ĐHĐN.công Nghệ Cán Và Thiết Kế Lỗ Hình Trục Cán - Lưu Đức Hòa, 229 Trang

7/24/2019 HN.cng Ngh Cn V Thit K L Hnh Trc Cn - Lu c Ha, 229 Trang

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Gio trnh: Cng ngh cn v thit k l hnh trc cn

Trng i hc Bch khoa nng - 2005

1

Chng 1

Tng quan v qu trnh

sn xut cn

1.1. Sn phm cn

Sn phm cn c s dng rt rng ri trong tt c cc ngnh kinh t qucdn nh: ngnh ch to my, cu ng, cng nghip t, my in, xy dng,quc phng... bao gm kim loi en v kim loi mu. Sn phm cn c th phnloi theo thnh phn ho hc, theo cng dng ca sn phm, theo vt liu... Tuynhin, ch yu ngi ta phn loi da vo hnh dng, tit din ngang ca sn phmv chng c chia thnh 4 loi chnh sau:

1.1.1. Thp hnh

L loi thp a hnh c s dng rt nhiu trong ngnh Ch to my, xydng, cu ng... v c phn thnh 2 nhm:

a/ Thp hnh c tit din n gin

Bao gm thp c tit din trn, vung, ch nht, dt, lc lng, tam gic, gc..

1Thp trn c ng knh = 8 200 mm, c khi n 350 mm.

2Thp dy c ng knh = 5 9 mm v c gi l dy thp, sn phmc cun thnh tng cun.

3Thp vung c cnh a = 5 250 mm.

4Thp dt c cnh ca tit din: h x b = (4 60) x (12 200) mm2.5Thp tam gic c 2 loi: cnh u v khng u:

- Loi cnh u: (20 x20 x 20) (200 x 200 x 200).

- Loi cnh khng u: (30 x 20 x 20) x (200 x 150 x 150)

b) Thp hnh c tit din phc tp: l cc loi thp c hnh ch I, U, T,thp ng ray, thp hnh c bit.

H.1.1. Cc loi thp hnh n gin.

H.1.2. Cc loi thp hnh phc tp


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1.1.2. Thp tm

c ng dng nhiu trong cc ngnh ch to tu thu, t, my ko, chto my bay, trong ngy dn dng. Chng c chia thnh 3 nhm:

a/ Thp tm dy:S = 4 60 mm; B = 600 5.000 mm; L = 4000 12.000 mm

b/ Thp tm mng:S = 0,2 4 mm; B = 600 2.200 mm.

c/ Thp tm rt mng (thp l cun): S = 0,001 0,2 mm; B = 200

1.500 mm; L = 4000 60.000 mm.

1.1.3. Thp ng

c s dng nhiu trong cc ngng cng nghip du kh, thu li, xydng... Chng c chia thnh 2 nhm:

a/ ng khng hn: l loi ng c cn ra t phi thi ban u c ng

knh = 200 350 mm; chiu di L = 2.000 4.000 mm.b/ ng cn c hn:c ch to bng cch cun tm thnh ng sau cn

hn gip mi vi nhau. Loi ny ng knh t n 4.000 8.000 mm; chiudy t n 14 mm.

1.1.4. Thp c hnh dng c bit

Thp c hnh dng c bit c cn theo phng php c bit: cn bi, cnbnh xe la, cn v t v cc loi c tit din thay i theo chu k.

H.1.3. Mt s loi sn phm cn c bit


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1.2. my cn

1.2.1. Cc b phn chnh ca my cn

My cn gm 3 b phn chnh dng thc hin qu trnh cng ngh cn.a/ Gi cn:l ni tin hnh qu trnh cn bao gm: cc trc cn, gi,

trc cn, h thng nng h trc, h thng cn bng trc,thn my, h thngdn phi, c cu lt tr phi ...

b/ H thng truyn ng: l ni truyn mmen cho trc cn, bao gm hpgim tc, khp ni, trc ni, bnh , hp phn lc.

c/ Ngun nng lng: l ni cung cp nng lng cho my, thng dngcc loi ng c in mt chiu v xoay chiu hoc cc my pht in.

1.2.2. Phn loi my cn

Cc loi my cn c phn loi theo cng dng, theo s lng v phngphp b tr trc cn, theo v tr trc cn.

a/ Phn loi theo cng dng:

1My cn ph: dng cn ph t thi thp c gm c my cn phi thiBlumin v my cn phi tm Slabin.

2My cn phi: t sau my cn ph v cung cp phi cho my cn hnh v

my cn khc.

H.1.4. S my cnI- nguin ng lc; II- H thng truyn ng; III- Gi cn

1: Trc cn; 2: Nn gi cn; 3: Trc truyn; 4: Khp ni trc truyn;5: Thn gi cn; 6: Bnh rng ch V; 7: Khp ni trc; 8:Gi cn; 9:

Hp phn lc; 10: Hp gim tc; 11: Khp ni; 12: ng c in


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3My cn hnh c ln: gm c my cn ray-dm v my cn hnh c ln.4My cn hnh c trung.5My cn hnh c nh (bao gm c my cn dy thp).6My cn tm (cn nng v cn ngui).

7My cn ng.8My cn c bit.

b/ Phn loi theo cch b tr gi cn

1My c mt gi cn (my cn n a): loi ny ch yu l my cn phithi Blumin hoc my cn phi 2 hoc 3 trc.

2My cn b tr mt hng (b) c b tr nhiu l hnh hn.3My cn b tr 2 hay nhiu hng (c, d) c u im l c th tng dn tc

cn cc gi sau cng vi s tng chiu di ca vt cn.4My cn bn lin tc (e): nhm gi cn th c b tr lin tc, nhm gi

cn tinh c b tr theo hng. Loi ny thng dng khi cn thp hnh c nh.5My cn lin tc (f): cc gi cn c b tr lin tc , mi gi ch thc

hin mt ln cn. y l loi my c hiu sut rt cao v ngy cng c s dngrng ri. B truyn ng ca my c th tp trung, tng nhm hay ring l.

Trong my cn lin tc phi lun lun m bo mi quan h:F1.v1= F2.v2= F3.v3= F4.v4.... = Fn.vn; trong F v v l tit din cavt cn v vn tc cn ca cc gi cn tng ng.

c) Phn loi theo s lng v s b tr trc cn

1My cn 2 trc o chiu: sau mt ln cn th chiu quay ca trc lic quay ngc li. Loi ny thng dng khi cn ph, cn phi, cn tm dy.

2My cn 2 trc khng o chiu: dng trong cn lin tc, cn tm mng.3My cn 3 trc: c loi 3 trc cn c ng knh bng nhau v loi 3 trc

th 2 trc bng nhau cn trc gia nh hn gi l my cn Layma.4My cn 4 trc: gm 2 trc nh lm vic v 2 trc ln dn ng c

a b.

d

c

e

f

H.1.5- Phn loi my cn theo cch b tr gi cna-my cn n, b-my cn mt hng, c-my cn hai cp, d-my cn nhiu cp,

e-my cn bn lin tc, f-my cn lin tc.


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dng nhiu khi cn tm nng v ngui.

5My cn nhiu trc: Dng cn ra cc loi thp tm mng v cc mng.My c 6 trc, 12 trc, 20 trc v.v... c nhng my ng knh cng tc nh

n 3,5 mm cn ra thp mng n 0,001 mm.6My cn hnh tinh: Loi ny c nhiu trc nh ta vo 2 trc to lm

bin dng kim loi. My ny c cng dng l cn ra thnh phm c chiu dy rtmng t phi dy; Mi mt cp trc nh sau mi ln quay lm chiu dy vt cnmng hn mt t. Vt cn i qua nhiu cp trc nh th chiu dy mng i rt

nhiu. Phi ban u c kch thc dy S = 50 125 mm, sau khi qua my cn hnh

tinh th chiu dy sn phm c th t ti 1 2 mm.

7 My cn vn nng: loi ny trc cn va b tr thng ng va nmngang. My dng khi cn dm ch I, my cn phi tm ...

8My cn trc nghing: dng khi cn ng khng hn v my p u ng

H.1.6. Cc loi gi cna: Gi cn 2 trc; b: gi cn 3 trc; c: Gi cn 3 trc lauta; d: Gi cn 4 trc

H.1.7. S my cn hnh tinh1: L nung lin tc; 2: Trc cn ph (ch ng); 3: My dn phi

(dn hng); 4: Trc cn hnh tinh; 5: Trc ta; 6: Trc l sn phm.


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1.3. Quy trnh chung ca qu trnh sn xut cn

Quy trnh cng ngh sn xut cn ph thuc vo nhiu yu t: hnh dng snphm, mc thp, iu kin k thut v nhng c trng ring ca my cn; ngoi racn ph thuc vo trng lng ca thi thp c, thit b hin c ca phn xngcn v.v...

1.3.1. Quy trnh cng ngh cn thp cc bon v thp hp kim thp

a/ S cng ngh hnh 1.8a:Dng cho quy trnh cng ngh cn thp hnhc ln, cn phi tm v phi thi. Theo s ny my cn ph v my cn phi

tm, phi thi phi c ng knh trc cn D = 1,100 1.150 mm; nng sut cn

rt ln n trn 2,5 triu tn/nm. Thi c c trng lng G = 4,5 10 tn, c khi

t ti 15 20 tn. Khi cn phi tng nhit 2 3 ln.

b/ S cng ngh hnh 1.8b:Dng cho quy trnh cng ngh cn thp hnhtrung bnh. Cng c th cn trn my cn ph hoc cn phi c ng knh trc D =

650 900 mm. Thi c trng lng nh. Khi cn phi tng nhit 2 3 ln.

Thi c

Nung nng thi c

Lm iu nhit

Cn ph hoc cn phitm

Cn trn my cn lintc

Lm ngui sn phm

Kim tra, tinh chnh

Thnh phm

Lm ngui

Ct u rt, lm sch bavia

Nung thi c

Cn phi

Ct, lm ngui,kim tra, lm sch

Nung phi

Cn ra sn hm

Lm ngui, tinh chnh

Kim tra, lm sch

Thnh hm

a/

b/

c/

H.1.8. S cng ngh cn thp cc bon v hp kim thp.


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c/ S cng ngh hnh 1.6c:Dng cho quy trnh cng ngh cn thp hnhc nh. Quy trnh ny ch c 1 ln nung phi, qu trnh sn xut ngn hn. Cc mycn c b tr hng. Tuy nhin cht lng sn phm khng cao.

1.3.2. Quy trnh cng ngh cn thp hp kim

Trong qu trnh cn thp hp kim, c mt s cng on trung gian v saukhi qua mt s ln cn b mt ca thp hp kim b bin cng ln cn phi lmmm kim loi li, gim ni lc, lm thnh phn ho hc ca cc nguyn t hp kimv t chc ht u.

Thi c

Thi c trng thi nng

Tng nhit trong l ging

Cn ph p rn

Ct phi, lm ngui

Lm ngui thi c

(nu cn)

Kim tra, lm sch

Nung thi c

Cn ph p rn

Ct phi, lm ngui

T ra axt (tm th c)

(nu cn)

Kim tra, lm s ch

Nun hi

Cn thnh sn hm

Ct, lm ngui

Nhit luyn, ty ra axt

Kim tra, lm sch

Thnh hmHnh 1.9. Quy trnh cng ngh cn thp hp kim.


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thi thp c thp ccbon c khi lng t 10 72 tn dng cho my cn phi thi;

my cn phi tm c thi c nng 40 45 tn. i vi phi tm dy th thi c

c khi n 60 100 tn.

1.4.2. Cc khuyt tt ca thi c

Khuyt tt ca thi c nh hng ln n cht lng ca sn phm cn, nl nguyn nhn gy nn ph phm v th phm. Ngoi ra khuyt tt thi c cnlm hao mn thit b v nhng hng hc trong qu trnh cng ngh. Mt s khuyttt thng gp l:

a/ Thin tch

Thin tch l s khng ng nht v thnh phn ho hc trong thi c khi

ng c. Khi ngui, thi c ngui t ngoi vo trong v vy mt ngoi v y baogi tp cht cng t hn cc ni khc sinh ra thin tch vng.

Thi c ln th thin tch nhiu, thp hp kim th d xy ra thin tch hn.Thin tch lm cho cc phn ca thi c c thnh phn khc nhau, tnh nng khngging nhau, c, l tnh gim c khi thnh ph phm.

b/ Tp cht - phi kim loi: Trong bt k thi c no cng tn ti tp chtphi kim loi do 3 nguyn nhn sau:

- X ln ln vo kim loi.

- Do lin kt ho hc trong qu trnh nu luyn. Cc tp cht Al2O3, SiO2, MnOv.v...c to ra khi kh xy trong gu rt.

a/ b/

H.1.11a. Khun c thp si; b. Khun c thp lng


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- Do cc lin kt ca nhng xyt trn m to ra cc cht d chy.Tc hi ca tp cht phi kim loi l lm cho c, l tnh ca kim loi gim.

c) Bt kh: nguyn nhn gy ra bt kh l do kh b ln vo kim loi lng khirt vo khun.

d/ R co: do thi c khi lm ngui bn ngoi ngui trc bn trong nn khbn trong thot ra ngoi to nn.

/ Cc vt nt n: do b mt thi c khng c bng phng; cc vt ntngang l do khe h gia m gi nhit v thn thi c to ra. Cc vt nt dc sinhra khi tc lm ngui qu ln. Hin tng ny thng xy ra g mp v cnhgc l ni tp trung ng sut v ngui nhanh. Cc vt nt ln c th gy nn phphm.

e) Mng cng mt ngoi: khi nc kim loi rt vo khun vi tc v lulng ln n s bn ln v dnh vo thnh khun v ngui trc to thnh mt mngcng bm vo mt ca thi c. Cn phi lm sch mng cng ny ri mi tinhnh cn.

f) Lm co mt ngoi: khi rt kim loi lng vo khun, cc b mt thi ctip xc thnh khun ngui trc, gia ngui sau nn khi ng c chng co rtko kim loi ph trong lp v ngoi ng c to nn nhng lm co gn lp vngoi. Hin tng ny khng nhng gy ra ph phm m cn lm hhng b phn

dn h

ng v l hnh trc cn.1.4.3. iu kin k thut ca thi c

iu kin k thut ca thi c phi c gim c thng qua trn c s quynh chung ca nh nc. iu kin bao gm nhng im tng qut sau:

1Bo m thnh phn ho hc, nu luyn ng mc thp quy nh.

2Kch thc v hnh dng thi c ng bn v k thut yu cu.

3B mt thi c phi sch, khng b khuyt tt nu c khuyt tt b mt

phi nm trong phm vi cho php. Vt nt phi < 3mm; nu >3 mm th phi khsch trc khi cn.

4Lp bt kh ca thp si khng c cch b mt ca thi c > 10 mm.

5Phi ng mc thp v m nu luyn tht r rng.

6 Phi c cc chng t thng k t nu luyn, c, kim nghim, thnghim, phn tch thnh phn ho hc, mc thp v.v...

7nh gi cht lng thi c v a ti a im vn chuyn sau khi nu

luyn.


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1.4.4. Kh b cc khuyt tt ca thi c v phi trc khi cn

y l mt vic v cng quan trng v n quyt nh ti cht lng snphm cn. Nm vng cc iu kin k thut v cc khuyt tt, ta tin hnh kh bcc khuyt tt c bit l b mt bn ngoi ca phi cn.

a/ trng thi b mt thi c v phi trc khi cn

Trong gia cng kim loi bng p lc th vic nghin cu s phn b ng sutng u hay khng c mt tm quan trng to ln, n quyt nh ti nng sut vcht lng sn phm. Trong sn xut cn cng vy ng sut phn b u th snphm khng b cong vnh, nt n, b mt nhn bng v.v...ng sut phn b khngu dn n ph phm.

1Nhng nguyn nhn phn b ng sut khng u bao gm:

- Do nhit nung khng ng u.

- T chc kim loi sau kt tinh ln nhiu tp cht phi kim loi, c s thintch, c s khc nhau v c l tnh trong tng tinh th.

- Do bin dng ca kim loi khng ng u.

- Hnh dng dng c gia cng, lc ma st v.v...

2Cc khuyt tt b mt gy ra s tp trung ng sut: ta bit rng cc khuyttt b mt gy ra s tp trung ng sut thi c v phi. Ngc li s phn b ng

sut khng u l nguyn nhn gy ra cc khuyt tt b mt. V vy khi kh bkhuyt tt b mt ta ht sc ch vi nhng bin php thch hp. Nu ta khngthc hin ng th ngay trong qu trnh kh b khuyt tt v ct b 2 u thi c,cc g mp v.v...li gy ra hin tng tp trung ng sut.

b/ Bin php t chc kh b khuyt tt

Tu thuc vo quy trnh cng ngh cn m ngi ta c th dng cc binphp t chc kh b khuyt tt khc nhau. Nhn chung c cc bin php t chcsau:

- Dng my chuyn dng b tr nm trong trong dy chuyn cng ngh snxut kh b cc khuyt tt.

- Kh b khuyt tt trc khi a vo sn xut. Qu trnh ny phi tin hnh trng thi ngui.

- Kh khuyt tt phi m phi ny c cn t thi c cha c khkhuyt tt.

Ngoi cc bin php trn ngi ta cn phi da vo yu cu ca sn phm

bao gm kch thc, cht lng v cc yu cu c bit m kh khuyt tt ca thic trc khi cn v phi sa khi cn.


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i vi thp hp kim vic kh b khuyt tt li cng quan trng v cp thit.Cc bin php phi thn trng v nhiu cng on hn, c th gm cc bin phpsau:

thi c kh cc ng sut dv lm mm kim loi.

Gt o khuyt tt bng ba hi.

Ct bng kh.

Tin bc hon ton v ngoi ca thi c (vi phi trn cho cn ng).

Mi, phay b mt cho t yu cu k thut.

c/ Mt s phng php kh b khuyt tt

s b

Vi mc ch gim v kh ng sut dcng vi lm mm kim loi giacng c kh c d dng v lm gim tr khng bin dng ca kim loi khi cn.Vic kim loi cng cn thit i vi thp hp kim, thp ccbon cao, thp gi, thpkhng g v.v...

Qu trnh ph thuc vo thnh phn ho hc v kch thc ca thi c mtin hnh theo nhng biu thi gian v nhit trong cc l nung (l bung, lging). Ngoi ra khi ly thi c ra khi khun ta nn lm ngui chm trnhshin tng rn nt v xut hin cc m trng.

lm cho thnh phn ho hc ca thi c c ng u nht l nhit

(1050 11500C) th thnh phn ca kim loi phn b rt u. Nhit nn ly tng AC1tr ln hoc khong gia AC1-AC3hoc AC1-ACmtrong gin trngthi Fe-C.

Tm thc (ty bng axt)

Tm thc kim loi thng dng i vi phi cn, i khi cng dng i vithi c. Ta bit cc lp vy st bm vo b mt phi cn ch yu l Fe2O3v FeO.

Theo ti liu nghin cu th Fe2O3 chim khong (20 50)%, cn FeO chimkhong (50 85)% vy st.

lm cc lp vy st ny bong ra khi b mt kim loi ngi ta dng ccaxt H2SO4, HCL, HNO3v cc mui ca n cho tc dng vi cc lp xt to nnvy st . Ta thng dng 2 phng php tm thc:

- Tm thc bng in phn nhit bnh thng.

- Bng axt c nung nng nhit t (80 85)0C

Phng php 1 c s dng trong vic ch to cc loi thp tm trng km,trng thic v.v...trc khi sang nguyn cng m.


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Phng php th 2 c s dng rng ri trong qu trnh tm thc vi ccphn ng sau:

Fe2O3 + 3H2SO4 = Fe2(SO4)3+ 3H2O

FeO + H2SO4 = FeSO4 + H2OSau : H2SO4 + Fe = FeSO4 + H2

Kh H2cng nhiu cng tt v khi n bay ra s lm tung cc lp vy st vcc xt khc cha kp tc dng ho hc ri xung y b tm thc.

Tm thc kim loi mang ngha c hc l ch kh H 2sinh ra lm bt tungcc lp vy st bc quanh mt kim loi. Tng t nhvy khi dng kh HCl vHNO3. Kim loi sau khi c tm thc phi ra bng nc l sau ra qua ncnng. Sau khi hon thnh phi tin hnh kim tra b mt. Nu b mt cha sch

phi tin hnh x l bng phng php khc.

Lm sch bng la

Qu trnh lm sch b mt thi c v phi bng la gm 2 nguyn cngchnh:

- Nung kim loi (nung cc b).- Lm sch khuyt tt.u tin ngi ta nung ch c khuyt tt ca thi c phi t ti nhit

khong (1050 1350)0C bng kh l cc v acetylen, sau ngi ta thi O2vo,oxi tc dng vi C2H2to ra nhit lng ln nhit ny lm cho kim loi ti ch ckhuyt tt chy ra:

2C2H2+ 5O2= 4CO2+ 2H2O + Q

Di nhit t 2500 30000C v p lc kh ca cc vi phun C2H2vi O2m kim loi lng v x c to thnh c y ra b mt kim loi.

- Tc t kim loi ti nhit chy c nh hng ti nng sut lm schb mt. Vi mc ch tng tc nng chy ch khuyt tt b mt kim loi ngi

ta dng nhng thanh thp trn cacbon thp = (4 5)mm t chy v cho cc gt

kim loi nng chy ri vo ch khuyt tt nh kh ri lm sch - lm nhvynng sut tng rt nhiu.

Do vic t nng ch tin hnh cc b nhng ni c khuyt tt, nn sau khikh xong phi ngui i rt nhanh v truyn nhit vo cc vng ln cn. V nhvyng sut ph tp trung c to thnh. ng sut ph tp trung l ng sut do tc nung khng u gy ra nn hoc do kh b khuyt tt gy nn. N khc ng sut d(l ng sut cn li sau khi gia cng). Do ng sut ph tp trung sinh ra m mt sloi thp c hin tng nt sau khi lm sch bng la.

Cc nh kim loi hc nghin cu hin tng v cho ta thy rng:- Nt sau khi lm sch bng la l cso s bin i ca t chc kim loi khi


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lm ngui. Nu t chc stenic cng bn th kh nng nt cng nhiu. Lng chacacbon trong thp cng cao th sau khi lm sch tc ngui cng nhanh.

- Mun gim bt cc vt nt y th phi nung kim loi nhit cao hnquy nh bnh thng nu l kim loi dn nhit km, ngha l thp hp kim cao phi

nung nhit cao hn thp hp kim thp v thp cacbon thng. Khi lm nguin cng phi tc chm hn ci thin t chc kim loi v ng sut ph tptrung khng th vt qu gii hn bn ca n.

- Tu thuc thnh phn ca kim loi ch yu l thnh phn cacbon v crom(Cr) m ngi ta chn nhit nung kim loi trc khi lm sch bng la. Theo

thc nghim th phi nung s b phi nhit t (200 450)0C sau nung t

ti (1050 1350)0C th c th tin hnh nung bt k l no. Thp cacbon c hmlng < 0,5% v thp hp kim c thnh phn nguyn t hp kim


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. Lm sch bng ba hi

Dng ba hi lm sch khuyt tt ca phi thi c l mt phn phphin vn ang s dng rng ri.

Hnh dng ch lm sch v t s H/b > 1/6 ( ni phn cng ngh cn

thp hp kim)- Ba hi c su mi ln t 2 3 mm.- Tc c khong 0,3 m/pht.- Khi c khng c c theo vng trn ca phi v thi c, khng c

to ra cnh g vung gc trnh cc ng sut dph gy ra nt n v cc khuyttt khc.

. Lm sch bng my bo

kh khuyt tt trn b mt thi c ngi ta dngmt loi my c cu toc bit nhmy bo.

Tc ct ca my t 3 5 m/pht n ph thuc cng ca thi c vtrng thi b mt. Phng php ny nng sut khng cao m gi thnh li t.

. Lm sch bng my phay v my tin

Vi thi trn c < 300mm v trng lng t (500 550)kg c th khkhuyt tt trn my tin. Phng php ny ch dng vi thp hp kim c bit ltrong sn xut ng khng hn. V khi cn ng t thp hp kim b mt phi ng phi

c lm sch ht sc cn thn v kh b ht khuyt tt, c khi ngi ta tin bc ihn mt lp v ngoi.

Ngi ta cng c nhng loi my tin chuyn dng cho thi c tit dinvung hoc ch nht. Vi nhng thi c khng cn thit tin ht m ch khkhuyt tt tng vng th ngi ta dng my phay kh khuyt tt ti vng .

. Lm sch bng phng php mi

L phng php kh khuyt tt t tin nht. Mi ch s dng vi nhng thic qu nh khng th kh bng cc phng php c, tin, ba hi ...Mi c

tin hnh trc tip bng mi vi nguyn tc:- Thp mm th dng mi cng v ngc li thp cng th dng mi

mm ( mi cng hay mm l do s lin kt cc ht tinh th mi c chc haykhng). Ta c th gii thch nguyn tc nhsau:

- Thp cng lm cho mi chng mn. mi mm th ht d vng ra, htvng ra to thnh lp mi lm mn thp. Ngc li thp mm dng mi cngv khi mi ht b lu mn ht phi lin kt vng chc vi nhau mi lu mib vng ra.

- Kch thc mi ph thuc vo kch thc ca phi v s vng quay catrc mi. i vi thp cacbon cao v thp hp kim c trng lng v kch thc


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16

ca thi c nh th mi c dng c = (400 500)mm, rng (chiu dy) t

(60 80)mm.- Tc quay ca mi ph thuc vo lin kt ca vt liu. V d cht

lin kt l gm th tc t (25 30)m/s. Nu lin kt l pa-k-lit th tc t t

(35 45)m/s.S c xt do lc ma st sinh ra khi mi rt ln dn n phi c nhit ln.

Vi nhit cao y b mt phi c th b chy v nt n. V vy phi chn mic vt liu ph hp vi mc thp, tc quay, cng p mi ln b mt kimloi. Khi mi sinh ra bi nhiu v vy cn trnh bi cho cng nhn khi thao tc

1.5. Nung phi khi cn

1.5.1. Mc ch ca vic nung nng

Nung nng kim loi trc khi cn ch yu l gim lc chng bindng, tng tnh do cho kim loi d gia cng. Ngoi ra, nung phi cn lmgim lc cn, gim tiu hao nng lng, tng tui th ca trc cn v thit b,tng nng sut, tng cht lng sn phm.

1.5.2. Mt s vn xy ra khi nung

a/ Nt n: hin tng nt n xut hin bn ngoi hoc bn trong kim loi.

Nguyn nhn:Do ng sut nhit sinh ra v s nung khng u, tc nung

khng hp l v.v...ng sut nhit ny cng vi ng sut dsn c ca phi (cn,c) khi vt qua gii hn bn ca kim loi s gy ra nt n. (i vi thp thng

xy ra nt n t0


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17

Fe3C + CO2= 3Fe + 2COFe3C + H2O = 3Fe + CO + H2Fe3C + 2H2 = 3Fe + CH4

Tc dng mnh nht l H2O ri n CO2, O2, H2...

Qu trnh mt C ngc vi qu trnh xy ho v xy ra trn b mt kim loicng mt lc vi xy ho. Tc ca hai qu trnh khc nhau. Bt u nung tc mt C nhanh sau gim dn, cn tc xy ho th ngc li. Khi tc xy holn hn tc mt C th lp mt C gim i. Hp l nht l cn to nn lp xy ho

mnh hn mt C . Lp mt C bt u pht trin khi t0 = 6008000C v tng khinhit tng. Lng mt C tng khi thi gian tng nhng tc mt C gim.

gim s mt C c th dng cht sn ph ln b mt vt nung. Hin nayhay dng cht sn sau y ho vi nc hoc vi cn tyl:60%SiO2+ 15%Al2O3+ 11,2%CaO + 4,4%MgO +5%(K2O+N2O) + 0,8%Fe2O3.

d/ Hin tng qu nhit: Nu nhit nung qu cao th ht stenit cng lnlm cho tnh do ca kim loi gim nhiu, c th to nn nt n khi gia cng hocgim tnh do ca chi tit sau ny. i vi thp ccbon nhit qu nhit d i

ng c khong 1500tr ln (t0qn>toc- 150

0C). Nu thi gian gi nhit qunhit cng lu ht stenit cng ln th kim loi cng km do. Hin tng ny c

khc phc bng phng php .V d: Thp ccbon 750 9000C, nhng vithp hp kim th rt kh khn.

/ Hin tng chy: Khi kim loi nung trn nhit qu nhit (gn ngc) vt nung b ph hu tinh gii ca cc ht do vng tinh gii b xy ho mnhlit. Kt qu lm mt tnh lin tc ca kim loi, dn n ph hu hon ton bnv do ca kim loi. Khi chy kim loi s pht sng v c nhiu tia la bn ra.Sau khi b chy th kim loi b vt i hoc cht ra tng khc nu li.

1.5.3. Ch nung kim loi

a/ Chn khong nhit nung

%c

t

0

C t0

Ctmax tmax

tmin tmin%c

O Oa) b)

H.1.13. Gin chn khong nhit gia cng i vi thp cc bon

a) Gin l thuyt b) Gin thc t

0,8 1,7 0,8 2,1


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18

Yu cu:m bo kim loi do nht. Kim loi bin dng tt v hao ph tnht. Cht lng vt nung phi c bo m. i vi thp ccbon da trn gin Fe-C chn khong nhit GCAL.

Trong thc t c th chn

nhit nung khi gia cng p lctheo phm vi nhit nhhnh sau.

Trong sn xut xc nhkhong nhit ca cc kim loi vhp kim thng dng bng. i vicng nhn trong iu kin thiudng c o c th xc nh nhit theo mu sc khi nung.

V d: i vi thp khi nung

mu s sng dn t mu xm(5000c) n sng trng (12500c).

b/ Thi gian nung

Ch nung hp l cn m bo nung kim loi n nhit cn thit trongmt thi gian cho php nh nht. Nhit phi phn b u trn ton b tit dinphi. Qu trnh nung c 3 hnh thc: i lu (khi t0< 6000c th i lu l ch yu),bc x (khi t0> 6000c th bc x l ch yu), truyn nhit (c qu trnh nung). Thigian nung t nhit bnh thng n nhit ban u gia cng c th chia thnh2 giai on:

Giai on nhit thp: Thi gian nung giai on ny cn di, tc nungchm, nu khng kim loi d nt n hoc bin dng. Tc nung ny gi l: tc nung cho php v c th tnh theo cng thc:

K =5 6

3

, .

.

r (oc/gi).

K - Tc nung - Gii hn bn.. - H s dn nhit.

E - Modul n hi. - H s n di. r - Bn knh phi hnh tr.

K ch yu ph thuc vo cn cc thng s kia khng ng k.Giai on nhit cao:(850oc n nhit bt u gia cng)Khi nhit vt nung trn 850oc tnh do tng, tc xy ho mnh. Tc

nung giai on ny khng ph thuc nhiu vo h s dn nhit na, v th c thtng nhanh tc nung nhm tng nng sut nung, gim lng oxy ho v chyccbon, hn ch s ln ln ca cc ht kim loi, gim hao ph nhin liu.vv...

Tc nung ca giai on ny gi l tc nung k thut, n ph thucvo cch xp phi, di phi v.v...

1350

1100

800

Vng GCAL

Vng qunhit

Vng chy

vng bin cng

% cacbon0,8 1,1

00c

H.1.14. Phm vi nhit gc. p lc


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19

Bng 1.1. Nhit kt thc v nhit chy ca mt s thpSTT Mc thp v thnh phn ho hc Nhit kt thc nung

(0C)

Nhit

chy

1

23456789

101112

Thp cc bon c %C = 1,5

Thp cc bon c %C = 1,1Thp cc bon c %C = 0,9Thp cc bon c %C = 0,8Thp cc bon c %C = 0,5Thp cc bon c %C = 0,2Thp cc bon c %C = 0,1Thp l xo C60, C65Thp hp kim niken (Ni = 3 %)Thp hp kim Ni - CrThp hp kim Cr - VanadiThp gi

1050

10801120118012501320135012501250125012501280

1140

11801280128013501470149013501370137013501380

1.5.4. Thit b nung

Trong sn xut cn, cc loi l nung c s dng nhiu l cc l nung v hnung (l ging), c th nhsau:

- Nung thi c cho my cn ph th dng l ging.- Nung phi thi cho my cn Ray-dm th dng l lin tc v l bung

- Nung phi v thi c cho my cn hnh v cn ng th dng l lin tc, l

c y xoay.- Nung phi, thi c thp hp kim th dng l lin tc c bung t s b

H.1.13. Cu to l nung lin tc


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20

v l bung.- Nung thi c, phi cho my cn tm th dng l bung, l lin tc v l

c y di ng.Vic b tr nung thi c v phi cn l no cn ph thuc vo my cn v

sn phm. V vy cn phi chn l nung cho hp l m bo tng nng sutnung v nhn c tt nht vi gi thnh h.

Mt s nhn xt:

- L ging ch nung thi c v phi c kch thc ln, n ph hp vi cngngh cn th, cn ph v cn khng lin tc.

- L lin tc thch hp khi nung cho tt c cc loi phi v thi c khc nhau,n ph hp vi cn lin tc nhcn thp hnh, thp tm v.v...

- m bo c nhit v nng sut nung chnh xc v cao, cn phi trangb cc dng c o nhit, my y phi, my lt phi v cc thit b khc t ng

ho qu trnh nung.

H. 1.14. L ging hon nhit1: bung tch nhit nung khng kh; 2: bung tch nhit nung kh; 3:xe ch x; 4: khu

vc t thit b o v kim tra; 5: vng cha my pht in; 6: ng ra ng khi;7: ni lp cp nhit xung; 8: van o kh -khng kh


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21

Chng 2

Cc khi nim c bnca qu trnh cng ngh cn thp

2.1. L hnh trc cn

2.1.1. Khi nim v l hnh trc cn

Tt c cc loi thp hnh c tit din n gin nhthp trn, vung, ch nhtv.v...v c bin dng phc tp nhthp ch I, U, thp ng ray v.v... u c cntrn cc trc c to cc rnh c bin dng tng ng. Bin dng rnh ca 2hay 3, 4 trc to thnh mt bin dng calip gi l l hnh trc cn.

Trong cng ngh cn thp tm th qu trnh cn c tin hnh trn trc

khng to rnh (trc phng) song vic xc nh ch p, phn b lng p v tnhton xc nh bin dng trc cn t c sn phm c chiu dy ng u cngc gi l thit k l hnh trc cn.

Ni chung trn mi l hnh ch cn mt ln, song cng c th cn nhiu lnbng cch thay i khe h gia 2 trc cn.

2.1.2. Cc thng s c bn ca mt l hnh

Thng s c bn ca l hnh chnh l cc i lng cn tnh ton to nnl hnh, n tu thuc vo hnh dng cc l hnh:

a) L hnh hp ch nht

h - chiu cao l hnhb - chiu rng y l hnhB - chiu rng ming l hnh

- nghing thnh bn l hnhh1- chiu su rnh l hnhr1- bn knh ln vnh trcr - bn knh ln y l hnht - khe h gia 2 trc cn

a) b) c)

Hnh 2.1. Rnh ca trc cn to thnh l hnh.a) 2 trc; b) 3 trc; c) 4 trc

B

r1 rh

h1 t

Hnh 2.2- L hnh hp ch nht.


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22

nghing thnh bn l hnh cn gi l lng thot phi khi cn v cbiu th bng t s gia hiu s chiu rng ming v y l hnh v chiu cao rnhl hnh tnh theo %.

nghing thnh bn l hnh khng nhng to cho phi ra vo l hnh d

dng m cn to iu kin phc hi li ng kch thc ban u khi phc hi litrc. nghing thnh bn l hnh c th chn t 1 10% hoc ln hn.

Bn knh gc ln r v r1nhm loi tr s tp trung ng sut trong trc cnng thi trnh gc nhn cho vt cn do trnh c bavia, nt rn do rch gckhi nhit thp v gim tnh do.

C th chn: r = (0,1 0,15)h; r1= t.

b) L hnh thoi

i vi l hnh thoi v l hnh vung th bn knh ln r1 ming l hnh cth ly ln hn mt t to iu kin cho gin rng thun li trnh to bavia. Bngcch chn bn knh ln c th iu chnh c chiu cao v chiu rng ca l hnh.

h - chiu cao l hnh khng c bn knh ln.h1- chiu cao l hnh c bn knh lnb - chiu rng hnh thoib1- chiu rng ming l hnhr v r1- cc bn knh lnt - khe h gia 2 trc cn

c/ L hnh vung

L hnh vung c s phn bit vi hp vung cch b tr l hnh trn trccn. L hnh vung b tr rnh theo hnh cho. L hnh hp vung b tr rnh theocnh a.

d/ L hnh van

L hnh van c nhiu cch cu to: van mt bn knh, van nhiu bnknh, van bng, van ng.

Tu theo yu cu cng ngh m khi thit k l hnh ta chn cho ph hp:

b

b1

h h1

trr1

Hnh 2.3- L hnh thoi.

bb1

r r1

hh1

t

Hnh 2.4- L hnh vung.


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23

/ L hnh trn

Thng thng l hnh trn c mt ngknh d, song cng c mt s trng hp khi cncc loi sn phm ln th l hnh trn c

thit k theo 2 ng knh: ng knh thngng d v ng knh nm ngang d1.

2.1.3. Cch phn loi l hnh

a/ Phn loi theo hnh dng

L hnh n gin: ch nht, trn, vung, van v.v...

L hnh phc tp: l hnh gc, ch I, ch U, v.v...

b/ Phn loi theo cng dng

L hnh gin di (cn ph): nhm gim nhanh tit din ca phi.

L hnh cn th: ng thi vi gim tit din ca phi phi to c dnhnh dng v gn vi hnh dng ca sn phm.

L hnh trc thnh phm: tc dng khng ch c kch thc ca thnh phm

L hnh tinh: cho ra kch thc v hnh dng ca sn phm trng thi nngv phi m bo c dung sai ca sn phm.

c/ Phn loi theo cch gia cng l hnh trn trc cn

L hnh h:phn ln gp l hnh n gin, chng c ng phn chia kheh gia 2 trc cn x-x nm trong phm vi rnh ca trc cn d cho rnh cgia cng trn mt hay 2 trc.

L hnh kn: c ng phn chia khe h gia 2 trc cn x-x nm ngoi phm

b

h

R

r

Hnh 2.5- Cc thng s c bn ca l hnh vana. van mt bn knh; b. van hai bn knh; c. van bng

b

R1Rr

h

b

Rr

h

d

d1

Hnh 2.6- L hnh trn.

a) b) c)

Hnh 2.6- L hnh h.

x x x x


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24

vi rnh l hnh c cu to bi mt phn li v mt phn rnh ca 2 trc cn.

L hnh na kn

loi l hnh ny trn trc cnva c phn li va c phn lm. Kheh gia hai trc cn c cu to thnh bn ca l hnh.

2.1.4. ng trung bnh ca trc, ng cn, ng trung tuyn ca lhnh trc cn

a. ng trung bnh ca trc cn

ng trung bnh ca trc cn l ng nm ngang chia i khong cchgia 2 tm trc cn.

b. ng cn

ng cnl ng trn phn b l hnh trc cn.

c. ng trung tuyn ca l hnh

ng trung tuyn ca l hnh l mt ng thng i qua trng tm ca lhnh ng thi phi m bo tng hp lc ca kim loi tc dng ln mt trn v mtdi ca l hnh bng nhau. Chiu su rnh ca 2 trc u bng nhau.

ng trung tuyn c th trng vi ng trung bnh v ng cn trongmt s trng hp.

Nu nhlc ma st v h s ma st c hai trc nhnhau th s cn bngtc c th thay bng s cn bng ng knh trung bnh 2Rtbcatrc cn.

Rtbt= Rtbd

2Rtbt- ng knh lm vic trung bnh ca trc trn.2Rtbd- ng knh lm vic trung bnh ca trc di.

x x

Hnh 2.7 L hnh kn.

Hnh 2.8. L hnh na kn.

Dtt

Dtt

Dtt

/2

Dtt

/2

Dtt

/2

Dtt

/2

Dtt/2+y

Dtt/2-y

ng trung bnh catruc cn v n cn

ng trung bnhca truc cn

ng cn

Tm ca truc cn trn Tm ca truc cn trn

Tm ca truc cn di Tm ca truc cn di

y

a/ b/


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25

Nu khng xc nh ng ng trung tuyn ca l hnh c th ph v khpni hoc trc ni, gy tn tht nng lng, lm mn nhanh cc thit b dn hngv l hnh, gy ng sut trong vt cn.

xc nh c ng trung tuyn ca l hnh c nhiu phng php.

i vi cc l hnh n gin:

i vi trc cn c l hnh phc tp (dm ch I, ng ray): loi l hnhny cng c trc i xng v vy ng trung tuyn chnh l trc i xngnm ngang.

i vi cc l hnh kn v cc l hnh nh hnh khcNi chung vi l hnh kn th ng trung tuyn khng trng vi trc i

xng, ng thi vi nhng l hnh khng c tnh i xng th ng trung tuyn sl ng i qua trng tm la l hnh. Phng php xc nh trng tm ca l hnhphc tp, trong thc t dng phng php chia l hnh phc tp thnh nhng l hnhn gin xc nh trng tm sau tng hp to trng tm cacc phn ngin thnh trng tm ca l hnh.

q1= d. B mm2; q2= h.2

ab mm2

Rtbt

Rtbt

Rtbt

Rtbd

Rtbd

Rtbd

Hnh 2.9- ng trung tuyn ca l hnh.

Hnh 2.10 ng trung tuyn ca l hnh.

N NH

B

h

d zy

y1

y2

q2 q2

q1

ng trung tuyn ca l hnh

Hnh 2.11- S xc nh to trng tm thp ch U.q1, q2- din tch tit din tng phn. y1- to trng tm

ca q1; y2- to trng tm ca q2

b


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26

ng thng N - N l ng trung tuyn gi thit ca l hnh c to y.

Ta c:21

2211

21

21

qq

yq2yq

qq

M2My

+

+=

+

+=

trong , M1v M2- mmen tnh ca tit din.

Phn thn ca thp ch U l mt hnh ch nht, c trng tm i qua ngchia u d.

2

dy1=

To trng tm phn chn ca ch (hnh thang) c cch tnh nhsau:

++

=ba

ba2

3

hy2

hnh 2.11 c Z = d +y l khong cch t cnh ngoi ca l hnh n ng

trung tuyn N - N.

2.1.5. ng knh lm vic trung bnh (tip xc) ca trc cn

ng knh tip xc Dtx(lm vic) ca trc cn l ng knh m n bocho vn tc ra ca vt cn khi trc cn khng c s vt trc (vn tc ca vt cnv vn tc di ca trc cn bng nhau).

Chng ta bit rng tc di ca trc cn v tc ca kim loi khi ra khil hnh lin h vi nhau theo cng thc:

Vvc= (1 + S)vtrVvc- vn tc ca vt cn, m/s.Vtr- vn tc di ca trc cn, m/s.S - lng vt trc.

60

n.D.v

1v

V

v

vVS

txtr

tr

vc

tr

trvc

=

=

=

Khi cn trong l hnh th tc cn tnh theo ng bao ca l hnh trn trc

cn. ng knh lm vic (tip xc) Dtxca trc cn khc nhau. Do , tc cavt cn theo chiu rng ca l hnh cng khc nhau. V th phi xc nh mtng knh lm vic trung bnh, trn c s xc nh tc ra phi v cc i

lng bin dng khc. =n

1

txtb n

DD

Phng php xc nh ng knh lm vic trung bnh theo cn bng lc mast l rt kh. C th xc nh chng theo 3 phng php n gin hn:

Theo b mt tip xc ca l hnh


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27

Ta c biu thc:F

k.DD txtb

=

Trong :

Dtx- tng cc ng knh lm vic ti tng im trn b mt tip xc

gia kim loi vi trc cn.

k - h s phc hi li trc cn,0

minmax

D

DDk

=

D0- ng knh ban u ca trc cn. D0thay i t Dmaxn Dmin.F - din tch b mt tip xc gia kim loi v trc cn.

ng knh trc cn chn trn c s cng ngh: iu kin n kim loi, bn, cng sut ng c, tc cn v.v...

H s phc hi trc cn ca tng my cn c th nhsau:

My cn ph, cn phi : k = 0,08 0,15 My cn hnh : k = 0,08 0,15

My cn dy thp : k = 0,05 0,09

My cn tm nng : k = 0,04 0,07

My cn tm ngui : k = 0,03 0,06Mc gim ng knh khi phc hi DPHrnh l hnh hay vnh trc trongphm vi cho php sau:

L hnh tinh

Khi cn sn phm n gin: 0PH D61

51D

.

Khi cn sn phm phc tp: 0PH D51

4

1D

L hnh th

Khi cn sn phm n gin: 0PH D51

4

1D

Khi cn sn phm phc tp: 0PH D4

1

3

1

D

Theo chiu cao trung bnh ca l hnh th ng knh lm vic trung

bnh ca trc cn c tnh:

b

FDhDD tttbtttb ==

Dtt- khong cch gia 2 tm trc cn.htb- chiu cao trung bnh ca l hnhF - din tch tit din ca phi khi ra khi l hnh (din tch ca l hnh)

b - chiu rng ca phi khi ra khi l hnh (chiu rng l hnh)Phng php ny n gin nhng vi cc l hnh phc tp th cho kt qu


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28

khng chnh xc, v khng tnh n nh hng thnh bn ca l hnh.Theo ng bao ca l hnh: thng dng khi cn trong l hnh phc

tp, l hnh vung h v kn.

Di y gii thiu cch xc nh ng knh lm vic trung bnh ca trc

cn mt s l hnh thng dng:L hnh van

+ Theo b mt tip xc ca l hnh

Dtb= D+ 2Rov(1-costb)D- ng knh lm vic y l hnh; Rov- bn knh van.

tb- gc xc nh v tr ng knh lm vic trung bnh.

+=

ovov

ovtb R2

Carcsin

R2

barcsin25,0

bov- chiu rng vt cn khi ra khi l hnhC - cnh vt cn vung a vo l hnh van+ Theo chiu cao trung bnh ca l hnh

ov

ovtttb b

FDD = ; ( ) bnovovov FmhbF += ; m.b.kF ovovbn=

y, kovly theo th hnh 2.13 chn theo t s bov/m

22ovov C25,0R2R2m =

n gin hnngi ta xc nh ng knh lm vic trung bnh ca trc cn

hnh thoi nhsau vi hnh van mt bn knh un: h3

2DD tttb = .

L hnh trn

+ Theo b mt tip xc ca l hnhDtb= Dtt- 0,785d

d - ng knh thp trn.+ Theo chiu cao trung bnh ca l hnh

d785,0Dd4

dDhDD 02

0tb0tb ===

bov

h Rov

r

B

t

m/2

m

m/2

Dtt

Dd

Hnh 2.12-L hnh van xc nh D tb.

0,780,770,760,750,740,730,720,710,700,69

0,680,671 1,4 1,8 2,2 2,6 3 3,4

B/h.t

Kov=Q

ov

/B(h-t)

Hnh 2.13. Xc nh h s van kov

d

d1

Dtt


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29

L hnh vung

+ Theo b mt tip xc ca l hnhDtb= Dtt- hv+ 0,35.hv= Dtt- 0,65hv.

+ Theo chiu cao trung bnh ca l hnh:

v

vtttb b

FDD =

L hnh thoi

+ Theo b mt tip xc ca l hnh

75,1h

b

t

t = ; = 1200

Dtb= Dtt- ht+ 0,2.bt

Dtb= Dtt- ht+ 0,35ht= Dtt- 0,65ht.+ Theo chiu cao trung bnh ca l hnh:

tttt

tttttb h5,0Db

hb5,0DD ==

L hnh lc gic

+ Theo chiu cao trung bnh ca l hnh:

b

qDD tttb =

q - din tch tit din l hnh lc gic.b - chiu rng vt cn khi ra khi l hnh+ Theo chu vi ng bao l hnh:

( )a2b

a.D2bhDD

d

ddtttb +

+= ;

2

htDD ttd =

L hnh phc tp

321

nn2211txtb l...ll

lD...lDlD

l

l.DD

+++

+++==

l1, l2, ..., ln-tng phn t

ng bao.D1, D2, ...Dn- ng knh lm vic trungbnh tng ng vi cc phn t ng bao.

V d:vi l hnh dm ch I th:- ng knh lm vic trung bnh chn h

cho trc trn:( )

hh

hdk.hh.k b2Bh2

b2BDD2D

+

+=

- ng knh lm vic trung bnh chn kn cho trc di:( )

( ) kkkkk.akdkk.hHk.k a2b2Bh2H2

a.D2b2BDh.D2H.D2D +++

+++=

B

b

r r

h

h1

t

Dd

Dtt

bb1

r r1

hh1

t

b

b1

h h1

trr1

D

d

Dtt

DHD

d'

Dhk

Dak

H

hk

Dah

Dhh

hh

ah

bh

bk

ak


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30

T nhng phn tch trn c th tham kho cch xc nh ng knh lmvic (tip xc) ca mt s l hnh thng gp (bng 2.1)

Tn l hnh Hnh minh ho ng knh lm vic

L hnh phng

L hnh hp

L hnh vung

L hnh trn

L hnh thoi

L hnh van

L hnh 6 cnh

L hnh phc tp

Dtx= Dtt- h

Dtx= Dtt- h

Dtx= Dtt- 0,76a

Dtx= Dtt- 0,8d

Dtx= Dtt- 0,55h

1.3

2 mhDD

tttx

+=

2.b

hhbDD

tttx

22,0. +

=

+=

b

C1

2

hDD tttx

b

FDD tttx

D

tt

Dtx

h

Dtt

Dtx

h

ng

Dtt D

tx a

Dtt D

tx d

Dtt

Dtx

h

Dtx

h

BngDtt

h

Dtx

D

tt

m

Dtt

Dtt

h

b

Dtt

Dtx

h

b

C

Dtt

Dtx

Dtx

BF

F

Dtt

Dtx

Dtx

b

F

F


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31

2.1.6. B tr l hnh trn trc cn

a. Kch thc trc cn v ng cn

b tr c l hnh trn trc cn phi xut pht t cc kch thc c bn

ca trc cn.ng knh trc cn Dttca my cn l khong cch gia 2 ng tm ca

trc cn trn v di v tr bnh thng khi cn.

2

DDD DTtt

+=

Khi 2 trc tip xc nhau hon tonkhng c khe h ta c ng knh banu ca trc cn:

DT= DVT+ t ; DD= DVD+ tt - khe h gia 2 vnh trcDVT, DVD- ng knh theo vnh trcng tip xc ca 2 ng knh ban u DT, DDgi l ng cn.

DTtb, DDtb-ng knh trung bnh ca trc trn v di.DTtx, DDtx- ng knh lm vic ca trc trn v di.

i vi cc l hnh n gin nhhp ch nht, vung, thoi, van, trn khi btr l hnh trn trc cn th trc i xng nm ngang ca l hnh (ng trungtuyn) lun trng vi ng cn. Trong thc t nhm n nh phi lc ra khi l

hnh i theo mt hng, ngi ta s dng s chnh lch ng knh ban u ca 2trc cn. Nu DT> DD: ta gi l c p lc trn (phi lun cong xung di lc ra

khi l hnh, nu DT< DDta gi l c p lc di. p lc trn DTv p lc di

DTc biu th:

DT= DT- DD; DD= DD- DT.Khi cn cc sn phm c bin dng phc tp th dng p lc i vi trc no

c cha phn rnh l hnh kn. Khi cn hnh ngi ta dng p lc trn. Ap lc trn

c tr s 1 3 mm cho my cn hnh c nh v 10 mm cho my cn hnh c ln.

p lc di thng dng cc my cn m phi cn c trng lng ln (cntm nng) vi tr s p lc 10 15 mm.

Trong trng hp b tr l hnh c s dng p lc trn vi mt i lng DT= DT-DDth ng cn phi l ng v tr thp hn ng tip xc gia 2 ng

knh ban u ca trc trn v trc di mt khong cch l DT/4 bi v:

4

D

2

D

2

D;

4

D

2

D

2

D DDTBDTTTBT +=

+=

V DTTB= DDTBcho nn biu thc trn c th vit:

DT- DD= DTTrong trng hp s dng p lc di th ng cn s nm v tr cao hn

t

DTtx

DT

Dtt

DD

D

Dtx

DVD

DVT

DT/4

Hnh 2.14- B tr l hnh trn trc cn.


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32

ng trung bnh ca trc cn mt i lng DT/4. V ng trung tuyn ca lhnh nm trng vi ng cn nn khng cn phi xy dng mt ng ph trungbnh ca trc cn m ta c th tnh ng knh trc trn v trc di nhsau:

2

DDD;

2

DDD D

D

T

T

=

+=

vi D l khong cch gia 2 ng tm trc, chnh l ng knh ca trc cn.

b. Sp xp v b tr l hnh trn gi cn 3 trc

Gi cn 3 trc thng gp nhiu my cn hnh b tr hng, n lm nhimv cn ph, cn th. H l hnh thng dng cc gi ny l h l hnh hp chnht - vung hoc thoi - vung ty theo kch thc phi.

B tr l hnh trn gi 3 trc c hai cch: xen k v ln xung.* B tr xen k:

Theo cch b tr ny th trn mt chiu di ca trc cn ch xp c t lhnh. Nhng nu dng mt b trc cn 4 trc: mt trc trn, mt trc di v haitrc gia phi l hnh th vn c th tit kim c trc cn. B tr xen k ththit k l hnh s n gin.

* B tr ln xung:

Trong cch b tr ny th trc gia c dng chung cho trc trn v trcdi, do b tr c nhiu l hnh, qu trnh lt thp c thc hin l hnhdi. S dng cch b tr ln xung th khi thit k l hnh s phc tp hn.

2.1.7. Cc i lng bin dng khi thit k l hnh

a. H s bin dng

Trong l thuyt cn trnh by v cc i lng bin dng khi thit k l hnh:

Ll;Bb;hH ===

1

2

3

4

2

1

4

3

6

5

8

7


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33

Nu h s gin di sau mt ln cn l:

==L

l

f

F

1

Nu qu trnh cn phi qua nhiu ln (n) cn, c c sn phm cui cng

th h s gin di gi l tng lng bin dngnfF= qua mi ln bin dng,

din tch tit din ln lt gim dn v ta ln lt c cc h s gin di tng ng:

n

1nn

2

12

11 f

F;......;

f

F;

f

F ===

Vy: n21n

1n

2

1

1n.....

f

f....

f

f.

f

F

f

F=== =

Ly gi tr trung bnh cho lng gin di ta c:

nTB

nTB

nhoc

f

F ===

Tr s h s gin di trung bnh l mt i lng c trng cho cng bindng, mc s dng ph ti ca thit b, s tiu hao nng lng ca tng gi cn,ng thi n c mi lin h vi cc thng s cng ngh khc v.v.. Cc h s gin

di tng phn (1, 2, 3..) ph thuc vo nhiu yu t ch yu l lc p, lng ginrng, nhit , tnh cht thnh phn ha hc ca vt cn, trng thi b mt trc cn,ma st...

Qu trnh thit k l hnh thng thng xut pht t iu kin cng nghnh: vt liu v kch thc cho trc (phi v sn phm cn c). V vy, chng ta cth tm c s ln bin dng tng phn n:

TB

nnTB

n lg

flgFlgn

f

F

==

b. S lin h gia cc i lng bin dng

T s cn nhhnh 2.15 c th xc nh c mi lin h gia cc i

lng bin dng.lg= R.; h.Rlg =

vi, : gc n kim loilg: di cung tip xc trn trc cnR: bn knh lm vic ca trc cn

Khi gc n kim loi nh, ta c:

R

hhoc

h.R.R

=


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34

T hnh ta c:

D

h1cos

=

hoc hmax= D(1 - cos)

vi, h: lng bin dngV gc n kim loi cc i max

xut pht t iu kin ma st trn b mttip xc:

2max

max2

max

f1

1cos

tg1

1cos

+

=

+=

Suy ra,

+=

2max

f1

11Dh

xc nh s b lng p cc ic th dng cc biu thc n gin hn

hmax= 2max.R = R.f

2Gi tr gc n kim loi tu theo my cn c th tham kho theo bng 2.2

Bng 2.2 Gc n ca kim loi khi cn thp

iu kin cn v vt liu trc cn Gc n kim loi () h/DKCn ngui thp v thp hp kim khng nntrc sau khi n kim loi:

C dung dch bi trnKhng c dung dch bi trn

Cn nng:

Thp tm

Thp hnhThp hnh trn trc c hn vt

Cn nng kim loi mu:

Nhm 3500Cng thau 8000CNiken 11000Cng 9000C

3 4

6 8

18 22

22 2427 34

20 22

21 242227

1/700 1/400

1/250 1/100

1/20 1/15

1/15 1/121/9 1/6

1/16 1/151/15

1/9

h/2

h/2hH

B

lg

b/2

b/2

b

Hnh 2.15- S qu trnh cn.


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35

2.1.8. Nhng nguyn tc c bn khi thit k l hnh trc cn

Qu trnh thit k l hnh trc cn ph thuc vo sn phm cn, kiu my, cim k thut ca my, cng sut ng c, cht lng kim loi v cc yu t khc.

Xc nh s ln cn (ch p) phi xut pht t kh nng trc cn n

c vo kim loi (gc n ). Trong trng hp bn trc, cng sut ng ckhng m bo phi tng s ln cn. i khi s ln cn cn ph thuc vo cch btr gi cn...

Xc nh lng p nhng ln cn u tin theo gc n cho php, ccln cn sau phi xem xt theo bn trc, cng sut ng c, cht lng sn phm.

Xc nh lng p l hnh tinh v trc tinh theo iu kin bin dngtrong l hnh t c chnh xc ca sn phm v iu kin mi mn l hnh.C th nhsau:

- Vi l hnh tinh: = 1,1 1,2- Vi l hnh trc tinh: = 1,25 1,35

Xc nh kch thc phi ban u trn c s dung sai m cho php vxc nh nhsau:

- Kch thc phi trng thi ngui ang:

ang= a - /2- Kch thc phi trng thi nng an:

an= (a - /2).(1,012 1,015) (mm)

Thit k l hnh trc cn phi xut pht t kch thc sn phm. Kchthc l hnh s l kch thc sn phm theo tiu chun c xt n h s n nng

ca thp. V d vi thp trn c ng knh d, ph thuc vo dung sai kch thc tnh kch thc sn phm dn trng thi nng.

Tnh ton lng gin rng b trong l hnh phi chnh xc. Khong trngca l hnh dng cho gin rng bao gi cng phi ln hn lng gin rng tnh ton

bKL= (0,95 1)bLHtrong , bKL: chiu rng kim loi sau cn; bLH: chiu rng ca l hnh

i vi cc sn phm c bin dng phc tp (thp ch , thp ch I, thpng ray) phi chia l hnh thnh cc phn t ring bit v tnh h s bin dngcho tng phn t . Do cn gim bt s l hnh phc tp. Qu trnh thit k lhnh bt buc theo hng ngc vi hng cn.

Vi my cn b tr gi cn theo hng phi ch phn b s ln cn ccgi cn hp l m bo nng sut cao v ph ti u trn cc gi. Vi my cnlin tc phi bo m tc cn ln.

Tnh n ti trng ng c. Yu t ny gip tit kim nng lng, gimgi thnh sn phm.

Tnh n tui bn ca trc. Yu t ny dn n trnh phi thay trc nhiuln, gim nng sut ca xng.


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36

2.1.9- Phn tch c im cc h thng l hnh gin di

Cc l hnh cn th (to s gin di) rt a dng, chng c dng gimnhanh tit din vt cn. Ty thuc vo tit din, h s gin di, ng knh trccn... m dng l hnh gin di c hnh dng khc nhau theo mt th t ni tip

nhau nht nh.Tp hp mt lot cc l hnh to s gin di cng mt kiu theo mt th t

gi l h thng l hnh. Cc h thng l hnh to s gin di thng dng l:- H thng l hnh hp ch nht - vung.- H thng l hnh hp - trc phng.- H thng l hnh van - vung.- H thng l hnh thoi - vung.- H thng l hnh van bng - van ng.

- H thng l hnh van - trn.- H thng l hnh thoi - thoi.- H thng l hnh vung - vung.- H thng l hnh vn nng- H thng l hnh hn hp.

a) H thng l hnh hp ch nht - vung

H thng ny tp hp mt lot cc l hnh ch nht v l hnh vung. Trongqu trnh cn c th c lt phi hoc khng lt phi. H thng ny c dng my

cn ph, cn phi lin tc, cc gi cn th ca my cn hnh, dng sn xut phicho my cn thp dm.

c im

- u im: su rnh nh, cho lng p ln v ng u, tiu hao nng

lng t, thun tin cho vic c gii ho khi a vt cn t l hnh n sang l hnhkia. Trong mt l hnh c th cn c nhiu ln bng cch thay i khong cch

tm trc cn, c gc n ln (= 200300).- Nhc im:D to ra b mt li nu qu in y, d b lch phi khi i

vo l hnh, kh nhn c phi vung chnh xc.2Xc nh kch thc v cc thng s ca h thng

- Chiu rng ca y l hnh bKc th chn theo chiu rng phi vo l hnh.

bK(0,95 1)BI

- Chiu rng ming ca l hnh BK:BK= BI

+ b

Hnh 2.16- H thng l hnh hp ch nht - vung.


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37

vi, b l lng gin rng

trong l hnh, b = 5 10mmNu cn nhiu ln trong l

hnh th phn tch tng lng

gin rng b ca cc lncn xc nh c chiurng ming l hnh BK:

BK= bK+ b

y l hnh c li 10

15%, nghing thnhbn ca l hnh chn trong

phm vi 20 25%, h s hn

ch gin rng kbc gi tr:kb=0,6 0,8

- Lng gin rng b c th xc nh theo biu thc:

( ) ( )

+++

=

nTB

bTB

.R

b11hH

k.h.b2b (2.1)

hoc:

=f2

hh.R

H2

h.15,1b (2.2)

trong , H, h: chiu cao vt cn trc v sau khi cn.bTB: chiu rng trung bnh vng bin dng.

: gc n kim loi, rad.

R.: di cung n.

n = 1 khi bTB< R.; n = 2 khi bTB> R.

h: lng p, h = H - ht: nhit cn.

f: h s ma st khi cn, f = n1.n2.n3(1,05 - 0,0005.t) (2.3)n1: h s xt n trng thi b mt v vt liu ch to trc cn;

n1= 1 i vi trc thp, n1= 0,8 i vi trc gang.n2: h s xt n nh hng ca tc cn (hnh 2.18).n3: h s xt n nh hng ca thnh phn ho hc ca vt liu

cn (bng 2.3).H s ma st f cn c th tnh theo biu thc Ghely:

+ Cn trn trc thp: f = 1,05 - 0,0005t - 0,056V+ Cn trn trc gang: f = 0,92 - 0,0005t - 0,056V

vi V: tc cn (V < 5m/s); t: nhit cn (t > 7000C).

BF

rr1

S

BK

bK

2025%

15%

hK

Hnh 2.17- Kch thc l hnh hp.


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38

Bng 2.3 H s n3ph thuc vo vt liuVt liu H s n3

Thp Cacbon (CT1)Thp P18 (ledeburit)

Thp Peclit - mactenxitThp ostenit (X13H49)Thp ferit (1X17A)Thp ostenit + ccbit (X15H60)

11,1

1,24 1,31,41,551,6

Vi h thng l hnh hp nhhnh 2.16, cn c vo kch thc phi v snphm, iu kin phn b h s gin di xc nh cc kch thc l hnh vungtrung gian, sau mi tnh cc kch thc l hnh ch nht. Qu trnh tnh ton

c thc hin bng cch cho trc mt lng gin rng b, da vo biu thc cho xc nh cc kch thc bK, BK, hK...

Trn c s cc biu thc ny, tnh cc i lng gin rng b theo cc biu

thc 1v 2, nu gi tr lng gin rng b tnh c c gi tr tng ng vi

lng gin rng b chn th khng tnh ton li. Song, nu khng ph hp th

phi cn c vo lng gin rng b tnh ton xc nh li cc kch thc ca lhnh hp ch nht.

xc nh s b cc kch thc ca l hnh hp ch nht trung gian dngcc biu thc:

h2= C3- k3(b2- C3)b2= C1- k2(C1- h2)

Nhvy, c chiu cao ca l hnh hp ch nht trung gian l:( ) ( )

32

1323332 kk1

CkkkCk1h

++

=

trong , k2, k3: cc h s hn ch gin rng trong l hnh, vi l hnh cn th

k2, k3thay i trong phm vi 0,25 0,3.

0,30,40,50,6

0,70,80,9

n

0 4 8 12 16 18 20

Hnh 2.18- Xc nh n2 theo tc cn.


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39

Nu chn k2= k3= 0,3 th ta c:

91,0

C39,0C3,1h 132

=

T h2ta tnh li b2.

Nu xc nh kch thc ca l hnh hp ch nht theo gc n , tnh lngp h theo gc n .

V d:vi = 260c h = Dk(1 - cos).

Vy, h = 0,1Dk, suy ra:C1- h2= 0,1(D - h) (D: ng knh ban u ca trc cn)

9,0

D1,0Ch 12

=

Tm c kch thc h2xc nh b2v b2cng vi cc thng s khc ca l

hnh theo cu to ca n. C th chnh li kch th

c sao cho ph hp vi iu kinn, gin rng trong l hnh vung th ba.

b) H thng l hnh hp - trc phng (khng rnh)

H thng ny c dng nhm gi cn th ca my cn lin tc, cc mycn b tr theo bn c v ch Z.

H thng ny vn nng hn v c th thay i lng p h bng cch thayi khong cch 2 trc cn, d kh b vy rn khi cn trn trc phng.

V h s gin di th h thng ny km hn h thng trc. Song vi 2 uim trn, ngi ta vn s dng. Phng php tnh ton vi h thng ny cngtng t nhh thng trc, tuy nhin cc ln cn trn trc phng c h s gin

rng t do v gc n cho php nh hn.

C1

C1

h2

b2

C3

C3

Hnh 2.19- S xc nh tit din l hnh hp trung gian.

Hnh 2.20-H thng l hnh hp - trc phng.

I II III IV


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40

c) H thng l hnh van - vung

1c im:

c im ca h thng van - vung l gim rt nhanh tit din vt cn vc ng dng nhiu cc my cn hnh c nh, my cn dy.

- u im: c h s gin di ln, khi cn trong l van v= 1,3 2; cn

trong l vung v= 1,2 1,6.Trong qu trnh cn, phi

c lt nhiu gc nn ctnh ca sn phm c cithin.

Phi tit din vung da vo l hnh van, d dng ccu dn hng vi l hnh van,t khot su vo trc, khi cnphi vung th cc cnh vungchuyn ra pha ngoi van, dloi b khuyt tt b mt.

- Nhc im: Phi van cn trong l hnh vung kh n nh. dnhng phi kp cht, kh t mng vng a phi t l hnh van sang l hnhvung, c bin dng khng ng u theo chiu rng ca l hnh nn lm gim chtlng thp sn phm.

Hnh 2.21-H thng l hnh van - vung.

12

34

5

6

Hnh 2.22- Lt thp khi cn trong h thngl hnh van - vung.

h

H=

bv

h

H=

bv

Hnh 2.23- Bin dng khng u trong l hnh van v vung.


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41

L hnh vung c khot su vo trc ln, c hin tng bin dng khngu phi van trong l hnh vung. khc phc nhc im ny ngi ta thay lhnh van bng l hnh su cnh.

Ngoi bin dng ng u, h su cnh - vung n nh trong l hnh vungv d n vo trc, h su cnh - vung c s dng gi cn ph ca cc my cn

hnh loi nh, my cn dy.T dng ca mt l hnh su cnh, khi s dng bn knh y l hnh r c th

chuyn l hnh su cnh thnh l hnh van bng v van c cnh phng.

H s gin di trong l hnh van ph thuc vo t s gia chiu rng v

chiu cao ca van trn trcov

ov

h

bv mc hn ch gin rng trong l hnh kb.

2Tnh ton, thit k l hnh van - vung

Gi thit h c 3 l hnh nhhnh sau:

Hnh 2.24- H thng l hnh su cnh - vung.

r

hov

bov

hov

bov

r

Hnh 2.25- H l hnh van bng (a) v van c cnh phng (b).

a) b)

bv

hv

1,41C

C1

1,41C

C2

Hnh 2.26- H l hnh van - vung.


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42

Lng p trung bnh trong l hnh van c th xc nh theo biu thc:

hTB= C1- 0,74hovTrn c s mc gin rng trong l hnh van ta c th c mt chiu rng

ca l hnh bov:

( )

( ) ovovb1ovbovov1

ovb1

ov1ov

hk74,0Ck1b

h74,0CkCbCb

+=

+=+=

Xc nh t s gia chiu rng v chiu cao ca l hnh van:

ah

kh74,0Ck1

h

b

ov

ovbov1

ovb

ov

ov =+

=

Suy ra, ovb

1ov

b

ov k74,0a

Ck1

h

+

+

=

Ta xc nh li bov: ovb

1ov

bovov

k74,0a

Ck1ah.ab

+

+==

Biu thc ny khng cha lng p hTB, trn c s kch thc v chiu caov chiu rng ca l hnh van, xc nh c din tch Fovca l hnh van:

( )2ovb

21

ovb

ovovovk74,0a

C.a.k174,0hb74,0F

+

+==

Din tch phi vung cn trong l hnh van:Fv= 0,98.C1

2Trn c s din tch tit din, tnh h s gin di trong l hnh van:

( )( )2ovb

21

2ovb

21

ov

vov

k74,0a

C.a.k174,0

C.98,0

F

F

+

+==

Sau khi bin i v rt gn, ta c:

( )2ovbov

ov

2ov

bov

ov

ov

k1h

b

k74,0h

b33,1

+

+

= (2.4)

T biu thc ny, ta nhn thy h s bin dng trong l hnh van ph thuc

vo mt bin s trong t s gia chiu rng v chiu cao ca l hnhov

ov

h

bv vo

mc hn ch gin rng trong l hnh ovbk . Theo l thuyt v gin rng th hai


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43

bin s ny cng ph thuc ln nhau.

T biu thc (2.4) ta c th hnh thnh mi quan h gia ov, cnh vung a

v ovbk nhbng sau:

Bng 2.4- H s gin di ovtheo a vov

bk Tr s ovtheo

ovbk

ov

ov

h

ba=

0,5 0,6 0,7 0,8 0,9 1,01,52,02,53,03,5

1,381,661,952,242,53

1,311,551,802,102,30

1,251,461,681,902,10

1,201,381,571,771,96

1,151,311,481,651,83

1,111,251,401,551,71

Trn c s biu thc (2.4) ta c th tnh ton thit k cu to l hnh vanmt cch ti u:

( )( )2ovbi

2ovbi

ovk1a

k74,0a33,1

+

+=

T kinh nghim thc tin, ngi ta nhn thy rng:

+ C1= (50 100) mm khiov

bk = 0,5 0,6; a = 2 th ov= 1,55 1,65.

+ C1= (20 50) mm khiov

bk = 0,6 0,8; a = 2 3 th ov= 1,44 2.+ C1= (7 20) mm khi

ovbk = 0,8 1; a = 2,5 3 th ov= 1,4 1,7.

(trng hp ny ovbk kh ln gn nhc gin rng t do).

i vi l hnh vung ca h van - vung th trn thc t h s gin di ovcng ph thuc vo din tch tit din van cn trong l hnh vung, c ngha l ph

thuc vo t s ah

b

ov

ov = v h s hn ch gin rng trong l hnh vungTB

vb h

bk

= ,

ta c:

( )

( ) ovv b2v bov2ov

vb2

vTB

ovTB

vbvov

b.k.74,0Ck.76,029,1h

C.76,0b.74,0kC.29,1

hbkbh

+=

=

=

(tr s bv= 1,29C2v ta xt n bn knh ln vnh trc r = 0,15C2).

Vi,( ) ovv b2ovb

ov

ov

ov

b.k.74,0Ck.76,029,1

ba

h

b

+==

T 2 biu thc trn c th tnh li cc kch th

c ca tit din l hnh van:


44/229



44

v

b

2v

bov

k.a.74,01

C.a.k.76,029,1b

+

+= ;

v b

2v

bovov

k.a.74,01

Ck.76,029,1

a

bh

+

+==

Chiu cao trung bnh ca l hnh vung:

22

22

v

vvTB C.76,0C.29,1

C.98,0

b

Fh === (c bn knh ln y)

Din tch tit din phi van cn trong l hnh vung:

( )( )

22v

2vb

22

2vb

ovovov

C.98,0F

k.a.74,01

C.a.k.76,029,174,0h.b.74,0F

=

+

+==

H s gin di trong l hnh vung v:

( )( )2v v

2vv

v

ovv

k.a.74,01

k.76,029,1.a.755,0

F

F

+

+== (2.5)

T biu thc ny, ta c nhn xt: h s gin di trong l hnh vung phthuc vo t s ca din tch van ca hai trc cn trong l hnh v ch s hn ch

gin rng ca l hnh vung. Tr s vtheo a vv

bk

cho trong bng 2.5.

Bng 2.5- H s gin di vtheo a vv

bk

Tr s vtheov

bk

ov

ov

h

ba=

0,4 0,5 0,6 0,71,52,02,53,0

3,5

1,381,521,581,61

1,62

1,311,391,421,42

1,40

1,241,291,291,27

1,23

1,191,211,191,15

1,11

Trn c s bng 2.4 v 2.5 ta thy h s gin di trong l hnh vung nhhn nhiu so vi trong l van. Do c th a ra cc kt lun:

- Ty theo tr s ovbk vv

bk m h s gin di trong l hnh van v vung

c mi quan h: ov= 1 + (1,2 1,8)(v-1) (2.6)

Thng thng, hay dng: ov= 1 + 1,5(v-1)

- Theo t s gia chiu rng v chiu cao 7,22,2

h

b

ov

ov = vic chn h s

gin di phi m bo c iu kin n nh ca phi khi cn


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45

3Cc bc thit k l hnh van - vung

- T gc n cho php, tm h s gin di vtrong l hnh vung (hnh 2.27 )

50 60 70 80 90 100 110 120 130 140 1508

10121416182022242628

3032

Cnh vung a(mm)

Gcn(

) = 1,6

= 1,5

= 1,4

= 1,3

= 1,2

a)

20 30 40 50 60 70 80 90 100 110 1201012

16

20

24

28

34

Cnh vung a(mm)

Gcn(

) = 1,6 = 1,5

= 1,4

= 1,3

= 1,2

32

b)

5 10 15 20 25 30 35 40 45 508

101214161820222426283032

Cnh vung a(mm)

Gcn

()

= 1,5

= 1,45

= 1,4 = 1,3

= 1,2c)

= 1,35

= 1,25


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46

- Trn c s biu thc (2.5) v (2.6), xc nh h s gin di ov l hnhvan ln (nu tnh ngc hng cn).

- Trn c s v v ov, xc nh c h s gin di t l hnh vung

trc n l hnh vung sau: = v. ov. Sau c th iu chnh li kch thcca l hnh van trung gian.

Chiu cao ca l hnh van c th tm theo th hoc tnh ton theo lnggin rng b trong l hnh vung v l hnh van. Qu trnh tnh ton phi m bos in y bnh thng, trnh bavia. Kch thc ca l hnh van sau khi xc nh

lng gin rng b phi chnh li.- Tnh din tch ca l hnh van v l hnh vung.

+ i vi l hnh van:

( ) SbShb4

F ovovovov +

=

Gi a0l t s gia chiu rng bovv chiu cao hovkhi khe h S = 0:

( )Shb

aov

ov0 =

khi , din tch l hnh l:q' = a0.bov(hov- S)

Nu nhl hnh khng c in y th din tch q' c dng:q' = a0.b'ov(hov- S')

vi, ( )'Sh'b

aov

ov0 =

0 10 20 30 40 50 60 70 80 901214161820222426283032

Cnh vung a(mm)

Gcn

(

) = 1,5

= 1,45= 1,4

= 1,3

= 1,2

d)

= 1,35

= 1,25

34

100

Hnh 2.27- Cc th xc nh vtheo v cnh vung a theo

h thng l hnh van - vung cho cc ng knh trc cn.a) D = 800mm; b) D = 500mm;c) D = 250mm; d) D = 350mm

b'ov

hv

Hnh 2.28- Cu to l hnh van.

bov

S m/2

m/2

S'

Rov


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47

Theo hnh 2.28, ta c:

m = hov- S' v2ov

2ovov b25,0R2R2m =

Bn knh van:ov

2ov

2ov

ov

'h4

'hbR

+= , trong : h'ov= hov- S'

+ i vi l hnh vung:bv= hv= 1,41C

Kch thc thc t c bn

knh ln:b'v= hv= 1,141C - Sh'v= 1,141C - 0,828r

vi r 0,15C th:

F'v= Fv- 0,858.r20,98C2

d) H thng l hnh thoi - vungH thng ny c dng nhiu cc my cn hnh c trung bnh v nh trn

my cn phi lin tc.

r

CS

h'v

hv=1,1

41

C

b'v= 1,141C - S

bv= 1,41C

Hnh 2.30- Xc nh kch thc l hnh vung.

I II III IV

Hnh 2.31- H thng l hnh thoi - vung.

Hnh 2.29- Xc nh din tch van.

1,0 1,4 1,8 2,2 2,6 3,0 3,4 3,8 4,2

0,680,70

0,72

0,74

0,76

0,785

( )Shb

ov

ov

bov

hov

Rov

S

a0=

q/b

ov

(hov-s)


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48

1c im:H s gin di trong l hnh thoi v vung gn ging nhau v c tr s:

vt= 1,15 1,4

vi, v: h s gin di trong l hnh vung.

t: h s gin di trong l hnh thoi.Gc n = 110 1150.- u im:

+ Cho sn phm vung chnh xc, sc cnh.+ c p t mi pha ca l hnh.+ Kh tt vy rn.

- Nhc im:+ Rnh ca l hnh khot su vo trc.

+ D hnh thnh bavia khi qu in y l hnh.2Tnh ton h s gin di v kch thc ca l hnh thoi - vung:

Gi thit c 3 l hnh v cc kch thc c ghi nhhnh 2.32.

- Kch thc ca l hnh thoi khng c bn knh ln:

( )

( )

2h.k

C41,1.2

k1b

hC41,12

kC41,1

......hkC41,1b

tt b1

t bt

t1

tb

1

tTBt

b1t

+=

+=

+=

trong , t bk : h s gin rng trong l hnh thoi.

Ta c:t

tt

b1

tb

t

t

h

2

h.kC41,1.

2

k1

ah

b

+

==

(2.7)

Suy ra: tb

1t

btk.5,0a

C41,1.k.5,01h

+

+= (2.8)

C1h1

b1

Hnh 2.32- Xc nh kch thc h thng l hnh thoi - vung.

bt

h C2

b2

h2


49/229



49

T hai biu thc (2.7) v (2.8) ta tm c bt:

tb

1t

btt

k.5,0a

C41,1.k.5,01.ah.ab

+

+==

Din tch l hnh thoi:

( )( )2t b

21

2tbtt

tk.5,0a

Ck.5,01a

2

b.hF

+

+==

H s gin di trong l hnh thoi:

( )( )2t b

2tb

t

21

t

vt

k.5,01a

k.5,0a

F

C

F

F

+

+===

T biu thc ny, ta c th vit di dng:

( )( )2t bii

2tbii

tik.5,01a

k.5,0a

+

+=

Trong bng 2.6 cho tr s ca titheo aivt

bik :

Bng 2.6- H s gin di titheot

bik

Tr s titheot

bik ()t

t

h

ba=

0 0,4 0,5 0,6 0,7 0,8

100105110115120125130

1,191,301,431,571,731,922,15

1,191,301,431,571,731.922.15

1,131,201,291,391,501,631,79

1,121,181,261,351,451,571,72

1,101,161,241,321,411,521,65

1,101,151,221,291,371,471,59

1,091,141,201,251,341,431,55

T bng ta thy khi a tng th titng, ng thi khit

bik tng th tigim.

T thc tin, ta nhn thy rng:

+ Khi C1= 10 50mm; = 11001200 vi t bik = 0,5 0,8 th ti= 1,2 1,45

+ Khi C1= 50 100mm; = 10501150 vi t bik = 0,4 0,6 th ti= 1,15 1,4

- Kch thc l hnh vung ca h thng thoi - vung:Theo hnh 2.32 ta c:

ht= 1,41.C2- 0,5.v

bk (bt- 1,41C2) = 1,41C2(1 + 0,5v

bk ) - 0,5.v

bk .bt

vi,

( ) tv bv b2t

t

t

bk5,0k5,01C41,1

ba

h

b

+==


50/229



50

Suy ra:a.k.5,01

C.41,1.k.5,01ab

vb

2v

bt

+

+=

Tnh li:

a.k.5,01

C.41,1.k.5,01

a

bh

v b

2v

btt

+

+== (2.9)

T hai biu thc trn, tnh din tch hnh thoi:

( )( )2v b

22

2vbtt

tk.a.5,01

C.k.5,01a

2

b.hF

+

+==

Tnh h s gin di trong l hnh vung tip theo:

( )

( )2v

b

2vb

2

2

t

v

tv

k.a.5,01

k.5,01a

C

F

F

F

+

+===

hoc tng t:( )

( )2v bii

2vbii

vik.a.5,01

k.5,01a

+

+=

T biu thc (2.9) v bng 2.7 cho ta mt s gi tr theo aivv

bik :

Bng 2.7- H s gin di vitheov

bik

Tr s vitheov

bik ()ti

ti

h

ba=

0 0,3 0,4 0,5 0,6 0,7100105110115120125130

1,191,301,431,571,731,922,15

1,191,301,431,571,731.922.15

1,131,211,281,361,441,531,62

1,111,181,251,311,371,441,52

1,101,161,211,271,321,371,42

1,091,141,191,231,271,311,34

1,081,121,161,191,221,251,27

T thc tin nhn thy rng:+ Khi C2= 5 30mm;

vbk = 0,4 0,65 th v= 1,19 1,37.

+ Khi C2= 40 120mm;v

bk = 0,3 0,4b th v= 1,18 1,44.

Ni chung khi tnh theo h thng l hnh thoi - vung nh nu trn ta

c th chn tv.Th t cc bc tnh ton v thit k h thng l hnh thoi - vung cng

tng t nhvi h thng van - vung.- i vi nhng l hnh thoi c gc ln th din tch hnh thoi:

F't= 0,98.b't.h't= 0,49.bt.ht


51/229



51

vi: rt= 0,15.ht; r = (0,15 0,2)Ct;

2cos.2

h

2sin.2

bC ttt

=

=

Ty theo gc n nh l hnh thoi, c th tham kho kch thc l hnh

khi c bn knh ln nh bng 2.8.Bng 2.8- Kch thc l hnh thoi khi c bn knh ln

()2

tgh

b

t

t = h't b't F't

100105110115120125130

1,191,301,431,571,731,922,15

ht- 0,61rht- 0,52rht- 0,44rht- 0,37rh

t- 0,31r

ht- 0,28rht- 0,21r

bt- 1,19Sbt- 1,30Sbt- 1,43Sbt- 1,57Sb

t- 1,73S

bt- 1,92Sbt- 2,15S

Ft- 0,29r2- 0,59S2

Ft- 0,23r2- 0,65S2

Ft- 0,18r2- 0,72S2

Ft- 0,14r2- 0,79S2

Ft- 0,112r2- 0,87S2

Ft- 0,085r2- 0,96S2

Ft- 0,052r2- 1,08S2

e) H thng l hnh van bng - van ng

H thng ny dng ch yu cc my cn hnh c nh lin tc c trc bngv trc ng xen k.

1c im:

H thng l hnhvan bng - van ng cu im l khng cn lt

phi, c t s 3,12,1b

h= ,

bo m n nh phi trongl hnh, cht lng b mtsn phm tt.

2Tnh ton h s gin di, kch thc trong h thng van bng - ng:

Gi thit c 3 l hnhvi cc k hiu nhhnh2.34.

Vi t s chiu rng

v chiu cao 2,1b

h

I

I = ; ta

c:

( )ovTBITBovbIovIov

bbkh

bhh

+=

+=

I II III IV

Hnh 2.33- H thng l hnh van bng - van ng.

Hnh 2.34- S cu to l hnh van bng - van ng.

I

hI

bI

II

bov

hov

III

bIII

hIII


52/229



52

( )

( ) ovovbIovbovovI

ovbI

ovTB

ITB

ovbI

bk74,0bk75,02,1h

b74,0b75,0kb2,1

bbkb2,1

+=

+=

+=

Mc gin rng:TB

ovb h

bk=

t:ov

ovII b

ha = , ta c:

ov

ovov

bIov

bII b

bk74,0bk75,02,1a

+=

T 2 biu thc trn, ta suy ra:

ovbII

Iov

bov

k74,0a

bk75,02,1b

+

+= ;

ovbII

Iov

bIIov

k74,0a

bk75,02,1ah

+

+=

Din tch l hnh van bng II:

( )2ovbII

2I

ovbII

ovovovk74,0a

bk75,02,1a.74,0h.b.74,0F

+

+=

Din tch l hnh van ng I:

2IIIIII b.9,0h.b.2,1.75,0h.b4

F

Tm h s gin di trong l hnh van bng II:

( )2ovbov

ov

2

ovbov

ov

ov

Iov

k74,02,1b

h

k.74,0bh215,1

F

F

+

+

==

Ta nhn thy, h s gin di ovph thuc vo aIIvov

bk , c th vit di dng:

( )( )2ov ibiII

2ovibiII

ovk74,02,1a

k.74,0a215,1

+

+=

Bng 2.8 cho ta tr s ca ovtheo aII ivov

ibk .

Bng 2.8- H s gin di ov itheoov

ibk v aII i

Tr s ov itheoov

ibk

ti

ti

h

ba=

0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,11,52,02,53,0

1,161,431,691,95

1,141,371,611,86

1,131,331,551,77

1,111,291,491,68

1,101,271,451,63

1,091,231,391,55

1,071,201,351,49

1,071,181,311,44


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53

Trong thc t, ngi ta thng chn 5,22b

h

ov

ov = v tng ng c h s gin

di ov= 1,61 1,18.i vi l hnh van ng III ta c:

( )( )

( ) ovIIIbIIIIIIbIIIovIIIov

IIIbIII

IIITB

ovTB

IIIb2

IIIov

h.k.74,0bk.9,01b

b.9,0h.74,0kb

hhkbb

+=

=

=

vi( ) ovIIIbIIIIIIb

ovI

h.k.74,0bk9,01

ha

+=

Bin i v rt ra c:

IIIbIIII

III

bovk.a.74,01 b.ak9,01h

++= ; IIIbI

III

III

bovk.a.74,01 bk9,01b

++=

Din tch tit din ca l hnh van bng II:

( )( )2IIIbI

2IIII

2IIIb

ovovovk.a.74,01

b.a.k.9,0174,0h.b.74,0F

+

+=

Bng cch tnh ton nh trn ta rt ra c h s gin di 2trong l hnhvan ng III:

( )( )2IIIbI

2IIIbI

2

ov2

k.a.74,01

k.9,01a.822,0FF

+

+=

hoc:( )

( )2III ibI

2IIIibiI

2k.a.74,01

k.9,01a.822,0

+

+=

Bng 2.10 cho mt vi tr s 2c tnh ton theo cng thc trn.

Bng 2.10- H s gin di 2theo aivIII

ibk

Tr s 2theo III ibk ov

ovi b

ha = 0,2 0,3 0,4 0,5 0,6

1,52,02,53,0

1,171,371,531,65

1,121,271,371,43

1,091,211,261,27

1,071,141,161,17

1,041,091,101,08

Trn thc t, ngi ta thng s dng aI= 2 2,5 vIII

bk =0,3 0,5 v cho

h s 2= 1,37 1,14.


54/229



54

Nu cho trc mt h s gin di 2c th xc nh h s gin di trong lhnh van bng theo biu thc:

ov1 + (1,2 1,3)(2- 1) (2.10)3Cc bc tnh ton v thit k theo h thng van bng - van ng:

- Tm mt h s gin di tng qut t l hnh van ng ny sang van ngkia: = ov.2= 1,4 1,9. y gi tr nh ng vi bI= 5 15 mm, gi tr ln

ng vi bI= 30 40 mm.- Xc nh mi kch thc ca l hnh van ng (theo nhim v thit k).- Xc nh din tch tit din v cc kch thc ca l hnh van bng trung

gian (trn c s v biu thc *). ng thi phi xc nh lng gin rng btrong l hnh van ng v van bng theo cc biu thc tnh lng gin rng.

f) H thng l hnh van - trn

H thng ny thng s dng cn thp trn cc my cn hnh c trung,c nh, cn dy thp, s dng cc h blc cn dy.

1c im:

- u im:khng c gc nhn, phi ngui ng u, m bo cht lng bmt sn phm cn, lng p ng u nn hn ch c ng sut d trong sn

phm. t mi mn l hnh hn so vi h thng van - vung.- Nhc im: phi van vo l hnh trn kh, hay vn vt cn trong l hnh,

phi dng dn hng nghim ngt (cng), h s gin di khng ln (2= 1,2 1,4).2Tnh ton h s gin di ca h thng:

Gi thit c 3 l hnh nhhnh sau:

Tng t nhh van - vung ta c:

( )

( ) ovovb0ovbovov0

ovb0

ov0

0ovb0

ov0ov

h.k.74,0dk.8,01b

h.74,0d.8,0kd

h.74,0d

Fkd

bdb

+=

+=

+=

+=

bov

hov

d0

I

d0

II III

Hnh 2.35- H thng l hnh van - trn.


55/229



55

t ah

b

ov

ov =

Ta c:( )

a

h

k.h.74,0dk.8,01

ov

ovbov0

ovb =

+

Suy ra:ov

b

0ov

bov

k74,0a

dk.8,01h

+

+= ;

ovb

0ov

bov

k74,0a

dk.8,01.ab

+

+=

Din tch tit din ca l hnh van:

( )( )2ovb

20

2ovb

ovovovk74,0a

d.k.8,01.a.74,0h.b.74,0F

+

+=

Din tch tit din phi trn vo l hnh van:

20

20O d.785,04

d.F =

H s gin di trong l hnh van:

( )( )2ovb

2ovb

ov

Oov

k.8,01.a

k74,0a06,1

F

F

+

+==

y h s gin di trong l hnh van cng ph thuc vo t s a v h s

hn ch gin rng ovbk , v vy:

( )( )2ov ibi

2ovibi

ovk.8,01.a

k74,0a06,1

+

+=

Xc nh h s gin di trong l hnh trn, ta c:

( )

( ) I bovII bovIov

IbI

ITB

ovTB

IbIov

k.b.74,0dk.8,01h

d.8,0b.74,0kd

hbkdh

+=

+=

+=

Vi t s ah

b

ov

ov = c th bin i tng t nhtrn v tm c cc gi tr

ca bovv hov:

Ib

II

bov

k.a.74,01

d.k.8,01h

+

+= ;

Ib

II

bov

k.a.74,01

dk.8,01.ab

+

+=

Din tch tit din ca phi van:

( )( )2I b

2I

2Ib

ovovov k.a.74,01

d.k.8,01.a.74,0

h.b.74,0F

+

+

=


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56

Din tch l tit din phi trn:

2I

2I

I d.785,04

d.F =

H s gin di trong l hnh trn:

( )( )2I b

2Ib

I

ovI

k.a.74,01

k8,01.a.94,0

F

F

+

+==

Vit di dng tng qut:

( )( )2I ibi

2Iibi

iIk.a.74,01

k8,01.a.94,0

+

+=

g) H thng l hnh thoi - thoi

H thng ny thng s dng nhiu cc my cn hnh c ln v trung bnhkhi cn thp cht lng, thp hp kim.

1c im:

- u im: sc cnh, d nhn c tit din vung chnh xc, lng ginrng trong l hnh nh.

- Nhc im: vt cn d b lt trong l hnh, d hnh thnh bavia, h s gin

di nh (= 1,1 1,3).2Quan h kch thc v s hnh thnh h s gin di bin dng:

Theo hnh ta c:

2

tg.hb 111

= ;

2

tg.hb 222

=

H s gin di trong l hnh II:

2tg.h

2tg.h

h.b.5,0

h.b.5,0

F

F

222

121

22

11

2

12

===

Nu nhb2= 0 th h1= b2. Vy:

2tg.2tgb2

tg.2

tg.h

h

b

b.h

b.h 21

2

211

2

1

22

11

2

=

=== (2.11)

1

b1

h1C1

I2

b2

h2

C23

b3

h3

C3

II III

Hnh 2.36- H thng l hnh thoi - thoi.


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57

Nu nhkhi cn c gin rng b th h s gin di s gim i cng vi s

tng ca t bk v mi s tnh ton s tng t nhh thng thoi - vung. Nh

bng 2.6, khi 0k t b= th h s gin di trong l hnh thoi bng h s gin di trong

l hnh vung.

2tg

h

b

1

1vt

===

Vi h thng thoi - thoi, nu hai gc ca hnh thoi lin k nhnhau 1= 2th t biu thc (2.11) ta c:

.22

21

22

21

22

21

2C

C

b

b

h

h===

V vy,221

.bb = ;221

.hh =

c ngha l khi bit c 2ta c th xc nh c kch thc ca l hnh trc (tc l hnh I).

Nu vi mt lng gin rng b = 0 (h1= b2) th:

2tg

h

brasuy

h

b

b

b

122

1

12

21

1

2

1

=

=

==

Khi mt l hnh thoi c gc nh 1= 1000th h s gin di 2trong lhnh thoi II:

( ) 416,119,12

100tg

2tg 22122 ===

=

Ta c nhn xt rng, vi h thng thoi - vung, khi kb= 0, ta c:

v= t= 1,19 (bng 2.6 v 2.7)Khi thit k theo h thng thoi - thoi phi xt n khong trng cho gin

rng v lng gin rng c th tnh theo biu thc:

( ) ( ) 121

212k21 hbb

hb.Rhb4,035,0b ==

Khi dng hthng thoi - thoi nhhnh 2.37 th Rk2l bnknh lm vic y lhnh tng ng c gc

= 24 260; r 0,2C;

a = C/3; R1= C.

B

HC

r

R

Hnh 2.37- Cu to l hnh thoi c bn knh m rng.


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58

h) H thng l hnh vn nng

H thng ny l tp hp l hnh cc loi nh: vung, trc phng, van ng,van bng, trn. H thng ny c s dng nhiu cn cc loi thp trn t thpc cht lng, thp hp kim trn cc my cn c ln, trung bnh.

1c im:

- u im:bng cch thay i khe h ca trc cn, trong l hnh c th cn

c nhiu kch thc thp trn. mt s l hnh c gin rng t do nn loi trc khuyt tt, d lm sch vy rn, m bo cht lng b mt sn phm.

- Nhc im: h s gin di trong l hnh van nh (ov= 1,2 1,25).2Kt cu ca cc l hnh van ng v van bng ca h thng l hnh

vn nng:

- van ng kiu hp vung (hnh 2.39a):

S = 3 10mm; = 20 25%; r = 0,2C; gc n cho php = 21 220vi

= 1,3 1,35.Khi cn trn trc phng th phi vung c p theo chiu cao vi h s gin

di = 1,1 1,2; h s hn ch gin rng trong l hnh vung kb= 0,7.- van ng kiu trn (hnh 2.39b):

L hnh ny cho php cn thp trn vi 4 5 loi kch thc. Tm bn knh

R thp hn

ng nm ngang khi khe h S nh nht.V d, cn thp trn 20 33 mm th van ng kiu trn c kch thc B

Hnh 2.38- H thng l hnh vn nng.

C

C'

Cr

S

RH

h S

B

H

B

h

a) b) c)

Hnh 2.39- Kt cu kch th

c l hnh van.


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59

= 36 mm, h = 15 mm, R = 18,3 mm. Vi = 27 mm th S = 3 mm, H = 33 mm. Vi

= 33 mm th S tng ln S = 6 mm.- van bng (hnh 2.39c):L hnh ny c khe h rt ln, thun li cho vic kh vy rn. Chn cc kch

thc nhsau: chiu rng B = 1,15d + 5 mm (d bng ng knh ngoi thp trnln nht trong nhm sn phm c cn ra), h = 25 7 mm, ph thuc vo ngknh ngoi ca thp trn.

l hnh ny = 1,18 1,24.Kch thc phi vung (hnh 2.39a): C0= 1,3d.


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60

Chng 3

Cc i lng c trng cho s bin dng cakim loi khi cn

3.1- Cc thng s hnh hc ca vng bin dng

Quan st m hnh cn vi hai trc cn c tm O1v O2 quay ngc chiunhau vi cc tc V1v V2. Bn knh trc cn l R1v R2, cc im tip xc gia

phi cn vi trc l A1B1B2A2, gc tm chn cc cung A1B1v B2A2l 1v 2.Vi cc k hiu nhtrn, ta c cc

khi nim v thng s hnh hc cavng bin dng khi cn nhsau:

- A1B1B2A2: vng bin dng hnh hc

- A1B1nB2A2m: vng bin dngthc t.

- m, n: bin dng ngoi vng bindng hnh hc.

- 1, 2: cc gc n.- A1B1, A2B2: cc cung tip xc.- lx: hnh chiu cung tip xc ln

phng nm ngang.

- H, h: chiu cao vt cn trc vsau khi cn.- B, b: chiu rng vt cn trc v

sau khi cn.- L, l: chiu di vt cn trc v

sau khi cn.

3.2- Mi quan h gia cc i lng hnh hc

H - h = h: lng p tuyt i.

Hh

Hh1

HhH == : lng p t i.

b - B = b: dn rng tuyt i.

B

b1

B

b

B

Bb ==

: dn rng t i.

3.3. Lng p

Khi cn, tit din ngang ca vt cn u gim xung khi qua cc l hnh trc

cn. S gim tit din ngang chnh l s gim chiu cao ca vt cn sau mi ln cnqua cc l hnh ta gi l lng p.

O1

V1

h1A1 1

O2

V2

R1

R2

2 B2A2

m n

K

h2

hH

Hnh 3.1- S cn gia hai trc.

lx

B b

b/2

b/2

E

B1


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61

Lng p trong mi ln cn phi da vo cc yu t sau y m phn chiahp l cho tng l hnh:

Thnh phn ha hc, c l tnh ca kim loi cn

H thng l hnh, tng loi l hnh c th.

Lc cn cho php ca trc, cng sut ng c, mmen cn v.v.. Thit b ph khc ca my cn cho php tin hnh quy trnh cng ngh.Ngoi cc yu t trn cn phi lu ti nng lc ca thit b trong khi cn m

iu chnh cho ph hp ng thi khng ngng ci tin v tm ra quy trnh cngngh mi hp l hn t c lng p ln nht.

Lng p ln nht (lng p cc i) c tnh theo cng thc sau:

2LMAX .2

Dh = (mm) (3.1)

trong : DL- ng knh lm vic ca trc cn.- gc n ca vt cn

Trong qu trnh cn, trc cn lun lunb mi mn v vy b mt lm vic ca trc dnkhng t c yu cu k thut. p kimloi c tt v m bo cht lng b mt casn phm chng ta phi tin hnh mi, tin litrc cn.

H s mi li trc cn c tnh nhsau:

H

MINMAX

DDDK = (3.2)

trong :K - h s mi liDMAX- ng knh trc cn mi ch toDMIN- ng knh trc cn c mi li ln cui cng.DH- ng knh trc cn danh ngha

Trong thc t K = 0,08 0,12 i vi trc cn phi;

K = 0,1 i vi trc cn hnhNgoi ra hMAXdng cho my cn ph 2 trc o chiu c tnh theo cng

thc sau:

+

=2LMAX f1

11Dh (mm) (3.3)

trong : f - h s ma st gia vt cn v trc cn (c tnh theo L thuyt cn).Trong trng hp ang cn m tnh do ca kim loi b gim buc ta phi

iu chnh li lng p th phi kim tra li mmen un ca trc v lc cn theo

cng thc sau:

=

8b

4aPMu (3.4)

DL

DH

h


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62

trong :( )ba2

.D.8,0P u

3

= (3.5)

Thay (3.5) vo (3.4) ta c:

u

3

u

.D.1,0M =

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