P1 = 3,5 Kw

So vong quay: n = 45 vg/ph

Thi gian phuc vu: a = 8 nam

Quay 1 chieu, lam viec 2 ca, tai va ap nhe (1 nam lam viec 300 ngay, 1ca lam 8 gi)

Gia i :

Pyc = P1 = 3,5 Kw

T 2 0,6tck 0,8T 2 0,4tck

Pt = Ptd = P1 . . + . = 3,24 KwT tck T tck

Chon (bang 2.3)

= 0,95 = 1 . olan4 . br2 . k = 0,95 . 0,994 . 0,972 . 0,99 =0,85

ol = 0,99 Pt 3,24br = 0,97 Pct = = = 3,8 KWk = 0,9 9 0,85

nlv = 45 vg/phChon u = 3 , uh = 10 ut =u . uh = 3 . 10 = 30

nsb = nlv . ut = 45 . 30 = 1350 vg/ph

Da vao P1.2: chon ong c DK51_2

Co Pc = 4,5 KW , nc = 1440 vg/ph

nc 1440

ut = = 32nlv 45

ut = u . uh Chon u = 3,15 32

uh = 103,15


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T bang 3.1 u1 = 3,58 ; u2 = 2,79

P1 = Pct . . ol = 3,8 . 0,95 . 0,99 = 3,6 KW

nc 2900n1 = = = 921vg/ph

u

3,15

P1 3,6T1 = 9,55 . 106 . = 9,55 . 106 . = 37329 Nmm

n1 921

P2 = P1 . br . ol = 3,8 . 0,97 . 0,99 = 3,5 KW

n1 921n2 = = = 257 vg/ph

u1 3,58

P2 3,5T2 = 9,55 . 106 . = 9,55 . 106 . = 130058 Nmm

n2 257Pc 4,5

Tc = 9,55 . 106 . = 9,55 . 106 . = 29844 Nmmnc 1440

P3 = P2 . br . ol = 3,5 . 0,97 . 0,99 = 3,4 KW

n2 257n3 = = = 92 vg/ph

u2 2,79P3 3,4

T3 = 9,55 . 106 . = 9,55 . 106 . = 352935 Nmmn3 92

TrucThong so

ong c 1 2 3

Cong suat P, KW 4,5 3,6 3,5 3,4T so truyen u 3,15 3,58 2,79So vong quay n vg/ph 1440 921 257 92Momen xoan T, Nmm 29844 37329 130058 352935


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Tnh toan bo truyen ai : V cong suat thc te ma bo truyen ai nhan cch la Pct nen tnh toan ta dung Pct :

Pct = 3,8 KW , n1 = 1440 vg/ph , u = 3,15.

Theo hnh 4.1: chon ai A

Chon d1 theo bang 4.13 => d1 = 140 mm=>V = . d1 . n1/60000 = 140 . 1440 . / 60000 = 10,6 m/sd1 . u chon = 0,01

d2 = 140 . 3,15(1 - ) => d2 = 445 mm

(1 0,01)

Chon theo tieu chuan: d2 = 450 mm

Tren thc te: d2 450ut = = = 3,25

d1 . (1 - ) 140 . (1 0,01)

ut u 3,25 3,15u = = = 0,032 < 0,4

u 3,15

Theo bang 4.14 chon s bo khoang cach truc a = d2 = 450 mm

Theo cong thc 4.4, ta co:Chieu dai ai:

L = 2a + 0,5 . (d2 + d1) + (d2 d1)2/ 4a

= 2 . 450 + 0,5 . . (450 + 140) + (450 140)2 / (4 . 450) = 1880 mm

Chon theo tieu chuan: l = 2000 mmv 10,6

So vong chay cua ai trong 1 giay: i = = = 5,3 < 10 m/s

Tnh lai khoan cach truc: l 2a = ( + 2 - 2 . 8 ) / 4

ma = l ((d1 + d2) / 2 = 2000 (140 + 450) . / 2 = 1313 mm = (d2 d1) / 2 = ( 450 140) / 2 = 155 mm

=> a = (1313 + 13132 8 . 1552 ) 638 mm


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Goc om 1 = 180 57 . (d2 d1) / a = 180 57 . (450 140) / 638 = 1520

=> 1 > mm = 1200

Theo 4.16 : P1 . KdChon z =

Kd = 1,25 (bang 4.7) ([P0] . C . Cl . Cu . Cz)C = 0,92 (bang 4.15) 3,8 . 1,25Cl = 1,04 (bang 4.16) = = 1,69 2Cu = 1,14 (bang 4.17) (2,71 . 0,92 . 1,04 . 1,14 . 0,95)Cz = 0,95 9bang 4.18)

Vi t , h0 , e tra bang 4.21

Chieu rong ai: B = (z 1)t + 2e = (2-1).15 + 2.10 = 35 mm

ng knh ngoai: da = d + 2h0 = 140 + 2.3,3 = 146,6 mm 147 mm

Lc cang ai: 780.P1.kd 780 . 3,8 . 1,25F0 = = = 201,76 N

V.C . z + qm.V2 10,6 . 0,92 . 2 + 0,105 . 10,62

Lc tac dung len truc:

Fr = 2 . F0 . z . sin(/2) = 2 . 201,76 . 2 . sin(152/2) = 783,1 N

Bang 4.8/ 55 1 d1 140

chon = => = = = 3,5 mmd1 40 40 40FV = qm . V2 = 0,105 . 10,62 = 11,978 N.


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BO TRUYEN BANH RANG

Chon vat lieu: thep 45 toi cai thien.Banh nho: o ran 250 HB co b = 850 MPa, ch = 580 MPa

Banh ln: o ran 220 HB co b = 750 MPa, ch = 450 MPa

Ty so truyen: uhs = 10 u1 = 3,58 (cap nhanh)u2 = 2,79 (cap cham)

Xac nh ng suat cho phep:

* ng suat tiep xuc: tra bang 6.2/ 94

Hlim0 = 2HB +70 => Hlim10 = 2 . 250 + 70 = 570 MPa

SH = 1,1 Hlim20 = 2 . 220 + 70 = 510 MPa

Flim0 = 2HB +70 => Flim10 = 2 . 250 + 70 = 570 MPa

SF = 1,1 Flim20 = 2 . 220 + 70 = 510 MPa

Tnh NHO , NHE : ong c lam viec vi chu k thay oi nen:

NHO = 30 . HHB2,4 => NHO1 = 30 . 250 2,4 = 1,71.107

NHO2 = 30 . 2202,4 = 1,26.107TI 3

NHE = 60 . c . . .ni . tITmax

=> NHE1 = 60 . 1 . 921 . 300 . 7 . 16 . (0,6 + 0,83. 0,4) = 1,49.109

NHE2 = 30 . 1 . 257 . 300 . 7 . 16 . (0,6 + 0,83. 0,4) = 0,42.109

Ta co : NHO1 < NHE1 KHL1 = 1

NHO1 < NHE2 KHL1 = 1KHL 1 [H] = Hlim0. [H ]1 = 570 . = 518,2 MPa

SH => 1,11

[H ]2 = 510 . = 463,6 MPa1,1


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Vi cap nhanh s dung rang nghieng:

([H ]1 + [H ]2 ) ( 518,2 + 463,6)[H ] = = = 490,9 < 1,25 . [H ]min

2 2 < 1,25 . 463,6 = 579,5 MPaVi cap cham s dung rang thang:

* ng suat uon: TI 6

NFE = 60 . c . . .ni . tITmax

=> NFE1 = 60 . 1 . 300 . 7 . 16 . 921 .(0,6 + 0,86. 0,4) = 1,3.109

NFE2 = 30 . 1 . 300 . 7 . 16 . 257 .(0,6 + 0,86. 0,4) = 0,36.109

Chon NFO1 = NFO2 = 5.106 < NFE1 , NFE2=> KFL1 = KFL2 = 1

V bo truyen quay 1 chieu nen KFC = 1KFL 1 [F] = Flim0. KFC . KFL . [F ]1 = 450 . 1 . 1 . = 257,14 MPa

SF => 1,751

[H ]2 = 396 . 1 . 1 . = 226,3 MPa1,75

ng suat qua tai cho phep oi vi cac banh rang toi cai thien:

[F]max = 2,8 . ch = 2,8 . 450 = 1260 MPa[F]1max = 0,8 . ch1 = 0,8 . 580 = 464 MPa

[F]2max = 0,8 . ch2 = 0,8 . 450 = 360 MPa

Tnh toan cap cham: ( banh rang thang)

. Xac nh s bo khoan cach truc: 3 T2 . KH

aw2 = Ka . (u 1).

[H]22

. u2 . ba(bang 6.6/ 97) , ( bang 6.5)

Chon: ba = 0,4 , Ka = 49,5 , T2 = 130058 Nmm

bd = 0m53 . ba .(u 1) = 0,53 . 0,4 .(2,79 + 1) = 0,8

(bang 6.7) KH = 1,02 [H]2 = 463,6 MPa


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3 130058 . 1,02

aw2 = 49,5 . (2,79 + 1). = 154 mm463,62 . 2,79 . 0,4

Chon aw2 = 154 mm.

. Xac nh cac thong so an khp:m = (0,01 . . . 0,02) , aw2 = (1,54 . . . 3,08)

2 . aw2 2 . 154Chon m = 3 mm => z1 = = = 27

[m . (u2 +1)] [3 . (2,79 + 1)]

ma z2 = u2 . z1 = 28 . 2,79 = 75

Ty so truyen thc te : z2 75ut2 = = = 2,78

z1 27Tnh (z1 + z2) (75 + 27)

aw2 = m . = 3. = 153 mm2 2

Chon aw2 = 154 mmGoc an khp: ( z1 + z2) . m . cos (27 + 75) . 3 . cos200

cos tw2 = = = 0,942. aw 2 . 154

tw2 = 210

Kiem nghiem rang ve o ben tiep xuc:

2 . T2 . KH.(ut2 + 1)H = ZM . ZH . Z .

bw . ut2 . dw2

Chon:ZM = 274 . MPa (bang 6.5)

2 . cosb2 2 . 1 (vi cosb2 = 1)ZH = = = 1,73

sin 2tw2 sin 2. 21

Vi banh rang thang dung 6.36a , tnh Z(4 - ) (4 1,72) = 1,88 3,2 .(1/ z1 + 1/ z2)

Z = = = 0,87 Vi = 1,88 3,2 . (1/ 27 + 1/ 75)3 3 = 1,72


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ng knh vong lan banh nho :2 . aw2 2 . 154

dw1 = d1 = = = 81,48 mm 81,5 mmut2 + 1 2,78 + 1

. dw1 . n2 . 81,5 . 257

V = = = 1,096 m/s chon cap chnh xac 960000 600000

Chon: bang 6.16 -> g0 = 73 , bang 6.15 -> H = 0,006

aw2 154 H = H . g0 . V. = 0,006 . 73 . 1,096. = 3,57

ut2 2,78

Chon: (tra bang 6.14) -> KH = 1,02 bw2 = ba . aw2K

H= 1,13 = 0,4 . 154 = 61,6 mm

H . bw2 . dw1 3,57 . 61,6 . 81,5KHV = 1 + = 1 + = 1,06

2 . T . KH . KH 2 . 130058 . 1,02 . 1,13

KH = KH . KHV . KH = 1,13 . 1,06 . 1,02 =1,22

2 . 130058 . 1,22 . (2,78 + 1)=> H = 274. 1,73 . 087 . = 423,5 MPa

61,6 . 2,78 . (81,5)2

Theo 6.1: V= 1,14 -> ZV = 1 Cap chnh xac ong c la 9, chon capchnh xac ve mc tiep xuc la 9 khi o gia cong o nham R2 = 10 40 m

Chon ZR = 0,9 va da < 700 KXH = 1

[H]2 = [H]2 . ZV . ZR . KXH = 463,6 . 1 . 0,9 . 1 = 417,24 MPaTa thay H > [H] nhng chenh lech nay nho nen ta co the tang chieu

rong vanh rang. H 2 423,5 2

bwz = bw2. = 61,6. 63,46 mm 63,5 mm[H] 417,24Cac thong so va kch tht bo truyen:

aw2 = 154 mm = 0 ng knh vong chia:m = 2 mm z1 = 27 d1 = 81 mmbw2 = 63,5 mm z2 = 75 d2 = 225 mmut2 = 2,78


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ng knh nh: ng knh ay:

da1 = d1 + 2 . m = 87 mm df1 = d1 + 2,5 . m = 79,5 mmda2 = d2 + 2 . m = 231 mm df2 = d2 + 2,5 . m = 223,5 mm

Tnh toan cap nhanh: (bo truyen banh rang nghieng V)

1) Xac nh s bo khoang cach truc:3 T1 . KH Chon ba = 0,3, => bd = 0,5 . ba .(u1 + 1)aw1 = Ka . (u 1). = 0,5 . 0,3 . (3,58 + 1)

[H]2 . u1 . ba = 0,6873 37329 . 1,09 bang 6.7: chon KH = 1,09 ; Ka = 43

= 43 . (3,58 +1). = 102,6 mm T1 = 37328 Nmm(518,2)2 . 3,58 . 0,3

Chon aw = 120 mm

2) Xac nh cac thong an khp:

m = ( 0,01 . . . 0,02) * aw1 = (1,2 . . . 1,4)

Chon m= 2 ; = 300 => cos = 0,866

Tnh z1: 2 . aw1 . cos 2 . 120 . 0,866z1 = = =23 rang

m . (u1 + 1 ) 2 . (3,58 + 1)T so truyen thc: ut1 = 82 / 23 = 3,57

m . (z1 + z2) 2. (23 + 82)cos = = = 0,875

2 . aw1 2 . 120=> = 28,95 290

c) Kiem nghiem rang ve o ben tiep xuc:

ng suat tiep xuc tren be mat rang lam viec:

2 . T1. KH . (ut +1)

H = zM . zH . z . bw . ut1 . dw12

T1 = 37329 / 2 = 18664,5 Nmm

ZM = 274 MPa


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Theo 6.35: tgb1 = cost . tg ( = 200)

oi vi banh rang nghieng khong dch chnh:

tg tg 200

tw1 = t = arctg = arctg = 22,590

cos cos 290

=> tgb1 = cos 22,50 . tg 290 = 0,51

=> b1 = 27,060

2 . cosb1 2 . cos 27,060

ZH = = = 1,58sin2. tw1 sin2.22,590

Theo 6.37: bw1 . sin 0,3 . 120 . sin290

= = = 2,78( . m) . 21 1

= 1,88 3,2. + . cosz1 z21 1

= 1,88 3,2. + . cos290 = 1,4923 82

1 1Z = = = 0,82

1,49

bw1 = ba . aw1 = 0,3 . 120 = 36mm

ng knh vong lan banh nho:

2 . aw1 2 . 120d1 dw1 = = = 52,52 mm

(ut1 + 1) (3,57 + 1)V = . dw1. n1 / 60000 = .52,52 . 921 / 60000 = 2,53 m/s

Theo bang 6.13: chon cap chnh xac la 9.


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Theo bang 6.14, 6.15, 6.16 chon H = 0,002 , g0 = 73 , kH= 1,113

aw1 120 H = H . g0 . v . = 0,002 . 73 . 2,53 . = 2,14

ut1 3,57

H . bw1 . dw1 2,14 . 36. 52,52KHV =1 + = 1 + = 1,04

2 . T1. KH .KH 2. 37329 . 1,13 . 1,09

KH = KHV . KH . KH = 1,13 . 1,04 . 1,09 = 1,29

(2 .37329)/2 . 1,29 .(3,57 + 1)H = 274 . 1,55 . 0,84 . = 281,08MPa

36. 3,57 . (52,52)2

YF2 3,6F2 = F1 . = 71,88 . = 68,09 MPa < [F2]

YF1 3,8

Cac thong so va kch tht bo truyen:

aW1 = 120 mm ng knh vong lan:m = 2 mm 2 . aW 2 . 120bW1 = 36 mm d1 dW1 = = = 52,52 mmut1 = 3,57 (ut + 1) (3,57 + 1)

z1 = 23 d2 dW2 = dW1 . u = 52,52 . 3,57 = 187,5 mmz2 = 82 ng knh vong chia:

m . z1 2 . 23d1 = = = 52,57 mm = 53 mm

cos cos290

m . z2 2 . 82d2 = = = 187,43 mm = 187 mm

cos cos290

ng knh ay rang:df1 = d1 2,5.m = 48 mmdf2 = d2 2,5.m = 182 mm

ng knh nh:

da1 = d1 + 2.m = 57 mm


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da2 = d2 + 2.m = 191 mm

Mc dau thap nhat cach tam cac banh rang la:

d2 187xmin = - 2,25.m = - 2,25 . 2 = 89 mm

2 2Mc dau cao nhat cach tam cac banh rang la:

d2 225xmax = = = 75 mm

3 3=> 10 mm < x = 89 75 = 14 mm < 15 mm

(Thoa ieu kien boi trn).


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TNH TOAN TRUC

1) Chon vat lieu che tao cac truc la: thep 45 co b = 600 MPa

ng suat xoan cho phep [ ] = 12 . . . 20 MPa2) Xac nh s bo ng knh truc:

Theo (10.9), ng knh truc th k vi k = 1, 2, 3 3 Tk Vi: T1 = 37329 Nmmdk T2 = 130058 Nmm

0,2 . [ ] T3 = 352935 Nmm[ ] = 15 MPa

d1 23,2 mm =23 mm=> d2 35,13 mm = 35 mm

d3 49 mm = 49 mm

Xac nh khoang cach gia cac goc va iem.

Ta chon cac hieu rong o lan: (bang 10,2) at lc:

Tnh theo truc 2 (d2 b0 = 21)

Chieu dai may banh ai: lm12 = ( 1,2 . . . 1,5 ).d1= ( 27,6 . . . 34,5 ) = 31 mm

Chieu dai may banh rang tru: lm13 = ( 27,6 . . . 34,5 )lm14 = ( 27,6 . . . 34,5 )

Chon lm13 = lm14 = 29 mm

oi vi truc 2: lm22 , lm23 , lm24 = (1,2 . . . 1,5). d2= (42 . . . 52,5)

Chon lm22 = lm24 = 47 mmlm23 = 49 mm

oi vi truc 3: lm32 = (1,2 . . . 1,5). d3 = (58,8 . . . 73,5)

(noi truc vong ai an hoi) lm33 = (1,4 . . . 2,5).d3 = (68,6 . . . 122,5)

Chon lm32 = 72 mmlm33 = 115 mm


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Khoang cach lki tren truc th k t goc O en chi tiet quay th i:Truc 2: l22 = [0,5 .(lm22 + b0) + K1 +K2] = [0,5 . (47 + 21 ) + 10 + 10] = 54 mm

l23 = l22 + 0,5 .( lm22 + lm23) + K1 = 54 + 0,5 . (47 + 49) +10 = 112 mml24 = 2. l23 - l22 = 2 . 112 54 = 170 mml21 = 2. l23 = 2 . 112 = 224 mm

Truc 3: l31 = l21 = 224 mm ; l32 = l23 = 112 mmlc33 = 0,5 .( lm33 + b0) + K1 + K2 = 0,5 . (115 + 21) + 10 + 10 = 88 mm

l33 = 2 . l32 + lc33 = 2 . 112 + 88 = 312 mmTruc 1:

lc33 = -[0,5 .( lm12 + b0) + K1 + K2 ]= -[0,5 . (31 + 21) + 10 + 10 = -56 mm

l13 = l22 = 54 mm; l14 = l24 = 170 mm; l11 = l21 = 224 mm.


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TNH TOAN CAC GIA TR LC LEN CACTRUC:

Truc 1: lc t banh ai tac dung len truc 1 l12

Fr1,2 = 744,3 N l14 l132. T1 2 . 37329 l11

Ft13 = Ft14 = = RX11 RX10 Ry10 y

dw1 52,5 1 Fr14 Fr13 O z =1422,1 N Ry11 Fa14 Fa13 Fr12 xFr13 = Fr14 = Ft1. tgw .cosw Ft14 Ft13

Vi w = 22,590

46564,2 20658,7 41680,8 w = 28,950 88810,5=> Fr13 = Fr14 = 1422,1 . cos28,95 . tg22,59 Mx

= 676,2 NFa13 = - Fa14 = F1. tgw

= 1422,1 . tg28,95 = 786,7 N My

Tnh cac phan lc len o truc: 76793,4Fx = - Rx10 + Ft13 + Ft14 Rx11 = 0

Mx0

= - Ft13. l13 Ft14. l14 + Rx11. l11 = 0Ft13. l13 + Ft14. l14=> Rx11 =

l11(54 + 170)

= 1422,1 . = 1422,1 N 18664,5224 37329

=> Rx11 = Rx10 = 1422,1 N

Fy = - Ry11 + Fr14 + Fr13 Ry10 Fr12 = 0My0 = - Fr12. l12 Fr13. l13 - Fr14 .l14 Fa13. dw1/2 + Fa14. dw1/2 + Ry11. l11 = 0

Fr12. l12 + Fr13 .(l13 + l14) 744,3 . 56 + 676,2 (54+170)=> Ry11 = = = 862,3 N

l11 224

Ry11 = - 2. Fr13 + Fr12 + Ry10 = - 2 . 676,2 + 744,3 + 862,3 = 254,2 N


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Truc 2:Ft22 = Ft24 = Ft13 = 1422,1 N

2 . T2 2 . 130058Ft23 = = = 3211,3 N

dw1 81

Fr22 = Fr24 = Fr13 = 676,2 N Ft24 Ft22Fa22 = Fa24 = Fa13 = 786,7 N Fa24 Fa22Fr23 = Fr23 . tg20 = 1168,8 N Fr24 Fr22 RY20 yFx = RX20 - Ft22 - Ft23 - Ft24 +RX2 =0 1 RX21 Fr23 O zMX0 = - Ft22 .l22 - Ft23 . l23 - Ft24 . l24 Ft23 RX20 x

+ RX21 .l21 = 0 l24 l23 l22Ft22 .(l22 + l24) + Ft23 . l23 l21

=> RX21 =l211422,1 .(54 + 170) + 3211,3 . 112

= 4957,2224 MX

= 3027,8 N=> RX20 = 2. Ft22 + Ft23 - RX21 65255,8

= 2. 1422,1 + 3211,3-3027,8= 3027,8 N

FY = RY20 Fr22 + Fr23 Fr24+RY21=0 102691MY0 = Fr22 .l22 Fr23 . l23 + Fr24 . l24 256631,8- RY21 .l21 Fa22. dw2 + Fa24. dw2 = 0

Fr22 .(l22 + l24) Fr23 . l23 163501,2=> RY21 =

l21 MY676,2 .(54 + 170) 1168,8 . 112

=224 130058

= 91,8 NRY20 = 2. Fr22 Fr23 RY21

= 2 . 676,2 1168,8 91,8 = 91,8 N


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Truc 3:Ft32 = Ft23 = 3211,3 NFr32 = Fr23 = 1168,8 N

2 . T3F33 = ( 0,2 . . . 0,3 ) .

D0Tra bang 16.10:T3 = 352935 Nmm D0 = 120 mm

F33 =(1176,5 . . . 1764,7) RY31 Fr32 Ft32 RY30Chon F33 = 1750 N F33

FY = RY30 Fr32 + RY31 = 0 1 O RX30M0Y = Fr32. l32 - RY31.l31 = 0 RX31

Fr32. l32 l32

=> RY31 = l31l31 l33

1168,8 . 112= = 584,4 N

224FX = RX30 Ft32 + RX31 F33 = 0 154000

= 1168,8 584,4 = 584,4 N 256827,2MX0 = - Ft32. l32RX31.l31+F33. l33=0

- Ft32. l32 + F33. l33 MY=> RX31 =l31

- 3211,3 . 112 + 1750 . 312=

224= 831,9 N MX

=> RX30 = Ft32 + RX31 F33 65452,8= 3211,3 + 831,9 1750= 2293,2 N

352935


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M2 = MX2 + MY2Mtd = M2 + 0,75 . T2

Mtd10 = 52748 Nmm Mtd20 = 0 Mtd30 = 0

Mtd11 = 0 Mtd21 = 0 Mtd31 = 342255 Nmm

Mtd12

= 32328 Nmm Mtd22

= 208991 Nmm Mtd32

= 404557 NmmMtd13 = 121777 Nmm Mtd23 = 298482 Nmm Mtd33 = 305651 Nmm

Mtd14 = 91251 Nmm Mtd24 = 208991 Nmm

3 MtdTnh d : d vi [ ] = 63MPa

0,1 [ ]

d10 = 20,3 d20 = 0 d30 = 0d11 = 0 d21 = 0 d31 = 37,9d12 = 17,2 d22 = 32,1 d32 = 40d13 = 26,3 d23 = 36,2 d33 = 36,5d14 = 24,4 d24 = 32,1

e u ben va lap ghep c, ta chon:

d12 = 20 mm d20 = 30 mm d30 = 40 mm

d10 = 25 mm d21 = 30 mm d31 = 40 mmd13 = 28 mm d22 = 35 mm d32 = 45 mmd14 = 28 mm d23 = 40 mm d33= 38 mmd11 = 25 mm d24 = 35 mm

Tnh kiem nghiem truc ve o ben moi :

a) Vi thep CT6 45 co b = 600 MPa

M = MX2 + MY2 -1 = 0,436 . b = 0,436 . 600 = 261,6 MPa

-1 = 0,58 . -1 = 0,58 . 261,6 = 151,7 MPa


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Theo bang 10.6 : = 0,05 , = 0

M13 . 32 133719 .32a13 = = = 70,76 MPa . d133 . 26,83

M10 . 32 41680,8 . 32a12 = 0 , a10 = = = 27,2 MPa

. d103 . 253

M22 . 32 176042,5 .32a22 = = = 41,8 MPa . d223 . 353

M23 . 32 276415 .32a23 = = = 43,99 MPa . d233 . 403

M32 . 32 265036 .32a32 = = = 29,63 MPa . d323 . 453

M31 . 32 154000 .32

a31 = = = 24,51 MPa . d313 . 403

a33 = 0

Chon kch thc then (tra bang 9.1a)

Tiet dien knh truc b*h t1 w (mm3) w0 (mm3)12 20 6*6 3,5 642 142713 (14) 28 8*7 4 1825 398122 (24) 35 10*8 5 3566 773523 40 12*8 5 5364 1164832 45 14*9 5,5 7611 1655733 38 10*8 5 4670 10057


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21

O LAN:

Chon o lan cho truc 1: lc doc truc Fa = 0

Fr10 = (Rx10)2 + (Ry10)2 = (1422,1)2 + (254,2)2 = 1445 N

Fr11 = (Rx11)2 + (Ry11)2 = (1422,1)2 + (862,3)2 = 1663 N

Tnh o mat cat 10 : Fa = 0

d10 = 25mmX = 1

Q10 = X.VFr10 + Y.Fa = 1445 N V = 1

mat cat 11:

d11 = 20mmX = 1

Q11 = X.VFr11 + Y.Fa = X . V . Fr11 = 1663 N vi V = 1

Nh vay ta ch can tnh o mat cat 11 la u:

106 . L 60 . n1 . Lh 60 . 2 . 921 . 8 . 300 . 8Lh = => L = = = 2122 N

60 . n1 106 106

oi vi hop giam toc th L = (10 . . . 25). 103 h nen ta phai chia oi thigian lam viec L:

L 2122=> L = = = 1061

2 2

V che o tai trong thay oi theo bac nen: (vi m = 10/3)

QIm

. LIm

Q0m

Lh1 O02m

Lh2QE = = Q11 . + + - LI O01 Lh Q01 LhLh = 33600

Lh1 = 0,6 . 33600 = 20160 ; Lh2 = 0,4 . 33600 = 13440

QE = 1663 . (0,6 + (0,8)10/3 . 0,4)3/10 = 1549,5 N


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22

Ta tnh: Ctt = QE . (L)3/10 = 1549,5 . (1061)3/10 = 12529 N

T d = 25 mmCtt = 12529 N Ctt = 22,6 KN

=> chon o bi ua tru ngan c trung hep kieu 305 co C0

= 14,3 KN

Kiem tra ieu kien tai tnh:X0 = 0,5

Qt = X0.Fr + Y0.Fa vi Y0 = 0,5

Fa = 0KCD con lan Cdai con lan d D B r

9 9 25 62 17 2,0

Qt = 0,5 . Fr = 0,5 . 1663 =831,5 N < C0

Thoa ieu kien tai tnh.

Tnh o lan truc 2:

Fr20 = Fr21 = (R20X)2 + (R20Y)2 = (3027,8)2 + (91,8)2 = 3029,2 NFa = 0

X = 1Q20 = X.VFr20 + Y.Fa = 3029,2 N V = 1

106 . L 60 . n2 . Lh

Lh = => L = = 592,13 N60 . n2 106

Ta chia oi thi gian lam viec:

L 592,13=> L = = = 296,06

2 2V tai trong thay oi theo bac :

QE = Q20m Qim . Li/LI = 3029,2 . (0,6 + (0,8)10/3 . 0,4)3/10 = 2823 N

Chon m = 10/3

Tnh Ctt :


23/24


23

Ctt = QE . (L)3/10 = 2823 . (296,06)3/10 = 15564 N

T d = 35 mmCtt = 15564 N Ctt = 22,0 KN

=> chon o bi c trung hep kieu 306 co C0 = 15,1 KN

Qt = X0.Fr + Y0.Fa = 0,6 .4085,2 = 2748,12knh con bi d D b r

12,3 30 72 19 2,0

Kiem tra ieu kien tai tnh:X0 = 0,6 . Y0 = 0,5

Qt = X0.Fr + Y0.Fa = X0 . Fr = 0,6 . 3029,2 = 1817,52 N

=> Qt < C0Thoa ieu kien tai tnh.

Tnh o lan truc 3:

Fr30 = (RX30)2 + (RY30)2 = (2293,2)2 + (584,4)2 = 2366,5 N

Fr31 = (RX31)2 + (RY31)2 = (831,9)2 + (584,4)2 = 1016,7 N

Chon F33 cung chieu Ft32 :

F33 Ft32

R31X R30X

FX = - R30X R31X + F33 + Ft32 = 0

MY0 = - Ft32 . l32 + R31X . l31 F33 . l33 = 0

Ft32 . l32 + Ft33 . l33 3211,3 . 112 + 1750 .312=> R31X = = = 4043,2 N

L31 224

=> R30X = F33 + Ft32 R31X = 3211,3 + 1750 4043,2 =918,1 N


24/24


V R31X > R30X nen ta ch can xet Fr31

Fr31 = (R31X)2 + (R31X)2 = (4043,2)2 + (584,4)2 = 4085,2 NX = 1

Q = X.VFr

31 + Y.Fa= 4085,2 N V = 1

Fa = 0

60 n3 . Lh 60 . 8 . 8 . 2 . 300 . 92L = = = 4043,2 N

106 106

L=>L = = 106

2

V tai trong thay oi theo bac :QE = Qm Qim . Li/LI = 4085,2 (0,6 + (0,8)10/3 . 0,4)3/10 = 3806,5 N

Ctt = QE . (L)3/10 = 3086,5 . (106)3/10 = 15421 N

Vi d = 45 mmCtt = 15421 N Ctt = 31,9 KN

=> chon o bi c trung kieu 308 co C0 = 21,7 KN

ieu kien tai tnh:

Vi o bi 1 day X0 = 0,6Y0 = 0,5

Qt = X0.Fr + Y0.Fa = 0,6 .4085,2 =2748,12

d D B r ng knh bi40 90 23 2,5 15,08

Thoa ieu kien tai tnh.

thiet ke he dan dong thung tron - deso5

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