CHNG 5NHNG VN C BN TRONG TNH TON V THIT K CHI TIT MY5.1. Cc vn chung 5.1.1. Chi tit my Khi chng ta tho ri mt my, mt b phn my s nhn c nhng phn t nh ca my. V d nh: bu lng, ai c, bnh rng, trc. Nu tip tc tch ri cc phn t ny th n khng cn cng dng na. Cc phn t nh ca my c gi l chi tit my. Chi tit my c th phn thnh 2 nhm: - Nhm chi tit my c cng dng chung: bao gm cc chi tit my c s dng trong nhiu loi my khc nhau. Trong cc loi my khc nhau, chi tit my c hnh dng v cng dng nh nhau. V d: bnh rng, khp ni, trc, bu lng, ln. - Nhm chi tit my c cng dng ring: bao gm cc chi tit my ch c s dng trong mt loi my nht nh. Trong cc loi my khc nhau, hnh dng hoc cng dng ca chi tit my l khc nhau. V d: trc khuu, tuabin, v hp gim tc, thn my. 5.1.2. Ti trng v ng sut a. Ti trng tc dng ln my v chi tit my - Ti trng tc dng ln my v chi tit my bao gm lc, mmen v p sut. Ti trng l i lng vct, c xc nh bi cc thng s: cng , phng, chiu, im t v c tnh ca ti trng. Trong : + Lc, c k hiu bng ch F, n v o l N, 1 N = 1 kg.m/s. + Mmen un, k hiu l M, n v o l Nmm. + Mmen xon, k hiu l T, n v o l Nmm. + p sut, k hiu l p, n v o l MPa, 1 MPa = 1 N/mm2. - Phn loi ti trng: chng ta lm quen vi mt s tn gi ca ti trng, v c im ca n: + Ti trng khng i: l ti trng c phng, chiu, cng khng thay i theo thi gian. S ca ti trng khng i biu din trn Hnh 1-3. +Ti trngthayi: lti trngct nht mt trongbai lng (phng, chiu, cng ) thay i theo thi gian. Trong thc t tnh ton chi tit my, thng gp loi ti trng c cng thay i; s ca ti trng thay i c biu din trn Hnh 1-4. + Ti trng tng ng: l ti trng khng i quy c, tng ng vi ch ti trng thay i tc dng ln chi tit my. Hay ni cch khc: khi tnh ton chi tit my chu ti trng thay i, chng ta phi s dng mt ch ti trng khng i tng ng vi ch ti thay i v mt sc bn v tui th ca chi tit my. + Ti trng c nh: l ti trng c im t khng thay i trong qu trnh chi tit my lm vic. + Ti trng di ng: l ti trng c im t di chuyn trn chi tit my, khi my lm vic. +Ti trngdanhngha: lti trngtcdnglnchi tit mytheol thuyt. + Ti trng tnh. Khi lm vic, chi tit my, hoc mt phn no ca chi tit my phi chu ti trng ln hn ti trng danh ngha. Ti trng tng thm c th do rung ng, hoc do ti trng tp trung vo mt phn ca chi tit my. Chi tit my phi c tnh ton thit k sao cho phn chu ti ln khng b thiu bn. Nh vy ta phi tnh chi tit my theo ti trng ln hn ti danh ngha, ti trng ny c gi l ti trng tnh. b. ng sut - ng sut l ng lc xut hin trong cc phn t ca chi tit my, khi chi tit my chu ti trng. - ng sut l i lng vct, n c xc nh bi phng, chiu, cng . n v o ca ng sut l MPa, 1 MPa = 1 N/mm2. - ng sut c phn ra lm hai nhm: +ng sutphp k hiu l . ng sutphp c phng trngvi phng php tuyn ca phn t c tch ra t chi tit my. + ng sut tip k hiu l . ng sut tip c phng trng mt phng ca phn t c tch ra t chi tit my. - Tng ng vi cc ti tc dng, ng sut c phn thnh cc loi: + ng sut ko, k hiu l k, + ng sut nn, k hiu l n, + ng sut un, k hiu l u, + ng sut tip xc, k hiu l tx, hoc H, + ng sut dp, k hiu l d, + ng sut xon, k hiu l x, + ng sut ct, k hiu l c. Ngoi ra, ng sut cn c phn thnh ng sut khng i v ng sut thay i: - ng sut khng i hay cn gi l ng sut tnh, l ng sut c phng, chiu, cng khng thay i theo thi gian. S ca ng sut tnh c th hin trn Hnh 1-5. - ng sut thay i l ng sut c t nht mt i lng (phng, chiu, cng ) thay i theo thi gian. ng sut c th thay i bt k, hoc thay i c chu k. Trong tnh ton thit k chi tit my, chng ta thng gp loi ng sut thay i c chu k tun hon, hoc gn nh l tun hon. S ca ng sut thay i tun hon bin din trn Hnh 1-6. Mt chu trnh ng sut c xc nh bi cc thng s: + ng sut ln nht max+ ng sut nh nht min+ ng sut trung bnh m; ( )2min maxm + + Bin ng sut a; ( )2min maxa + H s chu k ng sut r;minmaxr hoc maxminr, khi min = 0. Cn c vo gi tr ca h s chu k ng sut r, ngi ta chia ng sut thnh cc loi: + ng sut thay i mch ng, khi chu trnh ng sut c r 0. + ng sut thay i i xng, khi chu trnh ng sut c r < 0. + ng sut tnh l trng hp c bit ca ng sut thay i, c r = 1. Vi cng mt gi tr ng sut nh nhau, nhng r khc nhau th kh nng ph hy vt liu ca ng sut cng khc nhau. Chi tit my chu ng sut tnh c tui th cao hn chi tit my chu ng sut thay i mch ng, chi tit my chu ng sut thay i i xng c tui th thp nht. 5.1.3 bn mi ca chi tit my a. Hin tng ph hng do mi Khi chi tit my chu ng sut tnh b ph hng, gi l b ph hng do ng sut tnh. Hay ni cch khc, chi tit my khng sc bn tnh. Tnh ton chi tit my ngn chn dng hng ny c gi l tnh ton theo sc bn tnh. Khi chi tit my b ph hng bi ng sut thay i, gi l b ph hng do mi, hay chi tit my khng sc bn mi. Tnh ton chi tit my ngn chn dng hng ny, gi l tnh ton theo sc bn mi. Khi ng sut tnh vt qu gi tr ng sut gii hn, chi tit my b ph hng t ngt. Vt gy nhm v mi, quan st di knh hin vi thy r kt cu ht kim loi (Hnh 1-7). Qu trnh hng do mi xy ra t t, theo trnh t nh sau: - Sau mt s chu k ng sut nht nh, ti nhng ch c tp trung ng sut trn chi tit my s sut hin cc vt nt nh. - Vt nt ny pht trin ln dn ln, lm gim dn din tch tit din chu ti ca chi tit my, do lm tng gi tr ng sut. - Cho n khi chi tit my khng cn sc bn tnh th n b ph hng. Quan st vt gy thy r phn chi tit my b hng do mi - b mt c v nhn - v phn chi tit my b hng do khng sc bn tnh - b mt mi v nhm (Hnh 1-8). Chi tit my s b ph hng do mi, khi m ng sut sinh ra trong chi tit my (, ) ln hn ng sut cho php ([], []). Gi tr ng sut cho php c chn khng nhng ph thuc vo c tnh ca vt liu ch to chi tit my, m cn ph thuc vo s chu k cn lm vic ca chi tit my. S chu k cn lm vic cng t th gi tr ca ng sut cho php c th chn cng caoNgi ta lm cc th nghim xc nh mi quan h gia gi tr ng sut v s chu k lm vic cho n khi hng ca chi tit my, biu din trn Hnh 1-9. y chnh l ng cong mi ca chi tit my trong h ta cc ON. Trong : NO: l s chu k c s. r : gii hn mi ca vt liu. m : m ca ng cong mi. N : gii hn mi ngn hn: N=KNr . KN: h s tng gii hn mi ngn hn: m0NNNK b. Nhng nhn t nh hng n sc bn mi ca chi tit my * Vt liu Vt liu c nh hng ln n sc bn mi ca chi tit my. Chi tit my c ch to bng vt liu c c tnh cao, sc bn mi ca chi tit s cao. V vt liu c c tnh cao, th kh nng xut hin cc vt nt s kh khn hn. Ni chung: - Chi tit my ch to bng vt liu kim loi c bn mi cao hn bng vt liu phi kim loi. - Chi tit my c ch to bng kim loi en c bn mi cao hn so vi bng hp kim mu. - Chi tit my bng thp c bn mi cao hn bng gang. -Chi tit mybngthphpkimcbnmi caohnbngthp ccbon thng. Trong cc loi thp thng, chi tit my bng thp c hm lng ccbon cng cao, bn mi ca ca chi tit my cng cao. * Kt cu ca chi tit my Chi tit my c kt cu phc tp: c cc bc thay i kch thc t ngt, c cc l, cc rnh, nh trn Hnh 1-10, s lm gim bn mi ca chi tit my. L do: ti nhng ch ny c tp trung ng sut, vt nt sm xut hin v pht trin kh nhanh. Trong tnh ton, nh hng ca kt cu n sc bn mi ca chi tit my c k n bng h s iu chnh k, k , gi l h s tp trung ng sut. rtrk ; rtrkTrong rt , rt l gii hn mi ca mu c tp trung ng sut; cn r , r l gii hn mi ca mu khng c tp trung ng sut. Gi tr ca h s k v k c th tra cc bng s liu trong S tay thit k c kh hoc sch Bi tp chi tit my, theo hnh dng v kch thc c th ca nhng ch c tp trung ng sut, trn tng loi chi tit my khc nhau. * Kch thc ca chi tit my Qua th nghim ngi ta thy rng: vi vt liu nh nhau, khi tng kch thctuyticachititmythgiihnbnmicachititmygim xung. L do: kch thc ca chi tit my cng ln, vt liu cng khng ng u, khnngxut hincckhuyt tt tronglngchi tit mycngnhiu. Nhng vt nt, r x, r kh trong lng chi tit my l nhng im c tp trung ng sut, l nhng im bt u cho s ph hng v mi. k n nh hng ca kch thc tuyt i, trong tnh ton ngi ta a vo h s iu chnh , , gi l h s nh hng ca kch thc tuyt i. H s v c xc nh bng thc nghim, gi tr ca n c th tra trong cc s tay Thit k c kh hoc sch Bi tp Chi tit my, theo kch thc v trng thi chu ti ca chi tit my.

rrd , rrd Trong rd v rd l gii hn mi ca chi tit my, c kch thc khc vi kch thc ca mu chun. Mu chun c ng knh d = 7 .10 mm. * Cng ngh gia cng b mt chi tit my Cng ngh gia cng b mt chi tit my quyt nh trng thi b mt ca chi tit my. Lp b mt chi tit my thng l lp chu ng sut ln nht, cc vt nt u tin cng hay xy ra y. nh hng ca cng ngh gia cng lp b mt n sc bn mi ca chi tit my c th tm tt nh sau: - Nhng chi tit my qua nguyn cng gia cng tinh, c bng b mt cao s c bn mi cao. - Nhng chi tit my ch qua nguyn cng gia cng th, b mt nhm, y nhp nh l nhng ch tp trung ng sut, d xut hin cc vt nt, bn mi gim. - Cc b mt c gia cng tng bn nh phun bi, ln p s san bng cc nhp nh v lm chai cng b mt, bn mi ca chi tit my c nng cao. nh hng ca cng ngh gia cng lp b mt n bn mi ca chi tit my, c k n bng h s trng thi b mt . Gi tr ca c th tra trong cc S tay thit k c kh hoc sch Bi tp Chi tit my. C th ly gn ng nh sau: khi b mt chi tit c mi nhn ly =1, khi b mt c gia cng tng bn ly >1, b mt c gia cng bng cc phng php khc ly 0), trng thi ng sut va ko va nn (r < 1) c bn mi thp nht. 5.2. Nhng ch tiu ch yu v kh nng lm vic ca chi tit my5.2.1. Ch tiu bn a. Yu cu v bn bnlchtiuquantrngnhtcachititmy. Nuchititmy khng bn n s b hng do gy, v, t, cong, vnh, mn, dp, r b mt, vv.. chi tit my khng cn tip tc lm vic c na, n mt kh nng lm vic. Chi tit my c nh gi c bn, khi n tha mn cc iu kin bn. Cc iu kin bn c vit nh sau: [] [] S [S]. Trong : + v l ng sut sinh ra trong chi tit my khi chu ti. + [] v [] l ng sut cho php ca chi tit my. + S l h s an ton tnh ton ca chi tit my,b. Cch xc nh ng sut sinh ra trong chi tit my ng sut sinh ra trong chi tit my c xc nh theo l thuyt ca mn hc Sc bn vt liu v L thuyt n hi. Trn c s , mn hc Chi tit my tha k hoc xy dng cc cng thc tnh ton ng sut c th cho mi loi chi tit my. * i vi cc chi tit my chu ti trng khng i - Trng hp trong chi tit my c trng thi ng sut n (ch c , hoc ch c ), ng sut sinh ra trong chi tit my tnh theo cng thc ca Sc bn vt liu. V d, tnh ng sut ko sinh ra trong thanh chu chu lc F: AFK . - Trng hp chi tit my c ng sut phc tp (c c v ), lc ng sut sinh ra trong chi tit my c ly theo ng sut tng ng t , t tnh theo thuyt bn "Th nng bit i hnh dng" - Thuyt bn th t: 2 2td3 + hoctheothuyt"ngsut tiplnnht"-Thuyt bnthba: 2 2td4 + - Trng hp din tch tip xc gia hai b mt kh ln, ng sut sinh ra c tnh theo ng sut dp. - Nu din tch tip xc gia hai b mt rt nh (ban u tip xc theo ng, hoc theo im), ng sut sinh ra l ng sut tip xc cc i ti tm ca vng tip xc, c tnh theo cng thc Hc H . * i vi cc chi tit my chu ti trng thay i V d, xt mt chi tit my lm vic vi ch ti trng thay i: trong thi gian s dng tb, chi tit my lm vic vi n ch ti trng, mi ch ti trng Mi lm vic vi thi gian ti (Hnh 2-1). ng sutsinh ra trongchititmys c tnh theo chtitrng khng i tng ng. Ch ti trng tng ng thng c chn nh sau: Mt = M1 (M1 l ti trng ln nht trong ch ti trng thay i). Thi gian lm vic tng ng tbtca chi tit my c xc nh da trn nguyn l "Cng n gin tn tht mi". Tui bn tng ng ca chi tit my, trong a s cc trng hp, c tnh theo cng thc:

imn1 i 1ibtdtMMt

,_

Trong trng hp xc nh s chu k ng sut tip xc, th tbtc tnh theo cng thc:

in1 i 1ibtdtMMt2m

,_

Trong m l m ca ng cong mi. Gi tr ng sut c tnh theo ti trng Mt, hoc theo ti trng M1, s chu k ng sut s c tnh theo tbt. c. Cch xc nh ng sut cho php - Xc nh ng sut cho php bng cch tra bng. Trong S tay thit k c kh, v trong sch Bi tp chi tit my c cc bng s liu ghi ng sut cho php ca mt s loi chi tit my thng dng. Bng s liu ng sut cho php c thit lp bng cch th nghim, hoc bng nhng kinh nghim c kt trong qu trnh s dng chi tit my. Cch xc nh ny cho kt qu kh chnh xc. - Tnh ng sut cho php theo cng thc gn ng:

[ ]Slim , [ ]Slim Trong : lim v lim l ng sut gii hn. Ty theo tng trng hp c th ng sut gii hn c th l gii hn chy (ch , ch), gii hn bn (b , b), gii hn mi (r , r), gii hn mi ngn hn (rN , rN) ca vt liu ch to chi tit my. S l h s an ton, h s S c xc nh t cc h s an ton thnh phn: S = S1.S2.S3 Trong : + S1 l h s xt n mc chnh xc trong vic xc nh ti trng v ng sut, S1 c th chn trong khong 1,2 1,5. + S2 l h s xt n ng nht v c tnh ca vt liu. i vi cc chi tit my bng thp rn hoc cn ly S2= 1,5 , cc chi tit my bng gang c th ly S2 = 2 2,5.+ S3 l h s xt n nhng yu cu t bit v an ton, i vi cc chi tit my quan trng trong my, hoc c lin quan trc tip n an ton lao ng, ly S3 = 1,2 . 1,5. - ng sut cho php cng c th c tnh theo cng thc thc nghim. V d, khi tnh bnh ma st, ng sut tip xc cho php c ly theo rn b mt: [H] = (1,5 2,5) HB, hoc [H] = (13 18) HRC. 5.2.2. Ch tiu cng a. Yu cu v cng - Chi tit my c coi l khng cng, khi lng bin dng n hi ca n vt qu gi tr cho php. - Khi chi tit my khng cng, chnh xc lm vic ca n s gim, nhiu khi dn n hin tng kt khng chuyn ng c, hoc lm tng thm ti trng ph trong chi tit my hoc nh hng n cht lng lm vic ca cc chi tit my khc lp ghp vi n. -cngcnglchtiuquantrngcachititmy.Trongmts trng hp chi tit my bn nhng cha cng, lc phi tng kch thc ca chi tit my cho cng, chp nhn tha bn. b. Cch nh gi ch tiu cng ca chi tit my Chi tit my ch tiu cng, khi n tha mn cc iu kin cng sau: l [l], y [y], [], [ ], h [h]. Trong : + l l dn di hoc co ca chi tit my khi chu ti, + y l vng ca chi tit my b un, + l gc xoay ca tit din chi tit my b un, + l gc xon ca chi tit my b xon, + h l bin dng ca b mt tip xc. + [l], [y], [], [ ] v [h] l gi tr cho php ca cc bin dng. Gi tr ca l, y, , c tnh theo cng thc ca Sc bn vt liu. Gi tr h ca vt th tip xc ban u theo im hoc ng c xc nh theo l thuyt ca Hc-Beliaep, ca vt th c din tch tip xc ln c xc nh bng thc nghim. Gi tr ca [l], [y], [], [ ], [h] c chn theo iu kin lm vic c th ca chi tit my, c th tra trong cc S tay thit k c kh, hoc sch Bi tp Chi tit my. nh gi kh nngchngbindng ca chi tit my,ngi ta cn dng h s cng C, l t s gia bin dng v lc tc dng do chng gy nn. Chi tit my c h s cng cng cao th kh nng bin dng cng nh. H s C c xc nh theo cng thc ca Sc bn vt liu. tng cng cho chi tit my cn chn hnh dng tit din ca chi tit my hp l, c bit nn s dng tit din rng. Trng hp cn thit nn dng thm cc gn tng cng. i vi chi tit my cn cng cao, nn chn vt liu c c tnh thp, trnh d bn. 5.3.2. Ch tiu bn mn - Khi hai b mt tip xc c p p, c trt tng i vi nhau v c ma st, th bao gi cng c hin tng mn. p sut cng ln, vn tc trt tng i cng ln, h s ma st cng ln th tc mn cng nhanh. Gia p sut p v qung ng ma st s c lin h theo h thc sau: pmS = hng. S m m ph thuc vo h s ma st f ca cc b mt tip xc. Gi tr ca m ly nh sau: + Khi c ma st na t (f = 0,01 . 0,09) ly m = 3, + Ma st na kh (f = 0,1 . 0,3) ly m = 2, + Ma st kh hoc c ht mi gia hai b mt tip xc (f = 0,4 . 0,9) ly m = 1. - Mn lm mt i mt lng vt liu trn b mt chi tit, kch thc dng trc ca chi tit my gim xung, kch thc dng l tng ln, cc khe h tng ln, lm gim chnh xc, gim hiu sut ca my. Khi kch thc gim qu nhiu c th dn n chi tit my khng bn. Mn cng lm gim cht lng b mt chi tit my, gim kh nng lm vic ca my, ng thi y nhanh tc mn. - Chi tit my c coi l ch tiu bn mn, nu nh trong thi gian s dng lng mn cha vt qu gi tr cho php. - m bo bn mn, chi tit my c tnh theo cng thc thc nghim sau: p [p] hoc pv [pv]. Trong p l p sut trn b mt tip xc, v l vn tc trt tng i gia hai b mt. - nng cao bn mn ca chi tit my, cn thc hin bi trn b mt tip xc y , dng vt liu c h s ma st thp. Tng din tch b mt tip xc gim p sut. Chn hnh dng chi tit my v quy lut chuyn ng ca n hp l vn tc trt tng i l nh nht. Dng cc bin php nhit luyn b mt tng rn, lm tng p sut cho php ca b mt. - Ngoi ra trnh n mn in ha, nhng b mt khng lm vic ca chi tit my cn c bo v bng cch ph sn chng g hoc bng phng php m. 5.2.4. Ch tiu chu nhit a. Yu cu v ch tiu chu nhit Trong qu trnh my lm vic, cng sut tn hao do ma st bin thnh nhit nng t nng cc chi tit my. Nhit lm vic cao qu gi tr cho php, c th gy nn cc tc hi sau y: - Lm gim c tnh ca vt liu, dn n lm gim kh nng chu ti ca chi tit my. - Lm gim nht ca du, m bi trn, tng kh nng mi mn. - Chi tit my b bin dng nhit ln lm thay i khe h trong cc lin kt ng, c th dn n kt tc, hoc gy nn cong vnh. b. Cch nh gi ch tiu chu nhit ca my My hoc b phn my c coi l ch tiu chu nhit, khi n tha mn iu kin chu nhit: []Trong : + l nhit lm vic ca my, b phn my. + [] l nhit cho php ca my. Nhit lm vic c xc nh t phng trnh cn bng nhit: = 1 + 2 Trong : + l nhit lng sinh ra trong mt n v thi gian, khi my lm vic, = 860.(1 - ).P (kCal/h) + : hiu sut lm vic ca my, + P : cng sut lm vic ca my, kW. +1lnhit lngtarami trngtrongmt nv thi gian, kCal/h. + 1 = kt.At.( - 0) (kCal/h) + kt: h s ta nhit ra mi trng, c th ly kt = (7,5 . 15) kCal/m2h0C + At: din tch ta nhit ca my, tnh bng m2, + 0: nhit mi trng lm vic ca my, 0C. +2lnhitlngdothitblmmttirangoitrongmtgi, kCal/h. Thay vo phng trnh cn bng nhit, ta c cng thc tnh nhit lm vic nh sau: ( )0t t2A kP 1 860 + Nhit cho php [] tra trong cc S tay Thit k c kh, ty theo loi du bi trn, vt liu ca chi tit my v chc nng lm vic ca chi tit my. Khi chi tit my khng ch tiu chu nhit, c ngha l > [], lc cn tm bin php x l. C th chn li cht bi trn tng nhit cho php []. Hoc l gim nhit lm vic bng cch: - Tng din tch b mt ta nhit At, bng cch dng cc gn, cnh tn nhit. - Tng h s ta nhit kt, bng cch dng qut gi, hoc phun nc. - Dng cc thit b lm mt. 5.2.5. Ch tiu chu dao ng Trong kt cu ca my, mi chi tit my l mt h dao ng c tn s dao ng ring 0. Nu chi tit my dao ng qu mc cho php, s gy nn rung lc gim chnh xc lm vic ca chi tit my v cc chi tit my khc. ng thi gy nn ti trng ph, lm cho chi tit bin dng ln, c th dn n ph hng chi tit my hoc gy ting n ln, ting n kh chu. Khi khi ng my, cc chi tit my bt u dao ng t do. Trong qu trnh lm vic, nu nh khng c ngun dao ng tc ng vo chi tit my, th dao ng t do ca chi tit my s tt dn sau mt vi pht. Nu chi tit my chu tc dng ca mt ngun gy dao ng, th n s dao ng cng bc. Ngun gy dao ng thng thng l cc chi tit my quay c khi lng lch tm, cc chi tit my chuyn ng qua li c chu k hoc do cc my xung quanh truyn n. Bin dao ng ca ngun cng ln th chi tit my dao ng cng nhiu, c bit l khi tn s ca ngun bng hoc gn bng vi tn s ring 0, lc chi tit my dao ng rt mnh (hin tng cng hng). Chi tit my ch tiu chu dao ng, khi bin dao ng ca n nh hn bin cho php. Trong thc t, vic xc nh chnh xc bin dao ng ca mt chi tit my l rt kh khn. Do , vic tnh ton ch tiu chu dao ng c thay th bng vic tm cc bin php hn ch dao ng ca chi tit my. Cc bin php hn ch dao ng ca chi tit my, c th k n l: - Trit tiu cc ngun gy dao ng: bng cch cn bng my, hn ch s dng cc quy lut chuyn ng qua li trong my, cch bit my vi cc ngun rung ng xung quanh. - Cho chi tit my lm vic vi s vng quay khc xa vi s vng quay ti hn (ng vi tn s ring 0) trnh cng hng. - Thay i tnh cht ng lc hc ca h thng, lm thay i tn s ring 0. - Dng cc thit b gim rung. 5.3. tin cy ca my v chi tit my5.3.1. Nhng vn chung - tin cy l mc duy tr cc ch tiu kh nng lm vic ca my, chi tit my trong sut thi gian s dng theo quy nh. Ni cch khc, trong sut thi gian s dng, my v chi tit my t xy ra hng hc, tn t thi gian hiu chnh sa cha, th tin cy ca my, chi tit my c nh gi l cao. - tin cy l mt trong cc c trng quan trng nht ca my v chi tit my, n l thng s th hin cht lng ca my tt hay xu, c ngi s dng a chung hay khng. Trong nn sn xut c kh ha v t ng ha tin cy cng c ngha c bit quan trng. Mt chi tit, thit b no o c tin cy thp, hay xy ra hng hc, s lm nh tr c dy truyn sn xut, c phn xng sn xut, thm ch ca c x nghip. - tin cy ca my, chi tit my c nh gi qua cc ch tiu sau y: + Xc sut lm vic khng hng hc, k hiu l R. Gi tr R ca my v chi tit my cng ln, th my v chi tit my c tin cy cng cao. + Xc sut lm vic hng hc, k hiu l F. Gi tr F cng ln, th my v chi tit my c tin cy cng thp. + Cng hng hc, k hiu l (t). L xc sut lm vic hng c tnh ti mt thi im trong thi gian lm vic ca my. Ti nhng thi im c (t) thp, my v chi tit my lm vic c tin cy cao. + Thi gian lm vic trung bnh cho n ln hng u tin, k hiu l tH. Gi tr tH cng cao, th my v chi tit my c tin cy cng cao. + H s s dng ca my, k hiu l KSd. Gi tr Ksd cng cao, th my v chi tit my c tin cy cng cao. Ch tiu tH v Ksd thng dng nh gi tin cy cho cc my hay xy ra hng hc, nhng sau khi iu chnh hoc sa cha nh th li c th lm vic bnh thng. 5.3.2. Cch xc nh cc ch tiu nh gi tin cy a. Tnh xc sut lm vic khng hng R v hng F ca mt i tng Xc sut lm vic khng hng R v xc sut lm vic hng F c tnh theo l thuyt xc sut thng k. Mt cch gn ng c th xc nh R v F ca tng my, chi tit my nh sau: - Ly ngu nhin N chi tit my, hoc my (ta gi chung l i tng th nghim), cho lm vic theo mt ch quy nh, trong mt thi gian nh. S lng i tng em th nn ly ln hn hoc bng 60, N 60. - Sau thi gian th nghim ta m c s i tng hng l NH, s lng i tng cn lm vic tt l NT, lc NNRT, NNFHTa nhn thy rng: R 1, F 1, v R + F = (NT + NH) / N = 1. b. Tnh xc sut Rnt v Fnt ca mt h gm n i tng mc ni tip Xt h thnggmn itng clp ghp ni tipnhauthnh mt chui. V d: mt dy chuyn sn xut gm n my. Mi i tng c xc nh xc sut R v F, i tng th i c xc sut Ri v Fi (Hnh 2-1). Ta phi tnh Rnt v Fnt ca ton h.

Ta nhn thy rng: khi mt i tng b hng th c h thng b dng; nh vy khi s lng n i tng mc ni tip cng nhiu, th xc sut lm vic khng hng ca h ni tip Rnt cng gim, v xc sut lm vic hng Fnt s tng ln, tin cy ca h gim. ng thi, nu ta tng gi tr Ri ca mi phn t, th xc sut Rnt s tng, tin cy ca h tng. Vi nhn xt nh trn, ta c th tnh Rnt v Fnt theo cng thc sau: nt ntin1 in 2 1 ntR 1 FR R ... R . R R c. Tnh xc sut RS v FS ca mt h gm m i tng mc song song Xt mt dy chuyn sn xut, trong c mt khu yu hay xy ra hng hc. Khu ny c tng cng bng cch lp m i tng c cng chc nng song song nhau. Mi i tng trong h mc song song c xc sut Riv Fi (Hnh 2-2).Chng ta cn tnh xc sut khng hng RS v xc sut hng FS ca ton b m i tng thuc khu yu ny, y chnh l RS v FS ca h gm m i tng mc song song. Kho st dy chuyn sn xut trn, chng ta nhn thy rng: khi mt hoc mt s i tng b hng th khu yu vn cha b hng, h thng cn hat ng c; ch khi c m i tng b hng th khu yu mi b hng, h thng mi b dng. Nh vy, khi s lng m i tng mc song song cng nhiu, th xc sut lm vic hng ca khu yu FS cng nh, tin cy ca h ln, v ngc li. ng thi, nu tng tin cy ca mi phn t mc song song, c ngha l gim Fi, th tin cy ca ton h cng tng, FS gim. C ngha l RS s tng. T nhn xt trn, ta c th lp cng thc tnh xc sut RS v FS ca h mc song song nh sau:

S Sim1 im 2 1 sF 1 RF F ... F . F F d. Xc nh cng hng (t) Cng hng (t) l xc sut lm vic hng ca i tng c tnh ti thi im t no . Cng (t) cng c tnh theo l thuyt Xc sut. Mt cch gn ng ngi ta xc nh (t) nh sau: - Ly ngu nhin N i tng em th. Th nghim cho n khi tt c N i tng b hng, thi gian th nghim l tth. Nn ly N 60. - Chia thi gian th tth lm n phn, k hiu cc phn l t1 n tn. m s i tng b hng trong khong thi gian t1, k hiu l N1. S i tng b hng trong khong thi gian ti l Ni. - Cng hng (ti) ti mt thi im thuc khong thi gian tic tnh theo cng thc: ( )iiit . NNt Th nghim vi nhiu loi my khc nhau, ngi ta nhn thy quy lut phn b ca (t) theo thi gian ca cc loi my c dng gn ging nhau. Dng ph bin nht c biu din trn Hnh 3-3. Quan st th phn b (t) theo thi gian, ta c nhn xt nh sau: - Thi gian u s dng my, cng hng (t) tng i cao. Khong thi gian t1khng di lm. y l thi gian hng ban u ca my. Cc c s sn xut thng ly thi gian t1 lm thi gian bo hnh my. - Trong khong thi gian t2, cng hng (t) tng i thp v t thay i, thi gian t2 kh di, y l thi gian lm vic n nh ca my. - Trong khong thi gian t3, cng hng (t) rt cao. y l thi gian hng ph hy ca my. Thi gian t3 khng c tnh vo thi gian s dng my. C ngha l, tui bn ca my ch bao gm thi gian t1 v t2. Nhng nhn xt trn c gii thch nh sau: - Cc my b hng trong khong thi gian t1thng l nhng my ph phm, cn ln vo sn phm xut xng, do kim tra khng pht hin ra c, hoc v mt l do no ngi thit k c tnh a vo. V d: tng dung sai kch thc, h gi thnh sn phm, trong nhiu trng hp ngi thit k chn gii php "lp ln khng hon ton". C ngha l ngi thit k bit trc ckhong5%snphmkhngchtlng,nhngvnccoilchnh phm. Cc c s sn xut nn sn sng thu nhn nhng sn phm b hng trong thi giant1v, v nilicmn lchsvikhchhng "Qu ngi gip chng ti tm ra sn phm km cht lng b ln trong s hng ha bn ra". - Sau khi s sn phm km cht lng b hng ht, theo l thuyt trong khong thi gian t2 s khng cn sn phm no b hng, (t) = 0. Song thc t, vn c my b hng do nhng nguyn nhn ngu nhin m khi thit k cha lng ht c. Nhng sn phm b hng trong thi gian ny, s khng c bi thng. - Sau mt thi gian di s dng, cc chi tit b mn, mi, lo ha, nn trong khong thi gian t3 cng hng rt cao. Mt my c nh gi l c tnh kinh t cao, khi m khong thi gian t3 ca n ngn. e. Xc nh thi gian lm vic cho n ln hng u tin tH Chn ngu nhin N i tng em th, nn ly N 60. Cho N i tng lm vic trong iu kin quy nh. Ghi chp thi gian lm vic cho n ln hng u tin ca tng i tng. K hiu thi gian lm vic cho n ln hng u tin ca i tng th i l tHi. Lc thi gian lm vic cho n ln hng u tin ca lat sn phm kho st, tH c tnh nh sau:

( )+ + +N1 iHiHN 2 H 1 HHtN1Nt ... t ttf. Xc nh h s s dng Ksd H s s dng KSd ca mt lat sn phm no cng c xc nh theo phng php thng k. Ngi ta theo di mt s lng sn phm ln (thng chn N 60), trong mt khong thi gian quy nh. Ghi chp v tnh thi gian lm vic trung bnh, thi gian chm sc, thi gian sa cha trung bnh ca lot my th nghim. Sau tnh h s Ksd theo cng thc:

ph ch lvlvsdt t ttK+ +Trong : + tlv l thi gian lm vic thc t ca my, + tCh l thi gian sa cha my, + tPh l thi gian phc v chm sc my. + (tlv + tch + tph) l khong thi gian th nghim. 5.3.3. Cc bin php nng cao tin cy ca my nng cao tin cy ca my, c th thc hin cc bin php sau y: - Nng cao tin cy ca tng chi tit trong my, bng cch: + Xc nh chnh xc ti trng v ng sut trong chi tit my. + Dng phng php tnh c chnh xc cao, cng thc tnh ton thch hp xc nh kch thc ca chi tit my. + Chn phng php gia cng tin cy, m bo ng cc ch tiu k thut theo thit k. + Chn phng php kim tra thch hp, thit b kim tra c chnh xc cao, m bo loi c ht cc ph phm ra khi lot sn phm xut xng. + Tun th tuyt i cc quy nh v s dng chi tit my v my. + Chm sc, bo dng thng xuyn chi tit my v my. - Tm kt cu hp l gim bt s khu lp ni tip trong my. - Tng tin cy nhng khu yu, hay xy ra hng hc, bng cch lp song song mt s chi tit c cng chc nng. Chng 6: B truyn ai6.1. Khi nim chung6.1.1. Cu to v nguyn l lm vica. Cu to- B truyn ai thng dng truyn chuyn ng gia hai trc song song v quay cng chiu (Hnh 11-1),trong mt s trng hp c th truyn chuyn ng gia cc trc song song quay ngc chiu - truyn ng ai cho, hoc truyn gia hai trc cho nhau - truyn ng ai na cho (Hnh 11-2). - B truyn ai thng thng gm 4 b phn chnh: + Bnh ai dn s 1, c ng knh d1, c lp trn trc dn I, quay vi s vng quay n1, cng sut truyn ng P1, m men xon trn trc T1. + Bnh ai b dn s 2, c ng knh d2, c lp trn trc b dn II, quay vi s vng quay n2, cng sut truyn ng P2, m men xon trn trc T2. + Dy ai 3, mc vng qua hai bnh ai. + B phn cng ai, to lccngbanu 2F0ko cnghainhnh ai. to lccng F0, c th dngtrnglng ng c (hnh 11-3, a), dng vt y (hnh 11-3, b) hoc dng bnh cng ai.b. Nguynl lm vic- Nguyn l lm vic ca b truyn ai: dy ai mc cng trn hai bnh ai, trn b mt tip xc ca dy ai v bnh ai c p sut, c lc ma st Fms. Lc ma st cn tr chuyn ng trt tng i gia dy ai v bnh ai. Do khi bnh dn quay s ko dy ai chuyn ng v dy ai li ko bnh b dn quay. Nh vy chuyn ng c truyn t bnh dn sang bnh b dn nh lc ma st gia dy ai v cc bnh ai. 6.1.2. Phn loi b truyn aiTy theo hnh dng ca dy ai, b truyn ai c chia thnh cc loi: - ai dt, hay cn gi l ai phng. Tit din ai l hnh ch nht hp, bnh ai hnh tr trn, ng sinh thng hoc hnh tang trng, b mt lm vic l mt rng ca ai (Hnh 11- 4, a). + Kch thc b v h ca tit din ai c tiu chun ha. Gi tr chiu dy h thng dng l 3 ; 4,5 ; 6 ; 7,5 mm. Gi tr chiu rng b thng dng 20 ; 25 ; 32 40 ; 50 ; 63 ; 71 ; 80 ; 90 ; 100 ; ... . mm. + Vt liu ch to ai dt l: da, si bng, si len, si tng hp, vi cao su. Trong ai vi cao su c dng rng ri nht. ai vi cao su gm nhiu lp vi bng vcaosusunfua ha.Cclpvi chutitrng,caosu dng lin kt bo v cc lp vi vtnghsma st vi bnh ai. ai vi cao su c chtothnhcun,ngithitkct chiu dicnthitvni thnh vng kn. ai c ni bng cch may hoc dng bulng kp cht. ai si tng hp c ch to thnh vng kn, do chiu di ca ai cng c tiu chun ha. - ai thang, tit din ai hnh thang, bnh ai c rnh hnh thang, thng dngnhiudy ai trongmt b truyn(Hnh 11-4,b).Vtliuch to ai thang l vi cao su. Gm lp si xp hoc lp si bn chu ko, lp vi bc quanh pha ngoi ai, lp cao su chu nn v tng ma st. ai thang lm vic theo hai mt bn. +Hnhdngvdintchtit dinai thangctiuchunha. TCVN 2332-78 quy nh 6 loi ai thang thng Z, O, A, B, C, D. TCVN 3210-79 quy nh 3 loi ai thang hp SPZ, SPA, SPB. + ai thang c ch to thnh vng kn, chiu di ai cng c tiu chun ha. B truyn ai thang thng dng c chiu di: 400, 450, 500, 560, 630, 710, 800, 900, 1000, 1120, 1250, 1400, 1600, 1800, 2000, 2240, 2500, 2800, 3150, 3550, 4000, 4500, 5000,... mm. - ai trn, tit din ai hnh trn, bnh ai c rnh hnh trn tng ng cha dy ai (Hnh 11-4, c). ai trn thng dng truyn cng sut nh. - ai hnh lc, l trng hp c bit ca b truyn ai thang. Cc ai c lm lin nhau nh rng lc (Hnh 11-5, a).Mi rng lm vic nh mt ai thang. S rng thng dng 220, ti a l 50 rng. Tit din rng c tiu chun ha. ai hnh lc cng ch to thnh vng kn, tr s tiu chun ca chiu di tng t nh ai thang.- ai rng, l mt dng bin th ca b truyn ai. Dy ai c hnh dng gn ging nh thanh rng, bnh ai c rng gn ging nh bnh rng. B truyn ai rng lm vic theo nguyn tc n khp l chnh, ma st l ph, lc cng trn ai kh nh (Hnh 11-5, b). Cu to ca ai rng bao gm cc si thp bn chu ti, nn v rng bng cao su hoc cht do. Thng s c bn ca ai rng l mun m, mun c tiu chun ha, gi tr tiu chun ca m: 1 ; 1,5 ; 2 ; 3 ; 4 ; 5 ; 7 ; 10 mm. Dy ai rng c ch to thnh vng kn. Gi tr tiu chun ca chiu di ai tng t nh ai hnh thang. Trn thc t, b truyn ai dt v ai thang c dng nhiu hn c. V vy, trong chng ny ch yu trnh by b truyn ai dt v ai thang. 6.1.3. Thng s lm vic ch yu ca b truyn ai- S vng quay trn trc dn, k hiu l n1, trn trc b dn n2 (v/ph).- T s truyn, k hiu l u,21nnu.- Cng sut trn trc dn, k hiu l P1, cng sut trn trc b dn P2 (kW). - Hiu sut truyn ng , 12PP . - M men xon trn trc dn T1, trn trc b dn T2 (N.mm). - Vn tc vng ca bnh dn v1, bnh b dn v2, vn tc di ca dy ai vd (m/s).- H s trt , ( )12 1vv v . - Thi gian phc v ca b truyn, cn gi l tui bn ca b truyn tb (h).- Lc cng ai ban u trn mi nhnh ai F0 (N). - Lc vng tc dng ln ai, cn gi l lc cng c ch Ft (N). 11tdT 2F - H s ko , 0tF 2F . - Yu cu v mi trng lm vic ca b truyn. - Ch lm vic. 6.1.4. Cc thng s hnh hc ch yu ca b truyn ai- ng knh tnh ton ca bnh ai dn d1, ca bnh b dn d2 l ng knh ca vng trn tip xc vi lp trung ha ca dy ai. Lp trung ho ca ai llp khng b ko, m cng khng b nn khi dy ai vng qua cc bnh ai. d2 = d1.u.(1- ). - Khong cch trc a l khong cch gia tm bnh ai dn v bnh b dn.- Gc gia hai nhnh dy ai (). - Gc m ca dy ai trn bnh dn 1, trn bnh b dn 2 (). 1 = 1800 - ; 2 = 1800 + ; 570.(d2 d1)/a.- Chiu di dy ai L (mm). c o theo lp trung ha ca dy ai. Quan h gia chiu di dy ai v khong cch trc a c xc nh nh sau: ( ) ( )a 4d d2d da 2 L21 2 1 2++ + ( ) ( )( )11]1

1]1

+ ++ 21 221 2 1 2d d 22d dL2d dL41a- S dy ai trong b truyn ai hnh thang z. - Din tch tit din mt ct ngang ca dy ai A (mm2) (Hnh 11-4). + i vi ai dt A = b h . Vi b l chiu rng, h l chiu cao ca tit din. + i vi ai thang A = A0 z.Vi A0 l din tch tit din ca mt dy ai. - Chiu rng bnh ai B1, B2. Thng thng B1= B2bng chiu rng tnh ton B. + i vi bnh ai dt, ly B = 1,1.b + (1015) mm. + i vi bnh ai thang, ly B = (z - 1).pth + 2.e mm. 6.1.5. Lc tc dng trong b truyn ai- Khi cha lm vic, dy ai c ko cng bi lc ban u F0.- Khi chu ti trng T1 trn trc I v T2 trn trc II, xut hin lc vng Ft, lm mt nhnh ai cng thm gi l nhnh cng v mt bnh bt cng i (Hnh 11-6).Lcnylccngtrnnhnhcng:2F FFt 0c+, lccngtrnnhnh khng cng:2F FFt 0kh- Khi cc bnh ai quay, dy ai b ly tm tch xa khi bnh ai. Trn cc nhnh ai chu thm lc cng Fv= qm.v-2, vi qml khi lng ca 1 mt ai. Lc Fv cn c tc hi lm gim lc ma st gia dy ai v cc bnh ai. Lc ny trn nhnh ai cng c lc vt 0cF2F FF ++trnnhnhaikhng cng c lc vt 0khF2F FF +- Lctcdnglntrcv mang btruynaillc hngtm Fr cphng vung gc vi ng trc bnh ai, c chiu ko hai bnh ai li gn nhau. Gi tr ca Fr c tnh nh sau:

,_

2cos F . 2 F0 r6.1.6. ng sut trong ai-Di tc dng ca lc cng Fc:+ Trn nhnh ai cng c ng sut AFcc .+ Trn nhnh ai khng cng c AFkhkh . ng nhin kh < c. - Ngoi ra, khi dy ai vng qua bnh ai 1 n b un, trong ai c ng sut un1udh . E1 . Trong E l m un n hi ca vt liu ai. -Tngt, khi dyai vngquabnhai 2, trongai c 2udh . E2 . Ta nhn thy u2 < u1.S phn b ng sut trong dy ai, dc theo chiu di ca ai c trnh by trn Hnh 11-7 ta c nhn xt:- Khi b truyn lm vic, ng sut ti mt tit din ca ai s thay i t gi tr min = kh n gi tr max = c + u1. Nh vy dy ai s b hng do mi. - Khi dy ai chy mt vng, ng sut ti mi tit din ca ai thay i 4 ln. hn ch s chu k ng sut trong ai, ko di thi gian s dng b truyn ai c th khng ch s vng chy ca ai trong mt dy. - cho u1 v u2 khng qu ln, chng ta nn chn t l hd1 trong khong t 30 40. CHNG XIII B TRUYN BNH RNG 13.1. Nhng vn chung 13.1.1. Gii thiu b truyn bnh rng B truyn bnh rng thng dng truyn chuyn ng gia hai trc song song nhau hoc cho nhau - b truyn bnh rng tr (Hnh 13-1, 13-2). Cng d th truyn chuyn ng gia hai trc ct nhau - b truyn bnh rng nn (Hnh 13-3). B truyn bnh rng thng c 2 b phn chnh: + Bnh rng dn 1, c ng knh d1, c lp trn trc dn I, quay vi s vng quay n1, cng sut truyn ng P1, m men xon trn trc T1 + Bnh rng b dn 2, c ng knh d2, c lp trn trc b dn II, quay vi s vng quay n2, cng sut truyn ng P2, m men xon trn trc T2. 148 + Trn bnh rng c cc rng, khi truyn ng cc rng n khp vi nhau, tip xc v y nhau trn ng n khp (Hnh 13-4).rng tr rng thngNguyn l lm vic ca b truyn bnh rng c th tm tt nh sau: trc I quay vi s vng quay n1, thng qua mi ghp then lm cho bnh rng 1 quay. Rng ca bnh 1 n khp vi rng ca bnh 2, y rng bnh 2 chuyn ng, lm bnh2 quay, nh mi ghp then trc II quay vi s vng quay n2.Truyn chuyn ng bng n khp, nn trong b truyn bnh rng hu nh khng c trt (ch c hin tng trt bin dng phnHnh 13-2: B truyn bnh rng tr rng nghingHnh 13-3: B truyn bnhrng nn (cn)nh v chn rng), hiu sut truyn ng ca b truyn rt cao. Rng ca bnh rng c phn nh rng, phn chn rng, phn bin dng rng v on cong chuyn tip gia bin dng rng v chn rng (Hnh 13-5). Trong qu trnh truyn ng, cc cp bin dng i tip tip xc vi nhau trn ng n khp. 13.1.2.Phn loi b truyn bnh rng Tytheohnhdngbnhrng, phngrngvonbindngrng, ngi ta chia b truyn bnh rng thnh cc loi sau: - B truyn bnh rng tr: bnh rng l hnh tr trn xoay, ng sinh thng, thng dng truyn chuyn nggia haitrc song song vi nhau, quay ngc chiu nhau. B truyn bnh rng tr c cc loi: + B truyn bnh rng tr rng thng, phng ca rng trng vi ng sinh ca mt tr, s biu din b truyn bnh rng tr rng thng trn Hnh 13-1. + B truyn bnh rng tr rng nghing, phng ca rng nghing so vi ng sinh ca mt trmt gc , s biu dinb truyn bnh rng tr rng nghing trn Hnh 13-2. + B truyn bnh rng rng ch V, bnh rng c to thnh t hai bnh rng nghing c gc nghing nh nhau, chiu nghing ngc nhau, s biu din b truyn bnh rng tr rng ch V trn Hnh 13-6. - B truyn bnh rng nn, cn c gi l b truyn bnhrngcn:bnhrngcdnghnhnnct, thng dng truyn chuyn ng gia hai trc vung gc vi nhau. B truyn bnh rng nn c cc loi: + B truyn bnh rng nn rng thng: ng rng thng, trng vi ng sinh ca mt nn chia.+ B truyn bnh rng nn rng nghing: ng rng thng, nm nghing so vi ng sinh ca mt nn. + B truyn bnh rng nn rng cung trn: ng rng l mt cung trn. - Btruynbnhrngthnkhai: bindngrnglmtonca ng thn khaica vng trn. y l b truyn c dng ph bin, a s cc cp bnh rnggp trong thc t thuc loi ny.- BtruynbnhrngNovikov: bindngrnglmt phnca ng trn.- B truyn bnh rng xiclit: bin dng rng l mt an ca ng xiclit.- B truyn bnh rng - thanh rng: thanh rng l bnh rng c bit, c ngknh bng v cng, dng i chuyn ng quay thnh chuyn ng tnh tinv ngc li.- Btruynbnhrnghnhtinh: t nht mt bnhrngtrongb truyn c trcquay quanh tm ca bnh rng khc.- B truyn bnh rng n khp trong: tm ca hai bnh rng nm v cng mt pha so vi tm n khp, hai vng trn ln tip xc trong vi nhau. - B truyn bnh rng sng: rng ca bnh rng c dng sng lin tc, thng dng n khp trong thc hin mt t s truyn rt ln. Trong chng ny, ch yu trnh by b truyn bnh rng thn khai, n khp ngoi. Cc loi b truyn khc s c cp n trong sch chuyn kho v bnh rng. 13.1.3. Thng s hnh hc ca b truyn bnh rng tr rng thng Hnh dng v kch thc ca b truyn bnh rng tr rng thng c xc nh qua cc thng s hnh hc ch yu sau y (Hnh 13-4, 13-5, 13-7): - M un ca rng bnh rng, k hiu l m, n v o l mm. Cc bnh rng c cng m un s n khp c vi nhau. Gi tr ca m un m c ly theo dy s tiu chun, hn ch s lng dao gia cng bnh rng s dng trong thc t. Vd:1;1,25;(1,375);1,5;(1,75);2;(2,25);2,5;3; (3,5);4; (4,5);5; (5,5); 6; (7); 8; (9); 10; (11); .. - H s chiu cao nh rng ha*, h s ny quyt nh rng cao hay thp. Chiu cao ca rng thng ly h = 2,25.ha*.m. Cc bnh rng tiu chun c ha* = 1. H s khe h chn rng C*, h s ny quyt nh khe h gia vng nh rng vvng trn chn rng ca bnh rng n khp vi n. Cn c khe h ny hai bnh rng khng b chn nhau. Thng thng ly C* = 0,25.H s bn knh cung ln nh dao gia cng bnh rng *, h s ny lin quannancongchuyntipgiachnrngvbindngrng. Gitr thng dng * = 0,38.H s dch dao x1 ca bnh rng dn, v x2 cabnh rng b dn. Gi tr h s dch dao thngdng -1 # x # 1.Chiu rng vnh rng bnh rng dn B1 v vnhrng bnh b dn B2, mm. Thng dng B1 > B2.BMc ch: khi c sai lch do lp ghp, th btruyn vn tip xc chiu di tnh ton B.Hnh 13-7: Kt cu bnhrng tr rng thngS rng ca bnh dn z1, ca bnh b dn z2.Gc prfil thanh rng sinh , , cn c gi lgc p lc trn vng trn chia.Gc n khp w, . L gc lm bi ng tip tuyn chung ca hai vng ln vi ng n khp. Nu xt = x1 + x2 = 0, th w = .ng knh vng trn chia d1 v d2, mm. C quan h d1 = m.z1, d2 = m.z2. ng knh vng trn ln dw1 v dw2, mm. C quan h dw1 = d1.cos/cosw. ng knh vng trn c s db1 v db2, mm. L ng knh vng trn c ng thn khai c dng lm bin dng rng.db = d.cos.ng knh vng trn chn rng df1 v df2, mm.ng knh vng trn nh rng da1 v da2, mm.Chiu cao rng h, mm. C quan h h = (2.ha* + C*).m = (da - df) / 2. Khong cch trc aw, l khong cch gia tm bnh rng dn v bnh rng b dn; mm. C aw = (dw1 + dw2) / 2.Chiu dy nh rng Sa1, Sa2,mm. Thng dng Sa # 0,2.m.Chiu dy chn rng Sf1, Sf2 mm. Kch thc Sf lin quan trc tip n hin tng gy rng.- Bc rng trn vng trn chia p, mm. L khong cch o trn vng trn chia ca hai bin dng rng cng pha gn nhau nht. Bc rng trn vng trn c s pb, c o trn vng trn c s. Bc rng trn ng n khp pk, c o trn ng n khp, pk = pb. - H s trng khp . Gi tr ca cho bit kh nng c nhiu nht bao nhiu i rng cng n khp v t nht c my i rng cng n khp. H s trng khp c tnh: = AE / pb , trongAElchiudica on n khp thc.Cc cp bnh rng thng dng c # 1,1.- H s gim khong cch trc y. Trong b truyn bnh rng dch chnh gc, tng hsdchdaoxt#0, khongcchtrcaw=(z1+z2).m.cos/(2.cosw) - y. m. Thng s hnh hc ca b truyn bnh rng tr rng nghing- Btruyn bnh rng tr rng nghing cmt bthng stng tnh b truyn bnh rng tr rng thng, c o trn mt u ca bnh rng. Mt s kch thc thucb thng s ny c thm ch s t. V d, m un mt, khong cch trc awt, ng knh vng chia dwt1, dwt2, gc n khp wt, gcprofil sinh t vv.. (Hnh 13-8). B thng s ny dng o, kim tra kch thc ca b truyn bnh rng. mt v t trn mt phngmt khng phi ly theo dy s tiu chun.- Mt sthng s c xc nh trn mtphngphptuynn-n,vunggcvi phng ca rng. Cc kch thc trong mt phng ny c thm ch s n. V d, m unmn, gc profil n, gc n khp wn, vv.. Cc bnh rng trrng nghing thng s trong mt phng php tuyn c ly theo dy s tiu chun. Cc thng s ny dng tnh ton b truyn bnh rng. - Gc nghing , gc lm bi phng rng v ng sinh ca mt tr. Phng rng c th nghing tri hoc nghing phi, gi tr ca : 0 < # 450. - H s trng khp dc . H s c xc nh nh sau (Hnh 13-9): + Gi s trin khai mt tr c s bnh rng dn v b dn, t song song vi mt phng n khp AA-EE. ng thng ca on AA l ng vo khp v EE l ng ra khp ca cc cp bnh rng. + Cng nh b truyn bnh rng tr rng thng, h s trng khp =AE / p bt+ H s trng khp dc c tnh theo cng thcAA B tg p = =p p btTrong b truyn bnh rng nghing, nu > 1, th ngay c khi < 1 b truyn vn lm vic bnh thng, v lun c t nht 1 i rng tip xc trongAvng n khp.Cc thng s xc nh trn mt mt v trnmt php tuyn c mi lin quan nh sau:mn = mt.cos,tgn = tgt.cosEtgwn = tgwt.cosB13.1.5. Thng s hnh hc ca b truyn bnhrng nn rng thng - B truyn bnh rng nn rng thng c mt b thng s tng tnh cabnhrng trrng Hnh 13-9: Tnh h s trng khp v thng, xc nh trn mt nn ph ln nht ca bnh rng, trong khong cch trc aw c thay bng chiu di nn L. B thng s ny dng o kim tra kch thc ca bnh rng. Mt s kch thc ca b thng s ny c thm ch s e. V d m un me, ng knh vng chia de1, de2, ng knh vng nh rng dae1, dae2, vv.. (Hnh 13-10). - Mt s thng s c xc nh trn mt nn ph trung bnh. Cc thng s c thm ch s tb. V d, m un mtb,ng knh dtb, vv.. Cc thng s ny dng tnh ton kim tra bn v thit k b truyn bnh rng nn.- Gcmtnnchia cabnhdn1, ca bnh b dn 2; . Thng dng b truyn bnh rng nn c gc giahai trc = 1 + 2 = 900 (Hnh 13-11).- Gc mt nn chn rng f1, f2 v gc mt nn nh rng a1, a2. Cc thng s xc nh trn mt mt ln v mt trung bnh c mi lin h nh sau: 13.1.6. Thng s lm vic ch yu ca b truyn bnh rng - S vng quay trn trc dn, k hiu l n1, trn trc b dn n2; v/ph. - T s truyn, k hiu l u, u = n1/n2 = d2/d1 =z2/z1. - Cng sut trn trc dn, k hiu l P1, cng sut trn trc b dn P2; kW. - Hiu sut truyn ng ; = P2 / P1. - M men xon trn trc dn T1, trn trc b dn T2; Nmm. - Vn tc vng ca bnh dn v1, bnh b dn v2; m/s. - Thi gian phc v ca b truyn, cn gi l tui bn ca b truyn tb; h. - Ch lm vic, - Cc yu cu v mi trng lm vic ca b truyn. 13.1.7. chnh xc ca b truyn bnh rng chnh xc ca b truyn bnh rng c nh gi qua 3 chnh xc thnh phn, l: - chnh xc ng hc, c nh gi bi sai s gia gc quay thc v gc quay danh ngha ca bnh rng b dn. chnh xc ny cn cho cc c cu phn . - chnh xc n khp m, c nh gi qua ting n v s va p. Khi sai s bc rng, sai s prfil ln, th chnh xc n khp m thp. chnh xc ny quan trng i vi nhng b truyn lm vic vi s vng quay ln. - chnh xc tip xc, c xc nh qua din tch vt tip xc trn mt rng. Ngi ta bi sn ln mt mt bnh rng, cho b truyn lm vic, sau o vt sn trn mt rng ca bnh th hai. chnh xc ny quan trng i vi cc b truyn lm vic vi ch ti trng nng. Tiu chun quy nh 12 cp chnh xc cho mi chnh xc ni trn. Cp 1 lchnh xc cao nht, cp 12 l thp nht. Ty theo c tnh lm vic ca mi b truyn, m chn cp chnh xc thch hp cho tng chnh xc. Trong mt bnh rng cp chnh xc ca cc chnh xc khng chnh nhau qu 2 cp. V mi chnh xc c quyt nh bi sai lch ca mt s kch thc ca bnh rng. Trong mt bnh rng, chnh xc ca cc kch thc khng th sai lch nhau nhiu. B truyn bnh rng thng dng trong cc my thng dng c cp chnh xc t cp 6 n cp 9. i vi cc b truyn c bit quan trng, chu ti nng v lm vic vi tc cao c th chn cp chnh xc cao hn (cp 4, 5). Ngoi ra, trnh hin tng kt rng theo cnh bn, tiu chun c quy nh 6 kiu khe h cnh bn. l: A, B, C, D, E, H. Trong kiu A c khe h ln nht, kiu H c khe h cnh bn bng 0. Mi kiu khe h cn c dung sai, quy nh mc chnh xc ca khe h. Cc bnh rng c chnh xc thp, khng c chn kiu khe h nh. Cc bnh rng c chnh xc cao c th chn kiu khe h E v H. Cc b truyn bnh rng thng dng thng chn kiu khe h A, B, C. Cch ghi k hiu chnh xc ca b truyn bnh rng, V d: Ghi k hiu: 8 - 7 - 7 - Ba TCVN 1067-84 + B truyn bnh rng c chnh xc ng hc cp 8, + chnh xc n khp m cp 7, + chnh xc tip xc mt rng cp 7, + Dng khe h cnh rng B v dng dung sai ca khe h l a. Nu chnh xc ng hc, chnh xc n khp m v chnh xc tip xc cng cp th ch cn ghi mt s, nu dng dung sai trng vi dng khe h th khng cn ghi dng dung sai, v d: 7 - B TCVN 1067-84. 13.1.8.Ti trng v ng sut trong b truyn bnh rng Ti trng danh ngha ca b truyn bnh rng chnh l cng sut P hoc m men xon T1,T2 ghi trong nhim v thit k. T ta tnh c lc tip tuyn Ft trn vng trn ln, v lc php tuyn Fn tc dng trn mt rng (Hnh 13-12).2 T 12 T 22 T 12 T 2 Hnh 13-12: Lc tc dng trnmt rng bnh rng Ngoi ti trng danh ngha nu trn, khi b truyn lm vic, do va p, c thm ti trng ng tc dng ln rng. Ti trng ny t l vi vn tc lm vic, c k hiu l Fv. Tnh chnh xc Fv tng i kh khn, nn ngi ta k n n bng h s ti trng ng Kv. Khi c nhiu i rng cng n khp, ti trng phn b khng u trn cc i rng, s c mt i rng chu ti ln hn cc i khc. i rng ny bn, khi tnh ton ta phi tng ti trng danh ngha ln K ln, K # 1. K gi l h s k n s phn b ti khng u trn cc i rng. Trn tng i rng, do cng khc nhau ca cc im tip xc, ti trng phn b khng u dc theo chiu di rng (Hnh 13-13). Nh vy cho im chu tilnnhtcarngbn,khi tnh ton phi tng ti danh ngha ln K ln, K # 1. K gi l h s k n s phn Titrngtcdnglnrngsgy nn ng sut tip xc v ng sut un trn rng. Khi ng sut vt qu gi tr cho php th bnh rng b hng. S hng hc s bt u t nhng im nguy him ca rng. Qua thc ts dng v phn tch bin dng ca rng, ngi ta nhn thy ng sut tip xc H ti im C c gi tr ln nht; ti im F c tp trung ng sut, vt nt thng bt u y, pht trin dn ln v lm gy rng. Hnh 13-14: Chu trnh mch Hnh 13-15: Chu trnh i xng cang ca ng sut H v FF, khi b truyn lm vic 2 chiuKhi rng vo n khp, ng sut H v F c gi tr khc khng, khi ra khi vng n khp gi tr ca n bng khng. Nh vy ng sut trn rng l ng sut thay i, rng b hng do mi. ng sut H l ng sut thay i theo chu trnh mch ng (Hnh 13-14). ng sut F thay i theo chu trnh mch ng, khi b truyn lm vic mt chiu. V F c coi l thay i theo chu trnh i xng, khi b truyn lm vic hai chiu, b truyn o chiu quay nhiu ln trong qu trnh lm vic (Hnh 13-15). 13.1.9.Lc tc dng ln trc v mang b truyn bnh rng tnh ton trc v mang bnh rng, cn bit cc thnh phn lc tc dng ln n. Khi b truyn lm vic, trc v chu tc dng ca nhng lc sau: i vi b truyn bnh rng tr rngthng, gm c cc lc (Hnh 13-16):- Lc tip tuyn Ft1 tc dng ln trc dn I, lc Ft2 tc dng ln trc II. Phng ca Ft1 v Ft2 trngvingtiptuynchungcahai vng ln. Chiu ca Ft1 ngc vi chiu quay n1, chiu ca Ft2 cng vi chiu quay n2.Gi tr Ft1 = Ft2 = 2.T1/dw1. - Lc hng tm Fr1 tc dng ln trc I, vungHnh 13-16: Lc trong b truyn gc vi trc I v hng v pha trc I. Lc bnh rngtr rng thng hng tm Fr2 vung gc vi trc II v hng v pha trc II. Gi tr Fr1 = Fr2 = Ft1.tgw. i vi b truyn bnh rng tr rng nghing gm c cc lc tc dng sau (Hnh 13-17): - Lc tip tuyn Ft1 tc dng ln trc dn I, lc Ft2 tc dng ln trc II. Phng ca Ft1 v Ft2 trng vi ng tip tuyn chung cahai vng ln. Chiu ca Ft1 ngc vi chiu quay n1, chiu ca Ft2 cng vi chiu quay n2.Gi tr Ft1 = Ft2 = 2.T1/dwt1.- Lc hng tm Fr1 tc dng ln trc I, vung gc vitrcIvhng v pha trc I. Lc hng tm Fr2 vung gc vi trc II v hng v pha trc II.Gi tr Fr1 = Fr2 = Ft1.tgwt.- Lc dc trc Fa1 tc dng ln trc I, song song vi trc I. Lc dc trc Fa2 song song vi trc II. Chiu ca lc Fa1, Fa2 ph thuc vo chiu quay v chiu nghing ca ng rng.Gi tr Fa1 = Fa2 = Ft1.tg. Hnh 13-17: Lc trong b truyn bnh rng tr rng nghing i vi b truyn bnh rng nn rng thng c cc lc tc dng nh sau (Hnh 13-18): - Lc tip tuyn Ft1 tc dng ln trc dn I, lc Ft2 tc dng ln trc II. Phng ca Ft1 v Ft2 trng vi ng tip tuyn chung ca hai vng ln. Chiu ca Ft1 ngc vi chiu quay n1, chiu ca Ft2 cng vi chiu quay n2. Gi tr Ft1 = Ft2 = 2.T1/dtb1. - Lc hng tm Fr1 tc dng ln trc I, vung gc vi trc I v hng v pha trcI.Lchng tm Fr2 vung gc vi trc II v hng v pha trc II.Gi tr Fr1 = Ft1.tgw.cos1.Fr2 = Ft2.tgw.cos2.- Lc dc trc Fa1 tcdng ln trc I, song song vi trc I. Lc dc trc Fa2song song vi trc II. Chiu ca lc Fa1 hng v y ln ca bnh dn, chiu ca Fa2 lun lun hng v pha y ln ca bnh b dn. Gi tr Fa1 = Ft1.tgw.sin1 = Fr2 Fa2 = Ft2.tgw.sin2 = Fr1. 13.2.Tnh b truyn bnh rng 13.2.1. Cc dng hng v ch tiu tnh ton b truyn bnh rng Trong qu trnh lm vic, trn bnh rng c th xut hin cc dng hng sau: -Gy rng bnh rng, mt hoc vi rng tch ri khi bnh rng. Gy rng l dng hng nguy him nht, b truyn khng tip tc lm vic c na v cn gy nguy him cho cc chi tit my ln cn. Gy rng c th do qu ti, hoc do b mi, khi ng sut un trn tit din chn rng vt qu gi tr cho php. - Trc r mt rng, trn mt rng c nhng l nh v su, lm hng mt rng, b truyn lm vic khng tt na. Trc r thng xy ra nhng b truyn c rn mt rng cao, ng sut tip xc khng ln lm v c bi trn y . Nguyn nhn: do ng sut tip xc thay i, mt rng b mi, xut hin cc vt nt trn b mt. Vt nt ln dn ln, n mt mc no s lm trc ra mt mnh kim loi, li vt lm. - Mn rng, pha chn rng v nh rng c trt bin dng, nn rng b mi mn. Mn lm yu chn rng v lm nhn rng. Mn thng xy ra nhng b truyn c ng sut tip xc trung bnh v bi trn khng y . - Dnh xc mt rng, trn b mt rng c dnh cc mu kim loi, km theo nhng vt xc. Dnh xc lm mt rng b hng, b truyn lm vic khng tt na. Dnh xc thng xy ra cc b truyn c rn mt rng thp, ng sut ln, v vn tc lm vic cao. Nguyn nhn: do ng sut ln v nhit cao lm vt liu ti ch tip xc t n trng thi chy do. Kim loi b bt ra dnh ln mt rng i din, to thnh cc vu. Cc vu ny co xc mt rng trong nhng ln vo n khp tip theo. C nh th mt rng b ph hng. - Bin dng mt rng, trn bnh rng dn c rnh pha gia, cn trn bnh rng b dn c g pha gia rng, dng rng b thay i, b truyn n khp khng ttna. Dng hng ny thng xut hin cc b truyn c rn mt rng thp, ng sut tip xc ln, v vn tc lm vic thp. Nguyn nhn: do ng sut ln, lu li trn mt rng lu, lp mt rng mm ra, kim loi b x y t ch n sang ch kia. Do chiu ca lc ma st, trn rng bnh dn kim loi b y v pha chn rng v nh rng, cn trn bnh b dn kim loi dn v pha gia rng. - Bong mt rng, c nhng vy kim loi tch ra khi b mt rng, to nn nhng vt lm nng v rng. Bong mt rng lm thay i bin dng rng, gim cht lng b mt, b truyn lm vic khng tt na. Dng hng ny thng c nhng b truyn mt rng c ti, sau khi thm nit, thm than. Nguyn nhn: do nhit luyn v ha nhit luyn khng tt, t chc kim loi trn mt rng b ph hng, km bn vng. Di tc dng ca ng sut ln v thay i, mt lp mng kim loi b tch khi mt rng. trnh cc dng hng nu trn, ngi ta tnh ton b truyn bnh rng theo cc ch tiu: H # [H] (13-1)F # [F] (13-2)ng thi chn ch v phng php nhit luyn hp l. Trong H l ng sut tip xc ti im nguy him trn mt rng, [H] l ng sut tip xc cho php ca mt rng, tnh theo sc bn mi, F l ng sut un ti im nguy him trn tit din chn rng, [F] l ng sut un cho php ca rng, tnh theo sc bn mi. Tnh ton b truyn bnh rng theo ch tiu 13-1, gi l tnh theo sc bn tipxc.Tnh theo ch tiu 13-2, gi l tnh theo sc bn un. Nub truyn bnhrng chuti trng quti trong mtthi gianrt ngn, cn phi kim tra cc bnh rng theo sc bn tnh, gi l tnh b truyn bnh rng theo qu ti. 13.2.2. Tnh b truyn bnh rng tr rng thng theo sc bn tip xc ng sut tip xc sinh ra trn mt rng c xc nh theo cng thc Hc Trong ZM l h s k n c tnh ca vt liu ch to cc bnh rng, MPa1/2 E1, E2 l m un n hi ca vt liu bnh rng 1 v 2, 1, 2 l h s Pot xng ca vt liu bnh rng 1 v 2, qn l cng ti trng trn ng tip xc ca rng, N/mm q n = KHv l h s k n ti trng ng dng tnh ng sut tip xc, KH l h s k n phn b ti khng u trn chiu di rng, khi tnh ng sut tip xc, lH l chiu di tip xc ca cc i rng. Ly gn ng lH = B, Coi nh c mt i rng n khp. Thc t s i rng n khp c lc ln hn 1. k n s khc bit ny ngi ta a vo h s iu chnh Z. H s Z c tnh theo cng thc kinh nghim 4 l bn knh cong tng ng ca hai b mt ti im tip xc, 2 1 l bn knh cong ca imgia rng bnh dn, gn ng 1 = dw1.sinw/2, 2lbnknhcongcaimgiarngbnhb dn, c2= dw2.sinw/2. K n s khc bit gia mt thn khai v mt tr, ngi ta a vo h s iu chnh ZH. H s ZH c tnh theo cng thc kinh nghim Z H = Thay Fn = Ft/cosw, cng cc thng s khc vo cng thc Hc, ta c cng thc tnh ng sut tip xc:

ng sut tip xc cho php [H] c xc nh bng thc nghim, ph thuc vo vt liu ch to bnh rng, phng php nhit luyn mt rng, tm quan trng ca b truyn v s chu k ng sut trong sut thi gian s dng b truyn. C th tra trc tip t cc bng, hoc tnh theo cng thc kinh nghim. Bi ton kim tra bn b truyn bnh rng tr rng thng theo sc bn tip xc, c thc hin nh sau: - Tnh ng sut tip xc sinh ra trn im nguy him ca mt rng, im gia rng nm trn vng trn ln, theo cng thc (13-4). - Xc nh ng tip xc cho php ca bnh dn [H1], v ca bnh b dn [H2]. Ly [H] = min([H1], [H2]). - So snh gi tr H v [H], kt lun. Nu H # [H], b truyn sc bn tip xc. Bi ton thit k b truyn bnh rng tr rng thng theo sc bn tip xc, thc hin nhng ni dung chnh sau: - Chn vt liu v cch nhit luyn cc bnh rng. Xc nh ng sut cho php [H1] v [H2]. Ly [H] = min([H1], [H2]). - Gi s ch tiu H # [H] tha mn, s dng cng thc 13-4, vi cc ch : + Hai bnh rng thng bng thp, nn ly gn ng ZM = 275 MPa1/2, + Bnh rng tiu chun dng gc profil = 200, v h s dch dao khng ln, do c th ly gn ng ZH = 1,76, + Cc b truyn bnh rng thng dng c h s trng khp # 1,6, + t phng trnh ph a = B/aw, a c gi l h s chiu rng bnh rng theo khong cch trc. Hoc d = B/dw1, l h s chiu rng bnh rng theo ng knh bnh dn. C quan h d = a.(u+1)/2. Gi tr ca a c chn theo kinh nghim. Mt rng c rn cao, dng gi tr nh, v ngc li: Nu bnh rng t i xng so vi hai , ly a = 0,30,5. Nu bnh rng t khng i xng so vi hai , ly a = 0,25 0,4. Nu bnh rng t v mt pha so vi hai , ly a = 0,2 0,25. i vi b truyn bnh rng ch V, ly a = 0,4 0,6. Ta c cng thc tnh ng knh bnh rng dn, hoc khong cch trc nh sau:- i vi cc b truyn thng dng, c th ly m un m = (0,01 0,02).aw, chn gi tr ca m trong dy s tiu chun. Tnh cc thng s khc ca b truyn. V d, B = a.aw; dw2 = u.dw1;Z1 # dw1/m, vv.. 13.2.3. Tnh b truyn bnh rng tr rng thng theo sc bn un Trng hp nguy him nht i vi dng hng gy rng l ton b lc Fn tc dng ln mt i rng, t ti nh rng. Lc Fn c phn thnh hai phn, lc nn rng Fnn v lc un rng Fnu (Hnh 13-19). Fnn = Fn.sinaFnu = Fn.cosaa l gc p lc trn vng trn nh rng.Lc Fnn gy ng sut nn n trn tit din chn rng, cn Fnu to nn m men un Mu = Fnu.l gy ng sut un u trn tit din chn rng.n = Fnn / B.Sfu = 6.Fnu.l / (B.Sf2).Vt nt chn rng thng xut hin pha chu ko ca chn rng, nn gi tr ca ng sut tng F c tnh theo cng thc: F = u n t l = e.m, v Sf = g.m. Trong e v g lhng s tnh ton, m l m un rng.2 T 1 K Fv K F Hnh 13-19: ng sut trntit din chn rng V tnh lc php tuyn F n = d w 1 cos w KFv l h s k n ti trng ng, tnh cho sc bn un, KF l h s k n s phn b ti khng u dc theo chiu di rng. Thay cc gi tr cc thng s vo cng thc tnh ng sut F, ta c: F = gi l h s dng rng g cos w g cos wGi tr ca YF khng ph thuc m un m, m ch ph thuc vo cc thng s xc nh hnh dng ca rng. YF c gi l h s dng rng. Khi tnh bnh rng, xc nh gi tr ca YF t cc bng tra trong sch Bi tp Chi tit my, ph thuc vo s rng z v h s dch dao x ca bnh rng. F 1 =Gitr ca[F] cchnphthucvovt liuchtobnhrng, phng php nhit luyn th tch rng, s chu k ng sut un, tm quan trng ca bnh rng, kch thc ca rng. C th tra trong s tay thit k, sch Bi tp Chi tit my. Bi ton kim tra bn b truyn bnh rng theo sc bn un, c thc hin nh sau: - Xc nh ng sut cho php ca bnh rng dn [F1], v [F2] ca bnh rng b dn, t cc bng tra, hoc tnh theo cng thc kinh nghim. - Xc nh h s dng rng YF1 ca bnh dn, v YF2 ca bnh b dn. - Tnh ng sut un F1 trn tit din chn rng bnh dn, v F2 trn tit din chn rng bnh b dn, theo cng thc (13-7). - So snh F1 vi [F1], v F2 vi [F2], a ra kt lun: Nu F1 # [F1], bnh rng 1 bn. Nu F2 # [F2], bnh rng 2 bn. Bi ton thit k b truyn bnh rng theo sc bn un, thc hin nhng ni dung ch yu sau: - Chn vt liu, phng php nhit luyn th tch cho bnh rng 1 v 2. Xc nh ng sut cho php [F1] v [F2]. - Xc nh h s dng rng YF1 ca bnh dn, tra bng theo s rng z1 v x1; YF2 ca bnh b dn, tra bng theo s rng z2 v x2. - Gi s ch tiu F1 # [F1] tha mn, T cng thc (13-7) ta rt ra c:d l h s chiu rng bnh rng theo ng knh d, ly theo kinh nghim nh trong phn tnh bnh rng theo sc bn tip xc. Ly gi tr ca m theo dy s tiu chun. - Kim tra sc bn un ca bnh rng 2. nu khng bn th phi chn tng gi tr m un m ln. - Tnh cc thng s khc ca b truyn, v kt cu ca cc bnh rng. 13.2.4. Tnh b truyn bnh rng tr rng nghing v rng ch V Phng php tnh b truyn bnh rng nghing v bnh rng ch V tng t nh tnh b truyn bnh rng tr rng thng. Cng thc tnh b truyn bnh rng tr rng nghing c thit lp bng cch: phn tch nhng c im v sc bn ca bnh rng nghing so vi bnh rng thng, a vo cng thc tnh ton bnh rng tr rng thng cc h s iu chnh, k n s khc bit v sc bn gia bnh rng nghing v bnh rng thng. B truyn bnh rng ch V l dng c bit ca bnh rng nghing, mi kt qu tnh ton bnh rng nghing c th s dng tnh bnh rng ch V. a)c im v sc bn ca bnh rng nghing so vi bnh rng thngB truyn bnh rng nghing n khp m hn b truyn bnh rng thng, do ti trng ng nh hn, gi tr ca h s Kv nh hn so vi bnh rng thng. Khi tnh chiu di tip xc lH trong b truyn bnh rng nghing, ta k n tt c cc i rng trong vng n khp, nn cng ti trng trn ng tip xc qn nh hn so vi bnh rng thng. K n s khc bit ny, ngi ta dng h s Z. Z = 1 / . ng thi phi a h s k n s phn b ti khng u chocc i rng K vo cng thc tnh ton.ng tip xc ca mt i rng trong bnh rng nghing nm chch trn mt rng (Hnh 13-20). Do chiudi cnh tay n l = g.m ca mmem un Mu nh hn; ng thititdinnguyhimlchsovititdinchn rng mt gc, nnm men chng un ca tit dinnguy him ln hn so vi tit dinchn rng. Nh vy, ng sut unF trong bnh rng nghing nhhn so vi bnh rng thng.trn mt rng ca bnh rng nghingDng rng ca bnh rng nghingtrnmtphngvunggcviphng rng (mt php tuyn), ging dng rng ca mt bnh rng thng c thng s m un mt = mn, v zt = z/cos3. Bnh rng ny c gi l bnh rng thng tng ng ca bnh rng nghing. Kh nng ti ca bnh rng thng tng ng bng vi kh nng ti ca bnh rng nghing, ta c th tnh ton bnh rng nghing thng qua vic tnh ton bnh rng thng tng ng. Nh vy, vi kch thc nh nhau, bnh rng nghing c gc cng ln th kh nng ti cng ln.b)Tnh b truyn bnh rng tr rng nghing theo sc bn tip xc Xut pht t cng thc Hc, c k n nhng c im v sc bn ca bnh rng nghing, ta c cng thc tnh ng sut tip xc ca bnh rng tr rng nghingH sk n c nhiu i rng n khp H s kn hnh dng mt rng Z H = Gi tr ca cc h s KHv, KH, KH c ly t bng tra trong s tay thit k c kh, hoc sch Bi tp Chi tit my. ngsutchophp [H] cly tngtnh tnhbnh rngtrrng thng. Bi ton kim tra bn b truyn bnh rng tr rng nghing theo sc bn tip xc, c thc hin nh sau: - Tnh ng sut tip xc sinh ra trn im nguy him ca mt rng, im gia rng nm trn vng trn ln, theo cng thc (13-9). - Xc nh ng tip xc cho php ca bnh dn [H1], v ca bnh b dn [H2]. Ly [H] = min([H1], [H2]). - So snh gi tr H v [H], kt lun. Nu H # [H], b truyn sc bn tip xc. Bi ton thit k b truyn bnh rng tr rng nghing theo sc bn tip xc, thc hin nhng ni dung chnh sau: - Chn vt liu v cch nhit luyn cc bnh rng. Xc nh ng sut cho php [H1] v [H2]. Ly [H] = min([H1], [H2]). - Gi s ch tiu H # [H] tha mn, s dng cng thc 13-9, vi cc ch : + Hai bnh rng thng bng thp, nn ly gn ng ZM = 275 MPa1/2, + Bnh rng tiu chun dng gc profil = 200, v h s dch dao khng ln, do c th ly gn ng ZH = 1,76, + Cc b truyn bnh rng thng dng c h s trng khp # 1,6, + t phng trnh ph a = B/awt, a c gi l h s chiu rng bnh rng theo khong cch trc. Hoc d = B/dwt1, l h s chiu rng bnh rng theo ng knh bnh dn. Gi tr ca a c chn theo kinh nghim, tng t nh bnh rng tr rng thng. Ta c cng thc tnh ng knh bnh rng dn, hoc khong cch trc nh sau: i vi ccbtruynthngdng, cthlymunmn=(0,01 0,02).awt, chn gi tr ca mn trong dy s tiu chun. Tnh m un mt v cc thngskhccabtruyn. V d, B=a.awt; dwt2=u.dwt1; Z1# dwt1/mt, vv.. c)Tnh b truyn bnh rng tr rng nghing theo sc bn un Thc hin tnh ton tng t nh vi bnh rng tr rng thng, c k n nhng c im v sc bn, ta c cng thc tnh ng sut un ti tit din chn rng ca cc bnh rng nh sau: Trong : Gi tr ca h s dng rng YF1 tra bng theo s rng zt1 v x1; h s dng rng YF2 tra bng theo s rng zt2 v x2. Y l h s k ng tip xc nm chch trn mt rng, Y = 1 # Y l h s k n c nhiu i rng cng n khp, Y = 1/. Gi tr ca cc h s KFv, KF, KF c ly t bng tra trong S tay thit k, hoc sch Bi tp Chi tit my. Bi ton kim tra bn b truyn bnh rng tr rng nghing theo sc bn un, c thc hin nh sau: - Xc nh ng sut cho php ca bnh rng dn [F1], v [F2] ca bnh rng b dn, t cc bng tra, hoc tnh theo cng thc kinh nghim. - Xc nh h s dng rng YF1 ca bnh dn, v YF2 ca bnh b dn. - Tnh ng sut un trn tit din chn rng bnh dn F1, v F2 trn tit din chn rng bnh b dn, theo cng thc (13-12). - So snh F1 vi [F1], v F2 vi [F2], a ra kt lun: Nu F1 # [F1], bnh rng 1 bn. Nu F2 # [F2], bnh rng 2 bn. Bi ton thit k b truyn bnh rng theo sc bn un, thc hin nhng ni dung ch yu sau: - Chn vt liu, phng php nhit luyn th tch cho bnh rng 1 v 2. Xc nh ng sut cho php [F1] v [F2]. - Xc nh h s dng rng YF1 ca bnh dn, tra bng theo s rng zt1 v x1; YF2 ca bnh b dn, tra bng theo s rng zt2 v x2. - Gi s ch tiu F1 # [F1] tha mn, ta tnh c: d l h s chiu rng bnh rng theo ng knh d, ly theo kinh nghim nh trong phn tnh bnh rng theo sc bn tip xc. Ly gi tr ca mn theo dy s tiu chun. - Kim tra sc bn un ca bnh rng 2. nu khng bn th phi chn tng gi tr m un mn ln. - Tnh m un mt v cc thng s khc ca b truyn, v kt cu ca cc bnh rng. 13.2.5. Tnh b truyn bnh rng nn rng thng Tnh b truyn bnh rng nn c thc hin tng t nh tnh b truyn bnh rng tr rng thng. Cc cng thc tnh b truyn bnh rng nn c thit lp bng cch: phn tch nhng c im v sc bn ca bnh rng nn so vi bnh rng tr, a vo cng thc tnh ton bnh rng tr cc h s iu chnh, k n s khc bit v sc bn gia bnh rng nn v bnh rng tr. a)c im v sc bn ca bnh rng nn so vi bnh rng tr - Tit din rng ca bnh rng nn c kch thc thay i dc theo chiu di rng, cng v pha nh nn, kch thc cng nh. Song, ti trng phn b trn ng tip xc ca rng cng t l vi kch thc tit din rng, nn gi tr ng sut tip xc H v ng sut un F ti cc tit din khng thay i dc theo chiu di rng (Hnh13-21). Thng ngi ta tnh ton b truyn bnh rng nn theo tit din trung bnh ca rng. - Dng rng ca bnh rng nn rng thngtrn mt nn ph trung bnh, ging nhdng rng ca bnh rng tr rng thngc cc cc thng s mt = mtb, zt=z/cos.Bnh rng thng ny c gi l bnhrng tng ng. Kh nng ti ca btruynbnhrngnnbng 0,85khnngticabnhrngthngtngng. Do , c th tnh ton b truyn bnhrngnnquabnhrngthngtngng,vititrngtngln 1/0,85 ln.b)Tnh b truyn bnh rng nn rng thng theo sc bn tip xc Xut pht t cng thc Hc, c k n nhng c im v sc bn ca bnh rng nn, ta c cng thc tnh ng sut tip xc ca b truyn bnh rng nn: Trong : H s k n vt liu ZM ly tng t nh bnh rng tr rng thng. Gi tr ca h s k n c nhiu i rng n khp Z, v h s ZH c ly tng t nh bnh rng tr. Gi tr ca cc h s KHv, KH, c ly t bng tra trong s tay thit k, hoc sch Bi tp Chi tit my. ngsutchophp [H] cly tngtnh tnhbnh rngtrrng thng. Bi ton kim tra bn b truyn bnh rng nn rng thng theo sc bn tip xc, c thc hin nh sau: - Tnh ng sut tip xc sinh ra trn im nguy him ca mt rng, theo cng thc (13-13). - Xc nh ng tip xc cho php ca bnh dn [H1], v ca bnh b dn [H2]. Ly [H] = min([H1], [H2]). - So snh gi tr H v [H], kt lun. Nu H # [H], b truyn sc bn tip xc. Bi ton thit k b truyn bnh rng tr rng nghing theo sc bn tip xc, cn thc hin nhng ni dung chnh sau: - Chn vt liu v cch nhit luyn cc bnh rng. Xc nh ng sut cho php [H1] v [H2]. Ly [H] = min([H1], [H2]). - Gi s ch tiu H # [H] tha mn, s dng cng thc 13-13, vi cc ch : tphngtrnhphd=B/dtb1, lhschiurngbnhrngtheo ng knh bnh dn. Gi tr ca d c chn trong khong t 0,3 n 0,6 tu theo v tr ca bnh rng so vi hai gi . Ta c cng thc tnh ng knh trung bnh ca bnh rng dn nh sau: i vi ccbtruynthngdng, cthlymunmtb=(0,02 0,03).dtb1, c th chn gi tr ca mtb trong dy s tiu chun. Tnh m un me v cc thng s khc ca b truyn. V d, B = d.dtb1; dtb2 = u.dtb1; Z1 # dtb1/mtb, vv.. c)Tnh b truyn bnh rng nn theo sc bn un Thc hin tnh ton tng t nh vi bnh rng tr rng thng, c k n nhng c im v sc bn, ta c cng thc tnh ng sut un ti tit din chn rng ca cc bnh rng nh sau: Trong : Gi tr ca h s dng rng YF1 tra bng theo s rng zt1 = z1/cos1 v x1; h s dng rng YF2 tra bng theo s rng zt2 = z2/cos2 v x2. Gi tr ca cc h s KFv, KF c ly t bng tra trong s tay thit k, hoc sch Bi tp Chi tit my. Bi ton kim tra bn b truyn bnh rng nn rng thng theo sc bn un, c thc hin nh sau: - Xc nh ng sut [F1] cho php ca bnh rng dn, v [F2] ca bnh rng b dn, t cc bng tra, hoc tnh theo cng thc kinh nghim. - Xc nh h s dng rng YF1 ca bnh dn, v YF2 ca bnh b dn. - Tnh ng sut un F1 trn tit din chn rng bnh dn, v F2 trn tit din chn rng bnh b dn, theo cng thc (13-15).- So snh F1 vi [F1], v F2 vi [F2], a ra kt lun: Nu F1 # [F1], bnh rng 1 bn. Nu F2 # [F2], bnh rng 2 bn. 13.2.6.Kim tra bn b truyn bnh rng theo ti trng qu ti C mt s trng hp, trong khi lm vic,ti trng tc dng ln bnh rng tng t ngt trong mt khong thi gian ngn. Ti trng ny gi l ti trng qu ti,k hiu l T1max, T2max. Trong trng hp ny cn kim tra sc bn tnh ca b truyn bnh rng theo ti trng qu ti. Ch tiu tnh ton: Hqt # [Hqt] Fqt # [Fqt] Trong : Hqt v Fqt l ng sut tip xc v ng sut un sinh ra trn rng, tnh theo ti trng qu ti Tmax, [Hqt] v [Fqt] l ng sut tip xc v ng sut un cho php theo sc bn tnh. ng sut Hqt v Fqt c tnh theo cng thc: ng sut cho php [Hqt] v [Fqt] c xc nh bng cch tra bng theo sc bn tnh ca bnh rng, hoc tnh theo rn mt rng. [Hqt] # 2,2.HBMPa, [Fqt] # 2,7.HBMPa. Bi ton kim tra bn bnh rng theo ti trng qu ti, c thc hin nh sau: - Tnh ng sut Hqt trn mt rng theo cng thc (13-6) v Fqt1, Fqt2 ca cc rng theo cng thc 13-17. - Xc nh ng sut cho php [Hqt1], [Hq2], [Fqt1] v [Fqt2] ca cc bnh rng, - So snh gi tr ng sut sinh ra trn rng v ng sut cho php, kt lun, Nu Hqt # min([Hqt1], [Hq2]), cc bnh rng sc bn tip xc tnh, NuFqt1 # [Fqt1] rng ca bnh rng dn sc bn un tnh. Nu Fqt2 # [Fqt2], rng ca bnh rng b dn sc bn un tnh. 13.2.7.Vt liu ch to bnh rng v ng sut cho php Bnh rng ch yu c ch to bng thp, ngoi ra c th dng gang, hoc vt liu phi kim loi. Tu theo cch nhit luyn, v rn mt rng, c th chia bnh rng thp ra hai nhm chnh: - Nhm bnh rng c rn b mt BH # 350 Trc khi ct rng, ngi ta nhit luyn phi liu bng ti ci thin hoc thng ho. Saukhict rngkhngphitivsarng. Chiphchoctgt tng i thp. hn ch dnh xc rng, v m bo sc bn u cho hai bnh rng, v s chu k ng sut ca bnh 1 ln hn ca bnh 2, nn chn vt liu bnh rng nhkhcvt liubnhrngln. ThngchnbnhdncHB1=HB2+ (3050), HB2 l rn mt rng bnh b dn. i vi cc bnh rng chu ti trng nh v trung bnh nn chn thp C40, C45, C50Mn, ti ci thin. i vi cc bnh rng chu ti nh, dng trong cc c cu khng quan trng, c th chn thp CT51, CT61, C40, C45, thng ho. - Nhm bnh rng c rn b mt HB > 350 Cc bnh rng thuc nhm ny, c gia cng phc tp hn. Phi liu c cho n nh, sau em ct rng. Thc hin ti b mt: thng thm than, thm nit, thm xianua trc khi ti. Sau khi ti phi gia cng sa rng bng nguyn cng mi hoc nghin. Nn chn hai bnh rng bng cng mt loi vt liu, nhit luyn t rn b mt nh nhau. Thng dng cc thp c hm lng cc bon thp nh: thp C15, C20, 15Cr, 20Cr, b mt c thm than trc khi ti. Gi tr ca ng sut tip xc cho php [H], c th tra bng, hoc xc nh theo cng thc kinh nghim: [H] = Hlim.SH.ZR.ZV.ZXH Trong : Hlim l gii hn mi tip xc ca mt rng, tra bng c gi tr. SH l h s an ton khi tnh sc bn tip xc, c th ly SH = 1,1 1,2 ; ZR l h s k n nhm b mt, bnh rng thng thng ly ZR = 0,95.ZV l h s k n vn tc vng, bnh rng thng thng ly ZV = 1,1. ZXH l h s k n kch thc ca bnh rng, cc bnh rng da < 700 mm, ly ZXH = 1. Gi tr ca ng sut un cho php [F] c tra bng hoc tnh theo cng thc cng thc kinh nghim: Y R Y S Y XF Trong : Flim l gii hn mi un ca rng, tra bng c gi tr. SF l h s an ton khi tnh sc bn un, c th ly SF = 1,1 1,2 . YR l h s k n nhm mt ln chn rng, cc bnh rng thng thng ly YR = 1. Cc bnh rng c chn rng c nh bng, ly YR= 1,0 1,1. YS l h s k n kch thc ca rng, thng thng ly YS = 1,08. YXF l h s k n kch thc ca bnh rng, i vi bnh rng thng dng c da < 700 mm, ly KXF =1. 13.2.8. Trnh t thit k b truyn bnh rng Trong nhim v thit k b truyn bnh rng, thng cho s liu v cc thng s lm vic ch yu ca b truyn, yu cu xc nh cc thng s hnh hc, v kt cu ca b truyn, bn v ch to cc bnh rng. Phn ny trnh by cc bc tnh ton thit k b truyn bnh rng tr rng nghing. Trnh t thit k b truyn bnh rng tr rng thng, rng ch V, bnh rng nn cng c thc hin theo cc bc tng t nh bnh rng tr rng nghing. Cc bc thit k bao gm: 1-Chn vt liu ch to cc bnh rng, cch nhit luyn, tra c tnh ca vt liu. i vi cc bnh rng c rn b mt HB # 350, thng chn vt liu bnh 1 c c tnh cao hn bnh 2, HB1 = HB2 + (3050). i vi cc bnh rng c rn b mt HB > 350, thng chn vt liu hai bnh nh nhau. 2- Xc nh gi tr ng sut cho php, [H1], [H2], [F1], [F2]. Nu b truyn lm vic c qu ti trong thi gian ngn, cn xc nh thm ga tr ca [Hqt1], [Hq2], [Fqt1] v [Fqt2]. 3- Tnh ng knh dwt1 theo cng thc 13-10, hoc khong cch trc awt theo cngthc 13-11, sau khi chn h s d, hoc a, h s KHv, KH v KH. 4- Ly gi tr m un mn trong khong (0,01 0,02).awt, thuc dy s tiu chun. 5-Chn s b gi tr gc nghing trong khong 80 150 (i vi bnh rng ch V chn = 200 450). Tnh m un mt = mn/cos. Ly z1 # dwt1/mt, lm trn thnh s nguyn. Tnh z2 = u.z1. Tnh li gc nghing theo cng thc: = ar cos 6- Tnh chnh xc khong cch trc, ng knh cc bnh rng, theo s rng, m un rng v gc nghing chn. 7- Xc nh chiu rng vnh rng B=a.awt, Tnh h s trng khp dc , tnh h s trng khp . Kim tra iu kin hoc > 1, hoc > 1. Nu khng tho mn, phi iu chnh li kch thc ca b truyn. 8-Kim tra li sc bn tip xc v sc bn un ca cc bnh rng. Nu khng tho nm, phi iu chnh li kch thc ca cc bnh rng. 9-Kim tra sc bn tnh ca cc bnh rng, nu nh c ti trng qu ti trong thi gian ngn. Nu khng tho mn, phi iu chnh li kch thc ca cc bnh rng. 10-Xc nh cc kch thc khc, v kt cu ca cc bnh rng trong b truyn. 11-Tnh lc tc dng ln trc v . c s liu tnh ton thit k trc v mang b truyn bnh rng. Ch : Khi thit k b truyn bnh rng tr rng thng, bc th 5, tnh s rng z1 = dw1/m, lm trn z1, tnh z2 = z1.u. Lc ny gi tr ca dw1, dw2 v khong cch trc aw b thay i. Mun duy tr gi tr ng knh v khong cch trc nh, dng cp bnh rng dch chnh gc. Vi gc n khp w c tnh t cng Hoc vi tng h s dch dao (xt = x1 + x2) c tnh theo cng thc:CHNG XIV B TRUYN TRC VT 14.1. Nhng vn chung 14.1.1. Gii thiu b truyn trc vt B truyn trc vt - bnh vt thng dng truyn chuyn ng gia hai trc vung gc vi nhau trong khng gian (Hnh 14-1), hoc cho nhau. B truyn trc vt c 2 b phn chnh: + Trc vt dn 1, c ng knh d1, trc vt thng lm lin vi trc dn I, quay vi s vng quay n1, cng sut truyn ng P1, m men xon trn trc T1. + Bnh vt b dn 2, c ng knh d2, c lp trn trc b dn II, quay vi s vng quay n2, cng sut truyn ng P2, m men xon trn trc T2. + Trn trc vt c cc ng ren (cng cthgilrngcatrcvt),trnbnhvtcrngtng tnhbnh rng. Khi truyn ng ren trc vt n khp vi rng bnh vt, tng t nh b truyn bnh rng.Nguyn tclmviccab truyn trc vt c th tm tt nh sau: trcIquay vi s vng quay n1, ren ca trc vt n khp vi rng ca bnh vt, y rng bnh vt chuyn ng, lm bnh vt quay, ko trc II quay vi s vng quay n2. Tuy truyn chuyn ng bng n khp, nhng do vn tccahaiim tip xc c phng vung gc vi nhau, nn trong b truyn trc vt c vn tc trtrtln (Hnh 14-2), hiu sut truyn ng cab truyn rt thp.cbit, khi sdngbnhvtdn,hiu sutcabtruynnh hn 0,5. Do , hu nh trong thc t khng s dng b truyn c bnh vt dn ng. Trc vt c gia cng trn my tin ren, bng dao tin c li ct thng, tng t nh ct ren trn bu lng. Bnh vt c gia cng bng dao phay ln rng trn my phay. Dao gia cng c hnh dng v kch thc tng t nh trc vt n khp vi bnh vt. Dao ct khc trc vt ch: trn dao c cc li ct, v ren ca dao cao hn ren trc vt to khe h chn rng cho b truyn trc vt - bnh vt. Nh vy mi mt bnh vt (c m un m v s rng z) c s dng trong thc t, cn c mt con dao gia cng. 14.1.2.Phn loi b truyn trc vt Ty theo hnh dng trc vt, bin dng ren ca trc vt, ngi ta chia b truyn trc vt thnh cc loi sau: - B truyn trc vt tr: trc vt c dng hnh trtrn xoay, ng sinh thng. Trong thc t, ch yu dng b truyn trc vt tr, v c gi tt l b truyn trc vt (Hnh 14-3).- B truyn trc vt Glbit, trc vt hnh tr trn,ng sinh l mt cung trn. Loi ny cn c gi b truyn trc vt lm (Hnh 14-4).- BtruyntrcvtAcsimet:trongmtphngcha ng tm ca trc vt bin dng ren l mt onthng.Trongmtphngvung gc vi ng tm trc vt bin dng ren l ng xon csimt (Hnh 14-5). Trcvtcsimt,ctrencthchintrnmytinthng thng,daotinclict thng,gngangtm my. Nucnmi,phidngcbin dng ph hp vi dng ren, gia cng kh t chnh xc cao vt tin. Do loi b truyn ny thng dng khi trc vt c rn mt rng c HB < 350. Loiny c dng nhiu trong thc t - B truyn trc vt thn khai: trong mt phng tip tuyn vi mt tr c s bin dng ren l mt on thng. Trong mt phng vung gc vi ng tm trc vt, bindngrenlmt phncangthnkhai cavngtrn, tng t nh rng bnh rng. Trc vt thn khai c ct ren trn my tin, nhng phi g dao cao hn tm, sao cho mt trc ca dao tip tuyn vi mt tr c s ca ren. C th mi ren bng mi dt thng thng, t chnh xc cao. B truyn ny c dng khi yu cu trc vt c rn b mt cao, BH > 350. - B truyn tr vt Cnvlt: trong mt phng vung gc vi phng ca ren, bin dng ren l mt an thng. Khi ct ren trn my tin, phi g dao nghing cho trc dao trng vi phng ren. Khi mi loi trc vt ny cng phi dng mi c bin dng c bit. Loi trc vt Cnvlt hin nay t c dng. Trong chng ny, ch yu trnh by b truyn trc vt tr c dng ren csimt. Cc loi khc c trnh by trong gio trnh ring v b truyn trc vt. 14.1.3. Thng s hnh hc ch yu ca b truyn trc vt Hnh dng v kch thc ca b truyn trc vt - bnh vt c xc nh qua cc thng s hnh hc ch yu di y (Hnh 14-2, 14-5, 14-6). Cc thng s thuc bnh vt c xc nh trong mt phng chnh ca bnh vt. - M un ca rng bnh vt, k hiu l m, n v o l mm. Tng t nh bnh rng nghing, bnh vt c m un xc nh trn mt phng mt m, v trn mt phng php mn. Gi tr ca m un m c ly theo dy s tiu chun. M un php mn = m.cos. V d: m =1; 1,25; (1,5); 1,6; 2; 2,5; (3); (3,5); 4; 5; (6);6,3; (7); 8; 10;12,5; 16; (18); 20; 25. - H s ng knh ca trc vt, k hiu l q. Gi tr ca q cng c tiu chun quy nh. ng vi mi gi tr m c mt vi gi tr q, vi mc ch gim s lng dao s dng gia cng bnh vt. V d: q = 6,3; (7,1); 8; (9); 10; (11,2); 12,5; (14); 16; (18); 20;(22,4); 25. Cc gi tr ca m v q c dng trong thc t ghi trong bng di y: m 2 2,5 3 4 5 6q 16 12 12 14 9 10 12 14 16 9 1012 9 10 12 14m 8 10 12 16q 8 9 10 12 8 10 12 8 10 8 9- S mi ren ca trc vt z1 (cng c th gi l rng ca trc vt), s rng ca bnhvt z2. Gi tr ca z1 c tiu chun ha, thng dng cc gi tr z1 = 1,2, 4. Srng bnh vt nn ly z2 # 28.- H s chiu cao nh rng ha*, ly tng t nh bnh rng, thng dng ha* = 1.- H s khe h chn rng C*, ly tng t nh bnh rng, thng dng C* = 0,2. - H s dch dao ca trc vt x1, v ca bnh vt x2. Gi tr h s dch dao thngdng -1 # x1 # 1, v x2 = - x1.- Gc prfil thanh rng sinh , cn c gi l gc p lc trn vng trn chia,thng dng gi tr = 200.- Gc n khp w, . L gc lm bi ng tip tuyn chung ca hai vng lnvi ng n khp. Trong b truyn trc vt, thng dng xt = x1 + x2 = 0, nnw = .- ng knh vng trn chia d1 v d2, mm. C quan h d1 = m.q,d2 = m.z2.- ng knh vng trn ln dw1 v dw2, mm. Cc b truyn trc vt thng dng,c d1 = dw1 v d2 = dw2.- ng knh vng trn chn rng df1 v df2, mm.- ng knh vng trn nh ren da1 v vng nh rng da2, mm.- ng knh vng trn ln nht ca bnh vt, da2max.- Chiu cao rng h, mm. C quan h h = (2.ha* + C*).m = (da - df) / 2.- Khong cch trc aw, l khong cch gia tm trc vt v bnh vt; mm. Khongcch trc c tnh aw = (d1+d2)/2 = (q+z2).m/2. Khong cch trc aw c th ly theo dy s tiu chun. V d: 40; 50; 63; 80; 100; 125; 160; 180; 200; 225; 250; 280; 315; 355; 400; 450; 500. - ChiudynhrngSa2,mm. Thng dng Sa2 # 0,3.m. - Chiu dy chn rng Sf2, mm. KchthcSf2linquantrctipnhin tng gy rng. - Bc rng trn vng trn chia ca bnh vt p, mm. Bc ren ca trc vt pr. Trong mt b truyn trc vt phi c pr = p.- Bccangxonvt.C quan h = z1.pr. Hnh 14-6: Kch thc b truyn trc vt - Gc nng ca ren trc vt , , thng dng gi tr trong khong 50200. Gc nghing ca rng bnh vt . Thng dng b truyn c = . - Chiu di phn ct ren ca trc vt B1, cn c gi l chiu rng trc vt; chiu rng vnh rng ca bnh vt B2, mm. Khi z1 =1 hoc 2, ly B2 = 0,75.da1 Khi z1 =4, ly B2 = 0,67.da1 - Khong cch gia hai gi trc vt l1, mm. -Gcmcabnhvt trntrcvt 2, thngly2#2.B2/(da1- 0,5.m). Gi tr gc 2 thng dng trong khong 900 n 1200. 14.1.4.Thng s lm vic ch yu ca b truyn trc vt - S vng quay ca trc vt, k hiu l n1, ca bnh vt n2; v/ph. - T s truyn, k hiu l u, u = n1/n2 = z2/z1. - Cng sut trn trc dn, k hiu l P1, cng sut trn trc b dn P2; kW. - Hiu sut truyn ng ; = P2 / P1. Hiu sut truyn ng ca b truyn trc vt bnh vt rt thp. C th tnh ton theo cng thc sau: = vi # l gc ma st trn mt tip xc gia ren v rng.Nu k n tn hao cng sut do khuy du, th = 0 , 98- M men xon trn trc dn T1, trn trc b dn T2; Nmm. - Vn tc vng ca bnh dn v1, bnh b dn v2; m/s. Vn tc trt vtr. Trong b truyn trc vt vn tc trt rt ln (Hnh 14-2), vtr = v1/cos. Tn tht cng sut ln, sinh nhit lm nng b truyn. - Nhit lm vic, lv, 0C, l nhit n nh khi b truyn lm vic. - Thi gian phc v ca b truyn, cn gi l tui bn ca b truyn tb, h. - Ch lm vic, - Cc yu cu v mi trng lm vic ca b truyn. 14.1.5. chnh xc ca b truyn trc vt chnh xc ca b truyn trc vt c nh gi qua 3 chnh xc thnh phn, tng t nh b truyn bnh rng: - chnh xc ng hc, c nh gi bi sai s gia gc quay thc v gc quay danh ngha ca bnh vt - chnh xc lm vic m, c nh gi qua ting n v s va p. - chnh xc tip xc, c xc nh qua din tch vt tip xc trn mt rng bnh vt. Tiu chun quy nh 12 cp chnh xc cho mi chnh xc ni trn. Cp 1 lchnh xc cao nht, cp 12 l thp nht. Trong mt b truyn trc vt cp chnh xc ca cc chnh xc khng chnh nhau qu 2 cp. trnh hin tng kt rng theo cnh bn, tiu chun c quy nh 6 kiu khe h cnh bn, tng t nh b truyn bnh rng. l: A, B, C, D, E, H. Trong kiu A c khe h ln nht, kiu H c khe h cnh bn bng 0. Mi kiu khe h cn c dung sai, quy nh mc chnh xc ca khe h. Cc b truyn c chnh xc thp, khng c chn kiu khe h nh. Cc b truyn trc vt thng dng thng chn kiu khe h A, B, C. 14.1.6.Ti trng v ng sut trong b truyn trc vt Tng t nh trong b truyn bnh rng, b truyn trc vt cng c ti trng danh ngha, ti trng ng v s tp trung ti trng ln mt phn ca rng. Ti trng danh ngha ca b truyn trc vt l cng sut P hoc m men xon T1, T2 ghitrong nhim v thit k. T ta tnh c lc tiptuynFttrnvng trnln,vlcphp tuyn Fn tc dng trn mt rng (Hnh 14-7).2 T 12 T 2Ft 1 =,Ft 2 = mt rng bnh rng k n ti trng ng, ngi ta a vo cng thc tnh ton h s ti trng ng Kv. knstptrungtitrnglnmtphncarng, khitnhton ngi ta a vo h s K, gi l h s k n s phn b ti khng u trn chiu di rng. Ti trng tc dng ln rng s gy nn ng sut tip xc trn mt rng v ng sut un trn tit din chn rng. Cng nh b truyn bnh rng, ng sut tip xc H ti tm n khp C c gi tr ln nht. ng sut H v F l ng sut thay i, rng b hng do mi. ng sut H l ng sut thay i theo chu trnh mch ng. ng sut F thay i theo chu trnh mch ng, khi b truyn lm vic mt chiu. V F c coi l thay i theo chu trnh i xng, khi b truyn lm vic hai chiu. 14.1.7.Lc tc dng ln trc v mang b truyn trc vt Khi b truyn lm vic, trc v mang trc vt v bnh vt chu tc dng ca nhng lc sau (Hnh 14-8): - Lc tip tuyn Ft1 tc dng ln trc dn I, lc Ft2 tc dng ln trc II. Phng ca Ft1 tip tuyn vi vng ln trc vt, phng ca Ft2 tip tuyn vi vng ln ca bnhvt.Chiu caFt1 ngc vi chiuquayn1,chiu ca Ft2 cng vi chiu quay n2. Gi tr ca Ft1 v Ft2: Quan h gia Ft1 v Ft2 c xc nh: Ft1 =Ft2.tg(+#)Trong # l gc ma st trn b mt tip xcFr2ca ren trc vt v rng bnh vt.Ft2- Lc hng tm Fr1 tc dng ln trc I, vungFr1gc vi trc I v hng v pha trc I. Lc hng tm Fr2 vung gc vi trc II v hng v pha trc II. Gi tr trc v b truyn trc vtFr1 = Fr2 = Ft2.tg/cos - Lc dc trc Fa1 tc dng ln trc I, song song vi trc I. Lc dc trc Fa2 song song vi trc II. Chiu ca lc Fa1, Fa2 ph thuc vo chiu quay v chiu nghing ca ng ren. Gi tr ca lc dc trc: Fa1 = Ft2 = 2.T2/d2 Fa2 = Ft1 = 2.T1/d1 Lc Fa1 tc dng ln trc vt c gi tr rt ln, d lm trc vt mt n nh. 14.1.8.Kt cu ca trc vt, bnh vt Trc vt thng c lm bng thp, ch to lin vi trc dn. V ng knh chn ren ca trc vt tng i nh so vi trc, nn khng th lm tch ri. V d, trc vt csimt c trnh by trn Hnh 14-9. gim tc mn, vnh rng bnh vt lm bng vt liu c h s ma st vi thp nh, nh kim loi mu, hoc hp kim mu. m bo sc bn, bnhvt thng c lpghp t hai phn. Vnh rng bnh vt 1 bng hp kim ng gim ma st; v may 2 thng bng thp hoc gang chu ti trng (Hnh 14-10).Trong mt s trng hp c bit, v d nh bnh vt qu nh, ngi ta ch to bnh vt lin khi bng hp kim ng. Hoc bnh vt qu ln, c th ch to lin khi bng gang.14.2. Tnh b truyn trc vt14.2.1. Cc dng hng v ch tiu tnh ton btruyn trc vtTrong qu trnh lm vic, b truyn trc vt - bnh vt c th xut hin cc dng hng sau:- Dnh xc b mt, thng xy ra cc b truyn c psuttrnbmttipxcln,vntclmvic tng i ln. Trn b mt ren trc vt c dnh cc ht kim loi, b bt ra t bnh vt. Mt ren tr nn sn si. ng thi mt rng bnh vt b co xc. Cht lng b mt gim ng k, b truyn lm vic khng tt na. Nguyn nhn: do ng sut ln v nhit cao lm vt liu ca bnh vt ti ch tip xc t n trng thi chy do. Kim loi b bt ra dnh ln mt ren trc vt, to thnh cc vu, cc vu ny co xc mt rng bnh vt. - Mn rng bnh vt v ren trc vt, do vn tc trt rt ln, nn tc mn cao. Vt liu ca bnh vt c c tnh thp, bnh vt b mn nhiu hn. Mn lm yu chn rng v lm nhn rng bnh vt. Mn thng xy ra nhng b truyn c p sut trung bnh v bi trn khng y . 175 - Bin dng mt rng, trn rng bnh vt c nhng ch li lm, dng rng b thayi, b truyn n khp khng tt na. Dng hng ny thng xut hin cc b truyn c p sut trn mt tip xc ln, v vn tc lm vic thp. - Gy rng bnh vt, mt hoc vi rng tch ri khi bnh vt. Gy rng l dng hng nguy him. Gy rng c th do qu ti, hoc do b mi, khi ng sut un trn tit din chn rng vt qu gi tr cho php. - Trc r mt rng, trn mt ren trc vt v rng bnh vt c nhng l nh v su, lmhngmt rng, btruynlmvickhngtt na. Trcr thng xy ra nhng b truyn bnh vt lm bng ng thanh c bn chng dnh cao, ng sut tip xc nh v c bi trn y . - Nhit lm vic qu cao. Khi nhit vt qu gi tr cho php, s lm gim cht lng du bi trn. Lm thay i tnh cht cc mi ghp, c th dn n kt . Lm cc trc dn di, c th lm tng ti trng ph. - Trc vt b un cong, do mt n nh. i vi nhng b truyn c trc vt mnh, t l gia khong cch l1 v ng knh df1 qu ln. Lc dc trc Fa1 nn trc vt, lm trc vt mt n nh. trnh cc dng hng nu trn, ngi ta tnh ton b truyn trc vt theo cc ch tiu: [H2] l ng sut tip xc cho php ca mt rng bnh vt. F2 l ng sut un ti im nguy him trn tit din chn rng bnh vt, [F2] l ng sut un cho php ca rng bnh vt, tnh theo sc bn mi. lv l nhit lm vic ca b truyn trc vt. [] l nhit lm vic cho php ca b truyn. [Fa] l lc dc trc cho php ca trc vt. Tnh ton b truyn trc vt theo ch tiu 14-1, l tnh theo sc bn tip xc. Tnh theo ch tiu 14-2, gi l tnh theo sc bn un. Tnh theo ch tiu 14-3, gi l tnh theo iu kin chu nhit. Tnh theo ch tiu 14-4, gi l tnh theo n nh thn trc vt. Nu b truyn trc vt chu ti trng qu ti trong mt thi gian ngn, cn phi kim tra theo sc bn tnh, gi l kim tra b truyn theo ti trng qu ti. 14.2.2. Tnh b truyn trc vt theo sc bn tip xc ng sut tip xc sinh ra trn mt rng c xc nh theo cng thc HcTrong E l m un n hi tng ng ca vt liu trc vt v bnh vt, MPa. E = 2.E1.E2/(E1+E2) E1, E2 l m un n hi ca vt liu trc vt v bnh vt, qn l cng ti trng trn ng tip xc ca rng, N/mm, q K Hv K H KHv l h s k n ti trng ng dng tnh ng sut tip xc, KH l h s k n phn b ti khng u trn chiu di rng, lH l chiu di tip xc ca cc i rng. Ly gn ng lH # 1,2.d1/cos, l bn knh cong tng ng ca hai b mt ti im tip xc, 1 l bn knh cong ca bin dng ren trc vt, 1 = #, 2 l bn knh cong ca im gia rng bnh vt, c 2 = d2.sin/(2.cos). Thay Fn = Ft2/(cos.cos), cng cc thng s khc vo cng thc Hc. S dng cc gi tr thng dng, E1# 2,15.105 MPa; E2 # 0,9.105 MPa; = 200; v # 100; ta c cng thc tnh ng sut tip xc:

ng sut tip xc cho php [H] c xc nh bng thc nghim, ph thuc vo vt liu ch to bnh vt, phng php bi trn, tm quan trng ca b truyn v s chu k ng sut trong sut thi gian s dng b truyn. C th tra trc tip t cc bng, hoc tnh theo cng thc kinh nghim. Bi ton kim tra bn b truyn trc vt theo sc bn tip xc, c thc hin nh sau - Tnh ng sut tip xc sinh ra trn im nguy him ca mt rng bnh vt, im gia rng nm trn vng trn ln, theo cng thc (14-6). - Xc nh ng tip xc cho php [H2] ca bnh vt. - So snh gi tr H v [H2], kt lun. Nu H # [H2], b truyn sc bn tip xc. Bi ton thit k b truyn trc vt theo sc bn tip xc, thc hin nhng ni dung chnh sau: - Chn vt liu v cch nhit luyn. Xc nh ng sut cho php [H2]. - Gi s ch tiu H # [H2] tha mn, s dng cng thc 14-6, vi cc ch : d1 = m.q;d2 = z2.m;v m = 2.aw/(q+z2). Ta c cng thc tnh khong cch trc nh sau: 14.2.3. Tnh b truyn trc vt theo sc bn un Xc nh chnh xc ng sut F2 trn chn rng bnh vt tng i phc tp, v chn rng cong v tit din rng thay i dc theo chiu di rng. Ngi ta dng cch tnh gn ng, coi bnh vt nh bnh rng nghing vi gc nghing = . ng sut F2 c tnh theo cng thc ca bnh rng nghing. Vi gc thng dng bng 100, ta c cng thc tnh F2: Trong , m un php mn = m.cos ; h s dng rng YF2 c tra theo x2 v s rng tng ng z2t = z2/cos3. Gi tr ca [F] c chn ph thuc vo vt liu ch to bnh vt, s chu k ng sut un, kch thc ca rng. Bi ton kim tra bn b truyn trc vt theo sc bn un, c thc hin nh sau: - Xc nh ng sut cho php [F2] ca bnh vt, t cc bng tra, hoc tnh theo cng thc kinh nghim. - Xc nh h s dng rng YF2 ca bnh vt. - Tnh ng sut un F2 trn tit din chn rng bnh vt theo cng thc (14-8). - So snh F2 vi [F2], a ra kt lun: Nu F2 # [F2], bnh rng 2 bn. M un ca rng trn mt phng mt c tnh theo cng thcm = 2.aw/(q+z2), ly m theo dy s tiu chun. Sau tnh m un ca rng trn mt phng php tuyn mn = m.cos. 14.2.4. Tnh b truyn trc vt theo iu kin chu nhit Nhit lng sinh ra trong b truyn trc vt rt ln, do c trt trn b mt tip xc. Ton b cng sut tn hao s bin thnh nhit nng lm nng b truyn. Sau khi lm vic mt thi gian, khong 20' n 40', nhit ca b truyn trc vt n nh. Nhit ny gi l nhit lm vic lv, c tnh theo phng trnh cn bng nhit lng. V d, i vi b truyn trc vt trong hp gim tc, phng trnh cn bng nhit c vit nh sau: # = #1 + #2 Trong # l nhit lng sinh ra trong mt gi, kCal/h, # = 860.(1-).P1 #1 l nhit lng ta ra mi trng xung quanh trong mt gi, kCal/h, #1 = At.Kt.(lv-0) #2 l nhit lng ti ra bn ngoi qua thit b lm mt, kCal/h. Gi tr ca #2 c ghi trn thit b lm mt. At l din tch b mt thot nhit ra mi trng xung quanh, m2. Din tch b mt thot nhit bao gm din tch cc b mt tip xc vi khng kh lu thng v 25% din tch cc mt gip tng, mt y hp. Kt l h s ta nhit, kCal/(h.m2.0C). C th ly Kt = 7,5 15 ty theo tc lu thng ca khng kh. 0 l nhit mi trng xung quanh. C th ly 0 = 300C 400C. T phng trnh trn, rt ra cng thc:Gi tr nhit cho php [] c chn theo loi du bi trn b truyn, tnh cht lm vic ca b truyn. Bnh thng c th ly trong khong 750C 900C. Bi ton kim tra iu kin chu nhit c thc hin nh sau:- Tnh nhit lm vic ca b truyn lv, dng cng thc 14-9. - Xc nh nhit cho php []. - So snh lv v [], kt lun. Nu lv # [], b truyn tha mn iu kin chu nhit. Nu lv > [], th phi tm cch x l b truyn tha mn iu kin chu nhit. Cc cch x l c th dng: - Nu nhit chnh lch khng nhiu, c th chn li cht bi trn tng gi tr ca [] ln. - Lm cc cnh tn nhit tng din tch ta nhit At. - C th dng qut gi, phun nc tng gi tr h s ta nhit Kt. - Trng hp cn thit, th phi dng thit b lm mt ti nhit ra ngoi, tng gi tr #2. 14.2.5. Tnh trc vt theo iu kin n nh Trc vt thng c ch to lin trc, bn ca trc s c tnh ton chnh xc theo h s an ton (xem chng Trc). y ch trnh by cch kim tra cng ca trc theo cch tnh mt thanh chu nn dc trc. Thng ch tin hnh kim tra i vi cc trc mnh, c chiu di l1 # 25.df1. Lc nn trc vt Fa1 c xc nh theo cng thc:Lc dc trc cho php [Fa] c xc nh theo cng thc le:Trong : E l m un n hi ca vt liu trc, J l m men qun tnh ca tit din chnren trc vt,J =S l h s an ton v n nh. C th ly S = 2,5 4. l h s lin kt. Trc vt c hai gi , c th ly = 1. l1 l khong cch gia hai gi trc vt. kim tra iu kin n nh ca trc vt, ta so snh gi tr lc Fa1 v lc [Fa], rt ra kt lun. Nu Fa1 # [Fa], trc vt iu kin n nh. Nu Fa1 > [Fa], th phi tm cch x l. C th tng ng knh df1, hoc rt ngn khong cch l1.14.2.6. Kim tra b truyn trc vt theo ti trng qu ti Nu b truyn chu ti trng Pmax trong thi gian ngn, ta xc nh gi tr h s qu ti Kqt = Pmax/P. Kim tra b truyn theo sc bn tnh, da vo cc ch tiu: Hqt # [Hqt] Fqt # [Fqt] Trong , ng sut tip xc v ng sut un qu ti c tnh theo cng thc: Hqt = H 2 K qt , Fqt =F 2 K qt14.2.7. Chn vt liu v ng sut cho php Vt liu ch to trc vt, bnh vt c th chn nh sau: - Khi truyn cng sut nh (di 3kW), nn dng trc vt Acsimet hoc Covlut khng mi. Trc vt c lm bng thp C35, C45, C50, C35CrCu, ti ci thin c rn b mt di 350 HB. - Khi truyn cng sut trung bnh v ln, ngi ta dng trc vt thn khai c mi. Thngdngloi thpC40Cr, 40CrNi, 12CrNi3Al, 20CrNi3Al, 30CrMnPbAl, ti t rn b mt 45 50 HRC. Sau khi ct ren, ti b mt ren, sau mi ren v nh bng. Trc vt ti thng dng n khp vi bnh vt bng ng thanh. - Bnh vt trong cc b truyn kn c vn tc trt vtr # 5 m/s, c lm bng ng thanh khng thic, nh: BCuAl9Fe4, BCuAl10Fe4Ni4; hoc ng thau LCu66Al6Fe3Mg2, LCu58Mg2Pb2. Nu vn tc trt trong khong 5 12 m/s, bnh vt c ch to bng ng thanh t thic, nh: BCuSn6Zn6Pb3, BCuSn5Zn5Pb5. Nu vn tc trt ln hn na, c th dng ng thanh nhiu thic,nh: BCuSn10P1, BCuSn10NiP. - Trong cc b truyn quay tay, hoc cng sut nh, bnh vt c ch to bnggang, v d nh: GX10, GX15, GX18, GX20. Trng hp ny dng trc vtbng thp C35, C40, C45, ti ci thin t rn 300 HB 350 HB.ng sut tip xc cho php c th chn nh sau:- i vi cc bnh vt bng ng thanh thic, c b < 300 MPa,ly [H] = (0,750,9).b.KNH ,Trong KNH l h s k n s chu k ng sut. KNH =4 - i vi cc bnh vt bng ng thanh khng thic, c b > 300 MPa, ly [H] = 250 MPa, khi vn tc vtr = 0,5 m/s, [H] = 210 MPa, khi vn tc vtr = 2 m/s, [H] = 160 MPa, khi vn tc vtr = 4 m/s, [H] = 120 MPa, khi vn tc vtr = 6 m/s, - i vi bnh vt bng gang, ly [H] = 120 MPa, khi vn tc vtr = 0,5 m/s, [H] = 110 MPa, khi vn tc vtr = 1 m/s, ng sut un cho php c th ly nh sau:- i vi bnh vt bng ng thanh, quay mt chiu, ly[F] = (0,25.ch + 0,08.b).KNF quay hai chiu, ly [F] = 0,16.b.KNF KNF l h s k n s chu k ng sut KNF = 9 - i vi bnh vt bng gang,quay mt chiu, ly[F] = 0,12.buquay hai chiu, ly ly [F] = 0,075.bu ng sut tip xc v ng sut un cho php qu ti c th chn nh sau: Bnh vt bng ng thanh thic, ly [Hqt] = 4.ch, [Fqt] = 0,8.ch, Bnh vt bng ng thanh khng thic, ly [Hqt] = 4.ch, [Fqt] = 0,8.ch, Bnh vt bng gang, ly [Hqt] = 1,5.[H2], [Fqt] = 0,6.b. 14.2.8. Trnh t thit k b truyn trc vt Thit k b truyn trc vt c th thc hin theo trnh t sau: 1- Chn vt liu trc vt, cch nhit luyn. D on vn tc trt vsb, chn vt liu bnh vt. Chn phng php gia cng, chn cp chnh xc gia cng. 2- Xc nh ng sut cho php [H2], [F2], nu c ti trng qu ti cn xc nh thm [Hqt], [Fqt]. Xc nh [Fa] v [].3- Chn s mi ren z1, tnh s rng z2 = u.z1. Chn h s ng knh trc vt q theo tiu chun. Tnh gc nng = arctg(z1/q). Chn gi tr s b ca hiu sut sb. 4- Tnh khong cch trc aw theo cng thc 14-7. Tnh m un m = 2.aw/(z2+q), ly gi tr ca m theo tiu chun. Tnh m un php mn = m.cos. 5- Tnh cc kch thc ch yu ca b truyn:ng knh vng chia trc vt, d1 = m.d; ng knh vng chia bnh vt, d2 = m.z2; Chiu rng vnh bnh vt B2 = 0,75.da1, khi z1 = 1 hoc 2. B2 = 0,67.da1, khi z1 = 4. Chiu di phn gia cng ren ca trc vt c th ly: B1 # (11+0,07.z2).m,khi z1 = 1 hoc 2. B1 # (12,5+0,09.z2).m,khi z1 = 4. 6- Kim tra vn tc trt vtr, kim tra gi tr hiu sut . Nu sai khc so vi gitr s b ban u qu 5%, th phi chn li gi tr vsb, hoc chn li sb v tnhli.7- Kim tra sc bn un ca bnh vt. Nu khng tha mn, phi iu chnhkch thc ca b truyn.8- Kim tra iu kin n nh ca trc vt. Nu khng tha mn, phi iuchnh kch thc ca b truyn.9- Kim tra iu kin chu nhit ca b truyn. Nu khng tha mn, phi tmcch x l.10- V kt cu ca trc vt, bnh vt.11- Tnh lc tc dng ln trc v .CHNG XV B TRUYN XCH 15.1.Nhng vn chung 15.1.1.Gii thiu b truyn xch B truyn xch thng dng truyn chuyn ng gia hai trc song song vi nhau v cch xa nhau (Hnh 15-1), hoc truyn chuyn ng t mt trc dn n nhiu trc b dn (Hnh 15-2). B truyn xch c 3 b phn chnh: + a xch dn 1, c ng knh tnh ton l d1, lp trn trc I, quay vi s vng quay n1, cng sut truyn ng P1, m men xon trn trc T1. a xch c rng tng t nh bnh rng. Trong qu trnh truyn ng, rng a xch n khp vi cc mt xch, tng t nh bnh rng n khp vi thanh rng.+ a xch b dn 2, c ng knh d2, c lp trn trc b dn II, quay vi s vng quay n2, cng sut truynngP2,mmenxon trn trc T2.+ Dy xch 3lkhutrung gian, mc vng qua hai a xch. Dy xch gm nhiu mt xch c ni vi nhau. Cc mt xich xoay quanh khp bn l, khi vo n khp vi rng a xchNguyn l lm vic ca b truyn xch: dy xch n khp vi rng a xch gn ging nh thanh rng n khp vibnhrng.a xich dn quay, rng ca a xch y cc mt xch chuyn ng theo. Dy xch chuyn ng, cc mt xch y rngcaaxchbdnchuyn ng, a xch 2 quay.Nhvychuynng c truyn t bnh dn sang bnh b dn nh s n khp ca rng a Hnh 15-2: B truyn c 3 a b dn xch vi cc mt xch. Truyn ng bng n khp, nn trong b truyn xich hu nh khng c hin tng trt. Vn tc trung bnh ca bnh b dn v t s truyn trung bnh ca b truyn xch khng thay i. 15.1.2.Phn loi b truyn xch Ty theo cu to ca dy xch, b truyn xch c chia thnh cc loi: - Xch ng con ln (Hnh 15-3). Cc m xch c dp t thp tm, m xch 1 ghp vi ng lt 4 to thnh mt xch trong. Cc m xch 2 c ghp vi cht 3 to thnh mt xch ngoi. Cht v ng lt to thnh khp bn l, xch c th quay gp. Con ln 5 lp lng vi ng lt, gim mn cho rng a xch v ng lt. S 6 biu din tit din ngang ca rng a xch. Xch ng con ln c tiu chun ha cao. Xch c ch to trong nh my chuyn mn ha.- Xch ng, c kt cu tng tnh xch ng con ln, nhngkhng c con ln. Xch cch to vi chnh xc thp,gi tng i r.1- Xch rng (Hnh 15-4), khpbn l c to thnh do hainachthnhtrtipxc lonhau. Mi mt xch c nhium xch lpghptrncht.Kh nng ticaxchrngln hn nhiu so vi xch ng con ln c cng kch thc.Gi thnh ca xch rng cao hn xch ng con ln. Xch rng c tiu chun ha rt cao.Trong cc loi trn, xch ng con ln c dng nhiu hn c. Xch ng ch dng trong cc my n gin, lm vic vi tc thp.Xch rng cdngkhicntruyntitrng ln,yucukchthcnhgn. Hnh 15-4: B truyn xch rng Trong chng ny ch yu trnh by xch ng con ln. 15.1.3.Cc thng s hnh hc ch yu ca b truyn xch ng con ln - ng knh tnh ton ca a xch dn d1, ca a b dn d2; cng chnh l ng knh vng chia ca a xch, mm; l ng knh ca vng trn i qua tm cc cht (Hnh 15-5). - ng knh vng trn chn rng a xch df1, df2, mm. - ng kch vng trn nh rng d