Chng 2: Mng nng 1. c im v phn loi ca cc loi mng nng? 2. Trnh by cu to mng nng? 3. Xc nh sc khng ca t nn di mng nng theo TTGH cng ? 4. Tnh ton mng nng theo TTGH cng , s dng? 5. Thi cng mng nng ni khng c nc mt v c nc mt? Chng 3: Mng cc ng knh nh 6. c im mng cc? Phn loi cc? 7. Cu to b cc ca mng cc ng knh nh? 8. Cu to cc c b tng ct thp c sn (ng knh nh)? 9. Xc nh sc khng nn dc trc cc n v nhm cc ca cc ng theo tiu chun thit k cu 22TCN272-05? 10. 11. 12. 13. 14. 15. 16. 17. Xc nh sc chu ti dc trc ca cc n theo phng php th Trnh by phng php xc nh sc khng ngang ca cc n v Tnh ton mng cc theo TTGH cng , s dng? Trnh by thi cng mng cc (h bng phng php ng)? Cu to v phm vi s dng mng cc ng BTCT ng knh ln? Phm vi s dng v c im, cu to ca mng cc khoan nhi ng Thi cng mng cc khoan nhi ng knh ln? Xc nh sc khng nn dc trc cc n ca cc khoan theo tiu nghim hin trng (ti trng tnh v ti trng ng)? nhm?
Chng 4: mng cc ng knh ln
knh ln?
chun thit k cu 22TCN272-05? Chng 5: mng ging chm 18. 19. 20. 21. c im, phm vi s dng v cu to ca mng ging chm? Thi cng mng ging chm? c im nn t yu? X l nn t yu bng mt trong cc phng php cc ct, ging ct,
Chng 6 : xy dng trn nn t yu
bc thm?
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Cu1: c im v phn loi mng nng? Tr li: c im: Mng nng l nhng mng c thi cng trong h o trn c chiu su h 1.5B th Cw1=Cw2=1,0; N N qm h s sc khng c iu chnh. m b) Cc pp bn thc nghim: * Th nghim xuyn tiu chun SPT(xuyn tiu chun)qult = 3.2 10 5 N corr B (C w1 + C w 2 Df B ) Ri
4
Trong : Ncorr= gi tr s ba trung bnh SPT hiu chnh trong gii hn chiu su y mng n 1,5B di y mng(ba/300mm) B= chiu rng mng; mm Cw1, Cw2: h s hiu chnh xt ti nh hng nc ngm. Df= chiu su chn mng ly n mng,mm Ri= h s chit gim tnh n ah nghing ca ti trng c cho theo TC. H= ti trng ngang cha nhn h s V= ti trng ng cha nhn h s * Dng CPT(chy xuyn ng) skun danh nh vi cc mng t trn ct v si:qult = 8.2 10 5 qc B (Cw1 + Cw 2 Df B ) Ri
* Dng kt qu o p lc: * Th nghim bn nn
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4. Tnh ton mng nng theo TTGH cng , s dng? I-Kim ton theo TTGH cng 1. Kim taons sc khng nn t di y mng : Cng thc kt : V = iiVi RR =R n =(qn A' ) = q R . A'
Vi i
tng ti trng thng ng ti y mng nhn h s.
A: din tch c hiu y mng=B x L B=B-2e B L=D-2e D B,L chiu rng v di ca mng e B ,e L lch tm theo 2 phng q R : sc khng tnh ton nhn h s q R =B q L =B qult q ult sc khng danh nh; h s sc khng ly theo bng (ph thuc loi t nn) B q ult xc nh da theo l thuyt c hc t hoc cc phng php bn thc nghim 2. Ph hoi do trt: sinh ra nu cc hiu ng lc do cc thnh phn ti nm ngang vt qu tr s nguy him Cng thc kim ton: H = ii H i RR = RR Trong R R sc khng nhn h s, chng li s trt: R R = + Rep ep
Tng ti trng ngang gy trt nhn h s R n Sc khng trt danh nh
h s sc khng gia t v y mng R sc khng trt danh nh gia t v mng
h s sc khng i vi sc khng b ng. e pR ep h s sc khagns b n danh nh ca t tc dng trong sut i tui th thit k cng trnh. 3. Kim ton lt hay mt tip xc qu mc: k kim tra: -khi mng t trn mt t, v tr hp lc ca cc phn lc phi nm trong phm vi chiu rng mng (e < 1 / 4 B ) -khi mng t trn , v tr cc hp lc ca cc phn lc phi nm trong phm vi 3/8 chiu rng mng (e < 3 / 8 B ) II- Kim ton theo TTGH s dng: Thit k theo TTGH SD phi bao gm ln tng v chnh lch ln cng nh n nh tng th. n nh tng th phi xt n mt hay nhiu h n cc k sau: c ti trng ngang hay ti trng nghing;
6
mng t trn nn p; mng t gn hay trong phm vi mi dc; kh nng mt chng do xi; a tng chu ti c th nghing ln * Phn tch ln Tng luns gm: luns n hi, n c kt, ln th cp; mm S t =S e + S c + S s Ln tc thi: xy ra t ko dnh, t dnh ko bo ha; Ln c kt xy ra nhiu trong t dnh ht nh c bo ha ln hn khong 80% Ln th cp cn quan tm ch yu trong trm tch t do cao hoc t hu c. Cc yu t khc nh hng ti ln: ti trng cu nn p v ti trng ngang hay lch tm v i vi cc mng trn t dng ht, ti trng rung ng do cc hot ti ng hay ti trng ng t cng cn c xem xt khi thch hp. + ln ca mng trn nn t ko dnh: C th d tnh ln n hi ca mng nh sau:Se = q0 (1 v 2 ) A Es z
q o cng ti trng (Mpa) A= din tch mng E S m ung young ca t, ly theo qui nh
Z h s hnh dng ly theo qui nhV h s poisson theo qui inh . + ln ca mng trn nn t dnh: -Ln c kt: tnh theo qui trnh phn 10.6.2.2.3b - ln th cp: +ln ca mng trn nn . * n nh tng th: S dng pp cn bng gii hn nh gi n nh tng th ca m cu, tng chn, mi t c chn gi, v nn t hay nn . n nh tng th c xem xt vi t hp ti trng TTGH SD, vi h s sc khng ph hp
7
Cu 5. Thi cng mng nng ni khng c nc mt: a)Mng h o trn(khng s dng bin php gia c) - c to thnh bng cch o b cc lp t mt trn n cao y mng to khng gian thi cng b mng. Thc t phi o su hn CDDM v gia c bng lp b tong ngho v lp dm m cht ri mi t mng - c th o vi dc thay i khi gp nhiu lp a cht, lm chiu ngh ti ni mi taluy i dc
-u im: thi cng n gin; tn dng c nhn lc; gi thnh r -Nhc im: Khi lng o p ln; din tch mt bng thi cng ln, nn ko p dng c ni cht hp; d xy ra hin tng sp vch h. Ch p dng c nhng ni ko c nc mt v cao mc nc ngoi thp hn y mng. b) Chng vch h bng vn lt:Trng hp t vch h ri rc, d sp, mt bng thi cng cht hp, cao mc nc ngm thp hn cao y h, c th dung vn lt chng vch h mng to khng gian thi cng.
A 1 2
3
A
*Cu to vn lt: -Vn lt ngang(1): gm nhng thanh g c tit din b. (b=0.2-0.25m) (=4-8cm), s dng g tt( loi 3 tr ln) -Thanh chng ng (2)c tit din hnh vung hoc nh trn ( d=10-20cm) -Thnh chng ngang (3)trn hoc vung: d= 14-22cm *Thi cng: Trong qu trnh thi cng o t n u t vn lt n sau t thanh chng ngang v thanh chng ng. trong trng hp mng tng i ln, p lc xung quanh h mng ln, ng ta tin hnh ng xung quanh mng thanh st ch I sau dng nm ta vn lt ln cch thanh st to ra kt cu chc. -u im: Gim khi lng o p v din tch chim dng mt bng, ko gy ln st ct xung quanh -nhc im: khng ngn c nc chy vo h mng -pvi p dng: Ch dng trong cc TH chiu su h mng Hm 4m, mt bng cht hp, nc ngm thp hn C y h mng c) Chng vch h bng cc vn *Cu to: -Cc vn phi c cng chu c lc ca h mng vi n~ h mng su phi b tr tm chng ngang gim p lc t. Hai kiu cc vn: cc mng, cc cht to s kn nc
*Thi cng: 8
1:m
1:n
b
-Dng ba ng hoc ba rung h cc xung n su cn thit, trc khi h cc c th lin kt vi cc vi nhau tng cng v gim thi gian thi cng -Cc vn phi chn di lp t 1 khong d tha mn 3 k: m bo n nh; Cc vn ko b lch; Nc ko lun qua chn cc vo h mng Trc khi ng cc vn phi sd cc nh v xd chnh xc v tr ng cc v sd khung dn hng gip thi cng cc vn n ng cao thit k. -Trong th h mng qu su, m bo n nh cc vn phi sd cc vn nhiu tng.
*u im: Thi cng c n~ ni c nc mt, nc ngm gn st mt t v mt bng thi cng cht hp; t nn ri rc. c s ln s dng lun chuyn kh nhiu, kh nng ti s dng cao nn gim khu hao, gim gi thnh thi cng. C th s dng cho h mng su nu c thm chng ngang *Nhc im: chi ph thi cng cao, i hi trnh k thut c tay ngh, thit b chuyn dng Thi cng mng nng ni c nc mt a) Vng vy t: ngn cn nc mt v lm kh ni thi cng, ng ta s dng vng vy t.B1 m = -21:1
MN C T Hn< 2m:3 -1
1:2
1:2
-B rng B ln i li, nc ko thm qua( 1-2m), mc nc thi cng 0 .5 m
H n =2 - 3 m ?t
-Vng vy cc vn kp: 1 vng vy t + 2 vng vy cc vn. p dng cho h mng su(hv) c) vng vy cc vn thp: s dng khi chiu su Hn>4m. d) vng vy cc ng thp dung cho cc h mng rt su, cc ng thp 600-1000mm, c ng xung quanh h mng m bo n inh t vch h mngHn=3-4m
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Cau 6: c im mng cc? Phn loi cc? Tr li: Mng cc gm 2 b phn chnh l cc v b cc -Cc l c phn ch yu truyn ti trng ct xung lp t tt nht di mi cc thng qua b cc. -B cc l b phn lin kt gia cc cc v truyn ti trng ct bn trn xg mi cc. *u: mng c l mng su do k n nh tt , kt cu tng i n gin -k/n chu ti ln v mi cc thg t lp t tt -cng ngh thi cng ph bin *)nhc: Ko th ko di chiu di cc theo mun do k cc nh v hn ch v mnh : Lc/d =30-80 -Khong cch gia cc tim cc phi 2.5d hoc 750mm do kch thc b phi m rng dn n tn nguyn vt liu -vi cc BTCT g knh nh th ct thp b tr ch yu phc v vn chuyn, tn ngvl. -k/nng chu lc ngang km -Thi gian thi cng mng cc ch tnh c cc vn chuyn chim 60-80% thgian thi cng do ko di thgian thi cng ct b) Phn loi cc *Theo vt liu: -Cc tre: sd trong cc ct ti trng nh,ct tm thi hoc x l tng t yu trong th tng t yu nh -cc thp: sd trong cc ct thi cng, ct tm c dng tit din hnh vung hoc hnh trong, c k/n chu ko v nn, tuy nhin d b n mn, gi thnh t. -Cc btct l loi cc c sd ph bin nht c tit din hnh vung, trn hoc ng. Chu lc tt c th p dng vi nhiu loi a tn tuy nhin trong lng ln, k/n chu ko km *Theo bin php thi cng -Cc h bng ba ng: gi thnh r, chi ph thp tuy nhin gy ting n v a/h n cc ct xung quanh, d lm h hi bt u cc v gy cc do lc xung kch. -Cc h bng pp p tnh: ko gy chn ng ph hp vi vic thi cng mng sa cha cc ct, tc thi cng chm, gi thnh cao -Cc h bng pp xon -Cc h bng pp rung kt hp vi xi -Cc m rng chn *Theo chc nng lm vic ca cc -Cc ma st l cc m k/n chu ti ca cc ch yu do thnh phn ma st ca cc v cc lp t nn to nn -Cc chng: khi cc da vo trong tng th chuyn v ca cc rt nh v sc chu ti ca cc ch yu do sc chng mi to nn *Theo kch thc cc d=250-600mm; g knh nh d=600-3000mm: g knh ln d=3000-5000mm: gin v mng d>5000mm: mng ging chm *Theo nghing -Cc ng -Cc xin: ch dng cc xin trong tg sc chu ti ngang ca cc thng ng ko chu lc ngang.
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Cu 7: Cu to b cc mng cc ng knh nh a)Cao : -C nh b : nu c nc mt thng t di cao sau xi -C y b: ph thuc vo k a cht, kh nng chn dng chy, vn xi l, chiu di t do ca cc, mc nc thit k Cy b =Cnh b Hb (chiu dy b) -chiu dy b (Hb) ly theo kinh nghim: 0.5-1m (mng nh) ; 2-3m (mng tb hoc ln). Hb phi tha mn k cc c th ngm vo trong b t nht 1 khong 300mm
d e lta
b) Kch thc b cc -Kch thc nh b ph thuc kch thc y cng trnh, thong thng m rng 0.2-1m -Kch thc y b ph thuc s lng cc v b tr cc trong mng *B tr cc: theo 22TCN272-05 v AASHTO-2007: k/c gia cc tim cc lin nhau mt phng y b ko c nh hn 2.5d hay 750mm, k/c t mp cc ngoi cng ti mp b 225mm c)vt liu b cc -b tng, ch to b cc: M200-300 -Ct thp: +Li ct y b: vai tr chng ph hoi khi chu un, ng knh ct thp v k/c gia cc mt li. ph thuc vo ln momen un. Trong trng hp momen theo 2 ph chnh lch ko nhiu th g knh ct v k/c gia cc mt li ly = nhau. +Li ct cu to nh b: tng cng cho b cc, tng cg lien kt gia tr v b, ng ta thng dng ct ng knh ln +Li ct cto xung quanh b: thng c g knh =12-16mm k/c gia cc ct : 200-300mm. Thng khng tnh n trong kim ton cng +CT cc b u cc: gim kh nng u cc chc thng b(v hnh minh ha) d) Lin kt cc vo b: 2 cch -Khi ko p u cc: cc ngm trong b tt thiu 300mm( khng k phn u cc h hi) -Khi p u cc: thi cng xong cc, p 1 phn u cc, b BT, b tr li CT dc d tha ra, lm cc CT ai, lois CT cc b, u cc ngm vo tt thiu 150mm(phn BTCT nguyn vn) Phn ct thp ch c t l 0.005 lng ct thp phi chu c lc =1.25 Fy As
Hb
11
Cu 8 : Cu to cc c BTCT c sn ng knh nh tr li 1.Kch thc cc : mt ct ngang cc : + Hnh vung cnh a = 20 , 25 ,30,35,40 ,45 cm + Hnh trn ng ng knh d=< 55,60 cm Chiu di ton cc phi tho mn yu cu v mnh = 30 70 Chiu di t cc : cc btct ng knh nh c ch to thnh tng t cc, chiu di 515m, chiu di ti a ca t c ph thuc vo ng knh cc ( d = 30 --> 8 m, d4515m), cc t cc s c ni dn vs nhau trong wa trnh thi cng chiu di c thit k, thng s dng lien kt = mi ni hn. 2.Vt liu ch to cc : BT: thng c mc 25Mpa30Mpa vs cc ng knh nh c sn CT: CT ch to cc gm nhng loi sau : CT ai : + 6 8mm + bc ct ai a = 510 cm u cc,a=15 20 cm gia cc. + C th dung ct ai ri hoc ct ai xon c + Vi tr : L ct thp cu to c nhim v chng nt,chng ct, chu ng sut cc b khi thi cng CT dc ch : + 12 32mm + S lng : 8 n 12 thanh trn 1 mt ct ngang + Vi tr : CHu lc trong qu trnh hkai thc,vn chuyn v c bit khi g cc. CT mc cu : + 14 25mm + 2 mc t cch u cc L1=0.207Ld , v c th them 1 mc treo cch u c L2=0.294Ld xc nh k, s lng ct dc ch v v tr mc cu L1,L2, chng ta tnh ton theo 2 cch b tr sau : Cch 1 : B tr cu v treo c khc nhau : (Hnh v ) M max = M-max= ( -a)2 x a2 = (L-2a)2 - a2 a=0.207 Ld Cch 2: b tr vs treo cc ti 1 v tr: M+max= b2 M -max= ( )2 - = - CT mi cc : + 3240 mm + Di 60 90 cm ,on nh ra khi mi cc khong 5 10 cm + Vai tr: nh hng cc, ph v hoc y cc vt cng trong qu trnh h cc -li ct u cc: + 68mm + mt li a = 5cm + Vai tr : trnh ph hoi be tong do chu ng sut cc b trong qu trnh ng cc Vnh ai thp u cc + Thp bn dy, = 8 12mm + Vai tr : dung ni cc dt cc vs nhau Mi ni cc : 12
-
+ i vs cc c , vung, tat hg s dng mi ni hn + i vs cc trn, ng to tat hg s dng mi ni bulng + Mi ni phi m bo cng mi ni tng ng hoc ln hn cng cc ti tit din c mi ni ( hnh v ) 1.CT dc ch 2.CT ai 3. CT mc cu 4.Li ct u cc 5.Vnh ai ct u cc 6. vnh ai ct mi cc 7.Thanh ct cng mi cc
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1. 2.
Cu 9 :xc nh sc khng nn dc trc, cc n v nhm cc ca cc ng theo tiu chun 22TCN 272 - 05 Tr li : Kim ton sc khng nn dc trc ca cc n : Cthc kim ton : N max + N Ptt Ptt : sc khng tnh ton chu nn ca cc n N Trng lng bn than cc N = Vc.bt Nmax : Ni lc ln nht tc dng ln 1 cc, xc nh = pp FB_ pier Kim ton sc khng nn dc trc ca nhm cc Cthc kim ton : Vc < QR = gQg Vc : tng lc gy nn nhm cc nhn h s QR : Sc khng dc trc tnh ton ca nhm cc g : Cc h s sc khng nhm cc Qg Sc khng dc trc doanh nh cu nhm cc - Xc nh Qg Qg = ( 2X + 2Y )Z + XYNcSu Trong : + Khi 2.5 Nc = 5( 1 + )(1 + ) + Khi >2.5 Nc = 7.5(1 + ) :cng chu ct ko thot nc trung bnh dc thoe chiu su ca cc ( MPa) Su : Cg chu ct khng thot nyc ti y mng (MPa) X,Y,Z : CHiu rng , CHiu di v c cao ca nhm cc
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10. Xc nh sc chu ti dc trc ca cc n theo phng php th nghim hin trng (ti trng tnh v ti trng ng)? - Th nghim ti trng tnh : 2 loi + L1 : gia ti n khi cc b ph hoi xc nh sc chu ti gii hn ca cc. + L2 : l nn tnh ktra. Thit b th nghim + Thit b gia ti : h thng kch v cc neo; hoc cc khi b tng , bao ti ct. + Thit b o ln : my thy bnh, ng h o ln + ng h o thi gian Trnh t th nghim: + H cc n cao tke, cho cc ngh 1 time (10-15 ngy i vi t ct v 10-30 ngy i vi t st) + Cht ti trng tnh theo tng cp, mi cp c ln P= ( : ) P d tnh ( P:xc nh theo CT kinh n ) + Khi cc ngng ln : ln i vi t ct sau 1h < 0.2 mm; i vi t st < 0.1 mm. Cht ti n khi cc b ph hoi, ngha l khi nn t ko sc gi cc lm cc ln t ngt hay ko t c n nh v ln. Xut hin cc TH sau th ngng gia ti : - ln ca cc = 5 ln ln cp ti trng trc . - ln ca cc = 2 ln ln cp ti trng trc nhng qua 24h vn cha t c ln tng i. - Tng ln cng dn vt qu 80-100mm Xc nh sc chu ti gii hn + Da vo chuyn v gii hn : trn ng cong quan h P-S, sc chu ti gii hn P l ti trng quy c ng vi chuyn v gii hn S. Thng thng ti trng ph hoi l ti trng tng ng vi chuyn v ca u cc vt qu mt tr s nht nh no , thng l 10% ng knh mi cc. + Da vo phng php th: - TH 1 ng cong S-P c im un r rng: Sc chu ti gii hn c xc nh trc tip trn ng cong , l ti trng tng ng vi im ng cong bt u thay i dc t ngt hoc ng cong gn nh song song vi trc chuyn v. - TH ng cong S-P thay i rt chm, rt kh hoc khng th xc nh chnh xc cc im un: Sc chu ti g.han c xc nh theo cc phng php th khc nhauO Pgh
P
O
Pgh
P
S
S
TH1
TH2
u nhc im : Phn nh tng i chnh xc s lm vic ca cc trong thc t v vy kt qu Th nghim c tin cy cao. Tuy nhin, th nghim cng knh phc tp, tn thi gian v kinh ph.
Th nghim ti trng ng : Nguyn tc TN : Cc cng chiu di, ng knh, h n cao thit k, cc c h ti hin trng. Dng qu ba trng lng q c chiu cao ri l h. ng 1 nht vo u cc, di tc dng ca ti trng cc s c ln 1 on l e, e gi l chi ca cc ( chi ca cc l ln ca cc di tc dng ca 1 nht ba). Tnh ton : Mqh gia e- P - Theo Gxevanov : P = + n : h s kinh nghim 15
-
F : din tch mt ct ngang cc q : trng lng phn cc( trng lng m cc, mi cc, cc dn k : h s phc hi sau va chm chi gi : - ng 1 nht ba , lm cho cc b ln : xc nh c chi gi e - Sau mt thi gian khi ng 1 nht ba, xac nh c chi thc e Thc t : e e ; i vi t ct e < e ; t st e > e khc phc chi gi cho cc ngh 1 time. u nhc im : Th nghim n gin, tn t tin, nhng ko phn nh chnh xc iu kin lm vic thc t ca cc , kt qu th nghim ko ng tin cy, thng dng kt qu theo di qu trnh thi cng. Chn loi ba ng cc : d dng quan st chi ca cc hay m bo b tng u cc ko b ph hng khi ng cc do chn ba c nng lng xung kch qu ln. - Da vo nng lc xung kch ca ba : E 25 Ptt E : nng lc xung kch ca ba Nmm Ptt : sc chu ti tnh ton ca cc (kN) - Da vo h s xung kch K : K = ; Q : trng lng ba E : nng lc xung kch ca ba q : trng lng cc
16
11. Trnh by phng php xc nh sc khng ngang ca cc n v nhm cc? o Sc khng ngang ca cc n. Sc khng ngang ca cc n c tnh theo CT : P = .P ; P : sc khng ngang gii hn ca cc n (Mpa) P : h s sc khng ngang ca cc + PP Broms xc nh P - Sc khng ngang ti hn trong trng hp u cc t do P = ().(9.c.B) ; i vi t dnh = -> i vi t ri L : chiu di cc ngp trong t H : cnh tay n ca lc ngang ti mt t B : ng knh cc L : chiu di ngm ca cc tnh t su cch mt t 1,5B hay L = L - 1,5B L : chiu su ti tm quay v L = (H+23L)/(2H+L) L : chiu su ti tm quay, tnh t su cch mt t 1,5B, hay L = L - 1,5B - Sc khng ngang ti hn trong trng hp u cc ngm P = 9.c.B(L-1,5B) -> t dnh = 1,5 .BL K -> t ri + Phng php Meyerhof xc nh P PP Mey xc nh ct ngn v ct di theo K = : K < 0,008 cc ngn K 0,012 cc di K = 0,008-0,012 l cc trung bnh K : cng tng i ca cc ; EI : cng chng un ca tit cc E : modun bin dng ngang ca t trong phm vi ngm tnh ton ; L : chiu di cc t ri : Cc ngn, t ri P = 0,12 BL K v P 0,4p BL trng lng th tch c hiu ca t B : cnh hay ng knh cc K : h s sc khng ngang P : p lc ti hn thu c sau tn nn ngang, c th ly : p = 50tan (45+ /2)e .tg Cc di, t ri : tnh tng t nhng s dng chiu di quy i L thay chiu di thc L nh sau: L = 1,65KL v L L t dnh Cc ngn : P = 0,4SBLK v P 0,4pBL S : sc khng ct ko thot nc ; K : h s sc khng ngang P : p lc ti hn thu c sau tn nn ngang, c th ly : p = 7S Cc di : : tnh tng t nhng s dng chiu di quy i L thay chiu di thc L nh sau: L = 1,5KL v L L ( K : cng tng i ca cc) o Sc khng ti trng ngang ca nhm cc : P = . P = . .P P : sc khng danh nh ca 1 cc n (N) P : sc khng danh nh ca nhm cc N : h s sc khng ca nhm cc c quy nh ( tra bng) : h s hu hiu ca nhm cc c xc nh trong quy trnh. = 0,75 cho t ri ; = 0,85 cho t dnh Phi ly sc khng ngang ca tng s ca sc khng ca mi cc c h s trong nhm cc. 12.Tnh ton mng cc theo TTGH cng , s dng? 17
TT mng cc theo TTGH cng : - Kim ton sc chu nn v chu nh dc trc ca cc n + Tnh ni lc u cc + iu kin kim ton : V + P Q : vi cc chu nn V Q : vi cc chu ko V : ni lc u cc chu nn dc trc ln nht; V : ni lc u cc chu nh ln nht Q : sc khng chu nn dc trc ca cc, nhn h s sc khng. Q : sc khng chu ko(nh) dc trc nhn h s sc khng P : trng lng bn thn ca cc n c xt n lc y ni - Kim ton sc khng v sc khng nh ca nhm cc + Kim ton sc khng dc trc ca nhm cc V Q= Q = Q + Kim ton sc khng nh dc trc ca nhm cc V Q= Q = Q V: tng lc gy nn nhm cc nhn h s; V: tng lc gy nh nhm cc nhn h s Q : sc khng dc trc ca nhm cc; Q : sc khng nh dc trc ca nhm cc ; : cc h s sc khng v khng nh ca nhm cc - Kim tra kh nng chc thng ca cc vo lp t yu pha di y mng: Nu nhm cc c ng trong lp trm tch tt nm trn lp trm tch yu phi xt n kh nng ph hoi, chc thng ca mi cc trong tng lp yu hn. Nu tng t nm di bao gm t nn ln yu hn phi xt n kh nng ln ln hn trong lp t yu hn. Sc khng bt k chiu su no di cc mi cc phi c xc nh trn c s kch thc hnh chiu mng quy c . Kh nng chu lc phi cn c vo tiu chun ca mng m rng c quy nh trong quy trnh - Kim tra sc chu ti ngang trc ca cc v nhm cc + Kim tra sc chu ti ngang ca cc n : Q P = P Q : ti trng ngang tc dng ln cc n ; P : sc chu ti ngang tnh ton ca cc P = P : sc chu ti ngang gii hn ca cc n; : h s sc khng + Kim tra sc khng ngang ca nhm cc Q P = .P = . .P Q : ti trng ngang tc dng ln nhm cc ; P : sc chu ti ngang tnh ton ca cc P : sc khng ngang danh nh ca cc n ; P : sc khng danh nh ca nhm cc : h s sc khng ( tra bng) : h s hu hiu ca nhm cc c xc nh trong quy trnh. TT mng cc theo TTGH s dng : Thit k mng cc theo TTGH s dng bao gm nh gi : + Chuyn v ngang ca nhm cc : Gii hn chuyn v ngang ca mng cc ko c vt qu chuyn v ngang cho php l 38mm. Chuyn v ngang ca nhm cc phi c tnh bng cch dng pp c xt n tng tc t- kt cu, chuyn v ngang cho php ca cc da vo so snh chuyn v ca cc b phn kt cu. Cn xt n tc ng ca sc khng ngang do b mng c chn ngp vo t khi nh gi chuyn v ngang. Sc khng ngang ca cc n c th xc nh theo TN ti trng tnh. + Ln ca mng cc: Dng tt c cc t hp tc dng trong t hp ti trng s dng. - Vi mc ch tnh ton ln ca nhm cc, ti trng c gi nh tc ng ln mng tng ng t ti 2/3 su chn cc vo lp t chu lc.
18
- p lc tnh ln q : q = = vi F = (a+2.z/2)x(b+2.z/2) - Ln ca mng cc trong t dnh : pp tnh tng t mng nng - Ln ca mng cc trong t ri SPT ; = 30.q.I. NI ; s dng CPT : = I = 1- 0,125 0,5 q : p lc mng tnh tc dng ti 2D/3 B : chiu rng hay chiu nh nht ca nhm cc : ln ca nhm cc I : h s nh hng ca chiu su chn hu hiu ca nhm D : su hu hiu ly bng 2D/3 D : su chn cc trong lp chu lc q : sc khng xuyn hnh nn tnh trung bnh trn su X di mng tng ng N1 : gi tr trung bnh i din hiu chnh i vi ng sut tng v hiu sut c hiu ca ba N1 = C.N N =( ER/60%).N C = [0,77log (1,92/ )] v C < 2,0 N : s nht ba cha hiu chnh (Ba /300mm) N : s nht ba hiu chnh cho hiu sut ca ba (nht/300mm) ER ; hiu sut ca ba tnh theo % ; ng sut thng ng hu hiu
Db
2 3 Db
19
Cu 13: Thi cng mng cc nhng ni ko c nc mt a) ng cc trc tip trn t: -Chun b ng cc -X trng tm mng cc v vtr tng cc -Tin hnh ng cc, sd cc dn x cao mng cc -p b tng u cc, o t h mng, p dng vn khun *)Pvi sd: do tn nhin liu cho ng cc dn v cc chnh nn ch s dng trong n~ TH chiu dy b mng nh, slg cc mng t v mc nc ngm nm cao hn cao y b
c ? c d?n
c?c d?n
a) b) c) d) b) ng cc trong h o: -X v tr kt v cao o, tin hnh o t h mng n cao y b hoc trn mc nc ngm trong th mc nc ngm nm trn cao y b th cao o phi kt thc sm hn mc nc ngm t nht 0.5m -Chun b cc, ba ng cc, x v tr tm nhm cc v vi tr tng cc -Tin hnh ng cc n cao thit k -p u cc lm sch h mng, lp dng vn khun v thi cng b *pvi p dung: trong cc th h mng su, kin a cht n nh, mc nc ngm nm su di cao y b. *u nhc: -Tit kim nng lng v rt ngn c cc dn -C th c gii ha vic o t tuy nhin gp kh khn trong vic n nh vch h o c)ng cc trn sn tm pvi: s dng tng cc th h mng su, cao mc nc ngm thp v c th tit kim c cc dn tuy nhin tn vl lm sn tm d)ng cc trn sn di ng: -Sd trong cc tg b rng h mng ln Thi cng nhng ni c nc mt a)p o nhn to *thi cng: xnh v tr, kthc, cao o -Tin hnh ng cc trn o nh ng cc trn t -Lp dng vn khun tin hnh ht nc trong h mng, p u cc v thi cng b *pvi: trong n~ th mc nc mt = M300 va ct thep coc : c ch tao thanh tng lng ct thep co chiu dai 6 12m.mi lng CT gm cac loai sau : 1.ct chu : ng kinh ct doc 12-32mm ( 40mm),s lng thanh c b tri theo thit k nhng ko c it hn 3 thnah va k/c mep 2 thanh lin k ti thiula 10cm.chiu dai ct chu phu thuc vao chiu dai ct thep. Mi ni cac lng la mi ni han howcj dp ep ng , chi sd mi ni buc cho cac oan lng ghep trong coc co d < 1,2m .va chiu dai toan b ng > 25m . 2.ct ai : kinh ct ai 6-16mm c qun xung quanh CT doc thanh hinh lo xo hoc ct ai ngang. Bc ct ai c b tri theo thit k nhng ko c ln hn 55cm. 3.thep inh vi: kinh xp xi bng ct chu,thay th ct ai 1 vi tri va t cach nhau khoang 23m.co t/d gia ung c ly ct chu va cung vi ct chu tao thanh khung. 4.tai inh vi :hinh cung hoc hinh det.kinh gn bng ct chu.tao lp B tong bao v u x quanh lng CT va tranh lec tm khi ha long vao l khoan. 5.moc neo :dung nng ha lng CT khi ni.co kinh phu thuc vao lng Thep. 6.ng thm do :bng thep hoc nha phu thuc vao chiu dai coc. +cu tao b coc :tng t nh b coc ng
1
5
4
3
2
6
1
3 4
1
3
D
D
Hinh ve : 23
Cu 16 : thi cng mong coc khoan nhi kinh ln ? Tra li : *khoan tao l : -khoan bng may khoan xon : lp mi khoan v trc khoan vo gi ba, ly t dng phoi t, s dng khi g kinh coc D= 300-800mm(D30 35 40 Cc nh), chiu su khoan ti 35m, n cac tng t cui soi, c bit l cui si kt(kt gia cc tng khc) -khoan tao l = may khoan gu xoay D= 800-2000mm ,L=68m ,tc khoan tng ti nhanh 1015m/h ,khoan qua t set ,cat,soi,san khoan c tng t c tr s SPT 1 ,e > 1. -t hu c,than bun. -mt s loai cat hat nho,hat min,hat bui bao hoa nc.
28
Cu 21 : x ly nn t yu bng 1 trong cac phng phap coc cat, ging cat, bc thm. Tra li : 1. p2 coc cat : la p2 tin hanh tao l trong t yu sau o nhi cat vao l va m cht. *tac dung :giam h s rng cua t ,tham gia chiu lc cung nn t va thoat nc trong t. *u nhc im : thi cng n gian bng cac VL ph bin ( cat th,san soi ) .co tin cy cao ,thich hp vi t co rng ln, ngay khi lp t yu tng i su. Ko hiu qua /v t dinh bao hoa nc. *PVSD: vi nn t la t ri rac ( cat,cat pha,set pha ) co h s rng ln va b day tng i ln (h>= 2m ). *tinh toan nn coc cat : - len cht cua t c anh gia bng giam h s rng e e = eo etk vi eo h s rng t nhin cua t yu etk h s rng cua t sau khi a cai tao +tim eo :x theo m t nhin w,trong ln th tich hat h ,trong lng th tich nc n eo = w.h / n.100 +tim etk : phu thuc vao ngun gc t yu . -t co ngun gc cat : etk = emax ID ( emax emin ) ta ly ID >= 0,67 -t co ngun gc set bao hoa : etk = 0,01wtk. -t cat pha (ham lng set din tich nen coc : Fnc = 1,4b(1+0,4b). goi Fc la din tich coc cat trn phn nen coc thi = Fc/Fnc la ty l din tich coc cat vi phn c nen cht. Nu li coc cat ko lam mt t tri ln va lng cat cho vao gia c nn = lng giam l rng trong t thi ta co. = Fc/Fnc = e/1+eo -x s lng coc cat : n = Fc / fc = Fnc/fc vi fc la dtich 1 coc cat. -x k/c gia cac coc cat : thong thng cac coc cat c b tri trn inh cac tam giac u.g/s co 3 coc b tri trn inh tm giac u ABC canh c ( hinh ve )
c
c
db tri hnh tamgiac du
d
Vth= (c2.3)/4.(1+eo) = ( 2c23-d2 ) / 8.(1+etk ) = Vsh Ta tinh c : C = ((1+eo)..d2/(2.(3)(eo-etk))) = 0,952d ( (1+eo ) / ( eo-etk ) ) -chiu su coc cat ly bng chiu su lp t nen lun cua nn nhng ko nho hn 2b /v HCN va ( 34 )b /v mong hinh bng. +thi cng coc cat : -x vi tri,chun bi mt bng thi cng.Va thi cng bng 1 trong 2 p2 sau .dung coc g ong xung ri nh ln va cho cat vao ( p2 nay th s it sd ) . dung coc thep,dung may chuyn dung ong ng thep xung,cho cat vao ng ri rut ng thep ln -> sd ph bin hin nay. .tin hanh rai 1 lp m cat 10-50cm. .Tin hanh t t nn ng va thi cng cac hang muc khac.
29
2. p2 ging cat : giam t/g nn t at c kt ngi ta tin hanh khoan tao l n cao thit k ,ri p cat thoat nc va tin hanh gia tai nen trong nen nc. *tac dung : +thoat nc trong t -> tng c kt. +ging cat tham gia chiu lc cung nn t. *u nhc im : +p2 thi cng n gian,sdVL cat ia phng. +hiu qua /v t dinh bao hoa nc. thi cng gy chn ng n ctrinh xung quanh. -keo dai t/g gia tai va trong qua trinh thoat nc co th bi tc ng thm. *PVSD :/v t dinh bao hoa nc co chiu day tng i ln. *cu tao ging cat: gm 3 b phn chinh la m cat ( 2 ) ,ging cat ( 1 ) va t p gia tai ( 3 ) nh hinh ve.
3h d
2 1 d
3 -5 c 0 0m
L
-m cat :chiu day tng m cat h c ly theo no sau : h= 0,3-0,5m -ging cat: kinh 20-60cm. -chiu su bng chiu su chiu nen cua t di ay mong. *tinh toan ging cat : da trn cach tinh toan c kt cua nn t. + lun cua nn tai thi im t :St = Qt.S + c kt Qt cua nn co ging cat c tinh theo cng thc sau : Q = 1 ( (1-Qv).(1-Qh)) trong o : Q c kt Qv c kt theo phng thng ng .tra bang Qh c kt theo phng ngang .tra bang *thi cng ging cat : -khoan tao l n cao thit k -nhi cat vao l -thi cng tng m cat,p t gia tai.theo doi qua trinh bin dang lun cua nn i cho nn c kt at yu cu ta tin hanh thi cng cac cng trinh bn trn.
30
3. P2 bc thm : giam thi gian nn t at c kt ngi ta tin hanh khoan tao l n cao thit k,ri p cat thoat nc va tin hanh gia tai nen trong nen trc. *tac dung :thoat nc trong t ->tng c kt nn t. *u nhc im :+thi cng nhanh bng may chuyn dung -p2 cn thit bi thi cng chuyn dung. - vt liu nhp ngoai,gia thanh cao *cu tao va b tri bc thm. +cu tao bc thm : -mt ct ngang :hinh ve
b a
b a cua bchm t o t
-kinh thc :bc thm co b rng khoang 100-200mm,day 5-10mm. -vt liu :loi cua bc la 1 bng cht deo co nhiu ranh nho nc do mao dn va ap lc nc ua nc ln. Loi c boc bng 1 lp vai ia ky thut d thm nc va rt dai co tac dung ngn ko cho cac hat t thm qua. +b tri bc thm :-b tri mt ng :chiu su cm bc thm la ht phn ap lc chiu nesncuar nn t hoc bng chiu su lp t yu. -b tri mt bng :hinh hoa mai hoc hinh vung, vi k/c bc thm 0,6-2,2m. *tinh toan bc thm :tng t nh ging cat vi cng kinh tng ng la d = .2. (b+t)/ vi la h s quy i ly bng 0,75 *thi cng bc thm :-chun bi mt bng,inh vi cac vi tri bc thm cn cm. -cm bc thm n cao thit k bng may chuyn dung -rai lp vai ia ky thut,trn lp vai ia ky thut thi cng lp m cat thoat nc. -p t gia tai,theo doi tc c kt. -i cho nn c kt at yu cu ta tin hanh thi cng cac cng trinh bn trn.
31
