phn 1: NG HC PHN BNG TI

n chi tit my s 5: Thit k h dn ng bng ti

MC LC3PHN 1: NG HC PHN BNG TI

31. Chn ng c

31.1. Xc nh cng sut cn thit ca ng c

41. 2. Xc nh tc ng b ca ng c in

41.3. Chn ng c

52. Phn phi t s truyn

52.1. Xc nh t s truyn chung ca h dn ng

52.2. Phn phi t s truyn

63. Xc nh cc thng s trn cc trc

63.1. S vng quay trn cc trc

63.2. Cng sut tc dng ln cc trc

63.3.M men xon trn cc trc

74. Bng tng kt

8PHN 2: TNH CC B TRUYN

81. Thit k b truyn ngoi

81.1. Chn loi xch

81.2. Chn s rng a xch

81.3. Xc nh s bc xch

101.4. Xc nh khong cch trc v s mt xch

111.5. Kim nghim xch v bn mn

121.6. Xc nh cc thng s ca da xch

141.7. Xc nh lc tc dng ln trc

141.8. Cc thng s ca b truyn xch:

152. Thit k b truyn trc vt-bnh vt

162.1. Chn vt liu ch to trc vt-bnh vt

162.2. Xc nh ng sut cho php

182.3. Xc nh s b khong cch trc

192.4. Xc nh cc thng s

192.5. Kim nghim bnh vt

212.6. Xc nh cc kch thc b truyn

222.7. Tnh nhit truyn ng trc vt

242.8. Lc tc dng ln trc.

252.9. Cc thng s b truyn trc vt.

26PHN 3: TNH TON THIT K TRC

261. Chn khp ni

261.1. Tnh chn khp ni

271.2. Lc t khp ni tc dng ln trc.

271.3. Bng cc thng s khp ni:

282. Tnh trc

282.1. Chn vt liu ch to trc.

282.2. Xc nh s b ng knh trc.

292.3. Xc nh khong cch gia cc im t lc.

302.4. t lc tc dng ln cc on trc.

302.5. V biu mmen cho trc I.

322.5. V biu mmen.

332.6. Xc nh ng knh cc on trc.

342.7. Chn then.

362.8. Kim nghim trc.

41PHN 4: TNH CHN LN

411. Trc I.

411.1. Chn loi ln.

421.3. Xc nh ti trng tc dng ln cc .

441.4. Kim nghim kh nng ti ng ca

461.5. Kim nghim kh nng ti tnh ca

482. TrcII

49PHN 5. THIT K V HP GIM TC

491. Tnh kt cu ca v hp

491.2. Cc kch thc c bn ca v hp.

511.3. Cc chi tit khc c lin quan.

511.1.3. Vng mc.

522. Bi trn hp gim tc

522.1. Cng dng

532.2. Bi trn b truyn trc vt

532.3. Bi trn ln.

533. Bng thng k lp ghp, tr s sai lch gii hn v dung sai lp ghp.

PHN 1: NG HC PHN BNG TI

1. Chn ng c

1.1. Xc nh cng sut cn thit ca ng c

+ Cng sut tng ng xc nh theo cng thc: Pct =

EMBED Equation.DSMT4 Trong : Pct, Pt: cng sut cn thit trn trc ng c v cng sut tnh ton trn trc my cng tc.

Vi gi thit h dn ng bng ti lm vic n nh vi ti trng khng i. Theo cng thc 2.10[1] v 2.11[1] ta c:

+ Cng sut cng tc Pt:

KW Vi : v= 0,72 m/s - vn tc bng ti;

F= 5375 N - lc ko bng ti;

+ Hiu sut h dn ng (:

( = ( (nib

Theo s bi th : ( =(k.(3l.(tv.(x;

Tra bng( 2.3) Ttttkhdck tp1 , ta c cc hiu sut:

(k = 0,99 - hiu sut ni trc.

(l = 0,99 - hiu sut mt cp ln; (tv = 0,80 - hiu sut b truyn trc vt khng t hm vi Z1=2;

(x = 0,93 - hiu sut b truyn xch h ;

( = 0,99. 0,993.0,80. 0,93= 0,715 ; Vy :

Pct= =

1. 2. Xc nh tc ng b ca ng c in

+ S vng quay ca trc my cng tc l nlv tnh theo cng thc 2.16[1]:

nlv =(vg/ph)

Trong :

v: vn tc bng ti; v = 0,72 m/s ;

D: ng knh bng ti ; D=525 mm ;

+ Theo bng 2.4[1] ta c th chn c:

un(sb): t s truyn ngoi, y b truyn ngoi xch: un(sb)=ux(sb)=2

uh(sb) t s truyn ca hp gim tc, y l b truyn trc vt: uh(sb)=utv(sb)=20

Vy t s truyn s b ca h dn ng:

usb= un(sb).uh(sb)= 2.20= 40

S vng quay s b ca ng c c tnh theo cng thc 2.18[1]:

nsb= nlv.usb= 26,2.40= 1048 (vg/ph)1.3. Chn ng c Chn ng c phi tha mn iu kin : Pc Pct , nc ( nsb

Chn s vng quay ng b ca ng c: nc= 1000 (vg/ph)

ng thi c mmen m my tha mn:

Do h dn ng hot ng ch ti trng tnh nn mmen m my bng mmen xon ca ti tc l: Tmm= T Do :

Vy ta cn chn ng c c:

Theo bng P1.3[1] ph lc vi Pct= 5,78 (Kw) v nc= 1000 (vg/ph)

Ta chn c ng c: 4A132M6Y3 vi cc thng s k thut l:

Cng sut : 7,5 (Kw)

S vng quay : 968 (vg/ph)

ng knh trc: 38 (mm)

Khi lng : 93 (kg)

Kt lun ng c 4A132M6Y3 c kch thc ph hp vi yu cu thit k.

2. Phn phi t s truyn

2.1. Xc nh t s truyn chung ca h dn ng

Theo cng thc 3.23[1] ta c:

Trong :

nc: S vng quay ca ng c chn

nlv: S vng quay ca trc my cng tc

2.2. Phn phi t s truyn

Theo cng thc 3.24[1] ta c:

usb= un.uh

Trong :

un: T s truyn ca b truyn ngoi: un= ux uh: T s truyn ca b truyn trong hp gim tc. uh= utv

Chn t s truyn ca b truyn xch ux=2,5 th t s truyn ca b truyn trc vt - bnh vt l:

3. Xc nh cc thng s trn cc trc

3.1. S vng quay trn cc trc + Tc quay ca trc I: n1 = nc= 968 (vg/ph)

+ Tc quay ca trc II: n2 = = = 64,5 (vg/ph)

+ Tc quay ca trc cng tc: = = = 25,8 (vg/ph)

3.2. Cng sut tc dng ln cc trc

+ Cng sut trn trc cng tc: Pct= Pt= 4,13 (Kw) + Trc II: P2= (kw)

+ Trc I: P1= (kw)

+ Cng sut trn ng c: (Kw)

3.3. M men xon trn cc trc

Ti = 9,55..

+ Trc I : (N.mm)

+ Trc II : (N.mm)

+ Trc cng tc:

(N.mm)

+ Mmen xon trn trc ng c:

4. Bng tng kt Trc

Thng sng cIIICng tc

U114,782,5

P(Kw)5,795,674,494,13

n(vg/ph)96896864,525,8

T(Nmm)57122,4255938,53664798,451528740,31

PHN 2: TNH CC B TRUYN

1. Thit k b truyn ngoi

Ta c bng thng s ca b truyn:

P= P2= 4,49 (Kw)

T= T2= 664798,45 (Nmm)

n= n2= 64,5 (vg/ph)

u= ux= 2,5

= 10o1.1. Chn loi xch

V ti trng nh vn tc thp nn dung xch con ln.

1.2. Chn s rng a xch

Vi u = 2,5, theo bng 5.4[1] ta chn z1= 25 (rng) l s rng a xch nh

S rng a xch ln c xc nh theo cng thc5.1[1]:

Z2 = u.Z 1 = 2.25 = 62,5 (rng)

Do Z2 nguyn ln chn Z2= 62 (rng)

T s truyn thc:

Sai s:

1.3. Xc nh s bc xch

Bc xch p c tra bng 5.5[1] vi iu kin

Vi

Pt : cng sut tnh ton

P: cng sut cn truyn, P = 4,49(kW)

Chn b truyn th nghim l b truyn xch tiu chun, c s rng v vn tc vng a xch nh l:

Z01= 25 (rng) v n01= 50 (vg/ph)

Do vy ta tnh c:

kz : h s rng, ta c kz = ,

kn: h s vng quay kn =, vi n01 = 50 vng/pht

Theo cng thc 5.4[1]

K= k0.ka.kbt.k.kc.kcVi

k0 : h s k n nh hng v tr b truyn,tra bng 5.6[1] vi chn k0=1

ka: h s k n khong cch trc v ciu di xch, chn a=(30-50)p tra bng 5.6[1] c: ka = 1 (a = 50p)

kc: h s k n nh hng ca lc cng xch

chn cch iu chnh bng con ln cng xch kc=1

kbt: h s k n nh hng bi trn, dng cch bi trn nh git, mi trng lm vic c bi, chn kbt = 1,3

k: h s k n ti trng ng, b truyn lm va p nh, chn k = 1,2

kc: h s k n ch lm vic b truyn, b truyn lm vic 2 ca, kc=1,25

K = 1.1.1.1,3.1,2.1,25 = 1,95

Kd: h s phn b khng u ti trng cho cc dy vi 3 dy ta c: kd=2,5

Vy ta c:

Theo bng 5.5[1], vi Pt=2,7 v n01 = 50 vng/pht, chn b truyn xch 1 dy c:

Bc xch : p = 25,4 (mm) tho mn iu kin bn mn

ng knh cht : dc= 7,95 (mm)

Chiu di ng : B= 22,61 (mm)

Cng sut cho php: [P]= 3,20 (Kw)

Pt < [P] = 3,20 (kW).

1.4. Xc nh khong cch trc v s mt xch

Chn s b:

a= 35p = 35.25,4= 889 (mm)

Theo cng thc 5.12[1], s mt xch

x = 114,49

Ly s mt xch chn xc = 114

Theo cng thc 5.13[1], tnh li khong cch trc:

=883 (mm)

xch khng phi chu lc cng qu ln, gim a mt lng

a = 0,003a = 0,003.1262 = 3(mm)

Vy khong cch trc thc t: a = 880 (mm).

+ S ln va p ca xch

Theo ct 5.14[1], ta c s ln va p I ca bn l xch trong 1 giy:

i =

Theo bng 5.9[1], vi p = 25,4 th [i] = 30

Vy i < [i]

1.5. Kim nghim xch v bn mn

Theo cng thc 5.15[1] ta c S =

Trong :

Q: Ti trng ph hng c tra trong bng 5.2[1].

Theo bng 5.2[1], vi xch con ln 1 dy c p= 25,4 th ti trng ph hu Q = 170,1 (KN) =170100 (N), khi lng 1m xch q = 7,5 (kg)

k : h s ti trng ng. Do ch lm vic trung bnh k = 1,2.

v=

Ft: lc vng, Ft =

Fv: lc cng do lc li tm sinh ra

Ta c: Fv = q.v2 =7,5.0,682 =3,468 (N)

F0 : lc cng do trng lng nhnh xch b ng gy ra, c tnh theo cng thc 5.16[1] :

Fv= 9,81kf.q.a

Trong :

a: Khong cch trc, a=0,88 (m)

kf: H s ph thuc vng f ca xch, ly kf = 4

q: Khi lng 1m xch, q= 7,5 (kg)

F0 = 9,81.4.7,5.0,88 = 258,98 (N)

Vy

S =

Theo bng 5.10[1] vi n=64,5 (vg/ph) v p=25,4 c [s] = 8,2

Vy s > [s] : b truyn xch m bo bn.

1.6. Xc nh cc thng s ca da xch

Theo cng thc 5.17[1] ta c:

ng knh vng chia:

Vy ng knh vng chia ca a dn d1 =202,66 (mm), a b dn

d2 = 501,49 (mm).

Theo bng 14.4b[1] ta c:

ng knh vng nh rng ca:

+ a dn:

+ a b dn:

Bn knh y: r = 0,5025.dl + 0,05

Vi dl tra trong bng 5.2[1] ng knh con ln ta c: dl=15,08(mm)

Vy: r = 0,5025.dl + 0,05=0,5025.15,08+0,05= 7,63 (mm)

ng knh chn rng:

+ a dn: df1 = d1 2.r= 202,662.7,63= 187,4 (mm)

+ a b dn: df2 = d2 2.r=501,49 2.7,63= 486,23 (mm)

Kim nghim ng sut tip xc trn mt rng a xch:

Theo cng thc 5.18[1]:

EMBED Equation.3 Trong :

Ft : Lc vng, Ft = 6602,94 (N)

Fv : Lc va p trn m=3 dy xch, theo cng thc 5.19[1]:

Fv = 13.10-7.n.p3.m

Fv = 13.10-7. 64,5.25,43.3 = 4,12 (N)

E = . Vt liu dung lm con ln v rng a l thp c

E1= E2= E = 2,1.105 (Mpa)

k : H s ti trng ng, k = 1,2

kr : H s k n s rng a xch, vi z1 = 25 (rng) tra bng trang 87 sch tttkhd 1 ta ckr = 0,42

kd = 2,5 (do s dng 3 dy xch)

Theo bng 5.12[1], vi p = 25,4 v 3 dy xch c A= 450 mm2Vy:

Nh vy theo bng 5.11[1] dng thp 45 ti ci thin, t rn HB210, ng sut tip xc cho php l = 600 (MPa) m bo c bn tip xc cho rng a xch 1.

Tng t vi rng a xch 2 do chn cng vt liu v ch nhit luyn nn cng c:

.

1.7. Xc nh lc tc dng ln trc

Theo cng thc 5.20[1]:

Fr = kx.FtTrong :

kx: H s k n trng lng xch => b truyn t nm nghing gc < 40o, chn kx = 1,15

Ft: Lc vng, Ft=6602,94 (N)

Vy Fr = 1,15.6602,94 = 7593,33 (N)

1.8. Cc thng s ca b truyn xch:

Thng sK hiuGi tr

Loi xchXch ng con ln

Bc xch (mm)p25,4

S mt xchx114

Chiu di xch (mm)L

Khong cch trc (mm)a880

S rng a xch nhz125

S rng a xch lnz262

Vt liu a xch

ng knh vng chia a xch nh (mm)d1 202,66

ng knh vng chia a xch ln (mm)d2501,49

ng knh vng nh a xch nh (mm)da1 213,76

ng knh vng nh a xch ln (mm)da2 513,55

Bn knh y (mm)r 7,63

ng knh chn rng a xch nh (mm)df1 187,4

ng knh chn rng a xch ln (mm)df2 486,23

Lc tc dng ln trc (N)Fr7593,33

2. Thit k b truyn trc vt-bnh vt

Cc thng s ca b truyn trc vt:

P= P1= 5,67 (kw)

n1= 968 (vg/ph)

u= utv= 14,78

T1= 55938,53 (N)

T2= 664798,45 (N)

n2= 64,5 (vg/ph)

S ca lm vic: 2 (ca)

Thi gian phc v: lh= 12500 gi

2.1. Chn vt liu ch to trc vt-bnh vt

+ Tnh s b vn tc trt

Theo cng thc 7.1[1], ta tnh vn tc trt s b:

Trong :

n1: S vng quay ca trc vt

T1: Mmen xon trn trc bnh vt

vs= 3,8 (m/s) < 5 (m/s). S dng ng thanh khng thic pA 9-4 c ly tm c ch to bnh vt c b= 500(MPa), ch = 200 (MPa).

Theo bng 7.2[1] ta s dng thp 45 ch to trc vt l thp ti, rn mt ren t cng HRC45

2.2. Xc nh ng sut cho php

V bnh vt lm bng ng thanh khng thic c c tnh thp hn nhiu so vi trc vt bng thp nn ch cn xc nh ng xut tip xc cho php v ng sut un cho php i vi vt liu lm bnh vt.

2.2.1. ng sut tip xc cho php

Theo bng 7.2[1], vi bnh vt lm bng ng thanh khng thic

pA 9-4 ta c: vi vs=3,8 (m/s) => .

2.2.2. ng sut un cho php

Theo cng thc 7.6[1] ta c:

Trong : ng sut un cho php vi 106 chu k

B truyn quay 1 chiu, theo cng thc 7.7[1] ta c:

KFL: h s tui th. Theo cng thc 7.9[1] ta c:

Vi

V ti trng khng i nn:

NFE= 60.n2.t

Vi:

n2: l s vng quay ca bnh vt trong 1 pht c: n2=64,5 (vg/ph)

t: Tng s thi gian lm vic ca b truyn t=lh=12500 gi

=> NFE= 60.64,5.12500=4,84. (chu k)

Do :

Vy :

+ ng sut cho php khi qu ti

Theo cng thc 7.14[1], ta c:

2.3. Xc nh s b khong cch trc

Theo cng thc 7.16[1] ta c:

Trong :

z2: l s rng bnh vt. Vi u= utv = 14,78, chn z1 = 2 => z2 = u2z1 = 14,78.2 = 29,56 (rng), chn z2=30 (rng)

T s truyn thc:

Sai s:

q: H s ng knh trc vt. Chn s b q = 0,25.z2 = 0,27.30 = 8,1

Theo bng 7.3[1], chn q = 8

T2: Mmen xon trn trc bnh vt. T2 = 664798,45 Nmm

KH: H s ti trn. Chn s b KH = 1,2

Vy:

Chn aw= 190 (mm).

2.4. Xc nh cc thng s

Modun dc trc ca trc vt:

Theo cng thc 7.17[1]:

Chn m theo tr s tiu chun trong bng 7.3[1] ta c: m = 10 (mm)

Tnh chnh xc khong cch trc aw:

Ly aw = 190 mm, khi h s dch chnh l:

Tho mn iu kin -0,7 < x < 0,7

2.5. Kim nghim bnh vt

2.5.1. Kim nghim v bn tip xc

Theo cng thc 7.19[1] ta c iu kin bn tip xc ca rng bnh vt l:

+ Tnh li vn tc trt:

Theo cng thc 7.20[1]:

Trong :

dw1 = m(q + 2x) = 10.(8 - 2.0) = 80 (mm)

Do :

KH: H s ti trng:

KH= KH.KHvVi KH: H s phn b khng u ti trng trn chiu rng vnh rng. Do ti trng khng nn ta c KH=1

KHv: H s ti trng ng. Theo bng 7.6[1], vs=4,18 (m/s)5 (m/s) ta c cp chnh xc ch to b truyn l 8. Tra bng 7.7[1] ta c: KHv= 1,24

KH= 1.1,24= 1,24

Vy: => B truyn m bo iu kin bn tip xc 2.5.2. Kim nghim bn un

Theo cng thc 7.26[1]ta c iu kin m bo iu kin bn un ca bnh vt:

Trong

mn = m.cos= 10.cos= 9,7 : mdum php ca rng bnh vt.

b2 : chiu rng vnh rng bnh vt, mm, theo bng 7.9[1] ta c: vi z1= 2, b2 0,75.da1 = 0,75(q +2)m = 0,75.10.(8+2) = 75

Ly b2 = 75 (mm)

d2 ng knh vng chia bnh vt, mm, d2 = m.z2 = 10.30 = 300 (mm)

YF : h s dng rng. Theo bng 7.8[1] theo s rng tng ng:

Tra c YF = 1,71.

KF: H s ti trng.

KF = KFv.KF = KHv.KH = 1.1,24= 1,24

Vy:

Vy b truyn tho mn iu kin bn un.

2.6. Xc nh cc kch thc b truyn

Khong cch trc: aw=190 (mm)

M un : m=10

H s ng knh: q=8

T s truyn : u= 15

S ren trc vt : z1= 2

S rng bnh vt : z2= 30

H s dch chnh : x=0

Gc vt :

Chiu di phn ct ren trn trc vt:

b1 (11+ 0,06.z2).m= (11+0,06.30).10= 128 (mm). Chn

b1= 130(mm)

Chiu rng bnh vt: b2= 75 (mm)

ng knh ngoi bnh vt:

daM2 da2+1,5.m

Vi ng knh vng nh bnh vt:

da2= m(z2 +2+2x)= 10.(30+2+2.0)= 320 (mm) daM2 da2+1,5.m = 320+1,5.10= 335 (mm) => chn daM2= 300 (mm) ng knh vng chia: d1= q.m= 8.10= 80 (mm)

d2= m.z2= 10.30= 300 (mm)

ng knh vng y: df1= m.(q- 2,4)= 10.(8- 2,4)=56 (mm)

df2= m(z2- 2,4+2x)= 10.(30-2,4+2.0)=276 (mm)ng knh vng nh trc vt:

da1= d1+2m= 80+2.10= 100 (mm)2.7. Tnh nhit truyn ng trc vt

B truyn lp thm qut ngui u trc vt. Ta c nhit u trong hp phi tha mn iu kin 7.31[1] :

Trong :

+ t0: Nhit mi trng xung quanh, t0=

+ Aq: Din tch b mt hp c qut ngui, Aq=0,3A.

+ A :Din tch thot nhit cn thit.

+ Ktq: H s ta nhit ca phn b mt c qut, vi s vng qut n= 968 (vg/ph) => Ktq= 20,5

+ [td]: Nhit cao nht ca du. Trc vt t di bnh vt. [td] = 90o + Kt: H s ta nhit. Chn Kt = 13 W/(m2.oC)

+ : h s k n s thot nhit qua y hp xung b my, ly = 0,25

+ : hiu sut thc t ca b truyn. Theo cng thc 7.22[1]

Vi : Gc ma st c tra bng 7.4[1]theo vn tc trt vs=4,18(m/s) v vt liu lm bnh vt.=> = 2,18

=>

+ : h s k n s gim nhit sinh ra trong mt n v thi

gian do ti trng ngt qung. Do ti trng khng i => =1

T iu kin 7.31[1] ta c din tch thot nhit cn thit ca hp gim tc l:

2.8. Lc tc dng ln trc.

Theo cng thc 10.2[1] ta c:

Vi gc ma st: nn ta c:

Ft1= Fa2= Fa1.tg= 4431,99.tg14,033 = 1107,76 (N)

Fr1= Fr2= Fa1.tg= 4431,99.tg20= 1613,11 (N)2.9. Cc thng s b truyn trc vt.Thng sK hiuGi tr

Khong cch trc (mm)aw190

M un (mm)m10

H s ng knhq 8

T s truynu15

S ren trc vtz1 2

S ren bnh vtz230

H s dch chnhx 0

Gc vt ()

Chiu di phn ct ren trc vtb1130

Chiu rng bnh vtb275

ng knh vng nh bnh vt (mm)da2320

ng knh vng nh trc vt (mm)da1100

ng knh vng chia trc vt (mm)d180

ng knh vng chia bnh vt (mm)d2300

ng knh vng y trc vt (mm)df156

ng knh vng y bnh vt (mm)df2276

ng knh ngoi bnh vt (mm)daM2300

Lc tc dng ln trc (N)Fa1= Ft2Ft1= Fa2Fr1= Fr24431,99

1107,76

1631,11

PHN 3: TNH TON THIT K TRC

1. Chn khp ni

Mmen cn truyn l: T = Tc= 57122,42 (Nmm)

ng knh trc ng c: dc= 38 (mm)

1.1. Tnh chn khp ni

Ta s dng khp ni vng n hi ni trc. Ta chn khp ni theo iu kin:

Trong :

+ Tt: L mmen xon tnh ton Theo cng thc 16.1[2] ta c: Tt = k.T

Vi k: h s ch lm vic, ph thuc vo loi my. i vi bng ti theo bng 16.1[2] ly k = 1,2

T: Mmen xon danh ngha T=57122,42 (Nmm)

Vy Tt = 57122,42.1,2= 68546,9 (Nmm)=68,5469 (Nm)

+ dt: ng knh trc cn ni.

Chn = 16 (MPa)

=>

Theo bng 16.10a[2]. vi:

Ta c cc thng s:

Tra bng 16.10 b[2] vi 125 (Nm) ta c:

1.2. Lc t khp ni tc dng ln trc.

1.3. Bng cc thng s khp ni:Thng sGi tr

Mmen xon ln nht c th t c (Nm)125

ng knh ln nht ca trc ni (mm)28

S cht4

ng knh vng tm cht (mm)90

Chiu di phn t n hi (mm)28

Chiu di on congxon ca cht (mm)34

ng knh ca cht n hi (mm)14

2. TNH TRC

2.1. Chn vt liu ch to trc.

Chn vt liu ch to trc l thp 45 c ng sut xon cho php .

2.2. Xc nh s b ng knh trc.

Theo cng thc 10.9[1]/186 ta c ng trc th k (k=1,2) c xc nh:

.

Trong :

T l momen xon, Nmm

[] l ng sut xon cho php, Mpa. Chn [] = 15 Mpa Trc 1 c: T1= 55938,53 (N)

=>

Trc 2 c: T1= 664798,45 (N)

=>

Trc 1 l trc v ca hp gim tc lp bng khp ni vi trc ng c nn chn d1 theo tiu chun tha mn: d1= (0,81,2).dc= (0,81,2).38= (30,445,6) (mm)

-Trc 1: chn d1= 35 (mm)

-Trc 2 : chn d2= 65 (mm)

2.3. Xc nh khong cch gia cc im t lc.

Chiu di trc cng nh khong cch gia cc im t lc ph thuc vo s ng, chiu di may ca cc chi tit quay, chiu rng , khe h cn thit v cc yu t khc. Theo cng thc 10.10[1], 10.11[1], 10.13[1] ta c:

- Chiu di may na khp ni, ta chn ni trc vng n hi nn:

lm12= (1,42,5).d1= (1,42,5).35= (4987,5) (mm)

Chn lm12 = 50 (mm)

- Chiu di may a xch :

lm23= (1,21,5).d2= (1,21,5).65 = (7897,5) (mm)

Chn lm23 = 85 (mm)

Chiu di may bnh vt:

lm 22= (1,21,8).d2= (1,21,8).65 = (78117) (mm)

chn lm 22= 95 (mm)

Theo bng 10.3[1]ta chn c cc thng s sau:

k1= 10 (mm) ,k2= 8 (mm), k3= 15 (mm), hn= 20 (mm)

Theo bng 10.2[1] vi:

d1= 35 (mm) => Chn b01= 21 (mm)

d2= 65 (mm) => Chn b02= 33 (mm) Khong cch gia cc chi tit quay v cc gi . Theo hnh 10.11[1] ta c:

l12= 0,5.(lm12 + b01) + k3 + hn=0,5.(50+21) +15+ 20=70,5 (mm)

l11= (0,91).daM2= (0,91).300= (270300).

Chn l11= 280 (mm)

l22= 0,5.(lm22 + b02) + k1 + k2= 0,5.(95+33) +10 +8= 82 (mm)

l21= 2.l22= 2.82= 164 (mm)

lc23= 0,5.( b02+ lm23) + k3 + hn= 0,5(33+85) +15 +20= 94 (mm)

l23= l21 + lc23= 164 +94= 258 (mm)

2.4. t lc tc dng ln cc on trc.

Trc I:

Fk= 248,62 (N) Fa1= 4431,99 (N)

Fr1= 1631,11 (N) Ft1= 1107,76 (N)

Trc II:

Fx= 7593,33 (N) Fr2= 1631,11 (N)

Fa2= 1107,76 (N) Ft2= 4431,99 (N)

2.5. V biu mmen cho trc I.

2.5.1. Tnh cc phn lc tc dng ln .

Trong mt phng ta Oyz xt cc phn lc Fyo, Fy1 sinh ra bi cc lc Fr1, Fa1.

Ta c phng trnh cn bng:

Vi: Vi d1 l ng knh vng chia trc vt. T (2) ta c:

Thay vo 1 ta c : Fy0= 1448,7 (N)

Trong mt phng O xt cc phn lc Fx0, Fx1 xinh ra bi lc Ft1.

Ta c cc phng trnh cn bng:

Thay vo (3) ta c:

Fx0= 859,14 - 616,48= 242,66 (N)

2.5. V biu mmen.

2.6. Xc nh ng knh cc on trc.

2.6.1. Trc I.

Tnh mmen un tng Mj v mmen tng ng Mtdj ti cc tit din j trn chiu di trc. Theo cc cng thc 10.15[1], v 10.16[1] ta c:

Trong : Mxj, Myj: L mmen un trong mt phng Oyz v O ti cc tit din j.

Xt ti tit din 0. Ta c: Mx0= 17527,71 (Nmm), My0= 0,

T0= 55938,53 (Nmm).

=>

Xt ti tit din 2. Ta c: Mx2= My2= 0, T2= 55938,53 (Nmm).

=>

Xt ti tit din 3. Ta c: Mx3= 86307,20 (Nmm),

My3= 202818 (Nmm), T3= 55938,53 (Nmm).

=>

ng knh trc ti cc tit din j c xc nh theo cng thc 10.17[1] ta c:

Vi : ng sut cho php ca thp ch to trc tra bng 10.5[1] ta c = 63 (MPa).

Vy ng knh trc ti cc tit din :0, 2, 3 ln lt l:

Xut pht t yu cu v bn, lp ghp v cng ngh ta chn ng knh cc on trc:

d10= 40 (mm), d12= 35 (mm), d13= 45 (mm).2.6.2. Trc II.

Xut pht t yu cu v bn lp ghp v cng ngh ta chn ng knh cc on trc nh sau:

d20= d21 = 65 (mm), d22= 60 (mm), d23= 70 (mm).

2.7. Chn then.

2.7.1. Trc I. Ti tit din ni trc:

Kch thc tit din then:

lt= (0,80,9).lm12= (0,80,9).50 = (4045) (mm).

Chn lt= 40 (mm).

Da vo bng 9.1a[1] vi d12= 35 (mm) ta c:

b= 8 (mm), h= 7 (mm), t1= 4 (mm), t2= 2,8 (mm). Da vo bng 9.5[1] ng vi dng lp c nh, vt liu thp, c tnh ti trng va p nh ta c:

= 100 (MPa) Da vo cc cng thc 9.1[1] v 9.2[1] ta c:

iu kin bn dp:

iu kin bn ct:

Vi : T= T1= 55938,53 (Nmm)

: ng sut ct cho php (MPa), vi then bng thp C45 ti trng va p nh : = (2030) (MPa)

=>

=>

2.7.2. Trc II.

Ti tit din lp bnh vt :

Kch thc tit din then:

lt= (0,80,9).lm22= (0,80,9).95 = (7685,5) (mm).

Ta chn lt= 80 (mm).

Da vo bng 9.1a[1] vi d23= 70 (mm) ta c:

b= 20 (mm), h= 12 (mm), t1= 7,5 (mm), t2= 4,9 (mm).

Ti ch lp a xch:

Tng t ta cng c:Kch thc tit din then:

lt= (0,80,9).lm23= (0,80,9).85 = (6876,5) (mm).

Chn lt= 70 (mm).

Vi d22= 60 (mm) tr bng 9.1a[1] ta c :

b= 18 (mm), h= 11 (mm), t1= 7 (mm), t2= 44 (mm).

2.8. Kim nghim trc.

Theo 10.19[1] ta c iu kin bn ca trc I ti tit din j l:

Trong : [s]: L h s an ton cho php .[s]= 1,52,5. sj: L h s an ton ch xt ring n ng sut php

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

Theo cng thc 10.20[1], 10.21[1] ta c:

Vi -1 , -1 : gii hn mi un v mi xon ng vi chu k i xng

Trc lm bng thp 45 c b = 600 MPa. Do :

-1 = 0,436.b = 0,436.600 = 261,6 MPa

-1 = 0,58.-1 = 0,58.261,6 = 151,73 MPa

aj,aj : bin ca ng sut php v ng sut tip ti tit din j.

mj,mj : tr s ng sut trung bnh ca ng sut php v ng sut tip ti tit din j.

Do trc quay, theo cng thc 10.22[1] ta c:

Vi : Wj l mmen cn un ti tit din j:

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

Vi W0j: Mmen cn xon ti tit din j :

: h s k n nh hng ca tr s ng sut trung bnh n bn mi. Theo bng 10.7[1] tra c:

h s. theo cng thc 10.25[1], 10.26[1] ta c:

Vi: kx : H s tp trung ng sut do trng thi b mt, ph thuc vo trng thi gia cng v nhn b mt. Tra bng 10.8[1] ta c kx= 1,06 vi phng phpgia cng trn my tin t Ra= (2,50,63) ()

ky : H s tng bn b mt trc. Tra bng 10.9[1] vi phng php tng bn, ky= 1.

h s kch thc k n nh hng ca kch thc tit din trc. Theo bng 10.10.kj, kj : H s tp trung ng sut thc t khi un, xon.

+ Ti tit din nguy him 0. Ta xc nh c cc thng s:

Vi d0= 40 (mm). Tra bng 10.10[1] ta c:

Tra bng 10.11[1] vi kiu lp k6 ta c:

Vy ti tit din 0 ta c:

Vy c:

Vy ta c:

+ Ti tit din nguy him 3:

Tng t ta c:

Vi d3= 45 (mm) ta c:

Tra bng 10.13[1] vi kiu lp k6 ta c:

Vy ti tit din 3 ta c:

Vy c:

Vy ta c h s an ton ti tit din 3:

Vy trc I bn.PHN 4: TNH CHN LN

1. Trc I.

1.1. Chn loi ln.

Do c lc dc trc ln nn ta s dng a cn.Vn tc trt trn b truyn bnh vt trc vt ln, nhit sinh ra nhiu, trc b gin di trong qu trnh lm vic nn ta b tr s nh hnh v: Dng s ch O cho 1 u c nh u cn li ty ng bng bi .

1.2. Chn s b .

Da vo bng P2.11 chn a cn c trung rng k hiu 7608 vi cc thng s:K hiu7608

ng knh trong (mm)d40

ng knh ngoi (mm)D90

B rng vng trong (mm)B33

B rng vng ngoi (mm)C128,5

Gc tip xc ()

11,17

Kh nng ti ng (kN)C80

Kh nng ti tnh (kN)C067,2

Da vo bng P2.7 ta chn bi c nh k hiu l 208 vi cc thng s:

K hiu208

ng knh trong (mm)d40

ng knh ngoi (mm)D80

B rng vng trong (mm)B18

Kh nng ti ng (kN)C25,6

Kh nng ti tnh (kN)C018,10

1.3. Xc nh ti trng tc dng ln cc .

Ta c tng phn lc tc dng ln :

+ Khng i chiu lc t khp ni:

Ta c:

Fx0= 242,66 (N) Fy0= 1448,70 (N)

Fx1= 616,48 (N) Fy1= 182,41 (N)

+ i chiu lc t khp ni:

Ta c:

Ta s tnh c:

Ta c phng trnh cn bng:

Vi: Vi d1 l ng knh vng chia trc vt. T (2) ta c:

Thay vo 1 ta c : Fy0= 1448,7 (N)

Trong mt phng O xt cc phn lc Fx0, Fx1 xinh ra bi lc Ft1.

Ta c cc phng trnh cn bng:

Thay vo (3) ta c:

Fx0= 1356,37 491,28= 865,09 (N)

Fx0= 865,09 (N) Fy0= 1448,70 (N)

Fx1= 491,28 (N) Fy1= 182,41 (N)

Vy ta chn tnh chn ln trong nhng trng hp i chiu lc t khp ni. Tc l ta c:

Fr0= 1687,34 (N)

Fr1= 524,05 (N)

1.4. Kim nghim kh nng ti ng ca 1.4.1. bi a cn.Bn cnh lc dc trc ngoi, trong cn xut hin lc dc trc Fs do cc lc hng tm Fr tc dng ln sinh ra:

Ta c:

Fs00= 0,83.e.Fr0 .

Vi

Fs0= 0,83.e.Fr0= 0,83.0,3. 1687,34 = 420,15 (N)

Tng t c :

Fs01= 0,83.e.Fr0= 0,83.0,3. 1687,34 =420,15 (N)

Tng lc dc trc tc dng ln cc :

Ti trng quy c trn cc :

Q0=(X.V.Fr0+ Y.Fa00).k.kt Q1=(X.V.Fr0+ Y.Fa01).k.ktTrong :

+ V: H s k n vng no quay. Do vng trong quay => V=1.

+ kt : H s k n nh hng ca nhit kt=1 khi

+ k : H s k n c tnh ti trng. Tra bng 11.3 vi ti trng tnh => k= 1.

+ X,Y : H s ti trng hng tm, h s ti trng dc trc. Tra bng 11.4 da vo tr s :

Ta c:

Vy vi 0 ta c:X=0,4 ; Y=0,4.cotg11,17=2,03

Vi 1 ta c : X=1 ;Y =0

Q0= (0,4.1. 1687,34 +2,03.4852,14).1.1= 10524,78 (N)

Q1= (1.1.1687,34 +0.420,15).1.1= 1687,34 (N)

Do 0 chu lc ln hn nn ch cn tnh cho 0.

Kh nng ti ng ca 0 l:Theo cng thc 11.1 ta c:

Trong :

Q= Q0= 10524,78 (N)

L: Tui th ca tnh bng triu vng quay =>

Trong :

Lh: Tui th ca tnh bng gi Lh=12500 (gi)

n1; s vng quay trn trc I; n1= 968 (vg/ph)

m: Bc ca ng cong mi khi th v ln vi a cn c m=10/3

=>

= 75,94410 (kN)< C= 80 (kN).

Vy a cn chn t kh nng ti ng.1.4.2. bi .Theo cng thc 11.3 ta c ti trng quy c:

Q= X.V.Fr1.kt.kTra bng 11.4 ta c: X=1

Vy:

Q= 1.1. 524,05.1.1= 524,05 (N)

Kh nng ti ng :

Vi bi th m=3.

=>

=4,70997 (kN)< C= 25,6 (kN)

Vy bi m bo iu kin bn ng.1.5. Kim nghim kh nng ti tnh ca 1.5.1. bi a cnTheo cng thc 11.18 ta c iu kin:

Qt C0Trong :

Qt: Tr s ln hn trong ln hn trong 2 gi tr Qt tnh c theo cng thc 11.19 v 11.20 :

Qt= X0.Fr0+ Y0.Fa00.

Qt= Fr0Vi :

X0,Y0: H s ti trng hng tm v h s ti trng dc trc. Ta tra bng 11.6 ta c:

X0= 0,5 ; Y0= 0,22.cotg11,17=1,11

Qt= 0,5. 1687,34 + 1,11. 4852,14= 6229,55(N)> Fr0= 1468,88 (N)

Vy ta c: Qt= 6229,55 (N)= 6,22955 (kN)< C0= 67,2 (kN)

Vy a cn m bo iu kin bn tnh.1.5.2.. bi d

Tng t ta c vi Fa00=0 ;X=0,6

=> Qt= X0.Fr1= 0,6.524,05= 314,43(N)< 524,05 (N)=Fr1Do Qt= 524,05 (N)=0,52405 (kN)< C0= 18,10(kN)

Vy bi m bo iu kin bn tnh.2. TrcII Trc c lp bnh vt c lc dc trc nn ta dng a cn.

Da vo bng P.11 ta chn a cn c trung k hiu 7313 c cc thng s:

K hiu 7313

ng knh trong (mm)d65

ng knh ngoi (mm)D140

B rng vng trong (mm)B33

B rng vng ngoi (mm)C128

Gc tip xc ()

11,50

Kh nng ti ng (kN)C134

Kh nng ti tnh (kN)C0111

PHN 5. THIT K V HP GIM TC

1. Tnh kt cu ca v hp

V hp ca hp gim tc c nhim v m bo v tr tng i gia cc chi tit v b phn my, tip nhn ti trng do cc chi tit lp trn v truyn ti , ng du bi trn, bo v cc chi tit may trnh bi bm.

Ch tiu c bn ca hp gim tc l cng cao v khi lng nh, v vy vt liu nn dng ca hp gim tc l GX15-32.

1.2. Cc kch thc c bn ca v hp.

Cc kch thc c bn ca v hp c trnh by trong bng di y.Trong :

a: khong cch trc ,chn l khong cch trc b truyn trc vt.

a = 190 (mm)

L,B: Chiu di v chiu rng ca v hp.

Chiu dy: Thn hp, ( Np hp, (1 ( = 0,03.a + 3 = 0,03.190 + 3 = 8,7(mm).

( Chn ( = 8 > 6(mm)

(1 = 0,9. ( = 0,9. 8 = 7,2 (mm)

( Chn (1 = 7 (mm)

Gn tng cng: Chiu dy, e

Chiu cao, h

dc e =(0,8 ( 1)( = 6,4 ( 8, chn e = 7 (mm)

h < 58 => chn h= 55 (mm)

Khong 2o

ng knh:

Bulng nn, d1Bulng cnh , d2

Bulng ghp bch np v thn, d3 Vt ghp lp , d4 Vt ghp lp ca thm du, d5d1> 0,04.a +10 = 0,04. 190 + 10 =17,6( Chn d1 = 18 (mm)

d2 = (0,7 ( 0,8).d1 ( Chn d2 = 13 (mm)

d3 = (0,8 ( 0,9).d2 ( Chn d3 =11 (mm)

d4 = (0,6 ( 0,7).d2 ( Chn d4 = 8 (mm)

d5 = (0,5 ( 0,6).d2 ( Chn d5 = 6 (mm)

Mt bch ghp np v thn:

Chiu dy bch thn hp, S3Chiu dy bch np hp, S4B rng bch np hp, K3S3 =(1,4 ( 1,5) d3 , chn S3 = 17 (mm)

S4 = ( 0,9 ( 1) S3 => chn S4= 16 (mm)

K3 = K2-( 3(5 )mm => chn k3=38 (mm)

Kch thc gi trc:

ng knh ngoi v tm l vt,

D3, D2B rng mt ghp bulng cnh : K2

Tm l bulng cnh : E2

k l khong cch t tm bulng n mp l

Chiu cao h

nh theo kch thc np tra bng 18.2 ta c: D3= 125 (mm), D2= 100 (mm)

K2=E2+R2+(3(5)mm=>Chn K2=42(mm)

E2= 1,6.d2 = 1,6 . 13 = 20,8 (mm).

R2 = 1,3 . d2 = 1,3. 13 = 16,9 (mm)

C=D3/2=125/2=62,5 (mm)

k ( 1,2.13 =15,6 ( k = 16 (mm)

h: ph thuc tm l bulng v kch thc mt ta

Mt hp:

Chiu dy: Khi khng c phn li S1

Khi c phn li:Dd,S1 v S2

B rng mt hp, K1 v qS1 = (1,3 ( 1,5) d1 ( S1 = 24 (mm)

Dd xc nh theo ng knh dao khot

K1 ( 3.d1 ( 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 cc bnh rng vi nhau. ( ( (1 ( 1,2) ( ( ( = 9 (mm)

(1 ( (3 ( 5) ( ( (1 = 40 (mm)

(2 ( ( = 8 (mm)

S lng bulng nn ZZ = ( L + B ) / ( 200 ( 300)

Lv B : Chiu di v rng ca hp

Hnh 13: Mt s kt cu ca v hp gim tc c

1.3. Cc chi tit khc c lin quan.1.1.3. Vng mc. nng v vn chuyn hp gim tc khi gia cng, lp ghp trn np v thn thng lp thm vng mc.

Kch thc vng mc c xc nh:

Chiu dy vng mc:

S =(2 ( 3).( = 16 ( 24 ( Chn S = 20 (mm)

ng knh vng mc : d =(3 ( 4).(= 30 (mm)

1.3.2. Cht nh v

m bo v tr tng i ca np v thn trc v sau khi gia cng cng nh lp ghp dng 2 cht nh v.Chn cht nh v hnh cn : Tra bng 18.4b ta c hnh dng kch thc cht nh v hnh cn:

d = 6 (mm) ; c= 1,0 (mm) ; l = (20 ( 110) (mm)

1.3.3. Np quan st.

Theo bng 18-5 tp 2 TTTKHDCK ta c kch thc np quan st:

Bng kch thc np quan st.

ABA1B1CC1KRVtS lng

150100190140175-12012M8x224

1.3.4. Nt thng hi.

Khi lm vic , nhit trong hp tng ln. gim nhit ta chn kt cu hnh 18-11C trang 96 sch TTTKHDCK tp 2.

1.3.5. Que thm du

2. Bi trn hp gim tc

2.1. Cng dng gim mt mt cng sut v ma st, gim mi mn rng, m bo thot nhit tt.

phng cc chi tit my b han g cn phi bi trn lin tc cc b truyn trong hp gim tc.

2.2. Bi trn b truyn trc vt.

Phng php bi trn b truyn trc vt l phng php ngm du. Trc vt nm di nn mc du phi ngp ren trc vt nhng khng c vt qu ng ngang tm con ln di cng. Nu khng ngm ht chiu cao ren trc vt th np vng vung du trn trc vt du c bn ln bnh vt n bi trn ch n khp.

2.3. Bi trn ln.

Cp a cn tren trc II c bi trn bng m. trnh cn bn v du bi trn trong hp bn vo trong qu trnh lm vic ta phi lm np chn m.

Cp a cn v trn trc I bi trn bng ngm du.

3. Bng thng k lp ghp, tr s sai lch gii hn v dung sai lp ghp.

Vng trong ca ln c lp vi trc theo h l. Vng ngoi ca ln c lp ln v hp theo h trc.

Bng thng k dung sai v lp ghp.

TrcV tr lpKiu lpDung saiKhe h di

TrcL

INp bi v v hp

+25

0-100

-290-100

-315

Vng trong ca bi v trc

+18

+2

Vng ngoi ca bi v v hp

+25

0

IIMay bnh vt v trc

Bc-trc

+30

0+21

+2+21

-28

TI LIU THAM KHO

1. Trnh Cht L Vn Uyn: Tnh ton Thit k h dn ng c kh tp 1,2- NXBKH&KT,H Ni, 2007

2. Nguyn Trng Hip- Chi tit my tp 1,2-NXBGD, H Ni, 2005

3. Ninh c Tn- Dung sai v lp ghp- NXBGD, H Ni, 2004

PAGE GVHD: TS Nguyn Tin Dng

SVTH: Nguyn Vit Dng CTM2 K52

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