1 Phn th nht TNG LUN CU CHNG 1: KHI NIM CHUNG VCC CNG TRNH NHN TO TRN NG 1.1 CC CNG TRNH NHN TO TRN NG Khixydngtuynngsgpphinhiuchngngikhcnhaunhsng,sui, mng mng, ni cao v.v vt qua cc chng ngi , bo m tuyn ng lin tc vchuynngcanton,ngitaxydngcu,cng,hm,ngtrnvcccng trnh khc gi l cc cng trnh nhn to trn ng. Cu l cng trnh vt qua pha trn chng ngi vt nh sng, sui, khe ni, thung lng, hoc vt qua ng, qua nh my, ch v.v Cng l cng trnh nm trong nn p ca tuyn ng nhm gii quyt cho dng chy luthngkhigiaoctvituynng.Cngtdimtngtithiu0.5mivi ngtv1mivingxela, vvyquavtrcngtuynngvnlintc. Cng ch c kh nng thot mt lng nc nh v va, v vy ngi ta xy dng cng khi tuyn ng i qua dng nc nh v va, hoc dng nc khng thng xuyn, lu lng khng ln lm nh mng mng, khe rnh. Cng c dng lm phng n so snh vi cu nh. Hmdngdnngxuynquani,tronglngtvctrnghpxydng trong nc. ngtrncxydngkhituynngctngangdngchycmcnckhng ln, lu lngc th thot qua kt cu thn ng. Mt nmch c mt vi gi hoc hn hu mt vi ngy nc ngp v trn qua mt ng, song xe c vn qua li c. Ngoi ra cn c cc cng trnh khc nh cu trn, tng chn, bn ph v.v Trong s cc cng trnh nhn to trn, cu l cng trnh ph bin nht, va phc tp v cu to, thit k v thi cng, kinh ph u t tng i ln. 1.2 CNG TRNH CU I- Cc b phn v kch thc c bn cacu 1. Cc b phn c bn ca cu - Ktcunhp:Bphntrctipmanghottivvtquakhongcchchngngi vt.Bphnchulcchnhldm,dn,vmv.vKtcunhpcncphnxechy, 2 ng ngi i v lan can, cc b phn phn xe chy, truyn ti trng cho b phn chu lcchnh,giaccbphnchulcchnhcnchlinktmbonnhv cng ngang cu. Kt cu nhp c t ln gi cu. C kt cu nhp xe chy trn, chy gia, chy di. - M, tr: b phn k kt cu nhp, tip nhn ton b ti trng v truyn xung nn t thngquaktcumng.Mcxydnghaiucu,mcncnhimvnitip gia ng vi cu. Tr c xy dng pha ngoi b phn chia cu thnh cc nhp. Nh vy, vi cu mt nhp s khng c tr m ch c hai m, cng c trng hp cu khng c m m kt cu nhp c ko di mt on mt tha ni vo nn ng p u cu. - Ngoi ra cn c cc b phn: ng dn vo cu, cng trnh dn dng, cng trnh bo v tr khi b tu b hoc vt tri va p, gi cu v.v Hnh 1-1 Cc b phn c bn ca cng trnh cu 1- kt cu nhp; 2- tr; 3- m; 4- gi cu; 5- mng; 6- m t p nn; 7- nn ng u cu 2. Cc kch thc c bn ca cu - Chiu di ton cu L(m): l khong cch gia ui ca hai m hay khong cch gia cc u mt ca kt cu nhp tip xc trc tip vi t p u cu nu nh cu khng c m nh cu mt tha. - Chiu di nhp tnh ton ltt: l khong cch gia tim hai gi kt cu nhp- KhNucuLo:lchiurngthotncdicu,ltngchiudiccnhptnhLo= loi;loikhongcchgiahaimpidintmptrnytimptrkia(hocm)xc nh ti mc nc cao nht (MNCN). Trng hp cu c m vi th MNCN khng tip xc vi tng thn m, do thay v nhp tnh st m khNu thot nc s c ly trung bnh cng ca hai tr s tng ng mc nc cao nht v mc nc thp nht. - Chiu cao cu H: l khong cch t mt xe chy (hoc y ray) n mc nc thp nht hay n im thp nht ca mt t nu l cu vt hoc cu cn. 3 - Chiu cao kh gm cu Ho l khong cch t mc nc cao nht n y kt cu nhp, m bo cy tri khng va p v mc nghn. Nu l cu vt th c tnh t mt ng bn di n y kt cu nhp. Theo quy trnh 79: i vi cu t: nu khng c cy tri, ln Ho 0.5m; nu c cy tri, ln Ho 1m N goi ra nu sng c thng thng ng thy th chiu cao kh gm cu phi m bo chiu cao kh thng thuyn cho cc phng tin ng thy qua li. - Chiu cao kin trc hkt l khong cch t mt xe chy n im thp nht ca y kt cu nhp. - Chiu rngtnh tonca kt cu nhp Bl khongcch gia trc ca hai dmbin (hay dn bin) trn mt ct ngang. Khi nim v cc mc nc: - MNLS-Mcnclchs,lmcnclnnhtttrcnnaymngitaiutra c. - MNCN - Mc nc cao nht, l kt qu tnh ton ng vi tn sut c quy nh ty theo loi cng trnh (1% hay 2%). Nu ni MNCN ng vi tn sut thit k 1% c ngha l mc nc ca cn l m 100 nm mi xut hin mt ln. - MNTN-mcncthpnht,cotrongmacnvngvimttnsutquynh (1% hay 2%), cn c vo MNTN b tr nhp thng thuyn. - MNTT -mc nc thng thuyn,l mc nc cao nht chophp tu bqua li, thng lyvitnsut5%,tmcncnyxcnhchiucaokhgmcucanhpthng thuyn.II- Phn loi cng trnh cu C nhiu cch khc nhau phn loi cng trnh cu. Phn loi theo chng ngi vt cu phi vt qua: - Cu qua sng, qua sui l loi cu ph bin - Cu qua ng hay cu vt, nh cu vt N g T S, cu vt ng st v.v - Cu cn hay cu dn, l cu c xy dng ngay trn mt t nhm dn ln mt cu chnh hoc nng cao tuyn ng ln gii phng khng gian bn di nh cu dn hai u cu chnh Thng Long. - Cucao,lloicucchiucaotrrtlncbcquaccthunglngsunhcu Chu u o. Phn loi theo mc ch s dng: - Cu t (cu ng b) 4 - Cu xe la (cu ng st) - Cu ngi i b (cu b hnh) - Cu hn hp - Cu thnh ph - Cu tu (dng cc bn cng) - Cu c bit dng dn kh, dn du, dn nc, dn cp in v.v Phn loi theo vt liu lm kt cu nhp- Cu g - Cu - Cu b tng, cu BTCT- Cu thpPhn loi theo cao mt ng xe chy - Cu c ng xe chy trn - Cu c ng xe chy di - Cu c ng xe chy gia Phn loi theo s tnh hc: Theo s tnh hc ca kt cu chu lc chnh c th phn chia cng trnh cu thnh cc h thng sau: - Cudm:ditcdngcatitrngthngngktcunhplmvicchuunvch truyn p lc thng ng xung m tr. H thng cu dm bao gm dm gin n, dm lin tc v dm mt tha. Theo cu to ca kt cu chu lc chnh c th phn thnh cu dm c sn c v cu dn. - Cuvm:cimcbncahvmltivtrchnvmlunxuthinthnhphn phn lc theo phng nm ngang (lc x). - Cu khung: l loi cu m m, tr c ngm cng vi kt cu nhp to thnh khung cng tham gia chu lc. - Culinhp:lloicuckthptcchnginhochnginctng cng cc b phn chu lc. Bng cch ngi ta c th to ra nhng kt cu chu lc hp l v c hiu quv cc phng din knh t, kthut c bit trong cc trng hp nhp ln. - Cutreo:lloiktcutrongbphnchulcchnhlccdylmvicchuko. Di tc dng ca hot ti h dm mt cu v dy lm vic nh mt h lin hp. Cu treo gm c cu treo parabol cn gi l cu treo v cu dy vng Theo quy m cng trnh5 - Cu nh: L 25m - Cu trung L = 25-100m - Cu ln: L> 100m hoc c nhp l30m Theo c im cng trnh Cu phao, cu quay, cu nng Theo thi hn s dngCu tm di 5 nm, cu bn vnh cu 10-50 nm, cu vnh cu trn 50 nm 6 Hnh 1-2 Cc s cu a,b,c- cu dm gin n, lin tc, mt tha; d- cu dn; e- cu khung; f, g- cu vm c ng xe chy trn vxe chy gia; h- cu lin hp dm-vm; i- cu treo; k- cu dy vng III- Lch s tm tt v phng hng pht trin ngnh xy dng cu 1. Lch s tm tt Cu l cng trnh nhn to, v vy lch s pht trin ca n gn lin vi s pht trin ca x hi. Vo thi k khai s ca loi ngi, con ngi da vo t nhin vt qua cc con sui, khe su nh nhng thn cy vt ngang, nhng dn dy leo hoc nhng cy tri ni mcvoccvtchngngi.Cthcoil nhnghnhnhutincacNudm,cu treov cu phaongynay. Ngi c xa bt chc cc hin tng tnhin to racc phng tin vt qua cc dng sng, con sui. Di tch ca chic cu c xa nht l cu qua sng Euphrate Babylon c lm bng thn cy c, nhp di 9m vi tng chiu di l 300m (khong 2000 nm trc cng nguyn). Thi k chim hu n l, phc v cc cuc chin tranh gia cc b lc cn thit phi xy dng nhng tuyn ng vcc cy cu, h thng giao thng vn ti bt u pht trin. u tin l cu g xut hin, sau l cu . Rt nhiu nhng cng trnh cu c xa c xy dng cch y hng ngn nm cn tn ti cc khu vc Babylon, Iran, La M. 7 Hnh 1-3 Cu vm gang qua sng Severn-Anh nm 1776-1779 Trongxhiphongkin,giaionsaukhichnghatbnxuthin,nhucugiao lu bun bn ngy cng tng tr thnh ng lc thc Ny s pht trin mnh m ca giao thng vn ti. Tuy vy thi k ny vn ch c cu v cu g, l lun tnh ton cha c. Sang thi k t bn ch ngha giao thng vn ti pht trin. Cu cng khng ch tng v mt s lng m c nhiu thay i v dng kt cu v vt liu. Bn cnh cu , cu g bt u xut hin cu gang, cu thp v sau l cu BTCT. V hnh dng kt cu xut hin cu dm, cu dn v cu treo... Nh cc ngnh ton hc v c hc pht trin mnh, cc cng trnhcucxydngdatrncslluntnhtonktcuvthitkvititrng nnghn(umyhincrai)ngthivtnhplnhn.Nm1776mtks ngi Nga tn l Ku-li-bin thc hin n thit k chic cu vm g nhp 310m bc qua sng NvaP-tc-bua.Cng trong thikny, xut hin cu kim loi, u tinl chic cubnggangbcquasngSevern(Anh),nhp31mvonm1776-1779.Bngdyxch st, nm 1741 ngi Anh xy dng cu treo u tin nhp 22m qua sng Tess. Cu treo c u im l vt nhp ln v c hiu qu kinh t cao. Vo nm 1820 khi xy dngchiccutreoquasngTvid(Anh)nhpdi110mngitasosnhvthygi thnh r hn 4 ln so vi phng n cu . Do c nhiu u im nn sau khi ra i cu treo phttrinmnh,chtrongmtthigianngncngdngnhiuvttinhp hngtrmmt.Nm1834ThySxydngchiccutreodycpnhp265mvn nm 1848 cu Virginia (M) t c nhp 308m. Tuy nhin thi k ny v l thuyt tnh ton c bit v n nh ng hc cn nhiu vn cha c gii quyt, do xy ra mt s tai nn i vi cu treo, nh cu qua sng Meine (Php) b sp khi c mt on qun i u qua cu lm cht 226 ngi. Nhng tai nn cngvi s hn ch v kh nng chu hot ti nng so vi cc loi cu khc khin vic p dng cu treo c xu hng chng li v chuyn sang nhng h thng cu cng nh dm, dn, vm. 8 Hnh 1-4 Cu dn mt tha qua vnh Forth-nm 1890 Nhng nm cui ca th k XIX, u th k XX ngnh xy dng cu pht trin mnh m v phong ph v mi phng din, cu BTCT bt u xut hin vi nhng cng trnh c xy dng Php, c. Hng lot cc cu kim loi (ch yu l thp) c xy dng vi cc dng dn, vm v t nhp hng trm mt trong cuc chy ua rt si ng v chiu di nhp klc.VcudnthpccuquavnhForth(Scotland)dngmtthanhp521mc sydngnm1890vcuQuebec(Canada)vtnhp549m(1917),nm1931cuvm thpquavnhSydney(Australia)tnhp503mvbnthngsautiNewYorkngita thngxecuBayonnecnhp504m,tipnlcuKyll-wan-koul(NewYork)nhp 511m. Hnh 1-5 Cu vm qua vnh Sydney nm 1931 Trc nhu cu vt qua cc con sng rng, su v cc eo bin ln, vo nhng nm na u th k XX cu treo c chp nhn tr li trn c s hon thin hn nhng nghin cu l thuyt v thc nghim, chnh thi k ny c bc nhy vt v kh nng vt nhp ca cutreo:nm1929cutreocnhplnnhtthgiicuAmbassasdor(M)vtnhp 564m,3nmsaucuG.WashingtonNewYork(1932)vtnhpdi1067mvtip theo l Golden Gate nhp 1280m San Francisco (1937). 9 Nm1940MxyravtainnspcutreoTacomanhp853m(cngtrnhmi hon thnh c 6 thng). y l v tai nn gy nhiu ch v thu thp c nhiu s liu lin quan, c bit ngi ta quay phim c ton b din bin ca tai nn. Hnh 1-6 Tai nn cu Tacoma nm 1940 V cu Tacoma khng lm cc nh xy dng lng trnh cu treo m ngc li b sungchongnhxydngcunhngvncnnghincuhonthin.Viccphng hng chnh l tng cng cng cho dm ch v tin hnh nghin cu thc nghim tmraccdngtitdinthotgi,cccngtrnhcutreonhplnvnctiptcxy dngnhcuVerrazano-NarrowNewYorknhp1298.45m(1964),cuHumberAnh nhp1410(1981)vklcvnhpcuicngcathkXXthucvcutreoAkashi-Kaikyo (Nht Bn) vi nhp chnh di 1991m, cng trnh c hon thnh nm 1998. Hnh 1-7 Cu treo Tacoma sau khi xy dng li 10 Hnh 1-8 Cu treo Golden Gate Mt dng cu treo na l cu dy vng dm cng c p dng kh ph bin cc nc chu u bt u t gia th k XX v hin ang c a chung. Vo khong nhng nm 70 ca th k ny nhng nghin cu cho thy cu dy vng c cc ch tiu kinh t k thut rtttivinhp200-300.Tuynhinnnayvinhngthayivquannimktcu cng nh cng ngh thi cng, tnh u vit ca cu dy vng cn th hin vi c cc nhp c chiudilnhn:cuNormandiebcquasngSeine(Php)xydngxongnm1994, nhpchnhdi856m,cuTatara(NhtBn)nhp890mhonthnhnm1999lnhpln nht th k i vi cu dy vng. Hnh 1-9 Cu dy vng Normandie - Php Song song vi cu thp, cu b tng ct thp ng sut trc trong nhng nm na cui th k ny chim lnh mt v tr quan trng. Vi vic s dng vt liu c cng cao cng vi s pht trin ca cng ngh thi cng, kt cu nhp b tng ct thp ng sut trc n nay t c nhp hng trm mt. nc ta t nm 1960 tr li y, nht l sau khi gii phng min Nam 1975, tquc thng nht, nhiu cu thp v BTCT hin i c xy dng trong c nc. ng ch l 11 cu Thng Long qua sng Hng th H Ni di trn 5km gm kt cu nhp dn thp gia hai b sng xe la chy di, t chy trn, mt cu t rng 31m, h thng cu dn hai u cu l nhp BTCT d ng lc xy dng vo nm 70-80 ca th k XX ny. Nhng nm gn ycu BTCT c xydng vi nhiu cu ln nh cuQunHu Qung Bnh, cu Ph ng, cu Thanh Tr, cu Vnh Tuy H Ni v.v Cu dyvngcng cxy dng nhiu, cu MThun ni bc qua sng Tin Giang vi nhp chnh 350m c hai mt phng dy, cu Bi Chy Qung Ninh- cu dy vng mt mt phng dy nhp di nht th gii 435m. 2. Phng hng pht trin ca ngnh xy dng cu Chon nayngnh xydng cu t c nhngthnh tu to ln v nhiu phng din, t nhng vn v kt cu cng trnhn kthut cng ngh, s hon chnh ca l thuyt i i vi nhng nghin cu thc nghim m ra kh nng ng dng vo thc tin nhngcngtrnhtcccchtiukinhtkthuttt,khnngvtnhpngymt ln. Thc vy, mt trong nhng phng n vt eo bin Gibralta ni Ty Ban Nha vi Ma rclsdngcudyvngcs3100+8400+4700m chothynhngtinbvt bc trong lnh vc xy dng ni chung v ngnh cu ni ring. Phn tch cc cng trnh cu hin i xy dng trn th gii trong nhng nm gn y thy r cc khuynh hng: Vvtliu:sdngvtliucngcao(thpcngcao,thphpkim,btng mc cao) v vt liu nh (b tng ct liu nh, hp kim nhm), nhm mc ch gim khi lng vt liu v gim nh trng lng bn thn kt cu. Vktcu:sdngnhngktcuhplpdngccbinphpiuchnhngsut nhm tit kim vt liu. - Kt cu bn trc giao - Kt cu thp-b tng ct thp lin hp - Kt cu ng sut trc - Kt cu dm tit din hp - Cc s cu treo vi cc bin php tng cng cng, cu dy vng, cu khung dm b tng ct thp ng sut trc. V lin kt v ghp ni: s dng cc bin php lin kt ghp ni c cht lng cao, thc hin n gin, tit kim nh lin kt hn v bu lng cng cao cho kt cu thp, dn keo epoxy vi kt cu b tng. V cng ngh thi cng: c th ni nhng tin b v cng ngh thi cng ng mt vai tr c bit quan trng trong s pht trin ca ngnh xy dng cu trong thi gian gn y. Cc 12 cngnghthicngtintinnhlphng,chng,cNycngviccthitbcng ngh hin i mang li hiu qu cao v kinh t cng nh k thut. N goiralthuyttnhtonthitkvntiptccnghincuvhonchnh.Vi phng tin my tnh in t qu trnh tnh ton ngy cng t c chnh xc cao bng cchxt ti yhnccyu t nh hng (vt l, hnhhc). Bncnh , cc nghin cuthcnghimccaov tin hnhmtcchquym.Thctchothynhngkt qu thc nghim c ngha rt ln trong vic kim chng, b sung v hon thin l thuyt tnh ton. CHNG 2: NHNG VN C BN TRONG THIT K CU 2.1 CC GIAI ON THIT K CU Theoquynhhinhnh,trnhtutvxydngmtcngtrnhbaogm3giai on: - ChuNn b u t - Thc hin u t - Kt thc xy dng a d n vo khai thc s dng Tuy nhin trong qua trnh thc hin d n th vic thit k cng trnh cu cng phn theo giai on sau: 1. Giai on thit k s b: Thitksbthucbcnghincukhthi:laramtviphngncu,cc phngnkhcnhauvvtliuktcuv.vtrncsmtctacht,iukinthu lcthuvnvmtscctiliukhccxcnhttrc.Tchnramt phng n thch hp nht thit k k thut v xy dng. Ni dung cng vic thit k s b gm - V li bnh , trc dc v tr cu - Tnh ton lu lng, khNu cu, xc nh coa vai ng u cu - Phn chia nhp v lp cc phng n cu. C th dng kt cu nh hnh s b tnh khi lng, xc nh cc kch thc c bn. - Chn loi mng v m tr da vo tnh hnh a cht c th v s b xc nh cao y mng, kch thc c bn ca mng m tr. S b xc nh s lng cc v su chn cc.- Lp bng so snh khi lng v gi thnh cc b phn chnh ca cu - Khi so snh v ch tiu kinh t k thut cn phi ch nmc cng nghip ho, c gii ho thi cng, vic p dng cng ngh mi v c bit hin nay th rt cn ch n v p m quan ca cng trnh cu. 13 Nh vy sau khi so snh la chn c phng n thch hp, lc kt thc thit k s b v chuyn sang thit k k thut. 2. Thit k k thut: Phn thit k k thut gm thuyt minh tnh ton v bn v chi tit cc b phn. Phn thuyt minh tnh ton: Tin hnh tnh ton cc b phn cu nh kt cu nhp, m, tr theo cc trng thi gii hn, m bo cc cu kin kh nng chu ti c tt c cc ti trng tc dng.Phn bn v gm: Bn v b tr chung theo chnh din th hin cch b tr kt cu nhp c lan can tay vn, cch b tr m tr v mng. Theo ngang cu th hin b rng cu, b rng phn xe chy v ngngii,sdmchnhvthhinlinktngang.Cckchthccbncacu c ghi trn bn v b tr chung cng vi cc loi mc nc, cc loi cao nh cao vaingucu,caomtngxechy,caoydm,caonh,y mngBng vt liu ch yu cho cc cu kin. Bn v chi tit cu to cc b phn: Cu to mng, m, tr, cu to kt cu nhp, chi tit mt s kt cu lin quan. 3. Thit k t chc thi cng: Thitktchcthicnggm:Btrmtbngcngtrng,tnhtonbtrnhnlc, mymcthitbphcvthicng,tnhtonccktcuphtmphcvthicngnh: Chn my bm, tnh vng vy cc vn, tnh vn khun, chn ba ng cc. Trongphnnychyuccnvnhthuaraphngnthicng,tccnh thu t thit k ra cc bin php t chc thi cng cho tng hng mc ca cng trnh. 2.2 CC TI LIU KHO ST PHC V THIT K 1. La chn v tr cu i vi cu nh, cu trung ni chung v tr cu ph thuc vo tuyn ng, cn cu ln th v tr cu li quyt nh, cn la chn cNn thn gi thnh cng trnh r nht. Yu cu khi la chn v tr cu l: - Khng lm tng kinh ph xy dng ng qu mc - Cc ti liu a cht thy vn khu vc nh lm cu phi n nh - Ni dng sng hp nht v khng c kh nng i hng dng chy - Tim cu nn vung gc vi dng chy dng chy m thun - Bo m giao thng ng thy nu c - Phi phc v yu cu pht trin kinh t x hi hin ti v tng lai, phc v quc phng 14 2. Cc ti liu cn o c iu tra kho st v tr cu Sau khi la chn c v tr cu hp l ta cn tin hnh thu thp cc ti liu sau y ti v tr cu lm c s cho thit k. a) o v bnh khu vc k c khu vc d kin b tr cng trng v ng u cu, trc ngang sng ti v tr cu. Phm vi o c v pha thng lu bng hoc gp ri chiu rng sng v ma l, v pha thng lu bng chiu rng sng v ma l. b) iu tra thy vn: iu tra cc mc nc MNCN, MNTN, MNTT, MNTC v.vvn tc dng chy, dc lng sng, b rng dng sng, tnh hnh xi bi, vt tri cy tri vo mal. Cc ti liu nydng phc v tnh ton khNu cu, ngxi l, quyt nh chiu cao y dm, thit k cc cng trnh bo v v nn dng chy. N goi ra phi xc nh thng thng ng thy, cp sng, ti trng tu b, cc yu cu v lung lch.Nu sng nm trong khu vc gn bin cn iu tra v nh hng ca thy triu, kh nng xm thc v cc nh hng khc ca hi nc mn. c)iutraachtcngtrnh:baogmcngvicxcnhvtrcclkhoanv khoan thm d bit a cht ch xy dng, ly mu t v th nghim, xc nh cc ch tiu c l ca t, cui cng v c mt ct a cht ca sng ti v tr cu th hin chiu dycc lp t, loi t, cc tnh nng cl ca t v.vnhmphc v cho vic thit k nn mng, chn loi mng v chiu su t mng. Trong qu trnh iu tra cn pht hin c hintngctchy,ttrt,xingm,ncxmthc,phonghav.vkhng.Nhiu trng hp phi thay i thit k k thut v phng php thi cng do iu tra khng chnh xc khng y . d) iu tra kh tng: bao gm thi tit, kh hu, ma kh, ma ma bo, hng gi, tc gi, thi gian l, nhit cao nht, thp nht v.vCc ti liu ny rt cn cho vic b tr cng trng, vch tin thi cng v cng lin quan ti thit k k thut chng hn nh cn tnh lc gi, tnh nh hng bin dng do nhit v.v e)iutrakhnngcungcpnhnlc,nguynvtliuaphng,ccxnghip cng nghip c lin quan cung ng vt t, my mc, thit b, nng lng, phng tin vn chuyn, ng giao thng st, thy b, tnh hnh cung cp lng thc, thc phNm, cht t phc v sinh hot, tnh hnh an ninh chnh tr v phong tc tp qun a phng. Cc ti liu ny rt cn cho thit k t chc thi cng nhm rt ngn thi hn xy dng v h gi thnh cng trnh. 15 2.3 LA CHN CC CNG TRNH NHN TO- PHN CHIA HNP- KH THNG THUYN- KH CU 1. La chn cc cng trnh nhn to i vi dng nc nh, dng nc khng thng xuyn vi lu lng nh, chiu cao t p ln dng cng l hp l, cu to v thi cng n gin, tn t vt liu, mt ng li lin tc. Cng c th l cng n, cng i hay cng ba. ividngncthngxuynhaychuklulngtngilncngkhngth thot nc c, hoc so snh vi cng khng c hiu qu v kinh t k thut th nn xy dng cu nh.i vi dng nc lu lng ln hn (Q>100m3/s) ngi ta xy dng cu trung hay cu ln. i vi cu nh v cu trung thng lm cu BTCT . i vi cu ln thng lm cu thp, khi khng c iu kin xy dng tr c th lm cu treo.KhiachtlkhngblncthlmcuvmBTCT.ivicuvtng thng lm cu khung. 2. Phn chia nhp Vic phn chia nhp xut pht t yu cu kinh t k thut v thng thng. Trng hp c thng thng th nhp thng thng phi m bo quy nh ti thiu ph thuc vo cp sng. Cc nhp khc c th tham kho kt lun sau y: gi thnh mt nhp (khng k phn xe chy) bng gi thnh mt tr l kinh t nht Trng hp sng khng thng thng cng phi xc nh v tr ca hai tr nhp bc qua dng ch trc trnh lm tr gia dng ch. T xc nh v tr ca cc tr khc v v tr m. Khong cch cc m tr phi m bo iu kin thot nc. 3. Kh thng thuyn Nu trn sng c cc phng tin giao thng ng thy th phi b tr mt s nhp thng thuyn. Thng thng l hai nhp, nhp xui dng v nhp ngc dng. Nu iu kin dng sng khng chophp b trhai nhp thng thuyn th cth chtmt nhp vi khNu ly theo kch thc ca kh xui dng. N hp thng thuyn trc ht b tr dng ch. Kh thngthuynlkhongkhnggiantrngdigmcudnhchoccphngtingiao thng ng thy qua li mt cch an ton m khng mt kt cu no c php vi phm m bo an ton giao thng. Kh thng thuyn phi c t lt di gm cu cao MNTT. Cc kch thc ca kh thng thuyn quy nh ph thuc vo loi phng tin vn ti v cp sng. 16 Bng 2-1 Chiu rng B (m)Chiu cao (m) N hp xui N hp ngc phn gia nhp H gi hCp sng su m bo thng thuyn (m)Khng nh hnKhng nh hn I>2.014012013.55.0 II1.6-2.614010012.5(10.0)4.0 III1.1-2.01208010.03.5 IV0.8-1.4806010.0(7.0)2.5 V0.6-1.160407.02.0 VI0.45-0.840(30)203.51.5 VII=0.25m >=0.5mXp xe xch>=0.1mXp xe t Hnh 2-6 S xp xe ngang cu Cthdngtitrngtngngthaythchovicxpxe.Trstitrngtng ngngviccdngngnhhngvngvimiloihotticnutrongquy trnh 22TCN18-79 3. Hot ti thng ng ca on ngi Theo quy trnh 22TCN18-79 hot ti tiu chuNn ca on ngi l 300KG/m2. Khi kim tra vi xe xch vxe bnh nng cbit nng th khng tnh ti trng on ngi. Khi tnh vn ng ngi ibng g phi kim tra thm lc tp trung 180kG, lc Nyvo tay vn theo phng ngang v thng ng l 130kG. 4. Cc tc dng ca hot ti v ti trng khc Lc xung kch v h s xung kch Xt n vic tng ln ca ti trng do yu t lch tm ca ng c t, gh gh ca mt ngmgyravacham,chnng.Lctngthmgillcxungkch.Cuthp 22 trng lng bn thn nh, tnh n hi ln nn tc dng xung kch ln hn cu BTCT. H sxungkchphthucvoloivtliu,bphnktcu,sktcu,loititrngv chiu di t ti. i vi kt cu nhp cu thp trn ng t (tr cu treo): (1+) = 1+ + 5 . 3715 i vi kt cu nhp cu dm b tng ct thp trn ng t: (1+) =1.3 khi 5m; =1.0 khi 50m; ni suy khi 5=40cm(120-150)cm>=4m70-80 Hnh 3-5 Kch thc x m Thng thng x m gm hai loi: X m lp ghp:khi sn xut ngi ta v tr cc hnh chp ct, kch thc pha di rng hn kch thc cc l 5cm . X m ti ch: sau khi ng cc n cao thit k ngi ta p u coc khong 30-40cmun ct thp to ra bn pha ri dngct thp ai d=6mm qun li sau dng ct thp ch d=(2024)mm v ct thp xin dng vn khun v b tng 25516 Hnh 3-6 Cu to x m lp ghp 40 >40cm25 25Ct chi m men dngCt chi m men m Ct xinct ch ca ccct thp u ccct ai Hnh 3-7 Ct thpx m ti ch 2. Tr do dng ct S dng:Khi tr cao hn 6m, chiu di nhp l = 3040m, vn tc nc Vnc > 1m/s. Cu to: Tr c th c 1, 2 hay nhiu ct thu thuc vo kh cu v kch thc ct. Ct c th c titdinvung, ch nht hoc hnh vnh khn.Ctvnhkhnhayc s dng do gim c trng lng khi lp rp. 5m10-12cm120-140cm1m70cm(0.8-1)mB tng >30cm Hnh 3-8 Tr do dng ct Ct c th t trc tip ln mng chung hay ring hoc trc tip ln t nn nu tng t kh nng chu lc. II. Cu to m do Thng c dng cc, ct, tng mng. 1. M do dng cc Phm vi s dng: Khi chiu cao t p H 6m, chiu di nhp l < 40m l loi dng ph bin v n gin nht. Cu to: Cng ging nh tr do n gm cc, x m nhng khc tr l trn x m c tng nh v tng cnh c nhim v chn t. Khi chiu cao t p H 2m, l 20m c th ch dng 1 hng cc. 41 Khi H, l ln b tr thm 1 hng cc xin. X m c chiu cao h 40 cm, ton b thn cc nm trong t p nn m v c kch thc nh cc ca tr do. H < 6mH < 2m> 40cmTng nhTng cnh Tng cnhTng nh> 40cm Hnh 3-9 Cu to m do dng cc 2. M do dng ct p dng khi cu c chiu cao t p len ti 3m Ging nh m do dng cc nhng do ng knh ct ln hn nn c th m ch cn 2 ct ng. Khi l (1215)m dng 2 cc ng 0.8m ng su 8m. Khi l (1824)m dng 2 cc ng 1.0m ng su 12m. H < 3mTng nhTng cnh> 40cm Hnh 3-10 Cu to m do dng ct 3. M c dng tng chn Khichiucaotpvchiudinhpkhnglnlm,nhtlcuvtng,cu trong thnh ph c th dng m dng tng mng bng BTCT c sn tng cng tam gic. u im ca loi ny l t p pha trc khng ln vo phn khng gian di gm cucc cu vt ng gim c chiu di nhp. Khi H, l ln, m c cu to cc tng song song ring r gim p lc t ln tng, nhg nhc im l nu nn p khng tt t s chui ra pha trc. 42 b)Tng chnSn tng cngTng cnh(35-40)cmTng dcTng cnha)Tng nh Hnh 3-11 Cu to m dng tng chn a) Tng chn dc b) Tng chn ngang 3. CU TO M, TR CNG I. Cu to tr cng Tr cng gm 3 b phn chnh: M, thn v mng tr. Trn nhng sng c dng nc chyxithocckhnngvapcatub,cytricthtbphnchngvax cho tr. 1. M tr M tr chu ti trng trc tip t kt cu nhp v truyn xung thn tr. Kt cu nhp ta trn m tr thng qua gi cu. Ti ch t gi cu, m tr thng b tr li ct thp chu ng sut cc b c bc (5 5) cm. Mt trn ca m tr phi to dc t nht 1:10 thot nc. B tng m tr thng s dng M250 hoc M300. Cu to: 1:10 Hnh 3.12 43 M tr dng b tr gi cu phn phi p lc cho tr to cho nc ma khng chy trnthntr,cchiudytithiul40cm,trninhmttodcthotnc 1:10,phnftipgipgiamtrvthntr>10cmdicrnhncmakhng chy ln thn tr> 40cm(1-3)m> 40cm10-15cm> 1m0.6-1ma)b)c) Hnh 3-13 Mt s dng tr cng thng gp a) Tr c thn hp; b) Tr c thn rng; c) Tr thn ct Ct thp ca m tr c b tr ph thuc vo cu to thn tr +Trcthnrng:ctthpmtrttheocuto:gmhailictthptrnv di ng knh D10 b tr cch nhau 200-250mm chng ng sut cc b N1N2d=10@200-250N1@200-250d=10N2 Hnh 3.14 Ct thp m tr c thn rng 44 + Tr c thn hp: ct thp m tr phn hng phi c t theo tnh ton. S tnh: Dm ngm mt u Ti trng: Trng lng bn thn m tr Trng lng k gi Phn lc gi do tnh ti: Rt Phn lc gi do hot ti: Rh (c xt n h s phn b ngang) d=22-30d=8-10d=12-14d=14-22N1 ( CT chu ko)N2 ( CT ai)N3 ( CT dc ph)N4 ( CT chu nn)@120-200N211 - 11 Hnh 3.15 Ct thp m tr c thn hp + Trthnct:pdngtrongcudnthpcngxechydi,cudmnhpl= 2030m. Ct thp chu lc ca xm thng c ng knh d=20mm,c b tr nh sau: d=18-32cmId=6mmI - Id=12-14mmICtd=80-200(300)cm Hnh 3.16 Ct thp m tr thn ct k gi bng BTCT M300, c li ct thp theo tnh ton. Li ct thp thng c cc kch thc sau: d = (812)mm c khi n 14mm @ = (8080 120120)mm Khongcchccliphithomnyucucutotckhongcchccli (5070)mm. Kch thc c bn ca m tr cng : c mt ct HCN hai u vt trn 45 c2c1ac1c2n1n2bBAc c m c c m Hnh 3.17 Theochiudccu:kchthcmtrAphthucvokhongcchhaigiaph thuc vo kch y gi n1 ,n2 v c1 , c2 c1=15-20cm c2= 30-50cmA=a + n1/2 +n2/2 +2 c1 +2 c2 Theo phng dc cu: B ph thuc vo khong cnh hai tim gi 2 dm bin b ,y gi m ,kch thc c=15-20cm B=b+ m + 2c +A 2. Thn tr Thn tr lm nhim v truyn p lc t m tr xung mng v chu cc lc ngang theo phngdccuvngangcu.Mtctngangcatrtrongphmvilngsngphic dngrnctt.Thntrphichucvapdocytri,ccnhpctuthuyn qua li cn phi chu c va ca tu. Hnh dng mt ct ngang thn tr ph thuc vo iu kin dng chy di cu. - Gim xi l lng sng v h chiu cao nc dng thng lu cup dng cho cu cn, tr hai ctRb- Trnh to thnh cc dng soy ngm gn tr- Dng chy mnhR = b / 2D thi cng Thn tr ch nht p dng cho cu cn Hnh 3.18 Mt s mt ct ngang thn tr Sn bn c th nghing 46 140(20) Hnh 3.19 Sn nghing Trcuhinicsnbnthng.Titdintrcchntheotitdintrnnh mng Mt s loi thn tr khc cng c s dng: + Tr thn c rng ( B tng, xy hoc BTCT) 300cm(75-100)cm Hnh 3.20 Tr rng lng3. Mng tr - Mng tr c nhim v truyn ti trng t thn tr m xung t nn bn di v xung quanh. Ngoi ra mng tr cn c nhim v phn b lc t thn tr xung 1 din tch rng hn m bo chu lc cho t nnvn nh cho tr. su tmng cn phi m bocho tr khng bmt n nh, nghing lchhoc b phhoi do xi l gy ra. u trn ca cc phi c ngm vo trong b hay x m BTCT mt tr s theo tnh ton ng thi phi ngp su vo trong b mt on khng nh hn2 ln chiu dy thn cc, vi cc cc ng knh d 60cm th khng c nh hn 1.2m. Vi cc cc cho ct thp chn vo trong b th cc phi ngm vo b (1015)cm v ct thp nm trong b t nht l 20 ln ng knh ct thp g v 40 ln ng knh ct thp trn trn. - Kch thc: (HV) m bo s truyn ti trng ng u xung cc cc th chiu dy b phi 2m. khong cch t hng cc ngoi cng n mp ngoi ca b mng ti thiu l 25cm47 >25cm >25cm

2m Hnh 3.21 Cu to mng tr a)Mng cc ng b)Mng cc ng knh ln - Cao nh mng: Ph thuc vo iu kin a cht, a hnh, kinh nghim ca ngi thit k. +Numngnng:Caonhmngphinmnganghocdimtttnhin khong (0.51)m. + Nu l mng cc: B thp: y mng n ng xi l phi tho mn h hmin

( t xung quanh mng chu c lc ngang) B cao: Cao y b, cao nh mng nm v tr bt kMNTN MNTN0.5mhmina)b) Hnh 3.22 Cao nh mng-Cao y mng: + Nu mng nng: y mong phi nmg di ng xi l 2.5m. + Nu l mng cc: Cc phi cm vo tng t chu lc 4m. II. Cu to m cng Trongcngtrnhcu,mthucktcuphndicchntrongt,nmtrong vng Nm t chu xm thc ca xi l. M c cc chc nng c bn: 48 - kt cu nhp - Chu ti trng thng ng v nm ngang t kt cu nhp truyn xung - Chu p lc t Ny ngang - B phn chuyn tip v bo m xe chy m thun t ng vo cu - m bo chng xi l b sng Cu to chung m: PM mTng thnB mTng nh Tng cnhNn m Hnh 3.23 Cu to chung m 1.Cutomchnht:ldngmcunginnhtdngvtknhmng nh,dngkhichiucaotpH60cm Hnh 3.25 Cu to tng nh M m:BTCT hm 40cm hm(10-15)cm (10-20)cmHb101Bcu(10-15)cm (10-15)cm Hnh 3.26 Mt ct dc, ngang m Tng thn: 50 + Chiu dy timt ctnhmng: b = (0.350.4)H. Vim c chiu cao t p H > 8m tng trc c th nghing 10:1. + Chiu di (ngang cu) = [ Bcu 2(1015)cm ]. Tng cnh: - Theo phng dc cu: + Xc nh chiu di tng cnh cn c vo: dc taluy nn m 1:n ngp su ca tng cnh m vo nn ng (s) hm1 : nsLc >60cm80-100cm(0.35-0.4)Ha)b) Hnh 3.27 Tng cnh dc cu Theo Quy trnh:H 6m dc 1:1 ( cu t) H >(612)m dc 1:1.25 ( cu t)H 6m s =0.65m H > 6m s = (0.751)m C th xc nh chiu di tng cnh theo cng thc sau: Lc = n H + s + on thng tng cnh: Tu ngi thit k. C th ly = h1 hoc = (80100)cm. + on xin: (6:1) (4:1) - Theo phng ngang cu: Kch thc tng cnh nh hnh v B mng: Ging tr b) Cu to m U BTCT Cc kch thc c xc nh tng t nh trn. Tng mng hn do c b tr ct thp:Tng nh: b1 =(3050)cm Tng cnh: bc = (4050)cm Tng cnh c hng ln: (1:1) (1:1.5) 51 h11:1.5(3-4)m1:1(30-50)cmP Hnh 3.28 Cu to m U BTCT 2. Cu to m vi: MchUvncthgimkhilngbngcch:tungcchsdngtngcnh ngn bng BTCT,tung trc c thay bng cc ct ta c loi m vi , khi m s hon ton chn trong t .a) M vi b tng, xy S dng khi chiu cao t p H = (520)m Gi A l giao im tng trc v m m. QT 79 quy nh: Taluy nn m phi cch A 1 khong 30cm B l giao im tng trc v nn m: B phi cao hn MNCN 25cm 70-100A170-100101(0.35-0.4)HB3-675-100 >60 Hnh 3.29 Cu to m vi b tng, xy b) M vi BTCT M vi BTCT thng c 2 loi:M vi tng dc M vi tng ngang + Tng nh, tng cnh dy 30cm. + Tng thn:. Nhiu tng: bt = (3540)cm . 2 tng: bt = (70100)cm. 52 Tng dca)bt2.5130bt30b)a 0.3a Hnh 3.20 Cu to m vi BTCT a) Nhiu tng dc b) Hai tng dc 3.4 KHI NIM TNH TON M, TR CU 1. Khi nim chung Khithitkmtrcutrchttachnloimtrcu,sbxcnhkchthc cc tit din. Sau tin hnh theo cc trnh t sau: - Chn s tnh ton. - Xc nh cc loi ti trng i vi tit din cn tnh ton ca cc b phn m tr. - Lpccthptitrngnhmxcnhcctrsnilcbtlirtckhnngxut hin trong qu trnh xy dng v khai thc cng trnh. - Kim tra li cc tit din theo cc trng thi gii hn. 2. Cc ti trng tc dng ln m tr cu a) Trng lng bn thn Xc nh theo kch thc hnh bao ca cc bn v k thut. Khi tnh ton nn chia m trthnhcckhihnhhcngintnhthtch,trnglngvcnhtaynt trng tm ca cc khi ny n 1 trc no cn tnh mmen hoc theo s tay thit k, thit k nh hnhCng thc tnh ton:Q = V (T) Trong : - trng lng ring ca vt liu V - th tch m tr Khi b phn m tr nm di nc (k c nc ngm) khi tnh n nh phi xt n tc dng ca p lc thu tnh. Khi trng lng ring l: = - 1 (T/m3) 53 b) Phn lc gi di tc dng ca trng lng bn thn kt cu nhp Xc nh da vo thit k c th ca gi cu v kt cu nhp.c) Trng lng t p: Trng lng ca t p trn cc b mng v cc thnh nghing ca tr m: P = H(T/m2) Trong : - trng lng ring ca t, = 1.8 T/m3. H - chiu cao t p. Ch : trng lng t p nm di mc nc cng phi tnh gim do p lc thu tnhcng thc xc nh: =1 / (1 + e) * ( - 0 ) trong : e h s rng ca t : t trng ca t ly bng 2.7/m3 0 : trng lng ca nc ly bng 1 d) p lc ngang ca t Rtquantrngkhitnhm.ivitrthtuloi,cthtnhhockhngtnhtu theo mc nh hng. Theo QT 79 p lc Ny ngang tnh theo cng thc: ep = tc H Trong :H chiu cao tng t tnh ton. = 245 tgo 2 - h s p lc ngang ca t. , tc gc ma st trong, dung trng th tch ca t. Khi ymng tcchmt t t nhin 3m coi p lc Nyngang ca t phn b theo quy lut ng thng Hp lc Ny ngang tnh theo cng thc: HB e21Ep=Trong :ep v H - p lc nm ngang ca t v chiu cao tng t. B - chiu rng tnh i ca m B xc nh nh sau: b1 2b2B = b. b1 > 2b2B =2 b2. 54 Vi m cc (ct) nu chiu rng tng cng cc cc (ct) < 1/2 chiu rng m tr th B =2 b( b chiu rng cc hoc ct) Vi m cc (ct) nu chiu rng tng cng cc cc (ct) 1/2 chiu rng m tr th B ly bng khong cch mp ngoi ca cc (ct). EH/3H< 3mb2 b1 b2b Hnh 3.21 p lc ngang ln m Cnh tay n ca hp lc cch y mng 1 khong: e = H / 3 e) Phn lc gi do hot ti thng ng gy ra Tu theo hot tinthit k : ti trng t, xe bnh xch, nghi s tnh c phn lc gi do ngi gy ra Xcnhphnlcnybngcchxptitrctiphocdngtitrngtngng xp xe ln ng nh hng phn lc gi. i vi ti trng t v ngi i b xc nh phn lc gi theo cng thc sau: R = ko o (1+) +pn n Trong :ko - ti trng tng ng ca 1 ln xe t tiu chuNn o - h s phn b ngang ca t xung cc gi cu - h s ln xe1+ - h s xung kch pn - trng lng ngi i trn 1m2 l ng n - h s phn b ngang ca ngi - din tch ng nh hng phn lc gi f) p lc ngang ca t do hot ti thng ng trn lng th trt M khng c bn qu : 55 Khi tnh ton p lc ngang do hot ti tc dng ln m, trng lng ca 1 trc bnh xe t c coi nh phn b utrn 1din tch (sb) v c thay th bng trng lng ca ct t tng ng c chiu cao ho. = . b . sPho Trong : - trng lng ring ca t P tng ti trng trn din tch (sb) bs0.2m Hnh 3.22S tc dng 1 trc bnh xe t (2ln xe) M c bn qu : Lc ny hot ti s phn b qua bn qu xung nn t. Trong cng thc tnh ho: s kch thc bn theo chiu ngang b = lb / 2 P tng hot ti ln bn PLbBn qu S Hnh 3.23 M c bn qu 56 Tnh p lc t + Xc nh chiu di lng th trt gi nh: =245 Htg loo + T lo tin hnh xp bnh xe + Tm c s tnhtra bng (sch M tr Cu ) tnh c p lc t do hot ti gy ra. g) Lc lc ngang Coi nh phn b u, tc dng theo phng ngang cu t nh mt ng xe chy. S = 0.2 T/m-on xe H10, H13. S = 0.4 T/m-on xe H30. h) Lc ly tmi vi m tr ca cc loi cu nm trn ng cong, c bn knh R cn phi tnh lc lytmcoinhtitrngnmngang,phnbutcdngnhmtphnxechy, trong cu ng t ch tnh lc ly tm khi bn knh R 600 m lc ly tm ca mt ln xe c tnh nh sau: C=15/(100+R)*l/L trong :R bn knh (km) l: tng chiu di ng nh hng nhng khng ln hn chiu di nhp L i) Lc hmi vi cu ng t v cu thnh ph vi mt ln xe theomt hng th ly di dng lc tp trung, t cao nh mt ng xe chy bng : 0.3p, 0.6P, 0.9P khi chiu di t ti tng ng 25m ,25m 50m v 50m,trong P l trng lng chic t nng trong on xe. Khi c nhiu ln xe theo mt hng th lc hm tnh vi tt c cc ln. - i vi cu dm th lc hm phanh truyn ton b cho gi c nh v truyn cho gi di ng 50% vi gi tip tuyn v 25% vi gi con ln - trng hp trn tr t hai gi khc nhau th lc hm phanh truyn ln tr ly bng tt ccclctruynlnginhngkhngclnhnlchmphanhtrongtrnghp hot ti cht ln nhp ln c gi c nh t ln tr - im t lc hm phanh khi tnh tr ly trng tm gi j) Lc gi57 Theohngngangculybng180kg/m2(khikhngchotti)vlybng 50kg/m2 (khi c hot ti) Khng tnh lc gi tc dng ln t k) Lc va tu b Ly theo quy trnh ph thuc vo ti trng ton phn ca tu b (tham kho bnh 4-4 trang 119 gio trnh m tr cu Trng HGTVT) l) nh hng lc xung kch: khi tnh m tr cng khng tnh n nh hng ca lc xung kch nhng tnh m tr do phi xt n lc xung kch 3. Cc t hp ti trng a) Cc t hp chnh: bao gm cc ti trong thng xuyn tc dng ln cu nh tnh ti, hot ti, lc li tm. b) Cc t hp ph: bao gm cc ti trong thuc t hp ti chnh v mt s ti trng cn li tr i ti trng ng t v ti trng do thi cng. Khi tnh xt theo c phng dc cu v ngang cu. c) T hp c bit: bao gm lc ng t hoc ti trng trong thi cng cng xy ra vi cc loi ti trng khc. 4. S v ni dung tnh tr cng: Ni dung tnh tr cng bao gm: - tnh kh nng chu lc ca thn tr ti tit din y thn tr v tnh kh nng chu lc ca t nn di y mng theo ti trng tnh ton ( t chn loi mng) - Kimtralchtmcavtrhplccacclcthngngnhmngvy mng nhm m ba chnh lch ng sut khng ln- Kim tra n nh chng lt chng trt theo ti trng tnh tonTin hnh tnh ton: - Xc nh kch thc tr. - Chn s tnh: Xt vi hai s cht ti: - hot ti cht mt nhp - hot tin cht hai nhp - Xc nh cc ti trng tc dng ln tr 58 i iii iib2b1ll A1 A2G1G2+ +1 1ah A1 ah A2gk1cht ti 1 nhpgcht ti 2 nhpk2d dc1c2Tf f Hnh 3.24 S tnh phn lc gi do hot ti trn kt cu nhp - T hp ti trng tc dng ln mt ct y mng: + T hp ti trng chnh dc cu gm: -- hot ti tc dng ln 1 v 2 nhp (k1 v k2 l ti trng tng ng) - - tnh ti (g). i ig 1a 1Th 1neyL c h t m n g t mh 1ig 1ia 2T Hnh 3.25 S tnh tr G1: Tng tnh ti thng ng tc dng ln tit din I-I A1,A2 : Tng hot ti thng ng tc dng ln tit din I-I (theo s 1v 2) Ta c :A1=no* k*m + nng*png*d*2)* A2=no* k*m + nng*png*d*2)*2* Trong :n l h s vt ti ca t ly bng 1.4 K : ti trng tng ng ca t m: s on t trn cu Png : ti trng on ngi (thgn ly bng 0.4T/m2) d: b rng l ngii 59 : din tch ng nh hng (m2) Vi s 1: max=(G1+A1)/F + A1*f / W Rubt etco = (Atc1* f ) / (G1tc+A1tc) 0.5*y Vi s 2: max=(G1+A1)/F Rubt Trong : N tng cc ti trng thng ng tiu chuNne0 : lch tm ca N y: khong cch t trng tm tit din I-I n mp tit din F: din tch tit dinW : m men chng un ca rit din i vi trc trung ho + T hp ti trng ph: - gm c ti trng hm phanh (T) - hot ti tc dng ln 1 v 2 nhp (k1 v k2 l ti trng tng ng) - tnh ti (g). trong t hp ny luc hm phanh gy mt n nh lt v trt nn h s vt ti ly bn 0.8 n(bt li) Vi s 1: max=(G1+A1)/F + (A1*f +T*h1)/ W Rubt eotc = (Atc1* f +Ttc * h1) / (G1tc+A1tc) 0.6*y Vi s 2: max=(G1+A1)/F +T*h1/ W Rubt Trong : T=0.8*n*(0.3*p)*mP: l trng lng ca xe nng trong onTnh tr ti tit din II-II: Trnh tv cc bc tnh ton tng t nhtit din I-I ch kghc ti trng thm G3 l ti trng do trng lng t trn g mng + Duyt n nh lt: (duyt vi im mp ngoi cng ca mt ct II-II) Ml/Mg < m (m=0.7 h s iu kin lm vic) Trong Ml m men lt : Ml =T*h Mg m men gi : Mg =N(y-eo)(trong N l tng cc lc thng ng) + Duyt n nh chng trt: mTTghtr mPTii(m=0.8h s iu kin lm vic) 60 - h s ma st. 5. Tnh m cu - Xc nh kch thc m - S tnh ah Al 1+k1ATd dgHE1H/3E2GQ2Q1Q2Q5Q4b Hnh 3.27 S tnh m Bc 1: xc nh v tnh cc ti trng: - -ti trng thng ng:Hot ti : gm xe t + on ngi (hoc xe bnh c bit)hot ti t tiu chuNn : A 1tc = k*m*hot ti t tiu chuNn : A 1tt = 0.8*n* k*m*on ngi tiu chuNn: A2tt =2-*d*png* ng Tnh ti : Trng lng m Q1, trng lng t v nn ng trn m Q2,Trng lngt trn g mng Q4 v Q5, trng lng gi vkt cu nhp G -ti trng nm ngang: . Lc hm phanhT=0.8*n*(0.3*p)*m . p lc ngang ca tE1=0.5*tc*tc*H2 *b trong : tc= tg2(45o-/2) tc = tg2(45o-tc/2) . p lc ngang ca t do hot ti gy ra: E2 (xem li phn ti trng) bc 2: tnh kim tra m: kim tra lch tm ca hp lc thng ng ti ti tit din nh mng v y mng: 61 ti nh mng: eo= y e1 0.8 khi ch tnh vi tnh ti 0.8 khi tnh vi t hp ph ng sut ti tit din y mng : uRWMFN + = ti y mng: etco = y e1tc0.8 khi ch tnh vi tnh ti khi tnh vi t hp tI trng ph: etco = y e1tc1.2 khi ch tnh vi nhp nh khi ch tnh vi nhp trung v ln trong : =M/W kim tra ng sut ti tit din y mng :khi tnh vi t hp c bndRWMFN + = khi tnh vi t hp ph dRWMFN2 .. 1 + = kim tra chng lt v trt ti y h mng: tnh theo tI trng tnh tona.chng lt : Ml / Mg m=0.7 b.chng trt: 8 . 0* *= = =mNTPTTTiiltr Chng 4 GI CU DM 4.1. KHI NIM V GI CU 1. Khi nim chung v gi cu Gi cu lm nhim v truyn p lc tp trung t KCN xung m tr v m bo cho KCN quay hoc di ng t do di tc dng ca hot ti v nhit thay i, ngoi ra gi cu cn kcn c nh v tr nhp v phng ca phn lc62 Cn c vo tnh cht lm vic gid cu c hai loi : gi c nh v gi di ngGicnh:chophpkcnchcchuynvxoaykhngchochuynvtheohai phng ngang v ngGi di ng : cho php u kcn va cp th chuyn v xoay va c th chuyn v theo c hai phng nganggi c nhgi di ng b tr gi cu : trong cu ng t v cu ng thnh ph theo phng dc cu thng b tr mt u gi c nh u kia gi di ngtrong cu gin n nhiu nhp trn mi tr thng b tr gi di ng ca nhp ny v gi c nh ca nhp kia i vi tr c chiu cao ln thng b tr hai gi di ng gim lc Ny ngang, khng nn b tr gi c nh tr cao gim lc Ny ngang cho tr gy bt li i vi cu mt tha vic b tr gi phc tp hn ,gi c nh v gi di ng b tr sao cho gi di ng t trn mt tha khng gy chuyeenr v ln . oi khi trn dm mt tha cn b tr gi chuyn v do nhit trn ton cu truyn v mt pha ,khi dm eo b tr gi di ng c hai u cn trong mt nhp chnh t gi c nh trn m tt c cc gi cn li l gi di ngkhibrngcukhnglnlm(b12m)b tr gi nh hnh vA BD Cb12mDACB Hnh 4.1 B tr gi trn mt bng 63 4.2.CU TO CC LOI GI CU cu to gi cu ph thuc vo tr s p lc truyn ln gi , i vi gi di ng cn ph thuc vo dch chuyn ca kt cu nhp ,chiu di nhp cng ln cu to gi cu cn phi hon chnh m bo chuyn v v xoay t do cu u kt cu nhp 1. cu to gi cu dm BTCT: trong kt cu nhp cu bn v cu dm gin n nhp < 9m i vi cu t c th dng cc bn m n hi lm gi. Vi nhp t 9-18mvi cu t dng gi tip tuyn -cu to gi tip tuyn: Gm2bnthpcchiudy(3050)mmgilthtgi.Thttrnl1tmthp phng c hn vo cc thanh thp neo chn sn trong dm BTCT. Tht di c mt tip xc hnh tr c lin kt vi thanh thp chn sn trong k gi. Cu to gi c nh v gi di ng ch khc nhau ch: Gi c nh c cht hoc vu ngn cn chuyn v ca thttrnivithtdi.Gidingctbnnpsnbnngnkhngcho tht trn chuyn v theo phng ngang so vi tht di. Gi c th chu c phn lc gi n 300T 124 351 Hnh 4.2 Gi tip tuyn 1. Thp neo;2. Tht trn;3. Tht di;4. Thp bn;5. Cht thp (gi c nh) Khi chiu di nhp ln hn 18-20m m bo chuyn v t do ca KCN p dng gi di ng con ln BTCT, gi thp hn c con ln ct vt hc con ln trn - Cu to gi conln BTCT: Gm 2 tm thp b mt hnh tr, gia l khi BTCT c M300400, chiu cao gi khong (6070)cm.Gi c th chu c phn lc n 80T 64 BTCT hg Hnh 4.3 Gi con ln BTCT Cu to gi con ln thp ct vt: - p dng: Dng cho nhp L = (3033)m. - Cu to: Ging nh gi con ln BTCT nhng khi BTCT c thay bng con ln thp vt gc. Con ln thp ct vthg Hnh 4.4 Gi con ln thp ct vt HinnaytrongcuBTCTngtgicaosuocpdngrtrngridocu im sau: Titkimthp,chiucaonhchtovbodngddng,lmgimchnng gia cc mt tip xc v gi cu hu nh khng cn bo dngGicaosuhinnayangdngnctachailoichnh* Gi cao su phng -p dng: Dng cho nhp L = (3033)m. -Cuto:Cccbnthpdy5mmnmgiacclpcaosu.Ccbnthpctc dng nh cc ct thpngn cn v tng cng ca gi khi chu phn lc thng ng. Nh tnh cht n hi ca cao su, tit din u dm c th chuyn v xoay v chuyn v trt. Gi c dng hnh trn hoc hnh ch nht. Hinnayloigicaosuphngchnhtcsdngtrongdncitonngcp cc cu trn Quc l 1. Gi c th chu ti trng n 200T. * Gi cao su hnh chu - p dng: Dng cho nhp L =(40130)m. 65 - Cu to: Gm 1 tm cao su hnh trn (1) t trong 1 b phn bng thp hnh chu (2). Trong gi di ng chuyn v trt ca gi do tm teflon PTFE (polytetra fluoroethylene). Tm trt teflon PTFE c t trong khc lm ca bn thp. Trn mt tm trt PTFE l 1 l thp hp kim c chiu dy 1mm. Tm PTFE c chiu dy t (48)mm. gi di ng c theo 1phng ngi ta t thm 1 bn np d hng. Gi c nh c np y di. Chu c phn lc gi (1002600)T Giong cao suNp yChu thpL thp hp kimTm cao sua PTFEBn trt thp Hnh 4.5 Gi cao su hnh chu 2. Cu to gi cu dm (dn) thp: Trongcuthpcnhpdi25mthngpdngloigitiptuyngingnhcu dm BTCT Khi nhp >25m phn lc gi t(70-80) T n (250-300) T nn dng loi gi di ng con ln - Cu to: (hnh v) S con ln khng nn qu 4. a)Bu lng neoTht trnb) Hnh 4.6 Gi cu thp nhp ln a) Gi c nh b)Gi di ng 66 4.3. NI DUNG TNH TON GI BN PHNG V GI TIP TUYNTasnghincucchtnhgibnthpphngvgitiptuynlhailoicp dng nhiu cho cu BTCT nhp < 20m v cu thp nhp 20m. - Cc thanh g trong dn lin kt vi nhau bng cht g, cht thp hoc inh tn, c th lp rp trong xng 71 (6)- Tr (cc)(14)- KCN dng dn(6)(14)MNTN Hnh 5.3-S cu dn Gao Giurapxki 5.2.CU TO CU DM G GIN N NHNP NH A. Cu to chung I. Kt cu nhp Dm dc l b phn cu lc chnh ca cu, dm dc c tc dng vt nhp v mt xe chy (trong c tay vn, va h) C 2 cch b tr dm dc: dm dc t dy (hnh 5.4) v dm dc t tha (hnh 5.5) 1.Dm dc t dy C ly tim 2 dm dc gn nhau (k nhau) khng ln, t 0.56m. Kt cu phn xe chy n gin, chiu cao kin trc b. Khi chiudi nhp l=(56)m, thng lm 1 tng; Khi lnhp >(56)m, lm 2 tng. X m chu un Hnh 5.4-B b tr dm dc t dy 2.Dm dc t tha Clytim2dmdcgnnhau(knhau)tngilnln,t1.51.8m(thpnht 11.2m). Dm dc t trc tip ln x m ti v tr cc ca m tr. Do vy x m khng chu un, nhng kt cu phn xe chy phc tp, chiu cao kin trc ln. Trn dm dc l ngang (dm ngang) t dy, trn ngang l vn mt cu. Dm dc t tha thng lm t 2 hay 3 cy g chng ln nhau lin kt vi nhau bng bu lng v cht bn chng x dch gia chng. M men qun tnh ca dm dc bng tng m men qun tnh ca cc cy g tp nn dm Chiu cao dm dc t tha tng i ln v vy cn cu to lin kt ngang chc chn m bo n nh: dng 2 thanh g ng kp cc cy g chng ln nhau; giadm dc dng lin kt cho hoc then chng ngang. 72 Ch : Khi khNu cu6m nn lm cu 1 nhp; Khi >6m, lm cu nhiu nhp Hnh 5.5 -B b tr dm dc t tha II. M tr M tr cu g thng lm v gi cc hay v gi k 1.V gi cc c dng phbin nht. Gmmt s cc ng xung t, chiu du ng cc vo t do tnh tnh ton quyt nh nhng khng c nh hn 3.54m. Trn nh cc dng thanh ngang lin kt gi l x m, khong cch H t y x m n MNTN (mc nc thp nht) hay mt t thin nhin c gi l chiu cao v gi. Dm dc t trc tip ln x m S cc trong 1 v gi ph thuc b rng cu v ti trng tnh ton. C ly tim 2 dm cc lin nhau thng t 1.52m tng cng cng ca v gi cc, ngi ta dng kp ngang v kp cho. Ty theo chiu cao H m s dng cc loi kp Khi H 2m, khng cn s dng kp (hnh 5.6a) Khi H = 23m, s dng kp ngang (hnh 5.6b) Khi H = 34m, s dng kp ngang v kp cho (hnh 5.6c) Khi H 5m, s dng kp ngang, kp cho v chng xin ta ln trn cc ph (hnh5.6d).Trnghpnycththaychngxinvccxin,nhvys tng c cng v gim c s lng cc (hnh 5.6e) Khi khong cch gia 2 cc ng ngoi cng l B>H th cn dng chng xin (n nh ngang ca v gi tt) Kp ngang di cng phi cap hn MNTN mt khong ti thiu l 0.30.5m d thi cng v thay th. Cc thanh kp thng lm bng g x i v lin kt cc bng bu lng, lp rp kp vi cc ti ch. 73 a) (H5.0m)HKp choKp ngangChng xinCcX m e) MNTNHKp choCc xinKp ngangCcX m Hnh 5.6a,b,c,d,e-Cu to v gi cc 2.V gi k (hay gi l Pa l) c s dng khi nn t tt (nn ct cht). Gm c cc ct, x , x m v kp cho (hnh 5.7) X t trn cc thanh g ngn, cc thanh g ngn t trn nn t thin nhin (hoc t trn lp dm, si m Ct ta trn x v trc tip x m Dng kp cho lin kt cc ct tng cng cng cho v gi k. V gi k c th sn xut trong xng hay ti cng trng. 74 MNTNX mKp choCc xinKp ngangG ngnX Ct Hnh 5.7-Cu to v gi k 3.V gi n, v gi kp, tng chn t m V gi n l v gi ch c 1 hng cc. V gi kp l v gi c 2 hng cc, theo phng dc cu, v gi kp cng c kp ngang v kp cho Khi chiu cao H nh (H4m v ngp trong nc >1,5m th dng v gi kp. i vi cu nhiu nhp, c 35m nhp v khng qu 2025m phi lm mt tr kp tng cng dc cu v tng kh nng tip nhn lc hm phanh. M cng c th s dng v gi kp (hnh 5.8) uKCNvxmcahngccmphannngkhngtipxctrcjtieeps vi t p, m bo bn lu di, ngi ta cu to nn t (t nn) thp hn x m, pha sau c hng tng cc vn chn t. (2)(1)- Tng chn t(2)- Cc tng chn(3)- X m(4)- Dm dc(5)- V gi n(6)- V gi kp(7)- Ging ngang(8)- Ging cho(1)(3)(4)(5)(6) (7) (8) Hnh 5.8-S dng kt hp v gi n, v gi kp v tng chn t m B. Cu to chi tit cc b phn cu dm g gin n nhp nh Cc b phn ca cu dm g gin n nhp nh gm: Mt cu xe chy, ng ngi i, lan can tay vn, b chn bnh xe, dm dc, v gi, b phn ni tip cu vi ng I. Cu to mt cu xe chy, ng ngi i, lan can tay vn, b chn bnh xe 1. Mt cu xe chy 75 a) Mt cu 1 lp vn Dng g trn x i hoc nguyn t st nhau trc tip ln dm dc (dm dc t dy).Dngmtcyglmbchnbnhxe,kpchtvnmtcuvidmdcngoi cng nh bu lng (hnh 5.9a).+ u im: Cu to n gin, d thi cng, kt cu thong. + Nhc im: Xe chykhng m thun,mau hng, khi h hng phi thay th hng lot + Phm vi p dng: Thng dng cho cu ng nng thn, ng lm nghip t xe qua li, ti trng nh, cu 1 ln xe, thi gian s dng ngn, cu tm. to cho xe chy m thun hn (to chomt cu bng phng hn), p lc bnh xe phn b u cho vn ngang mt cu, ta ph ln mt cu 1 lp t cp phi (hoc si, dm) dy 1012cm, tip gip trn mt cu, di lp cp phi l lp st do (hnh 5.9b) + u im: Bng phng, m thun. + Nhc im: Tng trng lng bn thn, khi ng nc chng b mc vn mt cu +Phmvipdng:Thngdngchocucnhiumyko,ccbnhxenng,xe xch. Ngoi ra, c thdng vn vt rng t 0.81.2m, dy 35cm, va c tc dng chng bo mn v phn phi p lc cho vn ngang di, va c u im l trng lng mt cunh.Vnvtgm46tmvncgibngaithpvghimbnginhthuyn, bt bu lng xung vn ngang di. Mt cu 1 lp vn c th dng khi dm dc t tha (hnh 5.9.c) (4)- Lan cana)b)c)(6)- Dm ngang(0.5-0.6)mP P PP P P(1)- Dm dc(2)- Vn mt cu(3)- B chn bnh (5)- Cp phi dm(2) (3)(4)(4)(5) (3)(1)(1)(2)(0.5-0.6)mP P(2)(1)(6)(1)(2)(6)(1)(6)(1.5-1.8)m (1.5-1.8)mP Hnh 5.9a,b,c-Cu to mt cu 1 lp vn 76 b) Mt cu 2 lp vn ividmdctdy:Lpvnditngangbnggtrnhocgx,lpvn trn t dc bng g vn x (hnh 5.10a). Lin kt gia 2 lp vn bng inh m. Lp vn dichulcphitnhtonquytnhchiudyvnngang,lptrnctcdng chngbomnvphnphiuplcchonhiuvndi,lmchomtcubng phng, xe chy m thun. Lp trn ly theo cu to dy t 56cm (0.5-0.6)mP P(1)- Dm dc (2)- Lp vn di (3)- Lp vn trn(3)(1)(2) (2)(3)(1) Hnh 5.10a -Cu to mt cu 2 lp vn khi dm dc t dy ividmdcttha:Cutothmngang(dmngang)vnmtcu. Dm ngang l cc cy g t dylndmdc vi c ly2 ngang k nhau 0.50.6m (nguyn tc tnh ton ngang ging nh tnh dm dc t dy). Trn ngang t 2 lp vndc:ccvnlptrntlinkht,ccvnlpdittokheh23cmcho thong (hnh 5.10b). (0.5-0.6)mP(1)(4)(1)(4)(1.5-1.8)m(1)- Dm dc(2)- Lp vn di(3)- Lp vn trn(4)- Dm ngangP(2)(3) (3) Hnh 5.10b -Cu to mt cu 2 lp vn khi dm dc t tha uim:Tndngcvncchiudikhcnhau,tbx,khisachachcn thay th cc vn trong phm vi vt bnh xe chy dc. Trng hp cu 1 ln xe: vn trn ch b tr phm vi vt bnh xe. Vntrn l lp cu to, dy 56cm (khng phi tnh ton)ctcdngchngbomnvtruynlcchonhiuvndi,tkhivntrnt ngang. Phm vi p dng: Dng cho cu c nhiu xe chy qua, ti trng trc ln, mt cu rng, thigianphcvludi.Khicucchiudinhpl>810mmidngdmdct 77 thavchiucaokintrcln,cutolinktdmdcvcutophnxechyphc tp. Khi l 8m, dng dm dc t dy hp l hn. c) dc mt cu mbothotncmtcutt,cutodcdcmtcu23%(tial 3%) v dc ngang 1,52%. Trng hp mt cu c khe h thong th khng cn to dc Cch to dc ngang: Lp vn ngangt cy g t nhin c thun 1% c t ungnrangoibin,dccnlit0.51%ciuchnhbngvtxmt gia ra 2 u. 2. ng ngi i a) ng ngi i dng vn lt dc: Dng thanh g ngn t chng ln dm ngang (vn ngang),trnthanhgngntvnltdcdy5cm.Brngngngiil0.75m (tnh t mp trong ct lan can n mp b chn bnh). (hnh 5.11a) b) ng ngi i dng vn lt ngang: Vn t ln 2 thanh g song song (thanh g b chn bnh v 1 thanh t st chn ct lan can). Vn lt ngang ng ngi i c lin kt vi cc thnh g k v g chn bng inh m hoc bu lng (hnh 5.11b) (3)a) b)(1)(2)(6)(4)>25cm(1)- Dm dc(2)- Mt cu(3)- Lan can(4)- Dm g va(5)- Thanh g ngn(6)- Vn t dc(1)(2)(8)(4)>25cm(7)(7)- Dm g cnh lan can(8)- Vn t ngang(3)(5) Hnh 5.11a,b-Cu to l ngi i 3. Lan can tay vn (hnh 5.11a) Tc dng ca lan can tay vn v cc thanh chn l m bo an ton cho ngi b hnh qua cu. Ctlancancao1.1m(tnhtmtngngii),clygiaccctl1,52.5m. Tit din ct vung 14x14cm. Lin kt ca ct lan can vi vn ngang (dm ngang) v cc thanh g dc (k ng ngi i) bng bu lng. C khidng thanh chng xin tavo u mt tha ca thanh g ngang vn ngi i gi n nh (bt bin hnh) cho ct lan can, thanh chng xin c tit din 6x8cm. Tay vn c kch thc 5x10cm. Khong cch t nh ct lan can n thanh chn trn l 45cm, gia 2 thanh chn l 4cm, gia thanh chn di v mt vn ngi i l 20cm. 78 4. B chn bnh ngncchphnxechyvingngiivmboantonchoxequacu, lmbchnbnhxebnggtrnhayghp.Bchnbnhxecaohnmtxechy 25cm, rng25cm. 5. Ct phng h trnh cho xe khng va qut voct lan can, 2 u cu phi dng cc ct thng ng hay ct xin gi l ct phng h II. Cu tao chi tit dm dc 1. Dm dc t dy Thng lm bng cy g t nhin c thun 1%, mt trn bt cho bng phng sut chiu di vn mt c(vn ngang hay dm ngang) ta ln c n nh, chc chn v tng din tip xc vi vn, mt di ch bt hai u dm dc t ta ln x m d dng. Lu pha gc bt nhiu hn ngn to cho cao mt trn dm dc bng nhau (hnh 5.13a) Khinhptnhtonl(l)thchiudicyglmdml(l+0.5m).ngknhdm dc v tr X so vi tim x m k u ngn : dX=d0 + 0.01X (d0l ng knh ngn dm dc v tr tim x m) Dm dc t dy c b tr xen k tro ui cho tng din tch mt ct cc dm theo chiu di khng i Trn x m ca tr cu, dm dc 2 nhp lin nhau c th t so le hoc i u bng cch bt cho theo mt phng ng. Khi , cc u to i u vi nhau, cc u nh i u vi nhau v c lin kt bng bu lng, b tr sp xp i xng theo tim cu. L+0.5mL(1)- Tr(2)- X m(3)- Dm dc(1)(2)(3)(3) (4)(4)- Bu lngDm dc t so le(2)(3)(2)Dm dc t i uNgnGc Hnh 5.13a-Cu to v b tr dm dc t dy Phm vi p dng: Dm dc t dy 1 tng dng cho nhp l 6m; Vi nhp l>6m, do m men un ln nn c th ghp chng i 2 cy g t tro u ui, mt tip xc gia 2 cy g bt bng phng; lin kt gia 2 cy g chng i bng bu lng.79 Gia 2 b dm dc khng chng i theo hng ngang c lin kt bng then g m bo n nh (hnh 5.13b) (1)(2)(3)(6) (5)(5)- Vn mt cu (6)- Dm ngang (7)- G neo(7) (4)(1)(2)(3)(6)(7) (4) Hnh 5.13b-Dm dc t dy bng g ghp chng i 2. Dm dc t tha Ty theo chiu di nhp v ti trng tnh ton, dm dc t tha c th ghp chng i (tng t hnh 5.13b) hoc ghp chng 3 (hnh 5.14a) (0.5-0.6)m(1)- Tr(2)- X m(3)- Dm dc(4)- Dm ngang(1)(2) (3)(4) (5)(1)(2)(3)(4)(5)(5)- Vn mt cu(6)- G neo(7)- Np g(8)- Bu lng(6)(7) (8) Hnh 5.14a-Cu to dm dc ghp chng 3 Cccygdmdcttrouui,linktvinhaubngbulngvchtbn. Chtbnbngthphocgcngnhmvidmdc,tdctheothcadmdcv ngmvomicygmtnachiucaocacht,vichiusukhngqu1/5ng knh cy g. Chiu dy cht bn t 35cm.(0.5-0.6)m(1)(2) (3)(4) (5) (8)(7) (6)(1)- Tr(2)- X m(3)- Dm dc(4)- Dm ngang(5)- Vn mt cu(6)- Bu lng(7)- Np g(8)- Cht thp(0.5-0.6)m(1)(2) (6)(4) (5) (9)(7) (3)(9)- Chm g Hnh 5.14b-Lin kt b dm dc bng cht thp hoc chm g 80 Lin kt ngang dng then g i vi dm dc chng i, dng kiu ch thp i vi dn dc chng ba (hai bn dng g kp ng). Trong cu dm dc t tha, cc b dm t v tr cc, cc dm ngang t cch nhau t 0.50.6m., mt cu gm 2 lp vn t dc. (1.5-1.8)m(5)(6) (7)(1)- Tr(2)- X m(3)- Dm dc(4)- Dm ngang(5)- Vn mt cu(6)- ng ngi i(7)- Lan can(1)(2)(3)(4) Hnh 5.14c-Cu to v b tr dm dc t tha 3. Dm dc l thp hnh hay b ray Dng dm g vt nhp trn 6m thng kh khn. V vy i vi cu tm, cu cng tc ngi ta c th dng thp hnh I, [ hoc ray lm dm dc thay dm g. Dm dc loi ny t dy hoc tha (cu to chi tit xem cu thp dm c). Cc b phn khc cu to bng g nh nu trn. Hnh 5.15-Cu g dm dc bng thp hnh (I , [ ]) III. Cu tao chi tit v gi (m, tr) 1. X m Mttrnbtbngphngkdmdc,mtdibtphngtivtrucc(hnh5.16a,b). Trng hp x m vung c th dng thm ai thp lin kt vi cc v bt bu lng (hnh 5.16c). 2. Cc C th dng cu g hay g hp. Khi chiu di cc ln phi ni di cc. V tr mi ni phicaohnMNTNtithiu0.30.5m.Yucuminiphingin,chcchnv chu lc tt. Cutomini:Niiubngngthphockpbngthpgcbtbulng.. Trng hp ni tp dng thp v bu lng kp cht (hnh 5.16d) 81 Khi chiu su ng cc ln, c th ni di trong qu trnh ng cc. Trng hp ny mi ni phi chc chn hn, dng 3 ai thp v gi cht bng inh mc hoc ng inh ai khng x dch theo cc khi ng (hnh 5.16e) 3. Lin kt cc vi x m Dng mng hay cht thp (hnh 5.16a,b) Chiusulmnglnhnchiucaomngt0.51cmplctruyntxm xung cc khng truyn ln mng. Kch thc l mng v mng (hnh 5.16a) Loi cht thp hin nay thng dng l inh xuyn tm. Cht thp c th l inh ng qua x m xung cc hay t cht vo l khoan sn. badd1(2)h(2) (2)10-12cma) b)a=(1/4-:-1/3)db=(1/5-:-1/4)dh=(1/4-:-1/3)d1(1)(2)(1)(1)(1)(1)- Cc (2)- X m(3)(3)- inh thp c) 6cm(4)- ai thp dy =(0.5-:-1)cm(1)(2)(4)(5)(5)- Bu lng d) e)3dd(3-:-3.5)dd(6)(1)(5)(7)(6)- ai thp (7)- inh mc/ng Hnh 5.16a,b,c,d,e-Chi tit cc lin kt v gi 4. X kp Cc thanh x kp ngang v cho lin kt vi cc bng mng trn khot thanh kp.Chiu su khot khng nh hn 23cm. Kp c bt cht vi cc bng bu lng 82 (8)- Kp ngang (9)- Kp cho(1)(8)(9)(2) Hnh 5.16f-Chi tit cc lin kt kp ngang, kp cho v cc 5. Thanh chng Thanh chng xin lin kt vi cc u thanh bng kiu rng ca, lin kt gia thanh chng vi cc cng dng bu lng 6. Bu lng Cc bu lng lin kt u c rong en bng thp dy3mm trnh p mt dp g IV. Ni tip gia cu v ng 1. Hin tng ln g Khi t chy qua ch tip gip gia cu v ng thng to nn g do nn ng b ln. Nguyn nhn l do cng gia ng v cu khc nhau. Xe chy trn ng vi cng nh, n u cu gp phi cng ln hn lm cho xe b sc, chuyn ng mt m thun. Tc dng nn cht ca bnh xe i vi nn ng gy ra ln cc b (gi l g hnh 5.17). Mt khc, u kt cu nhp tip xc vi t trc tip chng b mc, hng. V vy, cn gii quyt tt ni tip gia cu v ng. Yu cu: Khi chn bin php ni tip l xe ra, vo cu phi m thun, khng phi hn ch tc , tip gip giau kt cu nhp vi t nn ng phi thong lu b mc, hng. Hnh 5.17-Hin tng ln g 2.Cc hnh thc ni tip gia cu v ng Cu to tng chn t: Gm hng cc ng knh D1820cm, nh x m ca hng ccnybngmttrncadmdc.Vncatngchntlgtrnngknh D1516cm hoc g trn x i. 83 Cc cc mi ta luy nn ng c x m dc theo mi Vn chn t t thp hn nh t nn trc m t 3050cm, phng t b p tri. Pha sau vn chn t l lp st do gi vn lu mc C 3 hnh thc ni tip ng cu:Trn on ng ucu di 2.5m dng lp hc t trn m ct, vn chn t ta trc tip vo cc m (hnh 5.18a) + u im: Cu to n gin + Nhc im: Di tc dng ca xe chy, lp ct b ln n hi lm cho dn dn tch rinhaugynng,ngthidovnchnttatrctipvoccmnnccm chu lc ngang ln v u dm dc khng c thong + Phm vi p dng: Cho cu tm, cu 1 nhp, cu c chiu cao v gi thp Dng bn dn g di 1m, t su 40cm, bn dn gm cc thanh g t dc di, cc thanhngangtrnbnggxi(trnghpcuc1lnxecthchtbndn di vt bnh xe). Bn dn t trong lp ct, mt u ta ln tng chn t (hnh 5.18b) + u im: Xe chy m thun, ln tuyvn cn nhng khng ln v khngt ngt thay i cng t ng vo cu m tng dn nn khng gy sc mnh + Nhc im: Kh thay th khi b mc hng + Phm vi p dng: Dng cho cu c ti trng ln, chiu cao v gi ln (do c hng cc tng chn nn cc m khng trc tip chu lc ngang ca nn ng, u kt cu nhp thong, phng mc tt hn. Nn cht t nn v dn lng th dm trn on di 1m, su 0.6m (hnh 5.18c) + u im: Khc phc nhc im ca bn dn g (hay mc hng, kh thay th) + Phm vi p dng: Dng cho cu c ti trng ln, chiu cao v gi ln (do c hng cc tng chn nn cc m khng trc tip chu lc ngang ca nn ng, u kt cu nhp thong, phng mc tt hn. a) 200-:-250cm1:1(1)(2)(3)(4)(6) (5)30-:-50cm(8)(7) b) (9)(2)(4)(3)100cm30-:-50cm1:1(8)(6) (5) (7)(10) 84 c) (9)- Cc tng chn d18-:-22(9)(2)(3)(6) (5) (7)1:1(11)(6)- Dm ngang(7)- Mt ng(8)- t st(10)- Bn dn g(11)- Lng th dm(12)- T nn100cm60cm30-:-50cm(1)- hc(2)-Tng chn (d15-:-16)(3)- V gi(4)- Dm dc(5)- Vn mt cu Hnh 5.18-Cc hinh thc ni tip ng cu a) Ni tip bng 1 lp hc trn m ct, tng chn t trc tip vo v gi b) Ni tip dng bn dn g; c) Ni tip bng lng th dm 85 CHNG 6 TNH TON CC B PHN CU DM G GIN N NHNP NH 6.1.TNH MT CU A. Tnh vn mt cu Nguyn tc tnh ton:Ch tnh vn v kh nng chu m men un vi hot ti t, b qua trng lng bn thn Coi vn nh dm gin n, b qua tnh lin tc ca vn Cng tnh ton ca vn c tng thm 20% (theo quy trnh 22TCN18-79) I. Trng hp mt cu ch c 1 lp vn ngang (cu dm dc t dy) + Khi tnh vn A, p lc bnh sau ca chic xe nng trong on xe tiu chuNn truyn ton b cho vn A l Ps (tn).+ Vi hot ti tiu chuNn H8, H10 (c P=8T, 10T) th: Ps = 0.95P/2(6.1) + Nhp tnh ton ca vn lv = lo (c ly tim ca 2 dm dc vn) LvPsboLvboq =b0Ps (Psau) Ps Hnh 6.1-S cu to v tnh ton vn mt cu 1 lp vn + M men un tnh ton ln nht ti mt ct gia nhp (mt ct nguy him nht) )2(4.00 maxblP nMs =(6.2) Trong :b0:B rng tip xc ca bnh sau (xe nng) ln vn mt cu theo phng ngang n:H s vt ti ca t,n = 1.4 + Kim tra cng v la chn tit din: uRWM2 . 13 . 1max = (6.3) hayuRMW2 . 13 . 1max (6.4) Trong :W:M men khng un ca tit din ngang ca vn Ru:Cng tnh ton chu un ca g vn 1.3:H s k n s gimyu do vn b bomn (bnh xe tip xc trc tip vi vn g) 1.2:H s tng cng ca vn 86 II. Trng hp mt cu c 2 lp vn + Lp vn trn lun lun t dctheoyu cu cu to chngbomn, chiu dy vncm 6 51 = + Lp vn di t ngang (nu l cu dm dc t dy) hoc t dc (nu l cu dm dc t tha) 1. Trng hp vn di t ngang (cu dm dc t dy) P1aoaP1lol2 11lvl= lo - d/3d/31d111d 1boblv = min {lo, (l+ )} 1 2b = bo + 2. 1lvq=P1/bb Hnh 6.2-S cu to v tnh ton vn mt cu 2 lp vn(lp vn di t ngang) + Nh lp vn trn, ti trng Ps truyn cho mv vn di chu:mv=2 (vn) + p lc ln 1 vn l: 21svsPmPP = =(6.5) + Chiu di phn b ti trng theo hng ngang cu:b = b0 + 21(6.6) + Nhp tnh ton l:lv = min{(l1+2), l0}(6.7) Trong :l0: C ly tim 2 dm dc k vn l1: Khong cch gia 2 mp trong ca dm dc m vn ta ln 1,2:Chiu dy ca 2 lp vn trn v di + Cc cng thc tnh ton tng t nh vn 1 lp, ring cng thc (6.3) v (6.4) khng c h s 1.3 v khng nh hng ca s bo mn (lp vn di khng tip xc trc tip vi bnh xe) )2(4.1maxblP nMv = (6.8) 87 uRWM2 . 1max = (6.9) hayuRMW2 . 1max (6.10) 2. Trng hp vn di t dc (cu dm dc t tha) Psboblol2 11lvl= lo - d/3d/31d222d 2P1aoalv = min {lo, (l+ )} 1 2a = ao + 2. 1lvq=P1/aa Hnh 6.3-S cu to v tnh ton vn mt cu 2 lp vn(lp vn di t dc) + Nh lp vn trn, ti trng Ps truyn cho mv vn di chu. Nu b rng tip xc ca bnh sau xe nng theo hng ngang cu l b0 = 40cm hoc 30cm (tng ng vi hot ti tiu chuNn H10 hoc H8) th tng ng c mv = 3 (vn) hoc mv = 2.5 (vn)+ p lc ln 1 vn l: 5 . 21svsPmPP = =vi hot ti t H8 (6.11) Hoc: 31svsPmPP = =vi hot ti t H10 (6.12) + Chiu di phn b ti trng theo hng dc cu:a= a0 + 21 (6.13) + Nhp tnh ton l:lv = min{(l1+2), l0}(6.14) Trong :l0: C ly tim 2 dm dc k vn l1: Khong cch gia 2 mp trong ca dm ngang m vn ta ln 1,2:Chiu dy ca 2 lp vn trn v di a0:Chiu di tip xc ca bnh xe ln vn theo hng dc cu (a0=0.2m) + M men tnh ton ln nht ti mt ct gia nhp vn: )2(4.1maxalP nMv = (6.15) + Kim tra cng v la chn tit din vn tng t nh cng thc (6.9) v (6.10) 88 3. Trng hp mt cu ph t cp phi hoc dm lvPsbolvbP1/bPsaoq1bahd/4(a = ao + 2.h ; b = bo + 2.h)hbv Hnh 6.4-S cu to v tnh ton vn mt cu ph t cp phi + Cu dm dc t dy 1 lp vn c ph cp phi dy h, vn c ng knh l d + Ti trng tnh ton ca vn v lp cp phi l q1 (T/m): q1 = n1.g1 + n2.d.h.cp (T/m)(6.16) Trong :g1:Trng lng vn (T/m) cp:Trng lng ring ca cp phi (T/m3) n1:H s vt ti ca g lm vn, n1 = 1.2 n2:H s vt ti ca cp phi hoc dm,n2 = 1.5 +NhccpphimtitrngPstruyncho1svnchutrnchiudia(theo phng dc cu) v trn chiu rng b (theo phng ngang cu): a = a0+2.h v b = b0+2.h (6.17) + S vn tham gia chu ti:dh adamv. 20 += = (6.18) +p lc tnh ton tc dng ln 1 vn: vsmPP=1(6.19) + M men tnh ton ln nht ti mt ct gia nhp vn: )2(4.8.121maxblP n l qMv v + = (6.20) + Kim tra cng v la chn tit din vn tng t nh cng thc (6.9) v (6.10) B. Tnh dm ngang (trong dm dc t tha) Nguyn tc tnh ton:Coi dm ngang nh dm gin n, nhp tnh ton ldng l c ly tim 2 dm dc 89 q2q1lv lvP1 P2 P2dm dcPlvq=P1/bbP1bob Hnh 6.5-S cu to v tnh dm ngang Nu mi ni vn lt dc trn nm phn tn th khi tnh dm ngang c k n s phn phi n hi ca p lc bnh xe nh lp vn di t trn dm ngang. Ngha l, di tc dng ca p lc P t ln dm ngang A ang xt th khng phi dm ngang A chu ton b m ch chu mt phn P1P2>P3> S dm ngang tham gia chu lc nhiu hay t ph thuc vo cng ca vn mt cu v dm ngang. S ph thuc c c trng bi h s phn phi n hi K xc nh nh sau: v v dngdng dng vJ E lJ E lK. .. .. 833= (6.21) Trong : + lV:Nhp tnh ton ca vn (c ly tim 2 dm ngang) + ldng:Nhp tnh ton ca dm ngang (c ly tim 2 dm dc) + EV, JV, Edng, Jdng: M un n hi v m men qun tnh ca cc vn v dm ngang nhn p lc bnh xe. Ch :- Nu b0 = 40cm, 2 lp vn t dc th trong cng thc trn JV = 3J1vn - Nu b0 = 30cm, 2 lp vn t dc th trong cng thc trn JV = 2.5J1vn Cc trng hp: + K 1/3:plcPphnphicho3dmngang,dmngangAvtrlcPchu phn lc ln l P1, hai dm ngang hai bn chu phn P2 nh hn, xc nh nh sau: 90 P1 = 1.P P2 = 2.P (P = P1 + 2.P2) KK. 2 3. 2 11++= K . 2 312+= + 1/3 > K 0.055:p lc P phn phi cho 5 dm ngang, dm ngang A v tr lc P chu phn lc ln l P1, hai dm ngang hai bn chu phn P2, hai dm hai bn tip theo chu lc P3, xc nh nh sau: P1 = 1.P P2 = 2.P P3 = 3.P (P = P1 + 2.P2 + 2.P3) DK K21. 7 . 18 1 + += DK . 11 12+= DK . 3 13= (vi D = 5 + 34.K + 7.K2) + K < 0.055:p lc P phn phi cho 7 dm ngang (tng t) P1 = 1.P P2 = 2.P P3 = 3.P P4 = 4.P (P = P1 + 2.P2 + 2.P3 + 2.P4)

13 21. 26 . 131 . 72 1DK K K + + +=

122DK . 46 K . 57 1 + +=

123DK . 18 K . 23 1 +=

124DK . 3 K . 18 1 + = (vi D1 = 7 + 196.K + 193.K2 + 26.K3) + M men un tnh ton ln nht gia nhp dm ngang A: )2(4.8) . . (122 2 1 1 maxblPnlq n q n Mdng dd + + =(6.22) Trong : q1, q2: Tnh ti ca hai lp vn v dm ngang n1 = n2 =1.2 l h s vt ti tng ng ca cc tnh ti q1, q2 P1 = 1.P l p lc ca bnh sau (hoc trc) phn phi cho dm ngang A b: chiu di phn phi p lc P1 trn dm A theo hng ngang cu:- Nu b0 = 40cm th b = 3b1vn - Nu b0 = 30cm th b = 2.5b1vn + Kim tra cng , la chn tit din dm ngang (tng t nh tnh vn mt cu) 91 + Ch : - Khi tnh dm ngang cha cn xp xe dc cu v c ly gia 2 trc t (ti thiu l 4m) ln hn nhiu so vi c ly gia 2 dm ngang (thng l 0.50.6m). V vy khi tnh dm A vi p lc P ca trc sau, p lc trc trc khng nh hng ti A. -Khitnhdmngangcngchacnxpxengangcuvnhptnhtoncadm ngang ldng (c ly tim ucar 2 dm dc) thung di 2m; c ly tim 2 bnh xe ca 1 trc theo hng ngang cn ti thiu l 1.7m. Mt khc, b qua tnh lin tc ca dm ngang (coi l dm gin n). V vy khi t 1 bnh xe gia nhp th bnh xe th 2 lun lun nm ngoi nhp tnh ton. 6.2. TNH DM DC Nguyn tc tnh ton: Tnh dm dc theo TTGH th nht v cng (kh nng chu m men un, lc ct) v TTGH th 2 v bin dng (kim tra vng) Tnh ton vi 2 loi hot ti t v xe xch, ly kt qu ln hn ca ni lc kim ton (trng hp bt li nht): +Khi tnh vi xe xch, ch tnh vi 1 xe v khng c on ngi i. + p lc bnh xe tt coi nh lc tp trung +Tnhdmdctinhnhxpxetheohngngangcuvdccu(c2hng) xc nh c tr s m men, lc ct ln nht i vi dm ang xt. Qu trnh tnh ton c tin hnh vi mt s dm tm ra dm nguy him nht a vo tnh ton Phn bit 2 trng hp tnh ton: dm dc t dy v dm dc t tha A. Tnh dm dc t dy I. iu kin tnh ton dm dc t dy C ly tim 2 dm dc0.8m; S mi ni ca vn mt cu trn 1 dm dc khng qu 30% tng s vn t trn dm dc v phi b tr phn tn.Khi tnh ton c xt n s phn phi n hi ca ti trng (tng t nh khi tnh ton dm ngang) Trnghpl 75mthvtrbtlinhtllcbnhxesautgianhp(khong cch 2 trc trc, sau ca xe ti thiu l 4m) II. Trnh t tnh ton Bc 1: Xc nh h s phn phi n hi K:v v dddd dd vJ E lJ E lK. .. .833=(6.23) Trong :ldd, Edd, Jdd,: Chiu di nhp tnh ton, m un n hi v m men qun tnh dm dc 92 lv, Ev, Jv,: Chiu di nhp tnh ton, m un n hi v m men qun tnh ca vn aoalvdm dcPbv Hnh 6.6-S xc nh s vn tham gia chu lc tnh dm dc t dy Ch :Jv c ly theo s vn tham gia chu lc:Jv = mv . J1vn (6.24) + i vi hot ti t:- Khi c 1 lp vn (ngang):mv = 1 - Khi c 2 lp vn (di ngang, trn dc): bamv =+ i vi hot ti xe xch:vxvblm= .(Vi lx l chiu di ai xch tip xc vi vn mt cu theo hng dc cu) Bc 2: Xc nh ti trng t v on ngi i truyn ln dm dc ang xt (xp xe ngang cu) P170T=75 T=75 G=600pngPT2PT2PT2PT2b)10 50 50 10 40 20P2 1 3 4 5 6 7 8 9 10 11 12 13c)q P3 1 q2 P4 P7 P8 P10 P11P2 1 3 4 5 6 7 8 9 10 11 12 13dP4 P7 P8752 1 3 4 5 6 7 8 9 10 11 12 1375 10x60=600P P170 110Ppnga)2 1 3 4 5 6 7 8 9 10 11 12 13 Hnh 6.7-S xc nh ti trng t v on ngi i truyn xung dm dc s 6 93 + Qu trnh xp xe ngang cu tin hnh nh sau Gi s c mt ct ngang cu kh 6, cn xc nh ti trng ln nht do hot ti t v on ngi i truyn ln dm dc s 6 chng hn, vi k=0.2 (trng hp 0.550.5mP P P>0.1m>1.1ma1 21y=y=32a2l 2y=1a1l 1AAl l lT T G1 2 3 4Pa1 2 3ah N Hnh 6.13-S xp xe t ngang cu v tnh h s phn b ngang K 98 + p lc ln nht ln dm A:Pmax = P (y1 + y2 +y3) = K . P ; (vi K = yi)(6.45) + Ch : - H s phn b ngang K ch ph thuc khong cch gia cc bnh xe theo hng ngang cu (k c trc trc v trc sau ca xe) - Tin hnh xc nh K cho 1 s dm tm ra dm nguy him nht. Kinh nghim cho thy i vi cu kh hp 1 ln xe, cc dm bin nguy him hn dm trong cn trong kh rng c th dm gn tim cu nguy him hn. b) i vi xe xch: + Khi l1 = l2 th v tr bt li nht l tim ai xch trng vi tim dm dc (AH NA i xng v c tung ln nht ti v tr tim dm dc A) + Khi l1l2 th v tr bt li nht tha mn iu kin 2211lblb=tc y1=y2 ah NAyy12A2.6m (X60)l l lT T G1 2 3 4a1 a21 2 31l l 1 2la1 a2 b2p =xb1 b2bx qxbxbxb1 Hnh 6.14- S xp xe xch ngang cu v tnh h s phn b ngang Kx+ p lc ln nht ln dm A do ti trng xe xch l: X XXXX Xq Kbqp q . .max= = = (6.46) Trong :Kx:H s phn b ngang ca xe xch: XXbK=qx:p lc phn b u theo chiu di ai xch: XXlQq. 2=Q: Trng lng xe xch 99 lx: Chiu di ai xch + Kt qu tnh ton Kx sau khi bin i: 2 1211l lbKXX+ = (6.47) Bc 2: Xc nh m men un tnh ton do hot ti (xp xe dc cu) v tnh ti a) i vi t: + Khi ldd 7.5m: V tr bt li nht khi bnh sau t gia nhp (hnh 6.15) v khong cch nh nht gia 2 trc xe trong on xe tiu chuNn l 4m. P s P trl /2 l /2 dd ddldd Hnh 6.15- S xp ti t bt li nht (theo hng dc) cu khi ldd 7.5m - M men un tnh ton ln nht ti mt ct gia nhp:4. . .(max)dd s l P K nM = (6.48) Trong :n = 1.4 l h s vt ti ca t. K:H s phn b ngang Ps:p lc bnh sau ca xe nng + Khi ldd > 7.5m v trn cu ch xp c 1 xe: V tr bt li nht khi bnh sau cch khong gia 1 khong c/2 (c l khong cch t bnh sau n hp lc R ca 2 lc Ps v Ptr). RMPs Ptrc/2 c/2l /2 l /2 dd ddl ddca Hnh 6.16- S xp ti t bt li nht (theo hng dc) cu khi ldd > 7.5m - M men un ln nht pht sinh ti mt ct v tr bnh sau: dddd lc l R K nM. 4) .( . .2(max)=(6.49) Trong :R = Ptr + Ps;vRPtr ac.=100 a:Khong cch gia 2 trc (trc sau v trc trc) ca t K:H s phn b ngang n:H s phn b ngang b) i vi xe xch: + Khi ldd lx:M men tnh ton ln nht ti mt ct gia nhp: 8. . .2(max)dd x x xxl q K nM =(6.50) + Khi ldd > lx: M men tnh ton ln nht ti mt ct gia nhp: )2(4. . .(max)xddx x x xxlll q K nM = (6.51) lddlx > l ddlddlx < l dda) b) Hnh 6.17- S xp ti xe xch gy bt li nht (theo hng dc) cu a) Khi ldd lx;b) Khi ldd > lx c) M men tnh ton do hot ti xe ( t hoc xe xch): tng t cng thc (6.30) d) i vi hot ti ngi i: +Tcdngchyulndmdcbin.plcdotitrngonngiitruyncho dm bin B (hnh 6.18), ly png = 400KG/m2 c phn b u theo chiu dc cu vi tr s: qng = png . T . y (6.52) TB1yT/2 T/2ah NB Hnh 6.18- S xp hot ti ngi + M men un do ti trng on ngi: 8. . .2dd x x xngl q K nM = (6.53) e) M men un tnh ton do tnh ti + q1, q2 l trng lng bn thn dm dc v vn vi h s vt ti tng ng l n1, n2. + M men un tnh ton do hot ti: 8) . . (22 2 1 1ddtlq n q n M + = (6.54) 101 f) Tng hp m men un tnh ton: Mtt = Mt + Mh + Mng(6.55) Bc 3: Xc nh lc ct tnh ton gi do hot ti (xp xe dc cu) v tnh ti a) i vi t: + Trn AH QB, t bnh sau xe nng ca on xe tiu chuNn ti B, cc bnh cn li xc nh theo s on xe, s bnh trn AH ph thuc vo ldd. Ptr Psl dd4mPtr Ps4mXenng8mB1ah QBy2y=1y3y4 Hnh 6.19- S xp hot ti t (dc cu) xc nh lc ct ti gi B + Lc ct tnh ton ln nht ti gi B: Q(B) = n.K.Pi.yi.(6.56) b) i vi xe xch: + S xp ti dc cu cho 2 trng hp lddlx v ldd>lx nh hnh 6.11 - Khi lX ldd: 2..) (dd Xx X B Xl qK n Q = (6.57) - Khi lX < ldd:X X x X B Xq K n Q . . .) (= (6.58) c) Lc ct tnh ton do hot ti + So snh Q(B) v Qx(B) ly tr s ln nht tnh ton: Qh(B) = max [Q(B),Qx(B)] (6.59) d) Lc ct tnh ton do tnh ti:Qt(B) = (n1.q1 + n2.q2).ldd / 2 (6.60) e) Lc ct tnh ton do hot ti ngiQt(B) = nng.qng.ldd / 2(6.61) f) Tng hp lc ct tnh ton: Qtt(B) = Qh(B) + Qt(B) + Qng(B) (6.62) Bc 4: Kim tra dm dc v cng v bin dng a) V cng Theo TTGH 1, kim tra kh nng chu m men un v lc ct (chu un v chu ct) theo cng thc (6.39), (6.40) 102 b) V bin dng Theo TTGH 2, kim tra vng ln nht gia nhp ( vng c tnh vi ti trng tiu chuNn) + i vi t: - Khi ldd7.5m:dd ghddddTCS l fJ El P Kf1801.. .4813max= = (6.58) - Khi ldd>7.5m:V AH vng ti mt ct gia nhp, dng ti trng ri u tng ng (tra bng) xc nh fmax + i vi xe xch: - Khi lX ldd: dd ghddddtcX Xl fJ El q Kf1501.. .38454max= =(6.59) - Khi lX < ldd:dd ghddddtcX Xl fJ El q Kf1501) 4 8 (.. .38413 24max= + = (6.60) Trong :1 < =ddXll 103 Phn th t CU , CU B TNG, CU B TNG CT THP Chng 1: CU V CU B TNG 1.1. KHI NIM CHUNG V CU V CU B TNG Cucxydngbngtnhinhaynhnto(nhgchnunghaygch btng),cncubtngcxydngbngbtnglinkhitich,khngcct thp. V v btngchu nn tt, chu kokm nn cu v btng thng c dng vm. Vnhvmlbphnchulcchnh.ngcongnitrngtmgiacctitdin vnh vm gi l trc vm. Trc vm c dng st vi ng cong p lc do ti trng tnh ton gy ra. Ch tip gip vnh vm v m tr l chn vm. Tit din trong mt ct i xng ca vnh vm gi l tit din nh vm. Khongcch(giatrngtmhaititdinchnvmgilnhptnhton.Khong cch f t trng tm tit din nh vm n ng thng ni trng tm hai chn vm gi l ng tn, lf gi l dc ca vnh vm. Vnh vm truynp lc chom tr ti chn vm ri tmtr quamng vo t nn. Ngoi thnh phn thng ng, p lc cn c thnh phn nm ngang, v vy m tr cu to t bit hn m tr cu dm. T s lf cng ln th lc Ny ngang cng nh. Thc t thng ly lf = 71 41. Cu vm v btng a s lm khng khp. Trn vnh vm l kt cu phn xe chy. 1.2. CC DNG C BN CA CU VM V B TNGCn c vo kt cu trn vm ngi ta phn cu v cu b tng thnh 3 dng c bn sau: 1. Dng vm p bng ct to, si hay dm: Thng dng cho cu 1 nhp khng ln lm ( Hnh 1.1 ). mp ngoi vnh vm xy tng chn theo dc cu, m cng xy tng cnh chn t to thnh on chuyn tip t 104 ng vo cu. nc khng thm c vo khi xy, cn ph lp phng nc ln mt trn vnh vm v mt trong ca cc tng chn t. dmVnh vmMTng chn t trn vnh vm Hnh 1.1. Dng trn vm m bng ct, si hay dm 2. Dng trn vm lt bng hc hay b tng Loi ny mt ngoi xy bng tt, c cng v cng cao lm lp o bo v v to dng p. i vm c dc ln v vm nhiu nhp dng dng ny l thch hp. Lp phngncphlnmtphnxyltcdc(1.52)%dnncvongthot nc.Mcudnglt.Trnvm,ngoitngcnhcnxytngtrclinvi tngcnhchuplcngangcat.Trnchnvnhvmlmkhebindng.Lp phng nc ph lin tc qua khe bin dng. 3. Dng kt cu trn vm rng Dng ny dng cho cu nhp ln v c bit ng tn f ln l hp l bng cch lm cc vm con theo dc hay theo ngang cu. Thng lm vm con theo dc cu ( tc l l vmhngngangcu)todngp.Vmcontalntngthngngxytrn vnhvmchnh.Giaccvmconchnbtngvckhebindng,trnphlp phng nc. Vi loi vm ny th kt cu trn vm nhgim c tnh ti vm. Vm conVnh vmTng Hnh 1.2.Dng trn vm rng 105 gimnhktcutrnvmngitacnlmvmchnhhpcbrngnhhn mt xe chy: b = l/15 v b khng nh hn 3m. Mt xe chy l bn BTCT hai mt tha . Cng c th lm hai vnh vm chnh hp t song song, khong cch gia chng t cc vm ln. Cu b tng c c im v cc dng c bn tng t nh cu , nhng phng php thi cng khc ch b tng ti ch trn gio. 1.3. CU TO CHI TIT CU VMVB TNG 1. Yu cu i vi vt liu: Vnhvmchulcvmtrxybngtnhinchnhdngtheothitk(hoc bngccvingchbtnghaygchnung)vivaximng.Ringvnhvmthng dng cc vin o hnh nm. tnhindngxyphiccng(khngnhhn400600kG/cm2), khng b nt n, khng b phong ho v n nh i vi nc bin khi xy dng cc cng trnh gn bin. Cc loi c cht lng tt hay dng l: Grant, iozt, Bazan, cng c th dng trm tch nh vi, lmt Va xy dng xi mng Pooclang, mc va M 100. Nu dng b tng th mc b tng M 200 i vi vnh vm v M 150 i vi m tr. 2. Cu to chi tit: Khi chiu dy vnh vm nh c th dng cc vin hnh nm c chiu cao vin bng chiu dy vnh vm, mch va hng tm. Khi chiu dy vnh vm ln, cc vin c xy thnh tng lp theo chiu cao v so le trnh trng mch. Mtrxytheocclpnmngang,phndicamnmtrnnnthngxy lp nghing. ch tip gip gia vnh vm v m tr, gia cc lp nghing vi cc lp nm ngang m hay gia tng chn trnvm vi vnh vm chnh ngi ta dng cc vin gia cng c bit theo mt ct c hnh nhiu cnh. Mt ngoi cu a s xy 1 lp o bo v khi xy khng b tc ng ca khng vtovp.Khixylpotinhnhngthivikhixychnhndnhcht 106 thnh1khivngchc.Chtlngphitt,mtngoiphiphng,chiudikhc nhau gia cc vin xen k khi xy gn chc. Lpphngnclmbngvtliukhngthm,bn,do,daigmlpvitNmnha cchncvlpphngncclithpnhcuBTCT.Ngitatlpcchnc ln trn lp va xi mng to dc cho b mt khi xy (1.52)cm. Vi cch nc t cho mp ngoi ln ln v cc lp xp chm ln nhau. Cu to ng thot nc nh trong cu BTCT. Chng 2: CU DM B TNG CT THP 2.1. KHI NIM CHUNG V CU BTCT I. c im v phm vi p dng: 1. c im a) Vt liu Khi xy dng cu BTCT thng dng cc vt liu a phng: ct, , xi mng l ch yu,phnctthpchchimtlnhsovitrnglngtonktcuvthngdng loi thp trn vi gi r hn loi thp cn dng lm kt cu. b) bn, cng Khi kt cu nhp cu BTCT c cng rt ln, c bn p ng miyu cu khai thc an ton, thun tin. Tui th cu BTCT cao. c) Hnh dng, h thng Kt cu nhp cu BTCT c hnh dng hp l vmt c hc, thmn ccyu cu v thuntinkhaithc,vpkintrc.Vdcucongtrnmtbng,curnhnhch Y d) Tnh lin khi Kt cu nhp c b tng lin khi ti ch cng nh kt cu nhp lp ghp hin i u m bo c tnh lin khi vng chc. e) Trong lng bn thn Do trng lng bn thn ln nn kt cu nhp cu BTCT khng vt qua c nhng chiu di nhp k lc nh cu thp. Nhp cu vm BTCT di nht th gii c L =360m, trongkhinhpcutreodinhtcL>1300m.Tuynhinchnhdonngnmcu BTCT t b nh hng xung kch ca hot ti qua cu hn so vi cu thp, ting n khi xe 107 qua cu cng nh hn, dao ng t hn. Do nhiu cu trong thnh ph c lm bng BTCT. f) Chi ph duy tu bo dng Ni chung chi ph ny rt thp, hu nh khng ng k so vi chi ph duy tu cu thp. g) Vt nt Ni chung kh trnh khi nhng vt nt nh trong cu BTCT d l BTCT d ng lc. Cc vt nt nh hn 0.3mm vng kh hu khng n mn c coi l cha nh hng xu n tui th kt cu. thin v an ton, Quy trnh 79 hn ch m rng vt nt khng qu 0.2mm trong kt cu BTCT thng. Khi thit k cng nh thi cng cn phi tm mi bin php cng ngh hin i v hp l gim nguy c xut hin v m rng vt nt. Trong cc kt cu BTCT d ng lc khn c php xut hin vt nt. 2. Phm vi p dng ca kt cu nhp BTCT + Trong cc cu BTCT trn ng t, thng dng nhiu dm gin n cc nhp L = 633m, c bit dm Super T c chiu di ln n 42m hin ang c dng rt nhiu lm nhp dn vo cc nhp chnh nh Cu M Thun, Cu Tn Cc nhp chnh ca ccculnnythngcdngdmlintc,dmhng,cukhungBTCTdnglc, cu vm. + Ni chung thng chn cc loi kt cu nh sau: - Dng cu dm gin n BTCT thng: dng vi cc nhp L = 624m - Dng cu dm gin n BTCT d ng lc: dng vi cc nhp L = 1242m - Dng cu dm kin tc hoc cu khung BTCT d ng lc: L = 33200m - Dng cu vm BTCT thng: L = 15 300m - Dng cu treo dy xin: L > 200m II. Phn loi cu dm BTCT 1. Phn loi theo v tr cu Tu theo loi chng ngi cn vt qua m c th gi l: - Cu qua sng, sui - Cu cn - Cu cao 2. Phn loi theo mc ch s dng Tu theo mc ch s dng c th phn thnh cc loi cu: - Cu ng st: ch cho tu ho chy qua. - Cu ng b: cho tt c cc loi phng tin giao thng trn ng t. 108 - Cu b hnh: ch cho ngi i b. - Cu thnh ph: cho t, tu in, ngi i b. - Cu hn hp: cho tu ho v t. - Cu cbit:dng phc v cho cc cngvic ring nh ng dn nc, ng hi t, ng dn du, cp in 3. Phn loi theo cao mt ng xe chy - Cu c ng xe chy trn: khi ng xe chy t trn nh kt cu nhp. - Cu c ng xe chy di: khi ng xe chy b tr dc theo bin di ca kt cu nhp. -Cucngxechygia:khingxechybtrtrongphmvichiucaoca kt cu nhp. 4. Phn loi theo s chu lc: Cu dm: b phn chu lc ch yu l dm, lm vic theo chu un.Loi cu ny gm cu dm ginn (L=1240m), lin tc (L=40150m) cu dm hng Cu vm: b phn chu lc ch yu l vm. Vm lm vic theo chu nn v chu un. L=60305m (BTCT) . Cu khung: tr v kt cu nhpc lin kt cng vi nhau chu lc,L=50150m L Hnh 2.1. Cu dm L Hnh 2.2. Cu vm L1 L2 L1 Hnh 2.3. Cu khung 109 Cu treo dy xin dm cng BTCT: l loi cu c dm cng ta trn cc gi cng l m tr v cc gi n hi l cc im treo dy vng. L=100890m 5. Phn loi theo hnh dng mt ct ngang kt cu nhp- Kt cu nhp bn - Kt cu nhp dm c sn - Kt cu nhp mt ct hnh hp 6. Phn loi theo phng php thi cng - Cu BTCT ti ch - Cu BTCT lp ghp III. Mt cu BTCT v nhng cu to chung khc. 1. Mt cu lp ph b tng atphan Mt cu gm cc lp sau: lp va m, lp cch nc, lp b tng bo v, lp b tng atphan. Lp va m bng va xi mng M150 200 dy (11.5)cm. Lp va m ny nhm to bng phng hoc to dc ngang cho cu. Lp cch nc gm 1 lp nha nng, 1 lp vi th tNm nha, trn ph tip 1 lp nha nng nhm bo v bn b tng mt cu khi b thm nc. Lpbtngbohcttrnlpcchnctnhnhnglctptrungnguy him hoc khi lp b tng atphan s lm hng lp cch nc. Lp ny c chiu dy t 34cm bng b tng M 200. tng tc dng bo v v bn ca lp ny thng t lictthpcngknhd=35mmvili(55)cmhoc(1010)cm.Lict thp ny nht thit phi t cc cu BTCT lp ghp c bn mt cu hng. Lp b tng atphan c t lm mt trn c chiu dy 45cm Loi mt cu ny thng c s dng v chng thm tt, d thi cng v sa cha. Hnh 2.4. Cu dy vng Bn BTCTLp va mLp BT bo v BT atphanLp cch ncBT xi mng110

2. Mt cu b tng xi mng Loi mt cu ny c lp va m v lp cch nc ging loi mt cu c lp ph bng btngatphan.Trnlpcchncllpbtngdy68cm,McM300,clict thp. Loi mt cu ny c cng tt, chng thm tt nhng sa ch kh hn. 3. Cc thit b khc trn cu a) L thot nc L thot nc c lm bng gang, nha PVC hoc tn un li thnh ng thng hoc cong. ng knhng ti thiu d =15cm, trnming phi c np yrc khi chui vo ng. u di ca ng thot nc phi nh ra khi b mt di ca b tng t nht 10cm trh nc chy tt vo mt bn. Quynh:c1m2bmtc1cm2dintchlthotnc.Khongcchgiaccl thot nc xa nht l 15m, thng b tr st l ngi i. b) Khe co gin Khe co gin trn cu m bo cho kt cu nhp c th chuyn v t do di tc dng ca hot ti, thay i nhit , t bin v co ngt ca b tng. Khe co gin phi m bo bn, chng nc r r xung gi cu v m tr, phi m bo xe chy qua li c m thun. Cc khe co gin c b tr theo hng ngang cu, trong cu gin n chng c b tr trn tt c cc m tr, trn cu lin tc c b tr trn 2 m. NG THU NC LI CHNd Hnh 2.6. ng thot nc 111 Hin nay, vi cc cugin n chuyn v nh t 12cmdng kheco gin h, 2 u dm bt thp gc ( Hnh 2.7a) v t mng thot nc bng cao su pha di; khe co gin knpdngchocccunhpnhctngphngnclintcvtngbtngboh gin on qua khe co gin ( Hnh 2.7b). Khe c b phn co gin bng ng thau hoc tn trng km. Cc cu L = 2033 thng dng khe co gin cao su bn thp (Hnh 2.7c) Cc cu ln nh cu dy vng thng dng khe co gin mun c) Lan can LancanthnglmbngvtliuthphocBTCT.Lancanthngcchiucao 11.2m 2.2. CU BTCT THNG GIN N B TNG TI CH I. Cu bn ti ch. 1. c im v iu kin p dng: Ktcunhpcubnnginmtnhphaynhiunhpcuimcutoktcu n gin, d thi cng, v c chiu cao kin trc ca bn thp. BT nhaTm tnu dmVa nha nga) b) Hnh 2.7. Khe co gin Dm Dm dc Di cao su c) 112 Vic c ti ch kt cu nhp cu bn c u im l thun tin cho vic to hnh dng, kin trc ca cu, iu ny c bit c ngha khi xy dng cc cu vt ng cc thnh ph yu cu c hnh dng kin trc p hay i vi cc cu nm trn tuyn ng cong. CubnBTCTthngthchhpvikhNunhpL6m.Mtrcthlmtr do hay m tr cng. Hin nay, kt cu m nh ang c dng nhiu thay th kt cu m tr nng bng xy hay b tng ti ch nh trc y.

2. Mt ct ngang v cc kch thc c bn Mt ct ngang kt cu nhp thng c dng tit din ch nht. ng ngi i lin khi vi bn chu lc v mt ng b hnh cao hn mt ng xe chy 25cm. Theo dc cu ngi ta ct gin on mt s v tr t mt bn b hnh n mt bn chu lc vi khe h 12cm bn mt tha khng tham gia chu un vi kt cu nhp. Chiucaobnhb= L .201151||

\| cnchiurngvchiudinhptnhtontuthuc vo nhim v thit k. Vt liu thng s dng b tng M250300, ct thp thng s dng ct thp trn c g hoc trn trn ( ct ai: 810, ct ch: 1632 ).TrongkhongB/6phibtrctthpdyhndophnbtitrngkhngu.Ct thpchulcphicunmtnalnlmctxin2vtrL/4vL/6,gcun thngt30o45o.Lpbtngbovkhngnhhn2cm.Ctthpphnbngang c b tr mt di ca bn. AA1/2A-A Hnh 2.9. Cu to cu bn BTCT ti ch L/4L/6AB/6A-AA20B/620113 II. Cu bn m nh. Cu bn m nh c cu to kt cu nhp ging nh cu bn ti ch. Tuy nhin, do kt cu m nh nn kiu cu ny cn c thm b phn thanh chng pha di chn m, nh vy to thnh mt kt cu 4 khp gm kt cu nhp bn BTCT nm ngang, hai tng m thng ng v cc thanh chng nm ngang pha di. Ton b h thng c gi n nh nh p lc t t sau hai m lm cn bng ln nhau. T c im chu lc ny m trong thi cng phi lun lun phi m bo cn bng sut trong qu trnh thi cng cng nh s dng. a) M nh: Hai m nh lm bng vt liu xy, b tng hoc BTCT mng c ti ch. Chiu dy tng m bng 1/6 n 1/7 chiu cao m. Cng c th cu to tng m nh c chiu dy thay i theo iu kin chu p lc ngang ca t. Ngoi ra hai bn tng m cn c hai tngcnhchnttrongphmvita-luycannngucu.Haitngcnh thng c b tr xin t 10o20o so vi phng ngang cu hng dng nc chy di cu c m thun. Mng haitng cnh thng c tch c lp ra khimng Thanh chngKt cu nhp bnM nhM nhTi trng thng ngBiu p lc ngang ca t Hnh 2.11. S tnh hc ca cu bn m nh 114 ca tng trc bng cc khe h rng khong 2cm trnh nh hng ln nhau do ln nn khng u. b) Thanh chng phngkhnngmbplctNyxvphadngchycnbtrthm thanhchngdcbngbtnghayBTCTctitdin4040cmhoc4050cmcti ch. Khong cch gia cc thanh chng t 35m theo phng ngang cu. V tr cc thanh chngphittrngmngmhaylinktvibmngmphngccthanh chng b ln su di y b mng. c) Cht thp Kt cu nhp c lin kt vi hai m bng cht thp 2832mm. Cc cht thp ny t cch nhau 0.51.0m theo phng ngang cu, chng c nh v v chn sn t khi b tng tng m. Khi b tng kt cu nhp bn, ti cc v tr cht thp cha cc l ngknh55cm.SaunycclnyclpybngMattit-bitumbovcht thp. III. Cu dm c sn b tng ti ch. Xut pht t nhc im ca kt cu nhp bn l tn ti vng b tng chu ko v vng b tng chu nn trc trung ho ln dn n trng lng bn thn nngkhng vt c nhp ln. khc phc nhc im ny ngi ta b bt vt liu vng trc trung hovvngchuko,ctthpchukocbtrthnhcmhnhthnhdngkt cu dm c sn trn c s m bo chu lc v cngcho php vt c nhp ln hn. 1. Phm vi p dng Dng cho cc cu giao thng min ni, vn chuyn cc khi lp ghp kh khn, khNu thng t 221m. 2. Cu to Mattit-bitum 5cmd=28-32cm Hnh 2.12. Cu to cht thp lien kt gia KCN v m m 115 a) Dm chnh Lbphnchulcchnh,chiucao 20171 =Lhtutheoclygiaccsndm, cp ti trng xe qua cu. Dm ch thng c dng mt ct ch T v c 2 xu hng thit k: Dng t dm ch, khong cch hai dm ch t 36m nhng dm ch li c chiu cao lnt p sau m ln. Dng nhiu dm ch, khong cch gia hai dm ch tng t 1.22.5m nhng li tn vt liu do c nhiu sn dm. B tng dng M250300 b) Dm ngang Dm ngang c tc dng lin kt cc dm chnh, tng cng ngang cu v phn phi lcchoccdmchnh.Chiucaodmngangh1=(0.60.8)h;brngdmngang thng bng 20cm. Ti hai u dm chiu cao dm ngang h1 c th cao bng h. Theo dc cu c cch 46m li b tr 1 dm ngang. c) Dm dc ph Dm dc ph c chiu cao h2 = (0.60.7)h1, b rng 15cm d) Bn mt cu Lbphnchlctrctipthottitruynxung,ckhNutnhtonbng khong cch gia cc dm ch hoc dm ch v dm dc ph, chiu dy bn chu lc 10cm Hnh 2.13. Kt cu nhp dm BTCT ti ch a) Mt ct ngang b) Li ct thp 116 3. Nguyn tc b tr ct thp : Cc nguyn tc chung b tr ct thp: +Pht huy ti a kh nng chu lc ca ct thp: - B tr ct thp tp trung vng chu ko -Btrctthpminchunntngcngkhnngchunnchobtngti vng chu nn. - Hn ch s n ngang v tng kh nng chu nn. - Chng co ngt ca b tng B tr ct thp cng xa trc trung ho cng tt +B tr ct thp khng cn tr vic b tng. - Theo QT79 thkhong cch tnh gia ccct thp 1.5 ng knh ct thp v 5cm. -TheoTiuchuNn2001thkhongcchtnhgiaccctthpsongsongphiln hn: 1.5 ln ng knh danh nh thanh ct thp 1.5 ln kch thc ti a cp phi th hoc 38mm. a) Ct thp ch Cc ct thp ch chu ko c t trong phn di cng ca sn dm. S lng v din tch ct thp cn c vo tnh ton. Ct thp thng dng CT5 c ng knh t 14 32mm vi cc dng b tr nh hnh 2.14. Kiurirc(Hnh2.14a)cdngtxa,ckhuytimlkchthcbudm phi ln m bo khong cch gia cc ct thp. Tuy nhin cch b tr ny li lm cho s truyn lc, dnh bm gia cc ct thp vi b tng xung quanh s tt hn. 3-5cm>5cm3cm3-5cma) b) c) d) Hnh 2.14. Mt s s t ct thp ch trong dm BTCT 117 t cm (Hnh 2.14b,c) c thun li l tit kim c ch cha ct thp tc l c th lm sn dm nh i. Tuy nhin vic un nghing mt s ct thp ch lm ct thp xin gp kh khn hn. Cc ct thp t chng v hn c u im l to ra khung ct thp vng chc v n gin thao tc lp c khung ct thp (hnh 2.14d). Khi tng khung ct thp chu ko nhiu hn 4 s to ra s ngn lp b tng pha trong v b tng pha ngoi trong cc tng ct thphncakhung.khcphcngitadngthanhctthpdi2030cmtcch qung. b) Ct thp xin Thng c un ln t ct thp ch chu ko cn c vo kt qu tnh ton- phi hp trn biu bao mmen v hnh bao vt liu nh trnh by trong mn Kt cu cng trnh. Gc nghing thng t 30o600 v on b tr ct thp sin phi m bo bt k mt ct thng gc no cng phi ct qua t nht 1 lp ct xin. c) Ct thp ai Ct thp ai cng than gia chu lc ct vi ct thp nghing v b tng, ngoi ra gp phn cng cc ct thp ch, ct dc ph to ra khung khng gian cc ct thp cng. Ct thp ai b tr theo tnh ton v theo yu cu cu to. QT 79 quy nh: 1 ct thp ai khng c vng qu 5 hng dc ct thp chu ko v 3 hng dc ct thp chu nn. Khi thit k tun theo cu to sau: Trn on L/4 hai u dm c ly ct thp ai a 30cm. Trn cc on cn li a < 50cm, ng thi 3h/4. ng knh ct thp ai ly bng (1/41/3) ng knh ct thp ch v khng c nh hn 6mm, nn ly 812mm. d) Ct thp dc ph gim m rng vt nt do co ngt v phn b chng u hn trn dc dm cn phi t ct thp dc ph trn ton chiu cao khu vc chu ko ca dm cho n tn st y bn cnh trn. Trong phm vi 1/3 chiu cao pha dica dm c t ct thp dc ph ng knh 814mm, cch nhau 1012 ln ng knh. Trn phm vi chiu cao cn li t cc thanh 610mm e) Ct thp trong bn mt cu Ct thp trong bn mt cu thng t thnh cc li nm ngang, s tu theo s tnh ton bn. ng knh ct thp chu lc ca bn c ly theo cc quy nh sau: 118 -Khng < 10mm i vi bn mt cu t -Khng < 6mm i vi bn va h. Ni chung trong cc n thit k thng ly d = 1214mm i vi cu ng b. S thanhctthptrn1mdibncxcnhtheotnhtonnhngkhngcqu20 thanh v cng khng nh hn 5 thanh trn 1m di. 2.3. CU BTCT THNG GIN N LP GHP I. Cu bn lp ghp 1. Mt ct ngang v cc kch thc c bn n gin cu to, d thi cng v chu lc hp l ngi ta thng lm cc khi bn dngmt ct ch nht vi cc rnh lm haimt bndnh cho cu tomi ni ngang cugiacckhibntheokiuminicht(khngtruyncmmen,chtruyn c lc ct ). i vi cu bn lp ghp hoc na lp ghp trn ng t thngdng nht cng l dng cu bn m nh ging nh cu bn ti ch. Cc b phn ca cu gm kt cu nhp bn, hai m, thanh chng c phn chia thnh nhng khi nh c trng lng v kch thc tu theo iu kin vn chuyn v nng lc ca thit b cNu lp v c ch to sn trong nh my hay xng b tng. Ra ti v tr xy dng, cc khi lp ghp c cNu lp vo v tr v tin hnh lin kt li nh cc mi ni thnh kt cu cu hon chnh. Kt cu nhp bn c phn thnh cc khi bn lp ghp bng bng cc mi ni dc theo chiu di nhp. Chiu rng ca mi khi tu iu kin vn chuyn v cNu lp, thng thng t khong 11.5m. S lng khi bn ph thuc vo kh rng ca cu. Hai tng m nh c th bng b tng c ti ch, nu dng cc khi b tng lp ghp cnlucutolinktgiachngbngccchtthp32.Tngmcngcth bng xy. Cc thanh chng theo hng dc cu thng c ch to sn bng BTCT, khi lp vo v tr cu m bo chnh xc vo chn tng ca hai m nh. n gin hn, cc thanh chng c th bng b tng ti ch. 30940980 119 Cc tng cnh xin ca m c th bng b tng hay xy. Pha sau m thng dng bn qu c chiu di 2.53m bng BTCT chuyn tip m thun cng gia cu v nn ng u cu. Pha trn cc khi bn lp ghp c cu to lp ph mt cu bng BTCT hay b tng nha. 2. Mi ni Mi ni cng: Sau khi cNu lp cc khi bn vo v tr ngi ta tin hnh hn ct thp ch c b tr sn cnh bn khi bn. Chiu di ct thp ch khong 20 ln ng knh ct thp ngang ca bnMi ni cht: Dc theo khe h ca mi ni c t vo cc on ct thp c ng knh t 35mm c un thnh hnh l xo. Sau ton b mi ni c lp y b tng M200. Cng c th khng cn dng ct thp l xo, sau khe h ca mi ni s c lp y bng b tng M300, c s dng ph gia trng n. Docutonhvynnminikiuchtchtruynclcctchkhngtruyn c mmen 3. u nhc im a) u im Soviphngphpctich,vicsdngktcubnlpghpkhcphcc kh khn v chi ph lm gio. Chi ph vn khun cng tit kim c do vic c ln lt tng khi nh v lun chuyn b vn khun nhiu ln. 6 5 52Ct thp l so d=5cm Hnh 2.16. Cu to mi ni cht 120 Cc khi lp ghp c ch to trong nh my hay cng xng m bo cht lng b tng hn. Thi gian thi cng c rt ngn b) Nhc im Kt cu thiu tnh lin khi do phi cu to mi ni gia cc khi lp ghp II. Cu dm c sn lp ghp 1. Mt ct ngang v cc kch thc c bn a) Mt ct ch T, bn hng, mi ni kh Cc khi dm c cu to hon chnh gm c na dm ngang. Khi lp vo th ch ni ti v tr dm ngang, khi bn mt cu lm vic theo s hng, chiu dy bn s gim dn t ch st nch dm ra n tn u mt hng. Nch dm c th vt hoc khng. Bn dy t 812cm. Sn dm dy 20cm, khong cch cc dm 90170cm. Phm vi s dng: L= 915m. b) Mt ct ch T, bn kiu dm, mi ni t Cckhiccutocctthpch,sautinhnhlinktcckhidmbng mi ni t, to lin kt h mng dm lin khi hon chnh. Do bn mt cu lm vic theo s ngm hai cnh hoc 4 cnh do chu mmen 2 dubtrhailictthp.Bnmtcudy1520cm.Khongcchgiaccsn dm 1.42.5cm. Phm vi s dng: L =921m. Hnh 2.17. Mt ct T c dm ngang (mi ni kh) Ct thp ch>30cm Hnh 2.18. Mt ct T mi ni t 121 c) Mt ct ch Cuimlnnhtrongthicng,cngngangln,miningin,cth chn va xi mng hay b tng vo khe h hoc dung bu lng cng cao lin kt hai sn dm st nhau. d) Mt ct I Mt ct nhp gm cc dm I lp ghp trc, sau tip tc lp ghp hoc c b tng ti bn mt cu to kt cu lin hp na lp ghp. Chiu cao dm t lp ghp thng trongkhongH/L=1/71/20vichiudinhpkhong12,15-24m.Chiucaodm ngang nn lybng 2/3 ca chiu cao dm ch, chiu dysn dm ngang vo khong 1215cm l hp l. 70-11080-120100-120 Hnh 2.19. Mt ct Tm BTCT c sn Mt cu BT atphan Hnh 2.20. Mt ct ngang cu c s dng dm mt ct I 122 2. Mi ni a) Mi ni phn bn * Loi c ct thp ch Cu to: Khi thi cng dm ngi ta s sn cc on ct thp ch ti v tr bn mt cu. thng thng khong t 2030 ln ng knh ct thp bn u im: Kt cu c tnh lin khi cao v truyn lc 1 cch ng u Nhc im: Thi gian thi cng lu do hn ni ct thp v b tng ti ch. * Loi dng bn thp ch kt hp vi bn thp hnh - Cu to: Sut dc chiu di mi ni ca c hai bn bn ngi ta chn sn cc bn thp ch c neo chc chn vob tng ca bn.Saukhi ra cng trngngi ta t cc bn thp hnh cch 5080cm 1 tm v hn bn thp cho c hai bn. - u im: Thi gian thi cng nhanh, chu c c mmen v lc ct - Nhc im: Tn thp, truyn lc thiu ng u.

b) Mi ni dm ngang Dm ngang c th dng ct thp ch nh mi ni bn hoc dng mi ni bn thp ch BT ti chBn thp hnhBn thp cha)b) Hnh 2.21. Mi ni bn Ct thp ch BT ti ch Hnh 2.22. Mi ni dm ngang 123 2.4. CU BTCT GIN N D NG LC I. Khi nim chung cu BTCT d ng lc. - Mc ch ca vic to d ng lc nhm iu chnh tr s ng sut ko trong b tng bng cch to ra ng sut nn trc trong n, nh m kim sot c kh nng chng nt ca kt cu .-Nguyntcchungcaccbinphptodnglcltmcchnotorang sut ko trong cc ctthp cng cao risau li dng tnh dnh bm ca cc ct thp vi b tng hoc dng mu neo truyn ng lc ko trong ct thp vo b tng to thnh d ng lc nn trc b tng. -C hai bin php to d ng lc ( ko trc, ko sau ) u i hi h thng thit b ng b: b cng cp, mu neo, kch, ct thp cng cao, thit b ph tr v cc buc cng ngh ng b . II. Cng ngh ch to v cu BTCT DL ct thp ko trc v sau khi b tng1. Ko cng ct thp trc khi b tng (ko cng trn b).

-Qu trnh cng ngh c gii thiu trn hnh v -Cc ct thp cng cao c th c ko cng trc bng bin php ckh hay bngphngphpnhit.Saukhickocngctthpcngcaoclinkt B c nh B kp gi u ct thpB kp nh v im unDm BTCTThn b c nh CT uc ko cng Hnh 2.23. S ko cng ct thp trc khi b tng 124 cht ch vo cc b c nh nh cc neo tm thi. Tip ngi ta lp t cc ct thp thng, dng vn khun ri c b tng dm . Khi b tng dm c bo dng cng th tho b cc neo ngoi tm thi. Khi cc ct thp cng cao khng cn b neo gi cht vo cc b c nh nn c xu hng co ngn li nh c .-Docccneongmbtrtrcnmtronglngkhibtngvdoclc dnh bm gia cc ct thp v b tng nn s co ngn ny b cn tr . ng thi trong b tng xut hin d ng lc nn tn ti lu di. Cc on ct thp tha nh ra khi u dm c ct b, cc neo ngoi tm thi c s dung li ch tao dm khc . -B c nh c th bng thp hoc BTCT xy trn mt t . Cng c th b cng c b tr ton b trn mt toa xe di ng theo ng ray i qua cc phn xng ca nh my sn xut BTCT, ph hp vi dy chuyn cng ngh- Phng php ny thch hp vi iu kin sn xut BTCT DUL trong nh my c th m bo cht lng cao ca dm. Do iu kin vn chuyn t nh my n cng trng theongst,ngthayngthuphctpnncccukinBTCTDULch to theo phng php ny phi hn ch v kch thc v trng lng. Chiu di ln nht cacukinxpx33m.Nhvyphngphpnychphhpchoktcudmhay bn gin n. -Nhc im ca phng php cng trn b l i hi nhiu thit b v ch ko cng ct thp c theo s thng hay s gy khc . 2. Ko cng ct thp sau khi b tng ( ko cng trn b tng ) -Trong qu trnh b tng dm, ngi ta to ra cc ng ng rng trong lng khi b tng theo cc dng ng cong hay ng thng d kin . Sau khi b tng cng cn thit, ngi ta lun ct thp cng cao vo cc ng rng ny ri dng kch thy lc ko cng ct thp, chn kch t trc tip ln b mt b tng u dm cn m cp ca kch kp cht ly neo hoc cc u ct thp m ko cng ra: Khi t d ngsutkocnctrongctthptheotnhtonthitkthtinhnhcnhccneo ngoi vnh c gi du ct thp vo b mt b tng u dm ri tho kch , on ct thp cng cao tha s c ct b. Tip theo ngi ta bm va b tng vo ng cha KchCT uc ko cng Hnh 2.24. S ko cng ct thp sau khi BT 125 cplpknphnrngcnlaigiactthpvccngng.Ccneongoicng c b tng bt kn chng g. -ui