hgt khai trien1

PhnI Chn ng c in- Lp bng thng s

n chi tit my Nguyn Quang Trung B CT 02 - K49

Mc lc

Trang

Phn I. Chn ng c v phn phi t s truyn

1. Chn ng c.3

2. Xc nh cng sut, momen v s vng quay trn cc trc.5

3. Bng thng s.6

Phn II. Thit k cc b truyn.A). B truyn ai. 1. Chn loi ai.82. Xc nh cc thng s ca ai v b truyn ai.83. Bng thng s b truyn ai.9B). B truyn trong hp.1. Chn vt liu.92. Xc nh cc loi ng sut cho php.103. Tnh ton cho cp nhanh.124. Tnh ton cho cp chm.205. Bng thng s cc b truyn bnh rng trong hp.27Phn III. Thit k trc v chn ln.

A). Thit k trc .

1. S phn tch lc ca h dn ng.292. Gi tr ca cc lc n khp.293. Tnh s b trc.304. Xc nh khong cch cc gi v im t lc.315. Xc nh ng knh v chiu di cc on trc.316. Tnh kim nghim trc v bn mi.317. Tnh kim nghim bn ca then.54B). Chn ln.1. Chn ln cho trc I.562. Chn ln cho trc II.583. Chn ln cho trc III.59Phn IV. Tnh ton cc yu t ca v hp v cc chi tit khc.

1. Tnh ton cc yu t ca v hp.632. Bng k cc kiu lp.66Ti liu tham kho v tra cu.

n c thit k da trn cc ti liu sau y :

1. Tnh ton thit k h thng dn ng c kh T1,T2 - Trnh Cht ,L Vn Uyn (2000) Cc s liu c tra trong qa trnh thit k v tnh ton da trn cc bng trong cun ( Tnh ton thit k h thng dn ng c kh T1,T2 - Trnh Cht ,L Vn Uyn (2000) 2. Bi tp k thut o PGS, TS Ninh c Tn

TS Nguyn Trng Hng

ThS Nguyn Th Cm TPhn I. chn ng c v phn phi t s truynI. Chn ng c in1. Chn loi, kiu ng cChn ng c in dn ng my mc hoc thit b cng nghip l giai on u tin trong qu trnh tnh ton thit k my.

Theo yu cu lm vic ca thit b cn c dn ng v phm vi s dng ca loi ng c, ta chn ng c in mt chiu.

Loi ng c ny c u im sau: m bo khi ng m, hm v o chiu d dng.Tuy nhin loi ng c ny cng c cc nhc im nh sau: Gi thnh cao, kh kim v phi tng thm vn u t t cc thit b chnh lu. 2. Cng sut trn trc ng c: - iu kin: Pc PycPyc =

Pct =

F = 7200N

v = 0,62m/s

=> Pct = 4,46 kW

- Hiu sut truyn ng:

( = ( . (k . (t . (2BR . (3lTrong : (k, (, (t, (l, (BR ln lt l hiu sut ca b truyn khp ni, ai, trt, ln v bnh rng. Tra bng (2.3) trang 19 ti liu I ta c:(k = 0,99

( = 0,95 (t = 0,99(l = 0,99(BR = 0,97( = 0,99. 0,95 . 099. (0,99)3 .(0,97)2 = 0.85- H s ti trng tng ng:

= = = 0,9(v thi gian m my l rt nh => b qua)=> Pyc = = 4,72 kW3. Xc nh tc ng b ca ng c in:- S vng quay ca trc my cng tc:

nct = 34,82 v/ph- Chn t s truyn s b ca h:

usb = usbh.usbn

Chn usbn = 4

usbh=12

usb = 4 . 12 = 48

S vng quay s b ca ng c l:

nsb = nct . usb = 34,82 . 45 = 1567 v/ph

- Chn s vng quay ng b ca ng c l: nb = 1500 v/ph

- Theo bng P.13 ta chn ng c in 4A112M4Y3 vi cc thng s:

Pc = 5,5 kW

nc = 1425 v/ph

= 2

= 1,54. Phn phi t s truyn: Xc nh t s truyn uch ca h dn ng

ndc - S vng quay ng c chn, ndc =1425v/ph

nlv - S vng quay ca trc my cng tc, nlv =34,82 v/ph

ung - T s truyn ca b truyn ngoi

Truyn ng ngoi bngai dt thng

Chn ung= 4uh - T s truyn ca hp gim tc

uh =uch/ung =40,92/4=10,23 Mt khc: uh = unh . uchunh - T s truyn ca b truyn cp nhanh

uch - T s truyn ca b truyn cp chm

Trong hp gim khai trin thng ly(Bng 3.1\43-TKHD-I):

unh= 3,89

uch =2,63

Tnh li

II. Tnh ton tc quay, momen, cng sut trn cc trc

a. Tnh cng sut P trn mi trc Trc3:

- Hiu sut ln, = 0,99

= (kW) Trc 2:

- Hiu sut b truyn bnh rng tr,

= (kW) Trc 1: - Hiu sut b truyn bnh rng tr, =0,97

= (kW) Trc ng c: - Hiu sut b truyn ai, =0,95

= (kW)b. Tnh s vng quay trn mi trc

Trc 1

nc - S vng quay ca ng c, nc =1425(v/ph)

ung t s truyn b truyn ngoi ca ai dt thng ,ung=4 n1=ndc/ung=1425/4=356,25(v/ph)

Trc 2

n2 = n1 / unh unh t s truyn ca cp nhanh ,unh=3,89n2 =356,25/3,89 =91,58 (v/ph)

Trc 3

n3 = n2 / uch

uch - T s truyn ca b truyn cp chm uch=2,63n3 =91,58/2,63 = 34,82(v/ph)

c. Tnh momen trn mi trc

Trc 1 = (Nmm) Trc 2 = (Nmm) Trc 3 = (Nmm) Trc cng tc = (Nmm) Trc ng c: = (Nmm)III. Bng thng sTrc

Thng sng cIIIIIICng tc

Cng sut P

(kW)5,254,944,74 4,554,46

T s truyn u

43,892,634

S vng quay n

(v/ph) 1425356,2591,5834,8234,82

Momen xonT

(N.mm) 3518413242749428913548821223234

Phn II. Thit k cc b truyn c khA. Thit k b truyn ngoi hp1.Chn loi ai:

Theo yu cu cho th ta chn b truyn ngoi l ai thang dt

Vt liu ai: vi cao su

2.Kch thc ai

- ng knh ai bnh nh : Theo cng thc 4.1 (trang 53 TKHD)

d1=(5,26,4).

T1:momen xon trn trc bnh ai nh

d1 :ng knh bnh ai nh

Ta bit momen xon trn bnh ai ln chnh bng trc 1, nn trn bnh nh l

T1=Tc=35184 (Nmm)

Thay s : d1=(5,26,4). =170,39209,71mm

Ta chn d1=180 mm

Vn tc ai: v=d1ndc/60000=3,14.180.1425/60000=13,43 m/s NHO2 KHL2= 1

S chu k chu ti tng ng ca bnh nh ln hn s chu k chu ti ca bnh ln u1 ln:

NHE1= 8,3.107.3,89 = 32,287.107 > NHO1 KHL1= 1

Theo cng thc 6.1a:

(MPa)

(MPa)

Cp nhanh s dng cp bnh rng tr rng thng do ng sut tip xc cho php l:

((H(n = ((H(2 = 481,8 (MPa)

NHE2= NHE3= 8,3.107> NHO3 KHL3= 1

NHE4= NHE3/u2 = 8,3.107/2,63 = 3,15.107 > NHO4 KHL4= 1

Theo cng thc 6.1a:

(MPa)

(MPa)

Cp chm s dng cp bnh rng tr rng nghing do ng sut tip xc cho php l:

((H(c =(((H(3 +((H(4)/2= (536,36+509)/2 = 522,68 (MPa) NFE2>NFO2 = 4.106 => NFE1>NFO1 = 4.106

Theo 6.2a vi b truyn quay 1 chiu KFC= 1

Theo cng thc 6.13 v 6.14 ta c: ng sut qu ti cho php:

((H(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) ((F(3max = 0,8.(ch3= 0,8.580 = 464(MPa) ((F(4max = 0,8.(ch4= 0,8.580 = 464(MPa)3. Tnh ton cho cp nhanh (b truyn bnh rng tr rng thng)

a.Xc nh s b khong cch trc:Khong cch trc aw c xc nh theo cng thc(6.15a):

(1)

Trong :

Ka H s ph thuc vo vt liu ca cp bnh rng v loi rng .

T Momen xon trn trc bnh ch ng (Nmm).

((H( ng sut tip xc cho php ca cp nhanh (MPa).

u T s truyn.

KH( H s xt n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v tip xc.

(ba H s quan h gia chiu rng vnh rng bw v khong cch trc aw.

(ba = bw/awTheo bng 6.5 (TR.96,TTTKHTDCK-T1) ng vi vt liu ca cp bnh rng (thp-thp) v loi rng thng ta c: Ka=49,5

Theo bng 6.6 (TR.97,TTTKHTDCK-T1) ng vi v tr bnh rng i vi cc trong hp gim tc khng i xng v rn mt rng lm vic H1 v H2 (350HB chn (ba1= 0,3. Vi u = u1= 3,89 => (bd1 = 0.53(ba1.(u1+1) = 0,53.0,3.(3,89+1) = 0,78theo bng 6.7 (TR.98,TTTKHTDCK-T1) ng vi (bd1 = 0,78 v s 3 chn KH( = 1,115.

T = T1 = 132427(Nmm)

((H( = ((H(n = 481,8(MPa)

Thay cc gi tr trn vo cng thc (1) ta c:

Ly aw1 = 198(mm).

b. Xc nh cc thng s n khp.

Chn moun.

T khong cch trc aw1 ta xc nh c moun theo cng thc (6.17):

m1 = (0,01(0,02).aw1 = (0,01(0,02).198 = (1,98(3,96) (mm)

Theo bng 6.8 (TR.99,TTTKHTDCK-T1) chn c moun theo tiu chun l: m1= 3 Xc nh s rng v t s truyn thc.

S rng z1, z2 :

(rng) ly z1 = 27

(rng) ly z2 = 105

Suy ra t s truyn thc l: um1 = z2/z1 = 105/27 ( 3,89 (ln)

Tng s rng:

Zt = z1+z2 = 27+105 = 132 aw = = 198(mm) ly aw = 200(mm)

Do cn dch chnh tng khong cch trc t 198 n 200(mm)

Tnh h s dch tm theo (6.22):

Y = aw/m 0,5.(z1+z2) = 200/3 0,5.(27+105) = 0,67Theo (6.23): ky = 1000.y/zt = 1000.0,67/132 = 5,08Theo (6.10a): kx = 0,197 (y = kx.zt/1000 = 0,197.132/1000 = 0,026

Tng h s dch chnh:

Xt = y+(y = 0,67+0,026 = 0,696(ly h s:

X1 = 0,5.(0,696 (105 - 27).0,67/132) = 0,15

X2 = Xt X1 = 0,696 0,15 = 0,546Gc n khp:

cos(tw = 132.3.cos200/(2.200) = 0,93

((tw = 21,570c. Kim nghim rng v bn tip xc.

Theo (6.33) ng sut tip xc xut hin trn mt rng ca b truyn phi tho mn iu kin sau:

(2)

Trong :

ZM H s k n c tnh vt liu ca cc bnh rng n khp. Theo bng 6.5 (TR.96,TTTKHTDCK-T1) ta tra c ZM = 274(MPa1/3)

ZH H s xt n hnh dng b mt tip xc.

Z( H s xt n s trng khp ca rng.

KH H s ti trng khi tnh v tip xc.

dw1 ng knh vng ln bnh ch ng(mm).

bw1 B rng vnh rng bnh ch ng(mm).

((H( ng sut tip xc cho php (Mpa)(tnh chnh xc)

T1 Momen xon trn trc T1= 132427(Nmm)

um1 T s truyn thc ca b truyn cp nhanh um1 = 3,89 (ln)

H s xt n hnh dng b mt tip xc ZH c xc nh theo cng thc (6.34):

Gc nghing ca rng trn hnh tr c s (b c xc nh theo cng thc :

(b = arctg(cos(t1.tg()

Gc prfin rng (t c xc nh nh sau:

(t1 = arctg(tg(/cos()

Trong :

( Gc prfin gc xc nh theo TCVN 1065 71, (=20(( Gc nghing rng (=0.

=>(t1 = ( = 20(,(b= ( = 0(Gc n khp (tw c xc nh theo cng thc:

(tw1 = arccos[(a1.cos(t1)/aw1] = 21,570Vy:

H s xt n s trng khp ca rng Z( c xc nh theo cng thc (6.36a):

v h s trng khp dc ((= bw1.sin(/(m1.() = 0

Vi h s trng khp ngang:

=>

ng knh vng ln bnh ch ng dw1:

B rng vnh rng bnh ch ng bw:

bw1= aw1.(ba1= 200.0,3 = 60(mm).

Vn tc vng v1:

H s ti trng khi tnh v tip xc KH c xc nh theo cng thc (6.39):

KH= KH(.KH(.KHvTrong :KH( H s xt n s phn b khng u ti trng cho cc i rng ng thi n khp khi tnh v tip xc. Vi bnh rng thng KH(=1,13KH( H s xt n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v tip xc. theo bng 6.7 (TR.98,TTTKHTDCK-T1) ng vi (bd1 = 0,53.(ba.(u+1) = 0,53.0,3.(3,89+1) = 0,78 v s 3 chn c KH( = 1,115.

KHv H s k n ti trng ng xut hin trong vng n khp khi tnh v tip xc

Xc nh h s KHv theo cng thc (6.41):

Trong :

(6.42)

Trong :

(H H s xt n nh hng ca cc sai s n khp. Theo bng 6.15 ng vi HB1,HB2

Vy:

=> KH= KH(.KH(.KHv= 1,07.1,115.1,13 = 1,35Tnh chnh xc ((H(:

Vi v1 = 1,53(m/s) v rn mt rng HB ZR=0,9

ng knh nh rng:

da1= dw1+2.m = 81,82+2.3 =87,82(mm) KxH = 1

Vy ng xut tip xc cho php l:

((H( = ((H(.ZR.Zv.KxH = 481,8.0,9.0,89.1 = 385,92(MPa)

Thay cc gi tr tnh c trn vo cng thc (2) ta c:

Kim nghim:

Tho mn.

bn tip xc c m bo.

Tnh li chiu rng vnh rng:

Ly bw= 57 mmd. Kim nghim rng v bn un.

m bo bn un cho rng ng sut un sinh ra ti chn rng khng c vt qu gi tr cho php (theo cng thc (6.43),(6.44)):

(3)

(4)

Trong :

Y( H s xt n nghing ca rng , vi rng thng Y(=1.

Y( H s xt n s trng khp ca rng.

YF1 H s dng rng ca bnh 1.


KF H s ti trng khi tnh v un.



((F1( ng sut un cho php ca bnh rng 1 (Mpa)(tnh chnh xc)



H s xt n s trng khp ca rng Y( c xc nh nh sau:

Y(=1/(( vi ((=1,73 (tnh c trn)

=> Y(=1/((=1/1,73=0,58H s dng rng ca cp bnh rng YF1, YF2:

S rng tng ng c xc nh theo cng thc sau:

Do y l cp bnh rng tr rng thng nn: zv1= z1= 27 ; zv2= z2= 105

Tra theo bng 6.18 (TR109.,TTTKHTDCK-T1) ng vi h s dch chnh bng x1 = 0,15; x2=0,546 v:

S rng tng ng zv1= 27 => YF1= 3,63S rng tng ng zv2= 105 => YF2=3,497H s ti trng khi tnh v un KF c xc nh theo cng thc(6.45):

KF= KF(.KF(.KFvTrong :

KF( H s xt n s phn b khng u ti trng cho cc i rng ng thi n khp khi tnh v un. Vi bnh rng thng KF(=1,37KF( H s xt n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v un. theo bng 6.7 (TR.98,TTTKHTDCK-T1) ng vi (bd1 = 0,78 v s 3 chn c KF( = 1,233.

KFv H s k n ti trng ng xut hin trong vng n khp khi tnh v un

Xc nh h s KFv theo cng thc(6.46):

Trong :

(6.47)

(F H s xt n nh hng ca cc sai s n khp. Theo bng 6.15 ng vi HB1, HB2

Vy:

KF= KF(.KF(.KFv = 1,233.1,37.1,11 = 1,88 (F1 = = 74,93 (MPa)

(F2 = 74,93.3,497/3,63 = 72,18 (MPa)

Tnh chnh xc ((F1(:

Vi m1 = 3(mm) h s xt n nhy ca vt liu i vi tp trung ng sut c xc nh:

=>Ys=1,08-0,0695ln(3)=1H s xt n nh hng ca nhm mt ln chn rng thng bnh rng phay th YR=1

ng knh nh rng da1 KxF = 1

Vy ng xut un cho php l:

((F1( = ((F1(.YR.Ys.KxF = 252.1.1.1 =252(MPa)

Tnh chnh xc ((F2(:

Vi m1 = 3(mm) h s xt n nhy ca vt liu i vi tp trung ng sut c xc nh:

=>Ys=1,08-0,0695ln(3)=1H s xt n nh hng ca nhm mt ln chn rng thng bnh rng phay th YR=1 KxF = 1


((F2( = ((F2(.YR.Ys.KxF = 236,57.1.1.1 =236,57(MPa)

ta thy:

(F1 < [(F1] (F2 < [(F2] Vy rng m bo v bn une. Kim nghim rng v qu ti.

Khi lm vic bnh rng c th b qu ti (lc m my, hm my,hoc c s c bt thng). V vy cn kim nghim rng v qu ti da vo ng sut tip xc v ng sut un cc i

Vi h s qu ti:

Trong :

T Momen xon danh ngha.

Tmax Momen xon qu ti.

Tmm Momen m my.

trnh bin dng d hoc gy dn lp b mt, ng sut tip xc cc i khng c vt qu mt gi tr cho php (6.48):

Vi cc gi tr c tnh trn:

(H = 376,68(MPa)

((H(max =1260(MPa)

=>

phng bin dng d hoc ph hng tnh mt ln chn rng, ng sut un cc i ti mt ln chn rng khng vt qu gi tr cho php(6.49):


(F1 = 74,93(MPa)

(F2 = 72,18(MPa)

((F(max = 464(MPa)

=>

Vy b truyn m bo v iu kin qu ti.4. Tnh ton cho cp chm (b truyn bnh rng tr rng nghing)

a. Xc nh s b khong cch trc.

Khong cch trc aw c xc nh theo cng thc:

(1)

Trong :

Ka H s ph thuc vo vt liu ca cp bnh rng v loi rng .

T Momen xon trn trc bnh ch ng (Nmm).

((H( ng sut tip xc cho php ca cp chm (MPa).

u T s truyn ca cp bnh rng cp chm.

KH( H s xt n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v tip xc.

(ba H s quan h gia chiu rng vnh rng bw v khong cch trc aw.

(ba=bw/awTheo bng 6.5 (TR.96,TTTKHTDCK-T1) ng vi vt liu ca cp bnh rng (thp-thp) v loi rng nghing ta c: Ka= 43

Theo bng 6.6 (TR.97,TTTKHTDCK-T1) ng vi v tr bnh rng i vi cc trong hp gim tc khng i xng v rn mt rng lm vic H1 v H2 (350HB chn (ba2= 0,3Vi u = u2= 2,63 => (bd2 = 0.5(ba2.(u2+1) = 0,5.0,3.(2,63+1) = 0,54theo bng 6.7 (TR.98,TTTKHTDCK-T1) ng vi (bd2 = 0,54 v s 5 chn KH( = 1,03.

T = T2 = 494289(Nmm)

((H( = ((H(C = 522,68(MPa)

Thay cc gi tr trn vo cng thc (1) ta c:

Ly aw2=210(mm).

b. Xc nh cc thng s n khp.

Chn moun.T khong cch trc aw2 ta xc nh c moun php mn theo cng thc sau:

mn = (0,01(0,02).aw2=(0,01(0,02).210=(2,1(4,2) (mm)

Theo bng 6.8 (TR.99,TTTKHTDCK-T1) v tnh thng nht ho chn c moun theo tiu chun l: m2= 3(mm)

Xc nh s rng, gc nghing v t s truyn thc.

Chn s b gc nghing ca rng ( = 10(=>cos( = cos10(= 0,9848

S rng bnh nh z1 c xc nh theo cng thc :

(rng)

Chn s rng bnh nh l: z3 = 38(rng)

Vy s rng bnh ln l: z4 = u2.z3 = 38.2,63 = 99,94(rng)

Chn s rng bnh ln l: z4 = 100(rng)

Suy ra t s truyn thc l: um2 = z4/z3 = 100/38 = 2,63 (ln)

Sai s t s truyn l: u = um2- u2 = 2,63 - 2,63 = 0

=> (u = u.100%/ u2 = 0%

Vy b truyn c m bo.

Xc nh chnh xc gc nghing (Ta c:

=> (=9,690c. Kim nghim rng v bn tip xc.

ng sut tip xc xut hin trn mt rng ca b truyn phi tho mn iu kin sau:

(2)

Trong :

ZM H s k n c tnh vt liu ca cc bnh rng n khp.

Theo bng 6.5 (TR.96,TTTKHTDCK-T1) ta tra c ZM = 274(MPa1/3)

ZH H s xt n hnh dng b mt tip xc.

Z( H s xt n s trng khp ca rng.

KH H s ti trng khi tnh v tip xc.



((H( ng sut tip xc cho php (Mpa)(tnh chnh xc)


um2 T s truyn thc ca b truyn cp chm um2 = 2,63 (ln)

H s xt n hnh dng b mt tip xc ZH c xc nh theo cng thc sau:

Gc nghing ca rng trn hnh tr c s (b c xc nh theo cng thc :

(b = arctg(cos(t2.tg()

Gc prfin rng (t c xc nh nh sau:

(t2 = arctg(tg(/cos()

Trong :

( Gc prfin gc xc nh theo TCVN 1065 71, (=20(( Gc nghing rng (=9,690=>(t2 = 20,270 => (b= 9,10

Gc n khp (tw c xc nh theo cng thc:

(tw2 = (t2 = 20,270Vy:

+) H s xt n s trng khp ca rng Z( c xc nh theo cng thc sau:

v h s trng khp dc >1B rng vnh rng bnh ch ng bw3:

bw3 = (ba2.aw2= 0,3.210 = 63 mmVi h s trng khp ngang:

Vy:

ng knh vng ln bnh ch ng dw3:

Vn tc vng v2:

H s ti trng khi tnh v tip xc KH c xc nh theo cng thc:

KH= KH(.KH(.KHvTrong :

KH( H s xt n s phn b khng u ti trng cho cc i rng ng thi n khp khi tnh v tip xc.

KH( H s xt n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v tip xc. theo bng 6.7 (TR.98,TTTKHTDCK-T1) ng vi (bd2 = 0,54 v s 5 chn c KH( = 1,03.

KHv H s k n ti trng ng xut hin trong vng n khp khi tnh v tip xc

Theo bng 6.14 (TR.107,TTTKHTDCK-T1) ng vi v2 = 0,55(m/s) cp chnh xcv mc lm vic m l 9 ta c KH(=1,13Xc nh h s KHv theo cng thc:

Trong :

(H H s xt n nh hng ca cc sai s n khp. Theo bng 6.15 ng vi HB1,HB2

Vy:

=>KH= KH(.KH(.KHv= 1,03.1,13.1,005 = 1,17Tnh chnh xc ((H(:

Vi v1 = 0,55 (m/s) ZR=0,95

ng knh nh rng: da3= dw3+2.m = 115,7+2.3 = 121,7(mm) KxH = 1

Vy ng xut tip xc cho php l:

((H( = ((H(.ZR.Zv.KxH = 522,68.0,95.1.1 =496,55 (MPa)

Thay cc gi tr tnh c trn vo cng thc (2) ta c:

bn tip xc c m bo.

Kim nghim tha bn:

Tho mn.

d. Kim nghim rng v bn un.

m bo bn un cho rng ng sut un sinh ra ti chn rng khng c vt qu gi tr cho php :

(3)

(4)

Trong :

Y( H s xt n nghing ca rng.

Y( H s xt n s trng khp ca rng.



KF H s ti trng khi tnh v un.





T2 Momen xon trn trc ch ng T2= 494289(Nmm)

H s xt n nghing ca rng Y(:

Y(= 1-((/140 = 1- 9,69/140 = 0,93 H s xt n s trng khp ca rng Y( c xc nh nh sau:

Y( = 1/(( vi (( = 1,74 (tnh c trn)

=>Y(=1/((=1/1,74 = 0,575H s dng rng ca cp bnh rng YF1, YF2:

S rng tng ng c xc nh theo cng thc sau:

Tra theo bng 6.18 (TR109.,TTTKHTDCK-T1) ng vi h s dch chnh bng x = 0 v:

S rng tng ng zv3= 40 => YF1= 3,74S rng tng ng zv4= 104 => YF2=3,61H s ti trng khi tnh v un KF c xc nh theo cng thc:

KF= KF(.KF(.KFvTrong :

KF( H s xt n s phn b khng u ti trng cho cc i rng ng thi n khp khi tnh v un. theo bng 6.14 (TR.107,TTTKHTDCK-T1) ng vi v2 KF(=1,37

KF( H s xt n s phn b khng u ti trng trn chiu rng vnh rng khi tnh v un. theo bng 6.7 (TR.98,TTTKHTDCK-T1) ng vi (bd2 = 0,54 v s 5 chn c KF( = 1,1.

KFv H s k n ti trng ng xut hin trong vng n khp khi tnh v un Xc nh h s KFv theo cng thc:

Trong :

(F H s xt n nh hng ca cc sai s n khp. Theo bng 6.15 ng vi HB1,HB2

Vy:

=>KF= KF(.KF(.KFv = 1,1.1,37.1,01 = 1,52Tnh chnh xc ((F1(:

Vi mn = 3(mm) h s xt n nhy ca vt liu i vi tp trung ng sut c xc nh:

=>Ys=1,08-0,0695ln(3)=1,004H s xt n nh hng ca nhm mt ln chn rng thng bnh rng phay th YR=1



((F3( = ((F3(.YR.Ys.KxF = 267,43.1.1,004.1 =268,5(MPa)

Tnh chnh xc ((F4(:

Vi mn = 3(mm) h s xt n nhy ca vt liu i vi tp trung ng sut c xc nh:

=>Ys=1,004H s xt n nh hng ca nhm mt ln chn rng thng bnh rng phay th YR=1



((F4( = ((F4(.YR.Ys.KxF = 252.1.1,004.1 =253(MPa)

Thay cc gi tr tnh c trn vo cng thc (3)&(4) ta c:

Vy rng m bo v bn un

e. Kim nghim rng v qu ti.

Khi lm vic bnh rng c th b qu ti (lc m my, hm my,hoc c s c bt thng). V vy cn kim nghim rng v qu ti da vo ng sut tip xc v ng sut un cc i

Vi h s qu ti:

Trong :

T Momen xon danh ngha.

Tmax Momen xon qu ti.

Tmm Momen m my.

trnh bin dng d hoc gy dn lp b mt, ng sut tip xc cc i khng c vt qu mt gi tr cho php:


(H = 466,45(MPa)

((H(max =1260(MPa)

=>

phng bin dng d hoc ph hng tnh mt ln chn rng, ng sut un cc i ti mt ln chn rng khng vt qu gi tr cho php:


(F3 = 137,43 (MPa)

(F4 = 132,65(MPa)

((F(max = 464(MPa)

=>

Vy b truyn m bo v qu ti.

T cc kt qu tnh ton trn ta c bng cc thng s ca cc b truyn trong hp:

Thng sGi tr

Cp nhanhCp chm

Mounm = 3(mm)mn = 3(mm)

Chiu rng vnh rng bw = 57(mm)bw = 63(mm)

H s dch chnhx1 = 0,15x2 = 0,546(y = 0,026x1 = 0

x2 = 0

(y = 0

Khong cch trc

(khong cch trc chia)aw1 = 200(mm)aw2 = 210(mm)

ng knh vng chiad1 = 81(mm)

d2 = 315(mm)d3 = 115,65(mm)

d4 = 304,34(mm)

ng knh vng lndw1 = 81,8(mm)

dw2 = 318,2(mm)dw3 = 115,7(mm)

dw4 = 304,29(mm)

ng knh nh rngda1 = 87,74(mm)

da2 = 324,12(mm)da3 = 121,65(mm)

da4 = 310,29(mm)

ng knh y rngdf1 = 74,4(mm)

df2 = 310,78(mm)df3 = 108,15(mm)

df4 = 296,84(mm)

ng knh c sdb1 = 76,12(mm)

db2 = 296(mm)db3 =108,68(mm)

db4 =285,99(mm)

S rng bnh rng z1 = 27z2 = 105z1 = 38z2 = 100

Gc prfin gc

(TCVN 105-71)( = 20(( = 20(

Gc nghing ca rng( = 0(( = 9,69(

Gc prfin rng (t1 = 20((t2 = 20,27(

Gc n khp(tw1 = 21,57((tw2 = 20,27(

H s trng khp ngang(( = 1,73(( = 1,74

Vn tc vng bnh rngv1 = 1,53(m/s)v2 = 0,55(m/s)

phn iii. thit k trc

I. S phn tch lc ca h dn ng.

II. Gi tr ca cc lc n khp.

1. Ti trng tc dng ln trc

a. B truyn cp nhanh: bnh rng tr rng thngLc vng:

Lc hng tm:

b. B truyn cp chm: bnh rng nghingLc vng:

Lc hng tm:

Lc dc trc:

c. Lc tc dng ln b truyn aiF = 797,69 (N)

Fx = F sin300 = 797,69. sin300 = 398,85 (N)

Fy = F cos300 = 797,69.cos300 = 690,82 (N)

d. Lc tc dng t ni trc n hi:

Vi Dt : ng knh vng trn qua tm cc cht.Theo bng 16.10a ta c Dt = 227,5 mmLc t khp ni tc dng ln trc

Fk = (0,2 0,3)Ft = (2382,21 3573.,32)

=> Fk = 3000 (N)

2. Tnh s b ng knh trcTheo cng thc 10.9:

Vi T: Mmen xon

: ng sut xon cho php. = 1530 MPaa. Trc 1

(mm)

Chn d1 = 40 (mm)

b. Trc 2

(mm)

Chn d1 = 55 (mm)

c. Trc 3

(mm)

Chn d3 = 65 (mm)

T cc ng knh trc ta xc nh gn ng chiu rng ln bo

Theo bng 10.2 (TR.189,TKHTDCK-T1):

3. Xc nh khong cch cc gi

a. Xc nh chiu di may

1. Chiu di may bnh ai:lm = (1,42,5)d1 = (1,21,5)40 = (4860)

chn lm = 60 (mm)

2. Chiu di may na khp ni:lmk = (1,42,5)d3 = (1,42,5)65 = (91162,5)

chn lm12 = 120 (mm) 3. Chiu di may bnh rng 1:

lm12= (1,21,5)d1 = (1,21,5).40 = (4860)

chn lm12= 574. Chiu di may bnh rng 2 v 3.

lm23 = (1,21,5)d2 = (1,21,5).55 = (6682,5)

chn lm22 = 70 (mm)

lm23 = 67 (mm)

5. Chiu di may bnh rng 4.

lm33 = (1,21,5)d3 = (1,21,5)65 = (7897,5)chn lm12 = 80 (mm)

b. Xc nh khong cch gia cc im t lc

+ Khong cch t mt mt chi tit quay n thnh trong hp

k1 = 12 (mm)

+ Khong cch t mt mt n thnh trong ca hp

k2 = 10 (mm)

+ Khong cch t mt mt chi tit quay n np

k3 = 15 (mm)

+ Chiu cao v u bulng

hn = 18 (mm)

* Xt trn trc 2

+ Khong cch t gi 0 n bnh rng 3

l22 = 0,5.(lm22 + b0) + k1 + k2 = 0,5.(70 + 27) + 12 + 10 = 75,5 (mm)


l23 = 0,5.(lm23 + b0) +lm22 + 2k1 + k2= 0,5.(67 + 27) + 70 + 2.12 + 10 = 151 (mm)

+ Khong cch t gi 0 n gi 1

l21= l23 + 0,5.(lm23 + b0) + k1 + k2= 151 + 0,5(27 + 67) + 12 + 10 = 220 (mm)

* Xt trn trc 1


l12 = l23 = 151 (mm)


l11 = l21 = 220 (mm)

+ Khong cch t gi 0 n ch lp bnh ail13= l11 + 0,5.(lm + b0) + hn + k3= 220 + 0,5(23 + 60) + 18 + 15 = 294,5 (mm)

* Xt trn trc 3


l33 = l22 = 75,5 (mm)


l31 = l21 = 220 (mm)

+ Khong cch t gi 0 n ch lp khp ni:l32 = 0,5.(lmk + b0) + hn + k3= 0,5(120 + 33) + 18 + 15 = 109,5 (mm)

4. Lc tc dng ln trc t cc gi - Biu mmen

a.Trc 1

* Lc tc dng ln trc t cc gi

* =>

=> =>

* =>

=> =>

* Momen trn trc y

Fy10.l12 = 635,39 . 151 = 95943,89 (Nmm)

Fy10.l11- Fr1.(l11- l12) = 635,39.220 1279,98.(220-151) = 51467,18 (Nmm)

* Momen trn trc x:

Fx10.l12 = 1150,56.151 = 173734,56 (Nmm)

Fx10.l11- Ft1.(l11- l12) = 1150,56. 220 3237,82.(220-151) = 29713,62 (Nmm)

* Biu momen trc 1

* Tnh chnh xc ng knh trc 1.

- Momen un tng ti v tr lp bnh rng 1:

Theo cng thc (10.15) & (10.16)

ng knh trc1 ti tit din lp bnh rng

theo cng thc(10.18)

Trn trc I c T1 = 132427(Nmm), do c bnh rng lin trc ln chn vt liu trc ging vt liu bnh rng l thp 45 ti ci thin t bn (b= 850MPa.Vy theo bng 10.5 (TR.195,TTTKHTDCK-T1) => ng sut cho php ca trc l [(] = 56,5 Mpa

Chn ng knh ti tit din nguy him d1 = 34 (mm)

- Momen tng ng v ng knh trc ti bnh ai

- ng knh ti v tr lp bnh ai:

Chn ng knh dk = 28 (mm)

- Momen tng ng v ng knh trc ti v tr lp ln

Mx = 29713,62 (Nmm) ; My= 51467,18 (Nmm); T = 132427 (Nmm)

=>

- ng knh ti v tr lp ln:

Chn ng knh d10= d11 = 30 (mm)

* Kim nghim trc v bn mi

Trc lm bng thp 45 c gii hn bn ko (b= 600Mpa.

gii hn bn un (-1=0,436.(b = 0,436. 600 = 261,6 (MPa)

gii hn mi xon (-1=0,58.(-1 = 0,58. 261,6 = 151,73 (MPa)

Kt cu trc c tit k m bo bn mi nu h s an ton ti tit din nguy him tho mn iu kin (10.19):

Trong :

[s] : H s an ton cho php [s]=(1,5(2,5)

s(i, s(i : H s an ton ch xt n ng sut php v ng sut tip ti tit din i.

+) H s an ton ch xt n ng sut php v ng sut tip ti tit din i c xc nh

theo cng thc (10.20) & (10.21):

Trong :

- (ai,(mi : L bin v tr s trung bnh ca ng sut php ti tit din i.

- (ai,(mi : L bin v tr s trung bnh ca ng sut tip ti tit din i.

i vi trc quay, ng sut un thay i theo chu k i xng, do :

(mi= 0 ; (ai= (maxi= Mi/Wi.

Trc quay 1 chiu ng sut xon thay i theo chu k mch ng, do :

(mi = (ai = (maxi/2 = Ti/Woi+ Mi,Ti : L momen un tng v momen xon ti tit din i.

+ Wi,Woi : L momen cn un v momen cn xon ti tit din i ca trc, c xc nh theo bng 10.6 (TR.196,TTTKHTDCK-T1)

- (( ,(( : H s k n nh hng ca tr s ng sut trung bnh n bn mi. Theo bng 10.7 (TR.197,TTTKHTDCK-T1) vi (b=600MPa => ((= 0,05, ((=0

- K(di , K(di : L cc h s c xc nh theo cc cng thc (10.25) & (10.26).

K(di =( K(/((+Kx -1)Ky

K(di =( K(/((+Kx -1)Ky+ Kx : H s tp trung ng sut do trng thi b mt. Theo bng 10.8 ng vi phng php tin t Ra(2,5 .. 0,63) v (b= 600Mpa => Kx=1,06.

+ Ky : H s tng bn b mt. Khng dng cc bin php tng bn => Ky=1

+ (( , (( H s kch thc k n nh hng ca kch thc ti tit din n gii hn mi theo bng 10.10 i vi thp cacbon

=>((= 0,88, ((= 0,81

K( , K( ; H s tp trung ng sut thc t khi un v xon ph thuc vo loi yu t gy tp trung ng sut

Tr s K(/(( , K(/(( tra theo bng 10.11 (TR.198,TTTKHTDCK-T1) i vi b mt trc lp c di. Chn kiu lp k6 vi (b= 600Mpa ,

=>K(/((=2,06 , K(/((=1,64

Cc h s K(d , K(d c xc nh.

K(d = ( K(/((+Kx -1)Ky = (2,06 + 1,06-1)/1 = 2,12

K(d = ( K(/((+Kx -1)Ky = (1,64 + 1,06-1)/1 = 1,7

Ti tit din lp bnh rng 1: d12= 34 mm.

Ta c momen un tng Mut= 198466,44(Nmm)

Momen xon T = 132427 (Nmm).

Vi tit din trn:

Momen cn un W:

W = = = 3240,28 (Nmm)

Momen cn xon Wo:

Wo = = = 7098,94 (Nmm)

Bin v tr s trung bnh ca ng sut php.

(m31= 0 ; (a1= (max= Mut/W= 198466,44/3240,28 = 61,25(MPa)

Bin v tr s trung bnh ca ng sut tip.

(m1 = (a1 = (max1/2 = T/2Wo= 132427/2.7098,94 = 9,33(MPa)

H s an ton ch xt n ng sut php c xc nh:

H s an ton ch xt n ng sut tip c xc nh:

10,29Vy h s an ton s c xc nh:

b.Trc 2

* Lc tc dng ln trc t cc gi - biu momen

=>

=> =>

=>

=> =>

- Momen trn trc y

Fy20.l22 = 2189,88 . 75,5 = 165335,94 (Nmm)

Fy21.(l21 - l23) = 108,8.(220 - 151) = 7507,2 (Nmm)

- Momen trn trc x:

Fx20. l22 = 6627,56.75,5 = 500380,78 (Nmm)

Fx21. (l21 - l23) = 5194,58.(220 - 151) = 355666,02 (Nmm)

Biu mmen trc 2:

*Tnh chnh xc ng knh trc 2.


Theo cng thc (10.15) & (10.16)

ng knh trc II ti tit din lp bnh rng 2

theo cng thc(10.18)

Trn trc 2 c T2 = 494289 (Nmm), theo bng 10.5 tr 195 tp I*vi d2 = 55 => ng sut cho php ca trc l [(] = 49,8 Mpa



Theo cng thc (10.15) & (10.16)

ng knh trc II ti tit din lp bnh rng 3

theo cng thc(10.18)

Trn trc I c T2 = 494289 (Nmm),theo bng 10.5 tr 195 tp I*

vi d2 = 55 => ng sut cho php ca trc l [(] = 49,8 Mpa



Chn ng knh d20= d21 = 45 (mm)

* Kim nghim trc v bn mi


+ (-1=0,436.(b = 0,436. 600 = 261,6 (MPa)

(-1=0,58.(-1 = 0,58. 261,6 = 151,73 (MPa)

+ (( , (( : H s kch thc k n nh hng ca kch thc ti tit din n gii hn mi

Ti thit din 1 , 2 c d22 = 50mm theo bng 10.10 i vi thp cacbon

+ K( , K( : H s tp trung ng sut thc t khi un v xon ph thuc vo loi yu t gy tp trung ng sut

=>K(/(( = 2,52 ; K(/(( = 2,03Cc h s K(d , K(d c xc nh.

K(d = ( K(/((+Kx -1)Ky = (2,52 + 1,06 - 1)/1 = 2,58K(d = ( K(/((+Kx -1)Ky = (2,03+1,06-1)/1 = 2,09Ti tit din lp bnh rng 3, d3= 50 (mm).

Ta c momen un tng Mut22 = 506881,26 (Nmm)

Momen xon T2 = 494289 (Nmm)

Vi tit din trc c 1 rnh then v ng knh trc ti ch lp bnh rng d22 = 50 mm

Theo bng 9.1a (TR.173,TTTKHTDCK-T1) ta c cc kch thc ca then:

b =16 (mm) ; t1 = 6 mm

Momen cn un W:

Momen cn xon Wo:


(m3= 0 ; (a3= (max= Mut/W= 506881,26/10413,29 = 48,68(MPa)


(m3 = (a3 = (max3/2 = T/2Wo=494289/2.22685,13 = 10,89 (MPa)



H s an ton s c xc nh:

Vy trc II m bo bn mi.

c.Trc 3

* Lc tc dng ln trc t cc gi - biu momen

=>

=> =>

=>

=> =>

- Momen trn trc y

Fy31.(l31 - l33) = 2163,27.(220 75,5) = 312592,52 (Nmm)

- Momen trn trc x:

Fk. l32 = 3000 . 109,5 = 328500 (Nmm)

Fx31. (l31 - l33) = 4425,44.(220 75,5) = 639476,08 (Nmm)

- Biu mmen trc 3:

*Tnh chnh xc ng knh trc 3.


Theo cng thc (10.15) & (10.16)

ng knh trc III ti tit din lp bnh rng 4

theo cng thc(10.18)

Trn trc 3 c T3 = 1354882 (Nmm),theo bng 10.5 tr195 tpI*

vi d4 = 65 => ng sut cho php ca trc l [(] = 49,4 (MPa)



- ng knh ti v tr lp ln:

Chn ng knh d30= d31 = 60 (mm)* Kim nghim trc v bn mi


theo cng thc (10.20) & (10.21):

+ (-1=0,436.(b = 0,436. 600 = 261,6 (MPa)

(-1=0,58.(-1 = 0,58. 261,6 = 151,73 (MPa)

+ Ti thit din d33 = 65mm theo bng 10.10 i vi thp cacbon

=>((= 0,77; ((= 0,74

+ K( , K( : H s tp trung ng sut thc t khi un v xon ph thuc vo loi yu t gy tp trung ng sut

Rnh then trn trc c ct bng dao phay ngn vi (b= 600Mpa

=> K( = 1,76, K( =1,54

=>K(/((= 1,76/0,77 = 2,28 ; K(/((= 1,54/0,74 = 2,08

+ Tr s K(/(( , K(/(( tra theo bng 10.11 (TR.198,TKHTDCK-T1) i vi b mt trc lp c di. Chn kiu lp k6 vi (b= 600Mpa

=>K(/((=2,52 , K(/((= 2,03

Vy gi tr K(/(( v K(/(( dng tnh K(di , K(di l:

K(/((= 2,52

K(/((= 2,03

Cc h s K(d , K(d c xc nh.

K(d = ( K(/((+Kx -1)Ky = (2,52 + 1,06 - 1)/1 = 2,58

K(d = ( K(/((+Kx -1)Ky = (2,03+1,06-1)/1 = 2,09

Ti tit din lp bnh rng 4, d 4 = 65 (mm).

Ta c momen un tng Mu33 = 711789,11 (Nmm)

Momen xon T3= 1354882 (Nmm)

Vi tit din trc c 1 rnh then v ng knh trc ti ch lp bnh rng d33 = 65 mm

Theo bng 9.1a (TR.173,TTTKHTDCK-T1) ta c cc kch thc ca then:

b = 20 (mm) ; t1 = 7,5 mm

Momen cn un W:

Momen cn xon Wo:


(m3= 0 ; (a3= (max= Mu33/W= 711789,11/ 23418,84 = 30,39(MPa)


(m3 = (a3 = (max3/2 = T/2Wo= 1354882/2.50107,59 = 13,52(MPa)



H s an ton s c xc nh:

Vy trc III m bo bn mi.

III. Tnh kim nghim bn ca then.Mi ghp then dng truyn momen xn t trc ti cc chi tit lp trn trc hay ngc li. Cc then dng trong b truyn l then bng .Trong qu trnh lm vic mi ghp then c th b hng do dp b mt lm vic, ngoi ra then c th hng do b ct.iu kin m bo bn dp v bn ct theo cng thc (9.1 & 9.2) :

Trong :

(d,(c L ng sut dp v ng sut ct tnh ton (Mpa).

[(d] L ng sut dp cho php (Mpa).

[(c] L ng sut ct cho php (Mpa).

d ng knh trc (mm)

T Momen xon trn trc (Nmm)

b, h, t Kch thc ca then(mm). Tra trong bng 9.1 (TR.173,TTTKHTDCK-T1)

lt Chiu di then c xc nh theo cng thc.

lt=(0,8(0,9)lmVi dng lp c nh, vt liu may l thp , ti trng tnh. Theo bng 9.5 (TR.178,TT-TKHTDCK-T1) ta c [(d]=150(Mpa) ; [(c]= 60(90(Mpa)

a) Kim nghim bn ca then trn trc I Trn trc 1, v tr lp then l ti bnh rng s 1:

d12= 34 mm

Theo bng (9.1a) ta chn then bng vi cc thng s sau:

b = 10 mm; t1= 5mm ; h = 8 mm; t2= 3,3 mm ; rmin= 0,25 mm; rmax= 0,4 mm.

Chiu di then:

lt= (0,80,9).57 = (45,651,3) mm. Chn lt = 50 mm.

Kim nghim bn dp v bn ct:

Theo cng thc (9.1), (9.2):

b) Kim nghim bn ca then trn trc II

Then ca bnh rng 2:

d23= 48mm theo bng 9.1 ta c :

Kch thc tit din then : b = 14mm ; h =9mm

Chiu su tit din then : t1=5,5mm ; t2=3,8mm.

Chiu di then : lt=(0,8(0,9)lm23=(0,8(0,9).67=(53,6(60,3)mm => Ly lt=60mm.

ng sut dp v ng sut ct tnh ton:

Then ca bnh rng 3:

d22= 50 mm theo bng 9.1 ta c :

Kch thc tit din then : b=14 mm ; h = 9 mm

Chiu su tit din then : t1=5,5 mm ; t2= 3,8 mm.

Chiu di then : lt=(0,8(0,9)lm22=(0,8(0,9).70=(56 ( 63)mm => Ly lt=60mm.


Then ca bnh rng 4:

d33=65mm theo bng 9.1 ta c :

Kch thc tit din then : b=18mm ; h=11mm

Chiu su tit din then : t1= 7 mm ; t2=4,4 mm.

Chiu di then : lt=(0,8(0,9)lm33=(0,8(0,9).80=(64(72)mm => Ly lt=70mm.


Nh vy ta phi lp thm mt then t cch nhau 1800, mi then s tip nhn 0,75T

Then ca bnh ai: d13= 28 mm theo bng 9.1 ta c :


Chiu su tit din then : t1=4mm ; t2=2,8mm.

Chiu di then : lt=(0,8(0,9)lm=(0,8(0,9).60=(48(54)mm => Ly lt=50mm.


Then ca ni trc n hi: d32= 58 mm theo bng 9.1 ta c :


Chiu su tit din then : t1=7mm ; t2=4,4mm.

Chiu di then : lt=(0,8(0,9)lm32=(0,8(0,9).120=(96(108)mm => Ly lt=100mm.


Nh vy ta phi lp thm mt then t cch nhau 1800, mi then s tip nhn 0,75T

Vy cc then m bo iu kin bn dp v iu kin bn ct.

b) Tnh ton chn ln.

1. Chn ln cho trc I.

a) Chn loi ln.

Ti trng hng tm nh, ch c lc hng tm, dng bi mt dy c nh cho cc gi 0 v 1.

b) Chn s b kch thc .

Vi ng knh ngng trc d=30mm v kt cu trc trn hnh v 3, chn s b c trung k hiu 206 c ng knh trong d=30mm, ng knh ngoi D=62mm, kh nng ti ng C=15,3(kN), kh nng ti tnh C0=10,2(kN) (bng P2.7 ph lc).

c) Tnh kim nghim kh nng ti ng ca .

Ti trng hng tm ca :

Vy ta tin hnh tnh kim nghim cho chu ti ln hn l 1Ti trng ng quy c Q i vi bi c xc nh theo cng thc (11.3):

Q= X.V.Fr.kt.kTrong :

X H s ti trng hng tm.

Y H s ti trng dc trc.

Theo bng11.4 (TR.215,TTTKHTDCK-T1) vi ln 1 dy iFa/Co = 0 v Fa/VFr = 0 < e

=> X=1,Y=0

Fr ,Fa Ti trng hng tm v ti trng dc trc. Fa=0(N)

V H s k n vng no quay. Vng trong quay V=1

kt H s k n nh hng ca nhit kt=1(nhit t(100(C)

kd H s k n c tnh ti trng. Theo bng 11.3 ta c kd=1 (ti trng va p nh)

Vy ti trng ng quy c :

Q= XVFr.kt.kd = 1.1.1689,04.1.1 = 1689,04 (N)Ti trng ng tng ng c xc nh theo cng thc (11.13)

Trong :

m Bc ca ng cong mi khi th v ln vi bi m=3

Li Thi hn khi chu ti trng Qi (triu vng quay)

Thi hn Li khi chu ti trng Qi c xc nh theo cng thc (11.13): Li= 60n.Lhi/106

Thi hn Lhi khi chu ti trng Qi (gi) c xc nh theo s ti trng thi hn phc v.

Vi tng thi hn phc v Lh =20000(gi)

T2 = 0,8.T1; t1 = 4 (h); t2 = 4(h); tck = 8 (h)

=> Q2 = 0,8.Q1 ; Lh1 = 10000 (h) ; Lh2 = 10000 (h)

Kh nng ti ng ca c xc nh theo cng thc (11.1):

Ti trng ng tng ng:

N

L = 60.n.Lh/106 = 60.356,25.20000/106 = 427,5 triu vng

=> < C =15,3(kN)

Vy chn l ph hp.d) Tnh kim nghim kh nng ti tnh ca .

Theo bng 11.6, vi bi 1 dy Xo= 0,6; Y0 = 0,5. Theo cng thc (11.19):

Qt = Xo.Fr+Yo.Fa = 0,6.1715,64 + 0,5.0 = 1029,38NTheo CT 11.20 Qt= Fr=> Qt < C0 = 10,2kN=10200N

Vy kh nng ti.

2. Chn ln cho trc II.

a) Chn loi ln.

Fa3/Fr3 = 1459,89/3361,06 = 0,43

Chn bi chn c trung- hpb) Chn s b kch thc .

Vi ng knh ngng trc d=45mm , chn s b c trung- hp k hiu 46309 c ng knh trong d=45mm, ng knh ngoi D=100mm, kh nng ti ng C= 48,1(kN), kh nng ti tnh Co=37,7(kN) (bng P2.12 ph lc)c) Tnh kim nghim kh nng ti ng ca .

Ti trng hng tm trn cc :

Theo bng 11.4 (TR.215,TTTKHTDCK-T1) vi bi chn gc tip xc (=12(i.Fa/C0 = 1.1459,89/37700= 0,039=> e = 0,35

Theo cng thc 11.8 lc dc trc do lc hng tm sinh ra (i vi bi chn) trn cc l:

Fs0= e.Fr0 = 0,35.6979,98 = 2442,99N

Fs1=e.Fr1 = 0,35.5155,73= 1804,51N

Theo bng 11.5 ng vi s b tr nh hnh bn ta c:

Tng lc dc trc (Fa0 tc dng vo 0:

(Fa0 = Fs1 + Fat = 1804,51+1459,89

= 3264,4N > Fs0 = 2442,99N

=> Fa0 = (Fa0 = 3264,4N


(Fa1=Fs0Fat=2442,99 1459,89

= 983,1N Fa1 = Fs1 = 1804,51N

Xc nh X v Y theo bng 11.4 vi bi chn 1 dy gc tip xc (=12(:

Fa0/(V. Fr0) = 3264,4/(1.6979,98) = 0,47>e = 0,35=> X=0,45; Y=1,53Fa1/(V. Fr1) = 1804,51/(1.5155,73) = 0,35 = e

=> X=1 , Y=0

Theo cng thc 11.3 ta c ti trng ng quy c trn 0 v 1 l:

Q0 =(X.V.Fr0 +Y.Fa0).kt. kd =(0,45.1.6979,98+1,53.3264,4).1.1=8135,52 N

Q1 = (X.V.Fr1 + Y.Fa1).kt. kd = (1.1.5155,73+1.1804,51).1.1 = 5155,73 N

T kt qu trn ta tnh cho 0 chu ti ln hn.

Ti trng ng tng ng c xc nh theo cng thc (11.13)


Vi : Q = QE = 7,41127kN

L = 60.10-6.n2.Lh = 60.10-6.91,58.20000 = 109,9(triu vng)

=> X0= 0,5, Y0 = 0,47.

Theo cng thc (11.19):

Qt = X0.Fr0+Y0.Fa0 = 0,5.7256,27 + 0,47.3181,96 =5123,66N

< Fr0 = 7256,27N

=> Qt = Fr0 = 7256,27N < C0 = 37,7kN=37700 N

Vy kh nng ti

3. Chn ln cho trc III.

a) Chn loi ln.

Fa4/Fr4 = 0.43 > 0,3

Do ta dng bi chn c trung hp mt dy gc tip xc (=12(cho cc gi 0 v 1.

b) Chn s b kch thc .

Vi ng knh ngng trc d=60mm v kt cu trc trn hnh v 5, chn s b c trung hp k hiu 46312 c ng knh trong d=60mm, ng knh ngoi D=130mm,

kh nng ti ng C= 78,8(kN), kh nng ti tnh C0=66,6(kN) (bng P2.12 ph lc)

c) Tnh kim nghim kh nng ti ng ca .V trn u ra ca trc III c lp ni trc vng n hi nn cn chn chiu ca Fk ngc vi chiu chn khi tnh trc .Khi phn lc trong mt phng z0x c tnh li nh sau:

* =>

=> =>

Vy, phn lc Fr0 > Fr0

Fr1 > Fr1Ly Fr0, Fr1 tnh.Theo bng 11.4 (TR.215,TTTKHTDCK-T1) vi bi

chn gc tip xc (=12(i.Fa/C0 = 1.1459,89/66600=0,022 => e = 0,32Theo cng thc 11.8 lc dc trc do lc hng tm sinh ra (i vi bi chn) trn cc l:

Fs0= e.Fr0 = 0,32.10175,99 = 3256,32N

Fs1=e.Fr1 = 0,32.2598,2=831,42N

Theo bng 11.5 ng vi s b tr nh hnh bn ta c:


(Fa0 = Fs1 - Fa4 = 831,42 1459,89= -628,46 N < Fs0 = 3256,32N

=> Fa0 = Fs0 = 3256,32N


(Fa1 = Fs0 + Fa4 = 3256,32+1459,89 = 4716,21N > Fs1= 831,42N

=> Fa1 = (Fa1 = 4716,21N

Xc nh X v Y theo bng 11.4 vi bi chn 1 dy gc tip xc (=12(:

Fa0/(V. Fr0) = 3256,32/(1.10175,99) = 0,32 = e

=> X=1 ,Y=0

Fa1/(V. Fr1) = 4716,21/(1.2598,2) = 1,82 > e

=> X=0,45 , Y=1,72

Theo cng thc 11.3 ta c ti trng ng quy c trn 0 v 1 l:

Q0 =(X.V.Fr0 +Y.Fa0).kt. kd =(1.1.10175,99+0.3256,32).1.1= 10175,99 N

Q1 = (X.V.Fr1 + Y.Fa1).kt. kd = (1.0,45.2598,2+1,72.4716,21).1.1 = 9281,07 N

Ti trng ng tng ng c xc nh theo cng thc (11.13)


Vi : Q = QE = 9,27kN

L = 60.10-6.n2.Lh = 60.10-6.34,82.20000 = 41,78(triu vng)

=> X0= 0,5, Y0 = 0,47.

Theo cng thc (11.19):

Qt = X0.Fr0+Y0.Fa0 = 0,5.10175,99 + 0,47.3256,32 = 6618,47N

< Fr0 = 10175,99N

=> Qt = Fr = 10175,99N < C0 = 66,6kN=66600N

Vy kh nng ti

Phn IV. Tnh ton cc yu t ca v hp v cc chi tit khc.

I.Cc kch thc c bn ca v hp gim tc.

Da vo bng(18.1) ta c:

Tn giBiu thc tnh ton

Chiu dy: Thn hp,( Np hp, (1

( =0,03a+3 =0,03.210+3 =9,3 (mm)

Ly ( =9 (mm)

(1=0,9.( =0,9.9 =8,1Ly (1= 8 (mm)

Gn tng cng: Chiu dy, e

Chiu cao,h

dce =(0,81).( =(7,29) ( e=8 (mm)

h 0,04a+10 =18,4(d1=20d2=(0,70,8)d1=(1416)( d2=16mm

d3=(0,8..0,9)d2=(12,814,4)(d3=14 (mm)

d4=(0,6...0,7)d2=(9,611,2)(d4=10(mm)

d5 =(0,5...0,6)d2 =(89,6) (d5=8 (mm)

Mt bch ghp np v thn:

+ Chiu dy bch thn hp,S3+ Chiu dy bch np hp,S4+ B rng bch np v thn,K3S3=(1,41,8)d3=(19,6..25,2)(S3=20 (mm)

S4=(0,91)S3=(18..20) (S4=19 (mm)

K3= K2 - (35) = (4547). Ly K3= 46 (mm)

Kch thc gi trc

+ ng knh ngoi v tm l vt D3 D2

+Tm l bulng cnh ,E2

ln trc I

ln trc II

+B rng mt ghp bulng cnh ,K2

tra bng(18-2)

*i vi ln trcI:

D = 62(D3 = 106, D2 = 80*i vi ln trcII:

D = 100 ( D3=144, D2 =118*i vi ln trcIII:

D = 130(D3=174, D2 = 148E2 =1,6.d2 =25,6 (E2= 25 (mm)

C1=D3/2= 106/2= 53(chnC1=53 (mm)

C2=D3/2=144/2=72(chnC2=72 (mm)

C3=D3/2=174/2= 87(chnC3= 87 (mm)

R2 =1,3d2 =20,8 (R2 = 20 (mm)

K2=E2 +R2+(3..5) = (4850)(K2=50mm

Mt hp:

+Chiu dy khi khng c phn li S1

+B rng mt hp,K1v qS1=(1,3..1,5)d1=(26..30)(S1=28 (mm)

K1=3.d1=60 (mm); q K1+2( =78 (mm)Ly q = 80 (mm)

Khe h gia cc chi tit:

+Gia bnh rng vi thnh trong hp

+Gia nh bnh rng vi y hp

+ Gia mt bn cc bnh rng vi nhau( (1..1,2)( = (910,8) (( =10(1 (3..5) ( = (2745) ((1=36( ( =10

S lng bulng nn ZZ=(L+B)/(200300)(3,455,6) L, B: Chiu di v rng ca hp.Ly Z = 4

*) Vng mc:+ Chiu dy vng mc: S=(23)( =(1827) ( S=24 (mm)

+ ng knh l: d=(3..4)(=(27..36) ( d=30 (mm)

*) Cht nh v cht tr:

+ ng knh d=6 (mm)

+ vt c=1,0 (mm)

+ Gc vt = 45o+ Chiu di l=(20160) (mm) ( l= 40 (mm)

*) Ca thm:

kim tra cc chi tit my trong hp khi lp ghp v tra du vo hp c ca thm c y bng np trn c th lp thm nt thng hi.

Theo bng (18.5):

ABA1B1CC1KRVtS lng

150100190140175-12012M8x224

*)Nt thng hi

Theo bng (18.6):

ABCDEGHIKLMNOPQRS

M27x21530154536326410822632183632

*)Nt tho du

dbmfLcqDSDo

10015103292,519,8322225,4

*)Kim tra mc du: Khi mun kim tra mc du, ta s dng que thm

6.3. Bng thng k kiu lp.

Kiu lp:

Kch thc danh ngha (mm)Sai lch Nmax , Nmin

Khp ni:

60+21+2

trc I - trc : 30+18+2

trc I - v :

62+300

trc II - trc :

45+18+2

trc II - v:

100+350

trc III - trc :

60+21

+2

trc III - v:

130+40

0

Bnh ai:

28+15+2

Bnh rng I - trc

34+18

-23

Bnh rng II - trc

48+18

-23

Bnh rng III - trc

50+18-23

Bnh rng IV - trc

65+21-28

EMBED Equation.3

EMBED Equation.3

Trc

I

II

III

D (mm)

40

55

65

b0 (mm)

23

27

33

EMBED PBrush

EMBED PBrush

6

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