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Chng II
Thit k mn hc : Kt cu tnh ton t.
nhim v thit k mn hc kt cu tnh ton t
Gio vin hng dn : TS . Nguyn c Ton.
Sinh vin thc hin : Nguyn c Minh.
Nhim v : Thit k truyn chnh vi sai ca t,vi cc s liu ban u
nh sau:
+) Loi t : Xe khch.
+).Ti trng cu trc :Ga1= 2538 (KG) +).Ti trng cu sau : Ga2= 5287 (KG)
+).Loi ng c : Xng
+).M men cc i ca ng c : Memax= 29/ 2250 (KG.m/nM)
+).Cng sut ln nht ca ng c : Nemax= 115/3200 (ml/nN)
+).Loi truyn lc chnh : n - Hypit
+).K hiu bnh xe : 8,25 - 20
+).T s truyn ca hp s cc tay s:6,55 ;3,09 ;1,71;1,00
+).T s truyn ca tryn lc chnh : i0 = 6,83
M u
Truyn lc chnh nm trong cu ch ng c cu to bi mt hoc vi cp bnh rng n khp. Truyn lc chnh gm c cc loi : Truyn lc trung ng vtuyn lc cnh hay cn gi l truyn lc cui cng.
Truyn lc chnh(TLC) c cng dng lm tng m men ca ng c v truyn mmen n cc bn trc di mt gc no i vi trc t.TLC ca t c th mt cp hoc hai cp.Loi hai cp c tr s tu thuc vo gi s TLC.TLC hai cp dng thay th cho s truyn tng ca hp s.
TLC c loi n hoc kp,loi n c mt cp bnh rng nn hoc hypit.Loi kp c mt cp bnh rng nn hoc hypit v mt cp bnh rng tr .Loi n thng dng cho t c ti trng trung bnh v nh.Loi kp thng dng cho t c ti trng ln
Loi n thng dng cho t c ti trng trung bnh v nh.Loi kp thng dng cho t c ti trng ln
Yu cu ca TLC l m bo c t s truyn cn thit ,m bm c hiu sut truyn cao khi nhit v s vng quay thay i ,ng thi m bo c tnh nng thng qua cao.
phn tch kt cu cc loi truyn lc chnh
1.Loi bnh rng nn rng thng
Loi ny c ng dng ph bin trn t ti v trn t but. . .Thong thng cp bnh rng c b tr di mt gc 90, nhiu lc n c b tr di mt gc khc 90.
Loi ny c nhc im l s rng nu ly t hn (8 - 9) rng th b ct chn rng,v vy mun c t s truyn ln th phi c kch thc ca bnh rng b ng ln ,n lm cho kch thc cu sau tng ln lm cho tnh nng thng qua gim.Mt khc khi truyn ng tc cao thng gy n.
2.Loi bnh rng nn rng xon
Loi bnh rng nn rng xon th s rng ca bnh rng b ng c th ly t (6 - 7) rng ,nhng vn m bo c iu kin n khp v iu kin bn.Nh m tng c t s truyn v gim c kch thc ca b tryun.ng thi TLC rng nn rng xon lm vic m du ngay c khi tc cao,v vy n thng dng cho ng c cao tc.Ngoi ra n cn c ch to n gin, v n c th gia cng trn my ct rng c cng sut cao,nhng n khng i hi chnh xc cao lm.Nhc im ca loi b truyn ny khi truyn ng thng sinh ra lc dc trc ln hng ca n ph thuc vo chiu quay ca trc ch ng
3.truyn ng hypit
Truyn ng Hypit c s dng rng ri cc loi xe ti c cng sut va v nh , Loi ny c nhng u v nhc im sau :
- u im :
+Dch c trc ca bnh rng b ng ln hoc xung,so vi bnh rng b ng mt khong E = (0,125 - 0,2)D2 (D2 l ng knh ngi ca bnh rng b ng).Nh m h thp c trng tm ca xe,lm tng n nh nn tng c tc cho php ca xe.
+ So vi loi truyn ng bng bnh rng nn ,khi c cng t s truyn v cng sut truyn nh nhau, th ng knh trc ca bnh rng ch ng ln hn khong (20 - 25)%.Do cng vng ca truyn ng hypit s ln hn,ng
thi n truyn ng bng cp bnh rng cn xon nn khi lm vic s m du hn.
+ Do c b mt tip xc ln nn p lc tng hp gim ,lm tng bn lm vic ca rng.
- Nhc im :
+ Do c s trt tng i theo c chiu di mt cnh ca rng, nn mun gim mi mn th phi s dng loi du bi trn c bit.Vn tc trt di c tnh theo cng thc:
- Truyn ng Hipoit i hi phi lp t chnh xc v bnh rng ch ng phi c im ta vng chc.
4.Truyn ng trc vt:
S dng truyn ng loi trc vt i hi t s truyn ln,kch thc b truyn b loi nay c nhng u v nhng nhc im sau :
- u im:
+ lm vic m du ,c kch thc b nhng truyn c t s truyn ln.
+ Cho php t vi sai ngay chnh gia cu sau,do n c kt cu i xng nn thun tin cho vic tho lp.
+ Trong t c ba cu ch ng th n c kh nng truyn mmen ln c hai cu m ch qua mt trc.
+ Khi t trc vt di bnh vt th c th h thp c ttng tm ca xe.
- Nhc im :
+ c hiu sut truyn thp ,i hi chnh xc lp rp cao.Nu chnh xc lp rp cng thp th n lm cho hiu sut truyn cng thp v trc vt cng chng mn.
+ Ch to cp bnh vt - trc vt kh khn phc tp.
+ Nu t trcvt di bnh vt th s lm gim khong sng gm xe ,v lm tng gc lch ca trc cc ng. Nhng nu t pha trn th lm kh khn cho cng tc bi trn.
pHN I--Thit k truyn lc chnh.
I- Thit k s b :
1.Chn ch ti trng tnh ton.
1.1.Mmen t ng c xung :
Mc = Memax.iI hs.(tl
Trong :
+ Memax : L mmen ln nht ca ng c : Memax = 29 (KG.m).
+ iihs : L t s truyn ca hp s ti tay s I : iihs = 6,55.
+ (tl : Hiu sut truyn ca h thng truyn lc ,tnh t ng c n bnh rng cn ch ng : (tl = (lh . (hs . (c = 0,96 . 0,96 . 0,98 = 0,90
Vy : Mc = 29 . 6,55 . 0,9 = 171,56 (KG.m)
1.2.M men truyn t mt ng ln b truyn lc chnh :
Trong : + ( :L h s bm ca bnh xe cu ch ng : ( = 0,7
+ Gb = 5287 (kg). L trng lng bm ca cu ch ng.
+ i0 : t s truyn ca TLC : i0 = 6,83.
+ rbx :bn knh bnh xe : rbx = 25,4 ( 8,25 + 20/2) = 463,55 (mm) + (tl : Hiu sut ca h thng truyn lc truyn t bnh xe ch ng ln TLC : (tl = 0,95
Vy:
So snh hai gi tr trn ta ly mmen c gi tr nh hn tnh ton.
2.Tnh ton cc thng s hnh hc ca b truyn lc chnh.
2.1 Xc nh s ca mi bnh rng trong truyn lc chnh.
Vi b truyn bnh rng nn hypit - n c t s truyn i0= 6,83 .Theo kinh nghim ta chn s rng ca bnh rng ch ng trong bng (6-2T99.KCTT):
Z1 = 6.
Vy s rng ca cp bnh rng b ng Z2 :
Z2 = i0 .Z1 = 6,83.6 = 41.
Kim tra li t s truyn:
[( i0 - Z2/ Z1 )/ i0 ] .100 % = [ (6.83 - 41/ 6)/ 6,83].100% = 0,05%
2.2 Xc nh ng sinh ca bnh rng cn theo cng thc thc nghim ta c:
2.3 Xc nh s b mun ca rng :
Theo tiu chun ta chn ms = 5 (mm).
2.4 Xc nh bn knh trung bnh ca cc bnh rng :
Mt khc r2 = E/ KE ( KE = E/r2
Trong : E : L khong cch dch chuyn trc ca bnh rng ch ng so vi bnh rng b ng ,i vi loi xe ti :
E = (0,125 - 0,2).D2= 2(0,125 - 0,2).102,5 = (25,63 - 41 ) (mm)
Chn E = 33 (mm).
Vy KE = 33/102,5 = 0,322.
Bn knh trung bnh ca bnh rng ch ng:
r1 = r2.k .Z1/Z2 = 102,5 . 1,3 . 6/41 = 19,5 (mm).
2.5. Gc cn ban u ca bnh rng b ng .
Trong : ( L gc truyn gia hai bnh rng ( = 9002.6 Gc gia ng php tuyn chung vi hai trc ca bnh rng b ng.
- Tnh gn ng:
-Tnh chnh xc :
2.7 Gc gia ng trung bnh khi thit k ca trc bnh rng b ng.
2.8. Gc cn ban u ca bnh rng cn ch ng.
2.9 .Chiu di ng sinh hnh cn ban u ca bnh rng ch ng.
2.10. Chiu di ng sinh hnh cn ban u ca bnh rng b ng.
2.11.Kim tra h s tng kch thc.
Sai lch vi chn ban u ch 1% nn ta chn k = 1,3.
2.12.Hiu s gc xon gia gc cn ca bnh rng ch ng v bnh rng b ng :
2.13.Gc xon ca rng trn bnh rng b ng.
2.14 .Gc xon ca rng trn bnh rng ch ng.
2.15. Mun php tuyn ti im gn gia rng.
2.16.Khong cch ca mt phng trung bnh ca ng knh bnh b ng vi trc.
2.17. Khong cch ca mt phng trung bnh ca ng knh bnh ch ng vi trc.
2.18.H s n khp khng i xng.
3.tnh ton nhng thng s hnh hc ca bnh rng hypit.
3.1. H s chiu cao chnca bnh rng b ng.
f2 = 0,8(1- ()Trong : ( L h s dch chnh u rng, tra theo bngI-8b(hdtkkctt) ng vi t s truyn i0 = 6,83 ta c : ( =0,7.
Vy f2= 0,8(1 - 0,7) = 0,24.
3.2. H s chiu cao chn rng ca bnh rng b ng.
fc2 = 0,8(2,25 - f2) = 0,8(2,25 - 0,24) = 1,68
3.3. Chiu rng vng rng tit din trung bnh ca bnh rng b ng.
Vi (0n L gc n khp lch ca bnh rng ch ng v bnh rng b ng.N ph thuc vo h s n khp khng i xng CE ,tra theo bng I - 8 Tr 44.Ta c tng ng viCE = 0,08 chn (0n= 200 Vy :
3.4.Gc chn rng ca bnh b ng.
Trong L2 L ng sinh ca mt nn bnh rng b ng.
3.5. Gc u rng ca bnh rng b ng.
3.6.Gc cn trong ca bnh rng b ng.
3.7. Gc cn ngoi ca bnh rng b ng.
3.8. Gc cn ph ca bnh rng b ng.
3.9. Chiu rng rng ca bnh b ng.
b2 ( 0,35.L2 = 0,35.103,7 =36,3 (mm).
3.10.Chiu di to bi hnh cn ngoi v cn trong ca bnh rng b ng:
+ Li2 = L2 - 0,5.b2 = 103,7 - 0,5.36,3 = 85,6 (mm)
+ Le2 = L2 + 0,5.b2 = 103,7 + 0,5.36,3 = 121,85 (mm)
3.11.Chiu cao chn rng v u rng ca bnh rng b ng ti im gia ca rng.
+ h'c2 = mn .fc2 = 4,35. 1,68 = 7,31 (mm)
+ h'd2 = mn .fd2 = 4,35.0,24 = 1,04 (mm)
3.12. .Chiu cao chn rng v u rng ca bnh rng b ng theo ng knh ngoi.
+ hc2 = h'c2 + 0,5.b2.tg(2 = 7,31 + 0,5.36,3.tg5,550 = 9,07 (mm)
+ hd2 = h'd2 + 0,5.b2.tg(2 = 1,04 + 0,5.36,3.tg0,790 = 1,29 (mm)
3.13.ng knh ngoi ca bnh rng b ng.
De2 = 2.(Le2.sin(2 + hd2cos(2) = 2.(121,85.sin81,260 + cos81,260) = 241,26 (mm)
3.14.Khong cch mt u n ng knh trung bnh ca n .
Gi tr ca n chn theo kt cu E2 = 23,56 (mm).
3.15.Khong cch nh rng ca bnh rng b ng knh ngoi t mt u c s ca n.
B'2 = E2 - 0,5.b2cos(2 + hd2.sin(2
B2'= 23,56 - 0,5.36,3.cos81,260 + 1,29.sin81,260 = 22,08 (mm).
3.16.Chiu rng ca bnh rng b ng.
B2 = E2 + 0,5.b2cos(2 + hd2.sin(2 Li2/Le2.
B2= 23,56 + 0,5.36,3.cos81,260 + (1,29.sin81,260 .85,6)/121,85 = 27,21 (mm).
3.17.Chiu di ca rng trn bnh rng ch ng dc theo ng sinh.
Chn b1 = 41 (mm).
3.18.Chiu cao chn rng v u rng ca bnh ch ng tnh ti im gia.
+ h'c1 = h'd2 + 0,25.mn = 1,04 + 0,25.4,35 = 2,13 (mm)
+ h'd1 = h'c2 - 0,25.mn = 1,04 - 0,25.4,35 = 0,0475 (mm)
3.19.Gc chn rng ca bnh ch ng.
3.20. Gc u rng ca bnh ch ng.
3.21.Chiu cao chn rng v u rng ch ng ng knh ngoi.
+ hc1 = hc1' + 0,5.tg(1 .b1 = 2,13 + 0,5.tg0,750.41 = 2,4 (mm)
+ h1 = hd1' + 0,5.tg(1.b1 = - 0,048 + 0,5.tg5,270.41 = 1,84 (mm)
3.22.Gc cn trong ca bnh rng ch ng.
(i1 = (1 - (1 = 8,30 - 0,730 = 7,570.
3.23.Gc cn ngoi ca bnh rng ch ng.
(e1 = (1 + (1 = 8,30 + 0,730 = 9,030.
3.24. Gc cn ph ca bnh rng ch ng.
((1 = 900 - (1 = 900 - 8,30 = 81,70
3.25.ng knh ngoi ca bnh rng ch ng.
3.26.Khong cch mt u c s ca bnh rng ch ng tnh t ng knh
trung bnh ca n.
Chn theo kt cu : E1 = 24,38 (mm).
3.27.Khong cch nh rng ca bnh rng ch ng tnh t ng knh
ngoi ca mt u c s.
3.28.B rng ca bnh rng ch ng.
3.29. Khong cch nh ca vng cn ban u bnh b ng theo mt u c
s ca n.
3.30. Khong cch nh ca vng cn ban u bnh ch ng theo mt u c s ca n.
3.31. Khong cch mt u c s ca bnh rng b ng theo trc ca rng.
A2 = C2 + E2 =19,478 +23,56 = 43,04 (mm).
3.32. Khong cch mt u c s ca bnh rng ch ng theo trc ca rng.
A1 = C1 + E1 = 97,38 + 24,38 = 121,76 (mm).
3.33.Khong cch nh ca hnh cn ban u ca bnh b ng tnh t tm trc ca bnh rng.
(A = A2 - A'2 = 43,04 - 39,5 = 3,54 (mm).
3.34.Chiu dy rng ti tit din trung bnh i vi thc o rng.
+ i vi bnh rng b ng:
S1 = Stb2 .cos2 n - Ch
Trong : Ch L h s ph thuc kt cu .
Do mn = 5 nn chn Ch = (0,05 - 0,2) chn Ch = 0,1
Vy S1 = 9,05.cos200 - 0,1 = 7,89 (mm).
+ Bnh rng b ng .
S2 = ( .mn) - Stb2 = 3,14.5 - 9,05 = 6,65 (mm).
3.35. Chiu cao rng , thc o rng .
+ Bnh ch ng .
h1 = h'1 - 0,5.tg(0n.S1 = 0,048 - 0,5.tg200.7,89 = 1,38 (mm).
+Bnh b ng.
h2 = h'2 = 1,04 (mm)
4.tnh ton bn.
4.1 Xc nh lc tc dng ln bnh rng .
xc nh c phng ca cc thnh phn lc tc dng,th ta phi chn chiu xon ca rng v chiu quay ca trc ,cch b tr truyn lc chnh.Ngoi ra trong truyn ng loi Hypit hng xon ca rng cn ph thuc vo hng dch trc ca bnh b ng ln pha trn hay pha d ng tm ca bnh b ng.i vi xe c cng sut va v ln th m bo khong sng gm xe,th ngi ta thng b tr trc ca bnh rng b ng ln pha trn.Nh vy ta c hng xon ca rng l hng xon phi(Nu nhn t y ln th rng i cng xa v pha tay tri).Nh vy
* Tnh lc vng tc dng ln bnh rng:
- i vi bnh rng ch ng :
+Lc vng : P1 = Me max .ih 1 .(tl / rt b = 29 . 6,55 . 0,89 . 1000 / 463,55
P1 = 368,8 (KG)
+Lc dc trc :
+Lc hng knh :
- i vi bnh rng b ng:
+Lc vng :
+Lc dc trc :
+ Lc hng knh :
4.2.Tnh ton bn ca bnh rng theo sc bn un v tip xc.
4.2.1. Tnh theo sc bn un :
Trong :
+ y : L h s dng rng tu thuc vo s lng rng quy i.
Vy h s dng rng ca bnh ch ng v b ng c gi tr nh sau:
Tra theo bng 42-T67 ta c : + y1 = 0,106
+ y2 = 0,136
+ts : Bc rng mt bn tnh t y ln ca hnh cn chia.
tsi = ( (.Dei )/Zi ( ts1 = (3,14.48,56)/6 = 25,43 (mm)
ts2 = (3,14.241,26)/41 = 18,48 (mm)
+ ri : Bn knh hnh cn chia tnh t mt y ln ca hnh cn chia.
r1 = 19,5 (mm)
r2 =102,5 (mm)
+ b : Chiu di rng theo ng sinh ca hnh cn .
b1 = 36,3 (mm)
b2 = 41 (mm)
+k : H s ti trng ng c chn theo kinh nghim, n c gi tr nm trong khong k = (1 - 1,5) .Chn k = 1,3
Vy gi tr ca ng sut ca hai bnh rng l :
Do bnh rng cn xon ca truyn lc chnh c ch to t thp t cc bon hoc thp hp kim cc bon trung bnh ,nn ng sut cho php nm trong khong :
[(u] = (700 ( 800) (KG/ mm2)
[(tx] = 1000 (KG/ mm2)
So snh vi hai gi tr tnh ton trn n tho mn iu kin bn un ca rng.
4.2.2.Kim tra bn theo iu kin bn tip xc.
Theo biu thc :
Trong :
+P : L lc vng tnh theo ch ti trng trung bnh.
+rtd :l bn knh tng ng ca bnh rng ch ng v bnh rng b ng gi tr ca n c tnh nh sau:
Vy lc vng trung bnh c tnh nh sau :
+ E : L mun n hi E = 2.106 daN/cm2
Vy ng sut tip xc ca cc bnh rng l :
So vi tiu chun (tx1,(tx2 < [(txphn II - Tnh ton bn .
Khi tnh ton bn vng ta phi tnh theo hai ch .
1. Ch tnh t ng c n bnh rng cn ch ng theo cng th :
Mtt 1= Me max .iH .H
iH : T s truyn tnh t ng c n bnh rng cn ch ng .
H : Hiu sut truyn lc t ng c xung chi tit tnh ton
H = 0,89 chn theo l thuyt t
=> Mtt 1 = Me max .ih1 .H = 19.3,5.0,89 = 59,18 (KG.m) = 591,8 (Nm)
2.Ch tnh t lc bm t mt ng n bnh rng cn ch ng
s bm . Chn Gb : Trng lng bm .
io : T s truyn ca truyn lc chnh .
+ Tnh theo bnh xe ca cu trc :
rbx = (8,25 + 20/2 ).25,4.0,93 = 431 (mm)
MTtt 2 = 0,7.870.431/4,1.0,89.1000 = 71,93 (KGm) = 719,34 (Nm)
+ Tnh theo bnh xe ca cu sau :
MStt 2 = 0,7.950.431.10-3 /(4,1.0,89) = 78,54(KGm) = 795,4 (Nm)
- So snh hai m men trn ta thy mmen tnh ng c xung nh hn mmen tnh t mt ng ln => chn ch tnh ton t ng c xung.
3. Tnh lc vng tc dng ln bnh rng:
+) i vi bnh rng ch ng :
Lc vng : P1 = Me max .ih 1 .H / rt b = 19.3,5.0,89.1000 /33,15 =1785,36(KG)
Lc dc trc : Q1 =P1 .(tg .sin ( + sin .cos() /cos Q1 = 1713,4(KG)
Lc hng knh : R1 =P1 .[( tg cos) - (sin1 .sin1)]/cos(1 = 649,1 (KG)
+) i vi bnh rng b ng :
Lc vng : P2 =(P1.cos2) / cos1 = 2312,7 (KG)
Lc dc trc : Q2 = P2.[( tg .sin ) + (sin .cos)] /cos
Q2 = 815.3 (KG)
Lc hng knh : R2 = P2.[( tg cos) - (sin2 .sin2) ] /cos(1
R2 = 795.4 (KG)
4. Tnh ton bn bnh rng
+)Tnh theo sc bn un :
H s dng rng y c tra theo bngtheo s lng rng quy dn
Zqd = Z /(cos3 .cos=> Z1qd =16,6; Z2qd =314,2
=> y1 = 0,526 ; y2 = 0,553.
tS= (2.r1) / Z1 = (2.3,14.33,15)/ 9 = 23,13 (mm)
- Tnh cho bnh ch ng:
u2 = (u1 .y1)/y2 = 998.0,526/0,553= 949,27(KG/cm2)
- Tnh ng sut tip xc .
Lc P l lc vng ly theo ch ti trng trung bnh P = 1785,36 KG
rt : Bn knh tng tng ca bnh rng ch ng v b ng
rt = r/(cos2(cos()
rt1 = 66,2 mm ; rt2 = 1288,2 mm
tx = 1101 (N/mm2) Tnh s b :
Mx : M men xon t ng c truyn xung : MX =171,56* 103(KG. mm)
ng sut un cho php . [(u] = (700 - 900) (KG/mm2)
chn d = 50 (mm) .Tra bng 14P sch chi tit my chn chiu rng B = 40 (mm)
+) L' : khong cch gia cc gi trc bnh rng nn nh
L' = (2,5 ( 3 )d ; chn L'=2,5d = 2,5 *50 =125 (mm)
Sau khi tnh ton theo chi tit my ta chn c cc khong cch sau :
a = 79 mm ; b = 75 mm ; c = 40 mm
-Xc nh phn lc ti cc gi ta A v B do lc vng P gy ra (trong mt phng XOY) .
Tnh ng knh trc ti hai tit din vi
[(] : ng sut cho php [(] = 80 (N/mm2) = 8(KG/mm2)
Mt : Mmen tng ng
Ti tit din I-I ta c :
Do ti tit din I-I c Mu = 0 ;Mx =RAx .80 + P1.130 = 795552,6 (KGmm)
( d ( 78,4 (mm) chn dI-I = 80 (mm)
Ti tit din II-II ta c :
chn dII-II = 68 (mm)
PHN IV - TNH TON VI SAI
1.Tnh s b L, Z1 , Z2 , ms
i vi bnh rng cn trong cm vi sai
MVS = Memax .ih1.i0 . = 29 * 6,55 * 6,83 * 0,89 = 1154,65 (KG.m)
S lng bnh rng hnh tinh : q = 4 ; T s truyn i = 1,5
Ta c L = 50 (mm) : ms =5 tra theo hnh I- 3 v bng I-1
=> sai s 3% < 5%
-Gc n khp : = 22030 => Z2 v Z1 tho mn Z1 > Zmin =10
Z2 =16 > Z2min tho mn
Tnh li
2. Chn gc n khp : n =200
3.Chn chiu rng vnh rng : b = L. L =( 0,250,35).L= (12,5
Chn b = 16(mm).
4.Chn h s chiu cao rng v khe h hng knh :
+ H s chiu cao rng : f0 = 1.000.
+ Khe h hng knh : C 'n =0,188.
5. Chn h s dch chnh chiu cao S v h s dch chnh tip tuyn :
tra bng S = 0,38 ; = f(i) =>2 = - 1 = 0,1 (trabng )
6. chn cp chnh xc : cp 8
III. Tnh ton kch thc hnh hc ca bnh rng cn thng
1. S rng ca bnh ch ng :Z1 = 11
+)S rng ca bnh b ng : Z2 =16
2. T s truyn :i = Z2/Z1 = 16/11 = 1,45
3. Gc n khp tit din php tuyn : (n = 200
4. Chiu rng bnh rng :b = (0,25 ( 0,35 )L = (12,5 ( 17,5) ; chn b = 16 mm
5. Gc xon gia vnh rng : ( = 0.
6.M un tit din mt u: ms =5
7. H s chiu cao rng : fos = 1
8. H s khe h hng knh : c'n = 0,188
9. Gc cn chia : (1 = 34030' vi ( = 900 khi (1 = arctgZ1/Z2
(2 =900 - (1 = 55030'10. Hm tam gic lng ca cc gc cn chia :
Sin(1 = 0,5665 ; cos(1 = 0,8240 ; sin(2 = 0,8211 ; cos(2 = 0,5693.
11.H s dch chnh chiu cao tit din u rng : (s = ( 0,38
12. H s dch chnh tip tuyn :( = 0,01
13.S rng ca bnh rng phng : Zc = Z2/ sin(2 = 16/0,8211 = 19,462.
14. Chiu di ng sinh hnh cn chia : L = Zc.ms/2 = 48,656 (mm)
15.Chiu cao lm vic ca rng : h1 = 2f0s .ms = 10,00
16.Khe h hng knh : c's = c'n.ms = 0,94.
17.Chiu cao ton b rng :h = h1 + c's = 10,94 (mm)
18. ng knh vng trn chia : d1 = ms.Z1 =5.11 = 55 (mm)
d2 = ms. Z2 = 5.16 = 80 (mm)
19. Chiu cao u rng : h'1= ms.(fos + (s ) = 6,9 (mm)
h'2 = h1 - h'1 = 3,1 (mm)
20. Chiu cao chn rng: h''1 = h - h'1 = 4,04 (mm)
h''2 = h - h'2 = 7,84 (mm)
21. Gc chn rng : (1 = arctg h''1/L = 4044'
(2 = arctg h''2/L = 909'
22. Gc cn ngoi : (e1 = (1 + (2 = 44039'
(e2 = (2 + (1 = 60014'
23.Gc cn trong : (i1 = (1 - (1 = 24046'
(i2 = (2 - (2 = 41021'
24. ng knh nh rng : De1 = d1 +2h'1cos(1 = 66,37 (mm)
De2 = d2 +2h'2cos(2
25. Khong cch t nh n mp ngoi ca vnh rng :
IV. Kim tra iu kin n khp v mt hnh hc ca b truyn
1. Kim tra iu kin khng b ct chn rng v chy rng
i vi nhng b truyn khi thit k ly cc gi tr (s v ( theo bng chn th khng cn kim tra na.
2. Xc nh gc n khp nh nht tit din u rng (smin m khng b ct chn rng
Trong : ( - gc chn rng
( - gc cn chia
i vi b truyn thng ( = 00 phi ly gc n khp (n ln hn (smin
Vy iu kin gc n khp nh nht tit din u rng c tho mn.
3. Kim tra khng b nhn rng l xc nh chiu rng pha cn ngoi tit din u rng.
Gi tr Se c tnh nn m bo quan h sau y
Khi
vy Se ( 0,5ms.cos( = 2,5
4. Kim tra h s trng khp
a. H s trng khp prphin mt u xc nh theo cng thc
( (s = 1,1245 vy tho mn iu kin (1 ( 1,25 )
b. H s trng khp theo chiu dc rng xc nh : (b = 0.
c.Tnh h s trng khp
PHN III - CHN LN :
- Chn ln cho trc ch ng ca truyn lc chnh :
+ Trc c tc vng quay n = nM / ih1 = 2250/6,55 = 343,5(v/p)
+ ng knh ngng trc ti 2 ln : dI-I = 90 (mm) ;dII-II = 76 (mm).
+ Phn lc ti cc gi :
+ Lc chiu trc trn bnh rng cn xon : Q1 = 5313,3 ( KG ).
+ Nhit lm vic di 1000C
+ Do trc c lc dc trc tc dng nn chn bi chn
+ D kin chn loi bi chn k hiu 36000 c gc +Vi dI-I = 90 chn loi bi chn k hiu 36218 c
D = 160 (mm) chiu rng ; B = 30 (mm) chiu cao ; D2 = 139 (mm) ;
d2 = 112,4 (mm) ; r1 = 1,5( mm) ;r = 3(mm).
+ Vi dII-II = 76 (mm) chn kiu c k hiu 36215 c cc thng s sau
D = 130 (mm) ; B = 25 (mm); D2 = 113 (mm) ; d2 = 92 (mm) ; r1 = 1,2( mm) ;
r = 2,5(mm).
Ta c s biu din cc lc tc dng sau:
Ti liu tham kho
1.Kt cu tnh ton t ( Trng i hc giao thng vn ti H Ni )
2.Hng dn thit k cu ch ng ( Trng i hc giao thng vn ti H Ni)
3.Hng dn thit k chi tit my ( Nguyn vn Lm )
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PAGE 23Gio vin hng dn : Nguyn c Ton.
SVTH : Nguyn c Minh - Lp c kh t B - k38.
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