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n chi tit my 1: Thit k h dn ng xch ti
N CHI TIT MY S 1: THIT K H DN NG XCH TIPHN 1: CHN NG C V PHN PHI T S TRUYN
Bng thng s u vo: Thit k h dn ng xch tiLc ko xch ti: F = 1500 (N)Vn tc xch ti: v = 2.88 (m/s)S rng a xch ti: z = 15 (rng)Bc xch ti: p = 105 (mm)Thi hn phc v: lh = 19000 (h)S ca lm vic: soca = 1 (ca)Gc nghing ng ni tm b truyn ngoi @ = 300c tnh lm vic: Va p va
1.1. Chn ng c in1.1.1 Cng sut trn trc my cng tc
(KW)1.1.2 Hiu sut chung ca ton h thng
Trong tra bng ta c: Hiu sut b truyn bnh rng: = 0,97 (1 cp bnh rng) Hiu sut b truyn xch h: = 0,92 Hiu sut ln: = 0,99 Hiu sut ni trc n hi: = 1 = 0,97.0,993.0,92.1= 0,86591.1.3 Cng sut yu cu trn trc ng c
= 4,99 (KW)1.1.4 S vng quay trn trc b phn cng tc
(v/ph)1.1.5 Chn t s truyn s b ca h thng
Tra bng chn s b: T s truyn b truyn xch: =3 T s truyn b truyn bnh rng = 4 = 3.4 = 121.1.6 S vng quay s b trn trc ng c
=109,71.12 = 1316,52 (v/ph)1.1.7 Xc nh s vng quay ng b ca ng c
Chn nbnsb nb=1500 (v/ph)1.1.8 Chn ng cTra bng ph lc ti liu 1.3[1], chn ng c tho mn: nbnsb v PcPycTa c ng c vi cc thng s sau: (vi mc v dc c tra t bng 1.7[1])K hiu ng cPc(KW)nc(v/ph)T/Tdnmc(kg)dc(mm)
4A112M4Y35,5142525632
1.2 Phn phi t s truyn
T s truyn chung ca h thng: Chn t s truyn ca hp gim tc ubr= 4
T s truyn ca b truyn ngoi: = 3,25
Vy ta c: 1.3 Tnh cc thng s trn trc Cng sut trn cc trcCng sut trn trc cng tc: Pct=Plv=4,32 (KW)Cng sut trn trc II:
= 4,74 (KW) Cng sut trn trc I:
= 4,94 (KW) Cng sut trn trc ng c:
= 4,99 (KW) Vn tcS vng quay trn trc ng c: nc= 1425 (v/ph)S vng quay trn trc I:
= 1425 (v/ph)S vng quay trn trc II:
=356,25 (v/ph)S vng quay trn trc cng tc:
= 109,62 (v/ph) M men xonMment xon trn trc ng c:
Mment xon trn trc I:
(N.mm)Mment xon trn trc II:
(N.mm)Mment xon trn trc cng tc:
(N.mm)1.4 Lp bng thng s ng hc TrcThng sTrc ng cTrc ITrc IIB phn cng tc
T s truyn-uuk=1ubr=4ux=3,25
Vn tc quay-n (v/ph)14251425356,25109,62
Cng sut-P (KW)4,994,944,744,32
M men xon-T (Nmm)33441,7533106,67127065,26376354,68
PHN 2 : TNH TON THIT K B TRUYN NGOI(B TRUYN XCH) Thng s yu cu:
2.1 Chn loi xch Chn loi xch ng con ln do ti trng khng qu ln v tc thp.2.2 Chn s rng a xchChn s rng a xch Z1 thng qua cng thc Z1 = 29 2u.
Thay gi tr u=3,25 vo ta c: Z1= 22,5 Chn s rng Z1 = 23
S rng Z2 = u.Z1 = 3,25.23 = 74,75 < zmax=120 ly Z2 = 75
T s truyn thc ca b truyn xch 2.3 Xc nh bc xchBc xch p c tra bng 5.5[1] vi iu kin Pt [P], trong :Pt Cng sut tnh ton: Pt = P.k.kz.knTa c:Chn b truyn xch th nghim l b truyn xch tiu chun, c s rng v vn tc vng a xch nh nht l:
Do vy ta tnh c:
kz H s h rng:
kn H s vng quay: k = k0kakckbtk.kc Tra bng 5.6[1] ta c:k0 H s nh hng ca v tr b truyn, vi @ = 300 ta c k0 = 1ka H s nh hng ca khong cch trc v chiu di xch:Chn a = (30 50)p => ta c ka = 1kc H s nh hng ca vic iu chnh lc cng xch c kc=1 (iu chnh bng mt trong cc a xch)kbt H s nh hng ca bi trn: kbt = 1,3 (b truyn ngoi lm vic trong mi trng c bi , bi trn t yu cu)
k H s ti trng ng-c tnh lm vic va p va k = 1,4
kc H s k n ch lm vic ca b truyn- lm vic 1 ca kc=1k = k0kakckbtkkc = 1.1.1.1,3.1,4.1 = 1,82Cng sut cn truyn P = 4,74 (KW)Do vy ta c:Pt = P.k.kz.kn = 4,74.1,82.1,09.1,12= 10,53 (KW)
Tra bng5.5[1] vi iu kin ta c: Bc xch: p = 25,4 (mm) ng knh cht: dc = 7,95 (mm) Chiu di ng: B = 22,61 (mm) Cng sut cho php: [P] = 19 (KW)
2.4 Xc nh khong cch trc v s mt xchChn s b: a= 40.p = 40.25,4 = 1016 (mm)S mt xch:
Chn s mt xch l chn: x = 130Tnh li khong cch trc:
xch khng qu cng cn gim a mt lng:
Do :
S ln va p ca xch i:Tra bng 5.9[1] vi loi xch ng con ln, bc xch p = 25,4 (mm) => S ln va p cho php ca xch: [i] = 30
2.5 Kim nghim xch v bnKim nghim v qu ti theo h s an ton:
trong :Q Ti trng ph hng, tra bng 5.2[1] vi p = 25,4 (mm) ta c: Q = 56,7 (kN) Khi lng 1m xch: q = 2,6 (kg).k H s ti trng ng: k=1,2Ft : lc vng.Ta c: vn tc trung bnh ca xch
(m/s) Lc vng:
Fv Lc cng do lc ly tm sinh ra: Fv=q.v2 =2,6.3,472 =31,31 (N)F0 Lc cng do trng lng nhnh xch b ng sinh ra:F0=9,81.kf.q.a trong :kf H s ph thuc vng ca xch: Do @ =300 => kf = 4
[s] H s an ton cho php: Tra bng 5.10[1] vi p = 25,4 (mm), n =400(v/ph) ta c [s] = 9,3
Do vy: b truyn m bo bn.
2.6 Xc nh thng s ca a xchng knh vng chia:
ng knh nh rng:
Bn knh y: vi tra theo bng 5.2[1] ta c:
15,88(mm) ng knh chn rng:
Kim nghim rng a xch v bn tip xc:
, trong :K H s ti trng ng: K=1,4A Din tch chiu ca bn l: Tra bng 5.12[1] vi p = 25,4 (mm);A = 180 (mm2)kr H s nh hng ca s rng a xch, tra bng trang 87[1], dng php ni suy, vi Z1=23 ta c kr= 0,44 k H s phn b ti trng khng u gia cc dy, vi xch 1 dy =>k=1Fv Lc va p trn m dy xch:
E Mun n hi:
do E1 = E2 = 2,1.105 MPa : C hai a xch cng lm bng thp.Do vy:
Tra bng 5.11[1] ta chn vt liu lm a xch l thp 45, vi cc c tnh ti ci thin t rn HB170 s t c ng sut tip xc cho php [H]= 500MPa, m bo c bn tip xc.2.7 Xc nh lc tc dng ln trc
trong :kx H s k n trng lng ca xch:kx =1,15 v b truyn nghing mt gc 300< 400.
=> 2.8 Tng hp cc thng s ca b truyn xch:
Thng sK hiuGi tr
Loi xch------Xch ng con ln
Bc xchp25,4 mm
S mt xch x130
Khong cch trca1004mm
S rng a xch nhZ123
S rng a xch lnZ275
Vt liu a xch------Thp C45(ti ci thin)
ng knh vng chia a xch nhd1186,54mm
ng knh vng chia a xch lnd2606,56mm
ng knh vng nh a xch nhda1197,49mm
ng knh vng nh a xch lnda2618,73mm
Bn knh yR8,03mm
ng knh chn rng a xch nhdf1170,48mm
ng knh chn rng a xch lndf2590,5mm
Lc tc dng ln trcFr1570,9N
Bng 2.1: Cc thng s ca b truyn xch PHN 3.TNH TON THIT K B TRUYN TRONG (B TRUYN BNH RNG TR RNG THNG)Thng s u vo:
3.1 Chn vt liu lm bnh rngTra bng 6.1[1] ta chn:-Vt liu bnh ln : Nhn hiu thp : C45 Ch nhit luyn: Ti ci thin rn : HB =192240 ; Chn :HB2=230 Gii hn bn : b2=750(MPa) Gii hn chy : ch2=450(MPa)-Vt liu lm bnh nh : Nhn hiu thp : C45 Ch nhit luyn: Ti ci thin rn : HB =241285 ; Chn :HB1=245 Gii hn bn : b1 =850(MPa) Gii hn chy : ch1 =580(MPa)3.2 Xc nh ng sut cho php:3.2.1 ng sut tip xc [H] v ng sut un cho php[F]:
Trong :
-Chn s b: -SH,SF H s an ton khi tnh v ng sut tip xc v ng sut un:Tra bng 6.2[1] vi : Bnh rng ch ng : SH1=1,1 ; SF1=1,75 Bnh rng b ng : SH2= 1,1; SF2=1,75-oH lim , oF lim - ng sut tip xc v ng sut un cho php vi chu k c s:
Bnh ch ng
Bnh b ng -KHL,KFL H s tui th,xt n nh hng ca thi gian phc v v ch ti trng ca b truyn:
, Trong :
+ mH,mF Bc ca ng cong mi khi th v ng sut tip xc.Do bnh rng c HB NH01ly NHE1= NH01 KHL1=1
NHE2>NH02ly NHE2= NH02 KHL2=1
NFE1>NF01 ly NFE1= NF01 KFL1 =1
NFE2>NF02 ly NFE2= NF02 KFL2=1Do vy ta c:
Do y l b truyn bnh rng tr rng thng
3.2.2 ng sut cho php khi qu ti:
3.3 Xc nh s b khong cch trc:
,vi:-Ka- H s ph thuc vt liu lm bnh rng ca cp bnh rng.Tra bng 6.5[1]
Ka=49,5 MPa1/3-T1-Mment xon trn trc ch ng: T1=33106,67 (N.mm)-[H]-ng sut tip xc cho php: [H]=481,82(MPa)-u- T s truyn: u=4-ba,bd -H s chiu rng vnh rng:Tra bng 6.6[1] vi b truyn i xng,HBta s dng bnh rng lin trc.
Ti v tr khp ni: lt=(0,80,9) lm12= 37,642,3 => chn lt=40 mm. 4.5.2 Tnh Trc IIa) Cc lc tc dng ln trc II:
Vi cc gi tr: Fx=1570,9(N); Ft2=1303,16(N); Fr2=534,91(N)
T h phng trnh cn bng lc: Trong :Fi Lc thnh phnMi Mmen unli Cnh tay nTa c :
b) Biu mmen
c) M men un tng, m men tng ng v ng knh:
Trong :+)Mj ,Mtj,dj -ln lt l mment un tng,mment tng ng,ng knh trc ti cc tit din j trn chiu di trc.+)Myj,Mxj-m men un trong mt phng yoz v zox ti cc thit din th j.+)[]-ng sut cho php ch to trc,tra bng 10.5[1] c []=65MPa
Tit din 2-0:
Tit din 2-1:
Tit din 2-2:
Tit din 2-3:
-ng knh trc ti cc tit din tng ng khi tnh s b.
vi=50N/mm2 tra bng 10.5/195-Ti tit din lp ln bn tri:-Ti tit din lp bnh rng: -Ti tit din lp ln bn phi: -Ti tit din lp bnh xch: Ta chn ng knh theo tiu chun v m bo iu kin lp ghp:d23 kt cu trc 2 nh hnh v
d) Chn v kim nghim then:Chn then bng tha mn iu kin:
Trong :-d,[ d]:ng sut dp v ng sut dp cho php; Tra bng 9.5[1] c [ d]=100MPa -c,[ c]:ng sut ct v ng sut ct cho php [ c]=4060MPa khi chu ti trng va p-T-M men xon trn trc-d-ng knh trc-lt,h,b,t kch thc then tra bng 9.1a[1]V trKch thc thenChiu su rnh thenBn knh gc ln ca rnh r
bhTrn trc t1Trn l t2Nh nhtLn nht
Lp bnh rng12853,30,250,4
Lp a xch10853,30,160,25
Chiu di then:Ti v tr lp bnh rng:
lt= (0,80,9)lm23=(0,80,9).45=(36-40.5)Chn lt=36(mm)
Tha mn. Ti v tr lp xch:
lt= (0,80,9)lm23=(0,80,9).45=(36-40.5)Chn lt=36(mm)
Tha mne) Kim nghim trc v bn mi:Kt cu trc cn m bo h s an ton ti cc tit din nguy him tha mn iu kin:
Trong :[s]-H s an ton cho php;[s]=2,5sjv sj-H s an ton ch xt ring ng sut php v h s an ton ch xt ring ng sut tip ti tip din j:
; Trong :-1 ,-1-gii hn mi un v mi xon ng vi chu k i xng.Vi thp 45 c b=600MPa ; -1=0,436 b=261,6MPa -1=0,58 -1=152MPaTheo bng 10.7[1] ta c h s k n nh hng ca tr s ng sut trung bnh n bn mi:=0,05 , =0.Cc trc ca hp gim tc u quay,ng sut un thay i theo chu k i xng,do :
; Cc tit din nguy him trn trc II: +Ti tit din lp bnh rng(tit din 2-1)Thng s ca then xc nh c: d21=40 th b=12, h=8, t1=5Thay vo cng thc trong bng 10.6[1] vi trc c 1 rnh then, ti v tr bnh rng:
+Ti v tr ln C ta c:
+Ti v tr lp xch
* Khi trc quay 1 chiu ng sut xon thay i theo chu k mch ng do :
Theo cng thc trong bng 10.6[1]v tra thng s ca then trong bng 9.1a[1]vi trc khng c rnh then ta c:+Ti v tr ln:
+Ti v tr bnh rng
Kdj v Kdj- H s xc nh theo cng thc:
; Trong :Kx-H s tp trung ng sut do trng thi b mt,ph thuc vo phng php gia cng v nhn b mt .Theo bng 10.8[1] vi yu cu cc trc c gia cng trn my tin,ti c tit din nguy him yu cu tRa=2,50,63 m,chn Kx=1,06Ky-H s tng bn b mt trc,khng dng phng php tng bn c Ky=1, H s kch thc,k n nh hng ca tit din trc n gii hn mi theo bng 10.10[1] ta c: =0,88 ; =0,81K , K-H s tp trung ng sut thc t khi un v xon.Tra bng 10.12[1] c: K=1,76 ; K=1,54
Do ta c: ; T cc kt qu tnh ton trn ta c:Ti v tr bnh rng:
Tha mn iu kin bn.
Ti vi tr ln:
Tha mn iu kin bn.PHN 5. TNH CHN V KIM NGHIM LN5.1. Chn ln cho trc I c kt cu n gin nht, gi thnh thp nht, ta chn bi mt dy .Chn kt cu ln theo kh nng ti ng, ng knh trc ti ch lp ln: d= 30 mm. * Chn ln l bi 1 dy 306( bng P2.7[1]) vi cc thng s:K hiu d(mm)D(mm)B(mm)r(mm)ng knh biC(kN)C0(kN)
306307219212.3022,0 15,1
5.2.Chn ln cho trc IIChn ln l loi bi 1 dy, s b tr: ng knh ngng trc d=35m => tra bng 2.7[1] c cc thng s :K hiu d(mm)D(mm)B(mm)r(mm)ng knh biC(kN)C0(kN)
3073580212,514,2926,2 17,9
*Tnh kim nghim kh nng ti ca :Phn lc hng tm tc dng ln cc
- bn tri bnh rng:
- bn phi bnh rng
Lc dc trc
Kim nghim theo kh nng ti ng: Vi bi : Q=(XVFr+YFa)kt,k-V h s k n vng no quay, vi vng trong quay c V=1-kt h s k n nh hng ca nhit ,c kt=1 khi nhit =1050-k-h s k n c tnh ti trng,tra bng 11.3[1] c k=1,4-X-h s ti trng hng tm, Y h s ti trng dc trc
Fa=0 => X=1, Y=0 => Q = D thy Q0 ly Q=Q1=1,4.Fr1=4432,29 N (Q0, Q1 ln lt l ti trng quy c tnh ti ln bn tri v ln bn phi bnh rng.
Vi m bc ca ng cong mi; bi c m=3 L:tui th (triu vng quay);
(triu vng)
>C=26,2Ta gim thi gian phc v ca xung mt na, tc L=203,065 triu vng, khi
Qt0=0,6.623,59=374,15 (N); Qt1=0,6.3165,92=1899,55 (N) *Qt=Fr: ta c Qt0=623,59; Qt1=3165,92
ti trng tnh quy c Qt1=3165,92 (N) =3,17 kN < C0=17,9kN tha mn.
PHN 6: KT CU V HP6.1. Kt cu v hp gim tcCh tiu ca hp gim tc l cng cao v khi lng nh.Chn vt liu c hp gim tc l gang xm c k hiu l GX15-32.6.1.1 Chn b mt np ghp v thnChn b mt ghp np v thn i qua tm trc v song song vi mt .6.1.2 Cc kch thc c bn ca v hpDng phng php c ch to np , vt liu l GX15-32.
Cc kch thc ca cc phn t cu to nn hp gim tc cTn giTnh ton
Chiu dy: Thn hp, Np hp, 1 = 0,03a + 3 = 0,03.125 + 3 = 6,75 (mm) Chn = 8 (mm)1= 0,9. = 0,9.8 = 7,2(mm) chn 1=8(mm)
Gn tng cng: Chiu dy, e Chiu cao, h dce = (0,81) = 6,4 8 mm Chn e = 8 (mm)
h < 5 mm = 5.=5.8=40 (mm)khong 20
ng knh: Bulng nn, d1
Bulng cnh , d2 Bulng ghp bch np v thn, d3 Vt ghp np , d4 Vt ghp np ca thm, d5d1 > 0,04a + 10 = 0,04.125 + 10 = 15 (mm)Chn d1 = 16 (mm)d2 = (0,70,8)d1=11,212,8 mm chn d2=12(mm)d3 = (0,80,9)d2 = 9,610,8 mm chn d3 = 10 (mm)d4 = (0,60,7)d2 = 7,28,4 chn d4 = 8 (mm)d5 = (0,50,6)d2 = 67,2 chn d5 = 6 (mm)
Mt bch ghp np v thn: Chiu dy bch thn hp, S3 Chiu dy bch np hp, S4 Chiu rng bch np v thn, K3S3 = (1,41,8)d3 = 1418 mm chn S3 = 16(mm)S4 = (0,91)S3 = 14,416 mm chn S4 = 16 (mm)K3 K2 - (35) = 37- (35)= 3234 mm chn K3 = 34 (mm)
Kch thc gi trc:ng knh ngoi v tm l vt, D3, D2
B rng mt ghp bulng cnh , K2 Tm l bulng cnh , E2 v C (k l khong cch t tm bulng n mp l) Chiu cao, h
Trc I: D2 = 88 (mm), D3 = 108(mm)Trc II: D2 = 96(mm), D3 = 115 (mm)K2 = E2+R2+(35)=18+16+3=37 (mm)E2 = 1,6d2 = 1,6.12=19,2(mm) chn E2 = 18 (mm)R2 = 1,3d2 =1,3.12=15,6 (mm) chn R2 = 16 (mm)Chn h = 45 (mm)
Mt hp:
Chiu dy: khi khng c phn li S1 khi c phn li: Dd, S1 v S2 B rng mt hp, K1 v q
S1 (1,31,5)d1 =(20,824) chn S1 =24(mm)
S1 (1,41,7)d1 =(22,427,2);
S2(1,01,1)d2=(1213,2) chn S2=13 (mm) K1 3d1 = 3.16=48 chn K1=50(mm), q K1 + 21 =50+2.8= 66 (mm)
Khe h gia cc chi tit:Gia bnh rng vi thnh trong hp
Gia nh bnh rng ln vi y hp
(11,2)1 = (11,2).7=(78,4) chn = 8 (mm)
1 (35)1 = (35).7=(2135) chn = 30 (mm)
S lng bulng nn, Z
Z=4
*Gi trc trn v hpng knh gi trc cng chnh l ng knh np , c xc nh theo cng thc:
Trong D l ng knh l lp ln.Ta cV tr
h
Trc I7288 10865M8410
Trc II809611575M8410
6.1.3 Mt s kt cu khc
a) Ca thm kim tra qua st cc chi tit my trong khi lp ghp v du vo hp, trn nh hp c lm ca thm.Da theo kch thc np hp ta chn c kch thc ca thm nh hnh v sau.A(mm)B(mm)
(mm)
(mm)C(mm)
(mm)K(mm)R(mm)Vt(mm)S lng
1004015065125-5012M64
b) Nt thng hiKhi lm vic, nhit trong hp tng ln. gim p sut v iu ha khng kh bn trong v ngoi hp, ngi ta dng nt thng hi.Nt thng hi thng c lp trn np ca thm. Tra bng 18.6[2] ta c kch thc nt thng hi
ABCDEGHIKLMNOPQRS
M2721530154536326410822632183632
c) Nt tho du Sau mt thi gian lm vic, du bi trn cha trong hp, b bn (do bi bm v do ht mi), hoc b bit cht, do cn phi thay du mi. thay du c, y hp c l tho du.Lc lm vic, l c bt kn bng nt tho du. Da vo bng 18.7[2] ta c kch thc nt tho du DbmfL cqDS
M2021593282,517,8302225,4
d) Kim tra mc du kim tra mc du trong hp ta dng que thm du c kt cu kch thc nh hnh v.3018126612
e) Cht nh v.Mt ghp gia np v thn nm trong mt phng cha ng tm cc trc.L tr lp thn hp & trn np c gia cng ng thi, m bo v tr tng i gia np v thn trc v sau khi gia cng cng nh khi lp ghp, ta dng 2 cht nh v. Nh cc cht nh v, khi xit bulong khng lm bin dng vng ngoi ca . Thng s k thut ca cht nh v l d=5; c=0,8; l=1690, chn l=35 f) Vng mcKch thc vng mc: S = (23) = 1624. Chn S=16 mmng knh d=15 mm
6.2. Bi trn hp gim tc5.2.1.Bi trn trong hp gim tc
Do b truyn bnh rng trong hp gim tc u c nn ta chn phng php bi trn ngm du. Vi vn tc vng v=3,79 m/s < 12 m/s tra bng 18.11[2] ta c nht ca du 80/11 ng vi 50oC Tra bng 18.13[2] ta chn c loi du l: AK-20.5.3.2.Bi trn ngoi hp Vi b truyn ngoi hp khi lm vic s dnh bi bm do hp khng c che kn nn ta dng phng php bi trn nh k bng m.Bng thng k dnh cho bi trnTn du hoc mThit b cn bi trnLng du hoc mThi giant hay du hoc m
Du t my koAK-20B truyn trong0,6 lt/KW5 thng
M TTt c cc v b truyn ngoi2/3 ch hng b phn1 nm
5.3.3. Bng dung sai v kch thc lp ghp
TrcV tr lpKiu lp ES es
EI ei
ITrc-vng trong bi30k6
+15
+2
Vng ngoi bi - V hp
+30
0
Trc-vng chn m
+98+15
+65+2
Np - v hp +46-100
0-290
Trc - Bc chn trc I
+33+15
0+2
IITrc - Bc chn trc II 30+33+15
0+2
Trc Vng trong bi35k6+18
+2
Trc-Bnh rng 40+25+18
0+2
Vng ngoi bi V hp80H7+30
0
Trc Vng chn m
35+119+18
+80+2
Np - v hp
+46-100
0-290
Mc lc:
PHN 1: CHN NG C V PHN PHI T S TRUYN11.1. Chn ng c in11.1.1 Cng sut trn trc my cng tc11.1.2 Hiu sut chung ca ton h thng11.1.3 Cng sut yu cu trn trc ng c21.1.4 S vng quay trn trc b phn cng tc21.1.5 Chn t s truyn s b ca h thng21.1.6 S vng quay s b trn trc ng c21.1.7 Xc nh s vng quay ng b ca ng c21.1.8 Chn ng c21.2 Phn phi t s truyn31.3 Tnh cc thng s trn trc31.4 Lp bng thng s ng hc4PHN 2 : TNH TON THIT K B TRUYN NGOI52.1 Chn loi xch52.2 Chn s rng a xch52.3 Xc nh bc xch52.4 Xc nh khong cch trc v s mt xch72.5 Kim nghim xch v bn72.6 Xc nh thng s ca a xch92.7 Xc nh lc tc dng ln trc102.8 Tng hp cc thng s ca b truyn xch:10PHN 3.TNH TON THIT K B TRUYN TRONG12(B TRUYN BNH RNG TR RNG THNG)123.1 Chn vt liu lm bnh rng123.2 Xc nh ng sut cho php:133.2.1 ng sut tip xc [H] v ng sut un cho php[F]:133.2.2 ng sut cho php khi qu ti:153.3 Xc nh s b khong cch trc:153.4 Xc nh thng s n khp:163.4.1 Mun php:163.4.2 Xc nh s rng:163.4.3 Xc nh gc n khp tw:173.4.4 Xc nh h s dch chnh173.5 Xc nh cc h s v mt s thng s ng hc:183.6 Kim nghim b truyn bnh rng:193.6.1 Kim nghim v bn tip xc:193.6.2 Kim nghim v bn un:203.6.3 Kim nghim v qu ti:213.7 Mt vi thng s hnh hc ca cp bnh rng:223.8 Bng tng kt cc thng s ca b truyn bnh rng:23PHN 4: TNH TON THIT K TRC V CHN LN244.1 Tnh chn khp ni.244.1.1 Chn khp ni:244.1.2 Kim nghim khp ni:264.1.3 Ti trng ph tc dng ln trc:264.1.4 Cc thng s c bn ca ni trc vng n hi:274.2 Thit k trc284.2.1 Chn vt liu ch to trc:284.2.2 S phn b lc284.2.3 Lc tc dng t xch ln trc:294.3 Xc nh s b ng knh trc:294.4 Xc nh khong cch gia cc gi v im t lc:294.5 Tnh thit k trc334.5.1 Tnh s b trc 1334.5.2 Tnh Trc II34PHN 5. TNH CHN V KIM NGHIM LN455.1. Chn ln cho trc I455.2.Chn ln cho trc II45PHN 6: KT CU V HP476.1. Kt cu v hp gim tc476.1.1 Chn b mt np ghp v thn476.1.2 Cc kch thc c bn ca v hp476.1.3 Mt s kt cu khc506.2. Bi trn hp gim tc545.2.1.Bi trn trong hp gim tc545.3.2.Bi trn ngoi hp545.3.3. Bng dung sai v kch thc lp ghp55
Ti liu tham kho1. Tnh ton thit k h dn ng c kh (tp 1 + 2) Trnh cht, L Vn Uyn2. Chi tit my (tp 1 + 2) Nguyn Trng Hip3. Dung sai v lp ghp Ninh c Tn
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