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Jun 26, 2015

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Automotive

jonhthien1

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  • 1. MC LC BN THUYT MINH GM NHNG PHN CHNH SAU PHN 1: Tnh chn ng c v phn phi t s truyn..... 1. Chn ng c... 2. Phn phi t s truyn v mmen xon trn cc trc....... 3. Tnh cc thng s trn cc trc......... PHN 2: Tnh ton b truyn ngoi...... 1. Chn loi xch 2. Tnh cc thng s ca b truyn xch 3. Kim nghim xch v bn. 4. Kch thc a xch... 5. Bng cc thng s ca b truyn xch. PHN 3: Tnh b truyn bnh rng..... 1.Tnh ton b truyn cp nhanh bnh rng tr rng thng... 2. Chn vt liu.. 3. Xc nh ng sut cho php... 6. Tnh cc thng s ca b truyn bnh rng tr rng thng. 7. Kim nghim rng.. 8. Bng cc thng s ca b truyn bnh rng tr rng thng... 9.Tnh ton b truyn cp chm bnh rng tr rng nghing... 10. Chn vt liu.. 11. Xc nh ng sut cho php... 12. Tnh cc thng s ca b truyn bnh rng tr rng nghing... 13. Kim nghim rng.. 14. Bng cc thng s ca b truyn bnh rng tr rng nghing... PHN 4: Tnh ton thit k trc.. 1. ....... .. 1 2 c 3.... .. . PHN 5: Tnh chn . 1. 1.. 2. 2..SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 1

2. 3. 3.. PHN 6: Thit k v hp gim tc ,bi trn v n khp 1. Thit k v hp 2. Bi trn trong hp gim tc.. 3. Lp bnh rng ln trc v iu chnh s n khp. 4. Thit k cc kt cu khc.. 5. Bng thng k cc kiu lp v dung saiSV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 2 3. PHN I TNH CHN NG C, PHN PHI T S TRUYN V M MEN XONTRN CC TRC.1324 F1.1.Chn ng c. Cng sut cn thit: - Cng sut lm vic trn trc my cng tc: Plv = F.V/1000 Vi F:L lc ko bng ti. V:L vn tc bng ti. Thay s ta c: =>Plv= 13000.0,32/1000 = 4,16 ( KW) Do ti trng thay i nhiu mc nn ta chn ng c theo cng sut tng ng. Ptd = Plv. (>1) Vi = Thay s ta c:= 0,816=SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 3 4. Vy cng sut tnh ton trn trc my cng tc l:Pt = Ptd = Plv.thay s ta c: Ptd = 4,16.0,816 = 3,4 (KW)-Hiu sut ca ton b h dn ng l: Ta gi ht l hiu sut ca ton b h thng v c xc nh theo cng thc: ht =k. ot.3. ol .2 x. brtTra bng 2.3 SGKTTTKHDCK tp 1 ta c: +k lhiu sut ca khp ni vi: k = 0,99.+ ot l hiu sut ca 1 cp trt: ot = 0,96. + ol l hiu sut ca 1 cp ln: ol = 0,99. + xl hiu sut ca b truyn xch: x = 0,98. + brtl hiu sut ca b truyn bnh rng tr: brt= 0,97. Thay s vo ta c: ht = 0,99.0,96.0,993.0,98.0,972 = 0,886 -Vy cng sut cn thit trn trc ng c:Pct= Pt/htthay s ta c: Pct = 3,4/0,886 = 3,84 (kW).-S vng quay ng b ca ng c: + S vng quay trn trc my cng tc: nlv= 60000.V/(Z.t) Vi V: l vn tc ca bng ti (m/s) Z: S rng a xch ti t: Bc xch ca xch ti (mm) SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 4 5. Thay s vo ta c:nlv== 24,4 (v/p)-Ta i chn s b t s truyn chung cho ton h dn ng: Ut= Ux.Ubrt Ta chn s b cc t s truyn nh sau. +T s truyn ca b truyn xch:Ux= 3 +T s truyn ca hp vi hp gim tc bnh rng:Ubrt = 20 Vy t s truyn ca ton h dn ng l: Ut = 3.20 = 60 - S vng quay trn trc ca ng c: nsb = nlv.Ut Thay s vo ta c: nsb = 60.24,4 = 1464 (v/p) Chn s vng quay ng b ca ng c : nb = 1500 ( v/p) chn ng c ta da vo bng P1.3 ph lc SGKTTTKHDCKtp1 Ta s dng loi ng c 4A112M4Y3 c cc thng s k thut nh sau:Bng thng s k thut ca ng cKiu ng cCng sut (kw)Vn tc quay(v/p)4A112M4Y35,51425cos0,85%85,52,22,0 m bo cho ng c lm vic c n nh ta cn i kim nghim li cc iu kin ca ng c khi lm vic +ndc= 1425(v/p) nsb= 1464(v/p) SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 5 6. +Pdc Pct = 3,84KW ng thi mmen m my phi tho mn iu kin: Tmm/T TK/Tdn Nh vy ng c chn ph hp vi yu cu t ra. 1.2:Phn phi t s truyn v mmen xon trn cc trc - Ta i tnh li t s truyn chung cho ton h dn ng: Vi: Ut = ndc/nlvthay s ta c: Ut = 1425/24,4 = 58,4 Ta i phn phi li t s truyn nh sau: chn Ubrt= 20 Trong : +T s truyn cp nhanh l: Ucn = 5,69 +T s truyn cp chm l: Ucc = 3,51 Ta c Ux= Ut/Ubrt = 58,4/20 = 2,92 1.3:Tnh cc thng s trn cc trc: -Tnh ton ton tc quay trn cc trc : +Trc ng c : ndc = 1425(v/p) +Trc s 1: = ndc/Uk thay s ta c: = 1425/1 = 1425(v/p) +Trc s 2: = n/Ucnthay s vo ta c: = 1425/5,69= 250(v/p) +Trc s 3: = /Ucc thay s vo ta c: = 250/3,51= 71(v/p) +Trc s 4: = /Ux thay s vo ta c: = 71/2,92= 24,3(v/p) - Tnh cng sut trn cc trc: =4,16 kw == 4,2 kw= 4,4 kw== 4,6 kw= 4,65 kw - Tnh mmen xon trn cc trc: +Tdc = 9,55.106.Pct/ndc= 9,55.106.4,65/1425= 31163(N.mm) + = 9,55.106. / = 9,55.106.4,6/1425= 308288(N.mm) + = 9,55.106. / = 9,55.106.4,4/250= 168080(N.mm) + = 9,55.106.P/n = 9,55.106.4,2/71= 564930(N.mm)SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 6 7. += 9,55.106.PIV/nV= 9,55.106.4,16/24,3=1634897(N.mm)Thng s Trc Trc ng c Trc s 1 Trc s 2 Trc s 3 Trc s 41,945Mmen xon (N.mm)4,653116314254,63082825020Cng sut(kw)1425T s truynTc quay(v/p)4,4168080714,256493024,34,161634897PHN : TNH TON THIT K B TRUYN NGOI 2.1: Cc s liu ban u + Cng sut: PI = 4,2kw + S vng quay ca trc dn: nI= 71 v/p +T s truyn: Ux= 2,92 + Gc nghing ni tm b truyn ngoi: 90o 2.2:Thit k b truyn xch 2.2.1:Chn loi xch V ti trng nh, vn tc thp nn ta chn loi xch ng con ln. 2.2.2: Xc nh cc thng s ca xch v b truyn - Chn s rng ca a xch dn theo cng thc: z1 = 29 2 u = 29 2 2,92 = 23,16 rng Chn z1 = 23 rng -Tnh s rng a xch ln theo cng thc: z2 = u z1 = 2,92 23=67,16rng Ly z2 =67 rng Ta c t s truyn thc t l Ux =SV thc hin: Phm Hu Luyn== 2,91Lp: HCNKT CK1_ K5 7 8. -Xc nh cc h s iu kin s dng xch K theo cng thc : K = Kd Ka Ko Kdc Kb Klv = 1 1 1,25 1 1,3 1,45 = 2,36 Trong : Kd = 1 (b truyn lm vic m) Ka = 1 (a = (30 50)p) Ko = 1,25 (ng ni hai tm a xch hp vi ng nm ngang mt gc ln hn 60o) Kdc = 1 (trc iu ch nh c) Kb =1,3 (bi trn t yu cu trong mi trng c bi) Klv = 1,45 (lm vic 3 ca) H s Kn = n01 / nIII = 200 /71 = 2,8 H s Kz = z01 / z1 = 25 / 25 = 1 Chn xch mt dy, Kx = 1. Cng sut tnh ton : Pt === 27,75 kwTheo bng 5.5SGKTTTKHDCK tp 1 theo ct n01 = 200 (vg/ph) ta chn bc xch p = 38,1mm. Theo bng 5.8SGKTTTKHDCK tp 1 s vng quay ti hn tng ng bc xch 38,1mm l nth = 500 vg/ph, nn iu kin n < nth c tha mn. -Vn tc trung bnh ca xch: V==== 1,13 m/sSV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 8 9. -Lc vng c ch: Ft === 3717 N-Kim nghim bc xch p Theo bng 5.8 SGKTTTKHDCK tp 1 ta c p < pmax- Chn khong cch trc s b a = (30 50) p = 40 38,1= 1524 mm. - S mt xch X X=== 126Chn X = 126mt xch. -Chiu di xch L = pX = 126.38,1= 4801mm.- Tnh chnh xc khong cch trc a = 0,25.p.= 0,25.38,1.= 1521,7 mm Ta chn a = 1517mm (gim khong cch trc (0,002 0,004).a) -S ln va p xch trong 1 giy:SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 9 10. =i== 0,86 *i+ = 20Theo bng 5.9SGKTTTKHDCK tp 1 vi bc xch p = 38,1 mm, ta chn [i] = 20. 2.2.3: Tnh kim nghim xch v bn. - Kim tra xch theo h s an ton Qs F1FvFo+ Ti trng ph hy Q tra theo bng 5.2SGKTTTKHDCK tp 1 vi bc xch p = 38,1 mm th Q = 127kN khi lng mt mt xch q = 5,5 kg/m + Lc trn nhnh cngF1 Ft = 3717N+ Lc cng do lc ly tm gy nn Fv = q v2 = 5,5.1,132 = 7 N + Lc cng ban u ca xch Fo Fo = Kf a q g = 4 1,522 5,5 9,81 = 328,5N s = 31,34> [s] = (7,3 7,6) Vy b truyn xch m bo bn - Kim nghim bn tip xc ca a xch theo cng thc: H = 0,47.= 0,47.SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 10 11. = 458 MpaTrong : +*h] l ng sut tip xc cho php + kr = 0,48 l h s nh hng n s rng a xch + Kd = 1 l h s ti trng ng + kd = 1 l h s phn b khng u ti trng cho cc dy + Fvd l lc va p trn m dy xch Fvd = 13.10-7n1.p3.m =13.10-7.71.38,13.1 = 5,1 N + E = 2,1.105 l mun n hi +Ft = 3717 N + A: l din tch chiu ca bn l tra theo bng 5.12SGKTTTKHDCK tp 1 ta cA = 395 mm2 Vy dng thp 45 ti rn HB210 s t c ng sut tip xc cho php *H+ = 600Mpa, m bo c bn cho rng a 1. Tng t vi rng a 2 cng tng t: H2 *H] (vi cng vt liu v nhit luyn) 2.2.4: Bn knh yvitra theo bng 5.2SGKTTTKHDCK tp 1ta c:=>SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 11 12. 2.2.5: Kch thc a xch d1 === 279 mmd2 === 813 mm-ng knh nh rng: da1 = d1 + 0,7.p = 305,67 mm da2 = d2 + 0,7.p = 839,67 mm -ng knh chn rng:2.2.6: Lc tc dng ln trc Fr = Kx Ft = 1,05 3717= 3903N Trong Kx = 1,05 do b truyn nghing mt gc ln hn 40o2.2.7:Cc thng s ca b truyn xch Thng s Loi xch Bc xch S mt xch Chiu di xch Khong cch trc S rng a xch nh S rng a xch ln Vt liu a xch ng knh vng chia a xch nh 10.ng knh vng chia a xch ln 1. 2. 3. 4. 5. 6. 7. 8. 9.SV thc hin: Phm Hu LuynK hiu ------p x L a----Gi tr Xch ng con ln 38,1(mm) 126 4801 (mm) 1517 (mm) 23 67 Thp 45(Ti,ram) 279(mm) 813(mm)Lp: HCNKT CK1_ K5 12 13. 11.ng knh vng nh a xch nh 12.ng knh vng nh a xch ln 13.Bn knh y 14.ng knh chn rng a xch nh 15.ng knh chn rng a xch ln 16.Lc tc dng ln trc 17.Xch mt dyPHN305,67(mm) 839,67(mm) r11,22(mm) 256,65(mm) 790,56(mm) 3903(N)Kx1,05. TNH TON THIT K B TRUYN TRONG3.1: Tnh ton b truyn cp nhanh bnh rng tr rng thng3.1.1: Cc s liu ban u = 4,6(kw) -= 5,69-ng n1 = 1425(vg/ph)-c lh = 160)Tmm = 1,4T1 T2 = 0,68T1 t1 = 3,2)SV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 13 14. t2 = 4,6 tck) )3.1.2:Thit k b truy 3.1.2.1: Chn vt liu ch to. i vi hp gim tc bnh rng tr2 cp chu cng sut khng ln lm (Pm = 7,5kw) ta nn s dng vt liu loi nhm I l loi vt liu c rn HB 350, bnh rng c thng ho hoc ti ci thin, nh c rn thp Nn c th ct rng mt cch chnh xc sau khi nhit luyn ng thi b truyn c kh nng chy mn ,hn na tng kh nng chy mn ca rng, Ta nn nht luyn bnh rng ln c rn thp hn bnh rng nh t 10 15 n v tc H1 H2 + (10.15)HB. Ta tra bng 6.1 SGKTTTKHDCK tp 1 ta chn -Vt li ch to bnh rng nh l: +Thp C45 ti ci thin; + rn:HB = (241.285); +Gii hn bn: = 850MPa; +Gii hn chy: = 580MPa; Chn rn ca bnh nh l: HB1 = 250. -Vt liu ch to bnh rng ln l: +Thp C45 ti ci thin; + rn HB = (192.240); +Gii hn bn = 750MPa;+Gii hn chy= 450MPa;Chn rn ca bnh ln:HB2 = 240. 3.1.2.2: Xc nh ng sut cho php - ng sut tip xc cho php [ vng sut un cho php xc nh theo cng thc sau:+[]=.ZR.ZV.KXH.KHL+[]=oc.YR.YS.KXF.KFC.KFLSV thc hin: Phm Hu LuynLp: HCNKT CK1_ K5 14 15. Trong : ZR: h s xt n nh hng nhm ca mt rng lm vic; ZV: h s xt n nh hng ca vn tc vng; KXH: h s xt n nh hng ca kch thc rng; YR: h s xt n nh hng ca nhm mt ln chn rng; YS: h s xt n nhy ca vt liu i vi tp trung ng sut; KXF: h s xt n kch thc rng nh hng n bn un ; Trong thit k s b ta ly: ZR.ZV.KXH = 1 v YR.YS.KXF = 1 Do cc cng thc trn ln lt tr thnh: ]==KHL(1-a) KFL.KF(2-a)Trong : ; . ln lt l cc ng sut tip xc cho php v ng sut un cho php vi s chu k c s,tra bng 6.12SGKTTTKHCK tp 1 vi thp C45 ti ci thin t rn HB = (180.350) Ta c: oHlim = 2HB + 70 ; SH = 1,1; oFlim = 1,8HB ; SF = 1,75; vi SH,SF l h s an ton khi tnh v bn tip xc v bn un; Thay cc kt qua trn vo cng thc ta c: oHlim1 =