bai giang may nang chuyen

8/6/2019 Bai Giang May Nang Chuyen

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MY

NNGCHUYN

Trnh ng Tnh

B mn C s Thit k my v Rbti hc Bch khoa H ni


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V tr v mc ch mn hc

Chuyn tip gia cc mn hc c s v

chuyn ngnh i tng l thit b tng th, khng cn l

cc chi tit ring l nh trong cc mn hcc s.

Cng c li cc kin thc hc nhSc bn VL, Nguyn l my, Chi tit my


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i tng nghin cu

Phng tin c gii ha vic nng/h

v vn chuyn vt nng. Cc thit b dng vn chuyn vt liu

vi s lng ln.


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Ni dung mn hc

Cc b phn v thit b my nng.

My chuyn lin tc.Yu cu v an ton thit b nng.


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Yu cu vi hc vin

Nm c cc ni dung sau:

Cu to, c im cu to ca mt s bphn v thit b my nng v my chuynlin tc.

Phng php tnh ton mt s b phn vthit b my nng v my chuyn lin tc.

Yu cu v an ton thit b nng.


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Ti liu tham kho chnh

[1]. o Trng Thng:My nng chuyn. HBK HN, 1993

[2]. Hunh Vn Hong, o Trng Thng:Tnh ton my trc. Nxb KHKT, HN, 1975

[3]. Trng Quc Thnh, Phm Quang Dng:

My v thit b nng. Nxb KHKT, HN, 2002[4]. Cc tiu chun lin quan.

next

Xem chi tit
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Ti liu tham kho1. o Trng Thng:My nng chuyn. HBK H Ni, 1993

2. Hunh Vn Hong, o Trng Thng: Tnh ton my trc.Nxb KHKT, H Ni, 1975

3. Trng Quc Thnh, Phm Quang Dng:My v thit b nng.Nxb KHKT, H Ni, 2002

4. TCVN 5864-1995. Thit b nng. Cp thp, tang, rng rc, xchv a xch. Yu cu an ton.

5. TCVN 5862-1995. Thit b nng. Ch lm vic.

6. TCVN 6395:1998. Thang my in. Yu cu an ton v cu to

v lp t.7. TCVN 4244-86. Thit b nng. Ch lm vic

8. TCVN 5744-1993. Ph lc 2: Tiu chun loi b cp thp.

9. OCT 1576-71.

Back


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M u

CC C TNH C BN

CA MY NNG


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0-2

1. Trng ti

Khi lng ln nht ca vt nng m

my c php vn hnh theo thit k. Trng ti Q (tn) thng c thit k

theo dy tiu chun.

Cm nng vt ti.
http://_solieu%20q%20tieu%20chuan.ppt/http://_solieu%20q%20tieu%20chuan.ppt/


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0-3

2. Vng phc v

Chiu cao nng H (m).

Khu v hnh trnh (vi cn trcdng cu) hoc tm vi v gc quay(vi cn trc quay).


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0-4

Chiu cao nng H (m)

L khong ccho t sn lmvic n tm mc v tr cao nht

Khu L


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0-5

Khu v hnh trnh (m)

Khu lkhong cch gia

2 ng ray dichuyn cu.

Hnh trnh l

qung ng cndi chuyn theophng dc ray.

Khu L

Ray


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0-6

Tm vi (m) v gc xoay

Tm vi l khongcch gia tm quay

v tm mc v trxa nht.

Gc xoay ca cn

quanh tm quay. Cntrc quay ngoi trithng c kh nngquay trn vng.Tm vi L

Ct

Cn


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0-8

Cc vn tc chuyn ng

Cc vn tc chuyn ng l vn tc cc c cutrn. Vi cn trc thng dng, vn tc ly trong

khong sau:Vn tc nng: vn = 6 12 m/ph

Vn tc di chuyn xe con: vx = 15 20 m/ph

Vn tc di chuyn cu: vc = 20 40 m/ph

Vn tc quay: nq = 0,5 3,0 v/ph


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0-9

4. Ch lm vic (CLV)

Phn nh c tnh lm vic c th ca loi thit b

ny: ng m nhiu ln v lm vic vi ti khc nhau. Cng trng ti v cc c tnh khc nhng mi my

nng c th c s dng vi thi gian v mc tinng nh khc nhau.

Do vy nu thit k nh nhau th hoc s tha an ton(lng ph) hoc s khng an ton.

CLV c phn nh trong tng bc tnh ton thitk cc b phn trong c cu v my nng.

CLV l c tnh ring, c a vo nhm mc chtit kim m vn m bo an ton khi s dng.


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0-10

Cch phn nhm CLV

Tiu chun quy nh cch phn nhm CLV.

Theo TCVN 4244-86, c cu nng c phn

thnh 5 nhm: Quay tay, Nh, Trung bnh, Nngv Rt nng da trn nhiu ch tiu khc nhau.

CLV ca my nng c ly theo CLV ca c cunng.

Cch phn nhm ny c mt s nhc im: Khng tng thch vi cc tiu chun khc

Qu nhiu ch tiu v phi hp khng nht qun


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0-11

Cch phn nhm CLVtheo 2 ch tiu

TCVN 5462-1995 phn loi c cu v my nngc lp vi cng phng php v ch da trn 2 chtiu: cp s dng (CSD) v cp ti (CT).

Cch phn nhm CLV ny tng thch ISO.

Cc ch tiu phn nh r nt hn mc ph hy(mi) ca cc chi tit

Nht qun trong cch phn nhm CLV Cc c cu phn thnh 8 nhm CLV: M1 M8

My nng phn thnh 8 nhm CLV: A1 A8

Xem chi tit


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0-12

Tm tt

Cc c tnh c bn ca my nng

Mc ch, ngha ca CLV

Cch phn nhm CLV theo 2 ch tiu(TCVN 5462-1995)

Vi CCN, CLV gm nhng nhm no? Vi MN

gm nhng nhm no? Cc ch tiu cp ti v cp s dng vi CCN v MN

Phi hp cc ch tiu ny c CLV.

next
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P0-13

Dy tiu chun v trng ti(tn)

- -- - - - - 0,05 - -

0,1 -- 0,2 0,25 0,32 0,4 0,5 0,63 0,8

1 1,25 1,6 2 2,5 3,2 4 5 6,3 8

10 12,5 16 20 25 32 40 50 63 80

100 125 160 200 250 320 400 500 630 800

140 180 225 280 360 450 550 710 900

1000

* Theo GOST 1575-61

Back


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P0-14

CLV TCVN 5462-1995Cc ch tiu phn nhm CLV cho cc c cu

* Ch tiu 1: Cp s dng - gm 10 cp T0T9 tu theo sgi lm vic trong c i my:

CSD T0 T1 T2 T3 T4 T5 T6 T7 T8 T9

t (h) < 200 400 800 1600 3200 6300 12500 25000 50000 100000

* Ch tiu 2: Cp ti - c 4 cp L1 L4 tu h s ph ti

CT L1 L2 L3 L4

Km < 0,125 0,25 0,50 1,0

Pi l cng sut ca c cu lm vic trong thi gian ti

t

t

P

PK

i

3

max

i

m

Next


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P0-15

CLV TCVN 5462-1995

Phn nhm CLV cho cc c cu

Next

CSD

CTT0 T1 T2 T3 T4 T5 T6 T7 T8 T9

L1 M1 M2 M3 M4 M5 M6 M7 M8L2 M1 M2 M3 M4 M5 M6 M7 M8

L3 M1 M2 M3 M4 M5 M6 M7 M8

L4 M2 M3 M4 M5 M6 M7 M8


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P0-16

CLV TCVN 5462-1995Cc ch tiu phn nhm CLV cho MN

* Ch tiu 1: Cp s dng - gm 10 cp U0U9 tu theo schu trnh lm vic trong c i my:

CSD U0 U1 U2 U3 U4 U5 U6 U7 U8 U9

c (x104) < 1,6 3,2 6,3 12,5 25 50 100 200 400 >400

* Ch tiu 2: Cp ti - c 4 cp Q1 Q4 tu h s ph ti

CT Q1 Q2 Q3 Q4

Km < 0,125 0,25 0,50 1,0

Pi l tng cng sut ca cc c cu lm vic trong chu trnh ci

C

C

P

PK

I

3

max

i

m

Next


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P0-17

CLV TCVN 5462-1995

Phn nhm CLV cho my nngCSD

CTU0 U1 U2 U3 U4 U5 U6 U7 U8 U9

Q1 A1 A2 A3 A4 A5 A6 A7 A8Q2 A1 A2 A3 A4 A5 A6 A7 A8

Q3 A1 A2 A3 A4 A5 A6 A7 A8

Q4 A2 A3 A4 A5 A6 A7 A8

Back


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Phn I

CC CHI TIT V THIT B

MY NNG


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Chng 1

S CU TO

C CU NNG


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1-3

1.1. C cu nng n gin

nng vt Q cn iukin:

Tp = TvTv = S.D0/ 2

Tp = P.R

S = QQ = S = 2.P.R / D0

Khng th nng ti ln !

Lm th no ?

PR

Do

S

Q

1

2

3

1. Tang

2. Tay quay

3. Dy

Tv

Tp


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1-4

Gii php tng trng ti Q

Thc t Tp


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1-5

1.2. C cu nng hin i

Cc b phn chnh: B phn mang ti

Palng

Tang cun cp

B truyn

B phn pht ng

B phn phanh hm.

a, p

Q

Dot

2

3

1

u1, 1

u2,


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1-6

V d


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1-7

1.3. Cc quan h tnh v ng hc

Cng sut ng c

T s truyn

M men xon trntrc khi nng vkhi h

n

yc

Q.v P , kW

60000.

c c 0

0

t n

n n . .Du

n v a

0 0

1

0 p t 0 0

Q.D QDT

2au 2au

0 p t 0' 0

1

0 0

Q.D QDT

2au 2au


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1-8

Tm tt

S pht trin ca CCN

Cc b phn chnh trong CCN hin i

Cc quan h tnh hc v ng hc Cng sut yu cu ca ng c

T s truyn

Mmen xon trn cc trc khi nng v khi h

next
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P1-10

V d v Palng (a = 4)

2 rng rcdi ng

2 rng rcc nh

Cc hnh nh ly twww.wikipedia.com

Q

S

S khai trin

End


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Chng 2

B PHN MANG TI


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2-2

Phn loi

Mc

B phn mang ti vn nng, c th s dng cho vt liu

bt k. Cp gi

B phn mang ti chuyn dng vi vt liu khi.Thng s dng vi loi vt liu c hnh dng v kch

thc nht nh. Gu ngom

B phn mang ti chuyn dng vi vt liu ri.


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2-3

2.1. Mc

Mc n: khi trngti nh v va

Mc 2 ngnh: khitrng ti va v ln

Vt liu: thp tccbon, thngdung thp 20.

Phng phpch to mc:

Rn

Dp

c


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2-5

Mc tm

Mc tm: khi trngti ln v rt ln

Khi trng ti ln v rt lnch to mc bng rn/dpkh v t nn thng dngmc tm.

Ch to mc bng cch ctcc tm thp thnh hnh

dng mc, sau lin ktcc tm bng inh tn.

C th thay th cc tm khicn thit.


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2-6

Tnh mc

Vi mc tiu chun khng cn tnh, ch cnla chn ng theo trng ti yu cu.

Vi mc khng tiu chun, cn tnh mc v bn ti cc tit din cung mc v thnmc.

Xem c th


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2-8

Tnh cp gi (loi ma st)

S

Fms Fms N

Q

N

a

S chu ti

S

Q/2

Na/2

c

b

Lc tc dng lntay n

Cn bng lc tc dngln tay n:

N.b Q.a/4 S.c = 0

S.cosg = Q/2 vt khng ri cn ma st: Fms > Q/2

hay (vi k > 1)N.f = k.Q/2

Thay th N v S, nhnc biu thc khng

ph thuc Q.


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2-9

2.3. Gu ngom

Loi 1 dy

1

2

4

35

Loi 2 dy

4

2

3

1

I II


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2-10

V d v kt cu


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2-11

V d (tip...)


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2-12

V d (tip...)


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2-13

V d (tip...)

next
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2-14

2.4. B phn mang ti khc


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2-15

B phn mang ti khc (tip)


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2-16

Tm tt

Phn loi b phn mang ti v phm vi sdng ca chng

Cc loi mc: Cu to chung, tnh mc khngtiu chun

Cp gi ma st: cu to chung, nguyn l hot

ng, tnh ton iu kin cp gi Gu ngom: cu to chung, nguyn l lm vic

Cc b phn mang ti khcnext
http://03-day.ppt/http://03-day.ppt/http://03-day.ppt/


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P2-17

Tnh mc khng tiu chun Tit din cung mc A-A:

tnh nh bulng chu ko,khng xit:

ng sut cho php ly 85MPa khi

dn ng tay hoc 40-50MPa khi

dn ng bng ng c.

Tit din thn mc: theo lthuyt thanh cong:

Next

d1

a

A A

B B

AA

B B

a/2 e1 e2

ydA

2

1

4

d

Q


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P2-18

Tnh mc khng tiu chun Tit din B-B:

Chu ko(th trong)

Chu nn(th ngoi)

Vi k h s ph thuc dng tit din

h = e1 + e2

r = a/2 + e1

A din tch tit din ng sut cho php ly 165 MPa khi dn ng

tay hoc 150 MPa khi dn ng bng /c.

Back

a

A A

B B

B B

a/2 e1 e2

ydA

a,e

k.A

Q

50

1

1

ha,

e

k.A

Q

50

2

2

dAyr

y

Ak

e

e

2

1

1


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1

Chng 3

DY TRONG CCN


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3-2

Khi nim chung

L chi tit mm lin kt b phn mang ti vtang hoc cc rng rc

Trong CCN s dng 2 loi dy chnh:

Cp thp bn bn t cc si thp c gii hnbn cao qua 2 thao tc bn.

Xch thng ch s dng 2 loi: xch hn tinh mtngn v xch tm.


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3-3

3.1. Cp thp bnCu to

Cc si thp c bn caob = 1400 2000 MPa (dothao tc tut si) bn vi

nhau thnh tao. Cc to bn vi nhau quanh

li thnh cp.

Cc si con c th cnghoc khc ng knh.

Li cp c th l ay, thphoc si tng hp.

Mt s loi cp khc
http://_uni-rope.htm/http://_uni-rope.htm/


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3-4

Phn loi v k hiu cp

Cp bn xui v cp bn cho (cpchng xon).

Theo dng tip xc gia cc si con:tip xc ng hoc tip xc im.

K hiu cp thng c dng nh sau:

K-P, 6x19 vi ngha:K-P - loi cp tip xc ng

6x19 - 6 tao, mi tao 19 si con.

Cp bn xui

Cp bn cho


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3-5

Tnh ton chn cp

Nhm m bo bn lu cho cp. bn lu cacp ph thuc 2 yu t: Smax v t s dc / Do

Tnh theo phng php thc dng, quy nh bi tiu

chun. Cp c chn cn m bo h s an ton:Zp = S / Smax Zp,min

Zp,min tra bng theo CLV M1---M8

xem TCVN 5864-1995 Lu : * Vi thit b ch ngi Zp,min = 9

* Vi thang my ch ngi Zp,min = 16 hoc 12 tus dy c lp treo cabin l 2 hay ln hn 2


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3-6

C nh u cp

Cn to khuyn u cp,sau khuyn ny s c

lin kt vi trc c nh. trnh cp ch st vi trc

c nh, cp c t tronglt cp.

Phng php khc

Vng lt cp1

>5dc 2 3

a) C nh bng kha cp1 - Vng lt cp 2 - Cp3 - Kho cp (s lng ti thiu 3)

>20.dc

b) C nh bng cch bn cp1 Vng lt cp 2 Cp3 Dy chng ri

1 2 3

Trc c nh


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3-7

C nh u cp

Chi tit ng cn hoc ng chm bngthp c, mt u gia cng sn l lin kt vi trc c nh cp.

c) C nh bngkha chm1 - Cp2 - ng chm

3 - Chm

1

2

3

d) C nh bng ng cn1. Cp 2. ng cn3. Kim loi nng chy ( y)

31 2


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3-8

Cc ch khi s dng cp

Cp phi c chng ch.

Dy cp phi l mt on nguyn.

Bi trn cp thng xuyn t ngoi bng m chuyndng.

Theo di cp v thay cp mi khi cp mn gim

ng knh 10%, t 1 tao hoc s si t trn mtbc bn ln hn gi tr cho php (TCVN 5744-1993).

Trnh cp ch st vi nhau v vi cc b phn khc.


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3-9

3.2. Xch hn v xch tmCu to

Xch hn: s dng loixch mt ngn: t2,6d;B3,5d. Loi th dng

cun vo tang trn,cn loi tinh n khpvi a xch.

Xch tm: c cu to

gn nh xch truynng nhng cc mxch lp trc tip lncht, khng qua bn l.

t

dB

t

t

Tm c dng

t

hoc dng


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3-10

Tnh ton chn xch

Tng t cp thp, xch c tnh theo phng phpthc dng, quy nh bi tiu chun. Xch c chn

cn m bo h s an ton:Zp = S / Smax Zp,min

Zp,min tra bng ty theo cch dn ng CCN.

xem TCVN 5864-1995


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3-11

3.3. So snh cp v xch

Nh Mm m => vn tc bt k bn lu tng i ln Lm vic an ton (ph hy

c bo trc qua s sit => khng t t ngt)

Yu cu ng knh tanghoc rng rc ln Phm vi s dng: a s cc

trng hp

Nng MmVa p, n => vn tc thp bn lu tng i ln Km an ton (mc ph hy khng

c bo trc => nguy c tt ngt)

Khng yu cu ng knh tang vrng rc ln Phm vi s dng: Khi vn tc thp,

yu cu nh gn hoc mi trngnhit cao

Cp Xch


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3-12

3.4. Cc bc tnh chncp v xch

Chn loi cp v cp bn thch hp hoc xch.

Tnh lc cng dy ln nht Smax.

T CLV cho, tra bng (tiu chun) c Zp,min.Tnh lc ko t yu cu:

S,yc = Smax . Zp,min

Tra bng chn cp (hoc xch) c ng knh (hocbc) thch hp sao cho:

S,bng S,yc


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3-13

Tm tt

Cu to chung, phn loi cp thp bnSi thp, tao, li Cp bn xui v cp bn cho

Cc yu t nh hng n bn lu ca cpCc ch khi s dng cp thp bn

Phng php tnh chn cp v xch

Mc ch v phng php tnh ngha ca h s an ton

So snh cp v xchnext

Gi tr ti thiu ca Zp i vi cp v xch ti
http://04-tang%20rongroc%20palang.ppt/http://04-tang%20rongroc%20palang.ppt/http://04-tang%20rongroc%20palang.ppt/


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P3-14

Gi tr ti thiu ca Zp i vi cp v xch ti(TCVN 5864-1995)Nhm CLV cac cu M1 M2 M3 M4 M5 M6 M7 M8

Zp,min 3,15 3,35 3,55 4,00 4,50 5,60 7,10 9,00GHI CH:1. Trong iu kin s dng nguy him (v d kim loi nng chy) th CLVkhng ly di M5 v khi t M5 tr ln, Zp,min ly tng thm 25%.2. Vi thit b ch ngi Zp,min ly bng 9, cn vi thang my ch ngi(TCVN 6395:1998) Zp,min = 16 hoc 12 tu theo s cp c lp treo cabin l2 hay ln hn. Lu , khng cho php treo cabin trn 1 dy cp duy nht.3. Vi xch dn ng bng ng c:+ xch hn cun ln tang trn: Zp,min = 6+ xch hn chnh xc n khp vi a xch: Zp,min = 8+ xch tm: Zp,min = 5Khi dn ng bng tay: Zp,min = 3 vi tt c cc loi xch

Back

S si t cho php trn 1 bc bn


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P3-15

S si t cho php trn 1 bc bnTCVN 5744-1993

H s an tonban uca cp

Cu to cp, s si

6x19=114 6x37=222

Bn cho Bn xui Bn cho Bn xui

9 14 7 23 12

9 10 16 8 26 13

10 12 18 9 29 14

12 14 20 10 32 1614 16 22 11 35 17

Back

Lift Rope 8x19+1 (KONE)


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P3-16

Lift Rope 8x19+1 (KONE)cp bn 1600(inner)/1300(outer) MPa

knh Athp

, mm2 S, kgf KL, kg/m

8 18,9 2 780 0,18

10 35,6 4 190 0,33

11 46,0 5 370 0,43

12 53,9 6 340 0,513 61,9 7 290 0,58

14 70,0 8 250 0,66

15 82,6 9 690 0,78

16 93,3 10 790 0,86

18 117,6 13 760 1,10

20 143,5 16 870 1,35

EndMore

Cp thp - 6x19+1 (GOST 2688-80)


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P3-17

Cp thp , 6x19+1 (GOST 2688 80)

knh S, N

b=1400MPa

S, N

b=1600MPa

KL, kg/m

8,3 - 34 800 0,256

9,1 - 41 550 0,305

9,9 - 48 850 0,357

11 - 62 850 0,461

12 - 71 750 0,527

13 71 050 81 250 0,597

14 86 700 98 950 0,728

15 100 000 114 500 0,844

16,5 121 500 139 000 1,025

18 145 000 166 000 1,220

19,5 167 000 191 000 1,405

EndMore

Cp thp -O, 6x19+1 (GOST 3077-80)


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P3-18

Cp thp O, 6x19+1 (GOST 3077 80)

knh S, N

b=1400MPa

S, N

b=1600MPa

KL, kg/m

7,8 - 29 900 0,221

8,8 - 39 800 0,294

10,5 - 53 650 0,388

11,5 - 66 150 0,487

12 - 72 000 0,530

13 - 81 000 0,597

14 - 97 750 0,719

15 - 115 500 0,853

16,5 118 000 135 000 0,997

17,5 136 500 156 000 1,155

19,5 162 500 183 000 1,370

EndMore

Cp thp -3, 6x25+1 (GOST 7665-80)


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P3-19

Cp thp 3, 6x25+1 (GOST 7665 80)

knh S, N

b=1400MPa

S, N

b=1600MPa

KL, kg/m

8,1 - 31 900 0,237

9,7 - 46 300 0,343

11,5 54 900 62 700 0,464

13 71 500 81 750 0,605

14,5 90 350 102 500 0,764

16 110 500 126 500 0,942

17,5 134 500 153 500 1,140

19,5 160 000 183 000 1,358

21 188 500 215 000 1,594

22,5 219 000 250 500 1,857

24 251 500 288 000 2,132

End

X h h h h h (GOST 2319 70)


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P3-20

Xch hn xch chnh xc (GOST 2319-70)

knhdy

Bc t(mm)

Chiu rngB (mm) S(kN)

KL(kg / m)

6 19 21 13,7 0,75

7 22 23 17,6 1,00

8 23 27 25,5 1,35

9 27 32 31,0 1,80

10 28 34 39,0 2,25

11 31 36 45,0 2,70

13 36 43 64,7 3,80

16 44 53 100,0 5,80

End


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Chng 4

B PHN CUN DY V

DN HNG DY


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Khi nim chung

Tang: b phn cun dy trong CCN, binchuyn ng quay thnh chuyn ng tnh

tin nng/h vt. Rng rc: b phn dn hng dy.

Palng: b phn gm cc rng rc, c nh

v di ng, lin kt vi nhau bng dy, dng gim lc cng dy hoc tng vn tc.

4 1 Tang cun cp


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4.1. Tang cun cpCu to chung

Tang thng c dng ng tr, hai u c moay lp vi trc, chuyn ng quay.

Vt liu tang: gang hoc thp.

B mt lm vic c th nhn (tang trn) hoc ctrnh dng ren trn c bc ln hn ng knhcp trnh cp ch xt vo nhau (tang x rnh).

Tang c th dng cun 1 lp hoc nhiu lpcp chng ln nhau.


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Tang trn

Khi cun nhiu lpcp, tang cn c g

chn. Chiu cao gtnh t lp cp trncng cn ti thiu 1,5ng knh cp trnh

cp tut khi tang.

Do

g

g = 1,5.dc t = dc

dc

d

L


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Tang x rnh

Kch thc rnh cp

t

dc D

d

DD1D

o

R = 0,55dct = dc+D

I

Do

I

L


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Cc kch thc c bn

ng knh danhngha Do.

Chiu di ti thiuphn cun cp trntang L.

Chiu dy thnhtang d.

Do

g

g = 1,5.dc t = dc

dc

d

L


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ng knh danh ngha

ng knh o theo tm lp cp di cng.

Xc nh t iu kin tng bn lu cho cp:

D0 h1.dcvi dc ng knh cp

h1 h s, tra trong tiu chun theo CLV cac cu nng.

TCVN 5864-1995 quy nh gi tr ti thiu ca h1. Lu : vi CCN dn ng bng /c, ng knh

tang cn tnh li, m bo vn tc nng cho trc.


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4-8

Chiu di cun cp

Tnh t s vng cp trn 1 lp (Z) v khong cchgia cc vng cp (bc cun cp - t): L Z.t

Bc cun cp t dc vi tang trn; t 1,1.dc vi tang x

rnh. S vng cp khi cun 1 lp tnh theo cng thc:

Z = Z1 + Z2 + Z3vi Z1 = a.H/(p.D0) s vng lm vic (H chiu cao nng;

D0 ng knh tang; a bi sut ca palng)

Z2 = 1,5..2 s vng cp d tr trn tang

Z3 = 0..2 s vng phc v c nh cp ln tang.

Khi cun n lp cp trn tang c th ly Z Z1/n.


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Chiu dy thnh tang

Chiu dy d thng chn trc theo vt liu tang: Thp: d = 0,001.D0 + 3 (mm) Gang: d = 0,002.D0 + (610) 12 (mm)

vi D0 ng knh tang, tnh bng mm. Kim tra tang vi kch thc chn v bn: Vi tang ngn (L/D0 3) ch cn kim nghim

bn nn: tang c tnh nh ng dy chu p sutngoi do dy vi lc cng Smax xit ln tang sinh ra.

Khi tang di (L/D0 > 3) cn tnh n nh hngca c un v xon.

Xem chi tit


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C nh cp ln tang

Bulng v tm kp

A

A - A

A

Cp

Vt chn

Cp


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4 2 Rng rc v a xch


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4.2. Rng rc v a xchCu to (tip)

Vi rng rc cho xch hn,ng knh danh ngha D0 xcnh theo ng knh dy thp

lm xch (d), bc xch (t) v srng (s hc) trn a xch (z):D0

d

z s hc, min = 5-6

D0= (

t

sin(90/z))

2

+(

d

cos(90/z))

2


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Lc cn v hiu sut rng rc

Khi cha quay: S2 = S1 Khi quay theo chiu trn

hnh v, do lc cn W

nn S2 > S1 hayS2 = S1 + W

Cc loi lc cn chnh:

Lc cn do cngdy (Wc)

Lc cn do ma st trong trc (Wo)

S1 2

S

n

W


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Lc cn do cng dy

Do cng nn khicun vo v khi nhkhi rng rc dy blch so vi trng hpl tng cc khong bv c nh trn hnh v

S2 = S1 + Wc

Kt hp phng trnhcn bng mmen tnhc lc cn do cng dy Wc = S1.j

S1(D0/2+b) = S2(D0/2-c)S1(D0/2+b) = (S1+Wc)(D0/2-c)

Wc = S1(b+c)/(0,5D0- c) = S1.j

b c

S1 S'2= S

1+Wc


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Lc cn do ma st trong

Gi s rng rc ngknh D0 lp trn trtc ng knh ngng d.

S2 = S1 + Wo vi Wo l

lc cn do ma st trong.

T mmen cn quay Tctnh c lc cn do mast trong Wo = Tc/ 0,5D0 = S1.xx = 2sin(a/2).f.d/D0

S''2

=S1+Wo

S1

Lc tc dng ln :

S = S1+ S''2 => [email protected]

Lc ma st trong : F = S.f

To mmen cn quay: Tc = F.d/2


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Hiu sut rng rc

Hiu sut = cng sut c ch / cng sut b ra

* Trng hp rng rcc nh:

C.s. c ch Pci = Q.vnC.s. b ra Pbr = S2.v0Lc cng dy S1 = Q

Vn tc dy v0

= vnHiu sut h = S1/S2

(l t s gia lc cng dy trnnhnh cun S1 v nhnh nh S2)

S

Q,

1

2S ,v0nv

n


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Hiu sut rng rc (tip...)

Hiu sut = cng sut c ch / cng sut b ra

* Trng hp rng rcdi ng:

C.s. c ch Pci = Q.vnC.s. b ra Pbr = S2.v0Lc cng dy S1+S2 = Q

Vn tc dy v0 = 2.vnHiu sut hd > S1/S2

S

Q,

12

S ,v0

nv

n

* Trong tnh ton thng ly:hd = h = 0,94...0,98 vi rng rc cp;h = 0,94...0,96 vi rng rc xch (a xch)

4 3 Palng


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4.3. PalngKhi nim chung

H thng rng rc c nh v di ng, lin kt vinhau bng dy. Tu cng dng, palng c phn lm 2 loi:

Palng li lc (hnh a) Palng li vn tc (hnh b)

Q

tang

S2......Sa S''1

S'1

S1

(a)

Q,vn

...

P,vP

S1 S2 Sa

(b)


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4.3.1. Palng li lc

Bi sut (a): s ln gim lccng dy so vi khi treo vt trctip trn 1 dy xt trng thing im (cc rng rc khngquay).

C th xc nh bi sut a quas nhnh dy treo vt. Trn hnh v l palng c bisut a = 4. Trong tnh ton, palng cth hin di dng khai trin...

h l l l


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Tnh ton palng li lc

Cho s khai trinpalng. Xc nh lc cngdy ln nht Smax=? nm u? Khi nng hay h? Hiu

sut ca c h thng hp=? Phng php: da vocc quan h lc cng dytrn cc nhnh ca rng rcv hiu sut h = Scun/SnhT , xt ln lt tngrng rc trong h thngpalng...

Q

tang

S2......Sa S''1

S '1

S1

(a)

T h ( i )


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Tnh ton... (tip)

Q

tang

S2......Sa-1Sa S''1

S'1

S1

Khi h th th no?

Khi nng vt Cc rng rc quay theo chiu nh

hnh v. Lc cng dy trn nhnhcun vo rng rc b hn trnnhnh nh ra nn suy ra Smax =S1 = Stang. Lc cng ln nht nm nhnh cun vo tang.

Tng lc cng dy cn bng vi Q:Q = S1 + S2 + ... + Sa

T quan h hiu sut rng rc:S1 = S1 = S1.1S2 = S1.h = S1.h

1

......Sa = Sa-1.h = S1.h

a-1

Q = Si = S1. (1+ h+ ... + ha-1

)

Smax = S1/ ht = Q.(1-h) / [(1-ha)ht] Hiu sut palng: hp = Q / (a.Smax)

P l k


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Palng kp

Bi sut palng kp khiu l "2a" v bng snhnh dy treo vt(trn s : 2a = 4)

Rng rc trung giankhng quay, ch ng vaitr cn bng nn trongtnh ton Smax c ththay th bng palngn vi bi sut

a' = 2a/2 v ti Q' =Q/2.

Hiu sut ca palnghp=Q' / (a'.Smax).

QDQ

Palng n Palng kp

D= 0

4 3 2 P l l i t


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4.3.2. Palng li vn tc

S1 = S1 = S1.1

S2 = S1.h = S1.h1

......

Sa = Sa-1.h = S1.ha-1

P = Si = S1. (1+ h+ ... + ha-1 ) (1)

Smax = S1; (2)

Sa = Q / h => Q = S1.ha (3)

T (1) (2) (3) tm c quan hgia P, Q, S

max

Q, vn

...

P,vP

S1 S2 Sa

C l h l


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Cc lu chung v palng

Lc cng cp

Palng kpBi sut k hiu l 2a. Rng rc cn bng khng quay.

Tnh ton coi nh palng n vi a = 2a/2 v Q=Q/s

S rng rc tCh tnh s rng rcpha tang cun cp

S c bit Trng hp gp s c bit cn thitlp cng thc tnhlc cng cp ln nht.

Q

S1

S1

S2

S

next

H s ng knh vi tang v rng rc(TCVN 5864 1995)
http://05-thietbi%20phanh%20ham.ppt/http://05-thietbi%20phanh%20ham.ppt/http://05-thietbi%20phanh%20ham.ppt/


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Nhm CLVca c cu M1 M2 M3 M4 M5 M6 M7 M8h1 11,2 12,5 14,0 16,0 18,0 20,0 22,4 25,0h2 12,5 14,0 16,0 18,0 20,0 22,4 25,0 28,0h3 11,2 12,5 12,5 14,0 14,0 16,0 16,8 18,0

GHI CH:1. ng knh danh ngha ca tang: D0 h1.dc

2. ng knh ca rng rc dn hng: D2 h2.dc

3. ng knh ca rng rc cn bng: D3 h3.dc

4. Vi cn trc t hnh: h1 = 16; h2 = 18; h3 = 14 vi CCN ti

h1

= 14; h2

= 16; h3

= 12,5 vi CCN cn

5. ng knh rng rc ma st trong thang my: D 40.dc(TCVN 6395:1998)

(TCVN 5864-1995)

Back

Kim tra tang cun cp v bn


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Kim tra tang cun cp v bn

Back

Vi tang ngn (L/D0

3) ch cn kim nghim bnnn: tang c tnh nh ng dy chu p sut ngoido dy vi lc cng Smax xit ln tang sinh ra:

sn = k.Smax/(t.d) [s]

k = 1; 1,28; 1,37; 1,45; 1,52; 1,53 ty s lp cp t 1..6[s] = 7090 MPa vi gang; 100120 MPa vi thp.

Khi tang di, cn tnh n un v xon:

u

u

t

ntn

W

TM22

22

75,0s

ssss

S khi h vt


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Smax khi h vt

Q

tang

S2......Sa-1Sa S''1

S'1

S1

Khi h vt, cc rng rcquay theo chiu ngc li.

Cc nhnh cun/nh ivai tr cho nhau. Lc cngln nht s n,f trnnhnh xa tang nht.

Tng lc cng dy vncn bng vi Q:

Q = S1 + S2 + ... + Sa

T d dng suy ra:

S*max = Sa = Q.(1-h) / (1-ha)

Back


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101/156

Khi nim chung

B phn khng th thiu trong ccu nng.

Cng dng:

Dng vt nng v tr mong mun.

Gi vt nng trng thi treo, khngri khi khng mong mun.


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5.1. Mmen phanh yu cu

Mmen phanh yu

cu khi h ln hnkhi nng

Chn phanh theoQPAT:Tph = n.T*t

HSAT n chn t1,5 2,5 theo CLV

ngha ca HSAT:

Tnh n ti ng

phng qu ti

Q

Tt

TTph

Tt

Q

TTph

Phanh khi nng

Tph= T- Tt Tph= T*t + T

*

Cn bng mmen trn trc t phanh

T =*t1 2auo

QDo

Phanh khi h

5 2 C cu bnh cc


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5.2. C cu bnh ccCc vn chung

Tnh ton c cu bnhcc: phng cc dnghng gy mt an ton: Gy con cc Gy rng bnh cc

Dp mp rng

Phng php tnh chung Chn trc s rng Tnh chn mun

Tnh kim nghim

Lso

Con cc

Bnh cc

Q

S cu to chung


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Tnh ton bnh cc

Ft

h

b

s

Tnh theo bn dp

q = Ft / b [q]

vi Ft = 2T / D = 2T / (m.z) ;

b = m.chn trc , z tnh munm, sau chn m tiu chun

Kim nghim bn un

= Mu / Wu= Ft.h / (b.s2 / 6) [ ]

vi bnh cc tiu chun:h = m; s =1,5m


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Cc thng s bnh cc

Vt liu bnh cc = b/m [q], N/mm [ ], MPa (*)+ Gang xm 1,6 - 6,0 150 30

+ Thp c 1,5 - 4,0 300 80

+ Thp CT3 rn 1,0 - 2,0 350 100+ Thp 45 rn 1,0 - 2,0 400 120

(*) ng sut un cho php ly thp i tnh n ti trng ng khic cu lm vic

(**) Ti trng ng xut hin do hin tng bnh cc b quay ngcli di tc dng ca trng lng vt nng trc khi n khp htvi con cc v b gi li. hn ch ti ng cn gim bt qungng ny: gim bc rng (do gim mun -> yu) hoc lpnhiu cc "lch pha" nhau


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Tnh ton con cc

Kim nghim v bnCon cc c tnh nhthanh chu nn lch tm bi

lc vng Ft:= n + u =

= Ft/ (cd) +

Ft.e /(dc2

/6) [ *]Con cc ch lm bng thp,[ *] = 65 MPa tnh nti trng ng.

Fte

d

c

5 3 Phanh m


107/156

5.3. Phanh mPhanh m n gin

Kh nng phanh tnh t iukin cn bng lc trn tayphanh v iu kin phanh:

N.a = F.c + K.l

Fms = k.F vi Fms = N.fSuy ra: K = (F / l ).(k.a / f - c)vi lc vng F = 2T / D.

* gim lc phanh yu cu K=> cc gii php:+ tng D, l, gim a: th sao?+ tng c: th sao? (K < 0 )

* Nu i chiu m men phanh ? bn lu: p = N / b.s [p]

KFms

N

a

cc'

a

c

FN

K

l

n

h h k l


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Phanh 2 m kiu l xo

Nguyn l lm vic

* lu cng dng ca cc

chi tit u nhc im v phm

vi s dng

Tnh ton phanh tng

t phanh 1 m* Kh nng phanh

* bn lu

a

l

F

F

K K KK

e

N N

21

4 3

5

9

8

7 6

10 11

5 4 Ph h i


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5.4. Phanh ai

Kh nng phanh:S1/ S2 = e

fa v

S1 - S2 = Ft = 2Tph/ D

=> K = S2.a / l =Ft.a / [l.(efa - 1)]

* Nu i chiu mmen:S2 - S1 = Ft v S2/ S1 = e

fa

=> K' = S2.a / l =Ft.a.efa/ [l.(efa - 1)]

bn lu:pmax = 2Smax / (D.b) [p][p] = 0,1-0,2 MPa vi aming

K

SS1

2

l

pmax

F

K

S1 (S )2S2 1(S )

l

a

a


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5.4. Phanh p trc


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5.4. Phanh p trcPhanh a

D1

D2

D

C th coi l trng hp cbit ca phanh nn ( = 90o)

K = 2Tph/ (D.f)

tng kh nng phanh: dngphanh nhiu a

K = 2Tph / (D.f.z)

5 6 Ph h t


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5.6. Phanh t ng

V sao gi l phanh t ng?

Lc trong c cu c s dng lm lc phanh

M men phanh t iu chnh theo tiPhn loi

Phanh t ng c mt ma st khng tch ri

Phanh t ng c mt ma st tch ri.u nhc im v PVSD tng loi phanh

Phanh t ng c mt ma st


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Phanh t ng c mt ma stkhng tch ri Cu to

c im cu to

Nguyn l hot ng

Tnh t ng ca

phanh:

* Lc phanh l lc dc

trc trn trc vt* Lc phanh c t l

thun vi lc phanhyu cu.

Fa2

Ft3


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Phanh t ng c mt ma st


115/156

Phanh t ng c mt ma sttch ri

Cu to

c im cu to

Nguyn l hot ng

Tnh t ng ca

phanh:

* Lc phanh l lc dc

trong b.t. vt - ai c* Lc phanh c t l

thun vi lc phanhyu cu.

3

4

1

2

5 6

Tnh t ng ca phanh t ng


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Tnh t ng ca phanh t ngc mt ma st tch ri

phanh hot ng tt cn tho mn iu kin: Kc Kyc,trong Kyc l lc phanh yu cu

Gi tr Kc v Kyc tnh nh sau: Kc = QD0 / [au0(d2tg( ) + f.D)] t l vi ti Q (xut phtt iu kin Tbr = Tr + TT - vn c ai c th m men trn

bnh rng cn thng ma st trn ren v ma st mt t) Kyc = ... = 2.n.QDo. / (2auo.D) t l thun vi ti Q

Khi tng ti Q th lc phanh yu cu Kyc tng, nhng lc phanh doc cu to ra Kc cng tng => loi phanh ny c kh nng t iuchnh lc phanh theo ti => khng s qu ti v v th HSAT phanhkiu ny thng ly b (n 1,2)

5 7 T t


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5.7. Tay quay an ton

L loi tay quay kt hp phanh hm,m bo gi vt an ton khng ri khi

khng c lc tc ng ln n. Tay quay an ton kiu I kt hp phanh t

ng c mt ma st tch ri.

Tay quay an ton kiu II kt hp phanh ai.

5 7 Ta q a an ton (tip)


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5.7. Tay quay an ton (tip)

next

Kiu I Kiu II

Cu toc im cu tou nhc im v PVSD.
http://06-co%20cau%20nang.ppt/http://06-co%20cau%20nang.ppt/http://06-co%20cau%20nang.ppt/


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Chng 6

C CU NNG

Khi nim chung


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6-2

Khi nim chung

L c cu khng th thiu trong my nng.

C yu cu cao v an ton.

Ty b phn pht ng phn ra:

CCN dn ng tay

CCN dn ng bng ng c

6 1 CCN dn ng tay


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6-3

6.1. CCN dn ng tay

Pht ng qua tay quay hoc bnh ko

Khi s dng sc ngi thng ly cng sut

N = P.v 0,1 kW. Khi s dng nhm cng nhn vn hnh,

tng lc tc ng P tnh theo: P = F.m.k

vi F lc do 1 ngi tc ngm s ngi tham gia vn hnh my

k h s tnh n s phn b khng u lc

6.1.1. S v c im


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6-4

cu to

Tnh cht chung ca cc ccu ny l t quan trng, thigian s dng ngn, tc

thp v khng c ti ng. c im cu to chung: n

gin, gn nh, gi thnh thp.

V vy thng dng tang trn,

cc b truyn h, trt v ts dng cc ni trc. Phanhthng dng kt hp vi tayquay (TQAT).Q

a, p

PR

6 1 2 c im tnh ton


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6-5

6.1.2. c im tnh ton

Tnh ton ng hc

T s truyn chung ca cc b truyn Uo xc nh t iukin v lc ch khng phi t yu cu v vn tc

Uo = Tv/(Tp. ) = QDo/(2.a.F.m.k.R. )

trong l hiu sut chung ca c cu.

m bo an ton vt khng ri

Cc b truyn bnh rng h tnh theo bn un,trnh hin tng hng gy mt an ton l gy rng.

Cc b phn khc: khi tnh ton thit k cc h s trabng theo CLV Quaytay

6.2. CCN dn ng bng


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6-6

g gng c

S dng ng c pht ng c cu.

C th gp nhiu loi ng c nh

ng c in, ng c t trong, ngc thy lc, kh nn, thm ch cndng c ng c hi nc.

ng c in c s dng rng rihn c.

6.2.1. S v c im


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6-7

cu to

y l cc c cu quantrng, nng sut v trng tiln, thi gian s dng ludi, tc tng i cao.

c im cu to chung:hiu sut cao, chc chn, tincy, tui th cao.

V vy thng dng tang xrnh, cc b truyn kn, ln v s dng cc ni trc.Phanh ai hoc phanh TK.

a, p

Q

Do

t 2

3

1

u1, 1

u2,

6 2 2 c im tnh ton


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6-8

6.2.2. c im tnh ton

Tnh ton ng hc

T s truyn chung ca cc b truyn Uo xc nh t iukin m bo vn tc nng cho trc:

u0 = nc/ntg = nc. .D0/(a.vn)

m bo an ton vt khng ri

Cc b truyn bnh rng che kn tnh theo bn tip

xc, kim nghim bn un v qu ti. Cc b phn khc: khi tnh ton thit k h s tra bng

theo CLV yu cu.

6 2 3 Qu trnh m my


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6-9

6.2.3. Qu trnh m my

Qu trnh m my xt khi nng vt, gi thit chuyn ngnhanh dn u trong sut qu trnh m my.Khi m my nng vt ng c cn pht mmen ngoi thng mmen

cn tnh do vt nng sinh ra Tt, cn thng thm qun tnh ca cc chitit trong h thng khi tng tc:

Tm = Tt + T = Tt + T1 + T2

Tt mmen tnh do trng lng vt nng sinh ra khi nng vt

T mmen do qun tnh

T1 do qun tnh cc chi tit chuyn ng thng (vt nng, mc)

T2 do qun tnh cc chi tit chuyn ng quay (rto, trc, )

Mmen tnh T


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6-10

Mmen tnh Tt

Mmen tnh khi nng (Nm) nh bittrong phn s cu to c cu nng:

vi Q trng lng vt nng, NDo ng knh tang, ma bi sut palngUo t s truyn ca cc b truyn

hiu sut chung ca c cu

0

0

2auQDTt

Mmen do qun tnh T1


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6-11

Mmen do qun tnh T1

Khi m my nng vt, vt nng v mc to ra lc quntnh Qqt. Lc ny ng vai tr ging nh ti Q, nnmmen do n gy ra trn trc ng c c tnh theo:

m

n

mqt t

v

g

Qj.mQ

60

0

0

1

2au

DQT

qt

m

c

tua

nQDT

2

0

2

2

0

1

375

0

000

au

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a

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Mmen do qun tnh T2


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6-12

Mmen do qun tnh T2

Khi m my, mmen cn do qun tnh ca mi chi titquay tnh trn trc ca n theo cng thc: Ti = J. i

Do , khi chuyn v trc /c (trc 1), ta c:

Trong ,

l mmen do qun tnh cc chi tit quay lp trn cctrc 2, 3 tnh quy i v trc 1.

...uu

TT

u

TT 3/1/

21

3

1

2

12

...TTT / 1212

Mmen do qun tnh T2 (tip )


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6-13

Mmen do qun tnh T2 (tip)

Do gia tc gc trn cc trc 2,3 kh nh so vi trc 1,li phi chia cho t s truyn u1, u1u2 nn cc thnhphn T2/1, T3/1 , khng ng k so vi trc 1. V vy,

mmen do qun tnh cc chi tit quay tnh theo:

Trong , Ti(I) l tng mmen do qun tnh cc chi titquay lp trn trc 1, cn kl h s tnh n nh hngca qun tnh cc chi tit quay lp trn cc trc khc.

)I(i TkT.kT 12



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6-14


Thay th

c

g

DGD

g

G.mJ

rad/s,t.

n

t

iiiiiii

2

m

c

m

42

60

2

222

1

1

Cui cng


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6-15

Cui cng

Vy trong qu trnh m my, ng c cn phtra mmen Tm = Tt + T = Tt + T1 + T2

m

cIii

m

c

mt

nDGk

tua

nQD

au

QDT

3753752

2

2

0

2

2

0

0

0

Cng thc ny c s dng tnh chn, kim trakh nng m my ca ng c hoc kim tra thigian m my, gia tc m my c ph hp hay khng.

6 2 4 Qu trnh phanh


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6-16

6.2.4. Qu trnh phanh

Qu trnh phanh xt khi h vt, gi thit chuyn ng chmdn u v phanh t trn trc 1.

Khi h vt mmen do phanh to ra cn thng mmen tnh do vt

nng sinh ra Tt* v mmen do qun tnh ca cc chi tit trong hthng khi gim tc:

Tph = Tt* + T

* = Tt* + T*1 + T

*2

Tt* mmen tnh do trng lng vt nng sinh ra khi h vt

T* mmen do qun tnh khi phanh (phanh khi ang h vt)

T*1 do qun tnh cc chi tit chuyn ng thng (vt nng, mc)

T*2 do qun tnh cc chi tit chuyn ng quay (rto, trc, )

Tng t qu trnh m my


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6-17

g t qu t y

Vy khi ang h, dng c c cu, phanhcn to mmen:

Cng thc ny c s dng tnh kim tra khnng phanh hoc kim tra thi gian phanh, gia tcphanh c ph hp hay khng.

Lu ii th h th


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6-18

v gii thch cc cng thc

Cn nu c: ngha v cc gi thit khi lp cng thc.

Cc thnh phn chnh trong cng thc: Tt, T1, T2nu ngha, vit cng thc tnh cc thnh phn ny.

Cc thng s trong cng thc v n v o: Q trnglng vt nng (N), Do ng knh tang (m), v.v

S dng cng thcnext

m

cIii

m

c

mt

nDGk

tua

nQD

au

QDT:duV

3753752

2

2

0

2

2

0

0

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Phn II

MY NNG

CNG DNG CHUNG


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Chng 7

THIT B NNG N GIN

7 1 Kch


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7-3

7.1. Kch

Loi TBN khng dng dy, khng ginchu ti.

Nng vt bng phng php y.

Cu to gn nh d di chuyn.

Chiu cao nng b, vn tc nng thp.

Phn loi kch


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7-4

Phn loi kch

Kch vt

Kch thanh rng

Kch thy lc

Q

Kch thanh rng


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7-5

Kch thanh rng

Cu to1. Thn kch 2. Thanh rng 2. Bnh rng

3. B truyn BR 4. Tay quay 5. u kch

Quan h gia cc i lngu = TV/ (TF. ) = Q.d1/(2. .m.F.l. )

d1 ng knh bnh rng 2

c im chung- Trng ti khng ln

- Cc bnh rng thng b tnh theo bn un

Q


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Ti xy dng


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7-7

Ti xy dng

S dng 2 t s truyn tng nng sut

u0 = z6/z5 . z2/z1u0 = z6/z5 . z4/z3

Thng ly u0 = 0,5.u0

Phanh t trn trc 2

PT

7.3. Palng


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7-8

7.3. Palng

Loi TBN dng dy - cp cun ln tang hoc xchn khp vi a xch.

Thng c treo trn cao, do vy yu cu kchthc nh gn.

Phn loi:

Palng tay: dn ng bng tay thng qua

xch ko Palng in: dn ng in, s dng cp hoc

xch hn.

Palng tay


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7-9

Palng tay

Dy c s dng l xch.

Dn ng tay bng cch koxch lm quay bnh ko an ton.

gim kch thc:

- Truyn cng sut thnh nhiudng

- Trc b dn lp lng khng trntrc dn

- S dng vt liu tt ch to

XchkoXch

nng

Bnh ko

an ton

Palng in


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7-10

Palng in

ng c in

tang khp ni hp s phanh a

I

II

IIIIV

cp

IIIIIIIV

s rng

z2/z1 = 50/14z4/z3 = 58/29z6/z5 = 42/15z8/z7 = 33/13

Dy c sdng l cphoc xch.

B truynbnh rngnhiu cphoc hnhtinh

Phanh thng dng phanh ma st nhiu a, loi thng ng.C th kt hp phanh t ng.

cn bng, ng c v phanh thng t 2 pha palng.

next
http://08-cau%20truc%20va%20can%20truc%20quay.ppt/http://08-cau%20truc%20va%20can%20truc%20quay.ppt/http://08-cau%20truc%20va%20can%20truc%20quay.ppt/http://08-cau%20truc%20va%20can%20truc%20quay.ppt/


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Chng 8

CU TRC V

CN TRC QUAY

8.1. Cu trcKhi i h


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2

Khi nim chung Loi TBN c s dng

gin chu ti, nng vtqua dy cun.

Cu to gm:

Dn chu ti t trncao: dm chnh vdm u

Cc c cu: CCN v 2CCDC

Phn loi: Cu trc 1dm v cu trc 2 dm

S dng nhiu trongcc phn xng.

8.1.1. S cu to


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3

8.1.1. S cu to

Cc b phn chnh1. Dm chnh2. Xe con

3. C cu nng

4. CCDC xe con5. CCDC cu

iu khinT mt sn hoc t cabin

Cc thng s chnhTrng tiKhu , chiu cao nngv hnh trnhCc vn tc chuyn ng

8.1.2. C cu di chuyn


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Lu Do khu lkca CCDC xecon v cu khc nhau nn cc

b phn ca chng cng b trtheo cc s khc nhau.

CCDC xe con1. ng c2. Phanh

3. Hp gim tc4. Ni trc5. Gi 6. Bnh xe



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5

y

CCDC cu (KCKL)1. ng c2. Phanh

3. Hp gim tc

4. Ni trc5. Gi 6. Bnh xe

Cng dng:

di chuyn tonb cu (kt cukim loi) dcphn xng.

8.2. Cn trc quay


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8.2. Cn trc quay

Loi TBN s dng tang v dy cun.

Thng t trn mt sn, s dng ko vt.

Phn loi

Ti tay

Ti in


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8.3. Kt cu kim loi


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8

Loi TBN dng dy - cp cun ln tang hoc xchn khp vi a xch.

Thng c treo trn cao, do vy yu cu kchthc nh gn.

Phn loi:

Palng tay: dn ng bng tay thng qua

xch ko Palng in: dn ng in, s dng cp hoc

xch hn.

Palng tay


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9

g y

Dy c s dng l xch.

Dn ng tay bng cch koxch lm quay bnh ko an ton.

gim kch thc:

- Truyn cng sut thnh nhiudng

- Trc b dn lp lng khng trntrc dn

- S dng vt liu tt ch to

XchkoXch

nng

Bnh ko

an ton

Palng in


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g

ng c in

tang khp ni hp s phanh a

I

II

IIIIV

cp

IIIIIIIV

s rng

z2/z1 = 50/14z4/z3 = 58/29z6/z5 = 42/15z8/z7 = 33/13

Dy c sdng l cphoc xch.

B truynbnh rngnhiu cphoc hnhtinh

Phanh thng dng phanh ma st nhiu a, loi thng ng.

bai giang may nang chuyen

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