CHNG 1
PAGE 17
B GIO DC V O TO B XY DNG
TRNG I HC KIN TRC H NI
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PHM TUN ANHTNH TON MNG B CC THEO M HNH H S NN C XT N TIN CY CA S LIU NN TLUN VN THC S XY DNG DN DNG V CNG NGHIP
H Ni Nm 2011
B GIO DC V O TO B XY DNG
TRNG I HC KIN TRC H NI
-----------------------
PHM TUN ANH
KHA: 2008-2011 LP: CH08-X
TNH TON MNG B CC THEO M HNH H S NN C XT N TIN CY CA S LIU NN TLUN VN THC SCHUYN NGNH: XY DNG DN DNG & CNG NGHIPM S: 60.58.20NGI HNG DN KHOA HCTS. NGUYN TNG LAI
H Ni Nm 2011
LI CM N
Trong qu trnh thc hin lun vn, ti gp nhiu kh khn trong vic tip cn nhng kin thc mi v hng gii quyt cho ti. Nh s hng dn tn tnh ca T.S Nguyn Tng Lai, ti nm bt c nhiu kin thc, do c th hon thnh ti. Ti xin gi li cm n su sc n thy.
Xin gi li cm n n cc thy c ca trng H kin trc H ni v Hc vin K Thut Qun S ch dy cho ti nhng kin thc b ch trong qu trnh hc tp ti trng v trong qu trnh hon thnh lun vn.
Xin cm n T.S Nguyn Vi gii thiu cc ti liu hu ch hon thin lun vn.
LI CAM OAN
Ti xin cam oan lun vn ny do ti t thc hin v cha tng c cng b di bt k hnh thc no.
MC LC1M U
3CHNG 1 : TNG QUAN
31.1. Cu to v ng dng ca mng b-cc
31.1.1. Cu to ca mng b cc
61.1.2. ng dng mng b cc
71.2. C ch lm vic ca mng b cc
101.3. Cc quan im thit k hin nay
101.3.1.Quan im cc chu ti hon ton
111.3.2. Quan im b chu ti hon ton
121.3.3. Quan im b - cc ng thi chu ti
131.4. Tng quan v cc phng php tnh ton mng b - cc
131.4.1. Cc phng php n gin
161.4.1. Cc phng php c k n s tng tc cc- t nn v b-t nn
191.5. Cc dng m hnh bin dng ca nn t
191.5.1. M hnh nn Winkler
211.5.2. M hnh bn khng gian n hi
231.6. Tnh ton cc lm vic ng thi vi nn
251.7. Tng quan v l thuyt tin cy
251.7.1. Cc m hnh tnh:
271.7.2. Cc phng php tnh
30CHNG 2 : XY DNG M HNH TNH MNG B - CC
302.1. Cc m hnh tnh ton
322.2. Xc nh cng l xo t
322.2.1. Phng php th nghim nn tnh ti hin trng
332.2.1. Phng php tra bng
352.2.2. Phng php s dng cc cng thc thc nghim
372.2.2. Phng php thc hnh xc nh h s nn
392.3. Xc nh cng l xo cc
392.3.1. Phng php nn tnh cc ti hin trng
402.3.2. Phng php tnh theo m un bin dng nn [7]
412.3.3. Phng php xc nh h s nn cc da theo ln cc n
452.4. Xy dng m hnh tnh mng b - cc
49CHNG 3 : V D MINH HA
493.1. Gii thiu cng trnh
493.1.1. c im cng trnh
493.1.2. iu kin a cht cng trnh
493.1.3. Ti trng tc dng ln mng
503.2. Tnh ton cc s liu u vo
503.2.1. Sc chu ti cc
513.2.2. Sc chu ti cc n xc nh theo cng thc ca Schmertmann SPT[1]
513.2.3. Xc nh cng l xo cc theo phng php truyn ti trng Gambin [6]:
543.2.4. Xc nh cng l xo cc theo mun bin dng nn[7]
543.2.5. Xc nh cng l xo t
563.3. Xy dng m hnh tnh
583.3.1. M hnh 1
633.3.2. M hnh 2
663.3.3. M hnh 3
70CHNG 4 : TNH TON MNG B CC C XT N TIN CY S LIU T NN
704.1. C s l thuyt
714.2. Cc bc tnh ton
734.3.1. Cc gi thit tnh ton v s liu u vo
734.3.2. Tnh ton tin cy
85KT LUN V KIN NGH
96TI LIU THAM KHO
DANH MC CC HNH V
4Hnh 11: Cu to mng b cc
6Hnh 12 : Mt bng kt cu mng ta nh 97- Lng H
8Hnh 13 : S lm vic ca mng b cc (Poulos, 2000)
9Hnh 14: Cc ng ng ng sut ca cc n v nhm cc [1]
13Hnh 15 : Biu quan h ti trng - ln theo cc quan im thit k
15Hnh 16: S tnh mng tuyt i cng
16Hnh 17: S tnh mng mm
18Hnh 18: M hnh tnh ton h mng b-cc theo phng php lp
19Hnh 19: M hnh nn Winkler
22Hnh 110: Mi quan h ln-ti trng trong m hnh nn bn khng gian n hi:
24Hnh112: ng cong P-Y v T-Z ca t [1]
26Hnh 113: M hnh tin nh
26Hnh 114:M hnh ngu nhin v hm khng ph hoi ca A.R. Rgianitsn [5]
30Hnh 21: M hnh 1
31Hnh 22: M hnh 2
33Hnh 23 : Quan h gia ng sut v ln thu c bng th nghim nn t hin trng
38Hnh 24 : Biu xc nh h s IF [9]
40Hnh 25: th S=f(P) theo kt qu th cc bng ti trng tnh
45Hnh 26: S phng php truyn ti trng Gambin [6]
58Hnh 31: S b tr cc trong i
59Hnh 32 : Biu bin dng b mng
59Hnh 33: Mmen M11
60Hnh 34: Mmen M22
60Hnh 35 : Phn lc gi ta l xo
63Hnh 36: M hnh mng 2
64Hnh 37: Bin dng ca b mng
64Hnh 38: Mmen M11
65Hnh 39: Mmen M22
66Hnh 310: Ti trng truyn xung cc
66Hnh 311: M hnh mng 3
67Hnh 312: M hnh mng 3 Phn lc u cc
68Hnh 313: M hnh mng vi s lng cc n = 35
80Hnh 41: Biu phn b sai s (M11)max
80Hnh 42: Biu phn b sai s (M11)min
81Hnh 43: Biu phn b sai s (M22)max
81Hnh 44: Biu phn b sai s (M22)min
82Hnh 45: Biu phn b sai s Pmax
82Hnh 46: Biu phn b sai s ((m)max
DANH MC CC BNG
34Bng 21: Bng tra h s nn theo K.X. Zavriev
34Bng 22: Bng tra gi tr Cz theo Terzaghi:
49Bng 31 : iu kin a cht cng trnh
50Bng 32: Bng gi tr ti trng tc dng ln mng
51Bng 33 : Bng tnh gi tr sc khng bn cc
53Bng 34: Bng tnh ln cc n theo phng php Gambin.
54Bng 35: Bng tnh cng l xo cc theo mun bin dng nn
57Bng 36: Bng thng k s liu u vo
61Bng 37: Kt qu tnh khi chiu dy b thay i
62Bng 38: Kt qu tnh khi khong cch cc thay i
62Bng 39: Kt qu tnh khi k n hiu ng nhm
68Bng 310 : Kt qu tnh khi tng s cc n = 35.
78Bng 41: Kt qu phn tch ni lc mng vi thng s u vo mang gi tr ngu nhin.
83Bng 42: tin cy ca ni lc vi n1 = 1,01
83Bng 43 : tin cy ca ni lc vi n2 = 1,03
M U
S cn thit ca ti
Mng cc ngy cng c s dng nhiu Vit Nam do nhu cu pht trin ca kinh t dn n nhu cu xy dng dn dng v h tng c m rng v pht trin khp cc vng min trn c nc.
Trong iu kin nc ta vic tnh ton thit k mng cc n nay vn cn s dng nhng m hnh tnh theo quan im c in cho rng cc ch c tc dng gim ln v gia c nn hoc cc chu ton b ti trng t b truyn xung.
Mc tiu v nhim v ca ti
Mc tiu ca ti l nghin cu phng php tnh ton mng b cc theo m hnh h s nn c k n tin cy ca s liu nn t. Vic tnh ton kt cu nn mng theo l thuyt tin cy v ang c quan tm nghin cu nhiu trn th gii nhng Vit Nam mi bt u c nghin cu trong thi gian gn y.
Vi mc tiu trn ti s cp n cc vn chnh nh sau:
- Nghin cu c s l thuyt v m hnh tnh mng b cc.
- Kho st tin cy gi tr ni lc trong kt cu mng, khi xem xt s liu nn t l cc bin ngu nhin.
Phng php v phm vi nghin cu
Phng php nghin cu ca lun vn l nghin cu l thuyt kt hp vi th nghim s trn m hnh ton.
Phm vi nghin cu ca ti: Mng cc cng trnh xy dng t trn nn t thin nhin. V d minh ha s dng s liu th nghim do m phng s hoc s liu th nghim t thc t.
Cu trc ca lun vn
Vi ni dung nh trn, bo co ca lun vn gm bn chng ni dung chi tit v phn kt lun.
+ Chng 1: Tng quan
+ Chng 2: Xy dng m hnh tnh mng b cc
+ Chng 3: V d minh ha
+ Chng 4: Tnh ton mng b cc c xt n tin cy ca s liu nn t.
+ Phn kt lun v kin ngh nh gi cc vn m lun vn gii quyt c, kh nng ng dng ca ti vo vic thit k cc cng trnh thc t, nhim v cn tip tc nghin cu trong giai on tip theo nhm xy dng hon chnh phng php tnh.
CHNG 1 : TNG QUAN1.1. Cu to v ng dng ca mng b-cc
1.1.1. Cu to ca mng b cc
Mng b cc l mt loi mng cc, cho php pht huy c ti a kh nng chu lc ca cc v tn dng c mt phn sc chu ti ca nn t di y b. Mng b - cc cn c gi l mng b trn nn cc. Mng b cc c rt nhiu u im so vi cc loi mng khc, nh tn dng c s lm vic ca t nn, pht huy ti a sc chu ti cc, chu c ti trng ln, cng ln, khng gian t do thng thong thun li cho vic b tr tng hm, lin kt gia b v kt cu chu lc bn trn nh vch, ct c cng ln ph hp s lm vic ca cng trnh. Mng b cc cu to gm hai phn: b v cc cc. - B hay i cc c nhim v lin kt v phn phi ti trng t chn kt cu cho cc cc, ng thi truyn mt phn ti trng xung t nn ti v tr tip xc gia y b v t nn. B c th lm dng bn phng hoc bn dm nhm tng cng chng un. - Cc cc lm nhim v truyn ti trng xung nn t di chn cc thng qua sc khng mi v vo nn t xung quanh cc thng qua sc khng bn. C th b tr cc trong i thnh nhm hay ring r, b tr theo ng li hay b tr bt k tu thuc vo mc ch ca ngi thit k, nhm iu chnh ln khng u, gim p lc ln nn y b hay gim ni lc trong b...
Cch b tr cc trong i thng theo nguyn tc trng tm nhm cc trng hoc gn vi trng tm ti trng cng trnh. Gii php ny c u im l ti trng xung cc c phn b hp l hn; tnh lm vic tng th ca nhm cc tt hn.
Hnh 11: Cu to mng b ccCc c th s dng cc ch sn hoc cc nhi.
- Cc ch sn thng gm hai loi:
+ Cc b tng ct thp (BTCT) c sn, c hoc khng c ng sut trc. Cc thng c dng hnh vung. Dng cc ny thng p dng cho c cng trnh c ti trng va v nh v chiu di cc hn ch, khong 30m. Cn cc ng sut trc c u im l sc chu ti ln, c th xuyn qua cc lp t ri c cht ln, tuy nhin loi cc ny cha ph bin nc ta.+ Cc thp (thp hnh ch H, hoc thp ng ch O). Do b dy tm thp mng, cc c th d dng xuyn qua cc lp t cng, ngi ta thng gia c thm mi cc. Vit Nam, ta thng h cc ch sn xung bng mt trong ba phng php:
+ Dng ba ng cc: thng gy chn ng v ting n ln. Hn na, kh c th ng cc qua lp t tt v cc thng b gy, v u cc. gim chn v gip qu trnh ng cc, ta c th khoan mi trc khi ng.
+ p cc bng kch thu lc v h i trng. c th p cc xung su thit k, ti trng p u cc phi vt qua hoc bng ti trng cc hn Pu ca t nn.
+ Rung: thng dng cho tng c, tng ngn.- Cc nhi:
Cc nhi l mt loi cc b tng c thi cng bng cch b tng ti vo mt h khoan trc.
So vi cc loi cc khc, cc nhi c lch s tng i mi. Nm 1908 n 1920, cc l khoan mi c ng knh nh 0,3m, di ch 6-12 m. Hin nay, ngi ta c th lm cc nhi m rng chn, s dng dung dch bentonite gi thnh h khoan. Cc nhi c s dng Vit Nam u nhng nm 1990. Kch thc ph bin ca cc nhi Vit Nam l : ng knh 1-2m, chiu di 40-70 m. Cc nhi thng p dng cho cc cng trnh c ti trng ln, nhng cng trnh xy chen khng th thi cng chn ng nh cc loi cc khc.1.1.2. ng dng mng b cc
Mng b cc thng c s dng tng i nhiu trong cc cng trnh xy dng. S d phi lm mng b cc v trng hp t yu rt dy, b tr cc theo i n hay bng trn cc khng . Cn phi b tr cc trn ton b din tch xy dng mi mang ti trng ca cng trnh. Hn na b cc s lm tng tnh cng tng th ca nn mng b p li s yu km ca nn t.
Nh dn dng: Ch yu l mng b trn cc nhi hoc barrette. Mng b cc thch hp vi kt cu ng, kt cu khung vch.
Hnh 12 : Mt bng kt cu mng ta nh 97- Lng H
Mt v d v cng trnh To nh 97 Lng H - ng a - H ni, mt bng 43,6 x 34,5m; kt cu khung-vch; s dng cc khoan nhi ng knh 1200. sc chu ti tnh ton cc n l 650 Tn; phn mng gm 65 cc c b tr khp nh. B mng dy 2.0 m.- Nh cng nghip: Ch yu l mng b trn cc ng hoc p. c im nh cng nghip l din tch mt bng ln, cu to a cht thng khng n nh; cc s dng trong cng trnh ny thng c tc dng gia c nn, gim ln lch v ln tuyt i.- Cng trnh cng, thu: Ch yu l mng b trn cc ng hoc p. c im ca cc cng trnh ny l chu ti trng nng, quy nh nghim ngt v ln tuyt i v ln lch. V d v cng trnh dng ny l cc u tu.1.2. C ch lm vic ca mng b ccc im ni bt ca mng b - cc l s nh hng tng h gia t v kt cu mng trong qu trnh chu ti theo bn nh hng sau:
( - S tng tc gia cc v t;
( - S tng tc gia cc v cc;
( - S tng tc gia t v mng b;
( - S tng tc gia cc v mng b;
Hnh 13 : S lm vic ca mng b cc (Poulos, 2000)
Nghin cu tc ng qua li khi k ti nh hng ca i cc, nn t di y i v cc cho thy c cu truyn ti trng nh sau:
+ S lm vic ca i cc: Ti trng t cng trnh truyn xung mng. i cc lin kt cc u cc thnh mt khi v phn phi ti trng tp trung ti cc v tr chn ct, tng cho cc cc. S phn phi ny ph thuc vo vic b tr cc cc v cng khng un ca i. mt mc nht nh n c kh nng iu chnh ln khng u (ln lch).
+ nh hng ca nn t di y i: Khi i cc chu tc ng ca ti trng mt phn c truyn xung cho cc cc chu v mt phn c phn phi cho nn t di y i. T l phn phi ny cn ph thuc vo cc yu t: cng ca nn t, chuyn v ca i, chuyn v ca cc v vic b tr cc cc.
+ nh hng ca cc: C ch lm vic ca cc l nh c h vo cc lp t tt pha di nn khi chu tc ng ca ti trng ng t i mng n s truyn ti ny xung lp t tt thng qua lc ma st gia cc vi t v lc khng mi cc lm cc chu ko hoc nn. Trong qu trnh lm vic cc cn chu thm cc tc ng phc tp khc nh: hiu ng nhm cc, lc ma st m ... Do c cng ln nn cc tip nhn phn ln ti trng t i xung, ch c mt phn nh do nn tip nhn. + S lm vic ca nhm cc:
S lm vic ca cc n khc vi s lm vic ca nhm cc. Khi khong cch cc cc kh ln (v d ln hn 6d) th cc lm vic nh cc n.
Xt cc v nhm cc trn hnh 1-4, cc ng cong trn hnh 1-4a th hin ng ng ng sut do cc n gy ra, cn hnh 1-4b, ta thy ng sut gia nhm cc s do ti trng truyn t nhiu cc ti, do ng sut di nhm cc ln hn ln. Nu mi cc trong nhm v cc n cng chu mt ti trng lm vic th ln ca nhm cc ln hn cc n.
a)
b)
Hnh 14: Cc ng ng ng sut ca cc n v nhm cc [1]Sc chu ti ca nhm cc cng nh hn cc n:
(1.1)
Trong :
( - h s nhm
N S lng cc trong nhm
Pnhu sc chu ti ca nhm cc Pu sc chu ti ca mt cc nKhi ng hoc p cc vo t ht th trng thi ri hoc cht va, t s cht ln, do ci thin c sc chu ti ca tng cc ( ( 1.Cn khi ng hoc p cc vo t dnh, cu trc t b xo trn, sc chu ti gim xung nhiu. Sau mt thi gian cc ngh, sc khng ct s phc hi dn nhng t khi phc hi c 100%. V vy, ( ( 0,8-0,9.Nhn xt: S lm vic ca h i cc - cc - nn t l mt h thng nht lm vic ng thi cng nhau v tng tc ln nhau rt phc tp. S tng tc d ph thuc vo cng khng un ca i cc, cng ca nn t (y i), cng ca cc (kh nng chu ti v b tr cc). Nh vo s tng tc m ti trng c phn phi xung nn t gy ra chuyn v ca nn, chuyn v ny phn phi li ti trng cho kt cu bn trn t c tc dng iu chnh chnh ln, gi c n nh khng gian cho mng. Tuy nhin, khng phi lc no gia t v kt cu mng cng c cc dng tng tc trn, do ty thuc vo s liu thc t ca mng v t m ta c th gi thit loi b mt dng tng tc no n gin cho tnh ton.1.3. Cc quan im thit k hin nayHin nay khi thit k cc loi mng dng bng cc, b cc thng c mt s quan im tnh ton nh sau:
1.3.1.Quan im cc chu ti hon ton
Theo quan im ny, cc cc c thit k nh mt nhm cc tip nhn hon ton ti trng ca cng trnh m khng k ti s tham gia chu ti ca nn t di i cc. Trong tnh ton, h mng cn tnh nh mng cc i thp vi nhiu gi thit gn ng nh:
Ti trng ngang do nn t trn mc y i tip thu
i mng tuyt i cng, ngm cng vi cc cc, ch truyn ti trng ng ln cc cc, do cc ch chu ko hoc nn Cc trong nhm cc lm vic nh cc n, v cc chu ton b ti trng t i mng.
Khi tnh ton tng th mng cc th coi h mng l mt khi mng quy c.
Tnh ton theo cch ny c u im l n gin, thin v an ton v c hng dn chi tit trong cc gio trnh v nn mng hin nay. ln ca mng tnh ton theo phng php ny nh, s dng nhiu cc v thng h s an ton cao, cha pht huy c ht sc chu ti ca cc. Nh vy, ta thy n c nhc im l qu thin v an ton v khng kinh t, o y l mt phng n lng ph.
Nhn xt: Quan im tnh ton ny ph hp cho nhng kt cu mng cc c chiu dy i ln, kch thc i nh, hoc nn t di y i yu, c tnh bin dng ln. Khi , ta c th b qua s lm vic ca t nn di y i v xem ton b ti trng cng trnh do cc chu.1.3.2. Quan im b chu ti hon ton
Theo quan im ny, b c thit k chu phn ln ti trng ln mng, cc cc ch nhn mt phn nh ti trng, c b tr hn ch c v s lng sc chu ti vi mc nh chnh l gia c nn, gim trung bnh v ln lch. ln ca mng trong quan im ny thng ln, vt qu ln cho php, ngoi ra vi ti trng cng trnh ln, tnh theo quan im ny thng khng m bo sc chu ti ca nn t di mng.
Nhn xt: Quan im thit k ny ph hp vi nhng cng trnh t trn nn t yu c chiu dy khng ln lm. Khi lin kt gia cc v i khng cn phc tp, v mc ch cc gia c nn v gim ln l chnh.1.3.3. Quan im b - cc ng thi chu ti
Theo quan im ny, h kt cu mng i - cc ng thi lm vic vi t nn theo mt th thng nht, xt n y s tng tc gia cc yu t t-b-cc. Trong quan im ny, cc cc ngoi tc dng gim ln cho cng trnh, cn pht huy ht c kh nng chu ti, do cn t cc hn, chiu di cc nh hn. Khi cc pht huy ht kh nng chu ti, th mt phn ti trng cn li s do phn b chu v lm vic nh mng b trn nn thin nhin.
Trong quan im ny, ln ca cng trnh thng ln hn so vi quan im cc chu ti hon ton nhng v tng th, n vn m bo nm trong quy nh vi mt h s an ton hp l, do quan im tnh ton ny cho hiu qu kinh t tt hn so vi quan im u. Tuy nhin, qu trnh tnh ton cn s dng cc m hnh phc tp hn, do hin nay quan im ny cha c ph bin rng ri.
Hnh 15 : Biu quan h ti trng - ln theo cc quan im thit kNhn xt:
Quan im thit k th nht thin v an ton, nhng khng kinh t, nn p dng khi cng trnh c yu cu cao v khng ch ln. Quan im thit k th hai, mng b trn nn thin nhin l phng n kinh t nhng ln ca b l rt ln v thng nn t khng sc chu ti vi cng trnh c ti trng ln. Quan im thit k th ba, dung ha c cc u, nhc im ca hai quan im trn, nn trng hp cng trnh khng c yu cu qu cao v ln, c th s dng tng tnh kinh t.1.4. Tng quan v cc phng php tnh ton mng b - cc
1.4.1. Cc phng php n gin
Phng php tnh ton nh mng cc i thp
Phng php ny tnh ton da trn quan nim tnh, xem ton b ti trng cng trnh do cc chu.
Chiu su chn mng hm phi tho mn iu kin ti trng ngang H c cn bng vi p lc t b ng ca t trong phm vi i cc, cho cc cc khng b tc dng ca lc ngang m ch hon ton lm vic chu nn.
Mmen ngoi lc c cn bng vi cc phn lc ti u cc vi cc ta (xi, yi) ca cc.
Ring i vi mng ch c mt cc t ng tm th cn phi xem l cc n chu mmen v ti trng ngang.
Do iu kin xem nh l mng cc i thp l cc phi c b tr trn 2 cc tr ln, chng li mmen.
Phn lc trn u cc c ta (xi,yi) l:
(1.2)Trong :
Mx mmen theo phng trc y
My mmen theo phng trc x
xi, yi - to ca cc th i so vi v tr ti trng
Phng php tnh ton nh mng b
Phng php ny tnh ton da trn quan nim tnh, xem ton b ti trng cng trnh do b chu lc, cc ch c tc dng gia c nn v gim ln.
Theo phng php ny, tu theo cng ca b m ta xem b nh mng cng tuyt i hoc mng mm.
Mng tuyt i cng
Khi xem mng l tuyt i cng, phn lc di y mng xem nh phn b u theo quy lut ng thng.
Khi , phn lc nn xc nh theo cng thc ca sc bn vt liu:
(1.3)
Hnh 16: S tnh mng tuyt i cng
Trong :Jx, Jy l mmen quan tnh ca tit din mng vi trc y, x.
eL, eB l lch tm ca trng tm mng v tm lc theo phng cnh L v cnh B.
Mng mm
Khi kch thc mng ln, cng ca mng gim, phn lc nn khng phn b theo quy lut bc nht, ta phi tnh mng nh mng mm.
tnh mng mm, ta c th dng phng php tnh ca dm trn nn n hi hoc n gin hn l s dng m hnh h s nn Winkler trong thay th t nn bng h l xo c lp, c cng l xo K = Cz.F vi cc l xo gia mng hoc K = Cz.F1 vi cc l xo bin mng. Trong Cz l h s nn ca t.
Hnh 17: S tnh mng mm
M hnh ny ch ng khi tnh ton phn lc t nn bn thn kt cu mng m khng dng tnh ln. tnh ln mng, ta phi dng cc phng php khc ca c hc t nh cng ln cc lp phn t hoc lp tng ng.1.4.1. Cc phng php c k n s tng tc cc- t nn v b-t nn
Phng php lp ca H.G. Poulos (1994)[2]Cc phng php thuc nhm ny c xt dn c im ni bt ca mng b - cc l s nh hng tng h gia t v kt cu mng theo bn nh hng sau:
- S tng tc gia cc v t;
- S tng tc gia cc v cc;
- S tng tc gia t v mng b;
- S tng tc gia cc v mng b;
S tnh mng b - cc: Mng b c m hnh bng phn t dm hoc bng phn t tm hoc c hai. Mng b lin kt vi cc l xo tng trng cho cc v cho t ti cc im nt. Cc l xo tng trng cho cc v t c nh hng tng h gia b, cc.Trnh t phn tch theo phng php ny:
Bc 1: Xc nh cng l xo cc c xt n tng tc cc-cc v nn-cc.
Bc 2: Xc nh cng l xo t c xt n tng tc cc-t v phn lc nn - t.
Bc 3: Tu vo sc chu ti cc v t, gi thit t l phn phi ti trng cho cc v b.
Bc 4: Sau khi bit phn lc cc v phn lc nn, xc nh cng l xo cc v t theo Bc 1 v 2.
Bc 5: Gn l xo vo m hnh mng b-cc, thm ti trng cng trnh.
Bc 6: Gii bi ton, xc nh li phn lc cc v nn.
Bc 7: Gii lp bi ton t bc 3 n khi phn lc cc v nn hi t.
Bc 8: Kim tra ln cho php.
Nhn xt: Phng php ca H.G.Poulos cho kt qu tng i hp l khi xt n cc qu trnh tng tc ln nhau ca h b-cc v nn t, phng php ny cng cho php s dng cc phn mm phn t hu hn trn my tnh gii bi ton. Tuy nhin, phng php ny cn cha xt n ln tng i ca b v cc. c bit, khi ln ca b qu ln so vi cc dn n ln tng th ca h khng tha mn.
T nhn nh trn, ta c th thay i li mt s bc trong phng php lp ny kt qu hp l hn v c xt n ln tng th ca h mng.
Hnh 18: M hnh tnh ton h mng b-cc theo phng php lpPhng php lp c chnh sa
Bc 1: Tnh ti tng ti trng cng trnh truyn v h mng b-cc Q.Bc 2: Gi thit ti trng do phn b chu: QbBc 3: Tnh ti trng truyn v h cc: Qc = Q- QbBc 4: Sau khi bit ti trng truyn v cc v nn, xc nh cng l xo cc v t theo Bc 1 v 2 ca phng php lp H.G.Poulos.
Bc 5: Cn c vo ti trng do b m nhn, tnh ln cho mng b Sb .
Bc 6: Cn c vo ti trng do cc nhn, tnh ln cho mng cc Sc .
Bc 7: Kim tra iu kin Sb < Sc.
Bc 8: Nu khng tha mn iu kin trn lp li bc 2 vi lng cc tng dn.Bc 9: Gn l xo vo m hnh mng b-cc, thm ti trng cng trnh, tnh kt cu mng.
1.5. Cc dng m hnh bin dng ca nn t
Hin nay c rt nhiu dng m hnh nn m phng s lm vic tip xc ca mng v t nn, khi tnh ton c th s dng cc m hnh nn khc nhau. Tuy nhin, khi p dng vo tnh ton, cn hiu r phm vi p dng ca tng m hnh nn vo tng trng hp c th. M hnh khc nhau th kt qu tnh ton cng khc nhau, nhiu khi s khc bit l rt ln. Vic s dng sai m hnh, sai quan im tnh ton c th mang li s c cho cng trnh.
1.5.1. M hnh nn Winkler
M hnh nn Winkler cn gi l m hnh nn bin dng cc b, l m hnh n gin v ph bin nht vi thng s duy nht ca t c a vo tnh ton l h s nn Cz.
c im ca m hnh ny l ch xt n bin dng n hi ngay ti ni c ti trng ngoi tc dng, m khng xt n bin dng n hi ca t vng ln cn, b qua c im t nh mt vt liu c tnh dnh v tnh ma st. M hnh bin dng tng ng vi l thuyt ny l mt nn n hi gm mt h l xo c bin dng lun lun t l vi p lc tc dng ln chng.
Hnh 19: M hnh nn Winkler
cng l xo k, vi k = Cz.F, trong F l din tch phn nh hng ca mt y mng vi nt ang xt, theo quy tc phn phi trung bnh.M hnh nn Winkler c u im l n gin, tin dng trong tnh ton, c th s dng nhng phn mm phn t hu hn c sn, thit k gn ng vi thc t, c bit l vi nhng nn t yu, c lc dnh v lc ma st nh, khi nh hng ca vng ln cn xung quanh vng chu ti nh, c th b qua.
Bn cnh , m hnh nn ny cng c nhng nhc im:
Khng phn nh c s lin h ca t nn, khi chu ti, t c th li ko hay gy ra nh hng ra cc vng ln cn.
Khi nn ng nht th ti trng phn b u lin tc trn dm, th theo m hnh ny, dm s ln u v khng bin dng, nhng thc t th dm vn b vng gia, nn nh hng ra xung quanh cng nh ln nhiu hn so vi u dm.
Khi mng tuyt i cng, t ti trng i xng th mng s ln u, ng sut y mng phn b u, nhng theo cc o c thc t th ng sut cng phn b khng u.
H s nn Cz c tnh cht quy c, khng phi l hng s vi ton b t nn di mng.Nhn xt: M hnh nn Winkler thng p dng tt cho t yu, th hin tnh bin dng ti ch, khi chu ti, khng lan truyn ra xung quanh. M hnh ny ch dng tnh bn thn kt cu mng, khng dng tnh ln, v bi ton tnh ln l bi ton phc tp, lin quan n nhiu qu trnh nh thot nc l rng, t bin, c kt v trong tnh ton phi s dng nhiu thng s c l ca t, ch khng th ch da vo h s nn Cz.
Ngoi ra, m hnh nn Winkler thay th t bng cc l xo c lp, tng i n gin cho tnh ton v cho php xy dng s tnh kt cu mng trong cc phn mm phn t hu hn thng dng hin nay.1.5.2. M hnh bn khng gian n hi
M hnh ny ng dng l thuyt n hi t li gii ca Boussinesq v Flamant, nn t c xem l mt bn khng gian bin dng tuyn tnh, c trng bi m un bin dng E0 v h s n hng (Bi ton Boussinesq
Bi ton Boussinesq xt cho trng hp nn nm trong trng thi ng sut- bin dng khi.Nh nhiu ti liu cho thy, khi c xt n chiu dy gii hn ca lp t, l thuyt tng bin dng n hi em li nhng kt qu ph hp vi thc t hn v bin dng ca mt t cc vng ln cn vng chu ti tt nhanh hn so vi khi xem nn t l mt na khng gian n hi c chiu dy v tn.Trong trng hp nn l mt na khng gian n hi th theo l lun ny, chuyn v thng ng Y ca mt im bt k trong t i vi ta x,y,z nm cch im t mt lc tp trung P trn b mt mt khong r c th xc nh theo biu thc sau y ca l thuyt n hi:
(1.4)Trong :
E0 - m un bin dng ca t;
- h s Poisson ca t;i vi cc im nm trn mt t th biu thc tnh ln c th rt ra t biu thc vi z=0:
(1.5)Hoc nu t c gi l h s na khng gian n hi, ta c:
(1.6)
Hnh 110: Mi quan h ln-ti trng trong m hnh nn bn khng gian n hi:
a. Bi ton Boussinesq
b. Bi ton Flamant Bi ton Flamant
Bi ton Flamant xt cho trng hp nn nm trong trng thi ng sut - bin dng phng, khi ln tng i Y ca mt im nm trn mt t, cch lc tp trung P mt khong r so vi mt im nm trn mt t, cch lc tp trung P mt khong d, tnh theo cng thc
(1.7)C th thy rng, theo biu thc, ln ca cc im trn mt t ti nhng vng ln cn quanh din tch chu ti khng phi bng khng, m c mt gi tr nht nh tc l ph hp vi thc t hn so vi kt qu tnh theo l thuyt bin dng n hi cc b Winkler. Rt tic l hin nay, m hnh ny cha th a vo cc phn mm tnh ton kt cu ph thng nh SAP, SAFE do mc p dng cn hn ch.1.6. Tnh ton cc lm vic ng thi vi nn
Hin nay, cng vi s pht trin ca my tnh v phng php phn t hu hn, ngi ta thng xy dng cc m hnh tnh ton cc lm vic ng thi vi nn.
Khi cc chu ti, di tc dng ca ti trng ng (ko hoc nn), nn t s tng tc vi cc qua nhng gi n hi theo phng ng. Quan h gia phn lc (k hiu l t ) v chuyn v ng ca cc gi (k hiu l z) l t = kz.z, vi kz l cng ca gi n hi theo phng ng. Biu quan h gia t v z gi l ng cong t-z.Di ti trng ngang, nn t s tng tc vi cc qua nhng gi n hi theo phng ngang. Quan h gia phn lc (k hiu l y) v chuyn v ngang ca cc gi n hi (k hiu l y) l p = ky.y, vi ky l cng ca gi n hi theo phng ngang. Biu quan h gia p v y gi l ng cong p-y [1]. Nh vy, phng php ny s dng phn t hu hn kt hp vi m hnh nn Winkler. Trong phng php ny, ngi ta chia cc thnh nhiu phn t, trn mi on, tng tc gia cc vi t c m t bng cc gi n hi.
Do khi lng tnh ton rt ln, ta cn phi s dng my tnh gii bi ton trn thng qua cc phn mm s dng phng php phn t hu hn.
ng cong P-Y v ng cong T-Z: Hnh dng v dc ca ng cong khng nhng ph thuc vo tnh cht ca t, m cn ph thuc vo su on cc ang xt, kch thc cc, mc nc ngm v dng ti trng (tnh hay ng).
Hnh112: ng cong P-Y v T-Z ca t [1]a) ng cong P-Y ca t st yu chu ti trng tnh
b) ng cong T-Z ca t st yu chu ti trng tnhNhn xt: Qua cc phn tch trn, ta thy mng b cc l mt h mng rt phc tp, s lm vic ca mng ph thuc vo s tng tc gia cc thnh phn: cc - t b, ch cn mt trong cc thnh phn ny thay i, s lm vic ca mng lp tc thay i theo. Trn thc t, cc thng s c trng cho cc thnh thn ny, c bit l s liu t nn, khng phi l mt gi tr c nh m c tnh cht ngu nhin, phn tn. Do , kt qu tnh chnh xc, ta cn xt n tnh cht ngu nhin ca cc gi tr ny trong tnh ton kt cu. Ni mt cch khc, l xt n tin cy khi tnh ton kt cu mng.1.7. Tng quan v l thuyt tin cy nh gi mc lm vic an ton ca kt cu cng trnh l mt trong nhng nhim v quan trng nht ca cng tc thit k v chn on k thut. Ni dung nh gi dn n dng bi ton so snh hai tp hp. Tp th nht S, cha cc thng tin c trng cho trng thi lm vic ca kt cu v tp th hai R cha cc thng tin c trng cho nng lc ca kt cu, c thit k theo mt tiu chun cht lng no .
1.7.1. Cc m hnh tnh
* M hnh tin nh:
Thc hin vic nh gi thng qua t s n = R/S hoc hiu s M = R-S.
iu kin an ton khi n>1 hoc M>0. Ngc li th khng an ton. Tn ti mt trng thi phn chia gia an ton v khng an ton khi n=1 hoc M=0, mang tnh l thuyt.Phn tch: m hnh ny n gin v tnh ton v gi tr R v S ly trung bnh thnh cc s c th, nhng cn nhc im l cha nh gi c chnh xc s lm vic an ton ca kt cu, v h s an ton n c th cao nhng cha chc kt cu an ton nu nh sai s ca R v S ln.
Hnh 113: M hnh tin nh
* M hnh ngu nhin:
Vi quan nim hai tp R v S mang bn cht ngu nhin, nn vic nh gi thc hin theo l thuyt xc sut, s liu bn trong v tc ng bn ngoi ln kt cu x l theo thng k ton hc. Kt qu ca nh gi th hin qua xc sut an ton Prob(M>0) hoc xc sut ph hoi Prob(M0,04m th tr s tiu chun ca sc chu ti Ptc, ly theo th trn ng vi ( =0,04m.
Nh vy, cng l xo mt cc c th xc nh theo kt qu nn tnh cc nh sau:
(2.10)
Hnh 25: th S=f(P) theo kt qu th cc bng ti trng tnh
Nhn xt: Phng php ny cho ta kt qu chnh xc, v n biu th quan h gia ng sut nn v ln cc thc t ti hin trng, ti chnh v tr t cc, khng b sai lch do cc nhn t khch quan. Tuy nhin, s lng cc nn tnh ti hin trng khng nhiu, ch chim 0,5% tng s cc. Ngoi ra, trong giai on thit k s b, thng thng ta cha c kt qu ca th nghim nn tnh cc.2.3.2. Phng php tnh theo m un bin dng nn [7]Phng php ny s dng kt qu ca th nghim xuyn SPT , cng l xo mi cc v thn cc c xc nh t m un bin dng ca nn t E0, gi tr ca E0 c xc nh t ch s SPT N ng vi tng v tr kho st.H s nn ti mi cc theo phng ng tnh nh sau:
- Cc ng: Kv = EoD-3/4
(2.11)- Cc khoan nhi: Kv = 0.2 EoD-3/4
(2.12)Trong :
Kv: H s nn mi cc theo phng ng (kgf/cm3)
: H s iu chnh mi cc, = 1
D: ng knh mi cc (cm).
Eo: M un bin dng nn (kgf/cm2)
Eo = 25N; (N: Gi tr xuyn tiu chun).
H s nn dc thn cc theo phng ng tnh nh sau:
- Cc ng trong t ri: ksv = 0.05 EoD-3/4
(2.13)- Cc ng trong t dnh: ksv = 0.1 EoD-3/4
(2.14)- Cc khoan nhi: ksv = 0.03 EoD-3/4
(2.15)
Trong :
ksv: h s nn thn cc theo phng ng (kgf/cm3)
H s nn ngang thn cc tnh nh sau:
kh = 0.2 EoD-3/4
(2.16)
kh: H s nn ngang thn cc (kgf/cm3).
Nhn xt: Phng php ny s dng kt qu ca th nghim xuyn tiu chun SPT nn c tin cy kh cao. Ngoi ra theo nghin cu ca Vin khoa hc cng ngh Giao thng vn ti, sai s tnh c t phng php ny so vi kt qu nn tnh l khng nhiu, t 10-12% v thin v an ton. Tuy nhin, cc h s a vo t cc cng thc trn cha c kim nghim trn quy m ln, s lng cc nhiu nn cn nhiu vn cha hp l. Ngoi ra, cng thc trn cng cha xt c nh hng cng ca cc n gi tr h s nn.2.3.3. Phng php xc nh h s nn cc da theo ln cc nNguyn tc ca phng php ny l xc nh ln ca cc S di ti trng P theo m hnh v phng php m ta xem l thch hp v n gin. Sau xc nh cng l xo tng ng ca cc theo cng thc bit .Xc nh ln ca cc n theo Phng php Vesic [11] ln ca cc n gm ba thnh phn nh sau:
(2.17)Trong : S1- Bin dng n hi ca bn thn cc
S2- ln ca cc do ti trng truyn ln t di mi cc
S3- ln ca cc do ti trng truyn ln t dc thn cc.
Bin dng n hi ca bn thn cc S1 (tnh ton nh thanh chu nn) c xc nh nh sau:
(2.18)Trong : Ap - din tch tit din cc
Ep-M un n hi ca vt liu ch to cc
L - Chiu di cc
Qb- Ti trng do mi cc chu
Qs - Ti trng do thn cc chu
( - H s ph thuc vo s phn b ma st bn, nu ma st bn phn b u th (=0,5; Nu cng xung su, ma st bn cng ln th (=0,67.Vy sc chu ti ca cc Qc=Qb + (Qs ln ca cc do ti trng truyn ln t di mi cc S2 tnh ton nh sau:
(2.19)Trong : qb- Sc khng mi n v ti trng lm vic qb.Ap=Qb
B - ng knh cc trn hoc cnh cc vung
( - H s Pot xng ca t di mi cc
Esb- Mun bin dng ca t di mi cc
( - H s tu thuc hnh dng cc,
(=0.79 vi cc vung;
(=0.88 vi cc trn;
hoc c th ly (=0.85 vi mi loi cc. ln ca cc do ti trng truyn ln t dc thn cc S3 tnh ton nh sau:
(2.20)Trong : qs- Sc khng bn n v ti trng lm vic, tnh trung bnh cho ton b cc: qs.u.L = Qs.
y: L-chiu di cc
u chu v cc
Is H s ph thuc mnh ca cc
( - H s Pot xng trung bnh ca t cc thn cc.
Ess- Mun bin dng trung bnh ca t dc thn cc.
Nhn xt: Phng php ca vesic ny sinh vn l phi xc nh c sc khng bn v khng mi thc ca cc ti trng lm vic, cn c cc s liu th nghim c th. Nu khng, ta cn s dng phng php tnh lp xc nh gn ng t l huy ng sc khng bn v mi so vi sc khng bn v mi cc i.Xc nh ln ca cc n theo Phng php Gambin [6]: Da theo nguyn l truyn ti trng.
Chia cc thnh n on. Tnh ton c bt u t mi cc, di 1 p lc tc dng vo t, gi thit ban u l 1 (to ra ln s1). Ta tnh ton chuyn dn t di ln trn n on cc th i, c cc thnh phn: + ng sut php tuyn i tc dng y on cc th i v nh on cc i -1.
+ ln si y on cc i.
+ ng sut ct cc t i thnh on cc th i, do ln si gy ra.
+ ng sut php tuyn i+1 tc ng ln u on cc th i, c tnh n ma st thnh on th i c xc nh theo biu thc:
(2.21)Trong : R l bn knh cc
Nu ta gi l bin dng ca vt liu on cc th i, th ln (Si+Hi ) chnh l ln chuyn ln y on th i +1. C nh th tip tc tnh ln cc on pha trn cho n nh cc s tm c gi tr ti v u cc Q tng ng.
So snh gi tr Q va tm c v gi tr ti trng lm vic theo thit k, tnh lp cho n khi hi t v gi tr Q th dng li.
Hnh 26: S phng php truyn ti trng Gambin [6] ln cc n c k n hiu ng nhm cc [6]:
Theo vesic, nhm cc d kin, da trn ln cc n tnh theo cng thc:
(2.22)Trong : Snh- ln nhm cc
Sd - ln cc n
B* - chiu rng tnh gia hai mp ngoi nhm cc
B - cnh cc vung hoc ng knh cc trn.Hoc c th tnh theo cng thc
Snh = Rs. Sc
Vi Sc l ln cc n
Rs l h s thc nghim
2.4. Xy dng m hnh tnh mng b - cc Sau khi xc nh c cng l xo thay th cho phn t t v cc di b, ta xy dng m hnh tnh mng b cc theo cc bc sau:Bc 1:
S b chn chiu di cc, chiu dy b v tnh sc chu ti cc.Tnh cng l xo ca cc v h s nn t.
Bc 2:Ti trng cng trnh truyn xung h cc theo nguyn tc:
Trong (QCT - Tng ti trng cng trnh
(Qcc - Tng ti trng truyn v h cc
(Qbe - Tng ti trng truyn v b.
Xc nh sc chu ti ca nn t di b: [(m]Ti trng truyn v b:
(Qbe = Am . (m
Trong Am - din tch mng b
(m - ng sut y mng, c th ly ~ 0.5.[(m]Bc 3: Tnh phn ti trng cng trnh truyn v h cc
(Qcc = (QCT - (Qbe
S b chn s cc cn b tr:
Trong : Qc - sc chu ti mt cc.
Bc 4: B tr cc iBc 5: M hnh ho h kt cu mng b - cc bng phng php phn t hu hn:
+Mng b c thay th bng phn t shell. i c chia thnh li vung hoc ch nht.
+Cc c thay th bng cc gi n hi spring c cng Kc tng ng theo m hnh 1 hoc phn t thanh gm nhiu on theo m hnh 2
+ Nn t c thay th bng cc gi n hi c cng Kd tng ng.
Bc 6: Gii bi ton, xc nh c ng sut y mng v phn lc u cc tng ng.
Kim tra cc iu kin :
(ttm < [(m]
Qc < [Qc]
Sbe < Scoc
Sbe < [S]
Trong : [(m] ng sut cho php ti y mng
[Qc] sc chu ti tnh ton ca cc
Sbe, ln ca mng b .Scoc, ln ca cc c xt n hiu ng nhm.
Nu mt trong cc iu kin trn khng t, ngha l s lng cc qu t, cn gi thit li s lng cc v tnh lp t bc 3.2.5. Phn mm SAP 2000 v9.03
SAP 2000 l mt phn mm phn t hu hn, c pht trin bi cng ty COMPUTER and STRUCTURE INC (CSI). T khi ra i t nm 1970 n nay, phn mm ny ngy cng hon thin, khng nhng phn tch kt cu tuyn tnh m c phi tuyn.
Kh nng ca phm mm SAP2000:
- Sap2000 cung cp nhiu tnh nng mnh m t cc bi ton kt cu ph bin trong thc t k thut nh: cu, p chn, bn cha, cng trnh nh..
- Phn mm c kh nng tnh ton cc phn t: thanh dn, dm, tm v, phn t khi.
- Vt liu tuyn tnh hoc phi tuyn.
- Lin kt bao gm: lin kt cng, lin kt l xo (spring), lin kt cc b kh bt cc thnh phn phn lc.
- Ti trng gm: lc tp trng, p lc, nh hng ca nhit , ti trng ph gia tc, ti trng di ng .
- Kh nng gii cc bi ton ln khng hn ch s n, tc gii nhanh v n nh.Nhn xt: Vi rt nhiu kh nng mnh v tnh n nh cao, phn mm SAP2000 ang l mt trong nhng phn mm tnh kt cu ph bin nht nc ta hin nay. Kh nng ca phn mm ny hon ton p ng c th m hnh ha v gii bi ton mng b cc theo phng php xy dng trn.CHNG 3 : V D MINH HA3.1. Gii thiu cng trnh
3.1.1. c im cng trnh
Cng trnh a vo v d minh ha l mt cng trnh nh chung c kt hp vn phng cao 18 tng vi 1 tng hm. Mt bng ch nht 16x33m, tng chiu cao 60,9m.Gii php kt cu, s dng h khung vch chu lc.
3.1.2. iu kin a cht cng trnhBng 31 : iu kin a cht cng trnhSTTTn tCao mt lp (m)Dy
(m)((kN/m3)W(%)(()c
(kN/m2)N30N60
1st pha1.51718.12515172015
2ct pha-15.524.519.52017122418
3Ct trung-40017.937.528143728
Ghi ch: Lp ct trung cha kt thc su kho st: -50 m.Ct y mng su -5 m so vi mt t t nhin.Mt bng cng trnh kh nh, ti trng li tng i ln nn s b chn phng n cc khoan nhi ng knh d = 0.8 m , chiu di cc d kin 30 m, cm su vo lp ct pha s 2 mt on 18 m.3.1.3. Ti trng tc dng ln mng
M hnh tnh cng trnh trong etab m phng cng trnh lin kt vi phn mng bn di thng qua cc gi ngm. Nh vy, ta s s dng ni lc chn ct, vch a vo tnh ton h mng b - cc.T hp ti trng nguy him nht a vo tnh ton :Bng 32: Bng gi tr ti trng tc dng ln mngTABLE: Joint Loads - Force
JointLoadCaseCoordSysF1F2F3M1M2M3
TextTextTextTonTonTonTon-mTon-mTon-m
455TH6GLOBAL00-867.67650.815771.42760
456TH6GLOBAL00-547.79150.407894.588720
457TH6GLOBAL00-869.20610.968731.42760
458TH6GLOBAL00-512.20340.509864.180840
459TH6GLOBAL00-818.73011.631551.42760
460TH6GLOBAL00-510.8778-1.223664.180840
461TH6GLOBAL00-807.20734.38478-0.152960
462TH6GLOBAL00-398.913-6.30592-5.812380
679TH6GLOBAL00-869.20610.968731.42760
680TH6GLOBAL00-512.20340.509864.180840
681TH6GLOBAL00-818.73011.631551.42760
682TH6GLOBAL00-510.8778-1.223664.180840
683TH6GLOBAL00-807.20734.38478-0.152960
684TH6GLOBAL00-398.913-6.30592-5.812380
787TH6GLOBAL00-547.79150.407894.588720
788TH6GLOBAL00-512.20340.509864.180840
789TH6GLOBAL00-510.8778-1.223664.180840
790TH6GLOBAL00-398.913-6.30592-5.812380
839TH6GLOBAL00-512.20340.509864.180840
840TH6GLOBAL00-510.8778-1.223664.180840
841TH6GLOBAL00-398.913-6.30592-5.812380
3.2. Tnh ton cc s liu u vo3.2.1. Sc chu ti cc Sc chu ti cc theo vt liu
Qv = ( (Rb.Fb + Ra.Fa)
(3.1)Trong : Qv: Sc chu ti cc theo vt liu
(: Hs un dc, (=1.
Rb,Fb: Cng b tng v din tch tit din cc, M300 c Rn=13000 kN/m2.
Ra,Fa : Cng thp v din tch thp, Thp dc dng thp nhm AII, c Ra=280000 kN/m2, hm lng thp (=0.8%.Vy Qv =1.(13000.3,142.1/4 + 280000.0.8.10-2.3,142.1/4 ) = 7656KN3.2.2. Sc chu ti cc n xc nh theo cng thc ca Schmertmann SPT[1]Sc khng bn:Bng 33 : Bng tnh gi tr sc khng bn ccSTTTn tN60Dy hi (m)Sc khng n v trong t ct:
fi = 1.82 x N60 (kpa)Sc khng n v trong t st, st pha:
fi = 2xN60x(110-N60)/47.86 (kpa)Sc khng bn: Qf=
Fi.hi.(.d
(kN)
1st pha151259.5492286.66945
2ct pha181869.2023986.02591
3Ct trung28050.9600
Tng cng Qf=6272.6953
Sc khng mi:
Mi cc nm trong lp t s 2 ct pha, sc khng mi xc nh theo cng thc: Qp = 153.N60.Fc = 1383.61 kNSc chu ti cc: < Qv3.2.3. Xc nh cng l xo cc theo phng php truyn ti trng Gambin [6]:
Xc nh cng l xo cc di sc chu ti cho php [Qc]
Chia cc lm n = 2 on, ng vi chiu di cc trong hai lp t.Phng php tin hnh nh sau
Bc 1:
Xc nh ln ca on cc u tin:- ng sut on mi cc, (1 gi thit trc, c th ly xp x gi tr:
(1 = Qp/Fc.Trong , Qp - ti trng truyn xung mi cc.
Fc - din tch tit din cc.- ln ca on th nht s1, xc nh theo cng thc:
Trong :
R = D/2. Vi D bn knh cc trn hoc cnh cc vung.Ep M un nn ngang ca t theo th nghim nn ngang PMT, nu khng c s liu c th ly theo tng quan vi kt qu th nghim SPT.
vi t st Ep = N/(0,8(1,1) , Mpa
st pha, ct pha Ep = N/3 , Mpa
ct Ep = N/(2(6) , Mpa( - H s hnh dng cc, vi cc trn ( = 1; vi cc vung ( = 1,12( - H s cu trc t, ( = Ep/E0 , vi E0 l m un bin dng ca tBc 2:Xc nh ln ca on cc th 2:
- Tnh ln vt liu ca on cc 1:
vi h1 l chiu di on cc th nht.
- Xc nh ln on cc th 2:
s2 = s1 + (h1- Xc nh ng sut tip nm bn thnh on cc 1:
Vi CL l h s tu thuc t s hi/R v loi ccvi cc on cc u c hi/R