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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.
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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...)
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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
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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
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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
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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/8/6/2019 Bai Giang May Nang Chuyen
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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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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
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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
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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.
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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
nD
a
nD
a
vv c
tg
n
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
Mmen do qun tnh T2 (tip)
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6-14
Mmen do qun tnh T2 (tip)
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
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
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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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4
8.1.2. C cu di chuyn
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
8.1.2. C cu di chuyn
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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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6
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.