TRNG I HC BCH KHOA TP. H CH MINH KHOA: KHOA HC NG DNG NGNH: C K THUT

N MN HC THIT K K THUT

Gio vin hng dn Sinh vin thc hin MSSV

: TS. V Cng Ha : L Phi H : K0304100

Thnh ph H CH MINH Thng 12 nm 2006

TRNG I HC BCH KHOA TP. H CH MINH KHOA: KHOA HC NG DNG NGNH: C K THUT

N MN HC THIT K K THUT

Gio vin hng dn Sinh vin thc hin MSSV Lp

: : : :

TS. V Cng Ha L Phi H K0304100 KU03BCKT

Thnh ph H CH MINH Thng 12 nm 2006

LI NI UH thng thng trn c s dng kh rng ri vi nhiu ng dng trong cng nghip, nng nghip, xy dng v sinh hot hng ngy. Mn hc n Thit K K Thut l c hi cho em tip xc, tm hiu v i vo thit k mt h thng dn ng thc tin, cng l c hi gip em nm r nhng kin thc hc v hc thm c rt nhiu v phng php lm vic khi thc hin cng vic thit k, ng thi cng tng bc s dng nhng kin thc hc vo thc t. Tp thuyt minh ny ch dng li giai on thit k, cha thc s ti u trong vic tnh ton cc chi tit my, cha mang tnh kinh t v cng ngh cao v gii hn v kin thc ca ngi thc hin. Em xin chn thnh cm n cc thy c gio trong B mn C K Thut cho em c hi c hc mn hc ny. Xin chn thnh cm n cc bn trong nhm cng tho lun v trao i nhng thng tin ht sc qu gi. Em xin chn thnh cm n thy gio Ts.V CNG HA tn tnh hng dn, gip em hon thnh cng vic thit k ny. Sinh vin L Phi H

TRNG I HC BCH KHOA TP.H CH MINH KHOA KHOA HC NG DNG B MN C K THUTPhng 106 B4, 268 L Thng Kit, Qun 10, Tp. HCM Tel: (84-8-) 8 660 586 Fax: (84-8-)8 651 211

N MN HC THIT K K THUTSinh vin thc hin Ngnh o to Gio vin hng dn Thi gian thc hin : : : : L Phi H MSSV: K0304100 C K Thut TS. V Cng Ha T 11/09/2006 n 11/12/2006

s 5: Thit K H Thng Dn ng Thng Trn Phng n s: 3

1 23

T T1 T2

45

tt1 t2

Hnh 1: S h thng dn ng

Hnh 2: S ti

H thng dn ng thng trn (Hnh 1) bao gm: 1: ng c in 3 pha khng ng b. 2: B truyn ai thang. 3: Hp gim tc bnh rng tr hai cp phn i cp nhanh. 4: Ni trc vng n hi. 5: Thng trn. S liu thit k: Cng sut trn trc thng trn, P ( kW ) S vng quay trn trc thng trn, n (v / p ) Thi gian phc v, L (nm) Quay 1 chiu lm vic 1 ca, ti va p nh Ch ti Ni dung: 1. Tm hiu h thng dn ng thng trn. 2. Xc nh cng sut ng c v phn phi t s truyn cho h thng truyn ng. 3. Tnh ton thit k cc chi tit my. - Tnh ton cc b truyn h. - Tnh ton cc b truyn trong hp gim tc. - V s lc tc ng ln cc b truyn v tnh cc lc. - Tnh ton thit k trc v then. - Chn ln v ni trc. - Chn thn my bulng v cc chi tit ph khc. 4. Chn du bi trn v bng dung sai lp ghp. Yu cu: 1. 01 bi thuyt minh. 2. 01 bn v lp A0 v mt bn v chi tit. Tin thc hin: : 3,5 : 45 : 8 : Mt nm lm vic 300 ngy, mt ca lm vic 8 gi. : (T ,34) ; (0, 7T , 30)

Tun1 2 36 78 9 12 13 14 15

Ni dung thc hinNhn bi, ni dung AMH Tm hiu h thng dn ng thng trn Xc nh cng sut ng c v phn phi t s truyn cho h thng truyn ng Tnh ton thit k cc chi tit my V phc tho v hon thnh kt cu trn bn v V hon thin bn v lp hp gim tc V 01 bn v chi tit, hon thnh ti liu thit k (thuyt minh v bn v) Gio vin hng dn duyt v k tn. Bo v.

n Thit K K Thut

Phn I

Phn I

TM HIU H THNG DN NG THNG TRN1. Khi nim:H thng thng trn l mt h thng chuyn dng trn, o cc nguyn vt liu vi nhau theo yu cu k thut v nhu cu ca con ngi, nhm to ra cc hn hp nguyn vt liu cn thit. Ngy nay, h thng thng trn c s dng trong rt nhiu lnh vc, t bit l trong cc ngnh cng nghip xy dng, ha thc phm ..

2. Kt cu h thng thng trn:H thng thng c rt nhiu loi v a dng tu theo mc ch s dng s c h thng tng ng, thch hp. Nhn chung, h thng c hnh thnh t 3 thnh phn c bn sau: - ng c: l ngun pht ng cho h thng. - Hp gim tc: chuyn cng sut t ng c sang thng trn theo cc ch tiu k thut v yu cu thit b. - Thng trn: cha v trn cc nguyn vt liu cn trn. Trong nhng ngnh s dng thng trn vi qui m v cng sut ln, ngi ta thng kt hp vi bng ti v cc thit b vn chuyn khc nhm nng cao nng sut lm vic, mang li hiu qu kinh t cao.

3. ng dng:Trong mt s lnh vc in hnh nh: - H thng thng trn nghin xi mng t, trong cng nghip khai khong. - H thng thng trn xi mng, ct, to va trong ngnh xy dng. H thng trn bt, cht lng , cht do, cc nguyn ph liu to cc hn hp ho cht - H thng thng trn s dng trong dy chuyn sn xut thc phm v thc n gia sc. S dng thng trn s c c nhiu u im: - Tit kim thi gian v chi ph nhn cng. - m bo cc yu cu k thut v thnh phn ca sn phm. - m bo v sinh an ton thc phm.

1

n Thit K K Thut

Phn I

MT S HNH NH MINH HO V THNG S CA MT S LOI THNG TRN

Tng cng sut: Cng sut trn:25(m3

18,5(kW ) 45(kW )h ) 75(m3

h

)

H thng thng trn b tng ca cng ty CITY NANHAI FOSHAN INCHINA JULONG CONSTRUCTION MECHINERY CO., LTD

2

n Thit K K Thut

Phn II

Phn II

XC NH CNG XUT NG C V PHN PHI T S TRUYN CHO H THNG TRUYN NG1. Xc nh ti trng tng ng:L trng hp ti trng thay i theo bc nn ta c:T Ti ti Pi ti 12 34 + 0, 7 2 30 Ptd = 1 n = P 1 n = 3,5 = 2, 66 (kW ) 34 + 30 ti tin 2 n 1 1 2

(3.10[2])

2. Xc nh cng sut cn thit ca ng c:4 Hiu sut chung ca h thng: = d k o l Vi: d = 0,96 : hiu sut b truyn ai. k = 0,98 : hiu sut khp ni n hi. ol = 0,99 : hiu sut mt cp ln. br = 0,98 : hiu sut b truyn bnh rng. 2 br

Vy: = 0,96 0,98 0,994 0,982 0,87 P 2, 66 = 3, 06 ( kW ) Cng sut cn thit ca ng c: Pct = td = 0,87

3. Phn phi t s truyn cho h thng: Chn t s truyn s b: Theo bng 2-2[1] ta chn t s truyn nh sau: o ai thang: ud = 3 o Hp gim tc hai cp: uh = 11 Nn t s trun s b ca h thng l: usb = 3 11 = 33 Vn tc s b ca ng c l: Vsb = usb n = 33 45 = 1485 (v / p) Chn ng c: Ta c: Pct = 3, 06 (kW ) & Vsb = 1485 (v / p ) nn chn ng c khng ng b 3 pha mang s hiu A02-41-4 (bng 2P[1]), c cc thng s k thut sau: o Cng sut: P = 4, 0 ( kW ) o Vn tc: V = 1450 (v / p ) Phn phi li t s truyn cho h thng: T s truyn thc u =V 1450 = = 32, 22 n 45

3

n Thit K K Thut

Phn II

Ta tin hnh chia t s truyn theo cc ch tiu: d bi trn, thun li cho vic ngm bnh rng trong du v trng lng nh nht Chn t s truyn ai: u d = 3 Khi t s truyn ca hp gim tc l: uh =32, 22 = 10, 74 3

Tip tc chn t s truyn qua b truyn bnh rng cp nhanh ( un ) v cp chm ( uc ) vi iu kin: uh = un uc & un = uc (1, 2......1, 4) . Chn:uc = 2, 77 un = 3,88

4. Xc nh cng sut trn cc trc: Trc I: P1 = Pdol = 4 0,96 0,99 3,8 (kW ) 2 Trc II: P2 = P dolbr = 4 0,96 0,99 2 0,98 3, 69 (kW ) 3 2 Trc III: P3 = P dolbr = 4 0,96 0,993 0,982 3,58 (kW )

5. Tnh s vng quay ca mi trc: Trc I:n1 = ndc 1485 = = 495 (v / p ) ud 3

TrcII:n2 = n1 495 = = 128 (v / p ) un 3,88 n2 127, 6 = = 46 (v / p ) uc 2, 77

Trc III:n3 =

6. Tnh moment xon trn trc v ng c:Theo cng thc sau:

T=Vi:

9,55 106 P n

(3.4[2])

P : cng sut (kW) n : s vng quay (vng/pht)

Moment xon trn trc ng c:Tdc = 9,55 106 Pdc 9,55 106 4 = = 25734 ( Nmm) ndc 1485

Moment xon trn trc I:T1 = 9,55 106 P 9,55 106 3,8 1 = = 72927 ( Nmm) n1 495

Moment xon trn trc II:T2 = 9,55 106 P2 9,55 106 3, 69 = = 274675 ( Nmm) n2 128

4

n Thit K K Thut Moment xon trn trc III:T3 = 9,55 106 P3 9,55 106 3,58 = = 739087 ( Nmm) n3 46

Phn II

Bng tng hp kt qu cc thng s cho hp gim tc v b truyn ai Thng s Cng sut (kW) T s truyn S vng quay (vng/pht) Moment xon (Nmm) ng c 4 3 1485 25734 Trc I 3,8 3,88 485 72927 128 274675 Trc II 3,69 Trc III 3,58 2,77 46 739087

Trc I

Trc III Trc II

5

n Thit K K Thut

Phn III

Phn III

THIT K B TRUYN AI THANGS liu u vo: Cng sut S vng quay T s truyn : Pdc = 4 (kW ) : n = 1485 (v / p ) : u =3

1. Da vo s liu trn v hnh 4.22[2] ta chn: ai thang loi A, c lm loi vt liu tng hp. Thng s ai thang loi A: Tn gi Chiu rng lp trung ha Chiu rng mt trn Chiu cao ai Khong cch t mt trung ha n th ngoi Din tch mt ct ngang ng knh bnh ai dn K hiubp

Gi tr 11 (mm) 13 (mm) 8 (mm) 2,8 (mm) 81 (mm)100 200

bo bo yoA d1

2. ng knh bnh ai nh s b: d1sb = 1, 2d min = 1, 2 100 = 120(mm) Theo tiu chun bng 5-15[1] ta chn ng knh bnh ai nh l: d1 = 125(mm) 3. Vn tc bnh ai:v1 = 3,14 125 1485 = 9, 7( m / s ) (5.18[1]) 60000 60000 v1 < vmax vi vmax = (30 m 35 m ) nn tha iu kin. s s =

d1n

4. Gi s chn h s trt tng i = 0, 01 . Ta c ng knh bnh ai ln: d 2 sb = ud1 (1 ) = 3 125(1 0, 01) = 371, 25(mm) Theo tiu chun tra bng 5-15[1] ta chn: d 2 = 360 (mm) Xc nh li t s truyn u: u =d2 360 = = 2,91 d1 (1 ) 125(1 0, 01)

Ch sai lch 3% so vi gi tr chn trc. 5. Gii hn khong cch trc c tnh:2(d1 + d 2 ) a 0,55(d1 + d 2 ) + h 2(125 + 360) a 0,55(125 + 360) + 8 970( mm) a 274, 75( mm) Chn s b a = d 2 = 360 (mm) khi u = 2, 91

6

n Thit K K Thut 6. Tnh chiu di ai:L = 2a + + 2 4a 3,14(125 + 360) (360 125) 2 = 2 360 + + = 1520 (mm) 2 4 360

Phn III

(d1 + d 2 ) (d 2 d1 ) 2

Theo bng 4-3[2] ta chn ai c chiu di L = 1600 (mm) = 1,6 (m). 7. S vng chy ca ai trong mt giy:i= v 9, 7 = = 6, 0625 ( s 1 ) , [i ] = 10 s 1 nn tha iu kin ny. L 1, 6(d1 + d 2 ) 125 + 360 = 1600 3,14 = 838, 2 (mm) 2 2 (d + d 2 ) 125 + 360 = 1 = = 242,5 2 2 k = L

8. Tnh li khong cch trc a:

Ma= a= k + k 2 8 2 4

838, 2 + 838, 22 8 242,52 = 330 (mm) 4

Vy gi tr a vn tha mn gii hn gi tr cho php. 9. Gc m bnh ai nh:1 = 180o 57(d 2 d1 ) 360 125 = 180 57 = 139, 4o = 2, 433 (rad ) a 3301

10. Cc h s s dng: - H s xt n nh hng ca gc m ai: C = 1, 24(1 e /110 ) = 0, 9 H s xt n nh hng ca vn tc:Cv = 1 0, 05(0, 01v 2 1) = 1 0, 05(0, 01 9, 7 2 1) = 1, 003

H s xt n nh hng ca dy ai, chn: C z = 1 H s xt n nh hng ca ch ti trng: Cr = 0,85 H s xt n nh hng ca chiu di dy ai: CL = 6L 6 1600 = = 1, 0085 Lo 1520

H s xt n nh hng ca t s truyn u: Cu = 1,14 v u > 2,5

11. Xc nh s dy ai: Theo th hnh 4.2a[2] ta chn: [Po ] = 2,4 (kW) . Ta xc nh s dy ai theo cng thc:z P 4 1 = = 1,84 [ Po ]C Cv C z Cr CL Cu 2, 4 0,9 1, 003 1 0,85 1, 0085 1,14

Chn s dy ai: z = 2

7

n Thit K K Thut 12. Xc nh lc cng ban u: Chn ng sut cng ai: o = 1,5 ( N / mm 2 ) Lc cng ban u: Fo = 2 A o = 2 811,5 = 243 ( N ) Lc cng trn mi dy ai: F =Fo = 121,5 ( N ) 2 1000 P 1000 4 1 Lc vng c ch: Ft = = = 412, 4 ( N ) 9, 7 v1 Lc vng trn mi dy ai: F=206 (N)

Phn III

13. H s ma st (ti liu 1 trang 159):f'= 2 F + Ft ' o ln o = 1, 03 f min = f sin(20 ) = 0,35 ( y ta chn gi tr = 1 ) 2 Fo Ft 1

H s ma st nh nht b truyn khng b trt trn l: 0,35 14. Lc tc dng ln trc: 1 Fr = 2 Fo sin = 456 ( N ) 2

15. Thit k bnh ai: Do vn tc lm vic v = 9, 7 ( m / s ) (nh hn 25m/s) nn ta chn bnh ai c bng gang CH12-28. Cc kch thc c chn bng cch tra bng 10-3[1]: - Chiu rng bnh ai: B = ( Z 1)t + S = 16 + 2 10 = 36 ( mm) - ng knh ngoi cng ca bnh ai dn:Dn1 = D1 + 2ho = 360 + 2 3,5 = 367 (mm)

-

ng knh ngoi cng ca bnh ai b dn:Dn 2 = D2 + 2ho = 125 + 2 3,5 = 132 (mm)

Cc kch thc ph khc lin quan theo tiu chun, tra bng 10-3[1].

8

n Thit K K Thut

Phn IV

Phn IV

THIT K B TRUYN BNH RNGChn vt liu: Do khng c yu cu no t bit nn ta chn vt liu cho cc cp bnh rng l ging nhau. Da vo bng 3-6[1] v 3-8[1] ta c bng sau:

Thng sTn thp Gii hn bn ko Gii hn chy rn 1. Thng s u vo: Cng sut T s truyn S vng quay Tui th

Bnh rng nhThp 45 (thng ha) b = 600 ( Nmm 2 ) ch = 300 ( Nmm 2 )HB = 200

Bnh rng lnThp 35 (thng ha) b = 500 ( Nmm 2 ) ch = 260 ( Nmm 2 )HB = 170

I. Tnh ton cp nhanh: bnh rng tr rng nghing phn i.:P = 3,8 (kW ) 1

: u1 = 3,88 : n1 = 485 (vng/pht) : 19200 (gi)

2. Xc nh ng sut tip v ng sut un cho php: ng sut tip xc cho php: S chu k tng ng ca bnh ln: (3.4[1]) Mi 7 2 2 N td 2 = 60u niTi = 60 1(1 0,53 + 0, 7 0, 47 )125 19200 = 11 10 M max n 485 Vi: n2 = 1 = = 125 (v / p ) u1 3,88' N td 2 > N o ; N o = 107 nn ta chn h s chu k ng sut k N = 1 cho c hai

2

bnh rng. ng sut tip xc cho php ca bnh ln (bng 3-9[1]) [ tx 2 ] = 2, 6 170 = 442 ( N / mm 2 ) ng sut tip xc cho php ca bnh nh (bng 3-9[1]) [ tx1 ] = 2, 6 200 = 520 ( N / mm 2 ) tnh sc bn ta dng tx 2 = 442 ( N / mm 2 ) ng sut un cho php: S chu k tng ng ca bnh ln: (3.8[1])N td 2 Mi 7 6 6 = 60u niTi = 60 1(1 0,53 + 0, 7 0, 47 )125 19200 = 8, 4 10 M max 6

9

n Thit K K Thut Vi: n2 =n1 485 = = 125 (v / p ) u1 3,88

Phn IV

' N td 2 > N o ; N o = 5 106 nn ta chn h s chu k ng sut k N = 1 cho c

nK nK 1,5 258 1 Bnh nh: [ ]u1 = = 143,3 ( N / mm 2 ) 1,5 1,8 1,5 215 1 = 119, 4 ( N / mm 2 ) Bnh ln: [ ]u 2 = 1,5 1,8 3. Chn s b h s ti trng: K = Ktt K d = 1,3 (3.19[1]) b 4. Chn s b h s chiu rng bnh rng: A = = 0, 4 A

hai bnh rng. Gii hn mi un ca thp 45: 1 = 0, 43 600 = 258 ( N / mm 2 ) Gii hn mi un ca thp 35: 1 = 0, 43 500 = 215 ( N / mm 2 ) Chn h s an ton: n = 1, 5 H s tp trung ng sut chn rng: K = 1,8 V ng sut un thay i theo chu k mch ng nn ta c: '' o kn'' (1, 4 1, 6 ) 1k N (3.5[1]) [ ]u =

Vi:b :chiu rng bnh rng A : khong cch trc

5. Tnh s b khong cch trc (3.10[1]), chn h s phn nh s tng kh nng ti ca bnh rng nghing so vi bnh rng thng: ' = 1, 25 1, 05 106 KN 3 A (i 1) [ ] i A ' n2 tx 2

Vi: -

[ ]tx = 422 ( N / mm2 ) : ng sut tip cho php.n2 = 485 = 125 (v / p) : s vng quay ca bnh rng b dn. 3,88 P 3,8 N= 1= = 1,9 ( kW ) 2 22

i : t s truyn, i = u1 = 3,88

1, 05 106 1,3 1,9 A (3,88 + 1) 3 125 (mm) 422 3,88 0, 4 1, 25 125 Chn khong cch trc: A = 144 ( mm)

6. Xc nh vn tc vng v chn cp chnh xc ch to bnh rng: - Vn tc vng: 2 An1 2 144 485v= 60 1000(i + 1) = 60 1000 (3,88 + 1) = 1,5 (m / s )

(3.17[1])

-

Vi vn tc ny ta chn cp chnh xc bnh rng l: 9 (bng 3-11[1]) 10

n Thit K K Thut

Phn IV

7. Tnh chnh xc h s ti trng K. Chiu rng bnh rng b = A A = 0, 4 144 = 57, 6 (mm) chn b = 60 ( mm) ng knh vng ln ca bnh rng nh: d1 = Do : d =2 A 2 144 = = 59 (mm) i + 1 3,88 + 1

b 60 = 1 , tra bng 3-12[1] ta c: K tt = 1,1 d1 59 Tra bng 3-14[1] ta c: K d = 1, 2

Vy ta c h s ti trng l: K = K tt K d = 1,11, 2 = 1,32 . Kt qu sai lch t so vi chn s b nn ta chn khong cch trc A chnh xc l: A = 144 ( mm) 8. Xc nh modul, s rng v gc nghing ca rng: Modul php: mn = (0, 01 0, 02)144 = (1, 44 2,88)mm . Chn: mn = 2mm Chn s b gc nghing: = 10o , cos = 0,985 2 A cos 2 144 0,985 Tng s rng ca hai bnh rng: Z t = Z1 + Z 2 = = = 142mn 2

S rng ca bnh nh: Z1 =

Zt 142 = = 29 i + 1 3,88 + 1

Theo bng 3.15[1], Z1 = 29 tha mn iu kin khng b ct chn rng. S rng bnh ln: Z 2 = iZ1 = 3,88 29 = 112,52 . Chn: Z 2 = 113 Tnh chnh xc gc nghing (3.28[1]):cos = Z t mn (29 + 113)2 = = 0,9861 = 9, 6o 2A 2 144

Chiu rng bnh rng b tha mn iu kin:b = 60 > 2,5mn 2,5 2 = 30 (mm) sin 0,167

9. Kim nghim sc bn un ca rng: S rng tng ng (3.37[1]): Bnh nh: Z td 1 = Bnh ln: Z td 229 = 30 (0,9861)3 113 = = 118 (0,9861)3

H s dng rng: (bng 3.18[1]): Bnh nh: y1 = 0, 451 Bnh ln: y2 = 0,517 Ly h s: h s phn nh s tng kh nng ti ca bnh rng nghing so vi bnh rng thng: ' = 1, 2519,1 106 KN [ ]u Kim tra ng sut un (3.34[1]): u = 2 ymn Znb ''

19,1106 1,32 1,9 = 25 ( N / mm 2 ) . Tha mn Bnh nh: u1 = 2 0, 451 2 29 485 60 1, 25 iu kin u1 < [ ]u1 = 143,3 ( N / mm2 )

11

n Thit K K Thut Bnh ln: u 2 = u1

Phn IVy1 25 0, 451 = = 21,53 ( N / mm 2 ) . Tha mn iu kin y2 0,517

u 2 < [ ]u 2 = 119, 4 ( N / mm2 )

10. Kim tra bn khi qu ti t ngt: ng sut tip xc cho php (3.34[1]): Bnh nh: [ ]txqt1 = 2,5 520 = 1300 ( N / mm 2 ) Bnh ln: [ ]txqt 2 = 2,5 442 = 1105 ( N / mm 2 ) ng sut un cho php (3.46[1]): Bnh nh: [ ]uqt1 = 0,8 300 = 240 ( N / mm 2 ) Bnh ln: [ ]uqt 2 = 0,8 260 = 208 ( N / mm 2 ) Kim tra qu ti tip xc (3.38 v 3.42[1]): tx =1, 05 106 Ai (i + 1)3 KN [ tx ] 'bn2

tx =

1, 05 106 (3,88 + 1)31,8 3,8 = 545 ( N / mm2 ) . 144 3,88 1, 25 60 125

Tha mn iu kin ng sut tip. Vi K = 1,8 l h s qu ti. Kim tra bn un: (3.38 v 3.42[1]): Bnh nh: uqt1 = 25 1,8 = 45 ( N / mm2 ) Bnh ln: uqt 2 = 21,52 1,8 = 38,8 ( N / mm2 ) u tha mn iu kin bn un. 11. Bng tng hp thng s bnh rng cp nhanh phn i:

Thng sModul S rng Gc n khp Gc nghing ng knh vng chia Khong cch trc Chiu rng bnh rng ng knh vng nh ng knh vng chn

Bnh rng nh (Bnh dn)m=2Z1 = 29

Bnh rng ln (Bnh b dn)Z 2 = 113

n = 20d1 = 59 (mm) b1 = 30 (mm) De1 = 63 (mm) Di1 = 54 ( mm)

o

= 9, 6od 2 = 229 (mm) b2 = 30 (mm) De 2 = 232 (mm) Di 2 = 224 (mm)A = 144( mm)

12. Tnh lc tc dng ln trc (3.50[1]): Lc vng: P =2 9,55 106 3,8 = 2523 ( N ) 59 485

12

n Thit K K Thut2523 tg 20o = 931 ( N ) cos 9, 6o Lc dc trc: Pa = 2523 tg 9, 6o = 427 ( N )

Phn IV

Lc hng tm: Pr =

II. Tnh ton cp chm: bnh rng tr rng thng.1. Thng s u vo: Cng sut T s truyn S vng quay Tui th : P1 = 3,67 (kW) : u2 = 2, 77 : n2 = 125 (vng/pht) : 19200 (gi)

2. Xc nh ng sut tip v ng sut un cho php: ng sut tip xc cho php: S chu k tng ng ca bnh ln: (3.4[1]) Mi 7 2 2 N td 2 = 60u niTi = 60 1(1 0,53 + 0, 7 0, 47 ) 46 19200 = 4 10 M max n 125 Vi: n2 = 1 = = 46 (v / p ) u1 2, 77' N td 2 > N o ; N o = 107 nn ta chn h s chu k ng sut k N = 1 cho c hai 2

bnh rng. ng sut tip xc cho php ca bnh ln (bng 3-9[1]) [ tx 2 ] = 2, 6 170 = 442 ( N / mm 2 ) ng sut tip xc cho php ca bnh nh (bng 3-9[1]) [ tx1 ] = 2, 6 200 = 520 ( N / mm 2 ) tnh sc bn ta dng tx 2 = 442 ( N / mm 2 ) ng sut un cho php: S chu k tng ng ca bnh ln: (3.8[1])N td 2 Mi 7 6 6 = 60u niTi = 60 1(1 0,53 + 0, 7 0, 47 ) 46 19200 = 3,1 10 M max n1 485 = = 125 (v / p ) u1 3,886

Vi: n2 =

' N td 2 > N o ; N o = 5 106 nn ta chn h s chu k ng sut k N = 1 cho c

hai bnh rng. Gii hn mi un ca thp 45: 1 = 0, 43 600 = 258 ( N / mm 2 ) Gii hn mi un ca thp 35: 1 = 0, 43 500 = 215 ( N / mm 2 ) Chn h s an ton: n = 1, 5 H s tp trung ng sut chn rng: K = 1,8 V ng sut un thay i theo chu k mch ng nn ta c: '' k '' (1, 4 1, 6 ) 1k N ]u = o n (3.5[1]) [nK nK

13

n Thit K K Thut Bnh nh: [ ]u1 = Bnh ln: [ ]u 21,5 258 1 = 143,3 ( N / mm 2 ) 1,5 1,8 1,5 215 1 = = 119, 4 ( N / mm 2 ) 1,5 1,8

Phn IV

3. Chn s b h s ti trng: K = Ktt K d = 1,3 4. Chn s b h s chiu rng bnh rng: A = Vi:b :chiu rng bnh rng A : khong cch trc.

(3.19[1])b = 0, 4 A

5. Tnh s b khong cch trc (3.9[1]): 1, 05 106 KN A (i 1) 3 [ ] i A n2 tx 2

Vi: -

[ ]tx = 422( N / mm2 ) : ng sut tip cho php.n2 = 125 = 46 (v / p ) : s vng quay ca bnh b dn. 2, 772

i : t s truyn, i = u2 = 2, 77

N = 3,58 ( kW )

1, 05 106 1,3 3, 67 3 A (2, 77 + 1) 222 (mm) . Chn: A = 266 ( mm) 422 2, 77 0, 4 46

6. Tnh vn tc vng ca bnh rng v chn cp chnh xc ch to bnh rng: - Vn tc vng: 2 An1 2 226 125v= 60 1000(i + 1) = 60 1000 (2, 77 + 1) = 0, 785 (m / s ) (3.17[1])

-

Vi vn tc vng ny ta chn cp chnh xc ch to bnh rng l 9 (bng 3-11[1]).

7. Tnh chnh xc h s ti trng ng K: Chiu rng bnh rng b: b = A A = 0, 4 226 = 90, 4 (mm) . Chn: b = 90 ( mm) ng knh vng ln bnh rng nh: d1 = Do : d =2 A 2 226 = = 120 (mm) i + 1 2, 77 + 1

b 90 = 0, 75 , tra bng 3-12[1] ta c: K tt = 1, 05 d1 120 Tra bng 3-13[1] ta c: K d = 1,1

Vy ta c h s ti trng l: K = K tt K d = 1, 05 1,1 = 1,12 sai lch nhiu so vi chn s b ban u nn ta tnh li khong cch trc (3.21[1]):A = Asb 3 K 1,12 = 226 3 = 215 (mm) 1,3 K sb

14

n Thit K K Thut 8. Xc nh modul, s rng v gc nghing ca rng: Modul php: mn = (0, 01 0, 02)215 = (2,15 4,3)mm . Chn: mn = 3mm .

Phn IV

2A 2 215 (3.24[1]) = = 38 m(i 1) 3(2, 77 + 1) S rng ca bnh rng ln: Z 2 = Z1u2 = 38 2, 77 = 105,3 . Chn: Z 2 = 105

S rng ca bnh rng nh: Z1 =

B rng bnh rng: b = A A = 0, 4 215 = 86 (mm) . 9. Kim nghim sc bn un ca rng: H s dng rng (bng 3-18[1]): Bnh nh: y1 = 0, 476 Bnh ln: y2 = 0,517 ng sut un ti chn rng bnh rng nh: u1 =19,1 106 KN 19,1 106 1,12 3, 67 = = 27, 7 ( N / mm 2 ) . 2 2 ym Zn2b 0, 764 3 38 125 86

Tha mn iu kin: u1 < [ ]u1 = 143,3 ( N / mm2 ) ng sut un ti chn rng bnh rng ln: u 2 = u1y1 27, 7 0, 476 = = 25,5 ( N / mm 2 ) . y2 0,517

Tha mn iu kin: u 2 < [ ]u 2 = 119, 4 ( N / mm2 ) 10. Bng tng hp thng s hnh hc ch yu ca b truyn cp chm:

Thng sModul S rng Gc n khp ng knh vng chia Khong cch trc Chiu rng bnh rng ng knh vng nh rng ng knh vng chn rng 11. Tnh lc tc dng ln trc: Lc vng: P =

Bnh rng dnm=3Z1 = 38

Bnh rng b dnZ 2 = 105

= 20od1 = 114 (mm) d 2 = 315 (mm) A = 214, 5 ( mm) b = 86 ( mm) De1 = 120 (mm) Di1 = 106,5 (mm) De 2 = 321 (mm) Di 2 = 307,5 (mm)

2M x 2 9,55 106 3,58 = = 4920 ( N ) d 114 125 Lc hng tm: Pr = P tg = 4920 tg 20o = 1790 ( N )

15

n Thit K K Thut

Phn V

Phn V

THIT K TRC V CHN THENChn vt liu cho trc: chn vt liu lm trc l thp 45, CT6. Theo bng s liu 7-1[1] ta c: - [ ] = 70 ( N / m 2 ) - b 600 ( N / mm 2 ) - 1 260 ( N / mm 2 ) - [ ] = 35 (MPa) : ng sut xon cho php (trang 115 [1]).

I. Thit k trc:Tnh s b ng knh ca cc trc (7.1[1]): Mx 72927 =3 22 (mm) - Trc I: d sb1 = 3 0, 2[ ]x 0, 2 35 Vi: moment xon: M x = 72927 ( N / mm 2 ) -

Mx 274675 =3 34 (mm) 0, 2[ ]x 0, 2 35 Vi: moment xon: M x = 274675 ( N / mm 2 )Trc II: d sb 2 = 3 Trc III: d sb3 = 3

Mx 739087 =3 47 (mm) 0, 2[ ]x 0, 2 35 Vi: moment xon: M x = 739087 ( N / mm 2 ) Ta chn tr s d sb 45 (mm) xc nh . y, ta chn loi bi mt dy c trung bnh, s hiu 309 c b rng B = 25 (mm), (bng 14P [1])Tnh gn ng chiu di trc: tnh chiu di ca trc ta chn cc kch thc sau: Khe h gia cc bnh rng Khe h gia bnh rng v thnh trong ca hp : 10 ( mm ) : 10 ( mm )

Khe h gia thnh trong ca hp v mt bn ca ln : 10 ( mm ) Chiu rng ln : 25 ( mm ) Chiu rng bch (bng 10.10a[1]), chn bulng M12 Khe h gia bulng v bnh ai : 23 ( mm ) : 10 ( mm )

Theo m hnh bn v phc hp gim tc trang sau ta xc nh s b chiu di mi trc theo cc kch thc: a, b, c, d, e, l. y, ta phi lm bc chn du bo v m trong b phn , chng ta khng th dng du bn to bi trn b phn v vn tc vng ca bnh rng thp hn 3 (m/s).

16

n Thit K K Thut

Phn V

Hnh v phc tho hp gim tc.

1. Trc I:Thng s u vo ca trc I: Cng sut S vng quay Moment xon o Tnh s b ng knh trc: : 3,8 ( kW ) : 485 ( vng/pht ) : 72927 ( N/mm )

Mx 72927 =3 22 (mm) (7.1[1]) 0, 2[ ]x 0, 2 35 Theo tiu chun (trang 133[1]) ta chn d = 25 (mm) , chn bi c trung bnh mang s hiu 308 (bng 14P[1]), vi b rng : B = 23( mm) . d sb1 = 3o Xc nh chiu di trc dng cho tnh ton: - l = 70 ( mm) - e = 55 ( mm) - c = 75 ( mm) o Cc lc tc dng: - Lc cng ai: Rd = 456 ( N ) Lc vng:P= 2 9,55 106 3, 78 P = 2523 ( N ) P1 = P 2 = = 1261,5 ( N ) 1 1 59 485 2

-

Lc hng tm:2523 tg 20o P Pr = = 931 ( N ) Pr11 = Pr2 = r = 465,5 ( N ) 1 o cos 9, 6 2

-

Lc dc trc:Pa = 2523 tg 9, 6o = 427 ( N ) Pa11 = Pa2 = 1

Pa = 213,5 ( N ) 2

17

n Thit K K ThutP1 1 A P B P1 r1 1 a1

Phn VP2 1 M1 P2 a1

x

zy

DC

E

P

2 r1

55

150

55

70

Rd

Mx ( Nmm)

25575 76190

29218

25575 94565 31920My ( Nmm)

3643,5 72927 S phn tch lc ca trc I

Mz ( Nmm)

o Xc nh moment tng ng theo thuyt bn IV v tnh li ng knh trc ti cc im A, B, C, D, E. Ti A: M td = 0 Ti B: M td = 292182 + 761902 + 0, 75 36463,52 = 87497 ( Nmm)dB 3

M td 87497 =3 = 23, 2 (mm) 0,1[ ] 0,1 70 M td 117410 =3 = 25, 6 (mm) 0,1[ ] 0,1 70 M td 70765 =3 = 21, 6 ( mm) 0,1[ ] 0,1 70 M td 63157 =3 = 20,8 (mm) 0,1[ ] 0,1 70

Ti C: M td = 292182 + 945652 + 0, 75 72927 2 = 117410 ( Nmm)dC 3

Ti D: M td = 02 + 319202 + 0,75 72927 2 = 70765 ( Nmm)dD 3

Ti E: M td = 02 + 02 + 0,75 72927 2 = 63157 ( Nmm)dE 3

Da vo cc iu kin trn v cng ngh ta chn cc thng s trc I theo tiu chun ( trang 133 [1]) nh sau: d A = 25 (mm); d B = 30 (mm)

dC = 30 (mm); d D = 25 (mm)d E = 22 ( mm)

18

n Thit K K Thut

Phn V

2. Trc II:Thng s u vo ca trc II: Cng sut S vng quay Moment xon : 3,69 (kW) : 128 (vng/pht) : 274675 (N/mm)

o Tnh s b ng knh trc: Mx 274675 d sb 2 = 3 =3 34 (mm) (7.1[1]) 0, 2[ ]x 0, 2 35 Theo tiu chun (trang 133[1]) ta chn d = 40 (mm) , chn bi c trung bnh mang s hiu 308 (bng 14P[1]), vi b rng : B = 23( mm) . o Xc nh chiu di trc dng tnh ton: - e = 55 ( mm) - c = 75 ( mm) - d = 2(e + c) = 2(55 + 75) = 260 ( mm) o Cc lc tc dng: - Lc tc dng ti v tr hai bnh rng ngoi:2523 0,98 = 1236 ( N ) 2 P 931 Lc hng tm: Pr12 = Pr22 = r br = 0,98 = 456 ( N ) 2 2 P 427 Lc dc trc: Pa12 = Pa22 = a br = 0,98 = 209 ( N ) 2 2

Lc vng: P21 = P22 = br =

P 2

- Lc tc dng ti v tr bnh rng gia: Lc vng: P =

2M x 2 9,55 106 3,58 = = 4920 ( N ) d 114 125 Lc hng tm: Pr = P tg = 4920 tg 20o = 1790 ( N )

o Tnh moment tng ng theo thuyt bn IV v tnh li ng knh trc ti cc im A, B, C, D, E. Theo s phn tch lc trang sau: - Ti A: M td = 0 - Ti B: M td = 763752 + 2032802 + 0, 75 1402202 = 248810 ( Nmm)dB 3

M td = 0,1[ ]

3

248810 = 32,8 (mm) 0,1 70 430937 = 39, 48 (mm) 0,1 70

- Ti C: M td = 1434822 + 3877802 + 0, 75 1402202 = 430937 ( Nmm)dC 3

M td = 0,1[ ]

3

- Ti D: M td = 763752 + 2032802 + 0, 75 1402202 = 248801 ( Nmm)dD 3

M td 248801 =3 = 32,8 (mm) 0,1[ ] 0,1 70

19

n Thit K K Thut - Ti E: M td = 0zx

Phn V

Pa12 A P21

Pr P1 r2

P22C

Pa22Pr22 E

y

B55

P

D55

150

143482 76375 24145

76375 24145Mx ( Nmm) My ( Nmm)

203280 387780

203280Mz ( Nmm)

140220 140220 S phn tch lc trc II Da vo cc s liu trn v xut pht t nhng yu cu v lp ghp v tnh cng ngh ta chn cc thng s trc II theo tiu chun (trang 133[1]), ring thng s ti A ta chn ng knh theo moment tng ng tnh ti mt ct pha bn tri im B nh trn s phn tch lc trc II. Ta c:M td = 241452 + 2032802 + 0, 75 02 = 204709 ( Nmm) dA 3

M td = 0,1[ ]

3

204709 = 30,8 (mm) 0,1 70

T ta chn ng knh nh sau:d A = 35 (mm) d B = 40 (mm) dC = 45 (mm) d D = 40 (mm)d E = 35 (mm)

20

n Thit K K Thut

Phn V

3. Trc III:Thng s u vo: Cng sut S vng quay Moment xon : 3,58 (kW) : 46 (vng/pht) : 739087 (N/mm)

o Tnh s b ng knh trc: Mx 739087 d sb3 = 3 =3 47 (mm) 0, 2[ ]x 0, 2 35 Theo tiu chun (trang 133[1]) ta chn d = 50 (mm) , chn bi c trung bnh mang s hiu 310 (bng 14P[1]), vi b rng : B = 27 (mm) o Xc nh chiu di trc dng cho tnh ton (theo s v phc): - a = 130 ( mm) - b = 130 ( mm) - d = 90 ( mm) o Cc lc tc dng: 2M x 2 9,55 106 3,58 = = 4920 ( N ) 114 125 d Lc hng tm: Pr = P tg = 4920 tg 20o = 1790 ( N )

Lc vng: P =

x

z

P M

y

A

B

Pr

C

D

130

130

90Mx ( Nmm)

116350 319800My ( Nmm)

759465 S phn tch lc trc III

Mz ( Nmm)

21

n Thit K K Thut

Phn V

o Tnh moment tng ng theo thuyt bn IV v tnh li ng knh trc ti cc v tr: A, B, C, D. - Ti A: M td = 0 Ti B: M td = 1163502 + 3198002 + 0, 75 7594652 = 740540 ( Nmm)dB 3

M td 740540 =3 = 47, 29 (mm) 0,1[ ] 0,1 70

-

Ti C: M td = 02 + 02 + 0, 75 7594652 = 657176 ( Nmm)dC 3

M td 657176 =3 = 45, 46 (mm) 0,1[ ] 0,1 70 M td 657176 =3 = 45, 46 (mm) 0,1[ ] 0,1 70

-

Ti D: M td = 02 + 02 + 0, 75 7594652 = 657176 ( Nmm)dD 3

o Theo s liu tnh ton, tiu chun lp ghp v tnh cng ngh, theo tiu chun (trang 133[1]) ta chn cc kch thc trc nh sau:d A = 55 (mm) d B = 60 (mm) dC = 55 (mm) d D = 50 (mm)

Kim nghim trc: H s an ton c tnh theo cng thc sau: n n n= [n] (7.5[1]) n + n Vi: o n : h s an ton ch xt ring ng sut php:

n =

n =

k

1 a + m 1 a + m

(7.6[1])

o n : h s an ton ch xt ring ng sut tip:

o 1 : gii hn mi un 1 (0, 4 0,5) b o 1 : gii hn mi xon 1 (0, 2 0,3) b o a : bin ng sut php

k

(7.6[1])

22

n Thit K K Thut

Phn V

2 o a : bin ng sut tip a = max min 2 o m : tr s trung bnh ca ng sut php + min m = max 2 o m : tr s trung bnh ca ng sut tip + m = max min 2

a =

max min

Do ng sut xon thay i theo chu k mch ng nn: a = m =o W : moment cn un. W

max2

=

Mx 2Wo

bt (d t ) (7.3a) 32 2d d 3 bt (d t )2 (7.3a) o Wo : moment cn xon. Wo 16 2d 2 o o : h s ng sut. = 1 o 2 o : h s sc bn mi. = 1 o o Da vo tiu chun ca vt liu ch to trc v cc bng tra ta chn cc s liu sau: 1 (0, 4 0,5) b = 0, 45 600 = 270( Nmm2 )3

d

2

1 (0, 2 0,3) b = 0, 25 600 = 150( Nmm 2 ) 0,1 0,05 =1 Theo bng 7.4[1] ta chn: = 0,86; = 0,75Theo bng 7.8[1] ta chn: k = 1,63; k = 1,5 Tp trung ng sut do lp cng, chn p sut l 30 ( Nmm 2 ) , tra bng 7-8[1] ta c: k = 2,5 k = 1 + 0,6 1 = 1,9 k

23

n Thit K K Thut

Phn V

Bng tng hp kt qu: Thng sMt ct nguy him M x (Nmm) M u (Nmm) W (bng 7.3b[1]) Wo (bng 7.3b[1])

Trc IC-C 72927 98976 2320 4970 40 7,3 2,7 10 2,6

Trc IIC-C 274675 413474 7800 16740 48 8,2 2,25 7,2 2,1

Trc IIIB-B 759465 340308 18760 40000 16 9,5 6,75 6,2 4,56

a ( Nmm2 ) a = m ( Nmm2 )n n n

Ta thy cc h s an ton u tha. Cc h s an ton cho php thng chn t 1,5 2,5

II. Thit k v chn then: y, chng ta chn then bng theo TCVN cho tt c cc mi ghp then. Chn tit din v rnh then theo TCVN 149 64 (bng 7-23[1]). Chn chiu di then theo TCVN 150 64 (bng 7-24[1]). iu kin bn dp trn mt cnh ca then khi lm vic c tnh theo cng thc:d =2M x [ ]d , ( N / mm 2 ) dtl 2M x [ ]c , ( N / mm 2 ) dbl (7.11[1])

iu kin bn ct ca then:c =(7.12[1])

Theo bng 7-20[1] v 7-21[1] ta chn:[ ]d = 100 ( N / mm2 ) [ ]c = 87 ( N / mm2 )

1. Trc I:

Moment xon cn truyn: M x = 72927 ( Nmm) Ta lp bng nh sau: V tr B C Ed (mm) b h (mm) 10 8 10 8 8 7

30 30 22

t (mm) 4,5 4,5 4, 0

l (mm) 28 28 32

d ( N / mm 2 )38, 6 38, 6 51,8

c ( N / mm 2 )17, 4 17, 4 25,9

Vy cc then trn trc I u m bo bn. 24

n Thit K K Thut2. Trc II:

Phn V

Moment xon cn truyn: M x = 274675( Nmm) Ta lp bng nh sau: V tr B C Dd (mm) b h (mm) 12 8 14 9 12 8

40 45 40

t (mm) 4,5 5, 0 4,5

l (mm) 32 70 32

d ( N / mm 2 )95, 4 34,9 95, 4

c ( N / mm 2 )35, 7

12,535, 7

Vy cc then trn trc II u m bo bn.3. Trc III:

Moment xon cn truyn: M x = 759465( Nmm) Ta lp bng nh sau: V tr B Dd (mm) b h (mm) 18 11 18 11

60 50

t (mm) 5,5 5,5

l (mm) 70 55

d ( N / mm 2 )65,8 98, 6

c ( N / mm 2 )20,1 30, 7

Vy cc then trn trc III u m bo bn.

25

n Thit K K Thut

Phn VI

Phn VI:

CHN LN V NI TRCI. Chn ln:Theo cu to ca hp gim tc, do khng c lc dc trc nn ta chn bi mt dy cho tt c cc trc. 1. Trc I: 465,5 (N) A 1384,3 (N) 683,7 (N) S chn cho trc I o Phn lc ti A: FA = 465,52 + 1383, 7 2 = 1460 ( N ) o Phn lc ti D: FD = 465,52 + 683, 7 2 = 827 ( N ) Tnh gi cho A v A c lc ln. Xc nh h s lm vic ca ln: C = Q(nh)0,3 Vi: Q : ti trng tng ng, (daN). Q = 146 (dnN ) n : s vng quay, (vng/pht). n = 485 h : thi gian phc v, (gi). h = 19200 C = 146(485 19200)0,3 = 17992

465,5 (N) D x y z

Tra bng 14P[1], ng vi ng knh 25(mm) chn bi c trung k hiu 305. C: ng knh ngoi : 62 (mm) B rng : 17 (mm) C : 27000 2. Trc II: 439 (N) A 3696 (N) 439 (N) E x 3696 (N) S chn cho trc II y z

26

n Thit K K Thut o Phn lc ti A: FA = 4392 + 36962 = 3722 ( N ) o Phn lc ti E: FE = 4392 + 36962 = 3722 ( N ) Tnh gi cho A v E. Xc nh h s lm vic ca ln: C = Q(nh)0,3 Vi: Q : ti trng tng ng, (daN). Q = 372, 2 (dnN ) n : s vng quay, (vng/pht). n = 128 h : thi gian phc v, (gi). h = 19200 C = 372, 2(128 19200)0,3 = 30756

Phn VI

Tra bng 14P[1], ng vi ng knh 35(mm) chn bi c trung k hiu 307. C: ng knh ngoi : 80 (mm) B rng : 21 (mm) C : 40000 3. Trc III: 877 (N) A 2411 (N) 2411 (N) S chn cho trc III o Phn lc ti A: FA = 877 2 + 24112 = 2566 ( N ) o Phn lc ti C: FC = 877 2 + 24112 = 2566 ( N ) Tnh gi cho A v C. Xc nh h s lm vic ca ln: C = Q(nh)0,3 Vi: Q : ti trng tng ng, (daN). Q = 256, 6 (dnN ) n : s vng quay, (vng/pht). n = 46 h : thi gian phc v, (gi). h = 19200 C = 256, 6(46 19200)0,3 = 15598

877 (N) C x y z

Tra bng 14P[1], ng vi ng knh 50(mm) chn bi c nh, rng va k hiu 110. C: ng knh ngoi B rng C : 80 (mm) : 15 (mm) : 25000

27

n Thit K K Thut

Phn VI

II. Ni trc n hi:Ni trc n hi gm hai na ni trc lp c nh vi hai trc v b phn n hi ghp hai na ni trc vi nhau. Ngoi kh nng b c cc sai lch ca trc nh bin dng ca cc chi tit n hi, ni trc n hi cn c th: Gim va p v chn ng. phng c cng hng do dao ng xon gy nn Trong mt s trng hp s dng ni trc n hi lm tng tui th ca c cu chu tc ng ca ti trng ng nhiu ln. Trong bi ny ta s dng ni trc vng n hi c cu to tng t nh ni trc a nhng c thay bng cht c bc vng n hi bng cao su. Vt liu lm ni trc: thp rn 35 Vt liu lm cht: thp 45 thng ha. Ni trc vng n hi n gin d ch to v gi r nn c dng rng ri. Tin hnh chn thng s ca b ni trc n hi theo bng 9.11[1] ta c: ng knh trc (d) ng knh a (D) Chiu di b ni ( l ) ng knh cht ( d c ) Chiu di cht ( lc ) Ren S lng cht (Z) ng knh ngoi ca vng n hi Chiu di ton b vng n hi ( lv )d =

: : : : : : : : :

48(mm) 190(mm) 112(mm) 18(mm) 42(mm) M 12 10 35(mm) 36(mm)

Kim nghim sc bn dp ca vng cao su (cng thc 9.22[1])2 KM x [ ]d ZDolv d c

Vi: - [ ]d = 2 ( N / mm2 ) - K = 1, 2 : h s ti trng ng (bng 9.1[1]) - M x = 739 ( Nm) - Do = D do = 190 36 = 154 (mm) d = 2 1, 25 739 103 = 1,85 ( N / mm 2 ) : tha iu kin sc bn dp. 10 154 18 36

Kim nghim sc bn un ca cht (cng thc 9.23[1])u =KM x lc [ ]u , chn [ ]u = (60 80) N / mm 2 0,1ZDo d c3

u =

1, 25 739 103 42 = 43, 2 ( N / mm 2 ) : tha iu kin sc bn un. 0,110 154 183

28

n Thit K K Thut

Phn VII

Phn VII

CHN THN MY, BULNG, CC CHI TIT MY, BI TRN V DUNG SAI LP GHPI. Cu to v hp:Chn v hp c, vt liu lm v hp l gang xm m bo cho cng ngh ch to v lp rp, ta chia v hp thnh hai phn: np v thn. Mt phn chia phi i qua ng tm ca cc trc, thn v np hp c ghp vi nhau bng bulng, mt bch lp ghp phi c gia cng phng. Phn np c ca quan st s n khp ca rng sau khi lp rp v rt du bi trn, c l thng hi v l lp bulng vng vic di chuyn hp gim tc c d dng. Phn thn c thc o du bi trn, phn y phi c c sao cho nghing 1o 2o vic tho du d dng v thun tin hn. Kch thc v hp: Bng 10.9[1] cho php ta xc nh c kch thc cc phn t cu to nn v hp nh sau: Chiu dy thnh thn hp Chiu dy thnh np hp Chiu dy mt bch di ca thn Chiu dy mt bch trn ca np Chiu dy y hp khng c phn li Chiu dy gn thn hp Chiu dy gn np hp Chiu dy phn bch y ng knh bulng nn ng knh bulng cnh ng knh bulng ghp np vo thn ng knh bulng ghp np ng knh bulng ghp np ca thm : : : : : : : : : : : : : 9 (mm) 8 (mm) 15 (mm) 13 (mm) 10 (mm) 8 (mm) 7 (mm) 20 (mm) 20 (mm) 16 (mm) 10 (mm) 8 (mm) 6 (mm)

Vi khong cch trc 144 x 214 tra bng 10.11a[1] v 10.11b[1] ta chn bulng vng M16 vi s lng l 2 ci. Chn s b chiu di hp L = 600 (mm) Chn s b chiu rng hp B = 350 (mm) S lng bulng nn n: n =L+B 600 + 350 = 3,8 . Chn n = 4 = 200 300 250

II. Bi trn hp gim tc:Bi trn b truyn bnh rng: do vn tc nh nn ta chn phng php ngm bnh rng trong hp du, mc du thp nht phi ngp chn rng ca bnh rng dn trc I, nh vy s c hao ph nng lng do lc cn ca du, tuy nhin do vn tc bnh rng trong hp nh (v = 0,785m/s) nn hao ph khng ng k. 29

n Thit K K Thut

Phn VII

Theo bng 10.17[1] chn nht ca du bi trn bnh rng l 116 centistc hoc 16 Engle v theo bng 10.20[1] chn loi du AK20. Bi trn ln: b phn c bi trn bng m, v vn tc b truyn bnh rng thp nn du khng th bn te ln . Ta c th dng m loi T ng vi nhit lm vic t 60o C 100o C v vn tc di 1500 vng/pht (bng 8.28). Lng m cha 2/3 ch trng ca b phn .

III. Dung sai lp ghp:Cn c vo yu cu lm vic ca tng chi tit trong hp gim tc ta chn cc kiu lp ghp sau: Dung sai ln: Vng trong ln chu ti tun hon, lp ghp theo h thng l. vng khng trt trn b mt trc khi lm vic nn ta chn mi ghp trung gian c di rt nh h6 theo h thng trc. Vng ngoi lp theo h thng l, vng ngoi khng quay nn chu ti cc b, c th dch chuyn dc trc mt lng nh khi lm vic, khi nhit tng trong qu trnh lm vic nn chn kiu lp trung gian H7. Dung sai lp ghp bnh rng v bnh ai: B truyn chu ti va p nh, mi lp khng yu cu phi tho lp thng xuyn nn chn kiu lp H7/m6. Lp ghp np v thn: Do mi lp cn tho lp d dng v c th iu chnh c nn chn kiu lp lng H7/e8. Lp ghp vng chn du trn trc: d dng tho lp ta chn kiu lp trung gian H7js6. Lp cht nh v: Chn kiu lp cht m bo ng tm cao v khng b ng sut: H7u8 Lp ghp then: Theo chiu rng chn kiu lp trn trc l P9/h9 Trn cc chi tit my nh bnh rng, khp ni ... chn kiu lp Js9/h9. Theo chiu cao sai lch gii hn kch thc then l h11 Theo chiu di sai lch gii hn kch thc then l h14.

IV. Chn cc chi tit my ph khc:1. Np ln: o Chn kch thc np : Da theo cc ng knh vng ngoi ca v bng 10-10b [1], ta chn kch thc ca np, b dy np xc nh theo b dy ca v hp gim tc. o Tc dng ca np ln: C nh trc theo phng dc trc: c nh trc theo phng dc trc c th dng np ln v iu chnh khe h ca bng cc tm kim loi gia np v thn hp gim tc, np lp vi thn hp gim tc bng vt, loi ny d ch to v d lp ghp. 30

n Thit K K Thut

Phn VII

Che kn ln: che kn cc u trc ra, trnh s xm nhp ca bi bm v tp cht vo , cng nh ngn m chy ra ngoi, y dng loi vng pht l n gin nht ( bng 8-29 trang 203 [1]) 2. Vng pht: C tc dng ngn khng cho du hoc m chy ra ngoi hp gim tc v ngn khng cho bi t bn ngoi vo bn trong hp gim tc. Chn vng pht theo tiu chun ISO 6194. 3. Vng chn du: C tc dng khng cho du v m tip xc nhau 4. Cht nh v: C tc dng nh v chnh xc v tr ca np v thn ca hp gim tc, dng 2 cht nh v m khi xit bu lng khng lm bin dng vng ngoi ca , do loi tr c mt trong nhng nguyn nhn lm chng hng v gip cho vic lp ghp hp gim tc c d dng hn. Cht nh v dng cht cn theo ISO 2339 c lm bng thp CT3 5. Np ca thm: C tc dng kim tra, quan st cc chi tit my trong hp gim tc khi lp ghp v khi du vo trong hp, quan st s an khp gia cc bnh rng. Ca thm c y bng np. Np ca thm c ch to kt hp vi b phn thng hi tay cm. Kch thc np ca thm c ly theo tiu chun bng 10-12[2] 6. Bulong tch np v thn: o Dng tch np v thn. o Chn bulong M10 x 40 tiu chun ISO 4014 7. Que thm du: o L chi tit dng kim tra mc du trong hp gim tc. o Kch thc c chn theo tiu chun (trang 289 [2]). 8. Nt tho du: o Dng tho du c, bn ra khi hp gim tc o Chn loi nt theo tiu chun DIN 910, kch thc M20 x 2.

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n Thit K K Thut

TI LIU THAM KHO1. Nguyn Trng Hip, Nguyn Vn Lm_ Thit K Chi Tit My _ Nh xut bn gio dc, 2005. 2. Nguyn Hu Lc _ C s Thit K My _ Nh xut bn i hc quc gia TP.H Ch Minh, 2004 3. Phan Th Bch Nga _ Bi tp C ng Dng _ Nh xut bn i hc quc gia TP.H Ch Minh, 2003. 4. Trn Hu Qu _ V K Thut C Kh, tp mt _ Nh xut bn gio dc, 2004. 5. Trn Hu Qu, ng Vn C, Nguyn Vn Tun _ V K Thut C Kh, tp hai _ Nh xut bn gio dc, 2004. 6. Nguyn B Dng, Nguyn Vn Lm, Hong Vn Ngc, L c Phong _,Tp Bn V Chi Tit My _ Nh xut bn Trung hc chuyn nghip, H Ni, 1978

n Thit K K Thut

MC LCTn Phn I. Phn II. Phn III. Phn IV. Phn V. Phn VI. Phn VII. Tm hiu h thng dn ng thng trn Xc nh cng sut ng c v phn phi t s truyn cho h thng Thit k b truyn ai thang Thit k b truyn bnh rng Thit k trc v chn then Chn ln v ni trc Chn thn my, bulng, cc chi tit my, bi trn v dung sai Trang 1 3 6 9 16 26 29