N CHI TIT MY S 4: THIT K H DN NG BNG TIThng s u vo : 1. Lc ko bng ti F =850 N 2. Vn tc bng ti v =1,95 m/s 3. ng kinh tangD = 370mm 4. Thi hn phc v Lh= 18000 gi 5. S ca lm vic: S ca = 2ca 6. Gc nghing ng ni tm ca b truyn ngoi: 1357. c tnh lm vic: Va p nh

PHN 1: CHN NG C IN V PHN PHI T S TRUYN1.1.Chn ng c in1.1.1.Xc nh cng sut yu cu ca trc ng c

Trong Pct : Cng sut trn mt trc cng tc Pyc : Cng sut trn trc ng c

Hiu sut ca b truyn:(1)

Tra bng ta c:

Hiu sut ca mt cp ln := 0,99Hiu sut ca b ai :0,95Hiu sut ca b truyn bnh rng :0,95

Hiu sut ca khp ni: 1Thay s vo (1) ta c:

= 0,993.0,95.0,95.1 = 0,876Vy cng sut yu cu trn trc ng c l :

1.1.2.Xc nh s vng quay ca ng c

Trn trc cng tc ta c:nct= =100,7 (v/ph)

Trong :(2)

Tra bng ta chn c t s truyn s b ca:Truyn ng ai: 3

Truyn ng bnh rng tr: 3,5 (hp gim tc mt cp)Thay s vo (2) ta c:

3.3,5= 10,5

Suy ra :100,7.10,5 =1057,35 (v/ph) Chn s vng quay ng b ca ng c: ndc =1500 (v/ph)1.1.3.Chn ng c T Pyc = 1,66 kW & ndc =1500 v/ph

Tra bng ph lc ta c ng c in + k hiu : 4AX90L4Y3

+ 2,2 (kW)

+ =1420 (v/ph)

+=24 (mm)

1.2.Phn phi t s truyn 1.2.1Xc nh t s truyn chung ca h thng Theo tnh ton trn ta c:

1429(v/p)nct = !00,7(v/ph) T s truyn chung ca h thng l :

1.2.2 Phnphi t s truyn cho h

Chn trc t s truyn ca b truyn trong = 3,5

1.3.Tnh cc thng s trn cc trc 1.3.1.S vng quay Theo tnh ton trn ta c: ndc = 1420(vg/ph) T s truyn t ng c sang trc I qua ai l:

S vng quay thc ca trc cng tc l:

1.3.2.Cng sut

Cng sut trn trc cng tc (tnh trn) l: Pct = 1,66()Cng sut trn trc II l :

Cng sut trn trc I l :

Cng sut thc ca ng c l:

1.3.3.Mmen xon trn cc trc

Mmen xon trn trc I l :

Mmen xon trn trc II l :

Mmen xon trn trc cng tc l:

Mmen xon thc trn trc ng c l :

1.3.4Bng thng s ng hc

Thng s/Trcng CIIICng Tc

U =3,5=1

n(v/ph)1420352,36100,7100,7

P(KW)1,901,791,681,66

T(N.mm)12778,2 48514,3159324,7157428

II.TNH TON THIT K B TRUYN NGOI.Tnh ton thit k b truyn aithang.

Cc thng s yu cu:

2.1.Chn loi ai v tit din ai.Chn ai thang thng.

Tra th vi cc thng s ta chn tit din ai: 0

2.2.Chn ng knh hai ai:

Chn theo tiu chun theo bng:d1=112 mm Kim tra vn tc ai:

tha mn.

Xc nh : d2 = u.d1.(1-) = 4,03.112. (1 0,03) = 437,8 mm

:H s trt, Chn = 0,03

Tra bng ta chn theo tiu chun: d2 = 450 mm

T s truyn thc:

Sai lch t s truyn:

Tha mn.

2.3.Xc nh khong cch trc a.

Da vo . Tra bng .. Ta chn

Vy:

Chiu di ai:

Da vo bngta chn L theo tiu chun:Chn

S vng chy ca ai trong.

Tha mn.

Tnh chnh xc khong cch trc:

Trong :

Vy:Xc nh gc m trn bnh ai nh:

Suy ra tha mn2.4.Tnh s ai Z.P : Cng sut trn bnh ai ch ng P=1,90(KW)

:Cng sut cho php.Tra bng theo tit din ai O,

V .Ta c: S ai Z c tnh theo cng thc:

:H s ti trng ng.Tra bng ta c :H s nh hng ca gc m.

Tra bngvi ta c:

:H s nh hng ca chiu di ai.

Tra bngvi ta c:

:H s nh hng ca t s truyn.

Tra bng vi ta c:

:H s k n s phn b khng u ti trng gia cc dy ai.

Tra bngtheota c:

Vy: Ly Z=22.5. Cc thng s c bn ca bnh aiChiu rng bnh ai : B = (Z-1).t + 2e

Tra bng ta c

B = (Z-1).t + 2e = (2-1).12+2.8 = 28 mm

Gc chm ca mi rnh ai : ng knh ngoi ca bnh ai :

ng knh y bnh ai:

2.6.Xc nh lc cng ban u v lc tc dng ln trc.

Lc cng ban u:

B truyn nh k iu chnh lc cng :

qm - Khi lng 1m ai, tra bng tit din ai O => qm = 0,061 (kg/m)

nn

Do :Lc tc dng ln trc bnh ai:

2.7.Tng kt cc thng s ca b truyn ai

Thng sK hiuGi tr

Tit din aiO

ng knh bnh ai nh

112

ng knh bnh ai ln

450

ng knh nh bnh ai nh

117

ng knh nh bnh ai ln

455

ng knh chn bnh ai nh

107

ng knh chn bnh ai ln

445

Gc chm rnh ai

38

S aiz2

Chiu rng ai

28

Khong cch trc

737,83

Gc m bnh ai nh

Lc cng ban u

100,4

Lc tc dng ln trc

391,31

Thit K B TRUYN BNH RNG CN RNG THNGThng s u vo: P = PI= 1,79(kW) T1= TI=48514,3(Nmm) n1= nI=352,36(vg/ph) u = ubr=3,5 Lh= 18000 (gi)3.1 Chn vt liu bnh rng

Tra bng , ta chn:Vt liu bnh rng ln: Nhn hiu thp: 45 Ch nhit luyn: Ti ci thin rn: Ta chn HB2=230 Gii hn bn b2=750 (MPa) Gii hn chy ch2=450 (MPa)Vt liu bnh rng nh: Nhn hiu thp: 45 Ch nhit luyn: Ti ci thin rn: HB=192240, ta chn HB1= 245 Gii hn bn b1=850 (MPa) Gii hn chy ch1=580 (MPa)

3.2.Xc nh ng sut cho php3.2.1.ng sut tip xc cho php[H] v ng sut un cho php [F]

, trong :Chn s b:

SH, SF H s an ton khi tnh ton v ng sut tip xc v ng sut un: Tra bng vi: Bnh rng ch ng: SH1= 1,1; SF1= 1,75 Bnh rng b ng: SH2= 1,1; SF2= 1,75

- ng sut tip xc v un cho php ng vi s chu k c s:

=>

Bnh ch ng:

Bnh b ng: KHL,KFL H s tui th, xt n nh hng ca thi gian phc v v ch ti trng ca b truyn:

, trong :mH, mF Bc ca ng cong mi khi th v ng sut tip xc. Do bnh rng c HB mH = 6 v mF = 6NHO, NFO S chu k thay i ng sutkhi th v ng sut tip xc v ng sut un:

NHE, NFE S chu k thay i ng sut tng ng: Do b truyn chu ti trng tnh => NHE= NFE= 60c.n.t , trong :c S ln n khp trong 1 vng quay: c=1n Vn tc vng ca bnh rngt tng s thi gian lm vic ca bnh rng

Ta c: NHE1> NHO1 => ly NHE1= NHO1 => KHL1= 1 NHE2> NHO2 => ly NHE2= NHO2 => KHL2= 1 NFE1> NFO1 => ly NFE1= NFO1 => KFL1= 1 NFE2> NFO2 => ly NFE2= NFO2 => KFL2= 1Do vy ta c:

Do y l b truyn bnh rng cn rng thng

=>(MPa)

3.2.3. Xc nh chiu di cn ngoi

Theo cng thc (6.15a):

Vi T1 l mmen xon trn trc ch ng. T1= TI=48514,3(N.mm)[H] - ng sut tip xc cho php; [H] = 481,8( MPa).

KR h s ph thuc vo vt liu lm bnh rng v loi bnh rng: i vi bnh rng cn rng thng lm bng thp =>U-T s truyn u=3,5

- H s chiu rng vnh rng : chn s b

=>

KH, KF H s k n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v ng sut tip xc v un: Tra bng vi

0,5-S b tr l s I- HB 0,04 Re + 10 = 0,04.136 + 10 = 16,2 (mm)Chn d1 = 18 (mm)d2 = (0,70,8)d1 = 12,614,4 mm chn d2 = 14mm)d3 = (0,80,9)d2 = 11,212,6 mm chn d3 = 12 (mm)d4 = (0,60,7)d2 = 8,49,8 chn d4 = 8 (mm)d5 = (0,50,6)d2 = 78,4 chn d5 = 8 (mm)

Mt bch ghp np v thn: Chiu dy bch thn hp, S3 Chiu dy bch np hp, S4 Chiu rng bch np v thn, K3S3 = (1,41,8)d3 = 16,821,6 mm chn S3 = 20(mm)S4 = (0,91)S3 = 1820 mm chn S4 = 20 (mm)K3 = K2 - (35) = 38- (35)= 3436 mm chn K3 = 35 (mm)

Kch thc gi trc: ng knh ngoi v tm l vt, D3, D2

B rng mt ghp bulng cnh , K2 Tm l bulng cnh , E2 v C (k l khong cch t tm bulng n mp l) Chiu cao, h

Trc II: D2 = 110(mm), D3 = 135 (mm)K2 = E2+R2+(35)=22+18+5=45 (mm)E2 = 1,6d2 = 1,6.14=22,4(mm) chn E2 = 22 (mm)R2 = 1,3d2 =1,3.14=18,2 (mm) chn R2 = 18 (mm)Chn h = 40 (mm)

Mt hp: Chiu dy: khi khng c phn li S1 khi c phn li: Dd, S1 v S2 B rng mt hp, K1 v q

Chn S1 = (1,31,8)d1 =(23,432,4) chn =S1 =28(mm)

S2=(1,01,1)d1=(1819,8) chn S2=18 (mm)K1 = 3d1 = 3.18=54 (mm), q K1 + 2 =54+2.8= 70 (mm)

Khe h gia cc chi tit: Gia bnh rng vi thnh trong hp Gia nh bnh rng ln vi y hp Gia mt bn ca cc bnh rng vi Nhau

(11,2) = (11,2).8=(89,6)chn = 10 (mm)

1 (35) = (35).8=(2440) chn = 35 (mm)

2=8 chn 2=8 (mm)

S lng bulng nn, ZZ=4

6.2.2 .Kt cu np v cc lt 6.2.2.1 Np ng knh np c xc nh theo cng thc:

Trong D l ng knh lp lnTrc II62759052M648

6.2.3.Ca thm

kim tra qua st cc chi tit my trong khi lp ghp v du vo hp, trn nh hp c lm ca thm.Da vo bng ta chn c kch thc ca thm nh hnh v sau:A(mm)B(mm)

(mm)

(mm)C(mm)

(mm)K(mm)R(mm)Vt(mm)S lng

10075150100125-8712M8224

`

6.2.4.Nt thng hi

Khi lm vic, nhit trong hp tng ln. gim p sut v iu ha khng kh bn trong v ngoi hp, ngi ta dng nt thng hi.Nt thng hi thng c lp trn np ca thm. Tra bng ta c kch thc nt thng hi

ABCDEGHIKLMNOPQRS

M2721530154536326410822632183632

6.2.5.Nt tho du

Sau mt thi gian lm vic, du bi trn cha trong hp, b bn (do bi bm v do ht mi), hoc b bit cht, do cn phi thay du mi. thay du c, y hp c l tho du.Lc lm vic, l c bt kn bng nt tho du. Da vo bng ta c kch thc nt tho du DbmfL cQDS

M2021593282,517,8302225,4

6.2.6.Kim tra mc du kim tra mc du trong hp ta dng que thm du c kt cu kch thc nh hnh v.3018126612

6.2.7.Cht nh v.Mt ghp gia np v thn nm trong mt phng cha ng tm cc trc.L tr lp thn hp & trn np c gia cng ng thi, m bo v tr tng i gia np v thn trc v sau khi gia cng cng nh khi lp ghp, ta dng 2 cht nh v, nh cccht nh v khi xit bulong khng lm bin dng vng ngoi ca .

Thng s k thut ca cht nh v l d=4c=0,6l=38

6.2.8.ng lt v lp ng lt c dng ln, thun tin khi lp v iu chnh b phn ng thi tri cho khi bi bm, cht bn ng lt c lm bng vt liu GX15-32 ta chn kch thc ca ng lt nh sau.

Chiu dy: , ta chn

Chiu dy vai v chiu dy bch

ng knh l lp ng lt

6.3.4. Bng thng k cc kiu lp v dung sai:Ti cc tit din lp bnh rng khng yu cu tho lp thng xuyn ta chn kiu lp H7/k6, tit din lp trc vi ln, khp ni, a xch c chn trong bng sau

TrcV tr lpKiu lp ES es

EI ei

I

Trc-vng trong bi35k6

+15

+2

V-lp

+30+21

0+2

Trc-bnh rng

40+21+15

0+2

Bc chn trc I

+98+15

+65+2

IIKhp ni38+21+15

0+2

ln 40k6+18

+2

Bnh rng 45+25+18

0+2

V v ln90H7+30

0

Then bnh rang

1400

-43-27

MC LC

PHN 1: CHN NG C IN V PHN PHI T S TRUYN .1.1.Chn ng c in........................1.1.1.Xc nh cng sut yu cu ca trc ng c1.1.2.Xc nh s vng quay ca ng c1.1.3.Chn ng c..1.2.Phn phi t s truyn.1.2.1Xc nh t s truyn chung ca h thng1.2.2 Phn phi t s truyn cho h..1.3.Tnh cc thng s trn cc trc1.3.1.S vng quay1.3.2.Cng sut........................1.3.3.Mmen xon trn cc trc1.3.4Bng thng s ng hc.PHN 2 : TNH TON THIT K B TRUYN AI2.1.chn vt liu bnh rng2.2.xc nh ng sut cho php...........................2.3.Xc nh s b khong cch trc..........................2.4.xc nh thng s n khp..2.5.xc nh cc h s v thng s hnh hc .2.6.kim nghim b truyn bnh rng .2.7.mt vi thng s hnh hc ca bnh rng.2.8.Tngkt cc thng s ca b truyn bnh rng.PHN 3 : TNH TON THIT K B TRUYN TRONG.3.1.Chn vt liu bnh rng...3.2.Xc nh ng sut cho php3.3.Xc nh chiu di cn ngoi theo cng thc sau..3.4.Xc nh cc thng s n khp.3.4.1. Xc nh m un php3.4.2. Xc nh s rng ....3.4.3. Xc nh gc nghing ca rng..3.5.Xc nh cc h s v mt s thng s ng hc... 3.6.Kim nghim b truyn bnh rng 3.6.1 Kim nghim v ng sut tip xc .3.6.1 Chiu rng vnh rng...3.6.2 Kim nghim bn un.3.6.3 Kim nghim v qu ti ..3.7.Mt vi thng s hnh hc ca cp bnh rng .3.8.Bng tng kt cc thng s ca b truyn bnh rngPHN 4 : CHN KHP NI & THIT K TRC4.1. Chn khp ni4.1.1.Chn khp ni .4.1.2.Chn vt liu .... 4.2.Tnh trc ........................................4.2.1S t lc .4.2.2.Chn vt liu ch to trc4.2.3.Xc nh s b ng knh trc ................................4.2.4Chn s b ln4.2.5.Xc nh khong cch gia cc gi & im t lc4.2.6Tnh phn lc v v biu m men ..4.2.7.Tnh m men tng tng ng4.2.8.Xc nh ng knh cc on trc 4.2.9.Chn then v kim nghim bn ca then.4.2.10.Kim nghim trc v bn mi... PHN 5 : TNH CHN V KIM NGHIM LN5.1.Chn loi ln ..5.2.Chn cp chnh xc ln 5.3.Chn kch thc theo kh nng ti ng.. PHN 6: TNH TON V HP V CC CHI TIT KHC6.1.Tnh ton v hp gim tc........6.2.Tnh ton kt cu cc chi tit khc6.2.1.Kt cu bnh rng6.2.2.Ca thm....6.2.3.Nt thng hi............................6.2.4.Nt tho du ..6.2.5.Kim tra mc du ..6.2.6.Cht nh v....6.2.7.ng lt v lp ..6.2.8.Bulng vng ..6.3.Bi trn v iu chnh n khp6.3.1.Bi trn trong hp gim tc.. .. 6.3.2.Bi trn ngoi hp 6.3.3.iu chnh n khp .. ...6.4. Bng thn k cc kiu lp v dung sai.Mc lc. .Ti liu tham kho

TI LIU THAM KHO1.Trnh Cht, L Vn Uyn Tnh ton thit k h dn ng c kh, tp 1,2 NXB KH&KT, H Ni,20072.Nguyn Trng Hip Chi tit my, tp 1,2 NXB GD, H Ni,20063.Ninh c Tn Dung sai v lp ghp NXB GD, H Ni, 2004