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2. GS. NGUYN NH CNGI ANH IIHN l[ HIENIEIIHIINHBIIIHIPSCH XUT BN K NIM 40 NM THNH LPTHNG I Hc XY NG 1966 - 2006NH XUT BN XY DNG H NI - 2006 3. LI NI UTllt kkt C btng C thp gm nhe^`tt cng on trong tnll ton tit din Ct l mt phz tng d qztan tI'ng v chla tmg mt SIv,l ` plllc rp nh Ct Clltt nlz lch tm xl'Iz; Ct c tt din t'n /zoc clit? T... Nhtng Vt? d` d tuy C llc ` C'p tt` tI'Ong Tlt clztlt thit k,CLng nh tl`0ng mt SI gl`0 tI'nh v Sclz tlzam kllo nhtfng I/lng ml` dc tI'nI1 by dng gllyn l Chiang m lt dlrc Cll tl` lla, CL_l tl1 [la q(` tlza vn dng tI'C tl^jJ. Ngay tl'ng hp n gn l tit din cht? nllt ClL4 nn lcll tm pllng, tlly dtrc g thll nh`u ti Ilt, c CL_t tllha bng cc Cng IIIIC IIHZ ton nhlmg Clng cn Clllta tmg mt vi vz d Cn lm Sng t hn.Tl'0ng klz tI11^t kl Cc Cng tl`t`nl1, nh`u k S v Snh vin tlzttng gp Cc vh d` vL`ra zll v yll cll tc g g p. `ll d thi tllc tc gi bn Son ti lbt ny n/lm gl thl'll mt SI vh ` v` tlnIl ton, [ly vng C tIzCLtngcJ dltc cc t/lng tin v phtrzg plzp Cn thit C/10 Cc Cn b nghin cmt v tl1e^tky l ti ltl tham kho m mt Stn dung vt 'a ngoi Cc gio tl`n/1 thllg tlllrng bc i hc. NI171g vtt ` tnh ton clz yl tlzeo st tll ClIull tllt k,IZl'n lnl C`a Vit Nanz TCQJVN 356 .' 2005. Tuy vy C mt SIVC/l ` dlrc m I'ng, g tllll tlleo nl1`LI qllan l klzc nllall nlzm gip c gi hl Sll v l`ng lln v l tlzuyt btng ct tllp.Tll Clttt TCQVN 356 - 2005 lrc ban lnlz v c lzu [l_rc L tllng 11 nm 2005, dlng dt[1ay hilll chut TCVN 5574 - 1991. TI'ong qtl trnh bn Son ti [l'lI ny (2004 - 2005) tc gi da vo tiu Chttt TCVN 5574. Kh TCDVN 356 dlrc' Czg b, th ti [ll ny Chbn xong v Chuql b em n. Tc gid d kp thi SLa Clzza bn tho tlleo ni dltng v k /liu ca TCVN 356. Cllc Chn I'ng nhL^71g vl d qltan tl'ng v c bn dlrc trnh by theo TCDVN 356. Tlty vy c mt vi v dl_1 dtng S,tItt Cl ca TCVN 5574 tc gi vn ngyn, v th l^ng n k/lng gy I'a Izllm [n v` nhn thc, khng nll lzzng dk mc d chnh Xc cla ti Ilt.Nm 2006 TI'lf'ng i hc fty drng k ml`m 40 Izm tlznl lp v 50 nm do to. Tc gi vlt ti Ill ny cng l gp plln vo l k nim .V thi gian c hn nn tc gi Chl mi Cp dh vic tlnh ton mt SII din ct m Chlta da tlzm cc vh d` khc nht xc nh n_0^l` lc, C to cht' tit. Hy vng c thq b7Sllng v hon Cl1lnh vo dp khc.Tc gi xn hoan nghnh v t lng bt 071 Cc bn c cht ra, gp kh cho nhlng Sal` St Cl-a ti LI d, tc gi c tl1h0n thin hn.Tc gi 4. Chng 1 I CNG V KHUNG V CT BTNG CT THP1.1. CC BC THIT K KT CU KHUNGThit k kt cu btng Ct thp ni Chung v kt cu khung ni ring thng theo th t Cc bc Sau:1. Gii thiu, m_t kt Cu.2. La Chn phng n, lp S kt Cu.3. Chn kch thc S b Cc tit din, chn Vt liu.4. Tnh ton cc ti trng, d kin Cc tc ng5. Xc nh n lc, t hp ni lc.6. Tnh ton tit din, kim tra cc iu kin s dng.7. Thit k, ch tit, Chn Cu to, th hin.Cc buc trn c quy V Cc giai on thit k gm: Thit k, C S (S b), thit k k thut v thit k bn V th cng.Vi Cc Cng trnh ln thit k theo ba giai on trong tht k C S gm ni dung Cc bc 1, 2, 3; thit k k thut gm ni dung Cc buc 4, 5, 6 v thit k bn V th Cng gm ni dung bc 7.Vi cng trnh va v nh thit k theo hai giai on hoc mt giai on (thit k' trc tip bn v thi Cng) tuy vy vn thc hin c 7 bc trong c mt s bc c th lm gn ng, n gin ha.H S thit k gm c bn thuyt minh v Cc bn v. Ni dung ca Cc bc C th c trnh by trong bn thuyt minh hoc trong Cc bn V. bc 1 Cn trnh by tn gi ca kt Cu, v tr (trn mt bng kt Cu ca Cng trnh), nhim v, c dim ca kt Cu.Bc 2 l bc kh quan trng trong d vic xut Cc phng n, phn tch v S0 Snh Chn c phng tn hp l c ngha ln n nhiu mt. Mt phng n hp l ca kt Cu l m bo c yu cu ca kin trc (yu Cu v s dng), bo m bn vng, S dng tit kim vt tiu v thun tin cho th Cng. 5. Vic Xut cc phong n c th theo hai cch:a) Da trn nhim v, c im ca kt cu m ra cc phong n c lp nhau (do mt s ngi hoc do mt ngi).b) Trc tin a ra mt phng n, phn tch u, nhc im ca phong n , trn C S tm cch khc phc nhc im m xut phng n khc.Vic lp S d kt cu mt cch ng n l rt Cn thit. i vi c ngi nh th l vic b tr kt cu tng th v v mt bng kt Cu. i vi kt cu khung th l Xc nh hnh dng, lin kt, cc kch thc c bn. Khi lp S kt Cu trc ht cn quan tm ti n nh tng th ca chng, sau mi Xem Xt n s lm vic ca tng b phn. Trong vic lp S khung th vn khung phng hoc khung khng gian l quan trng, vn ny c cp mc 1.2.Mt S Vn v t hp ni lc v chn kch thc tit din c trnh by mc 1.3 v 1.4.Ni dung chnh ca ti liu ny bao gm vic tnh ton tit din theo hai dng bi ton: tnh ton ct thp hoc kim tra.Tnh ton ct thp l khi bit ni lc v kch thc tit din cn Xc nh lng ct thp Cn thit, kh nng chu lc.Tnh ton kim tra l khi bit tit din v Ct thp cn kim tra Xem tit din c kh nng chu c ni lc cho trc hay khng.Vic chn kch thc bc 3 l S b, c th l hp l hoc Cha. nh gi kch thc tit din chn l hp l hay khng cn phi cn c vo kt qu tnh ton ct thp hoc kt qu kim tra. Nu kch thc chn l qu bt hp l (qu b hoc qu ln) th cn phi chn li v tnh ton li.1.2. S KT CU KHUNG Khung gm c cc Ct, cc dm lin kt vi nhau v lin kt vi mng. Trong S khung cc Ct v dm c thay bng ng trc ca n. V hnh hc v s lm Vic, phn bit khung phng v khung khng gian.Khung gi l phng khi truc cc b phn ca n cng nm trong mt mt phng v cc ti trng tc dng trong mt phng . Mt phng c gi l mt phng khung hoc mt phng un.Khung l khng gian khi trc cc b phn khng cng nm trong mt phng hoc tuy cng nm trong mt mt phng nhng c cilu ti trng tc dng ngoi mt phng khung.Trong kt cu nh, khung thng c cu to thnh h khng gian (khi `khung). H khung khng gian C th c Xem l gm Cc khung phng lin kt vi nhau bng cc dn, ngoi mt phng khung.6 6. Ty theo phong n kt cu chu lc chnh ca nh m h khung c th thuc v nh khung hoc nh kt hp.Vi nh khung, h khung chu ton b ti trng ng v ti trng ngang.Vi nh kt hp (Vi li cng, vch cng) khung chu phn ti trng ng trc tip truyn vo n v chu phn ti trng ngang c phn phi cho n.Tuy h khung l khng gian nhng v s lm A B vic v tnh ton c th theo s khng gian hoc theo s phng ty thuc vo ti trng tc dng v mc gn ng c th chp nhn c. phn bit trng hp lm vic ca khung l D C phng hay khng gian, xt trng hp h khung n gin gm 4 ct A, B, C, D v 4 dm lin kt cc u ct (hnh 1.1). Kho st cc trng hpkhung chu ti trng ng v ngang. Hnh l'1' H khung don gina) Khung Chu t trng ngTi trng trn Sn truyn vo khung ty thuc vo kt cu sn theo cc trng hp sau:Tl'ng hp I. Sn lp ghp dng panen t theo mt phng (hnh 1.2a), ti trng t panen chi truyn ln hai khung phng song song, hai khung ny lm vic theo khung phng, cc dm Vung gc vi cc khung ny chi ng vai tr lin kt, khng chu lc.)I, > 21, I l,2l, j Hnh 1.2. Cc trttg hp sn truy`n ti trng ng vo khung. 7. Trng hp 2. Khi bn Sn l ton khi k in 4 dm m t s gia Cc cnh bnI x , I, u A/ 0 'P . -2- > 2 , Xem gn ung ban chu uon mt phng, tai trng t ban truyen len hai khungIl . i din, mi khung lm vic theo khung phng (hnh 1.2b).Tlng hp 3. Bn k ln 4 dm m ty S cnh ban -- < 2 , ban chu un theo haill phng, truyn ti trng ln c 4 dm, h khung lm vic khng gian (hnh 1.20). Tl'ng /lp 4. Khi dng thm Cc dm ph (dm sn) bn sn, dm ph k ln dm khung. Ty theo S b tr dm ph m xt khung lm vic phng hoc khng gian.b) Khltng chu t trng ngang (gli)Ty theo phng ca ti trng. Khi xt ti trng gi theo phng ngang (hnh 1.3a) th cc khung AB v DC lm vic theo khung phng. Khi xt gi theo phng dc (hnh l.3b) th hai khung AD v BC lm vic theo khung phng, cn khi Xt gi theo phortg xin th h khung lm vic khng gian. Ch rng khi xt tc dng ca gi ngi ta Xem sn l cng V cng trong mt phng ca n nn sn lm c nhim v truyn ti trng gi vo cc khung.a) A B b) A B C) A B D CC D C TG / Hnh 1.3. Cc trng hp h khung Chu ti trng ngangVi h khung ca ton nh cng tin hnh phn tch nh trn Xem xt l khung lm vic theo phng hoc theo khng gian. T ch phn tch s lm vic ca Sn quyt nh cch truyn ti trng ng. Khi m c th Xem ton b ti trng ng trn sn chi truyn ln cc khung ngang (hoc khung dc) th Cc khung y c Xem l lm vic theo khung phng di tc dng ca ti trng ng. Ngc li th phi truyn ti trng ng ln c Cc khung dc v ngang v c khung khng gian. I`Vi ti trng ngang, thng ngi ta da vo mt bng kt cu nh xt trng hp bt li ca ti trng. Khi mt bng nh hp m di, cng tng th ca nh theo phng ngang l kh b so vi phng dc. Lc ny tc dng ca gi theo phng ngang s bt li hn do chi xt gi theo phng ngang (hnh 1.4a) v mi khung ngang c tnh theo khung phng, chu tc dng ca phn ti trng gi phn phi cho n.8 8. b) Gi __, --D Hnh 1.4. Cc trng hp bt li ca gi vi kt cn nh ~ / IKhi mt bng kt cu nh c dng gn vung, cng tng th ca nh theo hai phng gn bng nhau th phi xt tc dng ca gi theo ba trng hp: ngang, dc v Xin (hnh 1.4b). Vi gi dc v ngang nh cc khung lm vic phng Cn vi gi xin, khung lm vic khng gian.Tnh ton ni lc khung phng l bi ton kt cu thng thng, c th gii bng nhiu phng php khc nhau. Hin nay cc bi ton khung phng ch yu c gii nh vic s dng cc phn mm tnh ton trn my tnh.Tnh ton ni lc khung khng gian l kh phc tp v thng Chi c th gii nh cc chng trnh kh mnh. C th gii gn ng bi ton khung khng gian bng cch a V bi ton phng theo cch phn chia h khung thnh cc khung dc v khung ngang, trn mi khung Xp t cc tnh ti v hot ti tc dng ln khung . Gii ton b cc khung dc v khung ngang theo trng hp khung phng. Ni lc trong dln ca khung no l ca dm cn ni lc trong ct l bng tng ni lc trong ct y ca hai khung giao nhau.1.3. T HP NI LC KHUNG 1.3.1. i cng v t hp ni lcKhi tnh ton ni lc khung cn tnh ring ni lc do ti trng thng Xuyn (tnh ti) v ni lc o cc trng hp khc nhau ca ti trng tm thi (hot ti). Cui cng Cn t hp tim ra cc gi tr ni lc bt li.Vi cc khung phng thuc kit cu nh dn dng, trong t hp C bn cn Xt 6 trng hp ti trng Sau:1. Ti trong thng xuyn (tnh ti) (hnh 1.5a).2. Ti trng tm thi cch tng cch nhp trng hp 1 (hnh 1.5b).3. Ti trng tm thi cch tng cch nhp trng hp 2 (hnh 15.c).4. Ti trng tm thi trn ton b dm (hnh 1.5d).5. Ti trng gi t tri sang (hnh 1.5e).6. Ti trng gi t phi Sang (hnh 1.5g). 9. G P1 P2 d) et 9)-D -D _ 500 mm th dc theo cnh h cn t ct thp dc cu to c ng kinh t 12 +l6mm. Khong cch gia cc trc cc thanh ct thp o dc theo cnh h l S khng c ln hn 400mm. Din tch thanh ct dc cu to khng nh hn O,O0lSb1 vi b = min(0,5b v 200mm). blHnh 1.15. SC t tllp dc Cit lo 500 < h < 850850dc max (C tiu chun quy nh aj Z 0,3dc).B tr ct thp dai dc theo chiu cao ct ty thuc vo kt cu c yu cu chng ng t (khng chn) hay khng.Vi kt cu bnh thng (khng khng chn) khong cch ca ct thp ai trong ton b ct (tr on ni buc ct thp dc) l ad S otd dc min. Ly ud = 15 khi t s ct thp g S 0,03 v Ot 10, khi s > 0,03, ng thi ad S 400mm.Trong vng ni ct thp dc cn phi t Ct thp ai dy hn vi khong cch khng qu l04>dc min. Trong on ni buc ct thp dc phi c t nht 4 ct ai (hinh l.l6a).Vi kt cu c yu cu khng chn ct thp ai cn c t dy hn trong on gn St vi nt khung. Mc t dy ca ct ai ph thuc vo cp chng ng t ca cng trnh (hinh l.l6b). Ngoi ra cn c yu cu t ct ai cho ct trong phm vi nt khung khi nt khung c dm lin kt t 3 mt bn tr xung.V hnh dng, ct ai phi bao quanh ton b cc ct thp dc v t nht cch mt thanh ct thp dc c mt thanh dt vo gc ca ct ai (hnh 1.16). Trng hp cnh b S 400mm m trn t 4 thanh ct thp dc th c th khng tun theo quy nh va nu. (Tiu chun Php BAEL quy nh mi thanh ct thp dc c t) 2 20 u phi c neo gi bng ct thp ai khng th b un cong bt ra khi btng).I hS500 a) b) bS400 Hnh 1 .I 6. Bi tr ct thp ai a) Trong Ct bnh thng; b) Trang Ct c yu cu khng chn24 24. Ct ai lm vic chu ko do u mt phi c neo chc chn, thng lm mc neo gp ot S 45 vi on thng u mt S Z 3ai. Trng hp lm neo gp 90 th on thng S 2 Sa v cn dng dy thp buc u mt vo vi thanh ct ai, trnh cho khi ct ai chu ko mt bt ra ngoi (hnh l.l7). Khi dng thanh neo n, hai u phi cmc neo tiu chun vi S 2 3dai.Szd Hnh 1 .I 7. Neo ci thp ai1.5. NI LC V LCH TM1.5.1. Nn ng tm v nn lch tmCt chu lc nn N l ch yu. Ngoi ra ct cn c th b un theo mt phng hoc hai phng.Khi ct chi chu mt lc nn N t ng dc theo trc ca n, ct chu nn ng tm. Thc ra nn ng tm chi l trtrng hp l tng, trong thc t rt t khi gp. Kh nng chu lc ca ct chu nn ng tm l No c xc nh theo cng thc:NO = cp(RbAb + RSCASI) (l-6) Rb, RSC - cng tinh ton chu nn ca btng v ca ct thp. Ab, AS, - din tch tit din btng v ca ton b ct thp c. tp S l - h s gim kh nng chu lc do un dc (h s un dc). Xc nh (ptheo cng thc thc nghim (l.7), dng c khi 14 < ?L S l04. cp = 1,028 - 0,00002887t2 - 0,00l67 (1-7)tt = - - d mnh ca ct (Xem cng thc (1-2)).IminKhi tt S14, b qua nh hng un dc, ly tp = l. Ct va chu nn N Va chu un M c i thrih S lc nn t lch tm (hrlh M1.18) v c gi l ct chu nn lch tm. Gi tr ei = c gi l lch trn tnh hc.25 25. Khi mmen un M tc N dng trong mt phng cha trc i xng ca tit din lc trng hp nn lch tm , e:-l phng, khi M khng nmtrong mt phng i xng va nu, c trng hp nn lch tm xin_ Hnh 1.18. S ct chu ntt lch tm1.5.2. lch tm v lch tm ngu nhin v M , , ` x ' 1 -1 Ngoai lch tm el = , trong tinh toan con cn k n lch tm ngu nhinea gy ra bi nhng nhn t cha xt n c nh sai lch do thi cng, btng khngng nht v.v... Quy nh v vic xt lch tm ngu nhin ea trong cc tiu chun thit k l khc nhau.a) Theo tiu chun Vit Nam TCVN 5574 - 1991, (tiu chun c) lch tm ban u eo tnh ton l:e0=el+ea (l-8)i vi cc cu kin chu nn c S tnh nh hoc l b phn ca kt cu sieutnh nhng Chu lc nn trc tip dt ln n th gi tr ea ly khng nh hn 1/25 chiu cao tit din v khng nh hn cc tr s sau:20mm i vi ct v cc tm tng c chiu dy t 250mm tr ln. 15mm i vi cc tm c chiu dy 150 + 250 mm 10mm i vi cc tm c chiu d'y di 150mmi vi cc b phn ca kt cu Siu tnh khng chu lc nn trc tip cho php b qua lch tm ngu nhin (ea = 0).b) Theo tiu chun TCXDVN 356 Z 2005, lch tm ea trong mi trng hp ly khng nh hn l/600 chiu di cu kin v 1/30 chiu cao tit din. lch tm ban u eo ly nh Sau:- Vi cu kin ca kt cu siu tnh:eo = max (el ; ea) (l-9)- Vi cu kin tnh nh, Xc nh eo theo cng thc (1-8).c) Tiu chun Php BAEL - 99 ly ea = max (I/250; 20mm) v tinh eo theo (1-8).d) Tiu chun Anh BS 8110 quy nh lch tm eo khng nh hn lch tm ti thiu bng gi tr ln hn trong hai gi tr 1/20 chiu cao tit din v 20mm.26 26. 1.5.3. nh hng ca un dcCt c mnh ln c th b un dc lm cho n b cong (hinh l.l9). Lc ny lc nn N gy ra thm mt mmen th cp M2 = Nez vi ez l chuyn v tng i ca tit din ang xt So vi v tr t lc N.Mmen un t M tng ln thrlh M, = M + M2. Vic tng M nh vy l tng ng vi vic tng lch tm t eo thnh e'o = eo + e2.Tiu chun thit k ca cc nc al Xt vic tng lch tm ny theo cc cch khc nhau.a) Tiu chun ca Php BAEL v ca Anh BS 8110 a ra cng thc thc nghim xc nh ez.Tiu chun Php:12 e2 = 0,0003- (2+ot,p,)ot, v p, l cc h s k n nh hng ca tc dng di hn v t bin Hnh 1.19. nh hngca btng (ot, = O,7+lh; p, = 2 +1,5) ca "h dc ., o 1 lo 2 Tieu chun Anh: ez = Kh -- 2000 ck S l h s ph thuc mc chu nn ca tit din. C - cnh b ca tit din.b) Tiu chun ca Nga, Trung Quc, Vit Nam Xt vic tng d lch tm theo h s nhn.eg,=e0+e, =l+-2je0=neo (1-10) . 0 -T Z l f h s xt n un dc. Trong l thuyt n nh chng minh c cng thc Xc nh n 1 l = N1-: NCT(l-ll)NC, - lc nn ti hn. Cng trong l thuyt n nh chng minh cng thc le (Euler) i vi cu kin bng vt liu n hi, ng nht:27 27. _ TCZEJ lE - mun n hi ca vt liu;N(1-12)CYJ - mmen qun tnh ca tit din. EJ - cng chng un ca tit din. tnh ton cu kin btng ct thp ngi ta khng dng cng thc (l-12) m dng cc cng thc thc nghim. C kh nhiu cng thc nh vy vi cc mc gn ng khc nhau trong tnh ton NC, theo cng chu nn Rb hoc theo mun n hi Eb ca btng, c k hoc khng k n lch tm v s c mt ca ct thp. Cng thc theo cng chu nn Rb. _ 4800RbJ _ 400RbAbh2N ( l -1 3) CT Ab = bh - din tch tit din ch nht. Cng thc tnh theo mun n hi Eh: 2, 5E . NC, = 2 ' (1-14) lo Cng thc tnh theo Rb c k lch tm v ct thp: hz Ncr =CsRbAbI-2 (I-15) 0C _ 66000 1""R+350 h 5+ 200S + 1Trong : R - mc thit k ca btng theo cng chu nn trung bnh (kG/cm2); g - t l ct thp.Cng thc trlh tlieo Eb c k n lch tm, ct thp v tc dng di hn ca lc nn.6,4E V NCT: 12 bJ+uSJsJ (l-l6a)_ 0 I Ve-h s k n lch tm: Ve= O11e +0,1 (1-17a) O,l+-0 t h28 28. Kd - h s k n tc dng di hn ca lc nn: Mdh T Nhy M + Ny y - khong cch t trng tm tit din n mp chu ko (hoc chu nn t) ca tit din.Kd =1+ (l-l7b)Mdh, Ndh - phn ni lc do ti trng di hn gy ra.ES ot = ` , Ehvi ES l mun n hi ca ct thp.JS - mmen qun tnh ca tit din ct thp.Tiu chun TCXDVN 356 2005 cho cng thc tnh NC, trn C S ca cng thc (l-16a) vi cc h s chi tit hn.NCr=Cbb l 0*11 +0,1 +otsJ, (1-16b) l() l 0,1+ ..C. p Trong : e ly bng t s - nhng khng nh hn emin=0,5-0,011(-l-0,0lRb (Rb Z MPa)emin hJ tpp- h S Xt n nh hng ca ct thp ng lc truc n cng ca cukin. (tpp Z l - khi khng c ct thp ng lc trc (pp = l). Cb - h s. Vi btng nng v btng ht nh nhm A ly Cb = 6,4. Vi btng ht nh nhm B ly Cb = 5,6. (pi - h S, xc nh theo cng thc:E M B- h s, vi btng nng B = l; btng ht nh nhm A: B = 1,3; nhm B: B = 1,5.TI =1+BM - mmen ly i vi mp tiit din chu ko hoc chu nn t hn do tc dng ca ton b ti trng.M, - Nh trn, do ti trng thng xuyn v ti trng tm thi di hn. Tiu chun ca Trung Quc GBJ10 - 98 tuy cng Xt s tng lch tm bng h snhn T nhng khng dng cng thc (1-l l) m Xc nh T] bng cng thc thc nghim:29 29. 14005 h ho 0 l Q1 =;2 =1,15-0,01N hAb- din tch tit din.I I uu ^ .C r 'I I l 1 '7 'V 1 9 ' I Vi nhng ct ngan, Co manh be, 7 =S 14, co th bo qua anh hng cua un 1dc, ly e2 = 0 hoc T] = l. Qu trnh Xt s tng lch tm th hin trn hnh 1.20. Hnh 1.20. S tng d lch tmGi tr 11 tnh theo cng thc (1-1 l) l i vi tit din c chuyn v ln nht. Ty theo v tr tit din tinh ton m c th ly gi tr T] tng ng.Trn hnh l.l9a gi tr ez ln nht gia ct cn trn hnh l.l9b, C c ez ln nht chn ct, ti inh ct ez = 0 tng ng vi T = 1.Trong tnh ton thc t, thin v an ton c th Xem gn ng r l hng S trong ton ct. Tuy vy nu mun tnh ton chnh xc hon th cn da vo S bin dng bt li a ct ly gi tr Tl ng vi tng tit din.Sau khi k n lch tm ngu nhin ea v rlh hng ca un dc 11 th mmenun t gi tr bn u l M tng ln thnh M* M*=NneO (l-18)1.6. SLM VIC CA TIT DIN CT1.6.1. iu kin v bnTnh ton tit din ct theo phng php trng thi g hn. iu kin c bn m bo bn khi tnh theo trng thi gii hn v kh nng chu lc l:30 30. N = Ngh (1-19) Mu S Mg, (1-20) N - lc nn c Xc nh theo t hp ni lc. Ngh - kh nng chu nn ca tit din. Mu = Neu. Mmen un do lc nn N t lch tm gy ra i vi trc U (khng nm trong mt phng un). cu- khong cch t im t lc lch tm N n trc U chn (eu > 0). Mgh- kh nng chu un ca tit din ly i vi trc U chn. Xc nh Ngh, Mgh da vo s lm vic trng thi gii hn v kh nng chu lc. Trng thi ny c thit lp trn c s phn tch cc kt' qu nghin cu thc nghim,da trn cc quy lut chi phi s lm vic v cc gi thit c Xut. T lp ra cc biu thc ton hc tnh ton.Trong bi ton tnh ct thp hoc bi ton kim tra cn tha mn c hai iu kin nu trn trong c th ly mt iu kin theo du ng thc (=) cn iu kin kia theo du quy nh (S). Trng hp nn dng tm chi cn mt iu kin (l-19) trong Ngh = NO tnh theo cng thc (l-6).1.6.2. Cc kit qu thc nghim1.6.2.1. Quan h ng sut bh dngKt qu quan trng nht ca thc nghim l quan h gia ng Sut v bin dng t i 8 ca vt liu c gii thiu trn hnh 1.21. Rs l T i I : R l I I I I I l I l l I I 8 8 sa a) l bl b Hnh 1.21. Quan h gia Hnh 1.22. Quan h ng St - Sca vt lu.' - Edng trong tnh t0n.` a) Btftg,' b) Ci thp do. ` a) Btng,' b) Ct thp.31 31. dng vo tnh ton, cc quan h trn c n gin ha v sau khi a vo cc h s xt n an ton ( tin cy) th biu ng sut bin dng c ly theo hnh 1.22, trong :Rb, RS - cng tnh ton ca btng (v nn) v ca ct thp (v ko); sc, Sa - bin dng ca btng v ct thp trng thi gii hn;8,. z - gii hn bin dng n hi ca ct thp, trong E, l mun n hi. E .SVi btng, khi chu nn ng tm ngi ta cho rng khng nn cho btng c bin dng qu 2%o v nh vy ly sc = 2%o.. Khi trn tit din c mt vng chu ko mt vng chu nn, kh nng bin dng ca btng mp chu nn tng ln, tiu chun ca Php ly SC = 3,5%o, cn tiu chun ca M ly SC = 3%o.Vi ct thp, bin dng ca ct thp do khi b ko lt l kh ln (0,1+0,2). Tuy vy khi Xt s lm vic ca kt cu btng ct thp trng thi gii hn mt s tiu chunc quy nh hn ch gi tr ca Sa trong khong l0%o.1.6.2.2. S lm vic ca ci& kinThc nghim v s lm vic ca cu kin chu nn lch tm c tin hnh theo S trn hnh 1.23 vi lc nn N t cch trc mt on eo. Lm th nghim vi cc lch tm eg khc nhau v trong mi ln th nghim tng dn lc N cho n khi Cu kin b ph hoi.Kt qu thc nghim cho bit vi eo b ton b tit din chu nn v s ph hoi bt u t btng mp chu nn nhiu hn. Vi eo ln, mt phn tit din chu nn, phn cn li chu ko, btng chu ko c th b nt, s ph hoi c th bt u t vng btng chu nn hoc t ct thp chu ko.C hai quan im V ph hoi: ng sut v bin dng. Quan im v ng sut cho rng vt liu s b ph hoi khi ng Sut trong n t v `jt p Vt 1j Th Hnh 1.23. S tltc nghim quan im ny biu ng sut dng cu kin nn lch ,m32 32. cho tnh ton c thit lp t kt qu thc nghim v khng cn quan tm n cc gi tr SC, Sa.Quan im v bin dng cho rng S ph hoi c quyt nh bi bin dng ca vt liu. Theo quan im ny Xut pht lp S tnh ton l bin dng. T S biin dng, dng cc quan h hnh 1.22 Suy ra S ng Sut v dng S ng Sut lp cng thc.1.6.3. Gi thit tnh ton) lp cc cng thc tnh ton tit din btng ct thp ngi ta dng quy lut v cn bng lc, ngoi ra dn gin ha cng vic tnh ton ngi ta cn dng gi thit b qua s lm vic ca btng chu ko. Ring vi quan im v biin dng cn dng thm gi thit tit din phng v gi thit btng, ct thp c cng bin dng (ti mi v tr).1.6.4. S ng sut btng vng nnTuy xut pht ca hai quan im c khc nhau nhng S ng Sut trong btng vng nn c ly ging nhau v l mt ng cong nh trn hnh l.23c. ng cong c dng ca ng cong trn hnh l.2l hoc 1.22. Cn ch rng hnh 1.22 th hin quan h -8 theo hai trc vung gc cn hnh l.24b v l.24c th hin quan h - 8 theo hai trc Song song. Trc trung ha cchmp chu nn mt on XO, ti c S = 0 v = 0, trong on 8 < 2%o, quan h - 8 l blng cong, trong on 8 > 2%o, c lhng S, bng Rb (on AB trn biu ). Cl Nhm n gin ha vic tnh ton ngita thay biu c mt on cong OA vmt on thng AB trn hnh l.23c vi chiu dcao vng nn thc X0 bng biu hnh ch nht vi chiu cao vng nn tnh i X v ng Sut tnh di R'b. Xc nh X v R'b davo hailiu kin: gi tr ca hp lc Db v im t ca Db trong hai biu l trng Hnh 1-24- U718 Surng btng vng nn.' nhau. Trng hp tit din ch nht b rng ai S n lc" b Biu bn dng"_ p , I _ ` C) Bit d ng sut thc,' d) Bit d b th hai phng trinh Xac nh X Va R'b dn gin ha_33 33. l: Db = R'b bx v cb = (Xem hnh 1.24) trong Db v Cb c Xc nh theo Rhv X0. Trong trng hp chung c th biu in X theo XO v Rlb theo Rb nh Sau: X = 6x0 v R'b = [3bRb (l-21)Gi tr ca 9 v Bb ph thuc hirh dng vng btng chu nn v SC. Vi tit din ch nht c th ly Bb = l v 9 = 0,8 + 0,85 khi ec = 3%o, -:- 3,5%o . Vi tit din ch T c cnh trong vng nn ly Bb = 1 v 6 : 0,82 + 0,88 vi tit din trn B, z 0,9 + 0,95 v 6 = 0,8 -:~ 0,85.Gi tr 6 nh va nu dng cng thc (1-21) chi ng khi xo Sh. Trng hp nn lch tm m ton b tit din chu nn th trc trung ha nm ngoi tit din, X0 > h v XO c th tng n V cng trong lc gii hn ca X chi c th t in ti a l bng h. Khi m X0 > h v 6 = 0,85 c th dng tng qua gia X v xo theo biu thc (l-22) Sau y:` (xo - 0, 85h)hx = --- (1-22) xo - 0, 8235h1.6.5. ng sut trong ct thp1.6.5.1. Theo quan ditt ng sutVi trng hp ct thp c t tp trung trn cc cnh vung gc vi mt phng un l AS (chu ko hoc nn t hn) v A'S (chu nn nhiu hon) th cc ng Sut tng ng l os v 's c ly theo kt qa thc nghim nh Sau (hnh 1.25).a) Vi ct thp As. Khi As chu ko, os t gi tr ln khi ct thp t Xa trc trung ha v`a ngc li. Khong cch t As n trc trung ha l va = ho - X0 = ho - . Gi trln nht m os c th t n theo quy c tnh ton l Rs. t c iu ny th va phi ln hn mt phn no ca ho, tm t l ,lh0.Xva =hO -zotlho rt ra XS(l-(X1)6h0 =Rh0Vy iu kin os t n RS l:X Rho Ngi ta Xc nh gi tr R bng thc nghim, thy rng R ph thuc vo RS v Rb. c nhng cng thc thc nghim tinh ton R nhng thng c th tra bng ph lc 4.Khi X > R ho th os chu ko Cha t RS hoc khi X tng n mt mc no th os chuyn thnh chu nn. Tiu chun TCXDVN 356 : 2005 a ra cng thc thc nghimXc nh os:34 34. os =E2-1-)h--IJRS (l-23) 1-RCng thc (1-23) c dng cho btng c cp bng hoc nh hn B30, ct thpnhm Cl, AI, CII, AII, CIII, AIII. (Rs S 400) v chp nhn c khi X S ho cn khui CX > ho th ly os = -Rs. Cng thc (1-23a) do tc gi Xut cho kt qu dng c khi Rh0SXShVRsS40.os = [I - Rs (1-23a) h`Rh0Trong hai cng thc trn tnh c os > 0 l ng Sut ko cn os < 0 l ng Sut nn.b) Vi ct thpVi ct chu nt lch tm ct thp Ag lun lun l chu nn. lig sut trong Ag l og s ln khi khong cch t A'S n trc trung ha l v = XO - a' kh ln v ngc li. Gi tr ln nht m c th t ti l RSC. t c iu ny th v phi ln hn mt s ln no dy ca a' , c t l yl' ZX v =X0 -a'=-alzyla'Rt ra: X 2 (l + yl )6a' = la'Phn tch kt qu thc nghim thy rng 8, ph thuc vo RSC v thay i trong khong l,5+2 (6, tng ln khi RSC tng). n gin ha, chp nhn gi tr , = 2 cho mi loi ct thp (Vi R < 400MPa).SC '- Nh vy iu kin og t n R l:sc X 2 2a'Khi X < 2a' Xem l cha t nRg: Diu kin x 2 za' c ly theo tiuchun TCVN 5574 - 1991. Tiu chun TCXDVN 356 - 2005 khng a ra iukin cho og. Trong ti liu ny tc gi Vn gi li iu kin X 2 2a` nh l mt iu kin phn bit cc trng hp tnh ton. Vic ny chi c li l lm r rng hn s lm vic ca tit din v Hnh 1.25. ng sut cch tnh ton. trong c? thp q, 35 35. Vi ct thp c Rs S 400MPa ly RSC = Rs.Vi ct thp c Rs cao hn 400MPa th trong mt S tiu chun thit k cng chi ly RSC S 400MPa. L do l khi chu nn ng tm ly bin dng gii hn ca btng SC = 2%o v bin dng ca ct thp chu nn ti a cng bng khong y. Vi bin dng ng sut trong ct thp (c cng kh cao) khng vt qua gi tr.^ = SCES = 0,002 x200.000 = 400MPA(Ly mun n hi ca thp ER = 200.000MPa).Vi ct thp cng kh cao (AV, AVI, AVIII) TCXDVN 356 - 2005 cho php ly RSC = 500MPa trong mt s trng hp c bit.1.6.5.2. Theo quan rn bh dngXut pht t bin dng ca btng ti mp vng nn c quy nh, dng gi thiit tit din phng, khi bit v tr trc trung ha (bit X0) v v tr ca thanh hoc hng ct thp th i (hoi) s tnh ra c bin dng ca n l ci (hnh 1.26). IlllllIn,Hnh 1.26. ng sut tI'on g ct thp o',- c tnlz theo bn dng ici_Z hoi X0Si cc (l-24)X0 Khi: 8,l 2 8, th o, = Rs (1-Zsa) 18,] ST th o, = a,Es (1-25b)36 36. Trong : ST =SKhi tnh c Si > 0, i l ng Sut ko, khi Si < 0 c ng Sut nn.Theo quan im bin dng, cho i chu ko t gi tr Rs th:ei = hl -X0 cc 28T =-l- X0 ES Rt ra: X0 S T h()i (l-26a) 8 T = (1-26b) RS ec +Vi X = 6x0 th iu kin i t Rs l: X S Bi h,i (l-27a) B = e iu kin (l-27a) l tng t iu kin X0 S R ho trnh by trong mc 1.6.5.1. vi h(,l = ho (Xem hnh 1.25, 1.26).Vi ct thp chu nn, cho ng sut i t gi tr cng tnh ton v nnRSC th: lSiI=_Si = X0 -hoi ec 28T = RSC X0 ES ' . 80 Rut ra. X0 2 R hoiSC __ SCES v: X 2 82 h, (l-2821) 8, : 68 (1-28b)RSCac - Eiu kin (l-28a) l tng t iu kin X 2 2a' trong mc 1.6.5.1 (hnh l.26 v 1.25 cho thy h04 = a').1.6.6. Cc trng hp tnh tonT phn tch s lm vic ca tit din chu nn lch tm ngi ta a ra cc trng hp tnh ton. Trong vic ny cng c cc quan im khc nhau.37 37. Mt s nc u M phn chia ra hai trng hp da vo vng chu nn: Tit din chu nn ton b v tit din chu nn mt phn.Tiu chun thit k ca Nga, Trung Quc, Vit Nam phn chia ra hai trng hp nn lch tm ln v nn lch tm b da vo s lm vic ca ct thp AS, cng tc l da vo gi tr ca chiu cao vng nn X. Khi X < E_.Rh0 ct thp AS chu ko, ng Sut os t n Rs, Xy ra s ph hoi do, c trng hp nn lch tm ln. Khi X 2 Rh0 ct thp As c th chu ko hoc nn m ng sut trong n cha t n Rs hoc RSC, s ph hoi bt u t btng vng nn, c trng hp nn lch tm b. Tit din lm vic theo trng hp no l ph thuc vo tng quan gia M, N vi kch thc tit din v s b tr ct thp. Khi M tng i ln tiit din lm vic gn vi trng hp chu un, c vng nn v vng ko r rt. Nu ct thp chu ko AS khng qu ln th s ph hoi s bt u t vng ko, c trng hp nn lch tm ln.Khi N tng i ln, phn ln tit din chu nn, s ph hoi bt u t btng pha b nn nhiu, c trng hp nn lch tm b.Cn lu rng trong tnh ton thc hnh iu kin phn bit cc trng hp nn lch tm chi l tng di. C mt s trng hp, vi tit din v im t lc N cho, khi thay i ct thp c th chuyn s lm vic ca tit din t nn lch tm ln sang nn lch tm b v ngc li. Khi chuyn nh vy th gi tr lc dc gii hn m tit din chu c Ngh thay i theo.a eoN Hnh 1.27. Th d v` cc trng hp nn lch tmLy th d nh trn hnh 1.27. hnh l.27a, lch tm ca lc dc eo tng i b, thng l trng hp nn lch tm b vi kh nng chu lc N I. Nu gim ct thp AS tit din c th chuyn Sang lm vic theo nn lch tm ln vi kh nng N2 < N 1. hnh l.27b, lch tm eo tong i ln, thng tit in lm vic theo nn lch tm ln. Tuy Vy nu tng ct thp As th n mt lc no tit din s chuyn Sang lm vic theo nn lch tm b (vi ng Sut s trong As gim Xung khiAs tng) v lc dc N tit din chu c s cao hn.Vic phn bit trng hp nn lch tm ln hay b ch yu da vo gi tr chiu cao vng nn X, chi khi khng c cch no Xc nh c X th mi dng iu kin b tr, cn c vo lch tm eo.38 38. Chng 2 TIT DIN CH NHT CHU NN LCH TM PHNG2.1. S V CNG THIC C BN2.1.1. So v k hiu Xt tit din ch nht c cc cnh b, h. h - chiu cao tit din, l cnh song Song vi mt phng un. b- b rng, l cnh vung gc mt phng un. Trong nhng trtlg hp thng thng ct thp dc chu lc c t tp trung theo cnh b v k hiu l AS, Ag. A - din tch tit din ct thp pha gn vi lc dc t lch tm N, trong vng b nn nhiu. AS - din tch tit din ct thp pha i din vi Ag, ct thp AS c th b ko hoc nn t. a, a' khong cch t trong tm As, Ag n mp tit din gn nht. ho = h - a - chiu cao lm vic ca tit din. Za = ho - a' - khong cch gia trng tm AS v Ag. X - chiu cao Vng nn tnh i, gi tt l chiu cao vng nn; Rh - cng tnh ton v nn ca btng. Gi tr Rb c ly theo ph lc 2 nhn vi h s iu kin lm vic yh cho ph lc l. os, ; - ng sut trong ct thp As v Ag.RS, RSL. - cng tnh ton v ko v nn ca ct thp, ly theo ph lc 3.R - h s tnh ton gii hn vng nn, ly theo ph lc 4.S lc tc dng th hin trn hnh 2.la v 2.lb, vi trc U ly mmen l trc i qua trng tm ca As hoc Ag. Nh vy gi tr cu c ly bng e hoc e' (Xem cng thc lv20).e- khong cch t im t lc dc N n trng tm ct thp AS.39 39. e=ne0+0,5h-a (2-l) n Z l - h s Xt n nh hng un c, Xc nh theo cng thc (l-ll). Vitit din ch nht khi ?th = -ll- S Sc th b qua un dc, r = 1. X e' - khong cch t im t lc N n trng tm Ag . Ty trng hp im t N khong gia hay ~bn ngoi As Ag m c cch tnh khc nhau.C) dl u IHnh 2.1 . S tnh ton a, b - S d lc tc dng,' c - S d ng Su,' d- ti din2.1.2. iu kin v cng thc co bn iu kin v bn l cc iu kin (l-la) v (l-20) trong trc co bn ly mmen i qua trng tln ct thp As v nh vy Mu = Ne, iu kin (l-20) c vit thih: Ne S M, gh (2-2) Trong mt s trng hp c bit trc ly mmen c cho i qua trng tm Ag v iu kin s l: Ne' S Mzgh (2-3) M1 gh - mmen gii hn th hin kh nng chu lc ca tit din ly i vi trc i qua trng tm ct thp AS.40 40. Mzgh- mmen kh nng chu lc ca tit din ly i vi trc i qua trng tm Ag (ty trng hp tnh_ton s lp cng thc sau). Kh nng chu nn ca tit din Ngh c Xc nh bng tng hnh chiu cc lc:Ngh = RbbX+Ag -SAS (2-5)Trong cng thc (2-4) v (2-5) tnh ton vi gi tr tuyt i ca os v ; theo chiu ghi trn hnh 2.lc. Nu os l nn th cng thc (2-5) ly du cng trc SAS. Trng hp s c tnh theo cng thc vi du i S, quy c tig sut ko l dng th vn gi nguyn du ca cng thc (2-5) v lc os l nn s mang du m. Gi tr ca v os ly theo mc 1.6.5, cth l:Khi Z X Z 2a' th = RSC (2-6a) Khi : X < Rh0 th os = Rs (2-6b) Nh vy iu kin dng hit kh nng chu lc ca ct thp l:2a' S X < Rh0 (2-6c)Tnh ton ct thp hoc kim tra kh nng chu lc thng c tiin hnh theo iu kin (2-2) vi M 1 gh theo (2-4) trong X c Xc nh t iu kin N = Ngh. Vi gi thit l iu kin (2-6c) c tha mn th c phng trnh (2-7a):N = Ngh z Rbbx + R A' - RSAS (2-7a)SC S Khi Xy ra X > ,Rh0, gp trng hp nn lch tm b (mc nhin cng nhn X > 2a` do = RSL, ), Xc nh X cn gii ng thi hai phng trnh. Phng trnh th nht l iu kin cn bng lc nn:N = Ngh z Rnbx + R A' -o_,As (2-7b)SCSPhng trnh th hai l quan h gia ng Sut os v chiu cao vng nn X, ly theo mt trong cc cng thc (l-23) hoc (1-25).Cn ch l chi c th gii h hai phng trnh Va nu khi bit ct thp AS, Ag (bi ton kim tra) hoc bit quan h gia As v Ag (tnh ton ct thp i Xng). Khi cha bit AS v Ag (bi ton tinh ct thp khng i Xng) c th Xc nh X bng cng thc thc nghim, gn ng.Tiu chun thit k TCVN 5574 -1991 c a ra cc cng thc Sau:Khi eo < 0,2h0, tnh X theo cng thc (2.8a):4l 41. 0,5h hoX=h-[l,8+ -1,4RJeO (2-Sa)Khi 0,2h < co S cop, tnh X theo (2-8b):X = 1,8 (cop - eo) + Rh0 (2-8b) Trong : cop = 0,4 (l,25h - Rh0) (2-9) Khi eo > cop ly X = Rho (2-8C) Ngoi cc cng thc (2-8) cng cn c mt s cng thc khc: ITR = + h (2-10) X R 1+50s, O x=[E_,,,+(1-3e,,)(1-2,,)h,, (2-11)Trong hai cng thc trn th a, = -. Cng thc (2-10) dng cho mi 80 cn cngthc (2-l 1) chi dng c khi 80 - -, khi 80 > th ly X = Rh0.Cn ch^ rng gi tr gn ng ca X Xc nh theo cc cng thc (2-8a) n (2-l l) _chi c em dng tnh M 1 gh theo (2-4) m khng dng Xc nh gi tr os theo cc cng thc (l-23) hoc (l-25).Khi Xy ra X < 2a' dng iu kin (2-3) tinh ton s thun li hn (X < 221' mc nhin cng nhn X < Rh0 do os = RS). Tnh MzghzM,gh : RSASZ, + RbbXa'~(2-12a)Nhn Xt rng thnh phn th hai trong cng thc l kh b, v nu b qua th vic tnh ton thin v an ton hn, V vy thng ngi ta b qua tnh ton n gin.Gp trng hp a' kh ln, vic b qua thnh phn th hai cng thc (2-l2a) dn n vic gim ng k Mzgh hoc tng ng k ct thp AS th trong tnh ton c th b qua ct thp Ag hoc k thm thnh phn th hai.Gh ch quan trng: Cc bn thc lp dixc zh M I g,,,' M zgg v N g, cltl c gi t!' (c cltp rlhlz l rlg) klt din tclt ct thp As, AL `u dllg. Klz dllg Cc cng thc lp n tttlz c Ct thp m tll Chii c thikt lll l khng C`n z ct thp theo tllll toll (dt tllp tlteo Ct! to) cn cc kt qu tlzh torz (trng gian ltoc cui cng) l kllltg plzlz rlll dltg thc t42 42. 2.2. TNH TON CT THP I XNGBilt kch thc tit din b, h, chiu di tnh ton lo, ni lc M, N, chng loi vt liu. Yu cu tnh ton ct thp i Xng As = Ag.2.2.1. Chun b s liu- Xc nh cng tnh ton chu nn ca btng Rb theo ph lc 2 v khi cn th k thm iu kin lm vic yb theo ph lc l. Xc nh mun n hi Eb.- Tm cng tnh ton Rs; RSC ca ct thp theo ph lc 3.- Tm h s R theo ph lc 4.- Gi thit cc i lng a, a' tnh ho = h - a; Za = ho - a'.- Xt nh hng ca un dc, khi lil-S 4 ly T] = 1, khi -l> 4 cn tnh NC, v 1] theochi dn mc 1.5.2. Trng hp tnh ton NC, theo (l-16) th cn cn phi bit cc gi tr Mdh, Ndh Xc nh h s cp, v gi thit din tch ct thp hoc t l ct thp tnhI I 7 'V 1 1 I I I JS. Trong tnh ton thc t co th bo qua anh hng un dc khi - S 8.- Xt lch tm ngu nhin ea, tnh el, eo v e theo cng thc (2-l).Tnh ton ct thp bt u t vic Xc nh chiu cao vng nn X.2.2.2. Xc nh s b chiu cao Vng nn X1Trc ht cn Xc nh S b chiu cao vng nn ri cn c vo phn bit cc trng hp tnh ton.2.2.2.1. Xc nh x, khi RSC z RSVi nhiu loi ct thp thng dng c RS S 400MPa v nh vy RSC = Rs. Gi thit iu kin (2-6C) c tha mn, c th Xc nh X theo cng thc (2-13) rt ra t iu kin (2-7a) v t l Xl.N=- 2-13 Rbb X12.2.2.2. Xc nh x, khi RSC t RsTrng phi u M khng phn bit gi tr Cng tnh ton ca ct thp khi nn v khi ko v vy khng c trng hp RSC t Rs . Trng phi Nga v mt s nc khcc phn bit v khi Rs kh cao th RSC < Rs.43 43. tnh ton S b chiu cao cng nn X 1 cng tm gi thit X tha mn iu kin (2-6C). Ly gi tr N cng thc (2-7a) thay vo iu kin (2-2) vi du bng v dng biu thc M1 gh cng thc (2-4) vi og = RSC rt ra c phong trnh bc hai ca X.2N X2 -2(h0 +tS)X+(e+tS) z OI 2 Rscza 5 Rs _RSCGii phong trnh bc hai, ly nghim c ngha l X1. 2.2.3. Cc trng hp tnh tonGi tr Xl va tnh ton c chi mi l S b. Cn da vo X, phn bit cc trng hp tinh ton l nn lch tm ln thng thng, nn lch tm b hay l trng hp c bit.2.2.3.1. Nn lch tm ln thng thngKhi m 221' S X, S Rh0 iu kin gi thit l ng. Lc ny ly X = X 1 v ; = RSCthay vo cng thc (2-4), kt hp iu kin (2-2) v ch rng N = Rb bx rt ra c cng thc tnh A,H Z N(e+0,5x-ho)2-14 ., Rza ct thp i Xng, ly AS = Ag.2.2.3.2. Nn lch tm b. Kh xp > RhKhng dng c gi tr X1 v khng ph hp vi iu kin gi thit. Lc ny tim X cn phi gii ng thi ba phng trnh: phng trnh (2-7b) vi AS = Ag, phng trnh quan h gia os v X ly theo mt trong cc cng thc (1-23) hoc (1-25) v phngtrnh (2-2) kt hp (2-4). Kt qu rt gn li c mt phng trrih bc ba ca X.X3 + a2X2 + alx + ao = 0 (2-l5a)em t = X-, a phng trnh v dng khng th nguyn: 0.3+k,2+k,++k,,:O (2-15b)Cc h s k2, kl, ko c Xc nh ph thuc vo phng trnh quan h gia os v X, c cho trong bng Sau:44 44. Phng trnhk2 kl ko GS - X (1-23) -(R+2) 2(1+tpy+n8-2p) Zn (2pa-yo-8) (I-238) - (p + 2) 2(n8 + n) + Y(2 - Y - p) -n[Y]l (l-25b) E(Y-l,7Ba -1). Z. . N ;c=;=O,5(i-R);y=h-;B,z ~` ;C=O,5+0,6[3,Rhbh, h l,2R n:(7 SGii phng trnh bc ba c th bng cch gn ng vi ch bin thin ca trong khong gia R v l, c th dng phong php th hoc c th tham kho cch gii ph lc 5.Theo ngha vt l th chi bin thin trong khong R S STuy vy vi s t)cn thn cn thit c th hn ch trong khong Sau: R S S1 (2-16)Trong cc phng trnh quan h os - X c dng mt Vi iu gn ng, n gin ha do d c mt s trng hp gii phung trnh (2-15) c nghim khng nm trong gii hn nu (khi n v 8 u kh b thng tim c > l), lc ny cn dng iu kin (2-16) Xc nh .Sau khi c , tinh X = ho. trnh vic phi lp v gii phng trnh bc ba qu phc tp m kt qu cng cha bo m ng hon ton, trong tnh ton thc t c th dng cng thc thc nghim (2-8); (2-10) hoc (2-l l) Xc nh X. Sai s gia cc cng thc gn dng vi nhau v vi nghim ca phng trnh (1-15) l bnh thng, c mt vi trng hp hoi ln, tuy Vy kt qu ct thp tnh c chnh lch khng ng k v vn m bo iu kin an ton.Vi X c, dng iu kin (2-2) kt hpcng thc (2-4) rt ra cng thc trili Ag .A, z Ne- RbDX(hO -X/2). 2-17 S RSCZ ( )ct thp di Xng ly AS = Ag.2.2.3.3. Trng hp c bit. Khi x, 2a'Khi Xy ra Xl < 2a' th gi thit tnh X1 khng cn ng do cng khng dng c. Lc ny nu tnh ton chnh Xc th s c X > X, tuy vy X vn cn nh hon 2a'. tnh ton ct thp dng iu kin (2-3) kit hp cng thc (2-l2b), rt ra:45 45. A = Ne' =N(e-'Za) t RZ IgzaS B(2-18)Ct thp i Xng ly Ag = As. Khi a' l kh ln c th dng cng thc (2-26) tinh AS.2.2.4. S tnh .ct thp i XngBi ton tnh ct thp i Xng cu kin chu nn lch tm, tit din ch nht, c ct thp t tp trung theo cnh b c s ha nh trn hnh 2.2.S6 liu cho trc: b, h, lo, M, N, ea Chng loi vt liu: be tng, ct thpChun b s liu tnh ton:Tm Rh, Eb, Rs, Rgc , Es, h_ S6 R (theo cc ph lc) Gi thit a, a'. Tnh h, zaXt un dc, lnh N, 11Tnh e, eo. eI - Tnh ct thp xng Xc nh X bng cch gii phng trnh (2 -15) hoc dng cong thc gn ng (2-8)(2-10) X = X, Ai: = As = l2'18l `s = As = (2-17)A.=A',=2-14) nh g, x l kt quHnh 2.2. S tnh ct thp i xbrtg46 46. 2.2.5. nh gi v X l kt qu tnh ton Theo cc cng thc lp c th tnh ton c As, Ag l dng hoc m. Khi tnhc AS = Ag < 0 chng t kch thc tit din qu ln, khng cn n ct thp. Lcny nu c th c th rt bt kch thc tit din (hoc dng loi Vt liu c cng thp hn) tnh li. Khi khng th rt bt nh va nu th cn chn t ct thp theo yu cu ti thiu, gi l t ct thp theo yu cu cu to.Ch rng khi tnh ton c As, Ag m th cc kt qu trung gian tnh c hoc c chp nhn (chiu cao vng nn X1; ng Sut trong btng v trong ct thp...) lkhng chnh Xc, chng chi c tc dng nh l iu kin tinh ton ch khng phn nh ng s lm vic thc t ca tit din. Khi tnh c ct thp dng, tnh t l Ct thp: . '. . A', Z As +A hoc q% 2100(Ah + S) bh ` bho0sKim tra iu kin (l-5): min S s S maxKhi s < mjn chng t kch thc tit din l hi ln, cn X l nh khi tnh c ct thp m.Khi s > max chng t kch thc tit din qu b, cn phi tng kch thc tiit din hoc dng vt liu c cng cao hn (hoc dng c hai bin php) ri tinh tonli tha mn g S max.Chn v b tr ct thp cn tun theo quy nh v chiu dy lp bo v v khong h gia cc ct thp. Sau khi b tr ct thp cn Xc nh gi tr a, a' tnh li ho, Za, SO snh chng vi gi tr c dng trong tinh ton trc y. Khi gi tr ho v Za Va tnh ton c l ln hon hoc bng cc gi tr c dng th kt qu l thin v an ton. Nu gi tr ho v Za Va tnh ton c b hn cc gi tr c dng th kt qu nghing v pha thiu an ton, cn c X l thch ng. Khi mc b hn l khng ang k th chi cn chn ct thp tng ln so vi kt qu tnh c (mc tng ln c th bng hoc ln hn mc gim ca Za). Nu mc b hn l ng k th cn gi thit li a v tnh ton li.Mt vn rt quan trng trong khi dng cc cng thc tnh ton `l vic thng nht n V. Khi dng n v ca cng vt liu l MPa = N/mm2 th cn i n V chiu di (b, ho, Za...) thnh milimt v n v ca ni lc l.Niuton, Niutnxmm (k hiu l Niu v Nmm trnh nhm ln vi N dng k hiu lc nn). Khi s liu du vo c cho theo n v khc (v d kch thc tit din theo cm, ni lc theo kN, kNm) cn dng h s chuyn i n v thch hp, trnh s nhm ln lm sai kt qu.47 47. 2.2.6. Th dTh d I. Ct tng 5 ca khung nh mt nhp, sn ton khi, chiu di ct l = 3,8m, tit din ch nht b = 25cm; h = 40cm, btng mc 300 (theo tiu chun c) ct thp nhm CII. Yu cu tinh ton ct thp i xng. Khi ct chu cp ni lc M = l38kNm; N = 650kN, trong ni lc do ti trng thng Xuyn v ti trng tm thi di hn gy ra l Mdh = 80; Ndh = 500.Khung mt nhp, tng trn lo = 1,251 = 1,25x3,8 = 4,75m.Btng mc 300, khi khng Xt h s iu kin lm vic c:Rh = l3MPa, Eb = 29000MPa. Ct thp CII c RSC = RS = 280MPa Vi Rh = 13; Rs = 280 c R = 0,608 (ph lc 2, 3, 4)Gi thit a = a' = 4cm; ho = 40 - 4 = 36cm = 360mm; Za = 320.lo 4,75Xt un dc: - = -- = 11,8 > 8. Cn Xt un dc. h 0,4 lch tm tnh hc el = M = = 0,2l2m = 212mm. N 650 A , .A l h lech tam ngau nhien ea Z max -,-- = 13, 3mm. 600 30 lch tm ban u eo = maX(el, ea) = 2l2mm. Xc nh h s T] theo cng thc (1-l 1) trong NC, theo (l.l6b). Gi thit t l ct thp s = 1,5% = 0,015. JS = (As + A; )(0,5h - a)2 = ,bh,,(0,5h - a)2 = 0,015 x 250 x 360(200 -40)2 = 34,56 106 mm4_ b_113 _ 250x4003ot, =-Ei: 210000 =7,24; J- - =l333Xl06mm4 ` Eb 29000 12 l2 emin =0,5-0,0ll0-0,OlRb =O,5-0,01%-0,0lxl3=0,25l. 212 , , , e =e0=66=0,53 >cmin; (pp =l (khng co ng lc trc)cp, = l+B-i . Btng nng B = l. Trong cng thc tnh cp, th M v M, c ly i vi mp tit din chu ko:48 48. M=138+650X0-=268M,=80+5O0xQT4=18O180 =l --=1,67 ' +268_ 6,4E,, J 0,11+0,1 +(XSJs_6,4x290.000 1333x106( 0,11 47502 1,67 0,1+0,53= 3873100 = 3873kN.+0,1j+7,24x34,56x10l lTl: N 16501 .. ...._ .. N 3873=l,l8CTe =neO +-a =l,l8X2l2+i-i-9-40=4l0mmxl Rbb 13x250 Xl < RhO = 0,608 < 360 = 218mm. ng thi X, > 2a` = 80.A, = N(e+0,5X-ho) z 650Xl03(4l0+l00-360) " R802, 280x320 __ A,+A; _ l08822 c din tch: 1140 mm2. B tr nh trn hnh 2.3.Ly chiu dy lp bo v 25mm (2 tt) trih c chiu dy lp m a = 25 + = 36mmho = 400 - 36 = 364mm, ln hn gi tr dng trong tnh ton l 360mm. Khong h gia hai ct thp: _ 250-2x25 -3>22to = 67mm > 50 , t yu cu. o 249 49. Hnh 2.3. T dt ct - th d 1.Ct thp ai trong ct chn >6 2 l /4>dc max.Khong cch ct ai ad = 250 < l5dcmin = 330.Tlz d 2. Ct ca nh Cng nghip mt tng. Tnh ton cho phn ct di cu trc vi chiu cao H1 = 6,4m, dm cu trc khng lin tc. Tit din ch nht b = 40cm; h = 80cm, btng c cp bn 25, ct thp loi RB400. Yu cu tnh ct thp i Xtilig chu cp ni lc M = 480kNm. N = 500kN.Chiu di tnh ton lo = 1,5H, = l,5X6,4 = 9,6m.Btng cp 25 c Rb = l4,5MPa, Eh = 30000MPa.Ct thp RB400 c RSC = Rs = 365MPa.H s R = 0,558.Gi thit a = a' = 5cm; ho = 80 - 5 = 75cm = 750mm; Z, = 700mm. Xt un dc:-I=-gi-6-=l2>8 h 0,OO3 3 J =-L =--400x800 = 17060x l06mm412 12N _ 2,5E,,J _ 2,5x30.000x 17060x 10 l 96002Cl'= l3880000NiuNC, z 13880kN. 1 1= =---=l,O4 1-_l`L ,-_5fI.NC, 13880M 480 el==6=0,96=960mm50 50. lch tm ngu nhin ea Z maxv= 27mm 600 30Tinh ton ct nh cu kin tnh nh. eo = el + ea = 960 + 27 = 987mm.e=ne0 +-a=l,04x987+)-9-50=l377mmN _ 5001000= - = 86mm < 2a' = 100 Rab l4.5X 400XlTnh ton theo trng hp c bit: A, _ A _ N(e-Za) _ 500000(l376-700) S S RSZ 365 X 700A, +A; _ 1323x2 bh,, 400x750= l323mm2= 0,0096 = 0. 96%g=Chn ct thp: mi bn t 3>25, chn chiu dy lp bo v 35mm; a = 35 + t)/2 = 48lnm (hnh 2.421):X, = 90,6 < Za' = 96. Gh ch. G tlttilllzlrcllll [p bo t' 25mm, a = al = 25 + l2=38lntn, .r, = 86 > 2[lI = 76Ir1I1. Tinh li ct thp theo cng thc khc vi hg = 762, Z, = 724.A Z A, z N(e+O,5X-ho) Z 5000O0(1376+43-762) :1243mm2R,Z, 365 > 724T/l dll 3. Ct c chiu di tnh ton lo = 2,8m, tit din ch nllt b = 30cm; h = 50cm, btng cp bn 20, ct thp nhm CII. Ni lc tinh ton gm N = l320kN, M = 2] 8kNm. Yeu cu tnh ton ct thp i Xng.Btng cp 20 c Rh = ll,5MPa, Eh = 27000; ct thp CII c RSC = Rs = 280. H s u Z 0,61. Gi thit a = al = 4cm; ho = 50 - 4 = 46cm = 460mm; Za = 460 - 40 = 420mm.l=2-iEn=5,622 (hinh 2.4b)625Hnh 2.4. Tit din ct th d 2 v 3Ghi ch. Tl'ongtl1 d 3 nit muh tnh ton x bng cch lp v g phng t'l`nh th .`375 =0,51- =0,2; z=--=0,815(P ( R) 8 ho 460y='-=2=0,9l3;n= N =----1320000 =0,8696. ho 460 Rbbho i1,5x300x46Ok,=2p-3:-2,652 52. k, = 2(1+ oy + na - 2p) = 2(1+ 0,2 x 0,913 + 0,8696 x 0,815 n 0,4) = 2,9826.kg = 2n(2tps + yo - 8) z 2 x 0,8696(2 x 0,2 x 0,815 - 0,913 x 0,2 - 0,8l5)= -1,168. Phng trinh s l:=,3 - 2,62 + 2,9826,-l,l68 z 0 Gi c = 0,71; x = 0,71 >< 460 = 326.A, _ 1320000 >< 375 -l I, 5 X 300 X 326(460 - l63), : l368mm2. 280x4202.3. TNH TON CT THP KHNG I XNGTrong thc t chi c mt s t trng hp ngi ta mi t ct thp khng i Xng As : AL. Vi mt cp ni lc M, N cho trc th tnh ton ct thp khng i Xng cho tng lng ct thp As + Ag b hn trng hp ct thp di Xng (c bit l khi nnlch tm b). Tuy vy khi cu kin chu M di du m gi tr tuyt i gn bng nhau th tng lng ct thp trong trng hp t i Xng v khng i Xng chnh nhau khng ng k. Ct thp khng i Xfng tht s c hiu qu v tit kim vt liu chi khi tit din chu mmen khng i du hoc mmen theo chiu ny kh ln hn m men theo chiu ngc li. Trng hp c bit ca ct thp khng i Xmg l chi tnh ton ct thp mt phia, pha kia khng t ct thp hoc chi t theo cu to (khng k n trong tnh ton). tnh ton ct thp khng i Xng trc tin cng cn chun b s liu ging nh lm mc 2.2.1 i vi ct thp i Xng. 2.3.1. Trng hp tnh tonKhi t ct thp khng i Xng, ban u cha c cch g Xc nh c X da vo m phn bit trng hp tnh ton. Lc ny c th da vo lch tm:Khi neo > cop - tinh theo nn lch tm ln neo S cop - tnh theo nn lch tm b. Tiu chun TCVN 5574 cho cng thc thc nghim cop = 0,4 (l,25h -* RhO) c th ly gn ng cop = 0,3ho. 2.3.2. Nn lch tm lniu kin tnh ton l neo > cop v chiu cao vng nn X tha mn iu kin (2-6c). Lc ny c hai phng trnh l (2-2) v (2-7a) Xc nh ba n s l X, As vy l bi ton c nhiu nghim. Trong thc t khng cn tim c tt c cc53 53. nghim m chi cn mt nghim hp l l c. gii bi ton c th cho trc mt gi tr ca mt trong ba n s ri tm hai n cn li. Ch rng cc n s X, AS, Ag chi binthin trong mt khong nht nh, tng i hp nn gi tr cho trc hp l phi nm trong khong Xc nh va nu. Trong ba n th khong bin thin ca X l r rng hn c (2a` S X S E_,Rh0) V vy cho X mt gi tr tnh AS, Ag l thun li hn. Cng c thcho trc AL tinh X v As hoc cho trc As tinh X v Ag. Tuy vy vi khong binthin ca As l kh b v kh d on nn trong thc t tnh ton t ng cch cho trc As m thng thng chi cho trc X hoc 2.3.2.1. Chn x, tnh A; v AsCho X mt gi tr ty trong khong 2a' SX Skho. Thay X c vo biu thc (2-4) v dng iu kin (2-2) vi ch = RHC s rt ra cng thc tnh Ag . l cng thc c lp (2-17). Vit li:S R Z.SC&(2-19)Cng thc (2-17) v (2-19) c dng hon ton ging nhau, ci khc ch yu l gi tr ca X.Khi tnh c Ag > 0, em X v Ag thay vo phng trnh (2-7a) rt ra cng thc tnh As: = Rbbx + R A_'. - NA,(2-20)S ng vi mi gi tr ca X c AS v AL tng ng tuy vy tng lng ct thp A+A' thay i khng ln, vi khi tng X th Ag gim cn As tng. C th chng minh ch + IT ' I II I I il a thi tng ct thep As + As la be nht. Trong mt Sbng ton hc khi X = X A =ti liu ngi ta khuyn ly X = Rhg tnh ton vi ngha S dng ht kh nng vng btng chu nn v c c tng AS + Ag gn vi gi tr b nht.Trng hp tnh ton c Ag S 0 th chn li X b hn ri tnh li. Khi chn X b nht bng 2a' m vn tnh c Ag < 0 th chn A theo cu to v tnh As theo cng thc (2-26).2.3.2.2. Chn A; tnh x v As Khi bit trc hoc chn trc Ag cn tinh X t iu kin (2-2) vi du bng:54 54. Ne z Rbbx Lho --+ RScA_',Z.,, (2-21) trnh vic ii hng trnh bc hai, em t = L,ot = (l -0,5 )thay vo 8 P h m 0(2-2l) rt ra: z Ne - RscASZa (2_22) Rbbhoz 1 -,/1-Zom (2-23)Hoc t otm tra ra , theo bng ph lc 6.dmTinh X = ho v kim tra iu kin 2a' S X S Rh0 ( S R). Khi tha mn iu kin Va nu th thay X v A vo cng thc (2-20) tinh As.Khi , > R chng t A, bit l cha , cn tng A ri tnh li hoc Xc nh A theo cng thc (2-19).Khi X = ho < 2a', k c trng hpp otm < 0, chng t Ag l qu ln, niu c th c th gim bt Ag ri tnh li. Nu vn gi nguyn AL th tinh As theo trng hp c bit mc 2.3.2.4.2.3.2.3. Trng hp A = 0 l trng hp c bit khi khng cn n ct thp A (tnh c AL < 0, khng t ct thp chu nn hoc chi t theo cu to vi AL .>.0,0005bh0, khng k vo trong tnh ton).Lc ny tnh Otm theo cng thc (2-22) trong cho Ag -v 0. Cnh tay n nilc l Zb.Z, = ho - = (1 - 0, 5E_,)h0 z gho (2-24)g=1-O,5=0,5(1+,/1-zom (2-25) Din tch ct thp As c th c tnh theo cng thc (2-20) trong cho Ag = 0 hoc tnh theo cng thc (2`26):A`:N(e-Zb)2-26 ., Rszb 55 55. 2.3.2.4. Trng hp c bit x < 2a'Khi bit trc Ag , tnh om, m X = ho < 2a`, k c trng hp Otm < 0 th khng th dng X tnh tip. Lc ny tnh As theo cng thc (2-18).Trng hp a' l kh ln, dng (2-18) s tnh c A, kh ln. C th s tit kim hn nu b qua Ag trong tnh ton v lc ny khng cn iu kin X Z 2a'. Tnh ton As theo cng thc (2-26).Kit hp cng thc (2-18) v`a (2-26) c th viit thnh:Ap N(e-Z)2-27 , pRSZ < Trong : Z = max (Za; Zb). 2.3.2.5. Chn AS tnh x v AC th chn trc ct thp chu ko As tnh ton. Lc ny rt Ag t biu thc(2.7a) ri em thay vo (2-4) s a v c mt phng trnh cha X:Rbbx(---a' = RSAsZa -N(e-Z,,)Cng c th lp c phng trnh trn y bng cch ly mmen cc lc i vi trc i qua trng tm ct thp AL v vung gc vi mt phng un.Gii phng trnh, tm c X, khi X tha mn iu kin hn ch (2a' S X S Rh0) th thay X vo cng thc (2-l9) tinhNiu tm c X khng tha mn iu kin hn chi chng t gi tr As chn l khng hp l, cn chn li.Nh nhn Xt phn u ca mc 2.3.2, vic cho trc As tinh ton mang nng tnh cht l thuyt, thc t t dng n.2.3.3. Nn lch tm biu kin tnh ton l neo < cop. Gii hn ca chiu cao vng nn X l Rh0 S X S h.Nn lch tm b khi tha mn iu kin (2-28) th Xem l ring btng kh nng chu lc, ct thp hon ton t theo cu to.N S NB = Rbb (h - Ztco) (2-28) Khi N > NB cn tnh ton. Lc ny c 4 n s cn Xc nli l As, AL , X v S tronglc Chi c ba phng trinh. l phng trnh (2-2) kt hp cng thc (2-4), phng trnh (2~7b) v mt trong cc phng trnh quan h gia os v X.56 56. y l bi ton c nhiu nghim, tuy vy trong thit k thc t chi Cn mt nghim hp l l c. tm c nghim, v nguyn tc c th cho trc gi tr ca mt n s bt k ri gii h phng trnh tm ba n cn li. Ch rng cc n s X, As, AL, os chi bin thin trong mt khong Xc nh kh hep, nghim tim c chi hp l khi cho trc n s mt gi tr ph hp. Trong cc n th khong bin thin ca X l kh r rng v vy thng ngi ta chn trc X tnh cc n cn li. Tuy vy cng c th chn trc As tnh ton..Khng t vn chn trc os hoc A; v kh d on khong bin thin hp l ca chng, nu chn trc mt gi tr khng ph hp s c kt qu khng hp l v phi tnh li mt s ln.2.3.3.1. Chn trc x dtnh tonV nguyn tc ton hc c th chn truc cho X mt gi tr ty trong khong Xc nh ho S X 5 h. Tuy Vy nn Xc nh X theo cng thc thc nghim (2-8) hoc (2-10). Tnh ton ct thp AL theo cng thc (2-17).V phng din l thuyt, khi c X v Ag th c th tinh s v t (2-7b) rt ra cng thc tnh As:A : R,,bx+R,,A; ~NS(2-29a)SGi tr os tnh theo cng thc (l-23a) hoc (1-23).Tuy vy chi nn dng cng thc (2n29a) khi os l tng i ln cn khi os kh b th kh nng phm Sai s trong tnh ton l ln v rng cc cng thc Xc nh X v os u l cng thc thc nghim, gn ng. Hon na khi lch tm el l kh b, ct thp AS chu nn th vic tng lch tm t el thnh neo s lm gim din tich As. Lc ny S l bt li cho As nu gim lch tm. V nhng l do trn, ngoi yu cu v iu kin cu to, ct thp AS ca cu kin chu nn lch tm b cn cn tha mn iukin (2-29b) Sau: 6lA S As S Ai (2-29h)Yu cu A, SAL l phng khi tinh c , qu b, theo (2-28a) c th tinh ra As qu ln, khng ng vi thc t. Yu cu A, 2 6,A l phng khi lch tm qu b,kt qu tnh theo (2-29a) c tnh vi lch tm ln hn s cha an ton. H s a ly theo bng Sau:0,02 0,04 0,06 ` 0,08 0,10 57el/ho 0 57. 2.3.3.2. Chn trc As dtnh tonNn lch tm b c th chn trc ct thp As theo cu to. Lc ny c ba phng trnh Xc nh ba n S. Sau khi thc hin mt s bin i cn thit a v mt phng trnh cha X:0,5RbbdX2 + (2RSASZa - Rbbda`)X - (Ne`d + tRsAsZa) = 0 Trong : _ d=h-Rho;t=h+,Rh0;e'=Z,,-e. Gii phng trnh, kim tra iu kin ca X, em X thay vo cng thc (2-l7) tnhNu X khng tha mn iu kin hn ch nu chng t gi tr AS chn lkhng hp l. 2.3.3.3. Trng hp c bit As = 0 NNn lch tm b vi neo < cop c th thit k vi As = 0. Lc ny cn tnh ton btng v ct thp Ag chu ton bni lc. V khng c ct thp As, khngc iu kin gi cho os nn cng khng IIIIIIIIIIIIII IHnh 2.5. S d tnlt ton khi AS = 0cn iu kin X 2 Rh0 Chi cn iu kin 2 a' S X S h.Lp phoiig trnh Xc nh X bng iu kin (2~3), ly mmen i vi trc i qua trng tm Ag (hnh 2.5).Ne' : Mzgh = Rbbx-a'J (2-30)el = 0,5h-neo -al iu kin v kh nng chu lc l: N S Ngh = Rbbx + R; Ag (2e31) n gin vic gii phng trnh (2-30) t:oa z T =ot,(0,5o,, -1). Rt ra: T= Ne 2 (2-32a) Rbba'58 58. ot, z 1+ JH (2-32b) Hoc t T tra (xa bng ca ph lc 6. X = cta a iu kin l X S h Nu tnh c X.> h th bt buc phi t ct thp As, khng th b c. Sau khi c X, dem thay vo (2-31) rt ra cng thc tnh AL:- b A,:N RbxS R (2-33)SCKhi cn t ct thp AS theo cu to th ly AS 2 0,0005bh0.2.3.4. nh gi v x l kt quKt qu tnh ct thp khng i Xng c th dng hoc m. Vic X l tin hnh theo mc 2.2.5 nh i vi trng hp t ct thp i Xng.Trng hp nn lch tm b, nu cn tinh ct thp khng i Xng th cng chi nn tin hnh khi lch tm eo > 0,l5h0. Vi lch tm b hn, ton b tiit din chu nn, chi nn t ct thp i Xng.Khi t ct thp khng i Xng, trng hp nn lch tm b lun Xy ra AL Z As cn trng hp nn lch tm ln th AL c th ln hon hoc nh hon AS. Khi m Rbbx < Nth A ln hn A, v ngc li.2.3.5. Th dTh d l. Theo S liu ca th d l mc 2.2.6, yu cu tnh ct thp khng i Xng. S liu ( cho v tnh c):b = 250; h = 400; a = a' = 40; ho = 360; Za = 320mm.Rb = l3, Rg = RS = 280 MP8; R 0,608;N = 650kN; M = 138 kNm; eo = 212mm; T = 1,18;e = 410mm; Rh0 = 216mm.Tnh tip:cop = 0,4 (l,25h -R hg) = 0,4 (1,25 X 400 - 216) = 114mm neo = 1,18 X212 =250>e0p= 114.59 59. Tnh theo nn lch tm ln. Chn X = ghg = 2l6mm., Ne - RbbX(h0 - 0,5X) At = Z_TSL d.A, _ 65000041O-13x250x216(360-108)_ 280x320A __ R,,bx+R;A; -N _13x250x216+280x1000-650000 W R ` 280S= l000mm2z 1 l86mm2Tng lng ct thp A, +A; z 1186 + 1000 = 2l86mm2.Kt qu gn bng khi tinh theo i Xng. Ch thch. TI'ollg tlz d tl'It, Izt cllll X kltc ctllg dc, gi Sli cltll X = l 50nn (211, = 8(l < X < ,lI,, = 216). V X = [50 tlllz c AL = l423InIn2,' A_v = 843mm2, ttg lllg cttllp l." A_`. + A; = 22mm2. Kt qu gn blg 7 cc tI'`Ilg [rp d tlllt.Tltli Lll_t 2. Ct tiit din ch nht b = 40cm; h = 60cm, Chiu di tnh ton lo = 3,6m; btng mc 250 (c) ct thp RB400W. Ni lc tnh ton gm N = 2200kN, M = 352kNm. Yu cu tnh ct thp khng i Xng.S liu: Rh = ll,5, Eh = 27000; Rs = Ris = 365 MPa. R = 0,585. Gi thiit a = a' = 4cm; ho = 56cm = 560mm; Za = 520mm.Xt un dc: -l-==6 < 600 - 0, 585 >< 560) = 177neo = l60< cop = 177. Tinh ton theo nn lch tm b.6-Ozvl-6z0,267h600X: R+ 1_R ho: O,585+n---0415 2 560=364mm l+50X0,267cn llA, _ Ne - RbbX(h0 - 0,5X) _ 2200000X 420 - 1 l, 5 X 400 X 364(560 - 182)S = l802mm2 RSCZ, 365 x 52060 60. Tinh os theo cng thc (1-23):1_L 1_3646,: 2 -1 R- 2--1365=25OMPa ' l-R ` 1-0,585RbbX+R A_'. -N _ llX400X364+365X1802-2200000A--s-1----------:68mm ` ^ os 250 6-'==0,267;6, :0,3 ho 6000, 0025bh0 z 560mm2iu kin chn AS: AS 2 6,,A, = 0, 3 x 1802 = 600mm2Ly `A_, theo gi tr ln hn trong 3 gi tr tren, A, z 600. A, + Ag : 600 +1802 : 24022402=--=0,0107 z 1,07% 400x560s Cng vi bi ton trn th gii vi trng hp AS = 0. (Thc t chn AS theo cu to ti thiu As = 0,0005 X 400 X 560 = ll2mm2 v khng k vo tnh ton). e' = 0,5h - neo - a' = 300 - 160 - 40 = 100mmT_ Ne' _ 22O0.000x100- - z 31,25 Rbba'2 11,5x400x402Ota =1+li-l-2T =1+ 1+62,5 =8,97 X =cxaa' =8,97X40= 359mmA, z N -Rbbx 2 2200.000- 11x400x359 =1824mmz5 R 365SCA, +A; =1l2+1824 = l936mm22.4. TNH TON KH NNG CHU LC2.4.1. Cc loi bi tonKhi bit kch thc tit din v cu to ct thp c th tnh ton kh nng chu lc theo mt s bi ton khc nhau:- Kim tra Xem tit din c kh nng chu c mt cp ni lc M, N hay khng.61 61. - Vi lc nn N cho trc tnh Xem tit din chu c mt mmen M bng bao nhiu. - Vi lch tm eo cho trc tnh Xem tit din chu c lc nn N bng bao nhiu.- Vi M cho trc tnh Xem tit din chu c N bng bao nhiu. 2.4.2. Kim tra kh nng chu cp ni lc M, NTheo chiu tc dng ca M Xc nh v tr v tr s ca Ag, A (trng hp ctthp i Xng th khng cn). Chun b cc S liu nh trong mc 2.2.1. Ch rng y khng gi thiit a, a' m Xc nh trc tip t s liu cu to. bit c trng hp tnh ton cn tim gi tr ca chiu cao vng nn X. Trc ht gi thit iu kiri (2-6c) c tha mn t phng trnh (2-7a) rt ra X v dt l X2.N + RSA, - RSCA,2-34 Rbb )X2=Da vo gi tr X2 phn bit cc trng hp.2.4.2.1 . Nn lch tm ln thng thngKhi X2 nm trong phm v 2a' S X2 S kho, kt qu ng vi gi thit, ly X = X2 thay vo cng thc (2-4) vi , = Rg tnh M, Eh v kim tra theo iu kin (2-2).2.4.2.2. Nn lch tm bKhi tnh c X2 > `Rh0 khng dng c gi tr X2 v`l gi thit khng ng. Lc ny phi gii ng thi hai phng trnh Xc nh X. Phng trnh th nht l iu kin cn bng lc (2-7b), phng trnh th hai l quan h gia os v X (l-23) hoc ( l-25).Khi dng phng trnh (l-23a) kt hp (2-7b) rt ra c:X 2 (N -R A'.)(h-Rh0)+RsAS(h+Rh0)SC i't(2-35) iu kin ca X l phn < X Sh Sau khi c c X em thay vo cng thc (2-4) vi og = RSL. tnh Mlgh v kimtra iu kin (2-2). 2.4.2.3. Trng hp c bitKhi tnh c X2 < Za' cng khng dng c gi tr X2. Lc ny cn tin hnh kim tra theo iu kin (2-3) vi Mzgh Xc nh theo (2-12b).62 62. 2..4.2.4. Kiitl tra s chu nn ngoi mt phng unNgoi vic kim tra vi cp ni lc M, N, khi b < h cn cn kim tra s chu lc theo phng ngoi mt phng un. iu kin kim tra l N S NO vi No l kh nng chu nn ng tm, Xc nh theo cng thc (l-6).2.4.3. Xc nh M kh cho trc NTin hnh theo cc bc nh lp trong mc 2.4.2. Xc nh M1 gh. Thay biuthc ca e vo iu kin (2-2) tnh c eo.M,g,, -N(O,5h-a)= (2-36) eo YIN T eo v d lch tm ngu nhin ea tinh ra lch tm el: M = Ne, (2-37)Khi Xc nh M cn ch n du (chiu tc dng). Da vo du ca M bit ct thp nao l AS,Nu khng quy nh trc du ca M thi khi ct thp khng i Xng s Xc nh c 2 gi tr ca M theo hai chiu ng vi ct thp AS bn tri hoc bn phi.2.4.4. Xc nh N khi bit eoCho trc lch tm eo c ngha l cho trc im t ca N. Da vo im t phn nh ct thp As, AL (AL t gn N hn). Gi thit iu kin X Z 2a' c tha mn, em thay biu thc ca N (2~5) Vi ; = RSL. vo iu kin (2-2) rt ra phng trnh Xc nh X:2(R A'.e'- SASC) _SC S0 2-38 Rbb < >X2+2(e-h0)X+Cng c th lp c phng trnh (3-8) bng cch ly tng mmen cc lc i vi trc i qua im t ca N.Gi thit tip l X S E_,RhO ly os = RS thay vo phng trinh, gii ra, ly nghim c ngha v t l X312(RsAse - RSC Lei)2.38 Rbb ( alX3=(h0-e)+ (ho-e)2+Trong ny e' = e - Za c ly theo du i s tnh.Da vo X3 tim c phn bit cc trng hp.63 63. 2.4.4.1. Nn lch tm ln thng thngKhi tha mn c hai gi thit 2a' S X3 S Rh0 th ly X = X3 thay vo cng thc (2-7a) tnh N.2.4.4.2. Nn lch tm bKhi X3 > Rh0, phi tnh li X.Lc ny cn gii ng thi hai phng trnh tm X v os. Phng trnh th nht l phng trnh (3-38), phng trnh th hai l quan h os v X, c th chn mt trong cc cng thc lp (l-23); (l-23a) hoc (1-25).Sau khi tm c X, em thay vo cng thc (2-4) vi ; = RSC tnh M1 gh. T iu kin (2-2) rt ra:2 Mlgh 2 Rbbx(h0 -X/2)+R AZsc S a (2_39)N e e2.4.4.3. Trng hp dc bit Khi tnh c X3 < 2a' (k c trng hp X3 < 0) cn dng iu kin (2-3) v cng thc (2-l2b) Xc nh N:` = (2-39a)2.4.4.4. Gi thit trc 7]Trong biu thc Xc nh e c h s 11 cha tnh c, vy bn du phi gi thit mt gi tr T] 2 l tnh ton. Sau khi c c N th tnh li 11, so Snh vi gi tr gi thit, nu hai gi tr l gn bng nhau th chp nhn c.2.4.4.5. Kht tra ngoi mt phngGi tr N tnh c cn cn tha mn iu kin N S No nh trlih by trong mc 2.4.2.4. 2.4.5. Xc nh N khi cho trc M2.4.5.1. Tnh ton M0 theo trng hp chu unDa vo chiu ca mmen nh v ct thp chu ko As. Tnh ton M0 l mmen un tit din chu c khi N = 0.T phng trnh (2.7a) khi N = 0 rt ra X v t l X41= RSAS _ RSCAS X4 bRb64 64. Khi 2a' S X47 S 2Rh0 ly X = X4 tnh MO.M0 = RbbX(h(, -0,5X)+ R A.ZSC Sa(2-41) Cng thc (2-41) tng t nh cng thc (2-4). Khi X4 > kho vn dng cng thc (2-41) trong X = Rh0 Khi X4 < 2a' (k c khi X4 < 0) tnh M0 theo (2-42): Mg = Rs ANZ, (2-42) 2.4.5.1. Trng hp M > M0 Khi M > Mo cn Xc nh hai gi tr N I v N2 ng vi trng hp nn lch tm ln v nn lch tm b. Gi thit iu kin X 2 2a' c tha mn ( = RSC ). Thay biu thc e vo biu thc (2-5) vo iu kin (2v2) rt ra phng trnh bc 2 ca X.RbbX2 -Rbbhx - R A' (h -2a')-,A,(h -2a) + 2nM = O (2-43)SC S ,` Tm gi thit 11 tinh ton nh mc trc. Gi thiit tip X S kho c os = RS. Thay os = RS vo phng trnh (2-43) tim c hai nghim Xa, xb. ng vi mi nghim c ngha tnh c gi tr N tong ng. Vi X tha mn iu kin gi thit 2a' S X S Rho th thay X vo cng thc (2-7a) tnh N. Vi X > RhO cn gii ng thi hai phng trnh Xc nh s v X. Phuong trnh (2-43) v mt trong cc phng trnh quan h GS v X. C X s tnh c N. Vi X < 2a', bng cch bin i iu kin (2-3) rt ra:- TIM RSAsZa 0,5h-alN (2-44)2.4.5.2. Trng hp M M,,Lc ny Xy ra nn lch tm b v chi c mt gi tr N. Tnh ton bng cch gii ng thi hai phong trnh tm X v os Sau t X Xc nh N.2.4.5.3. Kin tra NGi tr N tinh c trong mc 2.4.5 u khng c ly ln hon gi tr No tnh theo cng thc (1-6).65 65. 2.5. BIU TNG TC2.5.1. Khi nim v biu tng tcTng tc y l tong tc gia kh nang chu mmen un M v kh nang chu lc nn N.Vi mt tiit din c ct thp bit biu tng tc th hin ton b kh nng chu lc ca n ng vi mi gi tr ca M v N.Nh trnh by trong mc 2.4 ng vi mi gi tr N tm c mt gia tr M, ng vi mi gi tr eg tm c mt gi tr N hoc ng vi mi gi tr M tm c mt hoc hai gi tr N. Lp biu vi hai trc l M v N. Mi cp gi tr nh Va neu cho mt im. Tp hp tt c cc im c c biu tong tc. Khi t ct thp i Xng biu c dng nh trn hnh 2.6. Trong hai trc c th ly trc ng biu din M hoc N ty theo s thun li khi th hin v khi dng.Hinh 2.6 th hin biu khi mmen M theo mt chiu. Khi Xt M theo hai chiu (dong v m) th biu c pht trin theo c hai pha nh trn hnh 2.7.N M C B D B N C D MHnh 2.6. Bt tng tc tlzilzn tlzeo la cclt l' M theo I Ch`uM N B C D C N B D` M D' DHnh 2.7. Bl`it tng tc vI` M la cllllKhi Xt c N theo hai chiu (nn v ko) th biu c pht trin thnh dng khp kin nh trn hnh 2.8.66 66. Hnh 2.8. Bt d tng tc vi M I N tlieo [lai clttl.Xt rieng gc mt phn t vi N nn, trn biu c ba im dc bit. im D ng vi N = 0 v M0 (cng thc 2-41, 2-42). im C ng vi M = 0 v No. Gi tr NO Xc nh theo cng thc (1-6). im B ng vi Mm,x v NB . C th chng minh c rng h, + ai2 tlih ton ct thp klng i Xng s c c tng lng ct thp AH + AL l nh nht.khi a = a`, im B ng vi trng hp X = XB = = 0,5h. Vi gi tr ny ca X, niuBiu tong tc chia lnt phng lm hai min: bn trong v bn ngoi. Vi mt cp ni lc M, N cho trc c mt im trong mt phng. Khi im thuc min trong (im I) tiit din kh nng Chu lc. Niu im thuc min ngoi (im K) tiit din khng kh nng chu lc (hinh 2.9).Trn biu , vng ln cn im B c th l nn lch tm ln hoc b, phn cn li trong on DB tng ng vi nn lch tm ln, trong on BC - nn lch tm b.Vi Cc im nm ngay trn biu (im P) kh nng chu lc Va ng bng ni lc m tiit din phi chu. Khi im nm trn on BC th mt trong hai ni lc M hoc N gim Xung s lm tng an ton v ngc li. Nu im nm trn on DB th khi M gim s tng an ton cn N giln s nguy him. Trong on M > M0 ng ll mi gi tr ca M c hai lc NI v N2. Khi N M thay i trong khong trn NI S N Mma, ----- S N2 th c c an ton (gi thit. I l I I 1 _-_,j--N2 > NI) cn niu N vt ra ngoi Mo D. . , .g Mk --------------------- -- plalin Vl tren la nguy hiem. ` Mp ----- `-l---'-Il- ------- -- -;,' I tl Il I I Diem B ng voi Mmax_ Vi mmen ny tit din chi kh I N nng chu lc khi N va bng NB NI NF NK NO ng N,, cn nu N tng ln hay gim Xung tit din u b nguy Hnh 2.9. Bit d v cc Cp 'tli ll_rC67 67. 'quy i din tch Ct thp rahim. Vi nhn Xt ny, khi thit k khng nn v`l mc ch tit kim ct thp m cho tit din lm vic im B nu cha c c tin cy cn thiit ca M v N. 2.5.2. Biu tng tc vi As khc AgVi tit din t ct thp khng i Xng, c bit l khi As v Ag khc nhau nhiuth biu tong tc c mt on li hi khc so vi cc biu v cho tit din t,ct thp i Xng. Xem Xt vn ny trc ht cn bn v trng tm hnh hc v trng tm vt liu ca tit din.Trng tm hnh hc O ca tit din ch nht cch u cc cnh, l giao im ca hai ng cho. Trng tm vt liu ca tit din, k hiu Ov c Xc nh c k n s khc nhau v kh nng bin dng ca btng V. ca ct thp thng qua gi tr mun n hi ca chng.Mt cch khc, trng tm vt liu Ov l trng tm ca tit din tng ng trong Hnh 2.10. Trng tm lznll hc 0v trng tm vt liu 0,, ca t din.din tich btng tng ng. H s quy i l t s ca Emun n hi ns = -SEb Ly trc i qua mp tit din pha c AL lm chun, khong cch t Ov n trc chun l yv c Xc nh nh Sau:_ 0, 513112 + ns(Ash0 + A;a')y bh+n,(A,+A;)Vi tit din c ct thp i Xng th yv = ,5h, im 0,, trng vi O.Khi lp S tnh ton chng ta ly trc i qua trng tm O Xc.nli lch tm eg v tnh M = Neo. Nhng nu ly trc i qua trong tm Ov th mmen MV S l (hnh 2.lla). ` T "_ M, = N(e0 - dv) = M- Nd, (2-46) Trong : dv - khong cch gia O v Oy; dv = 0,5h - yy. 68(2-45)) ' 68. Nhng nghin cu v s lm vic ca tit din btng ct thp cho rng tnh ton vi trc qua OV l ng hn, tuy vy trong nhiu trng hp sai lch khng ng k nn c b qua. Ngi ta chi nu ra vn ny khi mun th hin tht chnh Xc (theol thuyt) biu tng tc. b M B M0 ^'s ML L ` N Nic NLHnh 2.11. S v bil tng tc khi A. > AS Nh vy khi lc N t ng vo Ov tit din mi thc s chu nn trung tm v chu c lc nn ln nht NL = NO. Tuy vy nu so vi trc qua 0 th lc ny tit din vn cn chu mt mmen M = ML = Ndv. Khi v tr t lc N nm gia 0 v O, th thc cht, so vi 0,, M i chiu. Trn biu c mt on li t L (hnh 2. l lb).Khi As > Ag , im Ov gn vi As hon, S v biu tng tc nh trn hnh 2.12.M Mo IS N L M,Hnh 2.12. So d v bit d rng tc khi A, AA 2.5.3. Cc phng php v biu tong tcBiu tng tc c tnh ton theo tng im, ni cc im li thnh ng lin tc. Xc nh ta tng im c th dng mt trong nhng cch trinh by cc mc 2.4.3. n 2.4.5: Cho N tim M, cho M tim N hoc cho neo tm N. Dng bi ton bit neo tnh N trong vic v biu tng tc c ngha l t gc ta k ng Xin69 69. lp vi trc N mt gc cp m tgp = neo. im cn tm nm trn ng Xin (khi tnh c N). C th v nn dng kt hp cc phng php v mi phong php c ch mnh v ch yu ca n. Ba phng php trnh by l vi mt gi tr bit ca i lng ny tim gi tr tong ng ca i lng kia. V biu th chng ta t cho i lng ny cc gi tr khc nhau tim cc gi tr tng ng ca i lng kia v c c mt s im.Trong Cc phng php bit phng php no cng phi tnh ton thng qua mt biin trung gian l X. Vy c th Xem X l bin c lp t Xc nh cc gi tr ca M v N. V phong din Vt l, cho X bin i c ngha l S dng thay i mc chu nn ca btng t m Xc nh kh nng chu nn m kh nng chu mmen ca tiit din. V mt thc hnh ly X lm bin s l n gin hn c.Trc ht tinh X4 theo cng thc (2-40).Khi X4 2 2a' th ly X bin thin trong khong X4 S X S h.Khi X4 < 2a' (k c khi X4 < 0) ly 2a' S X S h.Nhn Xt rng, khi tnh ton tit din, niu k n lch tm ngu nhin ea v nh hng ca un dc T] th mmen t M tng ln thrlh M* = Nneo - Xem cng thc (l-18).Trong tnh ton thc hnh, trc mmen ngi ta khng t gi tr M m dt gi tr M*, lm nh vy vic lp v s dng biu n gin hn.Vi cc gi tr ca X trong khong 2a' S X S Rh() tnh gi tr N theo cng thc (2-7a), tinh Mlgh theo cng thc (2-4) trong = R ra t iu kin (2-2).v tnh Nreo theo cng thc (2-47) rtSCM z Nne,, = M,gh - N(0,5h -a) (2-47)Vi cc gi tr X trong khong Rh0 < X S h, dng cng thc (l-23a) Xc nh os, dng cng thc (2-7b) tinh N, dng cng thc (2-4)tr1h Mlgh v Xc nh Nheo theo (2-47).2.5.4. Nhn xt v biu tng tcBiu tng tc ca tit din t ct thp i Xng c hnh dng nh hnh 2.6. v 2.9. Khi dng biu ny Xem Xt kh nng chu lc ca tit din c mt s nhn Xt nh Sau (hnh 2.13):- im I nm bn trong min chu lc vi MI, NI. Cc im KI, K2, K3 nm bn ngoi min chu lc th hoc l c MKj, NKJ u ln hn MI, NI, hoc t nht cng c mt trong hai i lng ln hn M,, N,. l trong i th. - Trong Vng ln cn on DB tnh hnh c khc. Mt im Q min trong vi MQ; NQ. Mt im R min ngoi trong lc MR < MQ v NR < NQ. Nh vy tiit din70 70. chu c mt cp ni lc MQ, NQ trong lc khng chu c cp MR, NR u c gi tr nh hon. Tuy vy cp ni lc MR, NR c lch tm eo =14~ ln hn. iu nhn Xt va ri nh l mt nghch l, n c gii thch bng cch phn tch s lm vic ca tit din btng ct thp chu nn lch tm ln, s ph hoi bt u t vng ko. Khi M v N du gim nhng N gim nhiu hn th lc ko trong ct thp s tng ln. Cng thc (2-20) cho thy, khi N ln th ct thp chu ko As s b hon. Hnh 2.13. Nht! xt v biit d tI'tg tcHnh dng ng Cong ca biu tng tc ph thuc vo gi tr tuyt i v t l .- . , ty - ` -, - .zu tX gia cot thep hai phia, vao gia tr tng oi cua X4 la 4 = . 0Xt tit din t ct thp pha bn tri l A,, bn phi l Ap. Cho Ap mt gi tr c nh vo mc trung binh, cho A, thay i t O n mt gi tr kh ln. V cc biu vi c mmen dng (A, chu ko) v mmen m (Ap chu ko). Hnh dng cc biu thay i nh th hin trn hnh 2.14.Xt ring phn biu ng vi mmen dng (A, chu ko) thy nh sau:Khi A, = 0 c ng l, im Dl trng gc ta . Tuy A, = 0 nhng khi tng lc nn N kh nng chu mmen tng cho n im B. Tng A, nhng A, < Ap, c ng 2. Biu c im li L2 pha trn. Khi A, = Ap, c ng 3, i Xng. 71 71. Hnh 2.14. Cc dng dng colzg ca biit tng tcTng A, > Ap nhng vn gi cho 4 = -4 < E_,R (Xem X4 cng thc 2-40), c ng 0 4 vi im li L4 nm pha di. Khi gi tr 4 cng tng ln th on DB cng gim.Tng Al n mc 4 > R, on DB bin mt (ng 5).2.5.5. Biu tong tc khng th nguynLp biu vi kch thc tit din v ct thp bit nh phn trn trnh by chi thich hp cho mt trng hp c th no vi vic vn dng b hn ch. Lp biu vi cc thng s khng th nguyn, c bit l lp cc h biu s c c s vn. dng rng ri hon.Xt tit din t ct thp i Xng AS = AL v tha mn iu kin Rs = RSC.:t: n= N ' m= M = Nneo Rbbho Rbbh Rbbh 6-1-1. -lho ho' ho72 72. RSCAS __ RSAS (2_48) : - Rbbho RbbhoCng thc (2.7b) bin i thnh: n = , + ot (1 - (ps) (4-49)Trong p,=.KhisRthp,=1vn=.Vi > R v khi dng cng thc (l-23a) Xc nh os s c c:n 1_F_ E.>R) cps :: 1'l`RBin i cng thc (2-47) thnh: m=,(l-0,5E,)+(l-)(0t-0,5n) . (2-21) lp mt biu , cho , Rv ot mt gi tr chn sn, cho l thay i s tnh ra cc gi tr n v m. Mi cp n, m cho mt im ca biu . Ly t thay i t O n R, sau t ,, n h. Vi , R chn sn, li cho ot thay i (ot, OCZ... n) s c c mt h biu . Nh Vy mt h biu ng vi mt gi tr , R, nhiu gi tr Ot. Cho thay i s c nhiu h biu vi cc , ot khc nhau. Hnh 2.15 v mt h biu nh vy.0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2.0Hnh 2.15. H biu tzg tc kltng th ngttylt 73 73. Khi v biu cho mt tit din c th vi chiu di tnh ton lo cho trc, Xt c nh hng ca un dc theo phng yu nht (h s un dc (p) tnh gi tr No (kh nng chu nn ng tm, khi M = 0). Khi lp biu khng th nguyn cha Xt c iu va trnh by V vy gi tr n ng vi m = 0 chi mi th hin kh nng chu nn ln nht lc cha k un dc.2.5.6. Dng biu tng tcBiu tng tc ca tit din mt cu kin c th c dng ch yu kim tra kh nng chu lc theo cc bi ton trong mc 2.4. kim tra kh nng chu cp ni lc M, N th trc ht tnh eo, T] v M = Nneo. Vi M* v N c c mt im I. Khi im nm min trong ca biu th tit din kh nng chu lc. Xc nh M khi bit N th t N tm c im P l giao ca ng ging t N vi biu . T P tim ra M*, tnh c eo v M. Xc nh N khi bit eo, gi thit hs T. K ng Xin c gc 6 m tg 6 = ne.ng ny ct biu ti im P1. T P1 tm ra gi tr N (hnh 2.16). Ch tg9 y c don v chiu di, phi tnh ton theo t l trn hai trc ca biu . Hnh 2.16. Dng bit d tng tc dixc nh kll nng chu lcH biu khng th nguyn, ngoi vic dng kim tra kh nang chu lc nh trn (chi dng mt ng ng vi ot bit) cn dng tnh ton ct thp mt cch nhanh chng.T cc s liu M, N, b, h, Rs, Rh cho, gi thit a tnh ho v = -. Tnh n, 0neo, m. Dng h ng cong vi v R thch hp s t m, n tm c ot v :74 74. Ag z otRbbh0 RS(2-52)Tlt d. Ly s liu th d 3 mc 2.2.6: b = 300, h = 500mm; lo = 2,8m; Rb = 11,5; 37Rs = 280MPa; N = l320kN; M = 2l8kN. Gi thiit a = 37, ho = 463, =--1-6-3=0,08;R =0,61. tnh ton c eo = 165mm; T] = l._ N _ 1320x1000 R,,bh,, 1I,5x300x463n = 0, 864- Nne neo =0 864=O 308:n--m ... Rbbh h,, 463Vi n = 0,864 v m = 0,308 tm trn h biu c im K nm gia hai ng cong ng vi (1 = 0,2 v 0,3. Ni Suy c c ot = 0,28 (gn ng).A __ 0,28Xll,5X300X467S = l6l0mm2 ` 2802.5.7. Th d v biu tng tcTll dt . Cho tit din nh trn hnh 2.4b vi lo =- 2,8m; b = 30cm, h = 50cm, A, z A;= 2>22 + 225 = l7,4cm2, a = a' = 3,75cm, ly trn a = a' : 4cm, R, : 11,Rs = 260MPa; R = 0,64. Tnh ton v V biu tng tc. ho = 50 - 4 = 46cm = 460mm; Za = 420mm. RSAS - R A,SC 8:0X4 2 Rbb M0 = Rs A, Z, = 260Xl740 x420 = 19Ox106N,,,mm : l90kNm. No = p(R,,A,, + R,A,,) (cng thc 1-6).AS, = A, +A : 17,4 x2 : 34,8cm2 = 3480mm2Ab = 300x500 -3480 = l4650Omrn2.i = 0,288b z 0,288x300 = 8,6?t=l--gg)-=32,4>28,cn Xtundci T 86,475 75. tp = 1,028 - 0,O000288X32,42 - 0,00l6 X32,4 = 0,94 NO = 0,94 (l lx 146500 +260 X3480) = 2365000Niu.No = 2365kNCho X bin thin trong khong 2a' = 80 5 X S Rh0 = 276 `i Ly x = 80; N = Rbbx + CRSA- RSAS = ll X 300 X 80 + (0) = 264000 Niu = 264 kNMlgh = Rbbx (ho -0,5X) + RSCA Za =l1X300X80(460-40)+260xl740X420 = 300,9 >106Niu mm = 300,9kNm.M* = Mlgh - N (0,5h - a) = 300,9 - 264 (0,5 X 0,5 - 0,04) = 245,4 kNm.Tip tc ly cc gi tr khc ca X bng 120, 160, 200, 230, 250, 276, tnh c kt qu ghi trong bng: Tip tc cho X bin thin trong khong E_,RhO < X S h = 500, vi X = 300, (dng cng thc l.23a tnh os):=,-2n29g6o=zo4,gMp,h -Rh0 500-276N = Rbbx + R .A', - SAS.SL 5= ll X300 X 300 + 260 X1740 - 204,3 X1740 = l087Xl03Niu. Mlgh = Rbbx (ho - 0,5X) +RZaSL S=1l x300 x300 (460 - 150) +26O x1740 x42O z 496,9Xl06Nmm.M* = 496,9 - 1087 (0,5 X5 - 0,04) = 268,6.Tiip tc tnh ton vi cc gi tr ca X bng 350, 400, 450, 480. Kt qu ghi bng sau:76 76. X (mm) S (MPa) M* (kNm)300 204,3 350 88,2 400 -27,8 450 - l44 480 -2 l 3,6 Lc nn ti a ct chu c N = 2365 kN, khi tnh vi X = 480 c N = 2407 > NO, dng tnh ton. Kt qu V biu d ghi trong bng Sau:II l90 265,2 279,7 289 292,4 292T/l dtt 2. V biu tng tc khng thnguyn, ct thp i Xng vi = 1 =p0, 08 .hl) q , ,, , A. Ty l cot thep =l"-=0,0l, RS =260; Rh = 13; R =0,6 O oL= ASR =0,0lx2-6-q=0,2 Rbbho 13 Trong phm vi 0 < l < R = 0,6 c n = . Khi E_,= 0', n = 0; m = (1-) (ot - 0,5n) = 0,92 X0,2 = 0,184 vi =2=0,l6;n=0,l6 m = (l - 0,5) + (l- ) (CX - 0,5n) = 0,16 (l- 0,08) + 0,92 (0,2 - 0,08) m =0,2576.Vi cc gi tr khc ca S `R = 0,6, kt qu ghi trong bng Sau: , 0,32 0,329 0,328ViRAC, v Aga, Apcl) Xl tnh ct thp chu ko (A,a2) khi chn trc ct thp chu nn ALa2. Chn Aaz nh Sau:Al, Acl.Tnh ln hai cho cp C bng cch chn trc ct thp chu nn Acz = ,a2, tnh c ct thp chu ko APC2. So snh APC2 vi ALH2. S tinh ton mt vng th hintrn hnh 2.18. Hnh 2.18. S d mt vlg tlzll lpKhi Xy ra APC2 S Aaz th dng.Nu Apcz > A,a2 tiip tc tnh ton ln th ba vi Aag c chn trc trong khong A,.L`2 < Aas S Apbz . Vi Aas chn tip tc tnh ton mt Vng mi cho n khi tim c APC3 so Snh vi AJa3. Nu kit qu cha c nh mong mun th tinh tip vng th 4, th 5. Tnh nh th' n Vng th k (c th k = 2; 3 l ) khi Apck c gitr gn bng Apak th dng.82 82. 2.6.3. Nhn Xt v phong php tnhKhi cn tinh ton ct thp cho mt tieit din chu nhiu cp ni lc khc nhau th dng phng php tnh ct thp i Xng l on gin hn. Cng n gin hon na nu dng c h biu tong tc khng th nguyn.Vic tinh ct thp khng i Xng vi nhiu cp ni lc nhm S dng ct thp mt cch tit kim, hp l trong nhiu trng hp l kh phc tp v mang tnh l thuyit nhiu hn. c thivn dng trong thc t nn v cn lp cc chng trnh phn mm, S dng my tnh.2.7. TIT DIN C CT THP T THEO CHU VI2.7.1. i cng v vic t ct thp theo chu viTrong nhng phn trc y trnh by Cch tnh ton cho trng hp ct thp chu lc c t tp trung theo cnh b c din tch AL , Ag (hoc A,, Ap). Khi cnh h l kh ln, theo yu cu cu to, dc theo cnh h cn dt thm ct thp dc nhng chi Xem l Ct cu to m khng k vo trong tinh ton.Tit din c ct thp t theo chu vi l khi ct thp chu lc c t phn ra tng i u trn c cnh b v cnh l v thng t i Xng theo hai trc (hnh 2.19). Gi sl v S2 l khong cch gia cc trc thanh ct tlp theo Cnh b v canh h. Khi dng cc thanh cng ng knh () v sl = S2 c trng hp t ct thp u theo chu vi._y 1 ` Hnh 2.19. Ti' dll C` ci tllj Thng chl co the t ct thep u , lggo Clu ,ttrong tit din hnh vung hoc titdin ch nht m bl, hl l bi s ca khong cch S. Khi t khng u th nn to ra Int ct thp theo cnh b ln hon theo cnh h bng cch dng sl < S2 hoc chn ng kinh ct thp t theo cnh b ln hn.Xt v mt chu lc, khi nn lch tm phng th vic t thp theo chu Xll t hiu qu hn So 7i vic t thp tp trung dc cnh b. Tuy vy khi kch thc tit din kh ln, S lng ct thp kh nhiu th vic t ct thp theo chu 'i lm cho thi cng n gin hn v khng cn t them ct thp cu to. Hon na khi ct c th b un theo lai phorig th vic t thp theo chu vi tr nn cn thit.83 83. 2.7.2. S ct thp v S ng SutTit din c kich thc b x h trong h l cnh Song Song vi mt phng un. Ct thp c b tri thnh tng lp vung gc vi cnh h ln lt c din tch l Al, A2,. . .An trong Al v An l hai lp ngoi cng t theo cnh b vi A 1 l ct thp chu ko hoc nn t hn, An l ct thp chu nn nhiu. Khong cch t trng tm cc lp ct thp n trng tam tit din l yi. Ly du ca yi l dng khi ct thp khc pha vi lc nn N t lch tm, yi l m khi cng phia (so vi trong tm tit in). Gi hoi l khong cch t trng tm ct thp lp th i n mp vng nn. Mi ho, u dong (hnh 2.20). lp S ng Sut, dng cc C s v gi thit nu mc 1.6. S l thun tin hn khi theo quan im bin dng, dng gi thit tit din phng Xc nh bin dng Si ca cc lp ct thp, t ai suy ra ng sut i.Gi X0 - khong cch t trc trung ha n mp chu nn;X- chiu cao vng nn tnh i. Khi tit din c mt phn chu ko XO < h, ly X = 6h (6 = 0,85).Khi ton b tit din chu nn, X0 Z h, ly X theo cng thc (1-22) c th tnh ton Si v , theo cc cng thc (l-24) v (1-25).Tiu chun TCXDVN 356 - 2005 a ra cng thc thc nghim Xc nh i .C 0) o. z ` -- I+,p, (2-54) ,- i l,l spi - ng Sut trc trong ct thp, vi ct thp thong GSP, = 0. sc,, - ng Sut gii hn ca ct thp vng chu nn, vi cu kin lm t btng nng, btng ht nh, btng nh gi tr os c ly nh Sau: - Vi ti trng mc 2a ca ph lc l ly oscu = 50MPa - Vi ti trng mc 2b, ly Sc_u = 400MPa. 03 = ot - 0,008RbLy . = 0,85 i vi btng nng, CX = 0,8+0,75 i vi btng ht nh (Xem ph lc 4).. Chiu cao tlig i vng chu nn ca btng.i=X h .()lTheo cng thc (2-54) tnh c , > 0 l ng Sut ko, i < 0 l ng Sut nn. Gi tr ca 0', c ly trong gii hn - RSC S i S Rs.84 84. Theo TCXDVN 356 : 2005 nu gi tr i tnh theo cng thc (2-54) i vi ct thp nhm CIV, AIV, AV, AVI, AT VII m vt qu BRS thi phi tnh li theo chi dn ca tiu chun (cng thc 68, iu 6.2.2.l9).Hnh 2.20. S ng Si, bh dng v tit din c ci tlzp t tlleo Cllll v85 85. 2.7.3. Cng thc tnh ton co bnS ng Sut lp hnh 2.18 l tng qut do khng cn phn bit nn lch tm l ln hay b, khng cn cc iu kin X S RhO v X 2 2a' nh khi tnh vi tit din thng thng c ct thp t tp trung theo cnh b.Lp cng thc hnh chiu:N = Rh bx - oi Ai (2-55) Cng thc mmen: M* = Nneo = 0,5 RbbX(h - X) i'Zi Aiyi (2-56) Hai phong trnh (2-55) v (2-56) c dng ng thi vi phong trnh (2-54). Vic Vn dng cc cng thc, phng trnh va nu tnh ct thp l kh phc tp v`l phi chn trc V tr cc lp ct thp Xc nh yi, hoi, gi thit quy lut phn bdin tch cc lp ct thp v thng phi dng cch tinh gn ng dn. vn dng trong thc t th lp v dng h biu tng tc khng th nguyn l thun li hn c.2.7.4. Lp biu tng tcV nguyn tc c th lp biu cho mi trng hp t ct thp bt k, tuy vy thng dng hn c l trng hp Ct thp i Xng theo hai trc. Phng php dng bin S trung gian X l thun li.Lp biu cho mt cu kin c th khi bit kch thc tit din v b tr ct thp bng cch cho X thay i, ban u ly X = 0,lh ri tng dn tng Cp cho n X = h. Vi mi gi tr X tm c mt cp N v M*. Vi X kh b c th tnh c N < 0, ng 'i trng hp ko lch tm, khng dng cc gi tr . Chi ly s liu V biu khi N Z 0.Tinh thm mt gi tr na vi nn ng tm (M* = 0), lc ny N = Rbbh + RSCA. Khi c Xt n un dc cn a thm h s cp S l,Th dll. V biu tng tc cho tit din hnh 2.18 vi cc s liu sau:b = 400; h = 800mm; lo = 6m; btng c Rb = l4,5MPa; ct thp 16 t) 22 c Rs = 365MPa; Es = 210.000; a = 40mm. S liu v ct thp ghi trong bng: 86 86. Tnh i theo cng thc (2.54) trong ly sC_u = 400MPa.to = ot - 0,008 Rh = 0,85 -- 0,008 X 14,5 = 0,734... 400 0,734 0,734 i =[E3-n-l---,7----l)=l202[---l - i 1-0 34 i i l,l l,l ng thi - R_,C` S , S Rs, nh vy - 365 S , S 365. T A. A2 A3 A4 A5 A6 X h,,, = 760 I h,,z = 616 h,,_g = 472 h)4 = T h,,_ = 184 if h,,6 = 40 J l I z 2 s 3 4 4 s G5 680 X 365 >< 365 X 365 X 365 0,434 365 2 -365ir I 'J _TT IT -" -T- 160 X 365 X 365 X 365 0,488 365 0,869 -186 X -365 240 X 365 X 365 0,508 365 0,731 5 l,304 -365 X -365r ir + TL_ Tr P T_320 0,421 365 0,519 365 0,678 99 0,975 -297 l,739 -365 X -365 400 0,526 365 0,649_ l57 0,847 -160 l,2l9 L-365 X -365 X -365 . It . -i I- ,I 480 0,631 196 0,779 -69 1,017 -334 1,463 -365 X -365 X -365 560 0,737 -5 0,909 L_-231 l,l86 -365 X L -365 X -365 X -365 `I g g i g _ 640 _L0,842 LI54 1,039 I 352 X _L 365 >< [-365 X 365 X -i 365 720 0,947 L-270 l,l68 -365 X _J -365 i X -365 r X -365 X -365 800 l,052 -36L l,298_L-365 X I -365_ X L-365_L X -365 X -365Ch thch: N llttlg c dlllt dil X kltrlg cll tllll tollrn c tllily q = K, lzoc o',~ = -R,.,._ TI A.,=760 T A,=760 A,=1520y,,=-72 y,,=-216 y,=~360A Ayl GA Ay GA oAy1O" 106 103 106 lot 106277 -19,9 277 -59,8 199277 -19,9 -141 30,4 -554 19959,8 -554 199ci -i 59,8 -554 199 .-l . .T59,8 -554 19959,8JF-554_4 19959,8 _-554 19959,8 -554l 199Fi u i _ i l F T720 410 J 1471_ 277 19,9 59,8 554 199 800 -551 -198 -2771_-59,8 -277 -19,9 -277 19.9 -277 59,8J -554 19987 87. X R,,bx EA N 0,5R,bx(h-X) ZoAy M* iot 10* 10-* 10 106 Lu 106 r 80 464 E1108 -644 167 398 565 160 928 967 -39 297 488 785 240 1392 28l 1111 389 537 _L 926 _J 320 1856 -1.50 2006 445 539 984 400 495 _L 959 480 356 L_ 816 560 218 607 640 117 414 _ 720 52 _ 240 4 800 1 1 J Vi Mm = 0, ct chu nn ng tm, lc ny phi k n un dc theo phng yiu nht. Bn knh qun tnh b nht i = 0,288 X400 = l 15mm?L === 52,2 > 28 , cn k un dc lcp = 1,028 - 0,0000288 X52,22 - 0,00l6 x52,2 = 0,866 Tnh No theo cng thc (l-6): V NO =

8 h 600_pri _ 400X6003- =72Xl08mm2 12 12Jng vi R, = 9 c E, z 24000MPa.8 Z 2,5EbJ = 2,5X24000X72Xl0 =148O0X103Niul 54002Nth90 90. 1 1: :M-: ,15 ,-l ,-fPl 1 N,,, 14800 el = -M = = 0,17m = 170mm . lch tm ngu nhin ea. N 2000 ea Zsz= 20mm .Cu kin thuc kt cu Siu tnhe0=maX (e1,ea)= 170mm.ne()r'1g15XO,17 = 0,l95m = 195mmN _ 2000x1000: ---=0,926 Rbbh 9x400x600ll_ Nheo __ 200O000Xl95- - = 0, 301 Rbbhz 9X400X6002mD kin dng 14 thanh ct thp t theo chu vi nh trn hnh 2.21. Vi =1;Rs = 260; Rb = 9, tra biu vi n = 0,926; m = 0,301 c c im K nm gia hai biu vi g = 0,025 v 0,03. Ni Suy c s = 0,028. A,, : 0,028 x 400 x 600 z 6720mm2B tr 14 thanh, din tch mi thanh: -6-il-zg = 480mm2 . Chn dng ct thp )25 cdin tch 49lmm2 (hnh 2.23).Ch rng h biu hnh 2.21 mang nhiu tinh tng trng, cha chnh Xc dng cho thit k thc t. ph lc 9 cho mt s biu c th dng c.2.7.6. Phng php gn ng Xc nh NKhi bit lch tm neo cn Xc nh N m khng c biu tng tc ph hp dng th c th tnh ton nh sau:Tnh c chng chiu cao vng nn X theo cng thc (2-60):X=(0,5h-ne0)+ (0,5h-ne0)2 +(2-60) bLy hai gi tr X tnh ton l Xl v X2 vi X, = X + Otlh i X2 = X + Otzh. Gi tr Otl, (X2 ly ph thuc vo t s X/h theo bng Sau:91 91. 0,5+0,6 0,6+0,7 0,7+0,8 0,8+0,9 20,9 5EW -vI -0915 -~2 0h25 I5 EM v5 ng vi mi X chn, tnh ton i ca cc lp ct thp theo cng thc (2-54). T d tnh hai gi tr ca N l N v N2. Tinh N, theo cng thc (2-55) lp: N = Rbbx ZA Tnh N2 theo cng thc (2-61) rt ra t (2-56):*IEN _ M _0,5RbbX(h-X)+ZoiAiyi 2 'leo neo(2-61)Vi hai gi tr X tm c hai cp N 1, N2 nh Vy. Ghi kit qu vo bng sau: C th Xc nh N gn ng bng th hoc bng tinh ton. V th N I v N2 theo X. im ct nhau ca hai th cho bit gi tr ca N (hnh 2.22).Vi hai gi tr ca X nh t chn kh nng hai th N, v NZ ct nhau l rt ln. Nu chng vn cha ct nhau th cn phn on chn thm mt gi tr X mi ngoi hai gi tr c.Cng c th lp cng thc tnh ton N da trn cc s liu NI,N2 c. Trc ht cn bit khong ,XXj, xk m trong khong hai th .ct nhau. Tiu ch nhn bit l 2 Hinh 2.22. tiz xc nh gi tl- N Nzj < Nlj trong khi Nzk < Nlk (hocngc li).N: Nlj(N2kN2jlN2j(NlkNlj) (2_62)Nlj +N2k N2jN1k92 92. Thc cht ca phng php Va trnh by cng l phong php V biu tong tc nhng khng v ton b biu m chi tnh ton cho mt on vi hai im. Tnh cht gn ng cng thc (2-62) l Xem trong on ang Xt s thay i ca N 1 v N2 theo quy lut ng thng.Th d. Ly kt qu ca th d mc 2.7.5. Yeu cu Xc nh lc N khi cho neo = 195mm; Rh = 9; R, _-- 260MPa.Tit din th hin trn hnh 2.23. Cc s liu v ct thp dng tinh ton ghi bng Sau:K hiu Hnh 2.23. Ti din dt tlzp tlteo chu vi Tnh X theo cng thc (2-60):x : (300-l95)+ ,(300-195)2 +-----u-08X 260X6874X 520 = 570mm 9x 400 570-= g- =0,95 . Ly hai gi tr ca X l X1 v X293 93. X, = X + cxlh vi (xl = - 0,35; Xl = 570 - 0,35 X 600 = 360mm X2 = X + (l2h = 570 - 0,2 > 600 = 450mm Tnh i theo cng thc (2.54) vi ct) = 0,85 - 0,008X9 = 0,778= S' [-lJ=l0778-1J=13660778-IJ C i 1-07Z i il,l l,lX hii = .ng thi - Rs S i S Rs, tc l: - 260 S i S 260. Lp ct thp A5, hos = Kt qu tnh ton v lc:A, z 1964ys = -260 A Ay 103 10-510 132,6N _ M" _ 0,5R,,bx(h-x)+Zo,A,y, 2 ___ Nl(N2K N2jlN2j(NlK Nljl N `+N2K N2jNlKN94 94. _ l90l(l 193 - 2264)- 2264(2977 ~ l90l) 1901 + l 193-2264-2977Ghi ch: Tl'Ilg hp c bit tng tc tlt vic xc ilzh N s dll glill hrl. Bit plzN = 2083kN[lp Ct cc tltng SiSau.' Rh = 9.Rs = 260 MPa; = - =ct thp gm l4> c b tri thnh 5 hng trong h 600 15 hng th nht (Al) v hng th 5 (A5) c 4 t), cc hng khc c 24). H biu hnh 2.21 c cc thng s ph hp nh vy.Gi th biu hnh 2.21 c c chnh Xc cn thit th cch tim gi tr Nnh Sau:Tnh A,, z 1425 = 6874mm2;,:i=-E7-4-=O,0286:2,86% ' bh 400x600 tgozzl-9=0,325h 600K ng Xin gc vi tg6 = 0,325 gp cc biu d: Vi s = 0,025 tm c n = 0,87 g = 0,030 tm c n = 0,98 Ni Suy, vi g = 0,0286 c n = 0,94 N = n.Rbbh = 0,94 X 9 X 400 X 600 = 2030000 N = 2030kNKlt qu c Sai s So vi tinh ton (sai s do tnh ton gn ng v do chnh Xc ca biu ).95 95. Chng 3 TIT DIN CH T V CH I3. l. I CNG V TIT DIN CHT `b Tit din ch T gm c cnh v sn (hnh 3.1) vi k hiu cc kch thc nh trn hnh v. b - b rng Sn.: B - b rng cnh; c - b dy (chiu cao) cnh; h - chiu cao tit din (trong phong mt phng un);Aw, Ac - din tch ct thp t tp trung phn Sn (theo cnh b) v phn cnh (theo cnh B);aw, ac - khong cch t trong tm ct thp Aw, AC n mp tit din gn nht.Za = h - aw - ac - khong cch gia trng tm Aw v Ac.Hnh 3.1. Ti din Ch TI - im gia ca tit din, cch u hai mp mt on 0,5h; G - trng tm hnh hc ca tit din.Tit din c mt trc i Xng. Mmen un tc dng trong mt phng cha trc i Xng .Tiit din ch T thng gp l tit din ca vm vi cnh l phn V c lin khi vi Sn. Trong khung ca kt cu nh, ct c tit din ch T c th gp trong mt s trng hp c bit, hoc l ct c lp khi cn m rng ra hai bn, hoc ct c c lin vi tng ca vch cng, li cng.Trng hp ct c lp th b rng B c cu to trong mt phm vi gii hn no cn khi ct c c lin vi tng th b rng ca tng c th l kh ln, lc ny ly B a vo trong tnh ton cn c mt hn ch no .96 96. B=b+2vVi dm tit din ch T c nhng quy nh v vn ca cnh V c ghi trong cc tiu chun thit k. i vi ct c tit din ch T cc tiu chun cn t cp n. Sau y a ra mt vi gii hn c tnh cht tham kho, do tc gi ngh.. l . , , , ,r ,. Ly V S chiu cao cua ct, ong thi:Khi c z0,1h1yvs4c. 0,lh S c < 0,l5h ly v S 3c. 0,05h < C S 0,lh ly v S 2C. c S 0,05h ly v S C. Din tch tit din ch T l AT: Arr=bh+(B-b)cTrng tm hnh hc ca tit din G cch mp cnh mt on yc v cch mp Snmt don yw. 0,5bh2 +0,5(B - b)c2 yc = -- (3-l AT yw Z h _ yc Trng tm G cch im gia ca tiit din I mt on l d: d = 0,5h - yc (3-2)Mmen qun tnh ca tit din ly i vi trc i qua trng tm G v vung gc vi canh h l:(B - b)c312 +(B-b)c(yC ~0,5c)2 (3-3)J=(Yv+Y>+Bn kinh qun tnh i v mnh ?tri= l 7x=l (3-4) AT 13.2. NI LC V IU KIN TNH TON3.2.1. Ni lcNi lc tnh ton tit din ch T cng gm lc nn N v mmen un M nhng y cn ch mmen M c ly theo trc no, trc i qua im I hay trc i qua im G (hnh 3.2).97 97. Gi M1 - mmen un i vi trc qua I.MG - mmen un i vi trc qua G.MG = MI Nd (3-5) N MG =M, + Nd_ ml 05h 05hN MI ' Ye N N M, MG =M, - Nd b l IHnh 3.2: N lc 7ltlZ tit din cl1TTrong cng thc (3-5) ly du (+) khi mmen lm cho cnh chu ko, du (-) khi mmen lm cho cnh chu nn.Trong khi tnh ton kt cu v t hp ni lc cn ch l cc mmen c tnh theo trc qua I hoc qua G.Khi chuyn mmen t MI Sang MGt'th c th Xy ra l MI v Mc; cng chiu hoc khc chiu (MI l dng nhng MG l m). Thng thng, khi MI v MG cng chiu th trong tnh ton, d tinh vi MI hay MG cng khng c g khc bit, chi cn tinh ng gi tr lch tm e v e` (e v e' khng thay i khi tinh vi MI hoc MG). Khi MI v M khc chiu th phi o ngc php tnh vi nu ly theo chiu MI (gi th l mmendong) th cnh chu nn cn theo chiu MG (mmen m) cnh tr thnh chu ko. lch tm tnh hc: el = -BLNQ (3-6a) lch tm ban u eo:cu kin tnh dnh: eo = el + eacu kin Sieu tnh: eo = max (e, ea) (3-6b)Vi ea l lch tm ngu nhin (Xem mc l-5.2).Khi Xt nh hng ca un dc, lch tm t eo tng ln thnh neo vi Tl 2 l c Xc nh theo mc l-5-3.Khi ?L =S 28 b qua un dc, ly T] = l. Vi . > 28 cn tnh J theo cng thc (3-l 3) tnh Nth theo cng thc (l.l4) hoc (1.16) tnh T] theo (1.11). `98 98. 3.2.2. Cc trng hp tnh tonTnh ton tit din ch Tuc phn thnh hai trng hp chnh tu theo chiu ca MG lm cho cnh b nn hoc b ko. Trong mi trng hp chnh li phn thnh cc trng hp nn lch tm b, nn lch tm ln v trng hp c bit. tnh ton ct thp cn phn bit ct thp i Xng hoc khng i Xng.S pht trinca bi ton th hin hnh 3.3.Trn hnh 3.3 phn bit ra 14 trng hp tnh ton khc nhau, tuy vy Vn l cha . Khi cnh ch T nm trong vng chu nn cn cn phn bit trng hp trc trung ho nm trong cnh hoc ct qua Sn.chun b S6 liu: kch thc, vt liu nl lc N, MGi xng t Khng di Xng Khng i Xng t Di xng Ct thp ct thpB Ln B Ln Nn lch lm Hinh 3.3. S d bi ton tlzll t cllz CJLT99 99. 3.2.3. iu kin tnh tonTnh ton tit din btng ct thp theo trng thi gii hn v kh nng chu lc cn tun theo cc iu kin v bn (1.19) v (1.20) v cc chi dn mc 1.6.1. Tuong t nh i vi tit din ch nht, trc U ly m men thng c chn i qua trng tm ct thp Aw hoc Ac v nh vy M,, -"= Ne hoc Ne'. iu kin (1 .20) c c th ho thnh:Ne . 0,8yc ly X = E,Rh `3.3.3. Trng hp c bitNn lch tm ln, khi Xy ra X < 2ac (mc nhin cng nhn X < Rho c os = Rs). Lc ny dng iu kin (3-8) tnh ton, trong :M,g,, = RSAWZ, + MB (3-18) Trong : MB - mmen ca ni lc trong b tng vng nn. Trng hp trc trung ho nm trong cnh, X S c, tnh MB theo (3-18): MB = 0,8R,,Bxa,, - (3-l8a) Tuy vy trong nhiu trng hp, n gin ho tnh ton v thin v an ton c th ly MB = 0 v`l kh b. 3.3.4. Tnh ton ct thp i Xng Bit kch thc tit din, chiu di tnh ton lo, ni lc gm N v M (c chiu gy cho cnh chu nn). Bit c trtmg ca Vt liu (Rh, Eb, RS, RSC) h s R. Cn tnh ton ct thp i Xng Aw = Ac. 3.3.4.1. Chut b s' liu Gi thit aw, ac tinh ho = h - aw v Za = ho - ac. Tnh yc, yw, J; i, lx., Xt nh hng un dc, Xc nh 11.Khi 7=--l_S28 cth bqua undc,n= l.lKhi 7 > 28, tnh N,h, 11102 102. M Tnh el =- ;eo N Khi tinh lch tm tnh hc el cn ch l gi tr mmen M c ly i vi trc qua trng tm G. Nu M bit c ly i vi trc qua trung im O th cn tnhMG theo cng thc (3-5). Tnh lch tm.e theo cng thc (3w9).3.3.4.2. Lp cng thc tnh ton theo cc trng hp vi RSC = RSVi AS = Ac v RSC = Rs, gi thit X tho mn iu kin 2ac S X S E,Rh,, t iu kin (3-15) rt ra biu thc tnh X v tm t l XII_N-RB-mcal Rm ( xlDa vo X 1 phn bit cc trng hp tnh ton. a) Trng hp 1. Khi X .>. c ng thi Xl 2 2ac. Dng iu kin (3-7) trong Mlgh theo (3-13) vi ; = RSC._ Ne - RbbX(ho - 0,5X) - Rb(B - b)c(h0 - 0, 5c)A C RSCZ(3-20)Trong : khi X, S Rh,, - nn lch tm ll, ly X = Xl.Khi Xl > lRh,, - nn lch tm b, c th ly X theo cng thc thc nghim (3-16) hoc lp v gii h phng trnh (tng t nh i vi tit din ch nht) ng thi Xc nh X v os. Ct thp i Xng, ly Aw = AC.b) Trng hp 2. Khi X, 2 c m Xl < 2ac (mc nhin cng nhn X Rh,, tnh M0 theo cng thc (3-39) trong ly X = Rh,,. Tnh tiip vi cc gi tr khc ca X; X = X3 + AX. g vi mi gi tr ca X tim c N v M.3.4. TIT DIN C CNH B KO3.4.1. Cc trng hp tnh tonDa vo chiu ca mmen M = MG bit c trng hp cnh b ko do tc dng ca mmen un (khi k c lc nn N th cnh c th b ko hoc b nn t hn). Lc ny ct thp t pha sn Aw l chu nn, ct thp t phia cnh AC chu ko hoc nn it hon. Da Vao gi tr chiu cao Vng nn X phn bit nn lch tm ln (X S 2Rh,) v nnlch tm b (X >lRh,,).T`lrng ltp I . Khi X S h - c. Ton b cnh chu ko (hnh 3.5a). B qua s lm vic ca btng phn cnh. Lp cng thc tinh ton nh i vi tiit din ch nht b X h, ct thp A, = AC; A's = Aw..Hinh 3.5 : Cc tt'rzg hp cn/1 b ko109 109. Tt'`ng hp 2. Khi X > h - c, cnh c mt phn hoc ton b chu nn, thng chi c th Xy ra vi trng hp nn lch tm rt b. 3.4.2. iu kin tnh ton, cng thc c bn iu kin tnh ton l iu kin chung (3-7), trong : e=ne,, +yL. -ac (3-44) Trng hp c bit dng iu kin (3-8). Cng thc tnh Mlgh, Mzgh c thnh lp tu theo gi tr ca X. 3.4.2.1. Trng hp cnh chu ko ton bKhi cnh chu ko ton b; X < h - c; Xc nh M I gh theo cng thc (2-4), vi ccV chi dn v Xc nh X theo mc 2.1.2. Trng hp c bit, khi X S 2aw tnh Mzgh theocng thc (2-12). 3.4.2.2. Trng hp cnh c mt phn chu nnNn lch tm b, khi tinh c X > h - c th cnh c mt phn chu nn. t xc = X + C-h l chiu cao phn chu nn ca cnh. Tnh kh nng chu lc theo cng thc (3-45):M,g, = Rbbxho -+0,8Rb(B-b)XC(c-ac -0,5x,,)+R,CA,,Z._, (3-45)N = Ngh = Rbbx +0,8Rb(B-