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MC LC Trang

LI NI U ...............................................................................................................1 CHNG I

NHIM V -YU CU -PHNG N THIT K 1.1. TNG QUAN .....................................................................................................3 1.2. NHIM V - YU CU THIT K. ................................................................3

1.2.1. Nhim v thit k. ........................................................................................3 1.2.2. Yu cu thit k ............................................................................................4

1.2.2.1. Yu cu chung........................................................................................4 1.2.2.2. Yu cu c th trong tnh ton thit k cu trc ....................................5

1.3. PHNG N THIT K ..................................................................................6 1.3.1. c im, phn lai cu trc ........................................................................6

1.3.1.1. Mt s c im v cu trc ..................................................................6 1.3.1.2. Phn loi cu trc...................................................................................7

1.3.2. Chn phng n thit k.............................................................................14 1.3.2.1. Phn tch,chn phng n thit k.......................................................14 1.3.2.2. Chn cc thng s c bn ....................................................................17

CHNG II TNH CC C CU CHNH

2.1. CHN PHNG N V TNH TON C CU NNG.............................20 2.1.1. Chn phng n cho c cu nng ..............................................................20 2.1.2. Tnh c cu nng ........................................................................................21

2.1.2.1. Chn loi dy .......................................................................................21 2.1.2.2. palng gim lc....................................................................................21 2.1.2.3. Kch thc dy.....................................................................................22 2.1.2.4. Tnh cc kch thc c bn ca tang v rng rc ................................22 2.1.2.5. Tnh chn ng c in .......................................................................25 2.1.2.6. T s truyn chung ..............................................................................26 2.1.2.7. Kim tra ng c in v nhit............................................................26 2.1.2.8 .Tnh chn phanh ..................................................................................30 2.1.2.9.Thit k b truyn .................................................................................32

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2.1.2.10. Tnh cp lch tm...............................................................................34 2.1.2.11. Cc b phn khc ca c cu nng ....................................................37

2.2. TNH C CU DI CHUYN XE CON...........................................................44 2.2.1. Chn s tnh v cc thng s c bn .....................................................44 2.2.2. Tnh c cu di chuyn.................................................................................44

2.2.2.1. Tnh bnh xe ........................................................................................44 2.2.2.2. Chn ng c in ...............................................................................46 2.2.2.3. Xc nh t s truyn b truyn h......................................................50 2.2.2.4. Thit k b truyn h, bnh rng tr - thng .......................................50

2.3. TNH C CU DI CHUYN CU.................................................................56 2.3.1. Chn s tnh v cc thng s c bn .....................................................56 2.3.2. Tnh c cu di chuyn cu ..........................................................................57

2.3.2.1. Tnh bnh xe v ray .............................................................................57 2.3.2.2. Chn ng c in ...............................................................................58 2.3.2.3. T s truyn chung ..............................................................................59 2.3.2.4. Kim tra ng c in v mmen m my..........................................60 2.3.2.5. Tnh chn phanh ..................................................................................61 2.3.2.6. Thit k b truyn ................................................................................62 2.3.2.7. Tnh bnh xe v ray .............................................................................66 2.3.2.8. Tnh trc truyn....................................................................................68

CHNG III TNH KT CU THP CA CU TRC

3.1. TNH DM CHNH .........................................................................................70 3.1.1. Chn vt liu...............................................................................................70 3.1.2. Xc nh cc ti trng tc dng ln dm chnh ..........................................70 3.1.3. Chn kt cu dm chnh v kim tra bn ...................................................71

3.1.3.1. Chn kch thc tit din dm chnh...................................................71 3.1.3.2. Kim tra bn tit din chn .............................................................72

3.2. TNH DM CUI ............................................................................................75 3.2.1. Chn vt liu cho dm cui: .......................................................................75 3.2.2. Xc nh cc ti trng tc dng ln dm cui ............................................75 3.2.3. Chn kt cu dm cui v kim tra bn .....................................................75

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CHNG IV TNH CC THIT B PH V S B

TNH GI THNH 4.1. THIT K H THNG IU KHIN CHO CC C CU CNG TC ...79

4.1.1. Khi nim chung.........................................................................................79 4.1.2. H thng dy dn v cc thit b bo v .....................................................79

4.1.2.1. H thng dy dn cung cp in cho cu trc .....................................79 4.1.2.2. Cc thit b bo v ...............................................................................80

4.1.3. Thit k mch iu khin cho cc c cu cng tc ....................................80 4.2.3.1. Mch iu khin c cu nng: .............................................................80 4.2.3.2. Mch iu khin c cu di chuyn xe con v c cu di chuyn cu ...81

4.2. THIT K CC THIT B AN TON C - IN CHO CU TRC .........85 4.2.1. Thit b hn ch chiu cao nng .................................................................85 4.2.2. Thit b hn ch ti trng nng ...................................................................86 4.2.3. Thit b gii hn hnh trnh di chuyn v gim chn .................................86

4.3. S B TNH GI THNH SN PHM........................................................87

CHNG V KT LUN V XUT KIN

KT LUN ..............................................................................................................90 XUT KIN ..................................................................................................90

TI LIU THAM KHO............................................................................................91

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LI NI U

My nng ch yu dng nng vt nng phc v cc qu trnh xy lp, xp d v vn chuyn vt trong cc lnh vc cng nghip, xy dng, quc phng.v.v.. Vi vai tr quan trng trong vic c gii ha t ng ha, my nng l mt trong phng tin v cng cn thit, l iu kin c bn pht trin nn kinh t quc dn. V vy vic tm hiu cc my nng chuyn l khng th thiu i vi cc k s.

Nhm mc ch tm hiu nhiu hn v vn trn, do vy cui kha hc em nhn ti:Thit k cu trc 1 tn phc v cho vic di chuyn tn tm.

Ni dung thc hin bao gm cc phn chnh sau:

Chng I: Nhim v, yu cu v phng n thit k. Chng II: Tnh ton cc c cu chnh.

Chng III: Tnh kt cu thp. Chng IV: Tnh cc thit b ph v s b tnh gi thnh. Chng V: Kt lun v xut kin. Qua thi gian tm hiu thc t cng nhng ti liu c lin quan v c bit c

s gip tn tnh ca thy Nguyn Thi V, nay em hon thnh ti c

giao; mc d c gng nhiu, nhng khng th trnh khi nhng thiu st. Rt mong c s ng gp kin v gip ca cc thy cng bn c cun ti ny c hon thin hn.

Em xin chn thnh cm n thy Nguyn Thi V, cc thy trong khoa K Thut Tu Thy trng i Hc Nha Trang cng ton th cc bn nhit tnh gip em hon thnh cun ti ny.

Nha Trang, ngy 05 thng 7 nm 2008 Sinh vin thc hin:

Bi Mnh Linh

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CHNG I

NHIM V -YU CU -PHNG N THIT K

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1.1. TNG QUAN My nng chuyn l cc loi my cng tc dng thay i v tr ca i tng cng

tc nh thit b mang vt trc tip, s ra i v pht trin ca n gn lin vi yu cu v kinh t k thut ca ngnh cng nghip nhm gim ti a sc ngi trong lao ng.

c im lm vic ca cc c cu my nng l ngn hn, lp i lp li v c thi

gian dng. Chuyn ng chnh ca my l nng h vt theo phng thng ng, ngoi ra cn mt s cc chuyn ng khc dch chuyn vt trong mt phng ngang nh chuyn ng quay quanh trc my, di chuyn my, chuyn ng lc quanh trc ngang. Bng s phi hp gia cc chuyn ng, my c th dch chuyn vt n bt c v tr no trong khng gian lm vic ca n.

p ng yu cu v i hi ca cc ngnh cng nghip khc nhau, k thut

nng vn chuyn cng xut hin nhiu loi my nng vn chuyn mi, lun ci tin

v hp l ha phng php phc v, nng cao hn tin cy lm vic, t ng ha

cc khu iu khin, tin nghi v tha mn yu cu ca ngi s dng. Ty theo kt cu v cng dng, my nng chuyn c chia thnh cc loi: kch, bn ti, palng, cn trc, cu trc, cng trc, thang nng.v.v..

Cu trc l loi my trc kiu cu. Loi ny di chuyn trn ng ray t trn cao dc theo nh xng, xe con mang hng di chuyn trn kt cu thp kiu cu, cu trc c th nng h v vn chuyn hng theo yu cu ti bt k im no trong khng gian ca nh xng. Cu trc c s dng trong tt c cc lnh vc ca nn kinh t quc dn vi cc thit b mang vt rt a dng nh mc treo, thit b cp, nam chm in v.v.. c bit cu trc c s dng ph bin trong ngnh cng nghip ch to my v luyn kim vi cc thit b mang vt chuyn dng.

1.2. NHIM V - YU CU THIT K. 1.2.1. Nhim v thit k.

Thit k l mt qu trnh sng to, trong qu trnh ny ngi thit k phi tm

hiu, cp v gii quyt tho ng hng lot cc yu cu khc nhau v phng php

tnh ton, ch tiu kh nng lm vic, cng ngh ch to v quy trnh lp rp, s dung, sa cha theo nhiu phng php khc nhau. Nhim v chnh ca thit k l tm ra v

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c th ho cc gii php k thut t la chn ra phng php ti u, ph hp

vi nhim v th thit k. Cui cng l a ra nhng thng tin v i tng thit k

v t nhng thng tin c th to ra mt sn phm c th.

Vic thit k phi m bo kh nng thc hin c cc gii php k thut, ngha l phi c s ph hp gia cc c tnh k thut ca cc i tng mi vi cc

gii php k thut v mc pht trin ca khoa hc k thut cng nh thc t sn

xut. Trong ti ny, vic thit k c gii hn trong thit k cu trc 1 tn phc v cho vic di chuyn tn tm sao cho m bo c cc tnh nng k thut v yu cu t ra.

1.2.2. Yu cu thit k 1.2.2.1. Yu cu chung

Mi loi my nng c cu thnh t hai b phn c bn: kt cu thp v b phn c kh. Ngoi hai b phn trn cn c phn trang b in, cc b phn iu khin, cc c cu bo v an ton,

Phn kt cu thp c hnh dng, kch thc ngoi khc nhau, ph hp vi khng gian, tnh cht cng vic v i tng m chng phc v cng nh iu kin kinh t

k thut khc. Kt cu thp l xng sng, l b phn chu ti ca c my nng m trong qu trnh lm vic trng lng cc c cu c kh, ti trng nng chuyn n.

Cc c cu c kh c lp t trc tip trn b phn kt cu thp v thc hin chc nng nng h, di chuyn hoc quay my nng, thay i tm v. Ngi ta phi hp cc chc nng ca cc c cu trn nng h, di chuyn vt trong khng gian m my nng c th thao tc.

B phn c cu c kh l tp hp cc b truyn dn ng t ng c n b cng tc. Cc b phn ny c th l c kh, thu lc, kh nn hoc hn hp ca cc loi . i a s cc my nng s dng truyn ng c kh m kt cu ca chng l: ng c, hp gim tc, trong c cc trc, khp ni, bi, cc cp bnh rng, cp hoc xch truyn ng, tang cun cp, puli, phanh, c xp xp theo mt th t v

quy lut truyn ng nht nh. Tnh ton cc c cu truyn ng l tnh ton chc

nng ca my (ng hc, ng lc hc nh l s vng, tc , phng chiu chuyn

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ng, lc tc ng), sc bn cc c cu t nh ra kch thc hnh hc, cng sut ng c v cc thng s khc nhm lm cho my nng t c cc yu cu k

thut ph hp vi yu cu thc t i hi t ra.

i vi tnh ton sc bn nhm tm c kch thc ca cc c cu t cng vng v bn mn. Tnh ton bn thng tri qua hai giai on: trc tin l la chn s b sau l tnh chnh xc. La chn s b l mc ch xc nh nhanh nhng kch thc chnh theo phng php n gin v gn ng. Tnh ton chi tit hay tnh

chnh xc nhm mc ch kim tra v iu chnh li kch thc c cu la chn s

b. Cch tnh ny thng da vo tnh cht mi ca vt liu. H hng cc c cu my nng ch yu l do gy v mn. Vic tnh bn chi tit

l phi xc nh chnh xc kch thc c kh nng cng vng chng li cc ti

trng tc dng ln chng, bo m tui th ca chng ng thi bo m tnh kinh t khng qu lng ph vt liu. Mn ca cc chi tit c cu din ra t t v lu di. m bo mn cho php cn quan tm ti cht lng vt liu v phng php x l b mt cc vt liu ph hp iu kin lm vic theo yu cu ca tng chi tit, b phn v t c tui th ca c my xc nh trc.

1.2.2.2. Yu cu c th trong tnh ton thit k cu trc Trong tnh ton thit k cu trc 1T phc v cho vic di chuyn tn tm cn

tho mn cc yu cu sau:

- Phi phc v tt cho vic di chuyn tn tm trong phn xng c kh. - Hnh dng, kch thc ca cc kt cu phi ph hp loi vt mang v khng

gian nh xng.

- Phi t c tnh kinh t cao: ngha l thit b sau khi ch to v cc chi ph vn chuyn ca thit b phi l ti u nht.

- Kch thc cc chi tit kt cu ca cu trc phi nh gn m vn m bo c cc tnh nng ca n.

- Thit b phi d ch to hoc nm trong gii hn tiu chun v d lp t trong phn xng.

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- S dng n, lm vic phi c tin cy cao, t hng hc v b s c mi ch nng chuyn.

- Phi m bo cho vic bo dng v sa cha trang thit b c d dng trong nhng trng hp cn thit.

- Thit b phi t tui bn cn thit.

1.3. PHNG N THIT K 1.3.1. c im, phn lai cu trc 1.3.1.1. Mt s c im v cu trc

Cu trc l mt loi my trc c phn kt cu thp (dm chnh) lin kt vi hai dm ngang (dm cui), trn hai dm ngang ny c 4 bnh xe di chuyn trn hai ng ray song song t trn vai ct nh xng hay trn dn kt cu thp. Cu trc c s dng rt rng ri v tin dng nng h vt nng, hng ho trong cc nh xng, phn xng c kh, nh kho bn bi. Dm cu c gi l dm chnh thng c kt cu hp hoc dn, c th c mt hoc hai dm, trn c xe con v c cu nng di chuyn qua li dc theo dm chnh. Hai u ca dm chnh lin kt hn hoc inh tn vi hai dm cui, trn mi dm cui c hai cm bnh xe, cm bnh xe ch ng va cm bnh xe b ng. Nh c cu di chuyn cu v kt hp c cu di chuyn xe con (hoc palng) m cu trc c th nng h bt c v tr no trong khng gian pha di m cu trc bao qut.

Hnh 1.1. Cu trc dn ng in.

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Xt v tng th cu trc gm c phn kt cu thp (dm chnh, dm cui, sn cng tc, lan can), cc c cu c kh (c cu nng, c cu di chuyn cu v c cu di chuyn xe con) v cc thit b iu khin khc.

Dn ng cu trc c th bng tay hoc dn ng in. Dn ng bng tay ch yu dng trong cc phn xng sa cha, lp rp nh, nng h khng thng xuyn, khng i hi nng sut v tc cao. Dn ng bng in cho cc loi cu c ti trng nng v tc nng ln s dng trong cc phn xng lp rp v sa cha ln.

Cu trc c ch to vi ti trng nng t 1 n 500 t; khu dm cu n 32m; chiu cao nng n 16m; tc nng vt t 2 n 40 m/ph; tc di chuyn xe con n 60m/ph v tc di chuyn cu trc n 125 m/ph. Cu trc c ti trng nng thng c trang b hai hoc ba c cu nng vt: mt c cu nng chnh v mt hoc hai c cu nng ph.Ti trng nng ca loi cu trc ny thng c k hiu bng mt phn s vi ti trng nng chnh v ph, v d: 15/3 t; 20/5 t; 150/20/5 t; v.v..

1.3.1.2. Phn loi cu trc Cu trc c phn loi theo cc trng hp sau:

a. Theo cng dng Theo cng dng c cc loi cu trc c cng dng chung v cu trc chuyn

dng. - Cu trc c cng dng chung c kt cu tng t nh cc cu trc khc, im

khc bit c bn ca loi cu trc ny l thit b mang vt a dng, c th nng c nhiu loi hng ho khc nhau. Thit b mang vt ch yu ca loi cu trc ny l mc treo xp d, lp rp v sa cha my mc. Loi cu trc ny c ti trng nng khng ln v khi cn c th dng vi gu ngom, nam chm in hoc thit b cp xp d mt loi hng nht nh.

- Cu trc chuyn dng l loi cu trc m thit b mang vt ca n chuyn nng mt loi hng nht nh. Cu trc chuyn dng c s dng ch yu trong cng nghip luyn kim vi cc thit b mang vt chuyn dng v c ch lm vic rt nng.

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b. Theo k cu dm Theo kt cu dm cu c cc loi cu trc mt dm v cu trc hai dm. - Cu trc mt dm l loi my trc kiu cu thng ch c mt dm chy ch I

hoc t hp vi cc dn thp tng cng cho dm cu, xe con cheo palng di chuyn trn cnh di ca dm ch I hoc mang c cu nng di chuyn pha trn dm ch I, ton b cu trc c th di chuyn dc theo nh xng trn ng ray chuyn dng trn cao. Tt c cc cu trc mt dm u dng palng c ch to sn theo tiu chun lm c cu nng h hng. Nu n c trang b palng ko tay th gi l cu trc mt dm dn ng bng tay, nu c trang b palng in th gi l cu trc mt dm dn ng bng in.

Hnh 1.2. Cu trc mt dm. 1. B phn cp in li ba pha. 6. Palng in. 2. Trc truyn ng. 7. Dm chnh.

3. C cu di chuyn cu. 8. Khung gin thp. 4. Bnh xe di chuyn cu. 9. Mc cu. 5. Dm cui. 10. Cabin iu khin.

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Cu trc mt dm dn ng bng tay c kt cu n gin v r tin nht, chng c s dng trong cng vic phc v sa cha, lp t thit b vi khi lng cng vic t, sc nng ca cu trc loi ny thng khong 0,5 5 tn, tc lm vic chm.

Cu trc mt dm dn ng bng in c trang b palng in, sc nng c th ln ti 10 tn, khu n 30 m, gm c b phn cp in li ba pha.

Hnh 1.3. Cu trc hai dm.

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- Cu trc hai dm, kt cu tng th ca cu trc hai dm gm c: dm hoc dn ch 1, hai dm ch lin kt vi hai dm u 7, trn dm u lp cc cm bnh bnh xe di chuyn cu trc 6, b my dn ng 3, b my di chuyn hot ng s lm cho cc bnh xe quay v cu trc chuyn ng theo ng ray chuyn dng 5 t trn cao dc nh xng, hng chuyn ng ca cu trc chiu quay ca ng c in.

Xe con mang hng 11 di chuyn dc theo ng ray lp trn hai dm (dn) ch; trn xe con t cc b my ca ti chnh 10, ti ph 9 v b my di chuyn xe con 2, cc dy cp in 8 c th co dn ph hp vi v ch ca xe con v cp in cho cu trc nh h thanh dn in 12 t dc theo tng nh xng, cc qut in 3 pha t st trn cc thanh ny, lng thp lm cng tc kim tra 13 treo di dm cu trc. Cc b my ca cu trc thc hin 3 chc nng: nng h hng, di chuyn xe con v di chuyn cu trc. Sc nng ca cu trc 2 dm thng trong khong 5 30 tn, khi c yu cu ring c th n 500 tn. cu trc c sc nng trn 10 tn, thng c trang b hai ti nng cng vi hai mc cu chnh v ph, ti ph c sc nng thng bng mt phn t (0,25) sc nng ca ti chnh, nhng tc nng th ln hn.

Dm chnh ca cu trc hai dm c ch to di dng hp hoc dn khng gian. Dm gin khng gian tuy c nh hn dm hp song kh ch to v thng ch dng cho cu trc c ti trng nng v khu ln. Dm cui ca cu

trc hai dm thng c lm di dng hp v lin kt vi cc dm chnh bng bu lng hoc hn.

c. Theo cch ta ca dm chnh

Theo cch ta ca dm chnh c cc loi cu trc ta v cu trc

cheo.

- Cu trc ta l loi cu trc

m hai u ca dm chnh ta ln Hnh 1.4. Cu trc ta.

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cc dm cui, chng c lin kt vi nhau bi inh tn hoc hn. Loi cu trc ny c kt cu n gin nhng vn m bo c tin cy cao nn c s dung rt ph bin. Trn hnh 1.3 l hnh chung ca cu trc ta loi mt dm. phn kt cu thp ca gm dm cu 1 c hai u ta ln cc dm cui 5 vi cc bnh xe di chuyn dc theo nh xng. Loi cu trc ny thng dng phng n dn dn ng chung. Pha trn dm ch I l khung gin thp 4 dm bo cng vng theo phng ngang ca dm cu. Palng in 3 c th chy dc theo cnh thp pha di ca dm I nh c cu di chuyn palng . Ca bin iu khin 2 c treo vo phn kt cu chu lc ca cu trc.

Hnh 1.5. Cu trc treo. a) Loi hai ray treo; b) Loi ba ray treo.

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- Cu trc treo l loi cu trc m ton b phn kt cu thp c th chy dc theo nh xng nh hai ray treo hoc nh nhiu ray treo. Do lin kt treo ca cc ray

phc tp nn loi cu trc ny thng ch c dng trong cc trng hp c bit cn thit. So vi cu trc ta, cu trc treo c u im l c th lm dm cu di hn, do n c th phc v c phn ra mp ca nh xng, thm ch c th chuyn hng gia hai nh xng song song ng thi kt cu thp ca cu trc treo nh hn so vi

cu trc ta. Tuy nhin, cu trc treo c chiu cao nng thp hn cu trc ta.

d.Theo cch b ch c cu di chuyn Theo cch b ch c cu di chuyn cu trc c cc loi cu trc dn ng chung

v cu trc dn ng ring. - C cu di chuyn cu trc c th thc hin theo hai phng n dn ng chung v dn ng ring. Trong phng n dn ng chung, ng c dn ng c t gia dm cu v truyn chuyn ng ti cc bnh xe ch ng hai bn ray nh cc trc truyn. Trc truyn c

th l trc quay chm,

quay nhanh v quay trung

bnh (hnh 1.5, a, b, c). phng n dn ng ring (hnh 1.5, d) mi bnh xe hoc cm bnh xe ch ng c trang b mt c cu dn ng.

- C cu dn ng chung vi trc truyn quay

chm (hnh 1.6, a) gm ng c in 1, hp gim tc Hnh 1.6. Cc phng nh dn ng.

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2 v cc on trc truyn 3 ni vi nhau v ni vi trc ra ca hp gim tc bng cc khp ni 4. Trc truyn ta trn cc gi 5 bng bi. Do phi truyn momen xon ln nn trc truyn, khp ni v bi c kch thc rt ln, c bit khi cu trc c ti trng nng v khu dm ln. Cc on trc truyn c th l trc c hoc trc rng. So vi trc c tng ng, trc rng c trng lng nh hn 15 20%. Phng n ny c s dng tng i ph bin trong cc cu trc c cng dng chung c khu khng ln, c bit l cc cu trc c kt cu dm khng gian c th b tr d dng cc b phn ca c cu.

- C cu dn ng chung vi trc truyn quay trung bnh (hnh 1.6, b) c trc truyn 3 truyn chuyn ng n bnh xe di chuyn cu trc qua cp bnh rng h 4. V vy m mmen xon trn trc nh hn so vi trc truyn chm v kch thc ca

chng cng nh hn.

- C cu di chuyn dn ng chung vi trc truyn quay nhanh (hinh 1.6, c) c trc truyn 2 c ni trc tip vi trc ng c v v vy n c ng knh nh hn

2 3 ln v trng lng nh hn 4 6 ln so vi trc chuyn quay chm. Tuy nhin, do quay nhanh m n i hi ch to v lp rp chnh xc.

- C cu di chuyn dn ng ring (hnh 1.6, d) gm hai c cu nh nhau dn ng cho cc bnh xe ch ng mi bn ray c bit. Cng sut mi ng c thng ly bng 60% tng cng sut yu cu. Phng n ny tuy c s x lch dm cu khi di chuyn do lc cn hai bn ray khng u song do gn nh, d lp t, s dng v bo dng m ngy cng c s dng ph bin hn, c bit l trong nhng cu trc c khu trn 15m.

e. Theo ngun dn ng Theo ngun dn ng c cc loi cu trc dn ng tay v cu trc dn ng

my.

- Cu trc dn ng bng tay, (hnh 1.7) c dng ch yu trong sa cha, lp rp nh v cc cng vic nng - chuyn hng khng yu cu tc cao. C cu nng

ca loi cu trc ny thng l palng xch ko tay. C cu di chuyn palng xch v cu trc cng c dn ng bng cch ko xch t di ln. Tuy l thit b nng th

14

s song do gi thnh r v d s dng m cu trc dn ng bng tay vn c s dng c hiu qu trong cc phn xng nh.

- Cu trc dn ng bng ng c, (hnh 1.1) c dng ch trong cc phn xng sa ch, lp rp ln v cng vic nng - chuyn hng yu cu c tc v

khi ln. C cu nng ca loi cu trc ny l palng in. C cu di chuyn palng in, xe con v cu cng c dn ng t ng c in. Loi cu trc ny c dng ph bin nht do c nhiu u im ni bt l kh nng t ng ho, thun tin cho ngi s dung v c th s dung trong vic vn chuyn cc loi hng c khi lng ln.

Hnh1.7. Cu trc dn ng bng tay. a) Loi mt dm; b) Loi hai dm. f. Theo v tr iu khin Theo v tr iu khin c cc loi cu trc iu khin t cabin gn trn dm cu

(hnh 1.4) v cu trc iu khin t di nn nh hp nt bm (hnh 1.2). iu khin t di nn bng hp nt bm thng dng cho cc loi cu trc mt dm c ti trng nng nh.

1.3.2. Chn phng n thit k 1.3.2.1. Phn tch,chn phng n thit k

p ng yu cu v mc ch ca vic thit k mi cu trc 1 tn phc v

cho vic di chuyn tn tm, trc tin ta phi phn tch chn s kt cu cu trc

15

sao cho ph hp vi mc ch v c im sn xut ca ca phn xng sau tin

hnh chn phng n thit k cho ph hp, chnh xc v t hiu qu cao nht.

a. Chn m hnh thit k T cc lai cu trc trn, qua tm hiu thc t v c im kt cu v tnh nng

k thut ca cu trc phc v trong cc phn xng ti thy loi cu trc mt dm dng ch I c xe con treo palng di chuyn trn cnh di ca dm ch I l loi ph hp nht. Loi cu ny c u im hn c v c kt cu n gin v nh gn, thch

hp cho vic di chuyn tn tm trong cc phn xng cng nh yu cu v ti trng, lm vic tin cy, s dng n gin, thun tin cho vic bo dng thit b nu xy ra s c v t hiu qu kinh t cao. Chnh v vy ti chn loi cu ny thit k.

6

51

432

Hnh 1.8. Cu trc thit k.

Thit k cu trc mt dm vi ti trng nng 1 tn c kt cu nh sau: - Kt cu thp: dm chnh 1 ch I c hai u ta trn hai dm cui 5, kt cu

dm cui gm hai thanh thp ch I ghp song song. Pha trn dm ch I l khung gin thp 6 c tc dng lm gi cho c cu di chuyn cu ng thi m bo cng cn thit theo phng ngang.

- Xe con 3 mang palng in 4 di chuyn trn cnh di ca dm ch I, cu trc di chuyn dc theo nh xng nh c cu di chuyn 2.

16

- Phng n dn ng: mi c cu (c cu nng, c cu di chuyn xe con, c cu di chuyn cu) u c dn ng bng mt ng c in.

- Cu trc c trang b thit b mang vt l cp lch tm. - Cc c cu c iu khin bng hp nt bm t di nn nh.

b. Chn phng n thit k thit k bt c mt vn g vic u tin chng ta phi xc nh ta i thit k

ci g, n phc v mc ch g v ci ta thit k ra c nhng tnh u vit hn so vi

ci hin ang c hay khng. Mt vn v cng quan trng l ta thit k theo

phng php no, vn ny cn phi c xc nh ngay t u trc khi i thit k

bt c mt vn g. V nu khng xc nh c thit k theo phng php no th c th thit k khng c tnh kh thi v i khi l khng th thc hin c.

Hin nay i thit k mt vn no chng ta c 4 phng php c bn, l:

- Thit k theo mu.

- Thit k theo Quy Phm. - Thit k theo s liu thng k.

- Thit k theo tnh ton.

c im ca mi phng n thit k:

Thit k theo mu: u im ca phng php ny l cho php ta i thit k mt cch nhanh chng, chng ta ch cn da vo mu cu trc c sn hoc thit k mu i thit k ci gn ging vi ci ta cn thit k. Tuy nhin n cng c nhng

nhc im ca n l chng ta kh c th tm c mu cu trc thch hp hay

thit k mu gn nht vi ci ta cn thit k. Mt khc khi i thit k mt vn hon

ton mi th khng th p dng phng php ny c. Thit k theo Quy Phm: y l mt trong nhng phng php thit k cho ta i

thit k nhanh nht m bo d bn v cc Quy Phm t ra c da vo cc kinh nghim v cch tnh d bn. Tuy nhin phng php ny khng th p dng cho cc trng hp c bit c v cc chi tit thit k ra cho ta d bn.

17

Thit k theo s liu thng k: Chng ta thng k cc chi tit sn phm phn tch la chn xem chi tit no hot ng hiu qu v gn vi thit k mnh nht. T

cho ta thit k chi tit da vo kt qu va mi thng k c. Thit k theo tnh ton: y l mt trong bn phng php cho ta kt qu chnh

xc nht v c tnh kinh t cao, tuy nhin n c nhc im l kh khn trong cc

phng php tnh ton v i thit lp cc cng thc tnh ton.

Kt lun: mi phng n thit k u c nhng u nhc im khc nhau, do trong tnh ton thit k ta phi la chn phng n no cho ph hp nht theo yu

cu v mc ch ca vn cn gii quyt t hiu qu cao nht. Vy vi yu cu

v mc ch c th trong tnh ton thit k cu trc ti chn phng n thit k theo tnh ton v y l phng n cho ta kt qu chnh xc nht, tnh kinh t v hiu qu cao nht.

C th trong tnh ton Thit k cu trc mt dm vi ti trng nng 1 tn ta phi tnh cc c cu chnh sau:

- Tnh c cu nng.

- Tnh c cu di chuyn: c cu di chuyn cu v di chuyn palng in. - Tnh kt cu thp: tnh chn dm chnh va dm cui. - Tnh chn cc thit b ph: h thng iu khin, cc thit b an ton c

in.v.v..

1.3.2.2. Chn cc thng s c bn Vic la chn cc kch thc c bn ca cu trc phi cn c vo iu kin lm

vic, loi hng cn bc d v a hnh ca nh xng.v.v.. Qua kho st thc t cc loi cu trc phc v vic vn chuyn tn ti cc phn xng ta xc nh c cc

thng s c bn nh sau: - Ti trng nng: Q = 1T - Chiu cao nng: H = 5 m - Khu dm cu: L = 8 m - Vn tc nng :Vn= 10 m/ph

- Vn tc di chuyn cu: :Vc= 20 m/ph

18

- Vn tc di chuyn xe con: :Vx= 35 m/ph - Dng in xoay chiu 3 pha =/ 220/380 v, tn s 50 Hz - Ch lm vic: Nh.

Tng ng vi ch lm vic nh ta c:

Bng 1-1. Cc s liu v ch lm vic cc c cu ca cu trc.

Ch tiu Ch lm vic (T) - Cng lm vic, C%

- H s s dng trong ngy, kng - H s s dng trong nm, kn - H s s dng theo ti trng, kQ - S ln m my trong mt gi, m

- S cho k lm vic trong mt gi, ack - Nhit mi trng xung quanh, t0C

15 0,33

0,25 0,55 60

10 15 25

Thi gian phc v, nm

- ln

- Bnh rng

- Trc v cc chi tit khc

10

15 25

Thi gian lm vic trong

thi hn trn, h

- ln

- Bnh rng

- Trc v cc chi tit khc

1000

1500 2500

19

CHNG II

TNH CC C CU CHNH

20

2.1. CHN PHNG N V TNH TON C CU NNG 2.1.1. Chn phng n cho c cu nng

Theo yu cu cng ngh, c cu nng l mt b phn ca cu trc. Vic chn phng n cho c cu nng thit k cn phi m bo cc thng lm vic nh cng sut, tc , c tnh ng lc hc, phng php iu khin, mi trng sinh thi,

kh nng qu ti, kh nng tiu chun ha, kh nng lp t, vn hnh, an ton. Cc

ch tiu kinh t nh gi thnh, chi ph sn xut, khu hao, chi ph bo dng sa cha v.v..

i vi cu trc thit k phng n b tr cho c cu nng c chn c s nh hnh 2.1. Vi phng n ny c cu c kch thc tng i gn nh cho php

ch to tng cm c cu ring bit nn thun tin cho vic lp t v n gin trong vic ch to.

1. ng c in.

2. Khp ni vng n hi.

3. Phanh.

4. Hp gim tc.

5. Khp ni. 6. Tang.

Hnh 2.1. S c cu nng. y l loi c cu nng dy mm, c mt tang, truyn ng ca c cu l truyn

ng ring, nng lng s dng l nng lng in. Kt cu c bn gm ng c in 1, khp ni vng n hi 2, phanh 3, hp gim tc 4, khp ni 5, tang cun cp 6, ngoi ra cn c cc b phn khc nh dy cp, cp lch tm v rng rc cp (hnh 2.2).

6 5

4 3

2 1

21

Cc thng s ban u: - Ti trng nng: Q = 1T = 10000N. - Chiu cao nng: H = 5 m. - Tc nng vt: Vn =10 m/ph.

- Ch lm vic ca c cu: Nh.

- Trng lng b phn mang vt: Cp lch tm v palng thun, cp lch tm v palng thun c chn theo tiu chun ca Lin X, (atlat) c khi lng:

= QQm %25,0 25 kg = 250 N

2.1.2. Tnh c cu nng 2.1.2.1. Chn loi dy

C cu nng lm vic vi ng c in, vn tc cao, nn ta chn cp lm dy cho c cu, v cp l loi dy c nhiu u im hn so vi cc loi dy khc nh xch hn, xch tm v l loi dy thng dng nht trong ngnh my trc hin nay.

Trong cc kiu kt cu ca dy cp th kt cu kiu K-3 theo tiu chun ca Lin X c tip xc ng gia cc si thp cc lp k nhau, lm vic lu hng v

c s dng rng ri. Vt liu ch to l cc si thp c gii hn bn 1200 2100

N/ 2mm . Vy ta chn cp K-3 kt cu 6 x 25 (1+6; 6+12) + 1 li, gii hn bn cc si thp trong khong 1500 1700 N/ 2mm , d dng trong vic thay cp sau ny khi b mn, t.

2.1.2.2. palng gim lc Trn cc cu ln dy cp c cun trc tip ln tang; cu ln phc v trong

phn xng khi cn nng h vt theo chiu thng ng, tin li trong khi lm vic;

do ta chn palng n c mt nhnh dy chy ln tang. Tng ng vi ti trng cu trc, theo bng 2-6, [2- tr.25].

Chn bi sut palng a = 2. Palng gm mt rng rc di chuyn, s (hnh 2.2) Lc cng ln nht xut hin nhnh dy cp cun ln tang khi nng vt c

xc nh theo cng thc [2- tr.24].

kam

QSp.

0max =

22

Trong :

a = 2 bi sut palng. m = 1 s nhnh cp cun ln tang.

k = 1,5 h s ti trng ng.

=+= mQQQ0 10000 + 250 = 10250 N

p - hiu sut palng.

( )( ) 98,098,012

98,0.98,01)1(

).1( 2=

=

=

a

ta

p

Vi: t S rng rc i hng, t = 0

= 0,98 hiu sut ca rng rc t trn ln bi trn bnh thng. Hnh 2.2. s palng.

78445,1.98,0.2.1

10250max == S N

2.1.2.3. Kch thc dy Kch thc dy cp dc chn da vo cng thc (2-10) [tr.18]

392205.7844.max === kSS N

Trong :

S - lc ko t cp.

k = 5 - h s an ton bn ca cp, ly theo bng (2-2) [tr.19] ng vi ch lm vic nh.

Xut pht t iu kin bn theo cng thc (2-10), vi loi dy chn trn, vi gii hn bn ca si 1600=b N/ 2mm =160 kg/ 2mm . Theo tiu chun ca Lin X,

chn ng knh cp 1,8=cd mm c sc ko t NS 40350= xp x vi lc t cp

yu cu.

Trng lng 100 m cp = 23,40 kg = 234 N.

2.1.2.4. Tnh cc kch thc c bn ca tang v rng rc ng knh nh nht cho php i vi tang v rng rc xc nh theo cng thc

(2-12) [tr-20].

23

4,194)125(1,8)1( == edD ct mm Trong :

tD - ng knh tang n y rnh cp, mm.

1,8=cd mm - ng knh dy cp qun ln tang.

e = 25 h s thc nghim, tra theo bng (2-4) [tr.20] Ta chn ng knh tang 195=tD mm.

Rng rc lm vic, c th chn ng knh nh hn 20% so vi ng knh

tang.

156135.8,08,0 === tr DD mm

Chiu di ton b ca tang c xc nh theo cng thc (2-14) [tr.21]. 210 2LLLL ++=

Hnh 2.3. S xc nh chiu di tang Trong :

L chiu di ton b ca tang. L0 chiu di phn ct ren. L1 phn tang kp u cp.

L2 phn tang lm thnh bn. Chiu di mt nhnh cp cun ln tang khi lm vic vi chiu cao nng H = 5 m

v bi sut palng a = 2.

l = H.a = 5.2 = 10 m

S vng cp phi cun mt nhnh (2-tr.21)

L

L 2 L 0 L 1 L 2

24

1493,132)081,0195,0(10

)( 0 =++=++= pipi ZdDlZ

ct

vng.

Trong : Z0 = 2 s vng d ch khng s dng n ( 5,1 ). Vy chiu di phn ct ren l: L0 = Z.t Trong : t bc cp c xc nh theo cng thc kinh nghim.

t = dc + (2 3) = 8,1 + 2,4 = 10,5 mm L0 = 14.10,5 = 147 mm Chiu di L1, nu dng phng php cp thng thng th phi ct thm khong

3 vng rnh trn tang na, do : L1 = 3.10 = 30 mm V tang c ct rnh, cp cun mt lp, tuy nhin hai u tang trc khi vo

phn ct rnh ta tr li mt khong L2 = 20 mm lm thnh bn. L = L0 + L1 + 2L2 = 147 + 30 + 20.2 = 217 mm

thun li cho vic ch tao, chn chiu di tang: L = 220 mm. B dy thnh tang xc nh theo kinh nghim.

= 0,02Dt + (6 10) = 0,02.195 + 6,1 = 10 mm Kim tra sc bn ca tang theo cng thc (2-15) [tr.22].

Trong : Smax = 7844 N lc cng ln nht.

= 10 mm b dy thnh tang. t = 10,5 mm bc cun cp.

8,0= - h s gim ng sut i vi tang bng gang.

k =1 h s ph thuc lp cp cun ln tang.

Bng (2-1). H s k. S lp cun 1 2 3 4

k 1 1,4 1,8 2

76,595,10.10

7844.8,0.1==n N/mm

2

[ ]nn tSk

=.

.. max

25

Tang c c bng gang C 15 32 l loi vt liu thng thng ph bin

nht, c gii hn bn nn l: =bn 565 N/mm2. ng sut cho php xc nh theo gii

hn bn nn vi h s an ton k = 5.

1135

565===

kbn

n

N/mm2

Vy [ ]nn < 2.1.2.5. Tnh chn ng c in

Cng sut tnh khi nng vt bng ti trng xc nh theo cng thc (2-78).

.1000.60. nvQN =

Trong : Q = 10000 N ti trng nng ca cu trc. Vn = 10 m/ph vn tc nng.

- hiu sut ca c cu bao gm:

0.. tp=

Tong : p = 0,99 hiu sut palng tnh trn (mc 2). t = 0,96 hiu sut tang, bng (1-9). 0 = 0,85 hiu sut b truyn c k c khp ni, vi b truyn.

c ch to thnh hp gim tc hai cp bnh rng tr, bng(1-9). 807,085,0.96.0.99,0 ==

Vy 06,2807,0.1000.60

10.10000==N KW

Tng ng vi ch lm vic nh, s b chn ng c in.

Bng (2-2). Cc thng s ca ng c in.

Kiu

ng c

Cng

sut

(kw)

Vn

tc

(v/ph) cos

dm

k

MM

dmMM max

M men v

lng ca r to

GD2 (kgm2)

Trng

lng

(kg) K 41-4 1,7 1420 0,84 1,8 2,0 0,048 3,9

26

2.1.2.6. T s truyn chung T s truyn chung t trc ng c n trc tang c xc nh theo cng thc

(3-15) [tr.55].

Trong : nc = v/ph s vng quay danh ngha ca ng c. nt s vng quay ca tang m bo vn tc nng cho trc.

Vi : Vn =10 m/ph vn tc nng.

a = 2 bi sut palng. D0 ng knh tang tnh n tm cp.

D0 = Dt + dc = 195 +5,6 = 200,6 mm

322006,0.

2.10==

pitn v/ph

Vy i0 = 45321420

=

2.1.2.7. Kim tra ng c in v nhit. Do ng c in chn c cng sut danh ngha nh hn cng sut tnh yu

cu khi lm vic vi vt nng c trng lng bng trng ti (Nc= 1,7kW < N = 2,06kW), do phi c kim tra v nhit. Ta tin hnh kim tra ng c v nhit theo thi gian m my khi nng, h vi cc ti trng khc.

Q

t

Hnh 2.4. th gia ti trung bnh ca c cu my trc theo ch lm vic nh.

t

c

n

ni =0

0,2Q 0,75Q Q

t

0,2t 0,2t 0,5t

0.

.

DaV

n nt pi=

27

Chn s cho cc my trc lm vic vi ch nh v trung bnh theo s hnh 2.4. Theo s hnh 2.4 th c cu nng s lm vic vi cc trng lng vt

nng Q1 = Q; Q2 = 0,75Q; Q3 = 0,2Q v thi gian lm vic tng ng vi cc trng lng ny l 2 : 5 : 3.

Cc thng s cn xc nh l:

- Trng lng vt nng cng b phn mang. Q0 = Q + Qm = 10000 + 250 = 10250 N

- Lc cng dy trn tang khi nng vt, theo cng thc (2-19) [tr.24].

tan m

QS

).1(

)1(0

= 5176)98,01(1)98,01(10250

2 =

= N

- Hiu sut ca c cu khng tnh hiu sut palng khi lm vic vi vt nng

trng lng bng trng ti.

816,085,0.96,0. 0' === t

- M men trn trc ng c khi nng vt, theo cng thc (2-79) [tr.48].

2,14816,0.45.2

1.2006,0.5176..2

..

'

0

0===

imDS

M nn Nm

- Lc cng dy cp trn tang khi h vt, theo cng thc (2-2) [tr.25]

=

=

+

)1().1( 10

a

ta

hm

QS 5072)98,01(1

98,0).98,01(102502 =

N

- M men trn trc ng c khi h vt, theo cng thc (2-80) [tr.48].

3,945.2

816,0.1.2006,0.5072.2

...

0

'

0===

imDS

M hh Nm

- Thi gian m my khi nng vt, theo cng thc (3-3) - [tr.52].

..).(375..

)(375)(

20

21

2001

2

iaMMnDQ

MMnDG

tnmnm

liinm

+

=

Trong : = 1,1 h s k n nh hng qun tnh ca cc tit my quay trn cc trc sau trc I.

Iii DG )( 2 - tng mmen v lng ca cc tit my quay trn trc I,

Nm2(tra theo bng catalo ca chng).

28

Iii DG )( 2 khopiirotoii DGDG )()( 22 + = 0,48 + 0,216=0,686 Nm2

Mm mmen m my ca ng c, i vi ng c chn l ng c in

xoay chiu kiu dy cun, xc nh theo cng thc (2-75) [tr47].

dndndnmm

m MMMMM

M 5,12

1,19,12

minmax=

+=

+=

Mdn mmen danh ngha ca ng c.

Mdn = 9550 Nmn

N

c

c 4,111420

7,19550 ==

Mm = 1,5.11,4 = 17,1 Nm Vy khi Q1 = Q

=n

mt s06,1807,0.45.2).2,141,17(3751420.2006,0.10250

)2,141,17(3751420.686,0.1,1

22

2

=

+

Trong : 0.. tp= = 0,807 hiu sut nng ca c cu khi nng vt vi trng

lng bng trng ti. Gia tc khi m my vi ti trng Q1 = Q, c xc nh theo cng thc

2/157,006,1.60

1060

smt

vjn

m

n===

Thi gian m my khi h vt, xc nh theo cng thc (3-9) [tr.54].

..).(375..

)(375)(

20

21

2001

iaMMnDQ

MMnDG

thmhm

liihm

++

+=

=hmt s117,0807,0.45.2).3,91,17(375

1420.2006,0.10250)3,91,17(375

1420.686,0.1,122

2

=

++

+

Tnh tng t cho cc trng hp Q2 v Q3 theo cc cng thc dn trn.Kt qu tnh c ghi trong bng (2-3). Thi gian chuyn ng vi vn tc n nh l:

sv

Ht

n

v 30105.60.60

===

Mmen trung bnh bnh phng trn trc ng c, theo cng thc (2-37) - [tr.44].

29

+

=

t

tMtMM vtmmtb

).()(. 22

Trong : mt - tng thi gian m my trong cc thi k lm vic vi ti trng

khc nhau, s.

Mt mmen cn tnh tng ng vi cc ti trng nht nh trong thi gian

chuyn ng n nh vi ti trng , Nm.

tv thi gian chuyn ng vi vn tc n nh khi lm vic vi tng ti trng, s.

t - ton b thi gian ng c lm vic trong mt chu k, bao gm thi gian

lm vic trong cc thi k chuyn ng n nh v khng n nh, s.

Mm mmen m my ca ng c in, Nm.

Bng (2-3). Cc thng s tng ng vi cc trng hp ti trng.

Thay cc gi tr tng ng va tnh c vo cng trn ta c:

Mtb = 152,0.313,0.5117,0.2214,0.35,0.506,1.210.30)03,2.302,7.53,9.26,3.375,10.52,14.2(30

)152,0.313,0.5117,0.2214,0.35,0.506,1.2(1,17222222

2

++++++

+++++

++++++

= 11,26 Nm Cng sut trung bnh bnh phng ca ng c c pht ra theo cng thc (2-

76) [.47].

Thng s cn tnh Q1 = Q Q2 = 0,75Q Q3 =0,2Q Q0, N

, Sn, N

Mn, Nm

Sh, N

Mh, Nm

n

mt , s

hmt , s

10250 5176 0,807

14,14

5072 9,3

1,06 0,117

7750 3882

0,605 10,6 3804

7,02

0,50 0,13

2050 1035 0,75 3,2

1014

2,03

0,214

0,152

30

67,19550

1420.26,119550

.

===ctb

tbnMN kW

T kt qu tnh c ta thy, cng sut trung bnh bnh phng do ng c pht ra trong sut thi k lm vic vi ch ngt on lp i lp li nh hn cng sut

danh ngha ca n vi cng lm vic l 15% (Ntb = 1,67kW < Nc = 1,7kW). Vy ng c chn l K 41- 4 vi C 15% c cng sut danh ngha Nc = 1,7kW l hon ton tha mn yu cu trong khi lm vic.

2.1.2.8 .Tnh chn phanh phanh c nh gn, ta s t phanh trc th nht tc l trc ng c,

mmen phanh c xc nh theo cng thc (3-14) [tr.54].

Trong : k = 1,5 h s an ton phanh i vi ch nh, bng(3-2) [tr.54]. Q0 = 10250 N ti trng nng k c b phn mang vt. D0 = 0,2006 m ng knh tang tnh n tm cp.

= 0,807 hiu sut ca c cu.

i0 = 45 t s truyn chung. a = 2 bi sut palng.

Vy: NmM ph 1445.2.2807,0.2006,0.10250.5,1

==

Vic la chn phanh m bo iu kin lm vic bnh thng v s an ton trong qu trnh hot ng ca my nng l v cng quan trng. y l mt ch tiu

c TCVN 5863-1995 quy nh. i vi palng in, loai phanh c s dung l phanh a in t vi nhiu mt ma st v c kch thc nh gn, lm vic tin cy.

Phanh gm cc a ma st 5 khng quay v c th di chuyn dc theo cht dn hng 1. Trn cc a 5 c cc b mt ma st 6. Cc a thp 7 khng c b mt ma st lp bng then hoa vi trc phanh. Phanh ng nh lc l xo 4 p cc a 5 vo cc a 7, phanh m nh nam chm in 3 vi ngm ht 2 gn c nh trn a 5. Cc b mt ma st lm vic trong b du.

0

00

..2...

iaDQk

M ph

=

31

Hnh 2.5. Phanh a. Vi mmen phanh l thng s cho trc, cc thng s cn xc inh gm:

Rt bn knh trong ca b mt ma st chn nh nht, theo yu cu kt cu ca phanh ta chn Rt 4Dc = 4.0,035 = 0,14 m

Rn Bn knh ngoi, thng ly Rn = (1,25 2,5)Rt = 1,5.0,14 = 0,21 m Lc dc trc cn thit to m men phanh theo yu cu:

Trong : Mph = 14 Nm m men phanh.

z = 3 s i mt ma st. Rtb bn knh trung bnh. Coi cng do ma st mi im ca b mt tip xc nh nhau.

Rtb = 2nt RR +

= (0,21 +0,14) = 0,175 m

f = 0,12 h s ma st, tra theo bng (2-9) [tr.28].

22212,0.175,0.3

14==P N

p sut trn b mt kim tra treo cng thc:

fRzM

Ptb

ph

..

=

32

[ ]pRRPp

tn

+

=

1000)( 22pi

p lc cho php ca mt s loi vt liu trong phanh p trc tra theo bng(2-4). Bng (2-4). p sut cho php [p] i vi phanh p trc, N/mm2.

Vt liu ma st Khng bi trn Bi trn m Trong b du

Kim loi trn kim loi

Vt liu dt hay an trn kim loi Vt liu cn trn kim loi

0,2

0,3

0,6

0,4

0,6 1,0

0,8

0,8

1,2

Vy: 12,11000).14,021,0(

22222 =+

=

pip N/mm2

Vi p lc cho php [p] = 1,2 N/mm2, ta thy p sut tnh c l tha mn yu cu cho php v 2,112,1

33

Hp gim tc c u im: hiu sut cao, tui th ln, lm vic chc chn, s

dng n gin v c kh nng truyn cng sut nhng ch tc khc nhau.

Hnh 2.6. S hp gim tc. Cc thng s bit: I0 = 45 t s truyn chung. Nc =1,7kW cng sut ca ng c in.

nc = 1420 s vng quay trn trc ng c.

nt = 32 s vng quay trn trc tang.

S b ta chn hp gim tc c kt cu nh hnh (2-6). Ta lp bng phn phi t s truyn nh sau:

nc

nt

34

Bng (2-5). Bng phn phi t s truyn cho hp gim tc. Trc

Thng s

I II III

I I12 = 7,5 I23 = 6

n, v/ph 1420 190 32

N, kW 1,7 1,632 1,567

Mx, (N.m) 11,4 82,08 472,78

Vi:

n1 = nc =1420 v/ph ; n2 = 12

1

in

; 23

23 i

nn =

N1 =Nc =1,7 kW ; N2 = N1; N3 = N2

Mx = 1,14 (KG.m) ; M2 = i12. .Mx ; M3 = i23. .M2 Trong : Mx mmen xon trn trc ng c.

= (0,95 0,97) hiu sut b truyn bnh rng tr, chn = 0,96 Cn c vo yu cu vo cng sut phi truyn vi C15%, s vng quay trc

vo, t s truyn v yu cu v lp rp, ta chn phng n mua sn hp gim tc tiu

chun da vo bng phn phi t s truyn.

2.1.2.10. Tnh cp lch tm

Hnh 2.7. Thit b cp lch tm.

r

F 2

Q

N N

F 1 F 2 N

F 1

35

Trong thc t, nng cc dm thp hoc cc tm thp v tr thng ng ngi ta thng dng thit b cp lch tm.

Cp lch tm dng cp tn tm. Lc ma st gi vt nng trong c cu c to ra bi sc p ca cam lch tm khi nng vt hoc dng tay quay xit cht cam lch tm trc khi nng (hnh 2.7.). Khi cam lch tm p vo tm vt liu sinh ra phn lc N v lc ma st F:

F = F1 + F2 = N (f1 + f2) Q Trong : f1, f2 - h s ma st gia tm thp c nng vi cam lch tm v mt

trong ca m kp; chn f1 = 0,15; f2 = 0,1.

Vy: 400001,015,0

1000021

=

+=

+ ff

QN N

T iu kin cn bng ca tm thp trong cp lch tm, ta vit phng trnh mmen ti tm cam lch tm ta c:

( ) 0cos..cos..sin. 21 =+ rfNrfNrN ( ) 0cos.cos.sin 21 =+ rfrfr

++=

cos121

rfftg

m bo an ton cho vt khng ri ngi ta chn gc lch tm nh hn gc ma st nn ta c:

Trong : r bn knh cam lch tm.

- b dy ca tm thp (vt nng). Cn c vo cng thc trn xc nh kch thc cam lch tm m bo an

ton khng ri vt.

Vy tho mn cho vic nng vt vi ti trng 1tn, s b ta chn cc thng s ban u cho cp lch tm (theo cc cp ch to): = 100 gc lch tm

r = 100 mm bn knh cam lch tm.

= 10 16 mm - b dy cho php ca tm tn.

++ cos121

rfftg

36

Thay cc thng s vo biu thc trn ng vi hai trnh hp = 10 (16) mm. - Trng hp 1: khi = 10 mm 259,0176,0 < lun ng vi mi trng hp.

- Trng hp 2: khi = 16 mm 265,0176,0 < lun ng vi mi trng hp.

Vy cc thng s chn hon ton tho mn.

- Tnh cht cam lch tm. Khi kp tm tn vi ti trng nng 1 tn lm xut hin p lc P tc dng vo cht ca bnh lch tm. cht lm vic an ton ta phi tnh chn sao cho bn.

Tuy nhin khi nng tm tn vi b dy gim dn 10 16 mm thi p lc tc dng ln cht cng thay i theo chiu gim dn do gc lch tm thay i. vy trong thng hp ny ta ch cn tnh cho trng hp nguy him nht l khi cp tm tn vi

b dy l 10 mm, ng vi gc lch tm l = 14,50 (vi tg = 0,259). Khi p lc tc dng vo cht c xc nh theo cng thc:

P = N.cos = 40000.0,968 = 38720 N

Chn vt liu l thp 45CT c gii hn bn mi l ' 1 = 250 N/mm2. ng

knh cht c xc nh theo cng thc:

[ ]3 .1,0 uMd

Trong : Mu mmen un trn cht, xc nh theo cng thc:

2.

lRM u =

Vi: R phn lc ti gi cht; R = 193602

387202

==

P N

l = 40 mm chiu di cht, chn theo cc cp ch to. Vy: Mu = 19360.20 = 387200 N/mm

[ ] [ ] 67,665,2.5,1250

.

'

'

1===

kn

N/mm2

Thay cc thng s vo cng thc trn s b ta xc inh c ng knh cht

72,3867,66.1,0

3872003 =d mm

37

m bo cho cht lm vic tin cy ng thi tit kim vt liu, s b ta chn ng knh cht d = 40 mm.

Tnh cht chnh ca cp lch tm. Khi nng vt vi ti trng nng 1tn, cht s chu un, m bo cht lm vic an ton ta tin hnh tnh chn theo bn un.

Tnh tng t nh cht cam lch tm ta xc nh c ng knh cht, ng vi

l = 40 mm, P = 10000 N.

66,2467,66.1,0

1000003 =d mm

m bo cho cht lm vic tin cy ng thi tit kim vt liu, s b ta chn ng knh cht d = 25 mm.

Trong thc t tng tin cy ngi ta thng lm cam lch tm c b mt vi cc kha hnh rng ca.

m rng phm vi s dng ca cu trc, ta c th ch to cht chnh ca cp lch tm dng bulong ai c, thun tin cho vic tho lp, c th thay bng cc thit b mc vt khc nhau tu theo loi vt cn vn chuyn.

2.1.2.11. Cc b phn khc ca c cu nng b. Cp u cp ln tang Ta s dng phng php kp u cp

trn tang thng thng: mi u cp dng 3 tm cp, tng ng vi ng knh dy cp dc = 5,6 mm, bc ct rnh t = 8 mm, vt cy M10.

Lc tnh ton i vi cp cp xc nh theo

cng thc (2-16) (tr.22).

Nee

SS f 13527844

4.14,0max

0 === pi Hnh 2.8. Cp cp trn tang

Trong : Smax =7844 N

f = 0,14 h s ma st gia tang vi mt cp.

pi 4= - gc m ca cc vng cp d tr trn tang.

d1

lo

38

Lc ko cc vt cy

NfS

p 482814,0.2

13522

0===

Lc un cc vt cy

NfPP 67614,0.4828.0 === ng sut tng xut hin trong thn vt cy, theo cng thc (2-17) (tr.23)

31

002

1 ..1,04.

.

3,1dZ

lPd

Z

P+=

pi

Trong : d1 = 8 mm ng knh trong ca vt cy. l0 = 8 mm tay n t lc P0 (l0 l + c). Z = 3 s bu lng cp cp.

Vy: 232 /84,762,3564,418.3.1,08.676

48.

.3

4828.3,1mmN=+=+=

pi

Vy cc vt cy ny c th lm bng thp CT3 c ng sut cho php:

[ ] = 75 85 N/mm2 c. Tnh trc tang

S tnh nh hnh 2-9.

CD

R = 7844

RA =4047 RA = 3797 RARA

35 3015278,5

B

217

A

Hnh 2.9. S tnh trc tang. Xt trng hp v tr ca lc cng dy trn tang s khng thay i v nm

im gia tang.

Tr s ca hp lc ny bng:

39

R = Smax = 7844 N

T s tnh trc tang trn hnh (2-9) ta xc nh c ti trng tc dng ln may bn tri (im D).

RD = 7844 1525,78

= 4047 N

Ti trng tc dng ln may bn phi (im C) RC = R RD = 7844 4047 = 3797 N

Phn lc ti A bng:

RA = 391921730.3797)30152(4047

=

++ N

Phn lc ti B l:

RB = R RA = 7844 3919 = 3925 N Mmen un ti D

MD = 3919.35 = 137165 Nmm Mmen un ti C

MC = 3925.30 = 117750 Nmm Trc tang khng truyn mmen xon, ch chu un. ng thi trc quay cng

vi tang khi lm vic, nn n s chu ng sut un theo chu k i xng.

Vt liu trc tang dng thp 45 vi gii hn bn 2/610 mmNb= , gii hn

chy 2/430 mmNch = v gii hn mi 2' 1 /250 mmN= . Khi ng sut un cho

php vi chu k i xng trong php tnh s b c th xc nh theo cng thc (1-12) (tr.12).

[ ] [ ] 782.6,1250

.

'

'

1===

kn

N/mm2

Vi cc h s k v [ ]n tra theo bng 1-5 v 1-8 Vy ti im D trc phi c ng knh l:

[ ] 2678.1,0137165

.1,033 ==

DMd mm

40

Kt cu trc cng cc kch thc cho trn hnh 2-12. Trc cn c kim tra ti

cc tit din c kh nng chu ng sut ln nht: cc tit din cn kim tra l I-I, II- II, III-III v IV-IV.

- Ta kim tra tit din I-I, c ng knh d = 30 mm

ng sut un ln nht: 61,6530.1,0

176610.1,0 33

===

dM D

u N/mm

3015235

30

2830

28

IVIIIII

III IV III

I

Hnh 2-10. Kt cu trc tang. Xut pht t tui bn tnh ton A =15 nm, vi ch lm vic nh v s ti

trng hnh (2-4) ta s tnh s chu k lm vic nh sau: S gi lm vic tng cng

T = 24.365.A.kn.kng = 24.365.15.0,25.0,33 = 10573 h Trong : kn = 0,25; kng = 0,33 tra theo bng (1-1). S chu k lm vic tng cng

Z0 = 60Tnt(C) = 60.10573.32.0,15 = 3045024 Trong : nt = 32 v/ph s vng quay trc tang.

(C) = 0,15 cng lm vic ca c cu vi ch lm vic nh. S chu k lm vic tng ng vi cc ti trng Q1, Q2, Q3.

Z1 = 102 Z0 = 10

2 3045024 = 609005

Z2 = 105 Z0 = 10

5 3045024 = 1522512

Z3 = 103 Z0 = 10

3 3045024 = 913057

41

S chu k lm vic tng ng l:

Zt = 609005.18 + 1522512.0,758 + 913057.0,28 = 761258 H s ch lm vic

4,1761258

10108

7

87

===

tc Z

k

Gii hn mi tnh ton

3253,1.250.' 11 === ck N/mm2

H s cht lng b mt y ly = 0,9 b mt gia cng tinh. H s kch thc ly 88,0= (bng tnh chi tit my). H s tp trung ng sut 2=k .

H s an ton l

96,10.

61032561,65

9,0.88,02

325

.

.

1

1=

+=

+

=

m

bu

kn

H s an ton cho php ca trc trong iu kin lm vic bnh thng l: [n] = 1,5 2,6. Vy trc tang m bo an ton

i vi cc tit din II-II, III-III v IV-IV ta cng lm php kim tra tng t. d. Khp ni trc ng c vi hp gim tc Ta s dng khp ni vng n hi, l loi khp ni di ng c th lp v lm

vic khi hai trc khng ng trc tuyt i; ngoi ra loi khp ny cn gim c

chn ng v va p khi m my v khi phanh t ngt. Pha na khp bn hp gim tc kt hp lm bnh phanh.

Khi mmen ln nht m khp phi truyn c th xut hin trong hai trng

hp: khi m my nng vt v khi phanh hm vt ang nng.

- Khi m my nng vt, vi h s qu ti ln nht quy nh, s xut hin

mmen m my ln nht bng Mm max = 2,5.Mdn = 2,5.11,4 = 28,5 Nm

Phn d thng qun tnh ca c h thng. Md = Mm max Mn = 28,5 14,2 = 14,3 Nm

42

Trong : Mn = 14,2 mmen tnh khi nng vt ( tnh phn trc). Mt phn mmen Md ny tiu hao trong vic thng qun tnh cc tit my quay

bn pha trc ng c (rto ca ng c in k c na khp), cn li mi l phn truyn qua khp.

Mmen v lng ca na khp pha ng c ly bng 40%, mmen v lng ca c khp

( ) 2'2 0864,0216,0.4,0 NmDG kii == Trong : ( ) 22 216,0 NmDG kii = ( tnh phn trc). Mmen v lng cc tit my quay trn gi ng c

( ) ( ) ( ) 2'22'2 5664,00864,048,0 NmDGDGDG kiirotoiiIii =+=+= Momen v lng tng ng ca vt nng (c vn tc vn) chuyn v trc ng

c.

( ) 222

2

2

02 051,0

14201010250.1,0.1,0 Nm

n

vQDGc

n

tdii ===

Tng mmen v lng ca c h thng

( ) ( ) ( ) 8166,0051,0)216,048,0(1,1222 =++=+= tdiiIiiii DGDGDG Nm2 Tng mmen v lng ca phn c cu t na khp bn pha hp gim tc v sau

k c vt nng

( ) ( ) ( ) 2502,05664,08166,0'22'2 === Iiiiiii DGDGDG Nm2 Phn mmen d truyn qua khp

( )( ) 38,48166,0

2502,03,142

'2'

===

ii

iidd DG

DGMM Nm

Tng mmen truyn qua khp

Mk = Mn + Md = 14,2 + 4,38 = 18,58 Nm - Khi hm vt ang nng, mmen t trn phanh l Mph = 14Nm. Tng mmen

thng qun tnh ca cc h thng

Mqt = Mph + M*t = 14 + 9,3 = 23,3 Nm Trong : M*t = Mh = 9,3 Nm ( tnh phn trc)

43

Ta c th tnh c phn mmen truyn qua khp thng qun tnh cc tit

my quay trn pha ng c bng cch tng t nh trn. Mt khc cng c th tnh xut pht t thi gian phanh, theo cng thc (3-6) (tr.52).

20

2*1

200

*

1

.).(375..

)(375)(

iaMMnDQ

MMnDG

ttphtph

Iiinph

++

+=

= 1287,034,31.2.3,23.375

807,0.1420.1406,0.102503,23.375

1420.686,0.1,122

2

=+ s

Mmen truyn qua khp thng qun tnh s bng

( )663,16

1287,0.3751420.5664,0

.375. 1

'2'"

====

n

ph

Iiiqtk t

nDGMM Nm

Nh vy, khi phanh vt ang nng khp phi truyn mmen ln hn, do cn phi kim tra kh nng truyn ti ca khp theo mmen truyn yu cu l M =16,663 Nm. Kim tra iu kin an ton ca khp ni:

M.k1.k2 = 16,665.1,3.1,1 = 23,824 Nm < Mmax = 46,62 Nm Trong : k1, k2 l h s tnh n mc quan trng ca c cu v iu kin

lm vic ca c cu, gi tr c tra theo bng (2-7). Mmax = Rn.P = 46,26 Nm (da vo phn tnh chn phanh). Vy khp ni chn lm vic an ton.

Bng (2-7). H s k1 v k2. k2 i vi ch lm vic

Loi c cu k1 Nh TB N RN

C cu nng vt

C cu nng chuyn kim loi lng

C cu thay i tm vi

C cu di chuyn v c c vu quay

1,3

1,5

1,4

1,2

1,1

1,2

1,3

1,5

44

2.2. TNH C CU DI CHUYN XE CON 2.2.1. Chn s tnh v cc thng s c bn

S c cu di chuyn xe con Cch b tr c cu di chuyn xe con nh s hnh 2.11. C cu di chuyn b tr kiu treo, ray l dm thp ch I, cc bnh xe chy trn

cnh di ca dm ch I. C cu dn ng in, mmen xon c truyn t ng c in c gn phanh 1 n bnh xe 5 qua cc cp bnh rng (b truyn bnh rng h 4). Cc bnh xe c dn ng mt bn bng c hai bnh. Cc c cu c lin kt vi nhau bng tm chu lc 3.

Cc thng s c bn - Ti trng nng Q = 1 tn = 10000 N. - Trng lng xe ln k c b phn mang vt G0 = 4000 N (xc nh theo s liu cc cu trc ch to). - Vn tc di chuyn xe con Vx = 20 m/ph. - Ch lm vic ca c cu: nh.

1. ng c in c gn phanh.

2. Hp s.

3. Tm chu lc.

4. B truyn bnh rng h. 5. Bnh xe con.

Hnh 2.11. S c cu di chuyn xe con. 2.2.2. Tnh c cu di chuyn 2.2.2.1. Tnh bnh xe

Kt cu cm bnh xe c m t trn hnh 2.12.

5 4 3 2 1

45

40 130

Hnh 2.12. Trc v bnh xe ca c cu di chuyn. Ta chn loi bnh xe hnh cn c mt thnh bn vi cc kch thc theo OTC 3569 60. ng knh bnh xe s b chn Dbx = 130 mm, ng knh ngng

trc d = 40 mm. Ti trng tc dng ln bnh xe: Ti trng tc dng ln bnh xe gm c trng

lng bn thn xe ln G0 = 4000N v trng lng vt nng Q = 10000N. Trng lng xe xem nh phn b u cho cc bnh. Khi khng c vt nng cc bnh xe chu ti trng t nht

Pmin = 100044000

40

==

G N

Khi c vt nng ti trng tc dng ln bnh xe l u (vt nng t ti tm xe) Tng ti trng tc dng ln mi bnh xe l:

Pd = 2500410000

4==

Q N

Vy ti trng ln nht tc dng ln bnh xe l: Pmax = 1000 +2500 = 3500 N

Ti trng tc dng ln bnh xe theo cng thc

Pbx = kbx Pmax

Trong :

kbx = 1,1 h s tnh n ch lm vic ca c cu (bng 2-8) - h s tnh n s thay i ca ti trng xc nh theo cng thc

46

72,0

4000100001

1121

1

1121

33

0

=

+

+=

+

+=

GQ

Vy: Pbx = 0,72.1,1.3500 = 2772 N Sc bn dp ca bnh xe c kim tra nh sau: bnh xe c ch to bng thp

c 55. m bo lu mn vnh bnh c ti t cng HB = 300 320.

Sc bn dp c kim tra theo cng thc (2-67) [tr.42]

[ ]dbxdrb

P =

.

190

Trong : b = 30 mm chiu rng mt lm vic.

r = 65 mm bn knh bnh xe.

Vy: 22665.30

2772190 ==d N/mm2

ng sut dp cho php trong bng (2-19) [tr.44] l: [ ]d = 750 N/mm2 Vy kch thc bnh xe chn l an ton.

Bng (2-8). Tr s kbx.

2.2.2.2. Chn ng c in Phng php tnh lc cn chuyn ng cho kiu xe ln c 4 bnh xe, lp bng

ln trn khung cng chy trn hai cnh di ca dm ch I. Tng lc cn tnh, tnh theo cng thc (3-43) [tr.66].

Ch lm vic kbx My trc dn ng bng tay My trc dn ng bng my

Ch lm vic nh

Ch trung bnh Ch nng

Ch rt nng

1,0

1,1

1,2

1,4

1,6

47

Wt = W1 + W2 + W3 + W4 + W5 + W6

Trong : W1 lc cn do ma st, tnh theo cng thc:

W1 = DfdQGk t ++ 2)( 0

Vi: kt = 1,2 - h s tnh n ma st thnh bnh, ly theo bng (2-6) [tr.64]. = 0,3 0,5 h s ma st ln, chn = 0,3.

f = 0,015 h s ma st trong trc, tra bng (2-9). G0 = 4000 N trng lng xe con k c vt mang.

Q = 10000 N trng lng vt nng. D = 130 mm ng knh bnh xe. d = 40 mm ng knh ngng trc.

Bng (2-9). H s ma st trong trc bnh xe. Loi trt ln

Kt cu

h

C hp trc

bi du bi v thanh ln

nn

f 0,10 0,08 0,015 0,02

Vy: W1 = 15513040.015,03,0.2)100004000(2,1 =++ N

W2 lc cn theo dc, tnh theo cng thc (3-41) [tr.65]. W2 = (G0 + Q) = 0,002(4000 +10000) = 28 N

W3 lc cn ca gi, do thit b lm trong nh xng nn W3 = 0. W4 lc cn do ma st thnh bnh vo ray.

W4 = ( ) hrfQG 210 +

Vi: f1 0,17 h s ma st trt ca bnh xe trn ray h - khong cch t M n A (im tip xc ca thnh bnh xe n im ln ca bnh xe).

r bn knh trung bnh ca bnh xe, mm; thng thng r

h = 0,4 0,7

48

Vy: W4 = ( ) 2025,0.17,0100004000 2 =+ N W5 lc cn do trt ngang khi xe b xin lch so vi ng ray hnh (2-

14)

W5 = ( )rB

fQG+

+

10

Vi: - tng khe h hai bn thnh bnh v ray, mm.

= K k ; S b chn = 1mm. B khong cch trc gia hai bnh xe, s b chn B = 200mm.

r = 67,5mm bn knh trung bnh ca bnh xe.

Vy: W5 = ( ) 95,67200117,0.100004000 =+

+ N

W6 lc cn do trt hnh hc ca bnh xe hnh cn.

W6 = ( ) ( )2121

10 2 rrrrfQG

+

+

r1 = 75mm, r2 = 60mm bn knh ln nht v nh nht ca bnh xe (theo cc bnh xe ch to).

W6 = ( ) ( ) 13260752607517,0.100004000 =

+

+ N

Wt = 155 + 28 + 0 + 202 + 9 + 132 = 526 N

Hnh 2.13. S tnh lc cn do thnh bn.

KM

r

h

49

Hnh 2.14. Xe ln trn dm ch I. Cng sut tnh yu cu i vi ng c in, c xc nh theo (3-60) [tr.71].

21,085,0.1000.60

20.526.1000.60

===

c

xtt

vWN

kW

Trong : c = 0,85 kW hiu sut c cu di chuyn, theo bng (1-9) [tr.15] Tng ng vi ch lm vic ca c cu l nh c C 15%, ta chn ng c

in c tiu chun ha c km theo hp s v phanh.Vi phng n ny s gim

bt c cng vic tnh ton v tit kim c gi thnh m vn m bo c tin cy khi lm vic.

Hnh 2.15. ng c in VIGE.

4-12

131 1

50

165

4915 7

65165

250.8293.8

B K

50

Cn c vo cng sut tnh yu cu, ta chn ng c VIGE ca hng HITACHI

c cc thng s sau:

Kiu ng c: VIGE. Cng sut danh ngha: Nc = 0,3 kW. S vng quay u ra hp s: nhs = 150 v/ph. Khi lng: G = 9,9 kg.

2.2.2.3. Xc nh t s truyn b truyn h. S vng yu cu ca bnh xe m bo vn tc di chuyn xe con.

nbx = 50130,020

==

pipi bx

x

DV

v/ph

T s truyn cn c i vi b truyn h .

inh = 350150

==

bx

hs

n

n

2.2.2.4. Thit k b truyn h, bnh rng tr - thng Do c im dn ng ca c cu l dn ng mt pha bng c hai bnh s

truyn ng cho hai bnh cn li, nn khi thit k ta ch cn tnh ton cho mt bnh ch ng l .

T t s truyn va tnh c ta tn hnh thit k cp bnh rng tr thng, mt cp. Cc thng s tnh c l: s vng quay, mmen xon v cng sut trn trc ca

cp bnh rng tr thng c ghi trn bng sau:

Trc

Thng s I II

i 3

n (v/ph) N (kW) M (N.mm)

150 0,3

28650

50 0,288

82512

51

a. Chn vt liu v phng php ch to c im ca b truyn h bnh rng tr thng l chu ti ln, v vy i hi vt

liu ch to phi c c tnh cao.Tuy nhin chn vt liu phi tnh n gi thnh sn

xut v kh nng cng ngh cho php. Cn c vo nhng yu cu trn ta chn loi vt

liu cho cp bnh rng thit k nh sau:

- i vi bnh rng nh, chn thp C50 thng ha c cng HB = 220, b =

620 N/mm2, ch = 320 N/mm2, ng knh phi nh hn 100mm.

- i vi bnh rng ln, chn thp C45 thng ha, c cng HB = 190, b =

580 N/mm2, ch = 320 N/mm2, ng knh phi 100 300 mm.

b. Xc nh ng sut cho php - ng sut tip xc cho php:

[ ] Ntx KHB..5,2= Trong : KN h s chu k ng sut tip xc.

KN = 6 0tdN

N

Vi: N0 =106 - 107 s chu k ca ng cong mi tip xc, ta chn: N0 = 107.

Ntd s chu k ng sut tng ng.

N1 = 60 n1Tu = 60.100.10573.1 = 6,34.107 > 107 N2 = 60 n2Tu = 60.50.10573.1 = 3,17.107 > 107 Cho nn KN1 = KN2 = KN = 1

Vy ng sut tip xc cho php l:

[ ] 5501.220.5,21 ==tx N/mm2 [ ] 4751.190.5,22 ==tx N/mm2 - ng sut un cho php:

[ ] ( )

KnK N

u.

..6,14,1 '1=

Trong : 1 - gii hn mi un trong chu k i xng c th xc nh theo

cng thc:

52

1 = (0,45 0,5) b n = 1,1 h s an ton.

K = 1,8 h s tp trung ng sut chn rng.

m

tdN N

NK 0' = - h s chu k ng sut un.

Vi: m = 6 bc ng cong mi un. N0 = 5.106 s chu k c s ca ng cong mi un.

65,010.34,6

10.56

7

6'

1 ==NK

73,010.17,3

10.56

7

6'

2 ==NK

Vy: [ ] 1378,1.1,1

65,0.620.45,0.5,11 ==u N/mm

2

[ ] 1448,1.1,1

73,0.580.45,0.5,12 ==u N/mm

2

- ng sut tip xc cho php khi qu ti.

Bnh nh: [ ] [ ] 1373550.5,25,2 1max1 === txH N/mm2 Bnh ln: [ ] [ ] 1195478.5,25,2 2max2 === txH N/mm2 - ng sut un cho php khi qu ti.

Bnh nh: [ ] 484220.2,2.2,2max1 === HBF N/mm

2

Bnh ln: [ ] 418190.2,2.2,2max1 === HBF N/mm

2

c. S b chn h s ti trng Ksb, h s chiu rng m v s rng Z

Gi tr Ksb c th chn s b trong khong 1,2 1,6. Khi chn ch kh nng chy mn ca vt liu ch to, cch b tr b truyn v mi trng lm vic ca b truyn.

Gi tr h s m = b/m = 12 c chn theo bng 40 TKCTM.

S rng Z c chn theo kinh nghim tha mn iu kin:

Z1 > Z1min =17 rng, chn Z1 = 25 rng

53

Z2 = inh.Z1 = 3.25 = 75 rng d. Xc nh mun n khp theo sc bn tip xc v khong cch trc. - S b tnh ng knh vng ln bnh rng nh theo cng thc:

[ ]36

...

..10.1,19um

sbnzy

NKm

Trong : K = 1,5 h s qu ti. N = 0,3 cng sut ng c. y = 0,429 h s dng rng, xc nh theo (bng 36) da vo s rng tng ng.

[ ]u = 137 N/mm2 - ng sut un cho php. n = 150 s vng quay ca ng c.

Vy: 48,1137.12.150.25.429,03,0.5,1.10.1,19

36

=sbm

Theo tiu chun ta chn msb = 2.

Tnh khong cch trc A, s b

Asb =( ) ( ) 100

27525.2

221

=

+=

+ ZZmsb mm

e. Chn cp chnh xc ch to bnh rng chn cp chnh xc ch to bnh rng ta cn xc nh vn tc vng ca rng

V = ( ) 2,0)13(10.6150.50.2

110.6..2

10.6..

441

411

=

+=

+=

pipipi

inAdn

m/s

Theo gi tr vn tc vng tnh chn cp chnh xc ch to bnh rng l cp 9, theo bng (10-1-CTM).

f. Xc nh chnh xc khong cch trc A H s ti trng c xc nh theo cng thc:

K = Ktt.K Trong : Ktt h s tp trung ti trng.

21' +

=tt

tt

KK

54

Vi 1' =ttK , tra theo (bng 32) ng vi 4,0213

.2.02

1=

+=

+iA

Vy: 12

11=

+=ttK

Suy ra: K = 1.1,1 = 1,1

Vi K = 1,1- H s ti trng ng tra (bng 33 - TKCTM). Ta thy: K # Ktt hn 5% do ta phi i xc nh li khong cch trc A theo

cng thc:

972,11,1

.100 33 ===sb

sb KKAA

mm

Ta xc nh li gi tr, 94,12,11,1

.2 33 ===sb

sb KK

mm

V m < msb nn ta khng cn tnh li.

g. Kim nghim rng theo qu ti t ngt

b truyn c kh nng chu qu ti trong thi gian ngn cn kim tra b truyn qu ti theo iu kin :

[ ]max

. FqtuF K =

[ ]max

. HqttxH K =

Trong : u - ng sut un theo ti trng danh ngha

[ ]uunbzmyNK

=....

..10.1,192

6

Vi : K = 1,5 h s qu ti. N cng sut ca ng c trn trc tng ng.

y = 0,429 h s bin dng rng, xc nh da vo s rng. z s rng tng ng trn mi bnh. n s vng quay trn mi trc.

b = AA . = 0,2.97 = 19,4 mm, chiu rng bnh rng.

m =2 m un n khp.

55

Vy: 84,68150.4,19.25.2.429,03,0.5,1.10.1,19

2

6

1 ==u N/mm2

59,5650.4,19.75.2.501,0

288,0.5,1.10.1,192

6

2 ==u N/mm2

Ta thy: [ ]11 uu < , [ ]22 uu < Tha mn yu cu. Tng t ta cng tnh c gi tr ng sut tip xc theo cng thc:

( ) ( ) 60950.4,19

288,0.5,1.133.9710.05,1

.

.1.

10.05,1 36

2

36

=

+=

+=

nbNKi

iAtx N/mm2

Vi Kqt = 2,4 h s qu ti ca h thng, ta tnh c cc ng sut un qu ti

tng ng l:

[ ]max1

21 /288,1654,2.84,68 FF mmN ==

[ ]max2

22 /816,1354,2.59,56 FF mmN ==

[ ]max

2/46,9434,2.609 HH mmN ==

Kt qu tnh ton u tha mn.

h. Xc nh cc thng s hnh hc ch yu ca b truyn: - Khong cch trc A = 97mm - Chiu cao rng h = 2,25.m = 2,25.2 = 4,5 mm - h hng tm C = 0,25.m = 0,25.2 = 0,5 mm - ng knh vng chia dc1= m.Z1 = 2.25 = 50 mm dc2= m.Z2 = 2.75 = 150 mm - ng knh vng ln d1= dc1; d2 = dc2 - ng knh vng nh rng De1 = dc1 + 2m = 50 + 2.2 = 54 mm De2 = dc2 + 2m = 150 + 2.2 = 154 mm - ng knh vng chn rng Di1 = dc1 - 2m - 2C = 45 mm Di2 = dc2 - 2m - 2C = 145 mm thun tin cho vic lp rp ta c th ch to bnh rng gn lin vi bnh xe

ca c cu di chuyn (hnh 2-11).

56

i. Tnh lc Lc tc dng trong trong b truyn bnh rng c xc nh theo hai thnh

phn:

- Lc vng P1 = P1 = 14324028650.2.2

==

dM x

N

- Lc hng tm Pr1 = Pr2 = P.tg = 1432.tg200 = 521N 2.3. TNH C CU DI CHUYN CU 2.3.1. Chn s tnh v cc thng s c bn

M hnh c cu di chuyn cu: Nh bit, cu trc thit k l loi c ti trng nng v khu nh do yu

cu kt cu ca c cu di chuyn phi nh gn v lm vic tin cy. Vy ta chn phng n cho c cu di chuyn cu l: c cu di chuyn dn ng

chung vi trc truyn quay chm.

Kt cu ca c cu di chuyn cu bao gm: trc ng c in 1, hp gim tc 2 v cc on trc truyn 3 ni vi nhau v ni vi trc ra ca hp gim tc bng cc khp ni 4. Trc truyn ta trn cc gi 5 bng bi truyn chuyn ng ti bnh xe 7 di chuyn cu trc qua cp bnh rng h 6. Vi phng n ny mmen xon trn trc truyn s nh v kch thc ca n cng nh.

Hnh 2.16. S c cu di chuyn. Cc thng s c bn:

Ti trng: Q = 1t = 10000 N. Trng lng xe ln k c b phn mang vt: Gx = 4000 N.

1 2

4

5 3 6

7

57

Trng lng cu k c c cu di chuyn: G0 = 20000 N ( theo cc cu ch to). Vn tc di chuyn cu l : Vc = 35 m/ph. Ch lm vic ca c cu l : nh.

S c cu di chuyn - hnh (2.16.). 2.3.2. Tnh c cu di chuyn cu 2.3.2.1. Tnh bnh xe v ray

Ta chn loi bnh xe hnh tr c hai thnh bn vi cc kch thc theo OCT 3569-60. ng knh bnh xe chn Dbx= 320 mm, ng knh ngng trc lp d = 60mm. Cn c vo kch thc tng ng ca bnh xe chn chiu rng vnh bnh l 100 mm, chn ray cu trc KP80 lm ray cho cu ln.

Ti trng tc dng ln bnh xe: Bnh xe b tr vi khong cch bnh (khu dm) L = 8000 mm v khong cch trc B = 1200 mm. Khi ti trng tc dng ln cc bnh xe gm trng lng bn thn cu Gc, trng lng bn thn xe ln Gx v trng lng vt nng Q. Ti trng ln nht tc dng ln mi bnh xe l khi xe con mang c cu nng c vt nng ln nht ti mt u gii hn bn tri (phi) cu, hnh (2-17).

Hnh 2.17. S ti trng tc dng.

Pmax = P1 = P2 = ( ) cx GLlLQG

41

21

+

+ =

= ( ) 116502000041

84,08100004000

21

=+

+ N

Ti trng nh nht tc dng ln mi bnh xe khi xe ln khng c vt nng ti u gii hn bn tri (phi) cu.

l = 400

Gx + Q Gc

A

L = 8000

B

RB

4000

RA

58

Pmin (1;2) = =+ cx GLlLG

41

21

= 69002000041

84,08

.400021

=+ N

Ti trng tng ng tc dng ln bnh xe, xc nh theo cng thc

Pbx = kbxPmax

Trong : - h s tnh n s thay i ca ti trng, xc nh theo cng thc:

82,0

200004000100001

1121

1

1121

33

0

=

++

+=

+

+=

GQ

kbx = 1,1 h s tnh n ch lm vic ca c cu, bng (2-8). Vy: Pbx = 0,82.1,1.11650 = 10508 N

Bnh xe c ch to bng thp c 55 lm tt c rn HB = 300 320.

Sc bn dp c kim tra theo cng thc:

Trong : b = 60 mm chiu rng mt lm vic.

r = 160 mm bn knh bnh xe.

Vy: 199160.60

10508190 ==d N/mm2

ng sut dp cho php trong bng (2-19) [tr.44] l: [ ]d = 750 N/mm2 Vy kch thc bnh xe chn l an ton.

2.3.2.2. Chn ng c in Lc cn chuyn ng do ma st, xc nh theo cng thc:

( )bxD

dfQGW .201 ++=

Trong : = 0,5 H s ma st ln, bng (3-7) f = 0,02 h s ma st trong trc, bng (2-9).

[ ]dbxdrb

P =

.

190

59

Vy: ( ) 234320

60.02,05,0.210000240001 =+

+=W N

Lc cn do dc ng ray, xc nh theo cng thc:

( ) 34)1000024000.(001,002 =+=+= QGW N Trong : = 0,001 dc ng ray, tra bng(2-9) Tng lc cn tnh chuyn ng, xc nh theo cng thc:

Wt = ktW1 + W2 = 3,2.234 +34 = 783 N

Trong : kt = 3,2 h s k n lc cn do ma st thnh bnh v mt u moay bnh xe, tra bng (3-6) ng vi t s L/B =1.

Cng sut tnh yu cu i vi ng c in, xc nh theo cng thc:

537,085,0.1000.60

35.783.1000.60

.

===

c

ctt

VWN

kW

Trong : c = 0,85 hiu sut ca c cu di chuyn (hiu sut ca b truyn) Tng ng vi ch lm vic ca c cu l nh, s b ta chn ng c in

K 32-6 c cc c tnh sau: - Cng sut danh ngha Nc = 0,6 kW - S vng quay danh ngha nc = 930 v/ph

- H s qu ti dnM

M max = 1,9

- Mmen bnh ca roto ( )rotoii DG

2= 0,2 Nm2

- Khi lng 27 kg

2.3.2.3. T s truyn chung S vng quay yu cu ca bnh xe l:

3532,0.

35===

pipi bx

bxbx D

Vn

T s truyn chung yu cu i vi b truyn

5,2635

930===

bx

cc

n

ni

60

2.3.2.4. Kim tra ng c in v mmen m my - Gia tc ln nht m bo h s bm 2,1bk c tnh t cng thc:

+= 0

0max0 2,1 tbx

dd W

DdfGG

Ggj

Trong : g = 9,81 m/s2 gia tc trng trng. G0 = 24000 N trng lng cu k c c cu di chuyn. Gd = 6900 N tng p lc ln cc bnh dn khi khng c vt nng.

= 0,20 h s bm ca bnh xe vo ray khi lm vic trong nh.

f = 0,02 h s ma st trong trc, bng (2-9). 0

tW - Tng lc cn tnh chuyn ng ca c cu khi khng c vt nng

=0

tW =+

bxDdfG .20

160320

60.02,05,0.224000 =+ N

Vy: 41,01603,0

05,002,0.69002,1

2,0.690024000

81,9max0 =

+=j m/s2

- Mmen m my ti a cho php, xc nh theo cng thc

( )0

12

021

20

00

3753752 mIii

cmc

bx

cc

bxtm t

nDGti

nDGi

DWM ++=

Trong : ic = 25 - t s truyn chung i vi b truyn.

c = 0,85 hiu sut ca b truyn.

n1= 930 s vng quay ca trc ng c.

0mt - thi gian m my tng ng, 4,141,0.60

3560 max0

0=== j

vt cm s

= 1,1 h s k n nh hng qun tnh ca cc tit my quay trn cc trc sau trc ng c.

Iii DG )( 2 - tng mmen v lng ca cc tit my quay trn trc I, Nm2

Iii DG )( 2 [ khopiirotoii DGDG )()( 22 + ] = 1,1[0,2 + 0,13] = 0,363 Nm2

Vy: 43,264,1.375

930.363,085,0.4,15,26.375

930.32,0.2400085,0.5,26.2

32,0.1602

20

=++=mM Nm

61

- Mmen danh ngha ca ng c chn

16,6930

6,0.95509550 ===

c

cdn

n

NM Nm

- Mmen m my trung bnh ca ng c

( ) 5,1016,6.7,17,12

1,15,28,1)( ===

+= dn

dndncm M

MMM Nm

Ta thy: 0)( mcm MM < , mc d chn Mm max = 1,9Mdn ng c vn c mmen

m my nh hn so vi tr s cho php, vy ng c chn l hp l. 2.3.2.5. Tnh chn phanh

phanh c nh gn, ta s t phanh trc th nht tc l trc ng c. Khi

vic tnh chn phanh gm cc bc sau. - Mmen phanh c xc nh theo cng thc

( )0

12

021

20

0

3753752 phIii

phc

cbx

cc

bxtph t

nDGti

nDGi

DWM ++=

Trong : 0pht - thi gian phanh khi khng c vt nng:

42,141,0.60

3560 0

0===

ph

c

ph jv

t s

0phj = 0,75 gia tc hm, theo bng ( ) ty theo t l s bnh dn v h

s bm .

Vy: 97,642,1.375930.363,0

42,1.5,26.37585,0.930.32,0.24000

85,0.5,26.232,0.160

2

2

=++=phM Nm

Ta chn phanh cho c cu l phanh a in t vi nhiu mt ma st v c kch

thc nh gn, lm vic tin cy (hnh 2.5.). Kt cu v nguyn tc lm vic ca n c trnh by trong phn (2.1.2.8.).

n gin thi gian cho vic thit k v tit kim chi ph ng thi vn m

bo tin cy cho c cu trong qu trnh lm vic, ta chon phng n mua sn cn c vo mmen phanh yu cu.

62

2.3.2.6. Thit k b truyn B truyn ca c cu bao gm hp gim tc v b truyn bnh rng h, t s

truyn tng ng ca chng c tnh theo cng thc:

ic = inh.ih

Trong : ic t s truyn chung.

inh t s truyn ca b truyn bnh rng h. ih t s truyn ca hp gim tc.

T t s truyn ta tin hnh thit k cc cp bnh rng tr thng, cn c vo yu cu v t s truyn trung bnh ca b truyn bnh rng tr thng m ta c th chn: inh = 3 5, ih = 3 7. Vy s b ta chn inh = 5 th t s truyn ca hp gim tc s l:

ih = i/inh = 26,5/5 = 5,3 Tnh chn hp gim tc: Ta tin hnh thit k hp gim tc da vo cc thng s bit: Ih = 5,3 t s truyn chung. Nc = 0,6 kW cng sut ca ng c in. nc = 930 s vng quay trn trc ng c. S b ta chn hp gim tc c kt cu nh hnh (2-18). Ta lp bng phn phi cc gi tr thng s ng ng lc hc cc cp ca h

truyn dn nh sau:

Bng (2-11). Gi tr thng s ng ng lc hc cc cp ca h truyn dn. Trc

Thng s

I II III

I I12 = 2,65 I23 = 2

n, v/ph 930 351 175,5

N, kW 0,6 0,576 0,553

Mx, (N.mm) 6161 15673 30092

Vi: n1 = nc =930 v/ph ; n2 = 12

1

in ;

23

23 i

nn =

63

N1 = Nc =0,6kW; N2 = N1; N3 = N2

M1 = Mx = 6161(N.mm); M2 = i12. .M1; M3=i23. .M2 Trong : Mx m men xon trn trc

ng c ; = (0,95 0,97) hiu sut b truyn bnh rng tr, chn = 0,96

Cn c vo yu cu vo cng sut phi

truyn vi C15%, s vng quay trc vo, t s truyn v yu cu v lp rp, ta chn phng

n mua sn hp gim tc tiu chun da vo cc thng s ng ng lc hc ca cc cp

c ghi trn bng (2-11). Vi phng n ny s gim c chi ph v tnh ton.

Hnh 2.19. S hp gim tc. Thit k b truyn bnh rng h T t s truyn va tnh c ta ta tn hnh thit k cp bnh rng tr thng

rng thng, mt cp. Cc thng s tnh c l: s vng quay, mmen xon v cng

sut trn trc ca cp bnh rng tr thng.

a. Chn vt liu v phng php ch to c im ca b truyn h bnh rng tr thng l chu ti ln, v vy i hi vt

liu ch to phi c c tnh cao. Tuy nhin chn vt liu phi tnh n gi thnh sn

xut v kh nng cng ngh cho php. Cn c vo nhng yu cu trn ta chn loi vt

liu cho cp bnh rng thit k nh sau:

Trc

Thng s I II

i 5

n (v/ph) N (kW) Mx (N.mm)

175,5 0,553 30092

35 0,531

144441

nc

n tr

64

- i vi bnh rng nh, chn thp C50 thng ha c cng HB = 220, b =

620 N/mm2, ch = 320 N/mm2, ng knh phi 100 200mm.

- i vi bnh rng ln, chn thp C45 thng ha, c cng HB = 190, b =

580 N/mm2, ch = 320 N/mm2, ng knh phi 200 400 mm.

Cc bc tip theo lm tng t nh trnh by phn 2.2.2.4 ta xc nh c cc i lng chnh sau:

b. ng sut cho php - ng sut tip xc cho php:

[ ] 5501.220.5,21 ==tx N/mm2 [ ] 4751.190.5,22 ==tx N/mm2 - ng sut un cho php:

[ ] 1378,1.1,1

65,0.620.45,0.5,11 ==u N/mm

2

[ ] 1448,1.1,1

73,0.580.45,0.5,12 ==u N/mm

2

- ng sut tip xc cho php khi qu ti.

Bnh nh: [ ] [ ] 1373550.5,25,2 1max1 === txH N/mm2 Bnh ln: [ ] [ ] 1195478.5,25,2 2max2 === txH N/mm2 - ng sut un cho php khi qu ti.

Bnh nh: [ ] 484220.2,2.2,2max1 === HBF N/mm

2

Bnh ln: [ ] 418190.2,2.2,2max1 === HBF N/mm

2

c. H s ti trng Ksb, h s chiu rng m v s rng Z

Gi tr Ksb c th chn s b trong khong 1,2 1,6. Khi chn ch kh nng chy mn ca vt liu ch to, cch b tr b truyn v mi trng lm vic ca b truyn.

Gi tr h s m = b/m = 12 c chn theo bng 40 TKCTM.

S rng Z c chn theo kinh nghim tha mn iu kin: Z1 > Z1min =17 rng, chn Z1 = 30 rng

65

Z2 = inh.Z1 = 5.30 = 150 rng d. Xc nh mun n khp theo sc bn tip xc v khong cch trc.

56,1137.12.186.30.451,0

553,0.5,1.10.1,193

6

=sbm

Theo tiu chun ta chn msb = 2

Tnh khong cch trc A, s b

Asb =( ) ( ) 180

2150300.2

221

=

+=

+ ZZmsb mm

e. Cp chnh xc ch to bnh rng Cp chnh xc ch to bnh rng l cp 9. f. Xc nh chnh xc khong cch trc A khong cch trc:

1922,1

47,1.180 33 ===

sbsb K

KAA mm

M un n khp xc nh li c gi tr bng:

14,22,1

47,1.2 33 ===

sbsb K

Kmm

g. Kim nghim rng theo qu ti t ngt

b truyn c kh nng chu qu ti trong thi gian ngn cn kim tra b truyn qu ti theo iu kin :

[ ]max

. FqtuF K =

[ ]max

. HqttxH K =

Kt qu tnh c l:

[ ]max1

21 /92,854,2.799,35 FF mmN ==

[ ]max2

22 /32,614,2.55,25 FF mmN ==

[ ]36

...

..10.1,19um

sbnzy

NKm

66

[ ]max

2/62,5894,2.6,380 HH mmN ==

Kt qu tnh ton u tha mn.

h. Cc thng s hnh hc ch yu ca b truyn: - Khong cch trc A = 192mm - Chiu cao rng h = 2,25.m = 2,25.2,14 = 5,45 mm - h hng tm C = 0,25.m = 0,25.2,14 = 0,535 mm - ng knh vng chia dc1= m.Z1 = 2,14.30 = 64,2 mm dc2= m.Z2 = 2,14.150 = 321 mm - ng knh vng ln d1= dc1; d2 = dc2 - ng knh vng nh rng De1 = dc1 + 2m = 64,2 + 2.2,14 = 68,48 mm De2 = dc2 + 2m = 321 + 2.2,14 = 325,28 mm - ng knh vng chn rng Di1 = dc1 - 2m - 2C = 58,85 mm Di2 = dc2 - 2m - 2C = 315,56 mm thun tin cho vic lp rp ta c th ch to bnh rng gn lin vi bnh xe

ca c cu di chuyn (hnh 2-20). i. Tnh lc Lc tc dng trong trong b truyn bnh rng c xc nh theo hai thnh

phn:

- Lc vng P1 = P1 = 481560144441.2.2

==

dM x

N

- Lc hng tm Pr1 = Pr2 = P.tg = 4815.tg200 = 1753 N

2.3.2.7. Tnh bnh xe v ray Ta chn bnh xe hnh tr c hai thnh bn vi cc kch thc theo OCT 3569-

60 . Ta chn ng knh bnh xe Dbx = 320 mm, ng knh ngng trc lp d = 60 mm

Bnh xe c b tr vi khong cch bnh (khong cch hai ray) Lr = 8000 mm v khong cch trc 1200 mm. Ti trng tc dng ln bnh xe gm trng lng cng trc Gct, trng lng xe ln k c b phn mang vt Gx v trng lng vt nng Q. Vy ti trng ln nht tc dng ln bnh xe khi nng vt l:

67

( ) ( ) 85004

100004000200004max

=

++=

++=

QGGP xc N

Ti trng tng ng tc dng ln bnh xe tnh theo cng thc:

Pbx = .kbx.Pmax = 0,8 .1,1.8500 = 7480 N

Trong : = 0,8 h s tnh n s thay i ti trng, tng ng vi

1400

10000>=

xGQ

, tra bng 2-

kbx = 1,1 h s tnh n ch lm vic ca c cu, bng (2-8). ng sut dp c kim tra theo cng thc:

[ ]dd Pm = 3 2max..3600

Trong : max = 320 mm - bn knh tng ng ln nht, i vi ng knh

bnh xe l 320 mm, ta chn ray P15 c 146min = mm.

m - h s, ph thuc h s: 45,0320146

min

max==

, theo bng (2- )ta chn

m = 0,536.

Vy: 8073207480

.536,0.3600 3 2 ==d N/mm2

i vi bnh xe lm bng vt liu l thp 40 XH, ng sut dp cho php l:

[ ] 2200=d N/mm2, c cng HB = 300 400. So snh kt qu ta thy [ ]dd < , vy bnh xe v ray lm vic an ton.

Hnh 2.20. Bnh xe v ray.

b2

h2 h1

R r1

r2

r 4

b 1

320 60

20

100 60

68

Bng (2-12). H s m.

Bng (2-13). Cc thng s ca ray P15. Kiu

ray

h1

(mm) h2

(mm) b1

(mm) b2

(mm) r1

(mm) r2

(mm) r4

(mm) R

(mm) P15 120 24 50 76 36 52 45 146

2.3.2.8. Tnh trc truyn Trc truyn chuyn ng quay t ng c ti cc bnh xe nn trong trng hp

ny mmen tc dng ln chc ch l mmen xon, vi gi tr mmen xon cu truyn trn trc nn trong trng hp ny ta xc nh c ng knh trc theo mmen

xon.

Gi tr ng knh trc s b c xc nh theo cng thc:

[ ]3 .2,0 xx

sbMd

Trong : Mx mmen xon truyn trn trc, Mx = 30092 N.mm.

[ ]x - ng sut xon cho php, [ ]x = 10 N/mm2.

Vy: 68,2420.2,0

300923 =sbd mm

So b chn d = 30 mm, trc truyn l trc c.

mmin

0,05 0,1 0,15 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0

m 1,28 0,97 0,80 0,716 0,6 0,536 0,49 0,468 0,44 0,42 0,40 0,38

69

CHNG III

TNH KT CU THP CA CU TRC

70

Cu gm dm chnh kiu ch I ni cng vi hai dm cui. Trn dm cui t cc bnh xe cu di chuyn dc phn xng. Xe ln di chuyn trn cnh di ca dm ch I.

Kt cu kim loi ca cu c tnh theo cc s liu ban u sau: - Ti trng: Q = 1t = 10000 N - Trng lng xe ln k c b phn mang vt Gx = 4000 N - Trng lng cu vi c cu di chuyn Gc = 20000 N - Khu dm cu L = 8 m - Trng lng khung gin thp Gt = 1500 N

3.1. TNH DM CHNH 3.1.1. Chn vt liu

Trong cu trc kt cu dm chnh chim mt phn rt ln, khi lng kim loi dng cho dm chnh chim n 60 80% khi lng kim loi my trc. L k cu chu ti chnh nn i hi kt cu phi bn trong trng hp phi chu ti trng ln nht. Ngoi vic phi m bo bn khi lm vic, kt cu kim loi cn d dng ra cng, p c gi thnh thp, b mt ngoi ca kt cu cn phng d nh g v d sn. V th vic chn kim loi thch hp cho dm chnh s dng chng mt cch kinh t nht l rt quan trng.

Cn c vo yu cu trn ta chn loi thp cho dm l: thp CT3 vi lng lu hunh khng cha qu 0,05%, lng pht pho khng qu 0,045%. y l loi thp thng dng cho cc dm chu ti ca my trc.

3.1.2. Xc nh cc ti trng tc dng ln dm chnh Cc ti trng tc dng ln dm chnh gm: ti trng khng di ng, ti trng di

ng, lc qun tnh khi phanh cc c cu.

- Ti trng khng di ng: l ti trng ca khung gin thp v c cu di chuyn cu, ti tr