Đồ án máy tiện T6M16

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Thuyt minh n mn hc My ct kim loi

Trng i hoc KTCN Thi NguynLp K35MA 1

B gio dc v o to Cng ha x hi ch nghavit nam

---------------- c lp T doHnh phc

Trng i hc -------------o0o------------

Cng nghip thi nguynKhoa : c kh

B mn : My v T ng ho

ti

thit k n mn hc

Ngi thit k : Hong TunLp : K35MANgnh : C kh ch to myCn b hng dn : Nguyn ThunNgy giao ti : 10/03/2003

Ni dung ti : Thit k my tin ren vtvn nng

H=200 ; Z=22 ; = 1,26 ;nmax= 1600 (v/ph)

Ct -c ren quc t, mun, Anh, Pt

S lng v kch thc bn v : 02 bn A01 . Khai trin hp tc - A02 . Khai trin ng - A0

Thi Nguyn , Ngy 10 thng 3 nm 2003

T trng b mnCn b hng dn

(k tn)(k tn)

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Nguyn Thun

LI NI U

Trong giai on pht trin x hi nh hin

nay,vic xy dng mt nn cng nghip hin i l mt

nhim v trng tm trong thi k pht trin nn kinh

t th trng.Nhn r c nhim v quan trng

ng v nh nc ta rt ch trng n vic pht

trin nn cng nghip nng trong mi nhn lnghnh C KH CH TO MY.

Trnh k thut ca mt t nc trc ht c

nh gi bi s pht trin ca ngnh c kh ch to

my-Mt trong nhng ngnh ch o ca nn cng nghiptrong my ct kim loi l thit b ch yu ca

nghnh,chng dng bc i mt lng d no t

phi bin thnh nhng chi tit my theo

mun.Ngy nay cng ngh sn xut phi t nhng

thnh tu to ln trong vic to ra nhng phi c hnh

dng ging vi chi tit gia cng v lng d gia cng

bc i rt nh.Song khng v th m ngha ca my

ct kim loi trong nghnh c kh li gim m cn tngln v bi qa trnh gia cng trn my ct rt phc

tp v yu cu chnh xc rt cao m cc dng gia

cng khc khng th t c.

Sau thi gian hc tp ti trng i Hc K Thut

Cng Nghip n nay,em hon thnh chng trnh hc

ca nghnh c kh ch to my. c s tng hp cc

kin thc hc trong cc mn hc ca ngnh v c

c s khi qut chung v nhim v ca mt ngi

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thit k ,em c nhn ti thit k my Thit k

my tin ren vt chnh xc v chuyn thnh lp s

ng ct ren c bc ren thay i. c s ch bo

tn tnh ca thy gio hng dn

TS Nguyn ng Ho v tp th cc thy c gio trongb mn my v t ng ho cng vi s c gng ca bnthn,n nay em hon thnh n tt nghip camnh.Trong qu trnh lm n chc chn s khngtrnh khi nhng thiu st.Em rt mong c s chbo ca cc thy c em c iu kin hc hi thm.

Em xin chn thnh cm n.

Thi Nguyn, ngy thng

nm 2002

Sinh

vin thit k

Hong Tun

Nhn xt ca gio vin hng dn

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Ti liu tham kho

[1] : Thit k my ct kim loi.Tc gi : Mai Trng Nhn.

B mn my ct kim loi.Trng HKT CN Vit Bc.

[2] : Gio trnh my ct kim loi Tp I.Tc gi : GVC Hon Duy Khn.

B mn my ct kim loi.Trng HKT CN Thi Nguyn.

Thi Nguyn : 1996.[3] : H-ng dn thit k n mn hc dao ct.

Tc gi: Trnh Khc Nghim.

Trng: HKTCN Thi Nguyn.

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Bc Thi: 1991.[4] : Tnh ton thit k my ct kim loi.

Tc gi: Phm p Nguyn c LcPhm Th Trng Nguyn Tin Lng

Nh xut bn i Hc v trung hc chuynnghip.

H Ni : 1971[5] : Tnh ton h dn ng c kh Tp I,II

Tc gi: Trnh Cht L Vn UynNh xut bn gio dc.

[6] : K thut tin.Tc gi: NHEJNI CHIKIN TKHO.Ngi dch: Nguyn Quang Chu

Nh xut bn Thanh Nin 1999.

[7] : Gio trnh my ct kim loi Tp IV.Bin son: Dng Cng nh.Hiu nh: PTS Trn V Quc.

B mn my ct kim loi Trng HKTCNTNThi Nguyn: 1996.

[8] : Tp bn v s ng my ct kim loi.Trng HKTCN Vit Bc.

B mn my ct kim loi.Bc Thi 1980.

Mc lcPhn

Trang

Phn I: Tng hp cu trc ng hc my.1-8

Phn II: c trng k thut ca my.9-15

Phn III: Thit k ng hc my.16-56

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Phn IV: Thit k ng lc hc my.57-69

Phn V: Tnh ton chi tit my.70-80

Phn VI: Tnh ton h thng bi trn v lm mt.81-85

phn i :tng hp cu trc ng hc my

I. Cng dng ca my tin ren vt vn nng.

My tin ren vt vn nng l my cng c c

dng ph bin nht trong cc nh my, phn xng c

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kh ca cc x nghip. N c dng gia cng ccb mt trn xoay, b mt ren. Ph hp vi loi hnhsn xut n chic lot nh, thnh hp vi sacha,ch to cc chi tit thay th.

Ngy nay do tin b khng ngng ca khoa hc kthut, my tin ren vt vn nng c ci tin nhiucho ph hp vi nhu xu hng pht trin ca thi i.c bit l cc my c iu khin theo chng trnhs (CNC), ng dng cng ngh mi CAD/CAM/CNC.

Ngoi vic gia cng cc b mt trn xoay, b mtren. Nu s dng thm cc g chuyn dng th cth m rng thm kh nng cng ngh ca my thchin cc nguyn cng khc nh khoan, khot, doa, tincc b mt nh hnh, mt phng, ct t c chnh

xc cao.Nhng cng vic ch yu ca my tin ren vt vn

nng l tin trn v tin ren. My c th tinc cc loi ren h mt, ren h Anh, ren nhiu umi, ren khuch i, ren tiu chun v phi tiuchun, ren tri v ren phi. . .

II. To hnh b mt chi tit gia cng.

1. S gia cng.

My tin ren vt vn nng ch yu dng giacng cc b mt trn xoay (Tr trn) v b mt ren.Chn hai nguyn cng c trng ny ca my xcnh s gia cng.a. Nguyn cng tin tr trn.

B mt ny c hnh thnh nh hai chuyn ng:chuyn ng quay trn ca trc chnh mang phi Q1vchuyn ng tnh tin ca bn my mang dao T2nhmto ra lng chy dao.

Vy c hai chuyn ng to hnh l:+ S(Q1): Chuyn ng to hnh ng sinh 1.+ c(T2): Chuyn ng to hnh ng chun 2.

phng php to hnh b mt l vt ( qu tch)

b.Nguyn cng tin ren.

ng sinh (1) l prfin ren c hnh thnh tphng php chp hnh. ng chun (2) l ngxon vt tr c hnh thnh t phng php vt.

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to ra b mt ren th 2 chuyn ng thnh phnQ1, T2phi c mi quan h cht ch vi nhau m bokhi trc chnh mang phi quay c 1 vng th bn mamang dao phi dch chuyn mt lng bng bc ren thay bc xon H(i vi ren nhiu u mi). Vy chai chuyn ng to hnh l :

S(Q1) : Chuyn ng to hnh ng sinh 1.c(Q1, T2) : Chuyn ng to hnh ca ng chun2.

2. Cc chuyn ng cn thit ca my.

a. Chuyn ng to hnh (k hiu ) .Chuyn ng to hnh l chuyn ng cn thit to ra ng sinh v ng chun. S lng cc chuynng to hnh c xc nh qua biu thc:

N = Ns+ Nc- 1/2 NT

Trong :Ns : S lng thnh phn chuyn ng to hnhng sinh.Nc : S lng thnh phn chuyn ng to hnhng chun.NT: S chuyn ng trng.Trong : Ns=0 ; Nc =2 ; NT =0 (Tin ren)

Ns=1 ; Nc=1 ; NT =0(Tin trn)

Nh vy trong c hai trng hp tin th N=2 ngha lchuyn ng to hnh gm c 2 thnh phn (Q1,T2).

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b. Chuyn ng ct gt.Chuyn ng ct gt l chuyn ng cn thit

thc hin v duy tr qu trnh bc phoi, y chuynng ct gt trng vi chuyn ng to hnh do cutrc ng hc my n gin nhng n li hn ch cngsut ct gt. Ngoi cc chuyn ng chy dao dc cabn my, chuyn ng ph cn c chuyn ng chy daongang thc hin mt s nguyn cng khc nh : Xnmt u, tin ct t.

c. Chuyn ng phn .

Chuyn ng phn l chuyn ng cn thit dch chuyn tng i gia dao v phi sang v trmi, khi trn chi tit gia cng c nhiu b mt gia

cng cn bn ging nhau. V d nh : Tin ren nhiuu mi.d. Chuyn ng nh v.

Chuyn ng nh v l chuyn ng nhm khng chkch thc gia cng ca chi tit gia cng, n cnhim v xc nh hng ca ta phi v dao vinhau, tc l xc nh v tr tng i ca ng sinhv ng chun vi nhau trong cc trc to camy. Chuyn ng nh v c th l chuyn ng n daonu trong lc thc hin c tin hnh ct gt v cth l chuyn ng iu chnh nu trong lc thc hinkhng c qu trnh ct gt.

e. Chuyn ng iu khin.

L chuyn ng nhm m bo my hot ng theomt tin trnh cng ngh xc nh, chuyn ng nyca my l chuyn ng cn thit cho my tr thnhmy t ng hay bn t ng. V d cc chuyn ngthc hin ng m l hp hay khng ch hnh trnh.

g. Cc chuyn ng ph khc.L chuyn ng thc hin dch chuyn dao hay phi vitc ln m khng tham gia ct gt, cc chuyn ngny cn thit khi kt thc mt lt gia cng chuyn sang lt gia cng khc.

III . Thnh lp s cu trc ng hc my.

Tp hp mt hay vi nhm ng hc ni kt cu hayni ng hc vi nhau, to thnh cu trc ng hc

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ton my. V vy mun xy dng cu trc ng hc mycn nm vng nguyn tc ni ng v nguyn tc b trcc khu iu chnh. Bit rng ngoi chuyn ng chydao dc T2my cn c chuyn ng chy dao ngang T3tin mt phng, tin mt u v tin ct t. . . Do lin kt ca my phi c vt me ngang. Khi thchin tin trn chuyn ng tnh tin ca bn xe dao T2s do c cu bnh rng thanh rng m nhn khi ctren s dng vt me dc chy do dc.

Theo yu cu my ch to ra phi gia cng ccc loi phi c kch thc khc nhau nm trong phmvi cho php, nhm tho mn tnh cng ngh khi chn

ch ct hp l. V vy trc chnh phi c nhiutc tng ng vi ch ct. bo m iu ta phi thit k hp tc (iv) v c cu iu chnhtc cho trc chnh. to ra cc lng chy daokhc nhau (dc, ngang) trong my cn b tr hp chydao (is) khi ny s cu trc ng hc my v iuchnh ng hc my c th hin nh hnh v.

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Trc vt me ngang- T3

Trc vt me dc T2- Lin kt trong: Trc chnh Q1is

Trc trn (Br- Tr)- T2

- Lin kt ngoi: i/civTrc chnh.*. Xch tc :- T ng c M1n trc chnh mang phi

M11 2 iv3 4 Trc chnh.- Lng di ng tnh ton

n/c= nT/c (Vng/pht)- Phng trnh iu chnh.

n/ci12ivi34= nt/c- Cng thc ng hc.

iv= Cv. nt/c*. Xch chy dao tin trn.- T trc chnh mang phi n b truyn bnh rngthanh rng.Trc chnh 4 5 is6 8 BR TR- Lng di ng tnh ton.

1 vng trc chnh to ra Sd(mm) bn dao dc

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- Phng trnh iu chnh.1i45isi68.mn.Z =Sd(mn)

- Cng thc ng hc.is=Cs1.Sd

* Xch chy dao tin ren.- T trc chnh n b truyn vt me ai c dc(tvmd)- Trc chnh 4 5 is6 7 tvmd- Lng di ng tnh ton.

1 vng trc chnh to ra t(mm)bn dao.- Phng iu chnh.

1i45isi67tvmd= t(mn)- Cng thc ng hc

is= Cs2. t*. Xch chy dao ngang.- T trc chnh n b truyn vt me ai c ngang(tvmn)Trc chnh 4 5 is6 9 tvmn- Lng di ng tnh ton.

1 vng trc chnh to ra Sng(mm) bn dao ngang.- Phng trnh iu chnh

1i45isi69tvmn= Sng(mm)

- Cng thc ng hc.is= CS3.Sng*. xch chy dao nhanh.+. chy dao dc.- T ng c chy dao nhanh M2n b truyn BR/TR

M210 8 BR/TR- Phng trnh iu chnh

n/c(M2) i108.mn.z = Sd(mm)+. chy dao ngang.

- T ng c dao nhanh M2 n b truyn vt mengang.M210 9 tvmn

- Phng trnh iu chnh.n/c(M2)i109tvmn= Sng(mm)

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PHn II :c trng k thut ca my

I. c tr-ng cng ngh.

My tin ren vt vn nng c th gia cng c ccb mt tr trn xoay (trong, ngoi); mt u v ccb mt ren, tin ct t.Cc nguyn cng thc hin trn my tin l : Tin

tr trn, tin ren, khoan, doa, tar. . . Ngoi ranu b tr thm g th c th m rng thm phm vicng ngh ca my.Cc dng c ct c s dng trn my tin thng

l : Thp cacbon dng c, thp gi, thp hp kimdng c, hp kim cng . . .Phi c th gia cng c trn my l phi thanh,

phi rn hoc c. Vt liu phi ch y l thpcacbon, thp hp kim , gang. . . Ngoi ra cn c hpkim mu v vt liu phi kim loi.

Tuy theo phng php t chnh xc khi gia cngm chi tit gia cng c th t c chnh xc v bng b mt khc nhau.Cp chnh xc bng Rz(m) phng php giacng811 80 (m) khi tin th57 40 (m) khi tin bn tinh24 10 (m) khi tin tinh2 3,2 (m) khi tin mnh

My ny ph hp vi sn xut lot va v phc v sacha thay th.

II. c tr-ng kch th-c my.

c trng kch thc l kh nng thch ng ca myi vi vic gia cng cc chi tit v mt kch thc.Chiu cao tm my: H=200 (mm)ng knh chi tit ln nht c th gia cng ctrn bng my:

Dmax= 2H = 2.200 = 400 (mm)

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ng knh chi tit ln nht c th gia cng trn bndao, l ng knh gia cng hiu qu nht m ta dng tnh ton cc c trng k thut l:

D1max= (1,21.4) H . Chn D1max= 1,3H = 1,4.200 = 280(mm)

ng knh b nht ca phi c th gia cng c trnmy

max1min1

1D

RD

d

Trong Rdl phm vi thay i ng knh Rd= 810Chn Rd= 10

vy c : D1min= 28010

1 = 28 (mm)

ng kch phi ln nht c th lun qua trc chnh.dmax= ( 0,150,2 ) D1maxChn dmax=0,15.D1max = 0,15.280 = 42 (mm)Khong cch xa nht gia hai mi tm : L= (3,57).200=7001400 (mm)

Chn Lmax= 1000(mm)S tc quay ca trc chnh : 24 (22 cp khcnhau).Gii hn tc quay ca trc chnh:

nmin=12,5 (vng/pht) ; nmax= 1600 (vng/pht)

III. c tr-ng ng hc.

1. Xch tc .+ Vic tnh ton tc ct ln nht v b nht

ca my, bng cch phi hp nhng iu kin thun lihoc kh khn nht vi nhau, s dn ti tng rt lnphm vi iu chnh ca my v lm cho kt cu rtphc tp.Do chn cc tr s tc ct ti hn tt

nht l cn c vo cc ti liu thng k v s dngtc ct trn cc my khc nhau. C th tng tr stc ct ln nht ln 25% khi k n s tin b vmt kt cu v vt liu dng c ct.+ Chui s vng quay ti hn ca trc chnh

Theo ti thit k ta c : nmax= 1600 (v/ph)+ S cp tc l zn= 22+ Chn cng bi = 1,26Tra bng 4[1] tra c nmin = 12,5 (v/ph)

+ Phm vi iu chnh tc

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1285,12

1600

m in

m ax n

nRn

+Tnh s vng quay ca trc chnh:

Vi = 1,26 ta c s vng quay ca trc chnhnh sau:n1= 12,5 (v/ph) n9= 80 (v/ph) n16=400 (v/ph)n2= 16 (v/ph) n10= 100 (v/ph) n17=500(v/ph)n3= 20 (v/ph) n11= 125 (v/ph) n18=630(v/ph)n4= 25 (v/ph) n12= 160 (v/ph) n19=800(v/ph)n5= 31,5 (v/ph) n13= 200 (v/ph)

n20=1000 (v/ph)n6 = 40 (v/ph) n14= 250 (v/ph)

n21=1250 (v/ph)n7= 50 (v/ph) n15= 315 (v/ph) n22=1600 (v/ph)n8= 63 (v/ph)

2.Xch chy dao.

- Tc chy dao ca my ph thuc vo chiu suct khi gia cng v cht lng b mt, yu cu kthut ca chi tit cn gia cng.- Chiu su ct tmaxc lp bng lng d h thp,khi gia cng c, theo bng 6[1], vi kch thc phil: 2001000 (mm) c lng d 2 pha l a= 14(mm)

)(72

14

2max mm

at

Chiu su tmin c tnh gn ng theo biu thc

sau: m axm in4

1

2

1tt

Chn tmin= )(75,14

1mm

4

7tmax

Lng chy dao Smaxtra theo tmaxkhi tin th ngoi.Lng chy dao Smintra theo cht lng b mt gia cngc th.

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Bi v hp chy dao tin ren dng c tin trn.Nn phm vi iu chnh bc ren v lng chy daophi m bo ging nhau Rt = Rs

Chy dao dc: Sd=0,052,8 (mm/v)

Chy dao ngang: Sng= 0,0251,4 (mm/v)IV.c tr-ng ng lc hc my.

- thit k truyn dn nh gn kch thc ph hpm my vn bn khi lm vic mi tc v vyphi chn ch ct ph hp. c trng ng lc hcca my c xc nh theo ch ct tnh ton cti trng v cng sut ln nht.1. Ch ct tnh ton.Chiu su ct tnh ton: c xc nh theo biu

thc: 3 max1* 7,0 Dt (mm)

Vi D1max=280 (mm) thay s 5 7,42 8 07,0 3* t

(mm)Lng chy dao tnh ton:

c xc nh theo biuthc: 3,0.4,0 ** tS (mm/v)

532,13,0579,4.4,0* S (mm/v)

Tc ct tnh ton: c xc nh theo biu thc

sau:

yvxv

vv

St

KCV

**

*

.

. (m/ph)

Tra bng: 4-58 [3] vi dao l thp gi v vt liu

gia cng c b 750 (N/mm2)Ta c: Cv=50,2 ; Kv=1,09

xv=0,25 ; yv=0,66Thay s vo ta c:

23,28)532,1.()579,4(

09.1.2,50 66,025,0* V (m/ph)

2. Lc ct.Lc ct c tnh ton theo cng thc bng 9[1]

Px*= Cpx.t

*xpx.S*ypxPy

*=Cpy.t*xpy.S*ypy

P*z=Cpz.t*xpz.S*ypz

Vi t*= 4,226 (mm) ; S*= 1,39 (mm/v)

Cpx=650 xpx=1,2 ypx=0,65

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Cpy=1250 xpy=0,9 ypy=0,75Cpz=2000 xpz=1 ypz=0,75Thay s tm c:

Px*= 650.(4,579)1,2.(1,532)0,65 = 5324,22 (N)

Py*= 1250.(4,579)0,9.(1,532)0,75 = 6769,36 (N)

Pz*= 2000.(4,579)1.(1,532)0,75 =12610,86 (N)

3. M men xon ln nht.

4,17655202

28086,12610

2

m ax1m ax

*

DP

M zx (N.mm)

4. Cng sut ct.

933,510.60

23,2886,12610

10.60 33

**

VPN

z

c (kw)

5. Chn s b ng c.

chn la phng n truyn dn ta cn xc nhs b cng sut ng c v chn ng c cho my. Cngsut ng c truyn dn chung cho c xch tc vxch chy dao l

*

/ . c

scd

NKN

trong : Ks = (1,021,2) l h s k n cng sutchy dao chn Ks=1,2

)85,075,0( l h s hiu sut truyn dn;chn 85,0

thay s c: )(376,885,0

933,5.2,1/ KwN cd

Vy ta chn s b ng c khng ng b 3 pha ck

hiu DK62-4N= 10(Kw) ; n=1460(v/ph)

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Phn III :Thit k ng hc my

A: hp tc

Hp tc trong my ct kim loi dng truyn lcct cho cc chi tit gia cng, c kch thc, vt liukhc nhau vi nhng ch ct cn thit. Thit k hp tc yu cu phi m bo nhng ch tiu v k thut v kinht tt nht trong iu kin c th cho php. Hp tc phi c kch thc nh gn, hiu sut cao, tit kimnguyn vt liu, kt cu c tnh cng ngh cao, lm vic

chnh xc, s dng bo qun d dng, an ton khi lm vic...

I . Chn ph-ng n truyn dn.

1. Chn kiu truyn dn.Khi chn phng n truyn dn cn cn c vo phm vi

iu chnh, cng sut truyn, tr s trt, thun tiniu khin, thay i tc nhanh, tnh cng ngh tt.Vi my truyn ng chnh l quay c cng sut nhhn 100KW, theo ENIMS nn dng truyn dn iu chnh

tc c kh gm mt ng c xoay chiu v mt hptc bnh rng.2. B tr c cu truyn ng.

C hai phng n b tr truyn dn nh sau :+ Phng n 1: Hp tc v hp trc chnh chung

mt v+ Phng n 2 Hp tc tch ri hp trc

chnhTrong hai phng n trn, phng n mt thng p

dng vi cc my c trung v ln, nhng yu cu

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chnh xc khng cao ta chn phng n 1. N c cc uim sau: Kt cu gn nh, gi thnh h, d tp chungc cu iu khin to iu kin thun li cho victhc hin cc thao tc ca ngi ng my.Nhc im: C th truyn dung ng trong hp tc sang hp trc chnh, c th truyn nhit trong hptc sang hp trc chnh, kh dng truyn ng aicho trc chnh.

3. La chn b truyn cui cng.

B truyn cui cng c nh hng nhiu n ch ct, iu ho chuyn ng, bng b mt giacng.- Trc chnh quay vi tc 1600(v/ph) nn chn btruyn cui cng l b truyn bnh rng. cho trcchnh quay m vi tc vng ca bnh rng khngqu ln v ng knh bnh rng lp trn trc chnhkhng b hn ng knh phi ln nht. Nu gi v ltc vng cho php ca bnh rng th ng knh lnnht cho php ca bnh rng l:

)/(66,26)/(1600

)/(9000)/(9

)(5,10766,26.14,3

9000

.

max

max

max

svphvn

smmsmv

mmn

vD

thy m ax1m ax DD do vy ta s dng 2 bnh rngdn ng cho trc chnh trn hai dy tc thp vcao khc nhau.

II. Chn ph-ng n kt cu.

1. Chn dng kt cu.

Khi thit k my vic la chn kt cu n gin hayphc tp cn cn c vo phm vi iu chnh yu cu,cng dng ca my. Theo kinh nghim ca cc nh thitk my, ch ra rng cu trc n gin c s dngkhi phm vi iu chnh yu cu nh hn tr s tihn: RnR

*

n

i

n RR

R *

2

vi Ri= 8 ; = 1,26 ; R*n=50

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mt khc 1285,12

1600

m in

m ax n

nR

n

vy c Rn> Rn* nn chon kt cu phc tp, Z = Z1+ Z2

Cu trc phc tp c u im l:

M rng phm vi iu chnhRt ngn xnh truyn dn cc tc cao, dn

n gim c tn tht ma st.Nng cao hiu sut ca my, gim ti trng v kchthc b truyn, gim qun tnh quay.2. Chn ph-ng n kt cu.Phng n kt cu c biu din thng qua cng thckt cu:

k

m

k

pZ

1 trong : k- l trt t kt cu ca nhm ng hctheo xch truyn ng.

Pk-l b truyn trong nhm th t k.m -l s nhm truyn.

Theo yu cu thit k c Z = 22 nh phn tch trn ta s dng cu trc nhn phc tp Mt khc khngth phn tch Z = 22 = 211 (khng thc t) . tinphn tch ta ly Z = 24 cp tc vi s phng nkt cu k

!

!

q

mK

m: l s nhm truyn bnh rng trong hp s tc :m = 4q: l s nhm c cng s lng b truyn ging nhau.K: l phng n thay i v tr ca cc nhm truyn

4!3

!4K

(phng n)v b tr theo cc phng n sau:

Z = 2322 ; Z = 2223

Z = 2232 ; Z =3222Cc phng n trn c gi l phng n hp lv:

V s b truyn trong nhm Pk=2; 3 m bo s btruyn l nh nht. Trong bn phng php trn m

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bo iu kin trng lng truyn dn l nh nht P1 >P2> P3tc l cng v cui trc chnh s b truynl gim dn. Mt khc do tc quay gim dn lmm men xon tng ln cc trc dn nn khi tnh

ton kt cu ca trc cng nh trng lng ca ncng tng dn ln m bm s phn b ca nhmtruyn trong hp v trng lng. V vy chnphng n tt nht l: Z = 3 2 2 2

Theo my c trc 16K20 cng chng loi, btruyn tc thp b tr thm hai nhm truyn minhm c duy nht mt b truyn gim tc s dnh vy l kt cu khng gian ca my hp l vkhi ct ren khuych i ngi ta li dng on khc

nhau gia xch tc thp v xch tc cao to ranhm khuych i v thng trn trc u tin tronghp tc c b tr ly hp ma st o chiu quayca trc chnh, v vy gim kch thc chiu trc,trnh gy yu trc ngi ta b tr sao cho P1< P2tcl P1=2 , P2=3 ta c

Z= 2322Chn s trc ca phng n kt cu ST = m+1 , m = 4

ST = 4+1 =5 (trc)

3. Chn ph-ong n ng hc my :

Phng n ng hc my l phng n v trt t thayi cc b truyn trong nhm nhn c dy tc choTrong mt b truyn m c m nhm truyn th s c

m! phng n thay i tr s vng quay. i vi hptc ca my cng c th t s truyn nn chntrong gii hn: imini imax ;

imin= 1/4 ; imax= 2 ; 1/4 i 2vy ta dng cu trc nhn phc tp m bo truyndn tc cao, mt khc do my c nhiu cp tc nn ta tch lm 2 ng truyn:

+ ng truyn c tc cao: Z1=232+ ng truyn c tc chm: Z2=23211

phng n th t hp l nht s l : x1

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p - l s b truyn trong mi nhm.-Vi ng truyn tc cao : Z1= 232Trt t ng hc l :

IIIIIIZ 6211 232

kim tra li lng m ca nhm truyn m bo tr struyn 1/4i 2

Nhm I : x(p-1)= 1,261(2-1)= 1,26 < 8Nhm II : x(p-1)= 1,262(3-1)= 2,56 < 8

Nhm III : x(p-1)= 1,266(2-1)= 4 < 8- ng truyn tc thp trt t hp l l

11232 6212 IIIIII

Z

Vy ta c phng n Z = 24 . Vi yu cu thit kl Z = 22 ta thy tng ln 2 cp tc m boyu cu thit k ta lm trng 2 tc , ngtruyn tc cao do tin hnh gim c tnh canhm truyn cui cng t x = 6 xung x = 4. Do :

IIIIIIZ 4211 232 Vy ta c phng n ng hc ca hp tc l:

Zn = 23(2+211) =

22 = Z1+ Z2IIIIII

Z 4211 232

11232 6211 IIIIII

Z

4. L-i cu trc.

- T phng n kt cu biu din li cu trc theonguyn tc i xng cho ta bit:

+ S lng nhm truyn.+ S lng b truyn mi nhm.+ Th t thay i ng hc, c tnh x v mi

lin h t s truyn mi nhm.+ Phm vi iu chnh ca cc nhm truyn v b

truyn dn.+ S cp tc ca trc dn v b dn ca mi

nhm truyn.Tuy nhin thng qua li cu trc ny ta khng thxc nh c th ga tr ca cc i lng v vy m

qua ch nh gi s b truyn dn, trong qu trnh

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la chn phng n truyn dn khc phc nhc imny, ta i xy dng th vng quay ca my.

Ta c s li cu trc truyn ng nh hnh v(5)

5. th vng quay.

Mun xy dng c th vng quay ta phi xcnh c s vng quay ca trc dn n0, mc nh lto ra t s truyn gim dn v pha trc chnh nn

ta chn im u vo bt u t trc I.S vng quay n0 ca trc I xc nh nh sau: T ng

c c n = 1460(v/ph) qua b truyn ai 148/ 268 ct s truyn l 1,86 do tc ri trn trc I s

cn li l )/(80086,1

1460phvn ta ly y lm im n0.

Chn cc t s truyn: Trong mi nhm ch cn mt ts truyn c dc ca tia tu v phi m bo iukin t s truyn 1/4 < i < 2. Mt khc cc t s truyn

c tiu chun ho thun tin trong vic tnh tonthit k, chng ph thuc s b truyn p, c tnh x, canhm v cng bi ca chui vng quay vng quay n cdng:

i =E(E: nguyn v E > 0, E

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Xch truyn ng nhanh: Z1 = 21I32

II 24III ta chn

t s truyn nh sau:

+ Nhm I : chn i1= 1=1,26 =

45

+ Nhm II : chn i5= 41 = 426,1

1

= 52

+ Nhm III : chn i11= 31

= 326,1

1 =2

1

Xch truyn ng chm: Z2= 21I32

II26III111

+ Nhm I : chn i1= 1=1,26 =

45

+ Nhm II : chn i5= 41

= 426,11 =

52

+ Nhm III : chn i7= 61= 626,1

1 =4

1

T s truyn ca ng truyn n : i8= 61= 626,1

1

=4

1

i9= 31= 326,1

1 =2

1

Trong c hai xch truyn ng trn th cc t struyn t i1i5l dng chung cho c hai xch.

V th vng quay nh hnh v: Da vo tc

vng quay trc I nh chn n0 = 800 (v/ph) v cct s truyn cn li trong cc nhm ( Xc nh bngphng php gii theo li cu trc)

S cp tc chung cho c truyn dn Z = 22 (tn1=12,5 (v/ph) n n22= 1600(v/ph))

S cp tc trng z = 2 ( n17 = 500 v/ph v n8=630v/ph )- th vng quay cho phng n:

22)1122(32 6421 III

nZ IIIIIIZ 4211 232

11232 6212 IIIIII

Z th vng quay c v nh hnh 6

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III. Tnh ton ng hc bnh rng.

1. Ph-ng n tnh.C s tnh ton ng hc bnh rng l xc nh

s rng Z ca cc bnh rng sao cho m bo t struyn chn. Trong mt nhm truyn cc bnhrng n khp c th cc bnh rng phi c cngmodul vi nhau, trong nhng nhm truyn c lng m

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ln do chu lc ct ca bnh rng khc nhau nhiugia cc b truyn, nn c th s dng nhng gi trmodul khc nhau cho mt nhm truyn. Trong hp tc khi ta thay i tc ca trc chnh s dng khibnh rng di trt th dng bnh rng thng. C thtnh s rng ca tng nhm bng nhiu phng n nh :Phng n gii tch, tra bng hay tnh gn ng. Tnh s rng ca cc bnh rng thng trong mt nhm

truyn c cng mdul.

Zj=jj

j

ba

a

.E.K ; Zj' =

jj

j

ba

b

.E.K

Trong : Zjv Z/jl s rng ca cc bnh dn v b

dn ca cp th j trong nhm.

Sz- tng s rng ca b truyn. Sz=E.Kij - t s truyn ca b truyn th j

.aj , bjl cc s nguyn n gin.

c cu nh gn trong truyn dn chnh ngi ta gihn Sz100 120 rng. khi b ct chn rng s rng ti thiu ca bnhrng Zmin=18 20 rng. i khi ly Zmin=14 rng.

E l nhng s nguyn dng v cc Zjv Zj' bng hoc ln hn Zmin cho php . V vycn phi tnh Emin cho tng trng hp c th.+ Nu bnh rng nh nht Zminl bnh rng ch ng ( ij< 1 ) th:

Zj =jj

j

ba

a

.Emin.K zmin

Vy : Emin= min..

zKa

ba

j

jj

+ Nu bnh rng nh Zmin ng vai tr b ng ( ij> 1) th

Zj' =jj

j

ba

b

.Emin.K > Zmin V E

min min..

zKb

ba

j

jj

Cc tr s Emin v Emin tnh ra thng l s l cn phi

quy trn ln pha trnNu s rng ca cc bnh rng tnh ra c Sz > Sz maxcnphi iu chnh li, bng cch gim bt tr s K v

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chu sai s t s truyn i ( %2i ). C hai phngphp gim tr s K:

+ Phn tch li t s truyn c jj ba lm cho K

ln, sau tnh li K v Emin,chn li E , tnh li

Zj v Zj' .+ b bt tha s ca K ri tnh li nh trn songlc ny Zj v Zj' tnh ra thng b l ,sau khi quytrn th khong cch trc A ca cc b truyn ny sb thay i, do phi dch chnh cc cp bnh rng.

* Tnh s rng ca cc bnh rng thng trong cngmt nhm truyn c modul khc nhau.Trong trng hp ny :

jZ m

A

S j2

V : Zj =jjj

j

m

A

ba

a 2.

; Zj=

jjj

j

m

A

ba

b 2.

V 2A = const nn cc Zj v Zj khi tnh ra c th ls l , ta phi quy trn v dch chnh bnh rng.S rng ca cc bnh rng ch nguyn khi : 2A = E. mj

( jj ba )

Do 2A s l bi s chung nh nht ca cc mj

( jj ba ). Nu bi s chung nh nht ny qu ln th

ly bi s chung nh nht ca cc mj. Sau khi nhnbi s chung nh nht ny vi mt s nguyn ta lykt qu lm 2A .

j

zm

AS

2 l s nguyn , nhng s rng Zj =

jjj

j

m

A

ba

a 2.

; Zj=jjj

j

m

A

ba

b 2.

c th l s l phi quy trn v

dch chnh rng nu cn thit .

2. Tnh s rng cho php cc cu trc truyn dn

a. Tnh ton nhm truyn I (chung cho c hai xchtruyn ng chm v nhanh.

Nhm c hai t s truyn l :13

1726,1

1

111

b

ai

3011 ba

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17

2826,1

2

222

2 b

ai 4522 ba

Bi s chung nh nht K = 2.32.5 = 90

Ta thy i > 1 nn bnh nh ng vai tr b ngta c : minmin .

.Z

Kb

baE

j

jj mt khc thy i2c nghing

ln hn v l tng tc nn m in2

22m in .

.Z

Kb

baE

v Zmin= 18 (

rng )thay s c :

190.17

4518m in

E vy chn 1min E s rng ca cc

bnh rng l :5190.

30

171 Z ( rng ) ;

3990.30

131 Z (rng )

5690.45

282 Z (rng ) ;

3490.45

172 Z ( rng )

vy s rng l Z1= 51 ( rng ) ; Z1 = 39 (rng )

Z2=56 ( rng ) ; Z2 = 34 ( rng)b .Tnh ton cho nhm truyn II( chung cho c 2 xchtruyn ng chm v nhanh )

Nhm c 3 t s truyn :5

503

1

11

b

ai 21133 ba

4

4

224 47

29

26,1

11

b

a

i

76472944 ba

3

3

445 55

21

26,1

11

b

ai

76552155 ba

vy BSCNN, K = 76Ta thy i < 1 nn bnh nh ng vai tr ch ng ta

c : m inm in ..

ZKa

baE

j

jj mt khc thy i5c nghing ln

hn v l gim tc nn m in5

55min .

.Z

Kb

baE

v Zmin= 18 ( rng )

thay s c :

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7621

1876m in

E < 1vy chn 1m inE s rng ca cc bnh

rng l : 3876.2

13 Z (rng) ;

3876.

2

1

3

Z

(rng)2976.

76

294 Z (rng) ;

4776.76

47

4 Z (rng)

2176.76

21

5 Z (rng) ;

5576.76

55

5 Z (rng)

vy s rng ca cc bnh rng l :Z3= 29 ( rng ) ; Z4= 21 ( rng ) ; Z5= 38

( rng )Z3= 47 ( rng ) ; Z4= 55 ( rng ) ; Z5= 38

( rng )c . Tnh ton cho nhm truyn III ( ng truynnhanh ).

Nhm c hai t s truyn l :4

526,1110 i > 1

2

1

26,1

113311

i < 1

V hai t s truyn qu chnh lch trnh btruyn c kch thc ln ta dng b truyn c modulkhc nhau. T s truyn : i10c m10= 2,5 ; i11c m11= 3Phng n ny s c mt tc cao v mt tc thp, iu kin nhm truyn lm vic c l:

11

10

.2

.2

11

10

Z

Z

SmA

SmA

c :5

6

5,2

3

10

11

11

10 m

m

S

S

Z

Z

do c :70

35

60

30

32

16

30

15

2

111 i

chn 9011ZS (rng)

60

30

11

11

Z

Z (rng)

Chn SZ10 =5

6 SZ11 =5

6 90 = 108 (rng) Z10 = 60

(rng) ; Z10/= 48 (rng)

vy s rng ca bnh rng l:

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Z11= 30 (rng) ; Z11'= 60 (rng)

Z10= 60 (rng) ; Z10/= 48 (rng)

d . Tnh ton cho nhm truyn III ( ng truynchm)

nhm c 2 t s truyn :1

0

6 i

4

1

26,1

11667

i

V hai t s truyn qu chnh lch trnh btruyn c kch thc ln ta dng b truyn c modulkhc nhau. T s truyn : i6c m6= 2,5: i7c m7= 3t iu kin lp c cng khong cch trc A = const

7

6

.2

.2

7

6

Z

Z

SmA

SmA

t h thc trn ta c :6

5

3

5,2

7

6

6

7 m

m

S

S

Z

Z

mt khc c88

22

84

21

80

20

60

15

4

1167

i

chn60

157 i 757 ZS Z7= 15 (rng) ; Z7

/ = 60

(rng)

m : 905

6.75.

6

77

6

m

mSS

Z

Z chn : Z9= Z9/= 45 (rng)

e . Tnh ton cho -ng truyn n : Nhm c t s

truyn :72

18

68

17

60

15168

i

Chn Z8= 18 (rng) ; Z8/ =72 (rng) Vy c :

72

188 i

f . Nhm truyn n c t s truyn:

70

35

48

24

60

30

40

20

2

1139

i

chn : Z9= 30 ; z9/=60 vy c

60

309 i

3. Tnh ton ng hc b truyn aiB truyn ai thng truyn chuyn ng t ngc n hp tc ( trc cng tc). N c u im ltruyn ng gia hai trc xa nhau, c kh nng phngqu ti, kt cu n gin, r tin. C nhc im lcng knh, gy trt khi truyn ng . . .Chn kiu ai c thit din E = 138 mm2

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ng knh bnh nh : D1= 140 280 . Chn D1= 145mm

ng knh bnh ln : D2= D1(1- )i = D1(1- )1

/

n

n cd

Trong : = 0,02 thay s c

)(268800

1460)02,01(1452 mmD

Kim tra ai theo iu kin

)/)(3530(10.60

..ma x3

/1 smvnD

V cd

ma x3 07,1110.601460.145.14,3

vV tho mn

4. Kim tra sai s vng quay v iu kin lm vic.Trong qu trnh tnh ton s rng ca cc bnh rng

do phn tchj

j

jb

ai . Mt khc khi tnh ton Zjv Zj

/

cng c nhng sai s, nu s vng quay ca trc cui( trc chnh) c th c nhng sai lch so vi s vngquay tiu chun. V vy phi kim tra sai s vngquay ca trc chnh. kim tra trc ht phi tnh s vng quay thc tnttca trc cui cng ( trc chnh) t n1 nZbngcch vit phng trnh xch ng:

dv/cj ..inn jd (v/ph)njl tc th j ( j = 122)n/cl tc ca ng c (v/ph)ivjl t s truyn ca hp tc t ng c n trcchnhdl h s trt ca ai.

Sai s vng quay tnh theo biu thc:

%100./

/

ct

ctttn

n

nn

nttl s vng quay thc t ca trc cui (v/ph)nt/cl s vng quay tiu chun ly theo bng.Sai s nphi lm trong phm vi cho php

)%1(10 nn

%6,2)%126,1(10

n

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Sau khi tnh ton v kim tra ta c bng sai s vngquay nh sau:

STT Phng Trnh Xch ng ntt nt/c n%

n1 6030

7218

6015

5521

3951985,0

2681451460 12,41 12,

5-

0,72

n2 60

30

72

18

60

15

55

21

34

56985,0

268

1451460 15,92 16

-

0,5

n3 60

30

72

18

60

15

47

29

39

51985,0

268

1451460 19,62 20

-

1,9

n4 60

30

72

18

60

15

47

29

34

56985,0

268

1451460 24,71 25

-

1,16

n5 60

30

72

18

60

15

38

38

39

51985,0

268

1451460 31,79

31,

50,92

n6 60

30

72

18

60

15

38

38

34

56985,0

268

1451460

40,05 400,12

5

n7 60

30

72

18

45

45

55

21

39

51985,0

268

1451460 49,56 50

-

0,88

n8 60

30

72

18

45

45

55

21

34

56985,0

268

1451460 62,16 63

-

1,33

n960

30

72

18

45

45

47

29

34

51985,0

268

1451460

78,47 80

-

1,91

3

n1

060

30

72

18

45

45

47

29

34

56985,0

268

1451460 98,84 100

-

1,16

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n1

160

30

72

18

45

45

38

38

39

51985,0

268

1451460

127,1

8125

1,74

4

n1

260

30

72

18

45

45

38

38

34

56985,0

268

1451460

160,1

9 1600,11

9

n1

360

30

15

21

39

51985,0

268

1451460 199,2

5200

-

0,37

5

n1

4 60

30

55

21

34

56985,0

268

1451460 246,6

6 250

-

1,336

n1

5

60

30

47

29

39

51985,0

268

1451460

314 315

-

0,31

7

n16

60

30

47

29

34

56985,0

268

1451460

395,3 400

-

1,17

5

n1

7 60

30

38

38

39

51985,0

268

1451460 508,7 500 1,74

n1

860

30

38

38

34

56985,0

268

1451460 640,7 630

0,16

9

n1

948

60

47

29

39

51985,0

268

1451460 784,7

6800

-

1,90

5

n2

0 48

60

47

29

34

56985,0

268

1451460

988,4

2

100

0

-

1,15

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8

n2

1 48

60

38

38

39

51985,0

268

1451460

1271,

8

125

0

1,74

4

n2

2 48

60

38

38

34

56985,0

268

1451460 1602

160

0

0,12

5

5 . Kim tra iu kin di tr-t ca khi bnh rng 3

bc :

Hp tc ca my thit k c mt nhm truyn

gm 3 cp tc nhm truyn II. Dovy ta b tr 3

cp tc trn mt cp bnh rng di trt ba bc

v b tr chung hai trc II v III.

Khi bnh rng 3bc c bnh ln gia, bnh b

hai bn. di trt c th nh ca cc bnh bn

khng nh ra qu vng chn ca bnh rng gia mt tr

s ln hn khe h ca nh rng C = 0,25.m

Trong : m l muyn ca rng

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iu kin di trt: Da1 2fo + 2 (fo+ C0).

Vi : fol h s chiu cao rng, fo= 1

Co l h s khe h chn rng,

Co= 0.25 .. 0,35

Chn Co= 0,35 Z5 - Z1 > 5.

Ta c: Z5= 55 , Z1= 47

Z5 - Z1= 55 47 =8 > 5

Do vy tho mn iu kin di trt ca khi

bnh rng 3 bc.

B. Thit k hp chy daoI. Nhng la chn chung

1.Chn c tnh chy dao.So vi hp tc hp chy dao c c im sau y:+ Cng sut truyn b, thng ch bng t 5 10% cngsut truyn ng chnh.+Tc lm vic thp hn so vi hp tc .Do hai nguyn nhn trn trong hp chy dao c th

dng cc c cu gim tc nhiu v hiu sut thp nh

vt me- ai c, trc vt- bnh vt, bnh rng thanhrng .Theo yu cu cn thit k th my phi ct c cc

loi ren ( Quc t, modul, ren Anh, ren pt ) do hp chy dao phi m bo t s truyn chnh xc ct c phi chnh xc . c trng ca my l dng gia cng cc chi tit my v gia cng ren ctiu chun ho . Nh vy nu t s truyn thc t cahp chy dao m c sai s so vi tnh ton th n s

nh hng trc tip ti chnh xc ca hp xe dao .

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2.Chn c cu iu chnh .Trong my tin ren vt dng c cu Noc tng lmnhm c s bo m t s truyn chnh xc, n ginh thng iu khin nhng phi ch ti cc binphp kt cu lm tng cng vng. lm nhm c s ca my tin ren vt cng c thdng nhm bnh rng di trt , lc ny cng vngkh cao nh tnh ton n bo m t s truynchnh xc c kh khn hn v h thng iu khin cnnhiu cng gt . dng lm nhm khuch i ca my tin ren vt cth dng bnh rng di trt c cu Meanra, u iml iu khin bng mt tay gt , chiu di b, xonglp t kh khn, hiu sut thp nn ngi ta t dng

hn loi bnh rng di trtC cu then ko c u im l gn, iu khin ngin song cng vng km , kh nng truyn ti km.C cu bnh rng thay th c s dng nhiu trongcc hp chy dao ca cc my chuyn mn ho dngtrong sn xut hng lot v hng khiQua phn tch trn ta la chn nhm bnh rng ditrt cho nhm c s v khuych i cho my tin ren

vt cn thit k v n c cc u nhc im sau:+ u im : C cng vng cao, cng sut truyndn ln, hiu sut truyn dn cao, ch to n gin .+ Nhc im : Kch thc hng trc ln, kch thcng knh trc lnVi u im ta chn c cu iu chnh ca hp chydao l c cu bnh rng di trt khi thay i loiren ta thay i bnh rng thay th . gim bt slng bnh rng thay th ta dng bin php o hngtruyn trong nhm c s tin ren mt sang ren Anh.

II. Thit k hp chy dao.

Hp chy dao c hai cng dng l chy dao tin renv chy dao tin trn. Nhng khi thit k ta ch yuthit k cho chy dao tin ren ( Ren tiu chun ).Nhn chung cc bc tin trn kh dy c v vy chy dao tin ren ta dng c cu vt me-ai c, cn chy dao tin trn ta dng c cu bnh rng thanhrng.

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Nu gi : tvm: l bc ca vt met : l bc ren cn cti : l t s truyn gia trc chnh v

trc vt meVy ta c phng trnh ct ren l :

1vtc i tvm= t (mm) vmt

ti

Theo cng thc c bn thit k hp chy dao l :

i = icicsigb=vmt

t

Trong : ic: L t s truyn c nh b vo xchng

ics : L t s truyn ca nhm c s

igb : L t s truyn ca nhm gp bi+ Cc bc tin hnh tnh ton : Xp bng ren.

Thit k nhm c s.

Thit k nhm gp bi.

Thit k nhm b.

Kim tra li bc ren theo yu cu.Theo yu cu k thut th my thit k ra phi ctc cc loi ren: Ren quc t, ren modul, ren Anh,ren pt.Ren quc t v ren modul gi l ren h mt .+ Ren quc t c c trng bng chiu di bc rent (mm), gc nh ren. = 60on c dng kpcht.+ Ren modul c o bng s modul m, n c dng lm

ren ca trc vt+ Ren Anh v ren pt c gi l ren h AnhRen Anh c o bng s vng ren (n) trn mt tc Anh(1 inse = 25,4mm) loi ny dng kp cht gc nh = 55o.Ren pt (p) n c o bng s modul trong mt tcAnh n c dng truyn ng .

Cc cng thc tnh bc ren ca cc loi rentiu chun:

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Loi ren S o Bc ren (mm)

Ren quc t t t

Ren modul m t = . m

Ren Anh nn

t 4,25

Ren pt pp

t .4,25

Chn theo my c trc 16K20 c :Ren quc t : t = 0,5 112Ren modul : m = 0,5 112Ren Anh : n = 56 0,25Ren Pt P = 56 0,25

Thng thng ngi ta tiu chun ho ren theo cp bil 2, ngha l nu c mt bc ren t th c mt bcren tiu chun l 2t cn cc bc ren thit k l nhnhau, c tnh cht ri rc theo cp s cng vi cng

sai no . Vy truyn dn chy dao ca my tin cnc cc nhm c bn sau:- Nhm c s: Nhm ny c trng cho tnh cht ccbc ren tiu chun lin tip nhau phn b theo cps cng.- Nhm gp bi: Nhm ny c trng cho tnh cht ccbc ren tiu chun gp i nhau, v vy t s truynca nhm ny tun theo cp s nhn vi cng bi =2tr s c th ph thuc vo vic chn t s truyn cs .- Nhm b: l nhm dng iu khin li ngtruyn tc l b li sai s ca ng truyn khi ctcc loi ren khc nhau.- Nhm o chiu: L nhm dng ct ren phi vren tri .Trong thc t ct c cc bc ren khc nhau trncng mt my th ngi ta cn dng thm b truynbnh rng thay th. Khi ct ren bc ln th dngthm b truyn bnh rng thay th v dng thm nhm

truyn khuch i.

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Nh vy phng trnh ct ren tng ng l:+ Ren quc t :

1vtc ( itt )tic icsigbtvm= t

+ Ren modul:

1vtc itt ic icsigbtvm= .m+ Ren Anh:

1vtc ( itt )t( ic)

Acsi

1igbtvm=

n

4,25

+ Ren pt:

1vtc ( itt )t( ic)

Acsi

1igbtvm=

p

.4,25

T cc phng trng trn ta thy:+ Tr s ca ren lun t l vi ics+ Cc loi ren trong cng mt h th s dng mt

t s truyn c nh (i c )+ Ren modul v ren pt dng chung mt itt+ Ren quc t v ren Anh dng chung mt ( itt)

t

1.Xp bng ren.

Vi cc bc ren nh yu cu ca my ra ta tinhnh sp xp thnh bng gm cc hng v ct sao cho:

+ Cc bc ren theo hng to thnh cp s nhn ccng bi = 2.

+ Cc bc ren theo ct to thnh cp s cng vicng sai no .C hai kiu b tr xch truyn ct cc loi ren :

+ Khng i hng ng truyn trong nhm c s

khi i h ren.+ C i hng truyn trong nhm c khi i hren. tit kim bnh rng v kch thc my ta dng mtnhm c s trong cc xch ct ren khc nhau tc l tachn phng n c i ng truyn trong nhm c skhi i h ren.sp xp bng ren theo nguyn tc sau:Vi ren quc t v ren modul ga tr ca t,m c spxp theo t l thun vi ics v igb.

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Vi ren Anh vren pt gitr ch s can,p t lthun vi icsv igb.V trong nhmc s c ing truynkhi chuyn tt ren h mtsang ren hAnh, th renh mt c bc ren tmin t ( 1,1 ) cn ren h Anh

c tr s nmin t (1,n). Vy c s xp ren theobng :

Khi sp xp bng ren cn ch :+S hng khng ln hn 10.+S ct khng ln hn 4.

a.Bng ren quc t v ren modul.

Ren tiu chun Ren khuych i

0,5 1 2 4 8 16 32 64

_ 1,25 2,5 5 10 20 40 80

0,75 1,5 3 6 12 24 48 96

_ 1,75 3,5 7 14 28 56 112

b. Bng ren Anh v ren Pit.

Ren tiu chun Ren khuych i

i ng truyn trong nhm c s

H mt H Anh

(1,1)(1,n)

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32 16 8 4 2 1 0,5 0,25

40 20 10 5 2,5 1,25 _ _

48 24 12 6 3 1,5 0.75 _

56 28 14 7 3,5 1,75 _ _

2. Thit k nhm c s.Trong nhm c s ta dng khi bnh rng di trt

m bo thc hin ng t s truyn ca nhm , ta d

nh nhm gm hai trc mi trc cha nhiu nht 4bnh rng v nhm dng bnh rng di trt c kchthc hng trc ln b truyn nh gn hp l toiu kin gim ti thiu s lng bnh rng ta dngchung mt vi bnh rng chn cc gia tr modul khcnhau v dch chnh sao cho mt bnh rng c th nkhp vi nhiu bnh rng m b truyn vn hp l.a. Xc nh t s truyn.Da vo phng trnh iu chnh nhm c s:

iCS1: iCS2: iCS3: iCS4 = tCS1: tCS2: tCS3: tCS4chn ct s (4) lm ct c s ( v n c bc renlin tip nhau) .Chn iCS2= 1 ta c :

iCS1=4/5 ; iCS2= 5/5 ; iCS3= 6/5 ; iCS4= 7/5

Trong nhm c s s dng dch chnh rng mbo khong cch trc . Tuy nhin mun m bo t struyn hon ton ng vi iCS ta c th dng nhiuloi modul khc nhau to ra nhiu s rng khc

nhau sau ta chn s rng thnh hp.b. Xc nh s rng trong mt nhm c s.V t s truyn ca hp chy dao yu cu phi chnhxc nn y ta khng th dng cch tnh BSCNN nh tnh s rng ca hp tc . y ta dng cccp bnh rng thng mdul khc nhau.Chn khong cch trc A= 63 (mm) vy c SZ = 2A/m (1)Chn modul 31m (mm) . thay ln lt cc gi tr m1=

1,5 ; m2= 1,75 ;

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m3=2 ; m4= 2,25 vo phng trnh (1) c:vi m1= 1,5 Sz= 84

m2= 1,75 Sz= 72m3=2 Sz= 63

m4= 2,25 Sz= 56Tnh ton cc gi tr khc tng t. Vy ta lp

c nh sau:

Sz m ics 4/5 5/5 6/5 7/5

84 1,5 _ _ _ _

72 1,75 32/40 36/36 _ 42/30

63 2 28/35 _ _ _

56 2,25 _ 28/28 30/25 _

T bng trn ta chn nhng cp c cng modul v chungt s v mu s ghp thnh i sao cho i ct nht 3 bnh rng li to c cc t s truynkhc nhau.

Vy ta c cc cp sau :

35

28

5

41 CSi ;

25

30

5

63 CSi

28

28

5

52 CSi ;

30

42

5

74 CSi

Ngoi ra ta cn thit k thm cc li hp bo m ts truyn i = 1 (chn tu sao cho thch hp nht).3. Thit k nhm truyn gp bi.Nhm gp bi to ra 4 t s truyn c cng bi = 2.Nh bit tr s ca nhm gp bi ph thuc vovic nhm truyn t s truyn ca nhm c s do vic tnh ton nhm gp bi phi ph hp vi khongcch trc.

1

1:

2

1:

4

1:

8

18:4:3:2:1::: 4321 gbgbgbgb iiii

a. Chn ph-ng n kt cu v ng hc.Ta c phng n b tr truyn dn Z = 4, v hp chydao t pha trc ca my nn kch thc ngang cahp cng b cng tt. Ta cho tm trc ca hai nhm

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trng nhau, nng cao tnh cng ngh ca hp gim kch thc ca hp ta dng

phng n kt cu : Z = 2 2phng n ng hc : Z = 21

I 22II

V khong cch trc ph thuc vo icsnn trnh schnh lch qu ln gia hai cp bnh rng ta phi ct s truyn thay i t t.Theo phng n ny ta chn : igb max= 1

igb min=1/8

Do vy : 11:

21:

41:

81::: 4321 gbgbgbgb iiii

Suy ra v c li cu trc v th vng quay l:

Tham kho my c trc 16k20 c :

165

52

81

1 gbi ;45

52

21

3 gbi

16

5

5

4

4

12 gbi ;

4

5

5

4

1

14 gbi

b. Tnh s rng ca cc bnh rng trong nhm.+ S rng ca nhm th nht

1

1

5

41 b

aigbI 911 ba

2

2

5

22

b

aigbI 722 ba

C BSCNN : K =63

C 163.2

18.7minm in

K

Z

a

baE

i

ii

Chn Emin= 1

2854

63.41

Z (rng) 35

54

63.51

Z (rng)

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1852

63.22

Z (rng) 45

52

63.52

Z

(rng)S rng ca nhm th II tnh tng t nh i vi

nhm I vy c :Z3= 35 (rng) ; Z4= 15 (rng)Z3

/= 28 (rng) ; Z4/= 48 (rng)

So snh s rng Z1/v Z2

/vi Z3v Z3/ta thy: Z1

/=Z3=35 . Ta ly cp bnh rng ny dng chung cho haib truyn nn.

48

15

45

18

16

5

5

2

8

11 gbi

48

15

35

28

16

5

5

4

4

12 gbi

28

35

45

18

4

5

5

2

2

13 gbi

28

35

35

28

4

5

5

4

1

14 gbi

4. Thit k nhm b.Nhm b gm cc khu c nh do cc b truyn n,bnh rng thay th v nhm o chiu to ra, trn cs thit k nhm b l phng trnh lin kt ng

hc. 1VTC ib iCS igb tvm= tTrong :

+ ib: t s truyn ca nhm b+ iCS: t s truyn ca nhm c s+ igb: t s truyn ca nhm gp bi+ tvm: bc vt me+ t : bc ren cn ct

t phng trnh trn ta c: cdttvmttcs

b ii

tii

ti

Khi tin ren trong cng bng th ip= const, dovy tnh ibta chn trc mt bc ren no trongbng ct th. Khi ny th icsiv igbiv tvm cxc nh. V ng truyn khi tin ren h mt khcvi ng truyn khi tin ren h Anh nn.

vmgbicsi

ij

bMtii

ti

1 (ct ren quc t)

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vmgbicsi

ij

bMtii

mi

.2

(ct ren mdul)

vmijgbi

csibA

tni

ii

4,251 (ct ren Anh)

vmijgbi

csibA

tPi

ii

..4,252

(ct ren Pit)

Ta thy c bn t s truyn ibkhc nhau khi tin4 loi ren khc nhau. Lc ny cng thc kt cu canhm b l : Zb= Ztt ZcTrong : Ztt: t s truyn khu thay th

Zc: t s truyn khu c nh

Nh vy ta b tr s ca xnh chy dao nh sau:

Khi ct chnh xc ta ng cc li hp M1 v M4.itt1dng tin ren quc t v ren Anh.itt2dng tin ren modul v ren Pit ( n dng khs trong cng thc tnh ton)

tnh ta c cng thc : cdttb iii . Ta ct th ren no .

+ gi s ta ct th ren quc t c t = 4(mm)c tvm=12 ; tra bng ren quc t ta c : igb=1 ;

ics=28/35vy c:

12

5

1.35

28.12

4

..

gbcsvm

biit

ti

Tham kho my c trc 16K20 chn :

28

28

45

30cdi

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12

51 cdtt ii

28

28

45

3012

5

.12

5

1

cd

tti

i

64

40

8

51 tti

+ Khi ct ren Anh theo ng truyn khc nu phitnh li cc gi tr icd

Anh sau

ct th ren Anh vi: n = 8 175,38

4,254,25

nt

Tra bng ren Anh tm c :28

35

4

5csi ;

2

1gbi

vy c :

83512

64284,25

82

1

28

3512

4,254,251

niitiii gbcsvmcdttb

33

30

34

38

45

30

4043512

64284,25

1

tt

bA

cdi

ii

Hai gi tr trn khi ct ren quc t v ren Anhthy c cp 30/40 c th dng chung c. Nn ta dngchung cp ny.

Tnh ibkhi ct ren Pit v ren modul quc t nhng ch

thay i itt.Ct th ren modul c m=4 (mm)

Tra bng c :5

4

csi ; igb= 1 ;28

28

45

30cdi

C :1

5

412

42

gbcsvm

cdttbiit

miii

36

86

73

60

141230

45452

tti

+ Khi ct ren Pt ng truyn tng t nh ren Anhva c:

33

30

34

38

45

30 cdAcdpit ii

36

86

73

60m od ulttttpit ii

Nh vy ta tnh c cc gi tr itt, ic ca nhmb, cc s rng Z chn trn c s ph hp vi tiu

chun chung v thch hp vi my thit k.

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5. Kim tra sai s b-c ren.Trong khi tnh ton c mt s bnh rng chn s

rng gn ng nu xut hin cc sai s bc ren do phi kim tra sai s bc ren bng cch ct th ccren ri sau em so snh vi bc ren ct c vibc ren tiu chun.a. Ct ren quc t.Ct th ren vi bc ren t = 5 (mm)

Vi : tvm= 6 ; ics=5/5 ; igb= 1 ; ic= 30/45=2/3

8

5

64

401 tti

Phng trnh ct ren: 1VTCicicsigbitt1tvm= tThay s c:

5128

51

5

5

3

21

khng c sai s :b. Ct th ren modul.

ct th ren vi: m =5 (mm) ; igb=1 ; ics=1

3

2

45

30cdi ;

36

86

73

602 tti ; tvm=12

phng trnh ct ren: 1VTCicicsigbitt2tvm= m

thay s c:

70716,1551236

86

73

6011

3

21

c : t = m.= 5.3.14 = 15,70726Sai s bc ren: t = 0,0002

Trng hp ny sai s khng ng k c th chp nhnc.c. Ct ren Anh.

Ct th vi n = 5 t = 25,4/5 = 5,08 (mm)C tvm= 12 ; ics=5/5 ; igb= 1 ; itt1= 40/64 = 5/8

33

30

34

38

45

30Acdi

phng trnh ct ren l: ttiii

iVTCvmttgb

cs

A

cd 1

11

thay s c:

080213964,512

64

401

5

5

33

30

34

38

45

301

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vy sai s bc ren l :t = 5,0802139 5,08 = 0,0002139

(mm)Sai s c th chp nhn c.d. Ct th ren Pit.

Ct th vi p =5 95929068,155

.4,25

t

C : ics= 5/5 = 1 ; igb= 1 ; tvm=12

36

86

73

602 tti ;

33

30

34

38

45

30cdi

phng trnh ct ren : ttiii

iVTC vmttgbcs

p

cd 21

1

thay s :96975875,1512

36

86

73

6011

33

30

34

38

45

301

vy sai s bc ren l :t = 15,95975875 15,95929068 =

0,000461947c sai s bc ren.

Thnh lp s ng hp chy dao nh hnh v: hnh 7

C. Thit k hp xe dao.

Trong hp xe dao ch cha nhng b truyn n v,vy quy lut phn b lng chy dao S cng nh quylut phn b bc ren. Lc chng ta dng hp chydao va tin ren va tin trn. V vy phm vi iu

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chnh lng chy dao v bc ren phi ging nhau ;Rs=RtDa theo my c trc 16K20 ta c: Sd = ( 0,05 2,8 ) (mm)Cn c vo ta sp xp bng lng chy dao S tngt nh bng ren quc t nh sau:

Lng chy dao tin trn (S mm)

0,05 0,1 0,2 0,4 0,8 1,6

0,06 0,125 0,25 0,3 1 2

0,075 0,15 0,3 0,6 1,2 2,4

0,09 0,175 0,35 0,7 1,4 2,8

thc hin chuyn ng chy dao no th dngxnh ct ren tng t. Ly theo xch ct ren tmin, thc hin lng chy dao Sminth lng chy dao khccng m bo khi ct ren tmin= 0,05 (mm)Ta c phng trnh:

1VTCicitticsigbtvm= tmin (1)khi tin trn vi Smin ta c phng trnh.

1VTCicittigbicsimZ = Smin (2)Trong m v Z l modul v s rng ca bnh rng nkhp vi thanh rng chy dao dc.Tham kho my c trc 16K20 ta c : Z= 10 , m= 3vi : il t s truyn tng cng cn to ra hp xedao.

14,315

6,0

31014,45,0

05,012

...

.

min

m in

zmt

Sti vm

b tr kt cu khng gian hp l m bo tho mniu kin v mt ng lc hc, ng hc ca cc btruyn. Tham kho my chun v cc my tin ren vtvn nng khc ta c xch ni t trc vt me ngang nh

sau:

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- Li hp M5(m). Trc vt me:30

32

32

32

32

30

35

28

39

24

40

23 qua trc vt bnh vt 4/21

chy dao dc:Zm..

66

17

41

36

Sd

chy dao ngang: 516

29

29

55

55

34

36

36 Sn

- Lng chy dao ngang Sn= Sd / 2Da vo my chun ta dng kt cu cc cp bnh rng

theo my 16K20.

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Phn Iv:thit k ng lc hc my

I. Xc nh ch lm vic ca my.

Ch lm vic ca my bao gm ch ct gt,ch bi trn lm lnh, n ton . . . Mt my mi

thit k, ch to song phi quy nh r rng vch lm vic ca my trc khi a vo sn xut.Trong phm vi my c thit k y xc nh ch ct gt ti hn ca my lm c s tnh ton ng lchc ca my. Ch ct gt bao gm: Ch ct gtcc i, ch ct gt tnh ton, ch ct gt thmy.1.Ch ct gt cc i.Theo kinh nhim tnh ton S, v, t bng cc cng thc

sau:)(579,4280.7,0. 33 m ax1m ax mmDCt

trong : C= 0,7 ( chn vt liu gia cng bng thp)

ma xm in )4

1

2

1( tt chn : )(145,1579,4

4

1

4

1m axm in mmtt

)/()7

1

3

1( m axma x vmmtS chn )/(53,1579,4

3

1

3

1ma xma x vmmtS

ma xmin )

10

1

5

1( SS chn )/(153,053,1

10

1

10

1ma xmin vmmSS

S dng ch ct gt cc i s dn n ton b chitit my lm vic vi ti trng cc i. Thc tincho thy khng bao gi my lm vic ht ti trng. bng, chnh xc, trnh ngh nghip v nhngyu t khc l nguyn nhn hn ch kh nng s dngca my. tnh ton hp l hn c th s dng ch ct gt tnh ton.2.Ch ct gt tnh ton.

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Chui s vng quay n ca my bin i t nmin nnmaxvi z cp tc khc nhau. Chuyn lng chy daoS bin i t Sminn Smax, z cp khc nhau. Ticc tr s nmin, Sminmy lm vic vi Mxmax. V vyphi xc nh tr s ntnh, Stnhtheo cng thc

4

min

max

minn

nnn

t

Ta c bng sau:

Cc ch ct khc (s,t) v ng knh chi tit giacng (d) chn theo ch cng ngh bo m my lmvic ht cng sut.3.Ch th my.

Ch th my l do ngi thit k hoc nh snxut quy nh. Trc khi a my mi vo sn xut,nh my ch to phi nghim thu my theo mt ch

kim nghim nht nh.Th my theo ch ct gt vi mc ch kim tra

my lm vic c n nh v bn hay khng.Nu cho my lm vic ngoi phm vi cho php th chitit s nhanh hng do ta chn ch th tnhsc bn.Th vi chi tit D = 115(mm). cng HB207, dao P18, ch ct chn:

n = 400 (v/ph) ; S =1,4(mm/v) ; t = 4(mm)

)/(5,1441000

400.115.14,31000

.. phmnDV

Khi thit k cho php a Pmax, Mmaxqu ti n 25%trong thi gian cho php.

II. Tnh lc ct.Lc ct gm 3 thnh phn Px, Py, Pz l cc

thnh phn lc theo phng x,y,z v c xc nhtheo cng thc bng 9[1].

Px*

= Cpx.t

*xpx

.S

*ypx

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Py*=Cpy.t

*xpy.S*ypyP*z=Cpz.t

*xpz.S*ypzVi t*= 4,226 (mm) ; S*= 1,39 (mm/v)Cpx=650 xpx=1,2 ypx=0,65Cpy=1250 xpy=0,9 ypy=0,75Cpz=2000 xpz=1 ypz=0,75Thay s tm c:

Px*= 650.(4,579)1,2.(1,532)0,65 = 5324,22 (N)

Py*= 1250.(4,579)0,9.(1,532)0,75 = 6769,36 (N)

Pz*= 2000.(4,579)1.(1,532)0,75 =12610,86 (N)

III. Tnh cng sut ng c.Hin nay tnh chnh xc cng sut ng c in l

mt vn kh khn v kh xc nh ng iu kin

lm vic v hiu sut ca my, iu kin ch to cngnh nhng nh hng khc. C hai cch thng dng tnh cng sut ng c in. Xc nh cng sut ngc in gn ng theo hiu sut tng v tnh chnhxc song khi ch to my, bng thc nghim c th oc cng sut ng c ti cc s vng quay v ch ct gt khc nhau.1.Xc nh cng sut truyn dn chnh xc.Cng sut ng c gm c: Nc= Nc+ N0+ NpTrong : Nc: cng sut ct.

N0: cng sut chy khng.Np: cng sut ph tiu hao do hiu sut v

do nhng nguyn nhn nh hng ti s lm vic camy.a.Xc nh cng sut ct.

Cng sut ct c xc nh theo biu thc.

310.60

.vpN zc

Theo phn II c Nc= 5,933(kw)b.Xc nh cng sut chy khng.

)(10

60 tcitb

m nknd

kN

Trong : km: h s ph thuc cht lng ch to ccchi tit v iu kin bi trn, thng ly k = 3 6 ,ly k = 5

dtb: ng knh trung bnh ccngng trc ca my tr trc chnh.

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n : tng s vng quay ca cc trctr trc chnh.

ki : h s ph thuc trc chnhlp ln hay trt l ln ki= 1,5

ntc: s vng quay ca trc chnh

ti xc nh nghim.Ni, nil cng sut v s vng quay trc i

Ni= N/c.i (kw)

I= ai.= 0,97.0,99 = 0,9603

II=I.br= 0,9603.0,99.0,98 = 0,932

III= II.br= 0,932.0,99.0,98 = 0,904IV= III.br= 0,904.0,99.0,98 = 0,88

V= IV.br= 0,88.0,99.0,98 = 0,85VI= V.br= 0,826.0,99.0,98 = 0,826

NI= Nc.I= 10.0,9603 = 9,603 (Kw); n1= 800(v/ph)

NII= Nc.II= 10.0,932 = 9,32(Kw); n1= 1057(v/ph)

NIII= Nc.III= 10.0,904 = 9,04(Kw) ; n1= 532(v/ph)

NIV= Nc.IV= 10.0,88 = 8, 8(Kw) ; n1= 188 (v/ph)

NV= Nc.V= 10.0,85 = 8,5 (Kw) ; n1= 50 (v/ph)

NVI= Nc.VI= 10.0,826 = 8,2 6(Kw) ; n1= 42(v/ph)

Tnh m men xon trn cc trc ca hp tc .

n

NM i

x .10.55,9 6max (N.mm)

Ta c :

5,116987800

8,9.10.55,9 6 xIM (N.mm)

85444

1057

457,9.10.55,9 6 xIIM (N.mm)

163822532

126,9.10.55,9

6 xIIIM (N.mm)

447377188

807,8.10.55,9 6 xIVM (N.mm)

162311850

4498,8.10.55,9 6 xVM (N.mm)

1864524

42

2,8.10.55,9

6 xVIM (N.mm)

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)(.3 mmn

NCd

t

iSB

C: h s : C = 110 160 Chn C = 120

5VIVIIIIII

TB

dddddd

Thay s v tnh ton ta c bng sau:Bng ng lc hc my

Trcnmax

(v/ph)

Nmin

(v/ph)

Nt

(v/ph)

Ntrc

(Kw)

Mxmax

(N.mm)

dSB

(mm)

I 800 800 800 9,603

116987,

5 30

II 1250 1000 1057 9,32 854444 25

III 1250 400 532 9,04 163822 35

IV 1250 100 188 8,8 447377 45

V 315 25 50 8,5 1623118 65

VI 1600 12,5 42 8,26 1864524 70

Vy c:

)(415

7045352530mmdTB

KWN 55,0425,1262710

415

60

c.Tnh cng sut ph.Cng sut ph c tnh theo biu

thc: kkp ii

KNN

1

1

:k Hiu sut cc b truyn cng loic 9 b truyn

bnh rng v 1b truyn aiik:s lng b truyn cng loi

Np=Nc 91- 0,98+11- 0,97= 0.21Nc KWd.Tmh cng sut chy dao.

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Ns=81.910612 4

cd

sVQ

(KW)

Trong : c : h s chy dao c = 0,15 0,2 ly c= 0,15

Q : lc chy dao c tnh theo cng thc:Q = kpx+ f(P8+ G)Trong : k : h s tng lc ma st ly k = 1,5

f= 0,1 h s ma st gia sng trt v bngmy

G : trng lng bn my : G = 200 (kg)Px , Pz: lc ma stPx= 5324,22 (N) = 532,422 (KG)Pz= 12610,86 (N) = 1261,086 (KG)

Thay s c:Q = 1,5.532,422+ 0,1(1261,086 + 200) =944,7 (KG)VSl vn tc chy dao tnh theo cng thc

VS= S.nTheo ch ct th th : n = 400 (v/ph) ; S = 1,532

VS= 1,532.400 = 61,28 (m/ph)

Thay s c : )(63,015,0.10000.612

28,61.7,944kwNS

Nh vy cng sut ng c cn thit l :N/c = NC + N0 + NP + NS = 5,933 + 0,55 +

0,21.Nc + 0,63

)(68,863,021,01

55,0933,5/ kwN cd

Vy chn ng c N = 10 (kw) ; n = 1460 (v/ph) khiu DK32-4

2.Tnh cng sut chy dao nhanh. to iu kin thun li khi gia cng , gim

thi gian ph dn ti tng nng sut gia cng. Ta btr thm mt ng c khc chy dao nhanh.Theo kinh nhim vn tc chy dao c khng ch : 2

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