Chng 1

KT CU B TNG CT THEPPHN CU KIN C BAN

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M U 1

KT CU B TNG CT THEPPHN CU KIN C BANChng 1: M U. Chng 2: TNH NNG C LY CUA VT LIU.

Chng 1

Chng 3: NGUYN LY CU TAO & TNH TOAN BTCT. Chng 4: CU KIN CHU UN. Chng 5: SAN PHNG. Chng 6: CU KIN CHU NEN. Chng 7: CU KIN CHU KEO. Chng 8: CU KIN CHU UN XON. Chng 9: TNH TOAN CU KIN BTCT THEO TTGH TH II. Chng 10: B TNG CT THEP NG LC TRC.Tai liu: -Kt cu BTCT-phn cu kin c ban, Phan Quang Minh (chu bin). M U -Kt cu B tng v BTCT-Tiu chun thit k TCXDVN 356-2005. 2 P1.. P2.. P3.. P4.. P5..

1. BN CHT CA B TNG CT THP

M U.

Chng 1

1.1. BN CHT CA B TNG CT THP: B tng ct thep la vt liu xy dng phc hp do BT va ct thep cung cng tac chu lc: B tng la a nhn tao c ch tao t cac vt liu ri ( Cat, soi,...goi la ct liu) va cht kt dnh (Xi mng hoc cac cht deo).Chu nen tt, chu keo kem. Nen Ct thep: chu keo, nen u tt. Xt th nghim n gian sau: - Un mt dm b tng (khng ct thep):pha hoai kha sm do vt nt xut hin vung b tng chu keo. Nh vy kha nng chu lc cua BT vung nen cha c tn dung ht

Keo Nen

- Cung dm tng t nh vy nhng nu t mt lng ct thep thch hp vao vung b tng chu keo:Dm BTCT ch b pha hoai khi BT vung nen b ep v hoc ct thep chu keo b t. Mc khac thep chu keo va nen u tt nn co th t thep vao ca vung chu nen.

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Keo

Vy thc cht b tng ct thep la mt vt liu xy dng hn hp maChng 1 b trong o tng va ct thep a lin kt hp ly vi nhau cung lam vic trong mt kt cu.S d b tng va ct thep co th cung lam vic c la do:

- Lc dnh bam gia BT va ct thep: B tng khi ninh kt dnh cht vi ct thepnn ng lc co th truyn t BT sang ct thep va ngc lai, nh o co th khai thac ht kha nng chu lc cua ct thep, han ch b rng khe nt...

- Gia b tng va thep khng xay ra phan ng hoa hoc: B tng bao boc baov ct thep khng b han r va ngn nga tac dung co hai cua mi trng i vi thep.

- B tng va thep co h s gian n nhit gn bng nhau (ct= 1,2.10-5; b=10-51,5.10-5). Nn khi nhit thay i trong pham vi thng thng di 1000C thng sut ( ban u ) xay ra trong vt liu khng ang k.

1.2. PHN LOI BTCT: 1.2.1 Phn loi theo phng php ch to : a. B tng ct thp ton khi (BTCT ti ch):Tin hanh ghep van khun, t ct thep va BT ngay tai v tr thit k cua kt cu.

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2. PHN LOI BTCT

1.2. PHN LOI BTCT: a. B tng ct thp ton khi (BTCT ti ch):

Chng 1

1.2.1 Phn loi theo phng php ch to :Tin hanh ghep van khun, t ct thep va BT ngay tai v tr thit k cua kt cu.

* u im: - Cac cu kin lin kt toan khi nn kt cu co cng ln, chu tai trong ng tt. - Co th ch tao cac cu kin theo hnh dang tuy y. * Nhc im: - Tn vt liu lam van khun, a giao. - Thi cng chu anh hng thi tit.

b. B tng ct thp lp ghp (cu kin BTCT ch tao sn):Phn kt cu thanh cac cu kin ring bit co th ch tao sn ri em lp ghep lai thanh kt cu tai v tr thit k. PP nay khc phuc phn nao nhc im cua BT toan khi.

* u im: - Co iu kin Cng nghip hoa trong thi cng xy dng. - Tit kim vt liu lam van khun. - Rut ngn thi gian thi cng, am bao cht lng.. .. * Nhc im: - Cn co cac phng tin vn chuyn, cu lp. - X ly cac mi ni phc tap. - cng cua kt cu khng ln.

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M U 5

Chng 1

B TNG CT THP TON KHI P1.. P2.. P3.. P4.. P5..

B TNG CT THP LPU 6 M GHP

c. B tng ct thp na lp ghp :

Chng 1

Theo PP nay, ngi ta tin hanh lp ghep cac cu kin c ch tao sn cha hoan chnh, sau o t thm ct thep, ghep van khun va BT hoan chnh kt cu.

* u im: - cng cua kt cu ln. - Giam khi lng van khun, co th loai bo ct chng. * Nhc im: - Cn giai quyt tt lin kt ga BT cu va mi. - T chc thi cng phc tap.

1.2.2. Phn loi theo trng thi ng sut: a. B tng ct thp thng:Khi ch tao cu kin, ct thep trang thai khng co ng sut. Ngoai cac ni ng sut do co ngot va nhit , trong BT va ct thep ch xut hin ng sut khi co tai trong

b. B tng ct thp ng lc trc (BTCT d ng lc) :Khi ch tao cu kin, ct thep ban u c keo cng, lin kt cht vi BT, khi bung ra ct thep co lai gy nen trong BT.

1.2.3. Phn loi theo ct thp : - B tng co ct mm. (d40mm, thep hnh). 1.2.4. Phn loi theo trng lng th tch: - B tng nng co 1800 kg/ m3 (2500). - B tng nhe co < 1800 kg/ m3. P1.. P2.. P3.. P4.. P5..M U 7

3. U NHC IM CU BTCT

Chng 1 1,3. U NHC IM CU BTCT: 1.3.1 u im: S dung vt liu a phng (cat, soi, a..) tit kim thep. Re tin hn so vi thep

khi kt cu co nhp va va nho, cung chu tai nh nhau.

Chu lc tt hn kt cu g va gach a. Chu la tt hn g va thep. B tng bao v cho ct thep khng b nung nong sm. Tui tho cua cng trnh cao, chi ph bao dng t. BT co cng tng theothi gian, chu tac ng cua mi trng tt, ct thep c BT bao boc bao v khng b g.

Vic tao dang cho kt cu thc hin d dang. Va BT khi thi cng dang nhaoco th vao cac khun co hnh dang bt ky, ct thep u deo un theo hnh dang cua kt cu.

1.3.2 Nhc im : Trong lng ban thn ln nn gy kho khn cho vic xy dng kt cu co nhp ln bng BTCT thng. Khc phuc: Dung BT nhe, BTCT LT (2), kt cu vo mong (3),... B tng ct thep d co khe nt vung keo khi chu lc. Cn phai ngn ngahoc han ch khe nt kt cu trong mi trng xm thc, cac ng ng hay b cha cht long.. (Tnh toan han ch khe nt, s dung BTCT LT.. )

Cach m va cach nhit kem hn g va gach a. Co th s dung kt cu co lrng, kt cu nhiu lp, BT xp..

Thi cng phc tap, kho kim tra cht lng. Khc phuc: BTCT lp ghep.. Kho gia c va sa cha. Thit k cn phai phu hp yu cu s dung hin tai va d M U 8 P1.. P2.. P3.. P4.. P5.. kin phat trin m rng.

4. PHM VI NG DNG

1.4. PHM VI NG DNG CA B TNG CT THP:

Chng 1

Xy dng dn dung, cng nghip: Kt cu chu lc nha 1 tng va nhiu tng [1] [2], ng khoi, bun ke, xi l [3], mong may, hanh lang vn chuyn v.v.. Cng trnh cp thoat nc [4] [5]... (Hnh nh), (Hnh nh KC mi).. Xy dng cng trnh giao thng: Cu, ng, ta vet, u tau, cu tau, vo hm...(Hnh nh cng trnh ng), (Hnh nh cng trnh cu), (Hnh nh tunnel)..

Xy dng cng trnh thuy li: Tram bm, mang dn nc, p thuy in,...(Hnh nh cng trnh thy li), (Hnh nh cng trnh thy in, p)..

Xy dng cng trnh truyn thng, thng tin; Cac cng trnh c bit Xy dng cng trnh quc phong: Cng s kin c, doanh trai,...

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5. S LC LCH S PHT TRIN

1.5. S LC LCH S PHT TRIN: Qua trnh phat trin chia thanh 3 giai oan:

Chng 1

- Giai oan phat minh va mo mm trong thc tin, b tr ct thep theo cam tnh.1848: Lambot (Phap) ch tao chic tau bng li st ngoai trat vi thuy [1.5.1],... Va nm 1855 ng c trao bng sang ch cho cac ban ve dm BTCT va ct c gia c bng 4 thanh thep xung quanh. 1850: Monier (chu vn m Phap) co cac th nghim vi cac chu BT c gia c bng li thep. Va tip theo la cac bng sang ch vi cac ng va b cha c gia c, tm san, cu thang..

- Giai oan nghin cu l lun va s dung rng rai (sau 1880), nghin cu vcng cua BT va ct thep, lc dnh gia BT va ct thep, giai thch s lam vic chung gia chung. 1886: Koenen (Trng thanh tra XD cua Ph) xuat ban cun sach v phng phap tnh toan bn cua BTCT. T nm 1890 n 1920 cac ky s thc hanh a dn dn nm c kin thc v c hoc cua BTCT. Cac cun sach, bai bao, tiu chun a th hin cac ly thuyt tnh toan.

- Giai oan phat trin hin tai: XD cac phng phap tnh toan theo ng sut chophep da trn c s cua mn SBVL, tnh theo giai oan pha hoai co xet n tnh bin dang deo cua vt liu, tnh theo trang thai gii han. Nghin cu va ch tao thanh cng BTCT LT [5.2].

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Chng 1

Joseph-Louis Lambot (Sinh 22-051814, mt 02-08-1887), la ngi phat minh ra ximng li thep dn n s ra i cua BTCT ngay nay. ng a ch tao cac b cha dung va xi mng va ct thep. Nm 1848 ng a ch tao mt vo tau bng cach trn, va chic tau nay hin c trng bay tai Bao tang Brignoles.

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2.1. B tng

TNH NNG C LY CUA VT LIU.

Chng 2

2.1. B tng:- Tnh nng c hoc cua BT la ch cac c trng c hoc nh: cng va bin dang. - Tnh nng vt ly la tnh co ngot, t bin, kha nng chng thm, cach nhit, ... cua BT.

2.1.1. Thanh phn, cu truc va cac loai b tng: 2.1.1.1 Vt liu, thanh phn cua b tng: BT la loai a nhn tao c ch tao t cac vt liu ri (cat,a, soi) va cht kt dnh. Vt liu roi c goi la ct liu, gm cac c hat khac nhau, loai be la cat 15mm, loai ln la soi, a dm 5-40mm. Cht kt dnh thng la XM trn nc hoc cac cht deo khac. 2.1.1.2 Cu truc cua b tng: BT co cu truc khng ng nht v hnh dang, kch thc cac hat ct liu khac nhau, s phn b ct liu va cht kt dnh khng u, co cac l rng. 2.1.1.3 Cac loai b tng: Theo cu truc co: BT c, BT co l rng (dung t cat), BT t ong. Theo khi lng ring: BT nng thng co khi lng ring 22002500kG/m3; BTct liu be 18002200kG/m3; BT nhe 2500kG/m3; Theo thanh phn: BT thng thng, BT ct liu be, BT chen a hc. Theo pham vi s dung: BT lam kt cu chu lc, BT chu nong, BT cach nhit, BT TNH NNG C LY CUA VT LIU 1 chng xm thc.. P1.. P15.. P2.. P3..

2.1.2. Cng cua b tng

2.1.2. Cng cua b tng:

Chng 2

Cng la ch tiu c hoc quan trong, la mt c trng c ban cua BT, phan anh kha nng chu lc cua vt liu. Thng cn c vao cng phn bit cac loai b tng. 2.1.2.1 Cng chu nen: Chun b mu th: ah = 4a

P

Ban nenh =2D a D

a a

Mua

Mu khi lng tru.

Th nghim trn may nen, tng tai dn khi mu b pha hoai. Goi P la lc pha hoai mu. P ( 2 1) Cng nen cua mu: R = ; A A la din tch TD ngang cua mu. B tng thng co R=5 30MPa. BT co R>40MPa la loai cng cao. Ngi ta a ch tao c loai BT c bit co R80MPa. TNH NNG C LY CUA VT LIU 2 P1.. P15.. P2.. P3..

S pha hoai cua mu chu nen:

Chng 2

Khi b nen ngoai bin dang theo phng lc tac dung, mu con n ngang. Chnh s n ngang qua mc lam cho BT b pha v do ng sut keo (kha nng chu keo cua BT kem hn chu nen nhiu ln). Lc ma sat gia ban nen va mu th can tr s n ngang anh hng n cng BT khi nen. Bin dang ngang khng u Bi trn

Thp 1: Co ma sat trn mt tip xuc Kt qua cho thy trng hp 1 mu co cng ln hn. ln.

Thp 2: Khng co ma sat

Kch thc mu th: Mu kch thc nho co cng ln hn mu th co kch thc Hnh dang mu th: Mu lng tru co cng be hn mu khi vung co cung kchthc ay Rlt= (0.7-0.8)R...

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TNH NNG C LY CUA VT LIU 3

2.1.2.2 Cng chu keo: R(t). Mu chu keo trung tm: P R(t ) = ; ( 2 2a ) A Mu chu keo khi un: 3.5M R(t ) = ; (2 2b) 2 bh Mu tru tron chu nen che: 2.P R(t ) = ; ( 2 2c ) .l.D

P

a 4a

P

Chng 2a (10cm) a

Ph=b(=15cm) 4h b

P

Trong o: P, M: Lc va mmen un lam pha hoai mu. B tng thng co R(t)= 1040 kG/cm2.

2.1.2.3 Quan h gia cng chu keo R(t) va cng chu nen R: Cng thc dung quan h ng cong: R(t ) = t R ; (2 3a)Trong o t c ly phu thuc vao loai BT va n v cua R. Voi BT nng thng thng va n v cua R la MPa t = 0.280.30.

Cng thc dung quan h ng thng: R(t ) = 0.6 + 0.06 R; (2 3b)

Cng thc dung quan h ng cong theo h s Ct: R(t ) = Ct .R; (2 3c) R + 150 Ct = ; ( 2 4) Vi n v cua R la MPa, H s Ct: TNH NNG C LY CUA VT LIU 4 60 R + 1300 P1.. P15.. P2.. P3..

2.1.2.4 Cac nhn t anh hng n cng cua BT: - Cht lng va s lng xi mng: - cng, sach, cp phi cua ct liu: - T l N/X hp ly. - Cht lng cua vic trn va BT, m va bao dng BT. * Thi gian (tui cua BT):Cng cua b tng tng theo thi gian, luc u tng nhanh sau tng chm dn. Vi cng chu keo s tng cng theo thi gian nhanh hn so vi cng chu nen. R28 Rt R

Chng 2

* Thanh phn va cach ch tao BT: y la nhn t quyt nh cng BT.

t t 28

Cng b tng tng theo thi gian c xac nh theo cng thc thc nghim:

Rt=R1+(R10- R1).lgt. lgt 0.7 R28 lg t ; Cng thc cua Nga (1935), (Skramtaep): Rt = R28 lg28 (vi t = 7-300 ngay) Cng thc cua Vin nghin cu BT My ACI theo quy lut hyperbn: t Trong o h s a, b phu thuc loai XM. Thng thng Rt = R28 ; a + b.t a=4; b=0.85. Voi XM ng cng nhanh a=2.3; b=0.92. TNH NNG C LY CUA VT LIU 5 P1.. P15.. P2.. P3..

Cng thc cua Sec (1926):

* Tc gia tai va thi gian tac dung:

Chng 2

Khi tc gia tai chm cng at khoang 0.850.9 tr s thng thng va khi gia tai nhanh cng cua mu co th tng 1.151.2 ln. Khi th nghim phai tun theo quy trnh TN vi tc gia tai 0.2MPa/s.

2.1.3. Gia tr trung bnh va gia tr tiu chun cua cng : 2.1.3.1 Gia tr trung bnh:Th nghim n mu th cua cung mt loai BT thu c cac gia tr cng cua mu th la R1, R2.. Rn. Gia tr trung bnh cng cua cac mu th ky hiu la Rm, goi tt la cng trung bnh:

Rm = i=1 ; ( 2 6) n 2.1.3.2 lch qun phng, h s bin ng:

Ri

n

t i = |Ri - Rm| goi la lch. H s bin ng c tnh:

2 ; i Vi s lng mu u ln (n 15) tnh lch qun phng: == Rm ; (2 8)

n 1

( 2 7)

2.1.3.3 Gia tr c trng: Gia tr c trng cua cng cua BT (goi tt la cng c trng) c xac nh theo xac sut am bao la 95% va c tnh: Rch = Rm (1 S . ); ( 2 9)Trong o S la h s phu thuc xac sut bao am. Vi xac sut bao am 95% th S=1,64. TNH NNG C LY CUA VT LIU 6 P1.. P15.. P2.. P3..

2.1.3.4 Gia tr tiu chun:

Chng 2

Gia tr tiu chun cua cng cua BT (goi tt la cng tiu chun) c ly bng cng c trng cua mu th Rch nhn vi h s kt cu KC. Cng tiu chun v nen Rbn, v keo Rbtn: Rbn = KC .Rch ; (2 9a )H s KC xet n s lam vic cua BT trong kt cu co khac vi s lam vic cua mu th, c ly bng 0,7-0,8 tuy thuc vao Rch. .

Gia tr cua Rbn va Rbtn c cho TCXDVN 356:2005 (Bang 12 trang 35) 2.1.4. Cp bn va mac cua b tng:La tr s cua cac c trng c ban v cht lng cua BT. Tuy theo tnh cht va nhim vu cua kt cu ma quy nh mac hoc cp bn theo cac c trng khac nhau.

2.1.4.1 Mac theo cng chu nen: K hiu M B tng nng: M100, 150, 200, 250, 300, 350, 400, 500, 600. B tng nhe: M50, 75, 100, 150, 200, 250, 300. 2.1.4.2 Cp bn chu nen: K hiu B Cp bn chu nn ca b tng: k hiu bng ch B, l gi tr trung bnh thng k ca cng chu nn tc thi, tnh bng n v MPa, vi xc sut m bo khng di 95%, xc nh trn cc mu lp phng kch thc tiu chun (150 mm x 150 mm x 150 mm) c ch to, dng h trong iu kin tiu chun v th nghim nn tui 28 ngy. TNH NNG C LY CUA VT LIU 7 P1.. P15.. P2.. P3..TCXDVN 356:2005 (trang 4) quy nh phn bit cht lng BT theo cp bn chu nen:

Ch k ca Cp bn chu nn ca b tng: k hiu bng ch B, l gi tr trung bnh thng ng 2 cng chu nn tc thi, tnh bng n v MPa, vi xc sut m bo khng di 95%, xc nh trn cc mu lp phng kch thc tiu chun (150 mm x 150 mm x 150 mm) c ch to, dng h trong iu kin tiu chun v th nghim nn tui 28 ngy.

TCXDVN 356:2005 (Bang 9 trang 30) quy nh: BT nng co cp bn chu nen B3,5; B5; B7,5; B10; B12,5; B15; B20; B25; B30; B35; B40; B45; B50; B55; B60. BT nhe co cp bn chu nen B2,5; B3,5; B5; B7,5; B10; B12,5; B15; B20; B25; B30; B35; B40. Tng quan gia mac M va cp bn B cua cung mt loai BT th hin bng biu thc: B = . .M ; ( 2 10)Trong o la h s i n v t kG/cm2 sang MPa; co th ly bng 0,1. la h s chuyn i t cng trung bnh sang cng c trng (vi =0,135 th = (1-S.) = 0,778).

2.1.4.3 Mac theo cng chu keo: K hiu K. B tng nng: K10, 15, 20, 25, 30, 40. B tng nhe: K10, 15, 20, 25, 30. 2.1.4.4 Cp bn chu keo: K hiu Bt.Khi s chu lc cua kt cu c quyt nh chu yu bi kha nng chu keo cua BT, kt cu co yu cu chng nt. TCXDVN 356:2005 (trang 4) quy nh cp bn chu keo:

Cp bn chu ko ca b tng: k hiu bng ch Bt, l gi tr trung bnh thng k ca cng chu ko tc thi, tnh bng n v MPa, vi xc sut m bo khng di 95%, xc nh trn cc mu ko tiu chuNG Cc chCUo, VT LIh8 n LY t A dng U TNH N P1.. P15.. P2.. P3.. trong iu kin tiu chun v th nghim ko tui 28 ngy.

Ch th ng Cp bn chu ko ca b tng: k hiu bng ch Bt, l gi tr trung bnhng2 k ca cng chu ko tc thi, tnh bng n v MPa, vi xc sut m bo khng di 95%, xc nh trn cc mu ko tiu chun c ch to, dng h trong iu kin tiu chun v th nghim ko tui 28 ngy.

BT co cp bn chu keo doc truc Bt0,8; Bt1,2; Bt1,6; Bt2; Bt2,4; Bt2,8; Bt3,2. 2.1.4.5 Mac theo cac yu cu khac: Mac theo kha nng chng thm la con s ly bng ap sut ln nht (tnh bng atm) ma mu chu c nc khng thm qua. Cp chng thm cua BT: W2; W4; W6; W8; W10; W12.W cn quy nh cho cac kt cu co yu cu chng thm hoc c cht cua BT nh cac cng trnh thuy li, thuy in...

Mac theo khi lng ring trung bnh D (kha nng cach nhit): - BT nhe D800; D900; D1000; D1100; D1200; D1300; D1400; D1500; D1600; D1700; D1800; D1900; D2000. - BT t ong D500; D600; D700; D800; D900; D1000; D1100; D1200. - BT rng D800; D900; D1000; D1100; D1200; D1300; D1400.

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TNH NNG C LY CUA VT LIU 9

2.1.5. Bin dang cua b tng

2.1.5. Bin dang cua b tng: 2.1.5.1 Bin dang do co ngot: Co ngot la hin tng BT giam th tch khi ninh kt

Chng 2

Mc co ngot khi ng cng trong khng kh (3 - 5).10-4 . Khi ng cng trong nc BT n ra =1/5-1/2 mc co, mc tti a (6 - 15).10-5 . Cac nhn t anh hng n bin dang co ngot: - S lng va loai XM: lng XM co ngot , XM co hoat tnh cao co ngot . - T l N/X tng co ngot tng. - Cat nho hat, ct liu rng co ngot tng. - Cht phu gia lam BT ninh kt nhanh co ngot tng. Co ngot la mt hin tng co hai: - Lam thay i hnh dang va kch thc cu kin. - Gy mt mat ng sut trong ct thep ng lc trc. - Gy ra khe nt trn b mt BT, lam thay i cu truc cua BT, giam kha nng chu lc va tui tho cua cng trnh. Cac bin phap khc phuc: - Chon thanh phn ct liu, t l N/X hp ly. - m cht BT, bao dng BT thng xuyn m trong giai oan u. - Cac bin phap cu tao nh b tr khe co dan, t ct thep cu tao nhng ni cn thit chu ng sut do co ngot gy ra, mach ngng khi thi cng hp ly v.v.. TNH NNG C LY CUA VT LIU 10 P1.. P15.. P2.. P3..

2.1.5.2 Bin dang do tai trong tac dung ngn han:

Chng 2

Th nghim nen mu th hnh lng tru vi tc tng tai t t, o va lp c th gia ng sut va bin dang nh hnh ve. b 2 P el pl 1 D Rlt l C (1) b1 (2) Khi con be th t cong nhng khi th cong nhiu.

0O

*b pl el -im D ng vi luc mu b pha hoai: ng sut at Rlt va bin dang cc ai *b. Bin dang cua BT gm 2 phn: phn khi phuc c la bin dang an hi (1el), phn khng khi phuc la bin dang deo (2pl): b= el+ pl; (2-11) Do vy BT la vt liu an hi-deo. Mun an hi ban u Eb: b = Eb . el Eb = b = tg 0 ; ( 2 12) el ' ' (2 13) Mun bin dang deo cua BT Eb: b = Eb . b Eb = b = tg ; b ( NNG t = el goi laP1.. s an hi Eb = .Eb ; TNH2 14) C LY CUA VT LIU 11 h P15.. P2.. P3..' b

b

Ch Khi BT chu keo cung co bin dang an hi va bin dang deo: Eb= t.Eb. ng 2 Bin dang cc han khi keo kha be 0,00015. 2R Rt = t; Th nghim cho thy khi BT chu keo sp nt th t 0,5 nn *b = 0.5Eb Eb Eb 0,4 Eb Mun chng ct vi h s Posson = 0,2 vi moi loai BT: Gb = 2(1 + ) v1 b Tc gia tai khac nhau th cac v2 ng biu din quan h - khac v3nhau.

b

el pl 2 b

b

2.1.5.3 Bin dang do tai trong tac dung dai han - T bin: c P P b c b

ct

O

b

2

Hin tng bin dang deo tng theo thi gian goi la hin tng t bin cua BT. Phn bin dang deo tng ln do tai trong tac dung dai han goi la bin dang t bin. TNH NNG C LY CUA VT LIU 12 P1.. P15.. P2.. P3..

Chng Hin tng bin dang deo tng theo thi gian goi la hin tng t bin cua BT. 2 Phn bin dang deo tng ln do tai trong tac dung dai han goi la bin dang t bin. * Cac nhn t anh hng n bin dang t bin: - ng sut trong BT ln bin dang t bin ln. - Tui BT luc t tai ln bin dang t bin be. - m mi trng ln bin dang t bin be. - T l N/X ln, cng ct liu be bin dang t bin ln. - Lng X tng bin dang t bin tng.

Cac ch tiu , C u tng theo thi gian, va at n gii han n nh la 0, C0.

* Co th biu din t bin qua mt trong hai ch tiu sau: - c trng t bin: = C/ el. (la ai lng khng th nguyn) - Sut t bin: C= C/ b (MPa-1 hoc cm2/kG).

* Tac hai cua hin tng t bin: - Lam tng vong cua cu kin. - Lam tng un doc cua cu kin chu nen. - M rng khe nt trong BT. - Gy tn hao ng sut trong ct thep ng lc trc.

P1.. P15.. P2.. P3..

TNH NNG C LY CUA VT LIU 13

2.1.5.4 Bin dang do tai trong lp lai:

b bmax bmin b

Chng 2

b

b

Nu tai trong tac dung ln kt cu lp i lp lai nhiu ln (t vao ri d ra nhiu ln) th

bin dang deo se c tch luy dn, nu tai trong ln se gy hin tng moi cho kt cu . 2.1.5.5 Bin dang nhit: y la loai bin dang th tch khi nhit thay i, xac nh theo h s n v nhit cua BT t.

H s t phu thuc vao loai XM, ct liu, m co gia tr khoang (0,7-1,5)x10-5/. Thng thng khi nhit trong khoang t 0-1000 C ly t =1x10-5 tnh toan.

P1.. P15.. P2.. P3..

TNH NNG C LY CUA VT LIU 14

2.2. Ct thep

2.2. Ct thep:2.2.1. Yu cu i vi thep dung trong B tng Ct thep:

Chng 2

- am bao cng theo thit k. - Phai co tnh deo cn thit. - Phai dnh kt tt va cung chu lc c vi BT trong moi giai oan lam vic cua kt cu. - D gia cng: d un, ct, va han c ... - Tn dung ht kha nng chu lc cua ct thep khi kt cu b pha hoai. - Tit kim thep va tn t sc L.

2.2.2. Cac loai ct thep: Theo thanh phn hoa hoc cua thep: thng ch dung mt s mac thep cac bon thp va thep hp kim thp. Theo phng phap luyn thep: - Ct thep can nong: - Ct thep keo ngui: - Ct thep gia cng nhit: Theo hnh thc ct thep: Thep thanh tit din tron mt ngoai nhn (tron trn), hoc mt ngoai co g (cac g co tac dung tng dnh bam vi BT). Cung co th dung thanh thep hnh, o la ct cng co th chu lc c khi thi cng.

P1.. P15.. P2.. P3..

TNH NNG C LY CUA VT LIU 15

2.2.3. Cac tnh cht c ban cua ct thep:

Chng 2

2.2.3.1 Biu ng sut-bin dang: Biu ng sut-bin dang co phn thng ng vi giai oan an hi, phn cong ng vi giai oan co bin dang deo. ` Thep deo: B D Biu _ gm mt oan thng xin OA ng vi giai oan y lam vic an hi. C oan nm ngang c goi la thm chay, thep trang thai el A chay deo. Luc nay xac nh c gii han chay cua thep y. oan cong CD la giai oan cung c cua ct thep, ng sut *S O pl va bin dang tip tuc tng cho n khi ct thep b t, vi gii han bn B va bin dang cc han *S. B Thep rn (gion): y Biu _ gm mt oan thng xin OA ng vi giai C D A oan lam vic an hi, oan cong AD la giai oan ct el thep co bin dang deo. Khi keo t xac nh c gii han bn B va bin dang cc han *S. 2.2.3.2 Bin dang an hi va bin dang deo: 0,2% Nu keo thep trong giai oan an hi ri giam tai th toan O pl *S b bin dang c phuc hi.Khi keo thep vt qua gii han an hi (co bin dang deo) ri giam tai th biu khng TNH NNG C LY CUA VT LIU 16 P1.. P15.. dan d la v theo ng cu va co mt bin P2..g P3.. pl.

2.2.3.3 S cng ngui: o la hin tng tng gii han chay khi gia cng ngui ct thep.

Chng 2

Ly ct thep deo em keo ngui cho qua gii han chay ri giam tai se co c ct thep keo ngui. Ct keo ngui nay co gii han chay cao hn ct thep ban u. Sau vai ln keo hoc chut thm chay se bin mt, ct thep tr thanh rn vi cng tng cao va bin dang cc han giam.

2.2.3.4 Cng tiu chun cua ct thep: Gia tr tiu chun v cng cua ct thep c goi tt la cng tiu chun, k hiu Rsn c ly bng cng gii han chay (thc t hoc quy c) vi xac sut bao am khng di 95%. Vi my la gia tr trung bnh cua gii han chay khi th nghim mt s mu thep m th: Rsn = y (1 S . ); (2 15) 2.2.3.5 Mun an hi cua ct thep ES: Mun an hi cua ct thep, k hiu ES c ly bng dc cua oan OA trn biu ng sut bin dang.Gia tr cua ES phu thuc loai thep va khoang (17 - 21)x104 MPa, xem TCXDVN 356:2005 (Bang 28 trang 53)

2.2.3.6 deo cua ct thep: 2.2.3.7 Tnh han c:Tnh han c phu thuc vao thanh phn cua thep va cach ch tao. Cac thep can nong bng thep cha t cac bon va thep hp kim thp co tnh han c tt. Khng c phep TNH cng nhi LY P1.. a qua gia P3.. han i vi cac thepP15.. P2.. cng ngui hoc gia NNG Ct. CUA VT LIU 17

2.2.4. Phn loai (nhom) ct thep: 2.2.4.1 Phn loai ct thep theo TCVN:

Chng 2

Theo TCVN 1651:1985, c cc loi ct thp trn trn CI v ct thp c gn (ct thp vn) CII, CIII, CIV.

2.2.4.2 Phn loai ct thep theo mt s tiu chun khac: Theo TC Nga: - Cn nng: trn trn nhm A-I, c g nhm A-II v AC-II, A-III, A-IV, A-V, A-VI; - Gia cng bng nhit luyn v c nhit luyn: c g nhm AT-IIIC, AT-IV, ATIVC, AT-IVK, AT-VCK, AT-VI, AT-VIK v AT-VII. Ct thep cua Trung Quc chia thanh cac cp I, II, III, IV va cac loai si keo ngui. Ct thep cua Phap c ghi theo gii han chay nh: FeE230, FeE400, FeE500. 2.2.4.3 Tng quan gia mac thep va nhom (loai) thep:Mac thep c nh ra va k hiu chu yu da vao thanh phn hoa hoc va cach luyn, v du CT3, CT5, 182C, 25X2C.. Nhom ct thep c phn chia theo tnh nng c hoc. Hai cach phn chia nay la khac nhau nhng lin quan vi nhau v tnh nng cua thep la do thanh phn quyt nh. Ct thep nhom CI, A-I ch tao t thep cac bon mac CT3; Ct TNH NNG C LY CUA VT LIU 18 thep nhom CII, A-II P1..taP15.. P2.. cP3.. mac CT5; ch o t thep ca bon

2.3. B tng ct thep

2.3. B tng ct thep: 2.3.1. Lc dnh gia b tng & ct thep: 2.3.1.1 Cac nhn t tao nn lc dnh:

Chng 2

- Lc ma sat do co ngot khi ng cng BT ep cht vao ct thep. - S bam do b mt g gh cua ct thep. - Lc dan do keo xi mng co tac dung nh mt lp h dan BT vao ct thep 2.3.1.2 Th nghim xac nh lc dnh: Mu th nghim xac nh lc dnh: Cng trung bnh cua lc dnh: max P = ; (2 16) . .lmax ln C C ln

P

Trong o: P la lc keo (nen) tut ct thep. la ng knh ct thep. l la chiu dai oan ct thep chn vao BT. P 1 Lc dnh cc ai: max = = . ; (2 17) . . .l Trong o: la h s hoan chnh biu lc dnh. ( < 1)

P

Tr s lc dnh cc ai:

max =

.Rbnm

;

( 2 18)

Trong o: m la h s phu thuc b mt: thep trn m=56; thep co g m=33,5; la h s phu thuc trang thai chu lc: chu keo =1; chu nen =1,5; TNH NNG C LY CUA VT LIU 19 P1.. P15.. P2.. P3..

max =

.Rbnm

;

Tr s lc dnh cc ai:

max =

.Rbnm

;

( 2 18)

Chng 2

Trong o: m la h s phu thuc b mt: thep trn m=56; thep co g m=33,5; la h s phu thuc trang thai chu lc: chu keo =1; chu nen =1,5;

2.3.1.3 Cac yu t anh hng n lc dnh: Khi BT ct thep thng ng lc dnh ln hn ct thep nm ngang. Ct thep chu nen co lc dnh ln hn ct thep chu keo. Khi thay i chiu dai oan l tr s max khng i nhng tr s trung bnh thay i, khi tng l h s hoan chnh biu giam. Nu co cac bin phap han ch bin dang ngang cua BT lc dnh tng.

P1.. P15.. P2.. P3..

TNH NNG C LY CUA VT LIU 20

2.3.2. Anh hng cua ct thep n co ngot va t bin cua b tng:Chng 2 2.3.2.1 Anh hng n co ngot:Do s dnh kt gia BT va ct thep ma ct thep can tr bin dang co ngot cua BT: ct thep b nen lai con BT b keo ra, o la ng sut ban u do co ngot trong BTCT.

0

1 < 0 (=a) Ab AS

Nu khng co ct thep, BT t do co co ngot 0; Do ct thep can tr BT co co ngot 1< 0 bng bin dang cua ct thep a=1. (1) (2)

Bin dang cng bc trong BT: 0 - 1 Bin dang cng bc trong ct thep: 1. Hp lc trong BT: Nt = t.Ab;

t = (0 - 1).t.Eb. S = 1.ES.

trong ct thep: NS = S.AS.

V la lc ni tai nn chung cn bng nhau: Nt = NS; . .E E .E A t = t 0 b ; Suy ra: S = 0 S ; nSt = S ; = S ; 1 t Eb nSt . + 1 Ab 1+ nSt . ng sut keo do co ngot ln khi ham lng thep ln;ng sut keo do co ngot va ng sut keo do tai trong gy ra lam BT b nt sm hn so vi khi khng co anh hng cua co ngot, nhng khi a co khe nt th anh hng cua co ngot giam va n giai oan pha hoai th khng con anh hng n kha nng chu lc cua cu kin. Trong kt cu siu tnh lin kt tha ngn can co ngot cua BT nn xut hin ni lc TNH NNG C LY CUA VT LIU 21 P1.. P15.. P2.. P3.. phu.

2.3.2.2 Anh hng n t bin:Xet cu kin chu lc n gian nh hnh ve:

Chng 2 Ab AS N C1 NN

b = . .Eb ; = b ; Trong BT: .Eb Trong CT: S = .ES = b .ES = nS . b ; (2 19) .EbVi nS =

ES la h s tng ng ( 8-20). .Eb

T iu kin cn bng lc, co: N = b . Ab + S . AS = b .( Ab + nS . AS ); (2 20) t Ared = Ab + nS.AS goi la din tch cua tit din tng ng (qui i). N N n .N ng sut, bin dang cua cu kin: = ; b = ; S = S ; Ared Ared .Eb . AredKhi chu lc lu dai, BT co bin dang t bin. Ct thep cung anh hng n bin dang t bin cua BT.

Nu khng co ct thep, vi ng sut b bin dang t bin cua BT la C = C.b; Ct thep se can tr bin dang cua BT, va bin dang tng thm se la: 1 1: H s an toan cua kt cu. (k = 1.52.5)

Xac nh kha nng chu lc cua TD theo cac gia thit sau: - ng sut trong BT chu nen at Rb va phn b ch nht (u). - ng sut trong ct thep chu keo at gii han chay RS. u im: - a xet n s lam vic cua vt liu giai oan deo nn tit kim vt liu. - Cho khai nim ro rang hn v an toan cua kt cu . Nhc im: - H s an toan k= Sph / Sc gp chung lai nh vy la cha thoa ang v an toan cua kt cu phu thuc rt nhiu yu t nh tai trong, vt liu, iu kin lam vic v.v.. V vy khng th anh gia an toan bng mt h s duy nht c. - Cha xet n bin dang va nt cua kt cu la hai vn cung rt c quan tm. P1.. P2.. P3.. P3.5.. Picture NGUYN LY CU TAO & TNH TOAN 9

3.2. PHNG PHP TNH THEO TTGH

Ch 3.2. PHNG PHP TNH CU KIN THEO TRNG THI GII HN: ng 3

3.2.1 Cc trng thi gii hn (TTGH): - TTGH la trang thai ma t o tr i kt cu khng thoa man cac yu cu ra cho no (do chu lc qua sc,do mt n nh, do bin dang qua ln hoc do khent xut hin va m rng v.v..)

- Kt cu BTCT c tnh theo 2 nhom TTGH: TTGH th I (TTGH v cng ) va TTGH th II (TTGH v iu kin s dung) 3.2.2 Tnh theo TTGH v cng (TTGH I): TTGH th I c qui nh ng vi luc kt cu bt u b pha hoai, b mt n nh v hnh dang va v tr, b hong do moi do tac dung ng thi cua tai trong va mi trng. iu kin tnh toan: S Sgh. (3 - 6)S: La ni lc ln nht co th phat sinh tai TD do tai trong tnh toan gy ra. Sgh: La gii han be nht v kha nng chu lc cua TD xac nh theo cng tnh toan cua vt liu.

- Tnh theo TTGH th I la cn thit i vi moi kt cu cung nh cac b phn. - Tnh theo TTGH th I cho moi giai oan: ch tao, vn chuyn, cu lp, s dung,sa cha.. (mi giai oan vi s tnh phu hp).

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 10

3.2.3 TNH THEO TTGH 2

3.2.3 Tnh theo TTGH v iu kin s dng (TTGH II): Tnh theo TTGH th II v bin dang: f fgh. (3 7a)

Chng 3

f: bin dang hoc chuyn v do tai trong tiu chun fgh: bin dang hay chuyn v ti a ma qui pham cho phep:

Tnh theo TTGH th II v nt: - Nu kt cu c phep nt: acrc agh. acrc: b rng khe nt do tai trong tiu chun. (3 7b)

agh: b rng khe nt gii han ma qui pham cho phep; Gii han cho phep cua b rng khe nt va bin dang am bao iu kin lam vic bnh thng: agh ly theo Bang 1, Bang 2 trang12, fgh ly theo Bang 4 trang 14 TCXDVN 356:2005.

- Nu kt cu khng cho phep nt: Sc Scrc. (3 7c) Sc: ni lc do tai trong tiu chun. Scrc: Ni lc ti a ma TD co th chu c khi sp nt (kha nng chng nt).(Co th xem Sc la ng sut keo ln nht trong BT, Scrc la cng chu keo cua BT) TCXDVN 356:2005 iu 4.2.2 trang 10 qui inh: Tnh ton kt cu v tng th cng nh tnh ton tng cu kin ca n cn tin hnh i vi mi giai on: ch to, vn chuyn, thi cng, s dng v sa cha. S tnh ton ng vi mi giai on phi ph hp vi gii php cu to chn. Cho php khng cn tnh ton kim tra s m rng vt nt v bin dng nu qua thc nghim hoc thc t s dng cc kt cu tng t khng nh c: b rng vt nt mi giai on khng vt qu gi tr cho php v kt cu c cng giai on s dng. P1.. P2.. P3.. P3.5.. Picture NGUYN LY CU TAO & TNH TOAN 11

3.2.2 Cng tnh ton

3.2.2 Cng tnh ton ca vt liu :

Chng 3

3.2.2.1 Cng tnh ton ca BT: Cng tnh toan cua BT v nen Rb va v keo Rbt c xac nh nh sau: .R .R Rb = bi bn ; Rbt = bi btn ; (3 8)Trong o bc va bt la h s tin cy cua BT khi nen va khi keo. Khi tnh theo TTGH I bc=1,3-1,5 va bt=1,3-2,3 -TCXDVN 356:2005 Bang 11 trang 34. bi la h s iu kin lam vic cua BT (i=1, 2, .., 10) -TCXDVN 356:2005 Bang 15 trang 37.

bc

bt

3.2.2.2 Cng tnh toan cua ct thep: Cng tnh toan cua ct thep v keo RS c xac nh nh sau: .R RS = si sn ; (3 9)

Trong o S la h s tin cy cua ct thep. Khi tnh theo TTGH th nht s=1,05-1,2 stuy thuc loai thep. TCXDVN 356:2005 Bang 20 trang 46. si la h s iu kin lam vic cua ct thep (i=1, 2, .., 9). TCXDVN 356:2005 Bang 23 trang 49.

s

Rs =

Rsn

;

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 12

3.3. NGUYN L CU TO

3.3. NGUYN L CU TO:

Chng 3

3.3.1 Chon kch thc tit din cac cu kin: Khi thit k kt cu BTCT thng phai chon kch thc TD cac cu kin xac nh tai trong, tnh ni lc va ct thep. S hp ly cua TD chon theo yu cu chu lc c anh gia qua ham lng thep (mi loai cu kin co mt khoang hp ly cua ). Chon TD con phai quan tm n yu t thm my (yu cu tao dang cua kin truc) va iu kin ch tao (thng nht van khun, b tr thep va BT..). 3.3.2 Khung v li ct thp: Ct thep trong cu kin BTCT c lin kt thanh khung, li : - Gi v tr ct thep khi thi cng. - Cac ct thep cung chu lc, tranh cac pha hoai cuc b. - Chu cac ng sut phc tap ma tnh toan khng xet n.Lin kt cac ct thep bng cach buc hoc han.

Khung ct thep: gm ct doc, ct ngang, ct thi cng. Thng b tr cac cu kin dang thanh nh ct, dm. Li ct thep: Thng s dung cho cac cu kin dang ban nh ban san, ban mong.

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 13

Chu tai trong ng tt. (co deo cao) B tr ct thep linh ng. Khng cn thit b han. * Nhc im: Mc lin kt khng tt bng han. Thi cng chm, kho c gii hoa. 3.3.2.2 Khung li hn:Co th ch tao sn cac khung phng ri ghep thanh khung khng gian bng cac thanh ngang (ct thi cng)

Khung ct thep: gm ct doc, ct ngang, ct thi cng. Thng b tr cacChkin3 cu ng dang thanh nh ct, dm. Li ct thep: Thng s dung cho cac cu kin dang ban nh ban san, ban mong. Buc 3.3.2.1 Khung li buc: Buc Buc bng si thep 0,8 1. * u im:

Van khun

Thanh ni ngang (1-3 thanh/m).

Co th la li phng hoc cun nhng am bao mi cun G 500 kg phu hp cn cu thiu nhi khi thi cng.

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 14

3.3.3 Ct chu lc v ct cu to

3.3.3 Ct chu lc v ct cu to:

Chng 3

- Ct chu lc: Dung chu cac ng lc phat sinh do tai trong, c xac nh theo tnh toan. - Ct cu tao: Lin kt cac ct chu lc thanh khung hoc li, giam s co ngot khng u cua BT, chu ng sut do co ngot va thay i nhit , giam b rng khe nt, han ch bin dang (vong), phn b tac dung cua tai trong tp trung.. 3.3.4 L p BT bo v: Lp BT bao v ct doc chu lc TCXDVN 356:2005 qui inh: iu 8.3.2 trang 122: i vi ct thp dc chu lc (khng ng lc trc, ng lc trc, ng lc trc ko trn b), chiu dy lp b tng bo v cn c ly khng nh hn ng knh ct thp hoc dy cp v khng nh hn: - Trong bn v tng c chiu dy: c2 + t 100 mm tr xung: 10 mm (15 mm) + trn 100 mm: 15 mm (20 mm) - Trong dm v dm sn c chiu cao: + nh hn 250 mm: 15 mm (20 mm) 20 mm (25 mm) c1 c2 + ln hn hoc bng 250 mm: - Trong ct: 20 mm (25 mm) c2 30 mm - Trong dm mng: c1 - Trong mng: + lp ghp: 30 mm + ton khi khi c lp b tng lt: 35 mm P1.. P2.. P3.. P3.5.. b tng lt: 70 mm + ton khi khi khng c lp Picture NGUYN LY CU TAO & TNH TOAN 15

3.3.4 L p BT bo v

Chng 3 3.3.4 L p BT bo v: Lp BT bao v ct cu tao TCXDVN 356:2005 qui inh: iu 8.3.3 trang 122: Chiu dy lp b tng bo v cho ct thp ai, ct thp phn b v ct thp cu to cn c ly khng nh hn ng knh ca cc ct thp ny v khng nh hn: - khi chiu cao tit din cu kin nh hn 250 mm: 10 mm (15 mm) - khi chiu cao tit din cu kin bng 250 mm tr ln: 15 mm (20 mm)

ch thch: 1. Gi tr trong ngoc (...) p dng cho kt cu ngoi tri hoc nhng ni m t. 2. i vi kt cu trong vng chu nh hng ca mi trng bin, chiu dy lp b tng bo v ly theo quy nh ca tiu chun hin hnh TCXDVN 327 : 2004. Trong kt cu mt lp lm t b tng nh v b tng rng cp B7,5 v thp hn, chiu dy lp b tng bo v ct thp dc chu lc cn phi khng nh hn 20 mm, cn i vi cc panen tng ngoi (khng c lp trt) khng c nh hn 25 mm. i vi cc kt cu mt lp lm t b tng t ong, trong mi trng hp lp b tng bo v khng nh hn 25 mm. Trong cc cu kin lm t b tng nh, b tng rng c cp khng ln hn B7,5 v lm t b tng t ong, chiu dy lp b tng bo v cho ct thp ngang ly khng nh hn 15 mm, khng ph thuc chiu cao tit din.

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 16

3.3.5 B tr ct thp

3.3.5 B tr ct thp v khong cch gia cc ct thp: - Nu ct thep nm ngang hoc nghing khi BT:

Chng 3

TCXDVN 356:2005 qui inh: iu 8.4.1 trang 123: Khong cch thng thy gia cc thanh ct thp (hoc v ng t ct thp cng) theo chiu cao v chiu rng tit din cn m bo s lm vic ng thi gia ct thp vi b tng v c la chn c k n s thun tin khi v m va b tng. i vi kt cu ng lc trc cng cn tnh n mc nn cc b ca b tng, kch thc ca cc thit b ko (kch, kp). Trong cc cu kin s dng m bn hoc m di khi ch to cn m bo khong cch gia cc thanh ct thp cho php m i qua lm cht va b tng. 30 t0 d (t0 la khoang h gia cac mep cua cac ct thep-khoang cach thng thuy) t0 t0 50 t0 t0 25 d t0 1.5d 30 d

- Nu ct thep t ng khi BT: t0 50.

t0

25 d

Ngoai ra khoang cach gia cac ct thep cung khng nn qua ln nhm tranh cac vt nt do co ngot, thay i nhit , tranh s pha hoai cuc b va n nh cua khung (li) ct thep khi thi cng.. Trong moi trng hp t0 400. P1.. P2.. P3.. P3.5.. Picture NGUYN LY CU TAO & TNH TOAN 17

3.3.6 Neo ct thp

3.3.6 Neo ct thp:

Chng 3

Neo ct thep nhm bao am phat huy ht kha nng chu lc cua ct thep, tranh pha hoai 3d cuc b do tut. 2,5d

Neo co un moc u: Ct thep tron trn trong khung buc phai co moc neo hai u.TCXDVN 356:2005 qui inh: iu 8.5.1 trang 124: i vi nhng thanh ct thp c g, cng nh cc thanh ct thp trn trn dng trong cc khung thp hn v li hn th u mt thng, khng cn un mc. Nhng thanh ct thp trn trn chu ko dng trong khung, li buc cn c un mc u, mc dng ch L hoc ch U.

Un tay 6,25d 2,5d Un may 3,25d

45-900

Chiu dai oan neo: c xac nh theo kha nng truyn lc gia BT va ct thep (lc dnh): R lan = an s + an d ; (3 10) Rb nhng khng nh hn lan = an .d ; Trong an, an, an cng nh gi tr cho php ti thiu lan c xc nh theo TCXDVN 356:2005 Bng 36 (tr. 125).

Trng hp khi thanh cn neo c din tch tit din ln hn din tch yu cu theo tnh ton bn vi ton b cng tnh ton, chiu di lan theo cng thc (3-10) cho php gim xung bng cch nhn vi t s din tch cn thit theo tnh ton v din tch P1.. din P3.. P3.5.. Picture NGUYN LY CU TAO & TNH TOAN 18 thc t ca titP2.. ct thp.

Chng 3 3.3.6 Neo ct thp: Chiu dai oan neo: c xac nh theo kha nng truyn lc gia BT va ct thep (lc dnh): R lan = an s + an d ; (3 10) Rb

nhng khng nh hn lan = an .d ; Trong an, an, an cng nh gi tr cho php ti thiu lan c xc nh theo TCXDVN 356:2005 Bng 36 (tr. 125). Neo bng cach han cac thep neo u:2d 4d d d 4d 10mm 2.5d

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 19

3.3.7 Un ct thp

3.3.7 Un ct thp:

F F

F

Chng 3

Tai ch un cong, ct thep khi chu lc se ep cuc b vao BT. phn b lc nen ep u vao BT, ct thep c un cong vi ban knh cong r 10d. 3.3.8 Ni ct thp: 3.3.8.1 Ni chng (ni buc):

F r 10d d

Khi d 36. Khng nn ni buc ct thep trong vung keo tai TD tn dung ht kha nng chu lc. oan lan xac nh theo cng thc tnh oan neo va: Keo: lan 250 mm. Tuy thuc mac BT lan va loai ct thep Nen: lan 200 mm.Khng c ni buc khi d > 36 hoc khi cu kin chu keo 3.3.8.2 Ni hn : hoan toan (Thanh chu keo cua dan, thanh treo cua khung..). 0.85 d2 Han i u: d2 d1 10 mm

Han i u co nep: Han ghep (chng): Han trong mang:

d10 lh lh8d

d8 lh lh10d lh5d lh4d NGUYN LY CU TAO & TNH TOAN 20

P1.. P2.. P3.. P3.5..

Picture

Chng 3

Khung-Li thp

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 21

Chng 3

P1.. P2.. P3.. P3.5..

Picture NGUYN LY CU TAO & TNH TOAN 22

1. C IM CU TO

CU KIN CHU UN.P M&Q M&Q

Chng 4

Cu kin chu un la cu kin chu M hay ng thi M & Q.

P Q=0

Cu kin chu un la loai cu kin c ban rt quan trong c s dung rng rai va trong nhiu b phn cua cng trnh nh dm, san, cu thang, ... Co th quy v hai loai c ban: ban va dm.

4.1. C IM CU TO: 4.1.1 Bn: Ban la loai kt cu phng co chiu day kha be so vi chiu dai va chiu rng.

h

Trong kt cu nha ca ban co kch thc mt bng thng bng 26m. Chiu day ban chon theo yu cu chu lc va cng (bin dang, vong, goc xoay.. San nha thng khoang t 620 cm, trong cac kt cu khac cac kch thc o co th ln hn hoc be hn ). B tng cua ban thng chon co cp bn t B12.5 n B25.

Ct thep trong ban gm co ct chu lc va ct phn b (CI, CII i khi CIII ) a. Ct thp chu lc:Ct phn b

Chu m men un. B tr trong vung keo. ng Chon va P1..tr theo tnhP4.. nP412.. P42.. P5.. P54.. P56..UCt chu lU UN 1 b P2.. P3.. toa . C KIN CH c

Tai gi ct mu chu M(-) th a 100 tin BT; Tai ni co M ln th: a 200 khi chiu day ban h 150, a 1.5h khi chiu day ban h > 150, Tai ni co M be th ti thiu phai co 3 thanh/1m dai ban.

Ch ng knh d=612 mm, khoang cach gia cac ct thep a=720 cm. ng 4Ct phn b

b. Ct thp phn b (cu to):

Ct chu lc

Tac dung: gi v tr ct chu lc khi thi cng, chu ng lc do co ngot, thay i nhit , phn phi anh hng cua lc tp trung ra cac ct chu lc ln cn. ng knh d=48; Cach khoang a=200300 (a 350). Lng ct phn b khng t hn 10% s lng ct chu lc tai TD co m men un ln nht. 4.1.2 Dm:c c15 khi d10 c1,5d khi d>10

lan10d

Dm la cu kin chu un co kch thc TD (b rng va chiu cao) kha be so vi chiu dai (nhp). Hnh dang tit din:Tit din dm thng co dang ch nht, I, T, hp, khuyn, ... l h b

C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56..U KIN CHU UN 2

Chng 4

C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56..U KIN CHU UN 3

Hnh dang tit din:

b

Chng 4h

Kch thc tit din:1 1 - Bi s cua 50 khi h 600. Chiu cao h = nhp; - Bi s cua 100 khi h > 600. 8 20 1 1 Chiu rng b = h; - 100, 120, 150, 180, 200, ... - Bi s cua 50 khi b >250. 2 4Phai thun tin vi qui cach van khun va tiu chun hoa kch thc dm.

Ct thep: a. Ct dc chu lc: Ct xin Ct ai Ct doc ct

Ct doc ai 2 nhanh chu lc Chu M, xac nh theo tnh toan. ng knh d = 10-32 B tr trong vung keo, i khi co ca trong vung nen;

ai 4 nhanh

Co th b tr 1, 2 hay nhiu lp (khi b 150 phai co t nht 2 thanh)

C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56..U KIN CHU UN 4

Ct xin

Ct ai

Ct doc ct

Chng 4

Ct doc ai 4 nhanh ai 2 nhanh chu lc b. Ct dc cu to: ng knh 10 14. Ct gia: la ct doc t trong vung BT chu nen (tai cac goc cua ct ai). Ct thep phu: Khi dm co chiu cao ln h > 700. Lng thep 0.1% din tch BT sn. Co tac dung gi n nh ct ai, chu ng lc co ngot va nhit . c. Ct ai: chu lc ct, lin kt ct doc thanh khung, gn vung BT chu keo va vung BT chu nen vi nhau chu m men. Tnh toan theo lc ct. ng knh ct ai: d 6mm; khi h 800 d 8mm. d. Ct xin: chu lc ct (thng kt hp a ct doc ln chu M(-) mep trn). Goc nghing = 450 khi h 800; = 600 khi h > 800. C P1.. P2.. P3.. P4.. P412.. P42..ban. P54.. P56..U KIN CHU UN 5 = 300 khi dm thp va P5..

2. S LM VIC CA DM

4.2. S LM VIC CA DM:

Chng 4

Quan sat mt dm BTCT chu tai cho n luc b pha hoai:

Tai khu vc gia dm ni M ln co vt nt thng goc vi truc dm; Tai khu vc gn gi ta co Q ln th vt nt nghing.

C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56..U KIN CHU UN 6

3. TRNG THI US-BD

4.2. S LM VIC CA DM:

Chng 4

Quan sat mt dm BTCT chu tai cho n luc b pha hoai: Tai khu vc gia dm ni M ln co vt nt thng goc vi truc dm; Tai khu vc gn gi ta co Q ln th vt nt nghing. Khe nt nghing Khe nt thng goc

Nh vy vic tnh toan va cu tao cac cu kin chu un theo iu kin cng nhm:

- Khng b pha hoai trn TD thng goc: Tnh toan theo cng trn TD thng goc. - Khng b pha hoai trn TD nghing: Tnh toan theo cng trn TD nghing. 4.3. TRNG THI NG SUTBIN DNG TRN TIT DIN THNG GC:Din bin cua trang thai US - BD trn TD thng goc co th phn thanh 3 giai oan sau:

4.3.1 Giai on I: Khi tai trong be vt liu lam vic an hi.

I

b R th ly = R tc m = R tnh Mgh. C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 23

5. TNH TON THEO CNG TRN TIT DIN NGHING

4.5. TNH TON THEO CNG TRN TIT DIN NGHING: 4.5.1 c im ph hoi trn tit din nghing (TDng):

Chng 4

Trn TDng chu tac dung ng thi m men va lc ct. S pha hoai theo TD nghing thng theo 2 kiu:

Kiu 1: Hai phn dm nay quay xung quanh vung nen,vung nen thu hep lai cui cung b pha huy. Luc o ct thep chu keo at gii han chay hay b keo tut v neo khng u.

Kiu 2: Khi ct thep chu keo kha nhiu va neo cht th squay cua 2 phn dm b can tr. Dm b pha hoai khi min M BT chu nen b pha v do tac dung chung cua lc ct va lc ep. Hai phn dm co xu hng trt ln nhau va tut xung so vi gi ta.

Q M

Q

4.5.2 Tnh toan theo kha nng chu ng sut nen chnh (K han ch): Kt qua thc nghim chng to, BT khng b pha v v ng sut nen chnh cu kin cn phai thoa man iu kin: Q 0,3. w1. b1.Rb .b.h0 ; (4 39)Trong o w1 - H s k n anh hng cua ct ai t vung goc vi truc cu kin, c w1= 1+5..w 1,3; (4 - 40) xac nh theo cng thc: Vi

=

ES A ; w = sw ; Eb b.s

Asw- din tch tit din ngang cua cac nhanh ct ai t trong mt mt phng vung goc viP3.. cu kin va ct P42.. P5.. P54.. P56.. U KIN CHU UN 24 C P1.. P2.. truc P4.. P412.. qua TD nghing

4.5.2 Tnh toan theo kha nng chu ng sut nen chnh (K han ch)Chng 4 : Kt qua thc nghim chng to, BT khng b pha v v ng sut nen chnh cu kin cn phai thoa man iu kin: Q 0,3. w1. b1.Rb .b.h0 ; (4 39)Trong o w1 - H s k n anh hng cua ct ai t vung goc vi truc cu kin, c w1= 1+5..w 1,3; (4 - 40) xac nh theo cng thc: Vi

=

ES A ; w = sw ; b.s Eb

Asw- din tch tit din ngang cua cac nhanh ct ai t trong mt mt phng vung goc vi truc cu kin va ct qua TD nghing. b - Chiu rng cua TD ch nht; Chiu rng sn cua TD ch T va ch I. s - Khoang cach gia cac ct ai theo chiu doc cu kin. b1- H s xet n kha nng phn phi lai ni lc cua cac loai BT khac nhau:

b1 = 1 Rb ; (4 41)

- H s ly nh sau: + i vi b tng nng, b tng ht nh, b tng t ong: + i vi b tng nh: 0,02; Rb tnh bng MPa.

0,01;

iu kin (4 - 39) nu khng thoa man phai tng kch thc tit din hoc tng cp bn BT.C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 25

Chng 4.5.3 Tnh toan theo kha nng chu ct cua tit din BT (cu kin khng4 b tr ct ngang chu ct): Kt qua thc nghim chng to rng khi dm chu ct thun tuy, se khng xut hin khe nt nghing nu ng sut tip thoa iu kin: Q 2,5.Rbt ; = kc = b.h0 Tc kha nng chu ct ln nht cua tit din BT: Qb max = 2,5.Rbt .b.h0 ; (4 42)

Vy iu kin cng theo lc ct ln nht cua cu kin khng co ct thep ngang la: Qmax 2,5.Rbt .b.h0 ; (4 42a ) TCXDVN 356:2005 iu 6.2.3.4 trang 81 qui inh: i vi cu kin BTCT khng c ct thp ai chu lc ct, m bo bn trn vt nt xin cn tnh ton i vi vt nt xin nguy him nht theo iu kin: 2 b 4 .(1 + n ).Rbt .b.h0 Q ; ( 4 43) c V phi ca cng thc (4 - 43) ly theo khng ch: 2 b 4 .(1 + n ).Rbt .b.h0 2,5.Rbt .b.h0 b 3 .(1 + n ).Rbt .b.h0 ; ( 4 44 ) cTrong c chiu di hnh chiu ca TDng trn trc dc cu kin tnh t mp gi ta. c cmax = 2.h0; Q - Lc ct c xc nh t ngoi lc t mt pha ca TDng ang xt. H s b3, b4 ph thuc loi BT; C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 26

Chng 4.5.3 Tnh toan theo kha nng chu ct cua tit din BT (cu kin khng4 b tr ct ngang chu ct): TCXDVN 356:2005 iu 6.2.3.4 trang 81 qui inh:

i vi cu kin BTCT khng c ct thp ai chu lc ct, m bo bn trn vt nt xin cn tnh ton i vi vt nt xin nguy him nht theo iu kin: 2 b 4 .(1 + n ).Rbt .b.h0 Q ; ( 4 43) c V phi ca cng thc (4 - 43) ly theo khng ch: 2 b 4 .(1 + n ).Rbt .b.h0 2,5.Rbt .b.h0 b 3 .(1 + n ).Rbt .b.h0 ; ( 4 44 ) cTrong c chiu di hnh chiu ca TDng trn trc dc cu kin tnh t mp gi ta. c cmax = 2.h0; Q - Lc ct c xc nh t ngoi lc t mt pha ca TDng ang xt. H s b3, b4 ph thuc loi BT; H s n xt n nh hng lc dc; H s n xt n nh hng lc dc, c xc nh nh sau: Khi chu lc nn dc, xc nh theo cng thc: = 0,1n

N 0,5; (4 45) Rbt .b.h0 i vi cu kin ng lc trc, trong cng thc (4-45) thay N bng lc nn trc P; khng xt nu lc nn dc trc gy ra m men un cng du vi tc dng ca ti trng. N Khi chu lc ko dc trc, xc nh theo cng thc: = 0 ,2 ; ( 4 46 ) n R bt bh 0 C P1.. P2.. P3.. P4.. phi ca (4-46) P5.. P54.. P56.. nhng gi tr tuyt i ca v P412.. P42.. khng ln hn 0,8.U KIN CHU UN 27

5.4 iu kin cng TDng c ct ngang

4.5.4 iu kin cng trn tit din nghing ca dm c ct ngang: 4.5.4.1 S ng lc trn tit din nghing: Gia thit: Ni lc trong cac ct ngang la lc keo doc theo truc cua no. Qbs s s s s s Rsw Asw Rsw Asw Rsw As, inc Rsw Asw

Chng 4

c0

cGiai thch cac ai lng trong s ng lc: s: Khoang cach gia cac ct ai. RSW: Cng tnh toan cua ct thep ngang. ASW: Din tch tit din ngang cua cac nhanh ct ai t trong mt mt phng vung goc vi truc cu kin (goi la 1 lp). AS,inc(i): Din tch tit din ngang cua cac lp ct xin (i=1, 2,..). Zs: Khoang cach t trong tm ct thep doc chu keo n trong tm vung nen. Zsw(i): Khoang cach t cac lp ct ai n trong tm vung nen. Zs,inc(i): Khoang cach t cac lp ct xin n trong tm vung nen. CU KIN CHU UN 28

P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56..

Giai thch cac ai lng trong s : s: Khoang cach gia cac ct ai. RSW: Cng tnh toan cua ct ngang. ASW: Din tch tit din ngang cua cac nhanh ct ai t trong mt mt phng. AS,inc(i): Din tch tit din ngang cua cac lp ct xin (i=1, 2,..). Zs: Khoang cach t trong tm ct thep doc chu keo n trong tm vung nen. Zsw(i): Khoang cach t cac lp ct ai n trong tm vung nen. Zs,inc(i): Khoang cach t cac lp ct xin n trong tm vung nen.

4.5.4 iu kin cng trn tit din nghing ca dm c ct ngang: 4.5.4.1 S ng lc trn tit din nghing: Zs,inc2

Chng 4

Zs,inc1

Qb Nb+RSCAS RswAs,inc1 ZS RswAsw M x

s s

Q

RSAS+ RSPASP Z sw2 c0 c Zsw1

4.5.4.2 iu kin cng trn TDng theo lc ct: i vi cu kin chu un co t ct thep ngang, iu kin cng trn TDng Q Qb + Qsw + Qs ,inc ; ( 4 47 ) theo lc ct nh sau:Trong o: QSW: Kha nng chu lc ct cua ct ai. Qs,inc: Kha nng chu lc ct cua ct xin. Qb: Kha nng chu lc ct cua BT c xac nh theo cng thc thc nghim: C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 29

4.5.4.2 iu kin cng trn TDng theo lc ct:

Chng 4

i vi cu kin chu un co t ct thep ngang, iu kin cng trn TDng Q Qb + Qsw + Qs ,inc ; ( 4 47 ) theo lc ct nh sau:Trong o: QSW: Kha nng chu lc ct cua ct ai. Qs,inc: Kha nng chu lc ct cua ct xin. Qb: Kha nng chu lc ct cua BT c xac nh theo cng thc thc 2 nghim: b 2 . 1 + f + n Rbt .b.h0 Qb = ; ( 4 48 ) c

(

)

H s b2 xt n nh hng ca loi b tng. H s f xt n nh hng ca cnh chu nn trong tit din ch T, ch I. Trong mi trng hp phi khng ch gi tr: 1 + f + n 1,5;

(

)

Gi tr Qb tnh theo (4-48) ly khng nh hn: Qb Qb min = b3. 1 + f + n Rbt .b.h0 ; (4 50)

(

)

Kha nng chu ct cua BT phai am bao Qb Qbmin. T (4-48) va (4-50) suy ra:c

ng thi phai am bao Qb Qbmax. T (4-48) va (4-42) suy ra: c b 2 (1 + f + n ).h0; ( 4 50b)2 ,5

b 2 h ; ( 4 50a) b 3 0

C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 30

4.5.4.3 iu kin cng trn TDng theo m men:

Chng 4

am bao cng trn TDng theo m men, cn tnh toan vi TDng nguy him nht theo iu kin: M M s + M sw + M s ,inc ; ( 4 51)Trong o: M -M men cua tt ca ngoai lc t mt pha cua TDng i vi truc i qua hp lc cua vung nen va thng goc vi mt phng un. Ms, Msw, Ms,inc -tng m men i vi truc noi trn cua cac ni lc tng ng trong ct thep doc, ct ai va ct xin ct qua TDng.

4.5.5. Tnh toan ct ai khi khng t ct xin: 4.5.5.1 iu kin cng trn TD nghing khi khng t ct xin: Khi khng dung ct xin, iu kin (4-47) tr thanh: Qu Q Qb + Qsw ; ( 4 52)Trong o lc ct do ct ai chu co th vit lai nh sau: Qumin Qsw = Rsw. Asw = qsw.c; (4 53) R .A Vi: qsw = sw sw ; ( 4 54 ) s Cung vi (4-48), iu kin cng trn TDng (4-52) tr thanh:

c0

c

Q Qu =

2 b 2 .(1 + f + n ) Rbt .b.h0

4.5.5.2 Tit din nghing nguy him nht: Tr s c0 tng ng vi Qu nho nht (Qumin). tm c0: M dQu 2 = qsw 2b = 0; (4 56) Trong o: M b = b 2.(1 + f + n ) Rbt .b.h0 ; ( 4 57 ) dc c0 C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 31

c

+ qsw .c; ( 4 55 )

Chng 4 4.5.5.2 Tit din nghing nguy him nht: Tr s c0 tng ng vi Qu nho nht (Qumin). tm c0: M dQu 2 = qsw 2b = 0; (4 56) Trong o: M b = b 2.(1 + f + n ) Rbt .b.h0 ; ( 4 57 ) dc c0 Mb ; ( 4 58 ) Qu min = 2. M b .qsw ; Tm c: c0 = qsw

i vi cu kin ch t ct thp ai thng gc vi trc dc cu kin, c bc khng i trong khong tit din nghing ang xt, gi tr c0 ng vi cc tiu ca 2 biu thc (Qb + Qsw) xc nh theo cng thc: b 2 .(1 + n + f ).Rbt .b.h0c0 =

; q sw R .A trong : qsw ni lc trong ct thp ai trn mt n v chiu di cu kin: qsw = sw sw ;

i vi cc cu kin nh vy, gi tr Qsw Qsw = qsw .co ; ( 4 59) Khi , ct thp ai xc nh theo tnh ton phi tho mn iu kin: b3.(1 + n + f ).Rbt .b Qb min = ; (4 60) qsw 2 2.h0

s c xc nh theo cng thc:

C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 32

i vi cu kin ch t ct thp ai thng gc vi trc dc cu kin, Chng 4 c bc khng i trong khong tit din nghing ang xt, gi tr c0 ng vi cc tiu ca 2 biu thc (Qb + Qsw) xc nh theo cng thc: b 2 .(1 + n + f ).Rbt .b.h0; q sw R .A trong : qsw ni lc trong ct thp ai trn mt n v chiu di cu kin: qsw = sw sw ; c0 =

Khi tnh toan ct ai cn phai tnh vi nhiu TDng khac nhau vi gia tr c khng vt qua khoang cach t gi ta n TD co gia tr m men cc ai va khng vt qua gia tr tnh theo (4-50a). Qb q1 4.5.5.3 Tnh cu kin chu tai trong phn b u:

i vi cc cu kin nh vy, gi tr Qsw c xc nh theo cng thc: Qsw = qsw .co ; ( 4 59) Khi , ct thp ai xc nh theo tnh ton phi tho mn iu kin: b3.(1 + n + f ).Rbt .b Qb min = ; (4 60) qsw 2 2.h0

s

V phai cua (4-55) co thm q1:Qu =2 b 2 .(1 + f + n ) Rbt .b.h0

Nb+RSCAS x s s qsw ZS M

Qu =

Mb + (qsw + q1 ).c; ( 4 61 ) c

c

+ qsw .c + q1.c;

RswAsw Va TDng nguy him nht c0: qsw Mb c0 = ; ( 4 62 ) RSAS+ RSPASP Q qsw + q1 c C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 33

4.5.5.3 Tnh cu kin chu tai trong phn b u: V phai cua (4-55) co thm q1:Qu =2 b 2 .(1 + f + n ) Rbt .b.h0

q1

Qb Chng 4 Nb+RSCAS x

Qu =

Mb + (qsw + q1 ).c; ( 4 61 ) c

c

+ qsw .c + q1.c;s s qsw Q qsw RswAsw ZS

M

Va TDng nguy him nht c0: Mb c0 = ; ( 4 62 ) qsw + q1 iu kin cng theo lc ct se la:Qmax Qu =

RSAS+ RSPASP c

Mb + (qsw + q1 ).c; ( 4 63 ) c

Theo thc nghim: - Khi q1 0,5.qsw th c0 tnh theo cng thc: c0 = - Khi q1 > 0,5.qsw th c0 tnh theo cng thc (4-62);

Mb ; ( 4 64 ) q1

iu nay co ngha khi q1 be, gia tr c0 khng phu thuc vao s b tr ct ai.

Trong tnh toan thit k, tnh ct ai (qsw) nh sau:Q Qmax b1 ; ( 4 66 ) 0,6 Trong o: Qb1 = 2. M b .q1 ; ( 4 67 )

* Khi

Th

2 2 Qmax Qb1 qsw = ; ( 4 68 ) 4M b

2 Mb Q + Qb1 > Qmax > b1 ; ( 4 69 ) Th qsw = (Qmax Qb1 ) ; ( 4 70 ) * Khi C h0 0,6 P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 34 Mb

4.5.5.3 Tnh cu kin chu tai trong phn b u: Trong tnh toan thit k, tnh ct ai (qsw) nh sau:Q * Khi Qmax b1 ; ( 4 66 ) 0,6 Trong o: Qb1 = 2. M b .q1 ; ( 4 67 )Th

Chng 4

2 2 Qmax Qb1 qsw = ; ( 4 68 ) 4M b

Mb Qb1 (Qmax Qb1 )2 ; ( 4 70 ) + Qb1 > Qmax > ; ( 4 69 ) Th qsw = * Khi h0 0,6 Mb Qb1 Q ; Trong ca hai trng hp trn, ly qsw max 2.h0 M Qmax Qb1 Qmax b + Qb1; ( 4 71 ) ; ( 4 72) * Khi Th qsw = h0 h0 Q Nu tnh c qsw < b min th phai tnh lai qsw theo cng thc sau: 2.h0 2 2 Qmax b 2 Qmax Qmax b 2 qsw = + .q1 2.h + .q1 2.h ; (4 73) 2.h0 b3 0 0 b3

4.5.5.4 Trng hp ct ai t khng u (chu tai trong phn b u):q1 Vi tai trong phn b u, cang xa gi ta lc ct cang giam. Do o t khoang cach l1 nao o tnh t gi ta co th tng cach khoang ct ai (b tr ai tha hn). C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 35 l1 s1 s1 s1 s2 s2

4.5.5.4 Trng hp ct ai t khng u (chu tai trong phn b u): Chng 4 q1 Vic tnh toan tin hanh nh sau: - Tnh c01 va c02 theo (4-58): Mb c0i = ; qswi - Khi q1 > qsw1 - qsw2 th:Mb + qsw1.c01 Qmax + q1.c c l1 = c ; ( 4 74 ) qsw1 qsw2Trong o c = s1 s1 s1

s2

s2

l1

Qmax (Qb min + qsw 2 .c01 ) c01; ( 4 76 ) q1 Trong oan ct ai c giam (nm ngoai oan l1) gia tr qsw2 khng bt buc phai thoa iu kin (4-60). F F

Mb b 2 .h0 ; (4 75) q1 (qsw1 qsw 2 ) b3

- Khi q1 qsw1 - qsw2 th: l1 =

4.5.5.5 Tnh cu kin chu tai trong tp trung:

1

2

Trong trng hp nay cn tnh s s toan vi tt ca cac TDng ci xut phat t mep gi nhng khng vt qua TD co m men ln nht. C P1.. P2.. P3.. P4.. P412.. P42.. P5.. c1 P54.. P56.. U KIN CHU UN 36

4.5.5.5 Tnh cu kin chu tai trong tp trung:

F1

Chng 4 F2

Trong trng hp nay cn tnh toan vi tt ca cac TDng ci xut s s phat t mep gi nhng khng vt qua TD co m men ln c1 nht. c2 Gia tr qsw xac nh theo h s: Q1 Q Qbi i = i ; ( 4 77) Q2= Q1- F1 Qbi M Trong o Qbi = b ; ( 4 78) ci Q c i < 0i = b min . 0 th: qsw(i ) = Qi . 0i ; (4 79) * Nu Qbi 2.h0 c0 0i + 1 c Q Qbi 0i < i i ; (4 80) * Nu th: qsw(i ) = i c0 c0 ci c (Q Qbi )2 ; (4 81) < i i qsw(i ) = i * Nu th: Mb c0 h0 c Q Qbi i > i ; (4 82) * Nu th: qsw(i ) = i h0 h0 y: c0 ly bng ci nhng khng ln hn 2.h0; Cui cung ly gia tr qsw(i) ln nht xac nh ct ai. C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 37

Chng 4 4.5.5.6 Khoang cach ln nht gia cac ct ai: Co th xay ra trng hp pha hoai theo TD nghing nm gia 2 ct ai.

H s b4 xet n thiu chnh xac cua khoang cach ai do thi cng, s sai lch v phng cua khe nt nghing do BT khng ng nht.

T (4-43) ly c=smax, co: 2 b 4 .(1 + n ).Rbt .b.h0 smax = ; ( 4 83) Q

c=smax

4.5.5.7 Yu cu cu tao i vi ct ai trong dm va ban:l l

l/4

l/4

l/4

Khoang cach cu tao cua ct ai uct: h h oan u dm: 3 2 khi h 450 uct khi h > 450 uct 150 500 oan gia dm (co Q be co th khng cn tnh ct ai): 3h Vi dm co h 300 uct C P1.. P2.. P3.. P4.. 500 P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 38

5.6 Tnh ton ct ai v ct xin

4.5.6 Tnh ton cu kin c ct ai v ct xin:

Chng 4

Ct xin co nhim vu chu phn lc ct vt qua kha nng cua ai va BT trong vung co Q>Qu am bao cng trn TDng ct qua BT nm gia mep gi ta va u lp ct F xin th nht va gia cac lp ct xin tip theo.

iu kin cng (4-47) trn TDng c ct qua ca ct ai va ct xin: M Q b + qsw .c + Rsw . As ,inc .sin ; ( 4 84) cTrong o As,inc -Tng din tch cac lp ct xin ct qua mt ct nghing c. smax

smax

smax

Khi thit k cn kim tra cng cua tt ca cac mt ct nghing xut phat t mep gi va t cui cua cac lp ct xin. Ngoai ra cn kim tra cng cua cac TDng co im cui tai im t cua lc tp trung nm trong khu vc co ct xin. FKhi kim tra theo (4-84) cn lu y :

- Phai k n tai trong phn b q1.

- Khi tnh qsw.ci, nu ci>c0 th phai ly c2 ci=c0; khi ci>2.h0 th phai ly ci=2.h0. c1 c3 Mb c0 - Khi tnh phai thoa cac yu cu ci c khng ch (4-50a), (4-50b). C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 39

4.5.7 Kim tra cng trn TDng chu m men:Tit din nghing chu tc dng ca m men cn c tnh ton ti cc v tr ct hoc un ct thp dc, cng nh ti vng gn gi ta ca dm v u t do ca cng xn. Ngoi ra, tit din nghing chu tc dng ca m men cn c tnh ton ti cc v tr thay i t ngt hnh dng ca cu kin (ct mt phn tit din, v.v...).zs, inc

Chng 4

s

s

s Rsw Asw Rsw Asw

Nb zs

Rsw Asw Rs As

Rsw As, inc

zsw Ti cc v tr gn gi ta ca cu kin, m men zsw Ms chu bi cc ct thp dc ct qua vng chu ko ca tit din nghing c xc nh theo zsw cng thc: M s = Rs . As .z s ; c trong : As din tch ct thp dc ct qua tit din nghing; zs khong cch t hp lc trong ct thp dc n hp lc vng chu nn.

M M s + M sw + M s ,inc ; ( 4 51)

M men Msw c chu bi cc ct thp ai vung gc vi trc dc cu kin, c bc khng i trong phm vi vng chu ko ca TDng ang xt, c xc nh theo cng thc: M sw = 0 , 5 .q sw .c 2 ; ( 4 87 )

Khai trin iu kin (4-51) theo cac ct thep Ms, Msw va Ms,inc c: M Rs . As .zs + Rsw . Asw .zsw + Rsw . As,inc .zs ,inc ; (4 85)C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 40

4.5.7 Kim tra cng trn TDng chu m men: M men Ms chu bi cc ct thp dc: M s = Rs . As .z s ; M men Msw c chu bi cc ct thp ai: M sw = 0 , 5 .q sw .c 2 ;iu kin cng trn TDng chu m men :

Chng 4

( 4 87 )

M Rs . As .zs + Rsw . Asw .zsw + Rsw . As,inc .zs ,inc ; (4 85) Rb . Ab + Rsc . A's Rs . As Rsw . As,inc . cos = 0; (4 86)

Chiu cao vung nen x c xac nh t phng trnh cn bng:

X =0

Trong o Ab la din tch vung BT chu nen (nu la TD ch nht th Ab = b.x).

4.5.7.1 Kim tra neo ct thep doc chu keo tai gi ta t do: z s,inc = z s .cos * Trng hp dm co ca ct xin va lc tp trung + (c a1 ).sin ; ( 4 88 )iu kin (4-85) kim tra trn TDng nguy him nht co:

c=

Q Fi Rsw . As,inc .sin qsw + q

; (4 93)la M

Trong o Q -lc ct TD gi ta F, q -Tai trong tp trung va tai trong phn b trong pham vi TDng. Gia tr c tnh c theo (4-93) khng c ln hn chiu dai phn k gi ta ma pha ngoai oan o thoa man 2 iu kin: 0,8.b 4 .Rbt .b.h0

Q

ng vi gia tr c 0,8.cmax (cmax = 2,5.h0). C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 41

c

; ( 4 94 )

4.5.7.1 Kim tra neo ct thep doc chu keo tai gi ta t do: * Trng hp dm co ca ct xin va lc tp trung: Q Fi Rsw . As,inc .sin c= ; (4 93) qsw + qTrong o Q -lc ct TD gi ta F, q -Tai trong tp trung va tai trong phn b trong pham vi TDng. Gia tr c tnh c theo (4-93) khng c ln hn chiu dai phn k gi ta ma pha ngoai oan o thoa man 2 iu kin: 0,8.b 4 .Rbt .b.h0 la

Chng 4

M

Q

ng vi gia tr c 0,8.cmax (cmax = 2,5.h0).

c

; ( 4 94 )

* i vi dm chu tai trong phn b u, khng t ct xin va ct ai comt khng i, iu kin (4-85) c thay bng (4-95):

Q 2.( Rs . As .z s M 0 )(qsw + q ); ( 4 95) .

Trong o Q -Lc ct TD gi ta; man cac iu kin sau:

M0 -M men TD mep gi ta.

Cho phep khng phai kim tra cng trn TDng theo m men khi thoaQmax2 0,8.b 4 .Rbt .b.h0 2.Rbt .b.h0 ; Q ; ( 4 96 ) c

Trong o Q -Lc ct cui TDng; Qmax -Lc ct ln nht TD mep gi ta. c: chiu dai hnh chiu TDng xut phat t mep gi ta vi c 2.h0. C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 42

* Yu cu neo ct thep doc chu keo tai gi ta t do:C da d C da

Chng 4d

la

la 5d; da 0,5d; C 15mm khi d 10 mm; C 1,5d khi d > 10mm;

la

la 10d; Trn oan neo phai han hai ct ngang.

Nu thoa man iu kin (4-42a) va (4-43) (TD b tng u chu ct): la 5d; trongkhung va li han nu la ct trn th phai co mt ct ngang han vi ct doc cach u mut mt oan C. la 10d; trong khung va li han nu la ct trn th trn oan neo phai han hai ct ngang vi ct doc cach u mut mt oan C. q Fi

Nu iu kin (4-42a) va (4-43) khng thoa man (phai b tr ct ngang chu ct):

4.5.7.2 Kim tra cng trn TDng theo m men i vi cng xon: i vi cng xon chu tai trong tp trung khe nt nghing se xut phat t im t c l1 tai tp trung tai gn u mut cng xon. C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 43

Ch 4.5.7.2 Kim tra cng trn TDng theo m men i vi cng xon: ng 4 q Fi * i vi cng xon chu tai trong tp trung khe nt nghing se xut phat t im ttai tp trung tai gn u mut cng xon.

TDng nguy him nht: Q Rsw . As,inc .sin c= 1 ; (4 97) qsw

c

l1

Trong o Q1 la lc ct tai TD u khe nt nghing. Gia tr c tnh c theo (4-97) khng c ln hn khoang cach t khi im TDng n mep gi ta.

Kim tra cng trn TDng theo (4-85) vi:

l M = q.l1. c + 1 + Fi .c; (4 98) 2

* i vi cng xon chu tai trong phn b q TDng nguy him nht se kt thuc gi ta va co: Rs . As .z s c= ; (4 99) lan .(qsw + q )Trong o As la din tch ct doc c keo n u mut cng xon. zs c xac nh vi TD tai gi ta. Nu c < l-lan th vic kim tra theo cng trn TDng co th bo qua.

* i vi cng xon co chiu cao TD tng dn theo lut bc nht v pha gi ta,khi xac nh TDng nguy him theo (4-97) gia tr cua t s phai c giam i mt lng bng: Rs.As.tg khi mep chu nen nghing; Rs.As.sin khi mep chu keo nghing; ( la goc nghing so vi phng ngang) C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 44

4.5.7.3 Kim tra cng trn TDng theo m men khi giam ct doc chu keo: a) Biu bao vt liu: (BBVL) b) Un ct dc chu ko: am bao cng trn TDng N1-N1 th Zs,inc ZS. Vy khoang cach t im un ct thep doc trong vung keo (Tit din I-I) n TD ma tai o ct doc c s dung ht kha nng chu lc (Tit din tnh toan II-II) phai (h0/2). h0/2 N1 I II ZS A O W O W

Chng 4

B

N1 I Zs,inc ZS II

B

c) Ct ct dc chu ko:Tai gi lng thep doc mep trn nhiu. Xa gi M giam, tai tit din O-O theo tnh toan co th ct bt ct doc (Tai gi la AS, ct bt AS2, con lai AS1: AS = AS1+ AS2).

MA MO

Tit din O-O goi la mt ct ly thuyt.Xet trn TD nghing OA co MA>M0 nhng ct chu keo la AS1 va co thm mt s t ct ai se khng u chu m men un MA.

khng b pha hoai trn TD nghing do m men ta phai keo ct thep ct giam ra ngoai mt ct ly thuyt O-O mt oan W na (n im B). Q W= + 5.d 20d ; ( 4 100) Tit din B-B goi la mt ct thc t. 2.q swQ: Lc ct tai im ct ly thuyt, ly bng dc cua biu bao mmen. C P1.. t do c P4.. m. 5d: oa P5.. P54.. P56.. bt u chu U d: ng knh cP2..c P3.. ct giaP412.. P42..n cn thit ct docU KIN CHU lc. N 45

4.5.7.3 Kim tra cng trn TDng theo m men khi giam ct doc chu keo: a) Biu bao vt liu: (BBVL) b) Un ct dc chu ko: c) Ct ct dc chu ko:Tai gi lng thep doc mep trn nhiu. Xa gi M giam, tai tit din O-O theo tnh toan co th ct bt ct doc (Tai gi la Fa, ct bt AS2, con lai AS1: AS = AS1+ AS2). I Zs,inc Za W h0/2 W II Za A O O N1 I

Chng 4B

N1 II

B

TD O-O goi la mt ct ly thuyt.

MA MO

Xet trn TD nghing OA co MA>M0 ct chu keo la AS1 se khng u chu m men un MA.

khng b pha hoai trn TD nghing do m men ta phai keo ct thep ct giam ra ngoai mt ct ly thuyt O-O mt oan W na (n im B). Q W= + 5.d 20d ; ( 4 100) Tit din B-B goi la mt ct thc t. 2.q swQ: Lc ct tai im ct ly thuyt, ly bng dc cua biu bao mmen. d: ng knh ct doc c ct giam. 5d: oan cn thit ct doc bt u chu lc.

Khi trong vung ct thep co ct xin th: W =

Q - Qs,inc 2.q sw

+ 5.d

20d ; (4 101)

Trong o Qs,inc= Rsw.As,inc.sin vi As,inc la din tch cua ct xin trong vung ct thep. n gian va an toan co th ly As,inc la din tch ct xin trn oan:

Q - Qs,inc ; CU KI P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56..2.qsw N CHU UN 46

5.7 Tnh toan dm co tit din thay i: a. Dm co mep chu nen nghing:Chiu cao dm tng dn theo chiu tng cua m men, do o gia tr Qb c tnh vi chiu cao lam vic tai mut cua tit din nghing trong vung nen (ai lng nay thay i theo C). Trnh t tnh nh sau:

Chng 4Db Q RaFx1 M Z RaF RaFx2

u

- Kim tra K (4-39), (4-40) vih0=h01 la chiu cao lam vic cua TD thng goc i qua iim u TD nghing trong vung keo. Q

aFa C

- Chon cu tao ct ai, tnh q. - Chon mt gia tr cua h0 cui TDng tnh C0 theo (4-47) 2 Rk b h (1 + B ) ; Co th tnh h0 theo cng thc sau: h0 = 01 vi B = qd 1- B - Tnh lai h0 theo C0: h0 = h01 + C0.tg. So sanh vi h0 a gia thit. Tip tuc tnh toantheo cach ung dn xac nh c h0.

Co h0 tnh Qb ri kim tra vi Q, nu khng thoa man co th tng ct ai hoc b tr ct xin.C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 47

5.7 Tnh toan dm co tit din thay i: b. Dm co mep chu keo nghing:Mep chu keo nghing mt goc la , kha nng chu ct cua dm co tham gia cua ct doc chu keo la Qa (la hnh chiu cua ng lc trong ct doc ln phng lc ct Q). Zx2 Zx1

Chng 4

Q Db

u M RaFx1 RaF RaFx2 aFa C Zx2 h0

iu kin cng trn TD nghing theo lc ct la (khng dung ct xin): Q Qtd = Qb + Qa.Qa xac nh theo ng lc co th co trong ct doc chu keo aFa Trnh t tnh nh sau:

Q

- Chon cu tao ct ai, tnh q ri tnh C0. - Co C0 xac nh Qb;

- Co C0 bit c mut cui cua TDng, xac nh m men un M tai TD thng goc i qua mut o; - Tnh phn m men do ct doc chu la: Ma = M 0.5q.C02. M Tnh kha nng chu ct cua ct doc chu keo la: Qa = a tg ; Z (Z co th ly gn ung bng 0.9h )C P1.. P2.. P3.. P4.. P412.. P42.. P5.. P54.. P56.. U KIN CHU UN 480

5.1. Gii thiu chung

SAN PHNG.

Chng 5

5.1. Gii thiu chung:San BTCT co u im la kha nng chu lc ln, a nng, thit k va thi cng n gian.

5.1.1 Phn loai: a. Theo PP thi cng: co san toan khi, san lp ghep va san na lp ghep. b. Theo s kt cu: co san sn va san khng sn. - San sn toan khi co ban loai dm (ban san lam vic 1 phng). - San sn toan khi co ban k 4 canh. - San sn c. - San sn pa nen lp ghep. 5.1.2 Phn bit ban loai dm va ban k 4 canh: - Khi ban ch co lin kt 1 canh hoc 2 canh i din, tai trong truyn theo phng co lin kt. Ta goi la ban loai dm. q lKHOA XY DNG DD&CN

q l l

q

P1.. P2.. P3.. P4..

SAN PHNG 1

1 - Khi ban co lin kt 4 canh, tai trong c truyn vao lin kt theo ca 2 phng. Ta goi loai nay la ban k 4 canh. Xet ban k t do 4 canh chu tai trong phn b u; Ct cac dai ban theo 2 phng. Goi Ta co: q2 l2

Chng 5

1 q1 l1

q2

l2

q1 l1

tai trong truyn theo phng l1 la q1, tai trong truyn theo phng l2 la q2. (5 - 1) q = q1+ q2.

vong tai im gia cua mi dai: 4 5 q1.l1 5 q 2 .l4 2 . ; . ; + Theo phng l2: f 2 = + Theo phng l1: f1 = 384 E.J 384 E.J 4 4 Ta co: f1 = f 2 q1.l1 = q 2 .l 2 ; (5 - 2) 4 l2 l14 T (5 - 1) va (5 - 2): q1 = 4 4 .q va q2 = 4 4 .q; (5 - 3) l1 + l 2 l1 + l 2 4 l2 q1 = 4 .q2 ; (5 - 4) l1 Vy tai trong truyn theo phng canh ngn nhiu hn. Khi tnh toan thc P2.. u l2>2.l1 th KHOA XY DNG DD&CN P1.. t, nP3.. P4.. co th xem ban lam vic 1 phng. NG 2 SAN PH

5.1.3 Khai nim khp deo-S phn b lai ni lc do khp deo: 5.1.3.1 Khai nim khp deo:Xet 1 dm chu un cho n khi b pha hoai. Gia s dm c cu tao thep sao cho khi b pha hoai co:

Chng 5

Vung BT co bin dang deo ln

- ng sut trong ct thep chu keo at gii han chay;ct thep chu keo at - ng sut trong BT vung nen at gii han chu nen gii han chay va co bin dang deo ln; Tai TD pha hoai xut hin 1 khp deo co m men khp deo Mkd = Mgh. Vi kt cu tnh nh, s xut hin khp deo ng thi vi kt cu b pha hoai. Vi kt cu siu tnh xut hin khp deo lam giam 1 bc siu tnh cua h. S pha hoai cua kt cu khi s khp deo u h b bin hnh.

- Trang thai khi xut hin khp deo cui cung trc khi kt cu b pha hoai goi la trang thai cn bng gii han. - Phng phap tnh theo s deo (xet s hnh thanh khp deo cho n khi h sp b pha hoai) con goi la tnh theo trang thai cn bng gii han.

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 3

5.1.3.2 S phn b lai ni lc:Xet dm chu tai co s nh sau:

P Chng 5 a khng i. M A b MB

MA MB Nu tnh theo s an hi, ty s , M nh M nh

Mnh Tng P n P1 tai gi A xut hin khp deo trc. Luc nay nu P tng th m men tai gi A khng tng, con tai cac TD khac vn tng. Khi P tng n P2 gia s tai gi B xut hin khp deo. Nu P tng th m men tai cac gi A va B khng tng. (nh dm n gian) Khi P tng n P3 gia nhp hnh thanh khp deo, kt cu b pha hoai: y la TT cn MkdA MkdB bng gii han. P Nh vy khi hnh thanh khp deo, trong kt cu co s phn b lai ni lc, y la yu t co li tranh s pha hoai cuc b. Khi tai cac gi hnh thanh khp deo, t s trn co th thay ngam bng lin kt khp va mt m men khp deo. MkdA a b MkdB

a b Goi M0 la m men cua dm n gian vi P3: M 0 = M kd nh + .M kdA + .M kdB ; l l Kt hp vi M0 = M(P3) xac nh c tai trong TT cn bng gii han P3 theo cac Mkd.KHOA XY DNG DD&CN

Mkdnh

P1.. P2.. P3.. P4..

SAN PHNG 4

5.1.3.3 iu kin tnh theo s deo: hnh thanh khp deo, vt liu va h phai co cac tnh cht sau:

Chng 5

- Ct thep co thm chay ro rt (dung thep deo, dy thep keo ngui, khng dung thepdp ngui..)

- Tranh s pha hoai sm do BT vung nen hong do b ep v hoc b ct t (han ch lng thep chu keo han ch chiu cao vung nen x 0,3.h0). - han ch b rng khe nt tai TD co khp deo u tin cn han ch s cach bit gia tr m men un khi tnh vi s deo so vi s an hi :Theo qui pham cua Anh BS8110-1997: Mkd (0,8-0,9)Mh. Theo cac tai liu cua Nga th phai han ch Mkd 0,7Mh. Trong cac kt cu BTCT siu tnh vic tnh toan theo s deo co th tit kim 20-30% ct thep, PP tnh n gian, kt qua tnh lai kha phu hp vi trang thai lam vic thc. ng thi co th iu chnh hp ly vic b tr ct thep giai quyt tnh trang ct thep t qua day tai tit din nao o. Tuy vy PP tnh theo s deo khng c ap dung cho cac kt cu chu tai trong ng, chu tai trong lp, cung khng c ap dung vi kt cu co nhng han ch nghim khc i vi s phat trin khe nt.

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 5

5.2. San sn ban dm

5.2. San sn toan khi co ban loai dm: Cac b phn chnh cua san: 1. Ban, 2. Dm phu, 3. Dm chnh, 4. Ct, 5. Tng.

Chng 5

5.2.1 S kt cu:

1 2 3 5 4l2 l2 l2

3l1

3l1

3l1 l1 l1 l1 l1 l1 l1

l2 l2

San gm ban san va h dm (sn) uc lin khi: ban k ln dm phu, dm phu gi ln dm chnh, dm chnh gi ln ct va tng,

3l1

3l1

5

1

San co th co dm chnh t theo phng doc hoc theo phng ngang (tuy thuc s b tr chung cua cng trnh, yu cu thng gio, chiu sang..).

2l1 l1

3l1

4l1

Khoang cach dm phu l1 = (1-4)m, thng l1 = (1,7-2,8)m. Khoang cach dm chnh l= (4-10)m, thng l2 = (5-8)m. KHOA XY DNG DD&CN P1.. P2.. P3.. P4..

SAN PHNG 6

1 1 Chiu day ban hb = .l1 ; 35 25 5cm vi san mai; 6cm vi san nha dn dung; 7cm vi san nha CN;

Chng 5

1 1 Chiu cao dm phu hdp = nhp; 20 12

1 1 Chiu cao dm chnh hdc = nhp; 15 8

B rng dm

bd = (0,3 - 0,5)hd;

Nu chu vi san c k ln tng gach, oan k: (12cm va hb) vi ban; 22cm vi dm phu; 12cm 34cm vi dm chnh;

22cm

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 7

5.2.2. Tnh ni lc san: 5.2.2.1 Tnh ni lc ban theo s deo: - S tnh: Ct dai ban rng = n v (1m) theo phngcanh ngn; Xem cac dai ban lam vic c lp nh dm lin tuc ta ln dm phu va tng.

Chng 5

- Tai trong: Tnh tai g (trong lng ban thn ban BTCT va cac lp cutao..)

l2 l2 l1 l1 l1 3l1 3l1

Hoat tai p (tai trong s dung trn san) phn b u trnmt san c qui v phn b u trn dai ban.

- Nhp tnh toan: Nhp gia: l = l1 - bdp; b b h Nhp bin: l b = l1 dp t + b ; 2 2 2 - Ni lc: q.l 2 Nhp bin va gi th 2: M = b ; (5 5) 11 q.l 2 ; (5 6) Nhp gia va gi gia: M = 16 Trong o q = g + p;KHOA XY DNG DD&CN

bt

lb l1

l l1 g

l l1 p

l1

lb

l

l

l

P1.. P2.. P3.. P4..

SAN PHNG 8

- Ni lc: q.l 2 Nhp bin va gi th 2: M = b ; (5 5) 11 q.l 2 ; (5 6) Nhp gia va gi gia: M = 16 Trong o q = g + p; 5.2.2.2 Tnh dm phu theo s deo : - S tnh: nh dm lin tuc gi ln dmchnh va tng.

q.l2 b 11

q.l2 ng 5 Ch 16

a

- Tai trong: phn b u gm Tnh tai: gd = g.l1 + g0; Hoat tai pd = p.l1. - Nhp tnh toan: Nhp gia: l = l2 - bdc; b b a Nhp bin: lb = l2 dc t + ; 2 2 2 - Ni lc:

lb l2

bdc l l2 gd pd

l l2

bdc l2

lb

l

l

l

q.l 2 * M NG DD&CN gia P2.. P3.. P4.. = +0,0625.q.l 2 ; KHOA XY Dmen dngP1.. nhp gia: M = + 16

Co th t hp tai trong (vi cac dm bt ky) hoc dung cac cng thc va bang lp sn (dm u nhp) ve biu bao (BB) m men, lc ct. xet n anh hng chng xon cua dm chnh lam giam ni lc trong dm phu, khi tnh ni lc theo s deo hoat tai c tnh vi gia tr la p=p.3/4 va tnh tai la g=g+p/4. SAN PHNG 9

a

5.2.2.2 Tnh dm phu theo s deo : - S tnh: - Tai trong: - Nhp tnh toan: - Ni lc: Dung bang lp sn ve BB m men: Tung nhanh dng BB m men: M = 1.q.l2; (5 - 7) Tung nhanh m BB m men: M = 2.q.l2; (5 - 8)Cac gia tr 1, 2 tra bang.

Chng 5lb l2 bdc l l2 gd pd l l2 bdc l2

lb

l

l

l

Lc ct tai gi A:

QA = 0,4.q.l;

(5 - 9) (5 - 10)

tr tai mep trai gi B: Q B = 0,6.q.l;

tr ph Tai mep phai gi B va cac gi gia: Q ph = Q C = Q C = .. = 0,5.q.l; (5 - 11) B

Trong o q = gd + pd; l la nhp tnh toan.

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 10

5.2.2.3 Tnh dm chnh theo s an hi: - S tnh: nh dm lin tuc gi ta la ctva tng.

bt

Chng 5

- Tai trong: Tnh tai: Hoat tai

G = gd.l2 + G0. P = pd.l2.

l1

l1 P G l

bc l1

l1

l1 P G l

l1

l1 P G

- Nhp tnh toan: ly bng khoang cach trong tm cac gi l; - Ni lc:

+ Xac nh va ve B ni lc do tnh tai G (MG, QG) va do cac trng hp bt li cua hoat tai (MP1, QP1, MP2, QP2,..) + Cng B ni lc tnh tai MG, QG vi tng trng hp hoat tai MPi, QPi c B tng cng : Mi, Qi. + Tai mi TD chon trong cac B tng cng mt gia tr dng ln nht va mt gia tr m co tr tuyt i ln nht ve BB ni lc (co th xac nh BB ni lc bng cach ve cac B tng cng ln cung mt truc va cung t l, hnh bao se la cac oan ngoai cung). Minh hoaVi dm u nhp co th dung cac cng thc va bang lp sn ve BB ni lc:

Tung nhanh dng BB m men: M = (0.G + 1.P).l ; Tung nhanh m BB m men: M = (0.G - 2.P).l; KHOA XY DNG DD&CN P1.. P2.. P3.. P4..

(5 - 12) (5 - 13)SAN PHNG 11

Chng 5 Vi dm u nhp co th dung cac cng thc va bang lp sn ve BB ni lc: M = (0.G + 1.P).l ; (5 - 12) Tung nhanh dng BB m men: Tung nhanh m BB m men: M = (0.G - 2.P).l; (5 - 13) Tung nhanh dng BB lc ct: Q = 0.G + 1.P ; (5 - 14) Tung nhanh m BB lc ct: Q = 0.G - 2.P; (5 - 15)Cac gia tr 0, 1, 2, 0, 1, 2 tra bang.

5.2.3. Tnh ct thep: 5.2.3.1 Tnh ct thep ban: Tnh nh cu kin chu un TD ch nht t ct n co: b = 1m; h = hb; TD gia nhp bin va nhp gia vi m men dng ln nht. TD gi th 2 va gi gia vi m men m. Kim tra kha nng chu ct cua BT vung nen.Xet anh hng cua hiu ng vom trong ban,cac ban co canh u uc lin khi vi sn giam 10-20% lng thep tnh toan tuy theo s canh lin khi. % = (0,3-0,9) la hp ly; Va min (min = 0,05%, thng ly =0,1%)M men m M men dng

* i vi cac ban gia (co ca 4 canh lin khi vi sn) c giam 20% m men. * i vi cac ban bin (co 3 canh lin khi vi sn) c giam 20% m men tnh toan khi l2/l1 8cm nn un bt thep nhp ln gi (khoang 1/3 n 2/3 lng thep). (Minh hoa) 1/5.lb

lb l1 .lb .l

l l1 .l .l

l l1

lb l1 1/10.lb .lb 1/6.l

l l1 .l .l 1/6.l 1/6.l .l 1/6.l

l l1

Ct phn b: cung vi ct chu lc tao thanh li. 10% lng thep chu lc khi l2/l1 3; 20% lng thep chu lc khi l2/l1 < 3;

p/g 3: = 1/4 p/g P2.. P3.. KHOA XY DNG DD&CN P1.. > 3: = 1/3 P4..

lb

l

l SAN PHNG 15

chnh, gi ln tng. Lng thep nay 1/3 lng thep chu lc tai cac gi gia va 56/1m dai.

Ct thep mu cu tao: tai v tr ban gi ln dm

1/8 l0

Chngl0 l0 5

5.2.4.2 Cu tao ct thep dm: Ct thep dm tt nht la dung khung han: (Minh hoa)+ Gia nhp dung cac khung phng c keo dai n mep gi. + Trn gi dm phu b tr cac li thep chu m men m. + Trn gi dm chnh b tr cac khung han chu m men m. 1/6l Li thep Cu tao

6/a200 (56/1m dai)

Dm chnh

1/3l

1/3l Li thep Chu lc

Khung han nhpKhung thep cu tao

l

Khung thep dm chnh 15d

15d

Thanh ni

Nu dung khung buc:

+ Gia nhp b tr ct doc chu m men dng mep di, vao gn gi co th un 1 phn thep ln Khung han Ct chu m men m, thep con nhp lai keo vao gi 2 thanh. + Trn gi, ngoai cac thanh un t nhp ln, phai t thm mt s thanh u theo yu cu, ra xa gi tin hanh ct bt ct thep theo BB m men. Kha nng chu lc tng ng vi KHOA XYthep G DD&CNbng Hnh bao VP4..u SAN PHNG 16 b tr DNth hin P1.. P2.. P3.. t li

Khung thep trn gi

5.3. San sn ban k 4 canh

5.3. San sn toan khi co ban k 4 canh:San gm ban san va h sn uc lin khi,

Chng 5

5.3.1 S kt cu:l T l 2 2 (thng ly 1-1.5), l1 Kch thc cac canh l1, l2 = 4 - 6m. 1 Chiu day ban h b l1; 50 Bin dang cua ban:+ Mt di cua ban: Xut hin cac vt nt theo phng ng phn giac cac goc, con gia ban co cac vt nt theo phng canh dai. + Mt trn: Nu cac canh la ngam cng co cac vt nt chay vong theo chu vi, nu k t do cac goc ban se b vnh ln.

l1 l1 l1 l2 l2 l2 l2

l2

l1 Mt di Mt trn

5.3.2 B tr thep ban: B tr cac li thep, ct thep co th song song vi cac canh hoc theo phng xin (cheo vung goc vi cac vt nt), hiu quachu lc nh nhau, tuy nhin vi li co ct thep song song vi cac canh thi cng n gian hn.KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 17

5.3.2 B tr thep ban: Nn dung cac li han: + Gia nhp: s dung cac li co ct chu lc theo 2 phng. Co 2 cach b tr thep nay: t thep u (dung 1 li thep) t thep khng u (dung 1 li chnh cho toan ban va 1 li phu t gia ban).

Chng 5lk

l2

Canh ngam: lk=l1/4; Canh khp: lk=l1/8;

lk lk l1 0.5l1

+ Trn gi: dung li thep co ct chu lc theo phng vung goc vi cac sn.0.5l1

Nu dung li buc: + Gia nhp t theo tnh toan, vao gn gi (day bin lk) co th giam. + Trn gi: co th un 1/2 -> 2/3 lng thep nhp ln va t thm ct mu xen ke.KHOA XY DNG DD&CN

l2 0.5l1

l2 0.5l1

0.5l1

l1

0.5l1

0.5l1

l1

0.5l1

P1.. P2.. P3.. P4..

SAN PHNG 18

5.3.3 Tnh ban k 4 canh theo s deo: S tnh:Khi trang thai CBGH theo cac khe nt se hnh thanh khp deo. MI

MII

Chng 5 MII M2 f

M1 M2

MI MII

l2

- M men khp deo: Mkd = RS.AS.ZS;Mkd la m men khp deo trn 1 n v dai, AS din tch ct thep trn 1 n v dai, ZS la canh tay on ni lc (ZS 0.9h0). Nu canh k t do th m men trn canh o =0. MII MI f M1

M2

MI

Theo nguyn ly cn bng cng kha d: Wq = WM. (5 - 16)F F

M1 l1 2

Cng kha d cua ngoai lc: Wq = y.q.dF = q. y.dF = q.V ; (5 17)Vi V la th tch cua hnh khi tao bi mt phng ban ban u va cac ming cng trang 3.l 2 l1 thai CBGH,

6 Cng kha d cua ni lc: WM = i.Mi.li;

V = f .l1.

;

(5 - 18)

Khi b tr thep u: WM = i.Mi.li = (2.M1 + .MI + .MI).l2 + (2.M2 + .MII + .MII).l1 ;

WM = .[(2M1 +MI +MI).l2 + (2M2 +MII +MII).l1]; l1 2 3.l l TXY DNG q.l1 . 2 1 = (2M1 +MI +MI).l2 + (2M2 +MII +MII).l1; (5 - 19) (5 - 16) DD&CN KHOA SAN PHNG 19 P1.. 12 P2.. P3.. P4..

V tg =

2f l1

2f

Chng 5 Khi b tr thep khng u: M M WM = 2.M1.(l2 - 2lk) + 2 . 1 .2lk + (MI + MI)..l2 + 2.M2.(l1 - 2lk) + 2 . 2 .2lk + 2 2 + (M + M )..l ;II II 1

T (5 - 16) 2 3.l l q.l1 . 2 1 = (2M1 +MI +MI).l2 + (2M2 +MII +MII).l1 - 2.(M1 + M2).lk; (5 - 20) 12Trong cac phng trnh (5 - 19) & (5 - 20) co cha 6 m men cn tm, co th ly M1 lam n s, con cac m men con lai c biu din qua M1 va cac h s c chon:

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 20

* Tnh ban k 4 canh theo s an hi:

Chng 5

S tnh: Tuy theo lin kt (ngam, khi) chia ban thanh 11 loai co s nh hnh ve

1l1

l2

2

3

4

5

6

7

8

9

10

11

M men dng ln nht gia nhp: M men m tai gi: MI = ki1.P; MII = ki2.P;

M1 = mi1.l1.l2.q = mi1.P; M2 = mi2.P; Trong o i =1, 2, ..11 la ch s s , m, k cho trong bang tra.

Khi tnh ban lin tuc cn xet cac t hp bt li cua hoat tai:-M men dng gia nhp ln nht khi hoat tai t cach . -M men m ln nht khi hoat tai t trn cac k gi o.KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 21

5.3.4 Tnh va cu tao dm: Tai trong t ban truyn vao dm nh sau: Theo phng canh ngn dang tam giac, gia tr ln nht la q.l1; Theo phng canh dai dang hnh thang, gia tr ln nht la q.l1; Trong lng ban thn dm la g; Tnh ni lc theo s deo: l2

Chng 5

l1 q.l1 q.l1 l2 l2 l1 l1

g.l 2 ; (5 21) + M men nhp bin va gi th 2: M = 0,7M 0 + 11 g.l 2 ; (5 22) + M men nhp gia va gi gia: M = 0,5M 0 + 16 2 q.l .l Vi tai trong phn b tam giac: M 0 = 1 ; 12 q.l1.l 2 l Vi tai trong phn b hnh thang: M 0 = . 3 4. 2 ; = 1 . 24 2l2 + Lc ct trong dm: M M tr Tai gi th nht: Q A = Q0 B ; Tai bn trai gi th 2: QB = Q0 + B ; l l ph tr ph Tai cac gi gia: QB = QC = QC = .. = Q0 ;

(

)

0 KHOA XY DNG DD&CN

Q la lc ct cua dm n gian, MB la m men tai gi B (th 2);

P1.. P2.. P3.. P4..

SAN PHNG 22

5.4. San sn lp ghep

Theo s an hi:

Chng 5

Tnh nh dm an hi vi cac PP cua CKC. Co th qui i tai trong thanh phn b u n gian tnh toan:

Vi dang tam giac: Vi dang hnh thang:

qt = 5/8.qd; qt = (1 - 2.2 + 3)qd;

1

4 ld

5.4. San sn panen lp ghep: 5.4.1 S kt cu: Pa nen k ln dm hoc tng; Nhp cua panen lp = (2,8 -> 6,8)m; Nhp dm ld = (4 -> 7,2)m; (Hnh anh) 5.4.2 Cu tao panen: 5.4.2.1 Panen c:Co th 1 lp hoc nhiu lp. Chiu day h = 80->150. u im: D san xut, nhanh, lin kt n gian, chiu day san thp. Nhc im: Tn VL, cach m kem.

3 2 lp1. Panen

ld ld lp3. Ct

lp2. Dm100

lp4. Tng120

5.4.2.2 Panen co l:Co th 1 hoc nhiu l. L hnh thang, ch nht, tron, bu duc.. Chiu dai (nhp) = (2,5 -> 4,5)m. Chiu cao tuy thuc chiu dai (nhp).50

1000

40

1000 30

200 580

50 25 1180

200

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 23

5.4.2.2 Panen co l:Co th 1 hoc nhiu l. L hnh thang, ch nht, tron, bu duc.. Chiu dai (nhp) = (2,5 -> 4,5)m. Chiu cao tuy thuc chiu dai (nhp).50 580 200 50

Chng 530 200 25 1180

B rng = (45 -> 60)cm loai 1 l; (90 -> 120)cm loai nhiu l; B day canh = (2 -> 3)cm tuy thuc vung nen hay keo. B day sn = (2,5 -> 5)cm. u im: Tao c trn va san phng. Cach m, cach nhit tt, t tn VL. Nhc im: Kho ch tao.3 1 80 2 1490

5.4.2.3 Panen sn:

50 200 350 30

Gm ban va sn. Thng co 2 sn doc, cac sn ngang cach nhau (1,5 -> 2,5)m. Sn ngang be hn sn doc, sn co th pha trn hoc pha di. Chiu day canh 50 -> 60 khi sn pha di; 30 -> 35 khi sn pha trn;

1 Ban. 2 Sn ngang. 3 Sn doc.3 2

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 24

5.4.3 Tnh toan panen: 5.4.3.1 Tnh un tng th: S tnh: Xem panen nh 1 dm n gian k t do ln dm. Nhp tnh toan: Ly bng khoang cach trong tm cac gi.

Chng 5

Tai trong: Gm tnh tai va hoat tai phn b cua san trn din tch b mt panen. Tit din tnh toan: tnh kha nng chu un cua panen, qui i TD panen v cacdang n gian nh ch I, ch T. - Ct doc chu m men b tr trong vung keo. - Ct ai chu ct b tr trong sn (vi panen c tnh BT u chu ct).

5.4.3.2 Tnh un cuc b: (vi panen sn hoc panen co l) Tnh ban chu un: Xem ban lin kt an hi vi sn, tnh nh ban k 4 canh hocloai dm.

Tnh sn ngang: Nh dm n gian k t do ln cac sn doc.

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 25

Chng 5Khi thit k panen, co th chon chiu cao panen theo cng thc sau:

c.l0 .R a g c . + pc h= . ; Ea qcgc la tai trong tiu chun tac dung dai han (trn 1m2 san). pc la tai trong tiu chun tac dung ngn han. Tai trong toan phn qc = gc + pc ; la h s xet n s giam cng do tai trong dai han; ( = 2 vi panen co l, = 1,5 vi panen sn co canh trong vung nen). c la h s thc nghim c = 18 -> 20 vi panen co l, c = 30 -> 34 vi panen sn. (vi thep AII tr lai chon c ln, vi thep mac cao chon c be)

5.4.3.3 Kim tra vong: Tnh nh cu kin chu un (se c xet n trong phn tnh theo TTGH th 2). Tnh vi TD qui i thanh dang ch T, ch I tng ng, qui i theo qui tc sau: -Cac l tron i thanh l vung l bu duc i thanh l ch nht. -Gi nguyn v tr trong tm, din tch va m men quan tnh cua TD.

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 26

5.4.4 Cu tao ct thep cua panen: Dung khung va li han:

Ct chu lc

Chng 5 Khung thep

Ct cu tao Ct thep chu lc theo tnh un tng th la cac khung phng b tr trong sn. Trong ban (canh) t cac li thep. Khi chiu day ln t 2 lp, chiu day be t 1 lp gia.

Li thep Li thep

Khung thep (trong sn)

5.4.5 Cu tao va tnh toan dm: Tai trong: gm tai t panen truyn xung (vi panen c, panen hp la tai phn b,panen sn la tai trong tp trung tai v tr cac sn doc),

trong lng ban thn dm. Cu tao va tnh toan ct thep: nh dm cua san toan khi.Tuy yu cu chu lc, cach gac panen ma chon TD dm: ch nht, ch T canh di hay trn,.. Vi dm lp ghep cn kim tra kha nng chu lc khi vn chuyn, cu lp.

KHOA XY DNG DD&CN

P1.. P2.. P3.. P4..

SAN PHNG 27

6.1 Cu tao

CU KIN CHU NEN.

Chng 6

6.1. Cu tao:Cu kin chu nen thng gp la ct trong khung nha, trong thn vom, thanh dan, v.v.. Lc nen N tac dung theo phng truc doc cua cu kin. N e0 N N - Nen trung tm. M=N.e0 - Nen lch tm

6.1.1 Tit din ngang cu kin:i vi cu kin chu nen trung tm thng dung tit din vung, ch nht, tron, hay a giac u canh.. Cu kin chu nen lch tm thng dung tit din ch nht, ch T, ch I, ct rng hai nhanh, vanh khuyn... b h

b h

Chiu cao TD h (la canh // mt phng un). T s k.N Din tch TD co th chon s b: A = ; RbTrong o: - N: lc doc tnh toan. - k=0,91,1 khi nen trung tm. P1.. P2.. tm. - k=1,21,5 khi nen lch P3.. P4.. P4.3..

h = 1.5 - 3; b

CU KIN CHU NEN 1

n nh c c trng qua manh : l - Vi TD bt ky: = 0 0 ; r l - Vi TD ch nht: b = 0 0b ; b

Chng 6(r la ban knh quan tnh cua TD) (b la canh be cua TD)

Trong o: l0 la chiu dai tnh toan cua cu kin (l0 = .l). 0, 0b : manh gii han. i vi ct nha 0 =120, 0b =31; i vi cu kin khac 0 =200, 0b =52;

6.1.2 Cu tao ct thep: Ct thep doc chu lc: co 1240. Khi b >200 th nn dung 16. -Cu kin nen trung tm: ct doc chu lc c b tr u theo chu vi cua TD;Ast b AS AS

-Cu kin chu nen lch tm: AS la ct thep trong vung chu nen t hoc chu keo, AS la ct thep trong vung chu nen nhiu. A Ham lng ct thep: t t = st ; Ab Ab la din tch tnh toan cua tit din BT.

h

Cu kin chu nen lch tm ct thep b tr theo canh b th co: A A A st = A s + A ; A b = b.h 0 ; = s ; = s ; s CU KIN CHU NEN 2 b.h P1.. P2.. P3.. P4..0 P4.3.. b.h 0

6.1.2 Cu tao ct thep: Ct thep doc chu lc: co 1240. Khi b >200 th nn dung 16. -Cu kin nen trung tm: ct doc chu lc c b tr u theo chu vi cua TD; -Cu kin chu nen lch tm: AS la ct thep trong vung chu nen t hoc chu keo, AS la ct thep trong vung chu nen nhiu. A Ham lng ct thep: t ; t = st ; Ab Ab la din tch tnh toan cua tit din BT.

Chng 6

nhiu max = 3%, am bao s lam vic chung gia thep va BT th max = 6%. TCXDVN 356:2005 iu 8.6.1 trang 127: Hm lng ti thiu ca ct thp AS v AS trong cc cu kin chu nn lch tm m kh nng chu lc ca chng ng vi lch tm tnh ton c s dng khng qu 50% c ly bng 0,05 khng ph thuc vo mnh ca cu kin. CU KIN CHU NEN 3 P1.. P2.. P3.. P4.. P4.3..

Cu kin chu nen lch tm ct thep b tr theo canh b th co: A A A st = A s + A ; A b = b.h 0 ; = s ; = s ; s b.h 0 b.h 0 Nn han ch ham lng ct thep: min , ; min theo Bang 37 TCXDVN 356:2005 trang 127. 0 t max; 0 = 2.min, khi cn han ch lng thep s dung khng qua

Ch c ng 6 Ct doc cu tao: Khi chiu cao h > 500 th vi cu kin chu nen lch tm n b trct doc cu tao trn canh h: d 12 va khoang cach gia chung 500.h 400

Ct ai: Vai tro cua ct ai la n nh choct doc chu nen, nh v ct doc khi thi cng, chu lc ct, chu cac ng sut do co ngot va thay i nhit .. Han ch bin dang ngang cua BT.

h 500

b 400

b 400

Khoang cach cac ct ai: Hnh dang cua ct ai: ng knh ct ai:h >400 b >400 b >400 500 500

600 h 1000

b 400

b >400

h >1000

h >1000

Khi co yu cu bn cao hoc tnh deo cao, cac thanh ct doc chu lc c b tr trong mt ng tron va ct ai vung goc c thay bng ct ai un thanh hnh xon c vi nghing khoang 35-85mm.. Cac ct co ct ai xon thng co TD tron, cung co th vung hoc P1.. P2.. uP3.. h. a giac can P4.. P4.3.. CU KIN CHU NEN 4

6.2. 6.2 Nen trung