67

chng iV

xy dng cng trnh trn nn t yu. $.1 .Khi nim chung: + Nn t yu: y l khi nim ch l tng i V: : Trng thi vt l ca t

Tng quan gia kh nng ca t v ti trng CT (Cng trnh) v c th yu vi CT ny nhng tt vi CT khc. Thng t: C kh nng chu lc: 0,5 1,0 kg/cm2,

C tnh nn ln mnh. t yu hu nh:

Bo ho, h s rng e ln (e>1); M duyn tng bin dng b ( E 50 kg/cm2); Tr s sc chng ct c; nh: = 48o; c = 0,05 0,1 kg/cm2

Bao gm: St mm, cc loi ct ht nh, mn, ri rc, than bn Vic xy dng CT trn nn t yu cn thit phi nghin cu x l c 3 b phn ca CT: KCCT (Kt cu cng trnh), Mng, Nn. $.2 Cc bin php v KCCT: KCCT c th b ph hng ton b hoc tng b phn do:

cc /k v bin dng khng c tho mn hoc p lc tc dng ln mt nn qu ln.

Nhm mc ch: Gim p lc ln mng Tng kh nng chu lc ca KCCT

Ta dng bin php: Dng VL(Vt liu) nh; KC nh Lm tng mm ca CT Tng cng cho KC 1.Dng VL nh v KC nh:

gim trng lng KCCT Lu : Nhng CT thng xuyn chu ti trng ngang ln th lc phi c bin

php m bo tnh n nh trt.

68

2.Tng mm ca CT: Lm mm CT (k c mng) s kh c cc /s ph thm pht sinh trong KC khi ln khng u. Bng cch:

Ct cc b phn cng ca CT thnh nhiu phn tch bit bng cc khe ln. Dng KC tnh nh 1, Khe ln: + L 1 trong nhng bin php KC c hiu qu khi xy dng CT c ti khc nhau trn nn c tnh nn ln v khng u.

+ Cn c b tr m bo cc b phn ca CT c kh nng lm vic c lp, c cng v cng khi chu un, khng gy vt nt khi nn bin dng ln v khng u.

V tr khe ln: Da vo s phn b cc lp t. Hnh thc KC. Ti:

+ Chiu dy lp t thay i t bin v tnh nn ca t nn khc nhau ln + Khi t ng u th b tr ch thay i ln v chiu cao, hoc chnh lch ng k ti trng + Nu CT c hnh dng phc tp b tr ch c s thay i kch thc hnh hc

Chiu rng khe ln: : Tnh cht bin dng ca CT S phn b cc lp t yu.

69

Trng hp a: Khe ln nn cu to hp vo khong 1cm. Trng hp b: Khe ln ly ln hn khong 3-5cm. Chiu rng ti thiu cc b phn tch ra khng ta ln nhau:

= kh (tgph - tgtr) h = chiu cao ca khe ln ph; tr = gc nghing ca CT bn phi v bn tri khe ln. k: h s xt n tnh cht khng ng nht ca t, c th ly = 1,3 1,5.

Lu : khe ln l cn thit nhng khng phi lc no cng . N trnh c s truyn lc trc tip t cng trnh ny ln cng trnh khc nhng khng trit tiu c s chng cho nhau v /s trong nn t d gy ra ln ca CT c trc. Trong nn c tnh nn ln ln ring tc dng ca khe ln khng cn phi c nhng bin php c bit v d nh tng c cho php gim ng k s chng cho ny. 2. Dng KC tnh nh:

Bng cch thay mi ni cng bng khp hoc gi ta kh c ng sut ph thm pht sinh tuy nhin lm CT nng thm.

3. Tng thm cng cho KC: Mc ch: cc b phn sc chu thm ng lc sinh ra do ln khng u. Tng thm cng v cng khng gian cho KC. Bng cch: B tr ging lin tc bng gch ct thp hoc BTCT

dc theo tng chu lc v dc theo mng.

Xc nh v tr ging Thit k ging: Xc nh lng ct thp.

Tu CT b un cong xung hoc ln m ging c th b tr pha di hoc pha trn ca tng. Khi ct thp b tr 2 hng chiu dy ging > 150cm Thp = (6-12mm)

70

$.3 Cc bin php v Mng: 1. Thay i chiu su chn mng:

Chiu su chn mng tng Sc chu ti ca nn tng ln C t 2.Thay i kch thc mng:

Nu di mng l lp t yu c chiu dy thay i, m bo ln ca nn t ti mi im c gi tr nh nhau c th dng hai bin php sau: Cu to mng theo chiu su khc nhau chiu dy vng chu nn ca lp t di mng nh nhau. Hoc thit k mng c chiu rng thay i nhm mc ch to biu phn b ng sut tip xc c gi tr khc nhau ti mi im di mng.

3.Thay i loi mng v cng ca mng:

Chn loi mng quan trng khng nhng ch th hin v mt chu lc m cn c ngha kinh t ln. Da vo: Hnh thc KC cng nh tnh cht truyn ti trng. S b tr cng trnh ngm, cng trnh ln cn. Tnh hnh a cht.

iu kin xy dng mng: (phng tin thi cng, thi gian xy dng..) C th dng mt trong nhng loi mng nu trn. Tng cng ca mng lm chnh lch ln ca KC bn trn cng nh. tng cng c th lm: Tng chiu dy mng. Tng ct thp dc.

C th lm mng hp cng ln m nh.

71

$.4 Cc bin php x l nn: Mc ch:

Lm tng sc chu ti. Hn ch mc bin dng (c bit l bin dng khng ng u).

Ni dung: Lm tng cng lin kt gia cc ht t ( tng sc chu ti). Lm tng cht ca nn ( gim tnh nn ln v tnh thm nc).

Cc bin php: Loi c hc: lm cht bng m, bng chn ng, lm cht bng cc loi cc, phng php thay t, phng php nn trc, b phn p Loi Vt l: phng php h mc nc ngm, phng php dng ging ct, phng php in thm Loi ho hc: phng php keo kt bng xi mng, phng php silicat ho , phng php in ho

Sau khi gia c nn t khc nhiu so vi trc cn c nh gi thit k nn mng. 1.Phng php lm m: Nguyn l:

V ng sut gim dn theo chiu su (vi ti trng ngoi) nn khi gp lp t yu ngi ta thay n bng 1 lp t khc c tnh cht ph hp v m lu cht (ct trung, ct th, cui si hay t cp phi c m lu tng lp) Cng dng:

+ Lp m ng vai tr nh 1 lp chu lc, tip thu c ti trng CT v truyn xung lp t yu di. + Gim bt ln ton b v khng ng u, lm tng nhanh tc c kt v lp m c h s thm ln nn l ni nc trong t yu thot vo. + Tng kh nng n nh khi CT c ti trng ngang v lp m sau khi c m cht s c lc ma st ln. + Kch thc v chiu su chn mng s gim v cng ca lp m cao. + Thi cng n gin. Phm vi s dng:

Lp m dng hiu qu nht khi lp t yu trng thi bo ho nc v chiu dy 3m ( nu su hn khng kinh t)

Khi nc ngm c p lc tc dng trong phm vi lp m th khng dng bin php ny v ct trong lp m c kh nng di ng.

72

a, Thit k Xc nh kch thc m: Y/cu: + Di tc dng ca ti trng CT m phi n nh.

+ p lc do ti trng CT truyn ln mt lp t yu di lp m phi nh hn cng ca lp . + m bo ln ca nn nh hn ln cho php. * m bo n nh ca nn xung quanh lp m ct th chiu rng lp m phi c kch thc bin dng ngang do CT gy ra khng ln, nm trong gii hn cho php. Theo kinh nghim m bo y/cu trn ly = (gc ma st trong ca m) hoc = 30 45o Xc nh kch thc m: h = ???? gi thit mt chiu dy no xong kim tra v iu chnh cho ph hp. Khi c m ct th nn l mi trng phc tp nn trng thi / sut y hon ton khc vi cc trng hp nu trong c hc t v y kch thc m ct l gii hn. Tuy nhin n gin tnh ton ngi ta dng cc phng php gn ng sau: + Xem lp m nh b phn ca nn v vn dng cc quy lut phn b /sut trong c t tnh ton. + Xc nh kch thc m da v /k n nh v mt cng : xc nh Pgh ca nn th ta phi tnh ton trt su theo phng php cung trt trn v trt su theo mt tip xc y m ct v nh lp t yu v chiu dy ca tng m ct c xc nh ng dn theo yu cu trn. Tuy nhin gn ng c th lm nh sau: Gi nh h sau kim tra cc iu kin. iu kin: * V cng : Ti y mng:

p R; pmax 1,2.R p; pmax = p lc trung bnh v ln nht do ti trng tiu gy ra ti y mng.

R cng lp m. R= Pgh/Fs

Pgh = sc chu ti ca lp m ct tnh gn ng theo cng thc trong gio trnh C hc t.

b

tc

Qo

No

Mo

tc

tc

73

Ti y lp m: 1 + 2 Ry = Pghy /Fs 1 = ng sut thng xuyn do trng lng bn thn t nn v m ct tc dng

ln mt lp t yu di y m. 1 = h + hm

= trng lng th tch ca t v ca lp m. hm v h = chiu su t mng v chiu dy lp m.

2 = ng sut do ti trng ngoi gy ra ti b mt lp t yu tnh theo gio trnh C t.

tnh Pghy ta to ra mng quy c vi b rng mng khi quy c nh sau: b q = b +2 h.tg l q = l + 2 h.tg c th ly bng - gc ma st trong ca lp m ( thng ly 30o)

dc ta luy thnh h o (m) xc nh trn c s phn tnh n nh mi dc thc hin trong lp t yu.

* V bin dng: Tnh ln bng phng php cng ln tng lp. ( Lu : Kch thc m phi ng thi tho mn 2 iu kin trn. Nn gia c l gi nh nn ta phi kho st li nn gia c nh gi gia c v c cc s liu hiu chnh li mng thit k s b.) b. Vt liu m v bin php thi cng: Vt liu: Ct to hoc ct trung, theo kinh nghim 2 loi ny khi m c kh nng t n cht kh cao tip thu c ti trng ln ca cng trnh v khng di ng di tc dng ca nc ngm.

Ct phi tha mn mt s /k sau: i vi ct vng, hm lng SiO2 khng nn nh hn 70% v hm lng hu c khng vt qu 5%. Hm lng mica nn nh hn 1,5% v c ht c d>0,25mm chim trn 50%, cp phi ri u d=5-0,25mm. Ct en: Hm lng SiO2 khng nn nh hn 50% v hm lng hu c khng vt qu 2%. Hm lng mi ca v hm lng st nn nh hn 2% + tit kim vt liu c th trn 70% ct vng vi 30% ct en hay 3 phn si vi 2 phn ct vng (si nn chn c ht 20-30 mm) Bin php thi cng:

Khi thi cng phi m bo cht ln nht v khng lm ph hoi kt cu t thin nhin di tng m ct. Tu theo vt liu m v thit b m m chn bin php thi cng cho ph hp. Ri ct thnh tng lp chiu dy lp ri tu thuc thit b m nn: v d m th cng chiu dy =20cm; m bn rung =25cm...m rung c phun nc U20 =100150cm..

74

Nu t nn di m vn l lp t yu th nn ri mt lp vt liu ngn cch trnh ct b chm xung t yu to iu kin m cht lp ct theo yu cu. Vt liu ngn cch ph bin hin nay l vi a k thut: (geotextile) Nu thi cng trong nc c th dng bin php xa lc ct, y l bin php thi cng trong /k bo ho nc, nn mc nc ngm trong lp m phi cao hn mt lp ct ri khong 510cm. i vi m ct di mng n v mng bng, k/thc y m ct ln hn k/thc y mng, hn na vic o h thi cng m i hi phi c taluy m bo an ton thi cng do din tch o x l trong phn lp trng hp l ton b mt bng cng trnh.

Ch tiu nh gi cht lng m nn: a, cht m nn:

Da vo cht tng i D.

minmax

max

ee

eeD

= ; e = h s rng ca m ct.

31D ct ri;

32

31 < D cht va; 1

32 < D cht

Ngoi ra nh gi cht khi xy dng cc nn ng v sn trn lp m ct

ngi ta cn dng h s k: ktc

kk

=

k = dung trng kh ca ct sau khi m nn ktc = dung trng kh tiu chun trong th nghim m cht

thng thng k = 0,8-0,95 b, Xc nh Eo ca lp m c th dng th nghim bn nn. c, Kim tra cht lng m nn m ct: Mc ch: xc nh k v e ca lp m. Mt s phng php: Phng php cn: ly mu bng dao vng (c t) Phng php dng xuyn tiu chun Nu m c nhiu si: dng phng php o l ct tiu chun Xuyn tiu chun:

u im: xc nh c ngay v ton din cht lng m nn hin trng, c ng, rt ngn c thi gian Dng c gm: Qu ti nng 10,5kg Mt cn di 1,5m c mu khng ch chiu cao ri v u nhn ca chu xuyn c khc thc o n mm

Cch lm: Nng qu t ln n chiu cao quy nh th ri t do, nng ln th xung 3 ln ng thi ghi ln ca u nhn chu xuyn k t lc bt u th ti trng, sau c th tra biu hoc tnh ton tm ra e v k ( TK: XDCT trn nn t yu)

75

Phng php o l ct tiu chun: Vi nn khng dng dao vng ly mu th nghim c nh ln nhiu si lc ta lm nh sau: o l ly t cn xc nh c wQ t xc nh kQ Rt ct tiu chun bit trng lng vo h xc nh khi lng ct vo l

xc nh th tch l o:

cd

PV =

Pc = trng lng ct tiu chun vo trong l o = trng lng th tch ca ct tiu chun.

t xc nh c dung trng ca m.

Bi 1

Xc nh kch thc lp m ct di mng bng khi bit: b = 1,6m; hm= 1,0m. Vi t hp ti trng tiu chun mc mt t: Notc =10T/m; Motc = 2Tm/m; Qotc= 1T/m. Lp t di mng l lp st do nho c cc tnh cht nh sau:

1 = 1,8T/m3; c = 0,12 kg/cm2; = 5o Vt liu m: ct vng ht trung m n cht va: 2 = 1,9T/m3 Bi lm: Gi s chn chiu dy m ct: h = 2m Kim tra chiu dy lp m ct theo iu kin:

z + bt Ry trong s

udy F

PR =

bt = ng sut thng xuyn do trng lng bn thn t nn v m ct tc dng ln mt lp t yu di y m.

bt = 1 . hm + 2 . h 2/6,52.9,10,1.8,1 mTbt =+= z = /sut do ti trng ngoi gy ra ti b mt lp t yu tnh theo gio trnh C t. /s trung bnh ti y mng:

21 /45,68,1225,60,1.8,10,1.21.6,1

10.. mThh

FN

p mmtbtc

=+=+=+=

/s b mt lp t yu ti tm mng:

25,16,1

2==

bz

; 06,1

0==

bx

46,0 =p

z 2/345,6.46,0 mTz ==

Vy: 2/6,86,53 mTbtz =+=+ tnh Pu ti b mt lp t yu ta to ra mng quy c vi b rng mng khi quy c nh sau: b q = b +2 h .tg

c th ly bng - gc ma st trong ca lp m (vi ct vng ht trung m n cht va c th ly bng 30o)

76

2,7T/m2 6,55T/m2

5,6T/m2 3T/m2

P

400

1000

2000

3920

1600

max

tbP =6,45t/m2

Pmin

tt

Qo

tt

No

tt

Mo

0.000

-1.0

-3.0

3000

30 30

tg 30o = 0,58 bq = 1,6 +2. 2 .tg30

o = 1,6+2.2.0,58 =1,6+2,32=3,92m hq = hm + h = 1 + 2 = 3,0m Sc chu ti ca nn c tnh gn ng theo cng thc ca Terzaghi:

cNqNbNP cqu .....5,0 ++= Vi = 5o, tra bng ta c: N = 1; Nq= 1,56; Nc= 6,47 Thay s:

2/2076,774,853,32,1.47,6)9,1.28,1.0,1.(56,192,3.8,1.1.5,0

mT

Pu=++

=+++=

2/102

20mT

FP

Rs

u

dy ===

So snh: 22 /10/6,8 mTRmT dybtz =

77

- t c rng ln, ri, bo ho, tnh nn ln ln, hoc t c kt cu ph hoi km n nh, kinh nghim cho thy: Dng hiu qu vi t ct nh, ct bi ri trng thi bo ho nc, t ct xen k t bn mng, cc loi t dnh yu, t bn, than bn. a.Thit k: + Xc nh h s rng nn cht:

Nhiu kt qu nghin cu cho thy: * Vi t ri sau khi nn cht bng cc ct h s rng enc c xc nh:

enc = emax- D( ema x - emin) Kinh nghim cho thy c th chn D = 0,7 0,8 vi t ri

C th chn onc ee )75,065,0( eo l h s rng trung bnh ban u ca t

* Vi loi t st c th gn ng xc nh:

)5.0(100

+

= dn

nc we = ch s do Cng c ti liu (Lin x): cc loi t st enc c th ly tng ng vi tr s ep khi p = 0,5 1kg/cm2 da vo kt qu th nghim e - p

Lu : nu cc ct i qua nhiu lp t th nce c ly trung bnh ca cc lp t lc

s b ly chiu di cc ct bng 3b b: chiu rng mng (lc b cng c d kin)

21

2211.

lllele

e ncnctbnc

+

+=

Trong 11;lenc = h s rng v chiu dy lp 1 m cc ct i qua.

22 ;lenc = h s rng v chiu dy lp 2 m cc ct i qua. + B tr cc: B tr theo nh li tam gic u (c th b tr hnh vung) Khong cch cc ct xc nh da vo gi thit sau:

- m trong qu trnh nn khng i. - t c nn cht u trong khong cch gia

cc cc ct. - t khng tri ln trn mt t. - Th tch ca cc ht t trc v sau khi nn cht

xem nh khng i.

78

T /k:

Trng lng ca khi t, c y l tam gic u ABC s khng i sau khi c nn cht bng cc ct.

nco

o

cee

edL

+=

1952,0

eo = h s rng ca t trc khi c cc ct. enc = h s rng ca t sau khi ng cc ct. C th vit di dng:

nc

cdL

=

1

1952,0

nc = trng lng th tch ca nn sau khi nn cht.

bo m nn t c n nh v phng din bin dng cng nh kh nng chu ti, cc ct thng c b tr khng nhng di mng m cn phm vi ngoi mng Theo kinh nghim thit k s lng hng cc ct b tr theo hng dc v hng ngang

di mng thng ln hn 3 hng, trong trc ca hng cc ngoi cng ly rng hn kch thc mt bng mng mt khong cch ln hn 1,5 ln ng knh cc hoc 0,1 ln chiu di cc. C th b tr nh sau: Trn giy crki ta v li cc cc ct theo khong cch chn

trong thit k sau t giy can trn li v theo cng mt t l ta v s mng cng vi gii nn cht xung quanh. X dch giy can trn li ta s tm c s b tr hp l nht ca cc ct.

Cc ct ch c tc dng nn cht theo chiu su nu s lng ct cha y trong

cc p ng y/cu cht thit k. Trng lng ct cn thit trn 1md cc:

mw

e

fGnc

hc 1)100

1(1

1 ++

=

W1 = m tnh theo trng lng ca ct trong thi gian thi cng.

h = trng lng ring ca ct dng trong cc.

+ Xc nh chiu su nn cht ca cc: z 0,2 bt t tt

0,1 bt t yu 0,5 bt thu li

79

C th theo phng php lp tng ng ca Xtvich H = 2hs (hs= Ab) A = h s lp tng ng h s pot xng o; hnh dng mng v cng ca mng. Thng > 2 ln b mng ch nht 3 4 ln b mng bng. Tuy nhin: nu chiu dy lp t yu cn x l li nh hn chiu su nn cht nu trn th chiu di cc ct ly ht chiu dy ca lp t yu cn gia c. Theo kinh nghim nn t sau khi gia c cc ct sc chu ti:

c th tng t 2 3 t khi cha gia c v m duyn bin dng c th tng ln khong 2-3 ln.

Ta ly c s gi nh thit k b. Vt liu cc v bin php thi cng: +Vt liu: lm tng tc c kt v cht ca nn, ct dng lm vt liu cc thng l loi ct ht to hay ht trung, ct yu cu phi sch, hm lng bi v st ln vo khng qu 3% ng thi khng ln nhng hn to c kch thc ln qu 60mm. +Bin php thi cng:

ng thp thng c ng knh 40 50 cm nh b phn chn ng, my n ng xung n ct thit k. Sau nhc b phn chn ng ra, nhi ct vo v cao chng 1m. Ri li t my chn ng vo v rung trong khong 15 20 giy. Tip theo b my chn ng ra v rt ng ln khong 0,5m ri li t my chn ng rung khong 10 15 giy cho mi ca ng m ra v ct tt xung. Sau rt ng ln dn dn vi tc u, va rt ng va rung cho ct c lm cht. c, Kim tra cht lng: Sau khi thi cng kim tra li bng cc phng php sau: Khoan ly mu gia cc cc ct th nghim trong phng. Dng xuyn kim tra cht ca ct v cc ct. t yu c th dng th nghim ct cnh. Th bn nn tnh ti hin trng, trn mt nn cc ct. Din tch bn nn thng phi ln hn 4m2 trm c t nht 3 cc ct.

80

Bi 2: Thit k mng di ct tit din (30. 40) cm2; ti trng tiu chun tc dng ti ln mng ti cos 0,0: Notc = 110T; Motc= 8Tm; Qotc= 1,2T Mng t trn lp t ct bi c chiu dy 20m; di lp ct bi l lp st pha nho, mc nc ngm nm su cch mt t 1,5m. c trng ca lp ct bi nh sau:

= 1,80g/cm3; = 2,65; w = 30%; = 20o ; c = 0,0; emax = 0,96; emin = 0,56; qc = 30kg/cm

2

Bi lm:

- Xc nh trng thi ca lp ct bi da vo cht: 1hk

e

=

Trong : Wk +

=

1 ; 3k /38,1

100301

8,1cmG=

+= ; 92,01

38,165,2

e ==

1,056,096,092,096,0

minmax

max=

=

=

ee

eeD ct trng thi ri.

Xc nh mc m ca ct:

864,01.92,0.100

65,2.30e.

===

n

WG

G = 0,864 > 0,8 nn ct bi trng thi bo ho nc. t yu nn dng bin php gia c nn y dng bin php gia c nn bng cc ct

Gi thit mng c kch thc: b = 2m; hm=1m; l = .b;

trong : = 1+2e; e: l lch tm

NM

e = ; TmhQMM mo 2,91.2,18. =+=+= 1,01102,9

==

NM

e

l = .b = (1+2. 0,1) . 2m = 1,2 . 2m =2,4m. p lc di y mng:

2/251.24.2.2

110)( mThab

Np m

otctb +=+=

22max /8,294,2.2

6.2,925 mTW

Mpp otctb =+=+=

22min /2,204,2.2

6.2,925 mTW

Mpp otctb ===

/k kim tra: ptb R pmax 1,2R

Trong : R c tnh gn ng theo cng thc ca Terzaghi: cNisqNisbNisP cccqqqu ++=

81

cq NNN ;; cc h s sc chu ti ca nn ph thuc vo tra bng. H s hnh dng

s = 1- 0,2/ = 1- 0,2/1,2 = 0,83 sq = 1 sc = 1+ 0,2/ = 1+0,2/1,2 = 1,17

H s iu chnh nghing ca ti trng: 2

1

=

i 2

21

==

pi

cq ii

(gn ng coi l ti ng tc l = 0 nn 1=== cq iii ) Vi = 20o, tra bng ta c N = 4,97; Nq= 6,40; Nc= 14,8 { } 2/37,260.8,14.17,11.8,1.40,6.12.8,1.97,4.83,0.5,0 mTPu =++=

Chn Fs = 2 ; 2/2,13

237,26

mTFP

Rs

u===

Gi nh: Sau khi gia c Rgc = 2R cha gia c vy: Rgc = 2 .13,2 = 26,4T/m

2 So snh:

ptb= 25T/m2 < R = 26,4 T/m2

pmax= 29,8 T/m2 < 1,2 R = 1,2 . 26,4 = 31,68 T/m2

iu kin hp l v kch thc: [ ]{ } [ ]RpR %102,1 max

1,2R - pmax = 31,68 29,8 = 1,88T/m2 < 10%R = 2,64T/m2

Vy la chn s b b = 2,0m v a = 2,4m l hp l. Kch thc ca mng l: a b = (2,0 2,4) m2

(Nu so snh trn lch nhau qu nhiu chn li.) Xc nh enc khi dng cc ct:

)( minmaxmax eeDeenc = chn D = 0,75 ta s c: 66,0)56,096,0(75,096,0 ==nce

+Xc nh trng lng th tch ca t nn cht theo cng thc:

)01,01(1

h wenc

nc ++

=

W = m thin nhin ca t trc khi nn cht = trng lng th tch ca t thin nhin trc khi c nn cht.

82

3/08,2)30.01,01(66,01

65,2 cmGnc =++

=

Xc nh khong cch gia cc cc ct:

nc

cdL

=

1

1952,0

mL 04,180,108,2

08,24,0.952,0 =

=

Cn c vo mt bng mng ta b tr: 19 cc. Xc nh trng lng ct trn 1m di:

Theo kt qu th nghim, ct trong cc c c tnh sau:

h = 2,65G/cm3; W1= 12% mW

e

fGnc

hc 1)100

1(1

1 ++

=

W1= m ca ct trong thi gian thi cng.

h = trng lng ring ca ct dng trong cc.

tG 224,0100121

66,01

65,24

4,0.14,3 2

=

+

+=

Xc nh chiu su nn cht:

Chiu su nn cht y ly bng chiu dy vng chu nn, p dng phng php lp tng ng ta c: i vi ct v vi mng tuyt i cng tit din hnh ch nht tra bng:

l/b = 1,2; = 0,25 Aconst = 1,08 Vy chiu dy lp tng ng l:

Hs =1,08.2 =2,16m Chiu dy vng chu nn k t y mng:

H = 2.2,16 = 4,32m 4,5m. Tnh ton ln d tnh ca nn t sau khi nn cht bng cc ct. ln d tnh ca nn t sau khi nn cht bng cc ct c th xc nh theo cng thc:

;)1( 2

o

o

EpbS = v lc hc= 4,5m > 2.b (b = 2m)

p lc tiu chun ti y mng 2110

2.1 25 /. 2, 4.2otc

tb mNP h T ma b

= + = +

p lc gy ln: 2/2,231.8,125. mthpp mtb ===

5000

1000

1000

400

1000

2000

300

500

500

900 900 900 900

6000

1000

500

500

2000

2400

83

Eo= .qc; chn = 2 Eo= 2. 30= 60kg/cm

2 Sau khi gia c gi thit Egc = 2Eo = 2.60 =120kg/cm

2

1200)25,01(96,0.2.2,23 2

=S = 0,035m = 3,5cm

ln tnh ra nh hn ln cho php. 3. Phng php nn trc: Nn t c tnh nn ln, bin dng khng ng u: st, st pha ct trng thi chy; ct nh, ct bi bo ho nc nn cht n th ti trng phi tc dng thng xuyn v trong thi gian di th mi c hiu qu. Trong nhiu trng hp nu ln d tnh ln, vt qu ch dn cho php, CT c th s dng ngay sau khi thi cng th mt trong nhng bin php hay dng l nn trc bng ti trng tnh. a, Nguyn l:

Trc khi xy dng CT hoc sau khi xy xong mng dng cc loi vt liu nh ct, si, .v.v. cht ng ln mt t trong phm vi xy dng mng gy ra mt p lc nn (gi l p lc nn trc) lm nn ln xung t c cht li. Khi t nn t cht yu cu ngi ta d p lc nn trc ri tin hnh xy dng CT Lc CT va c cng t yu cu va c tnh nn ln nh. Ni dung:

Xc nh ln ca p lc nn trc. Thi gian nn trc. p lc nn trc: c hiu qu tt, p lc nn phi tng ng p lc CT

rt ngn thi gian nn trc c khi ngi ta dng p lc nn > p lc CT (thng =1,2 ln).

Do t nn yu nn p lc phi tng dn tng cp v khng ch tc tng nn khng b ph hoi.

Thi gian nn trc:

Do yu cu thit k. Do nu khi thit k, qua tnh ton nhn thy ln tnh ton ng vi thi gian quy nh vn nh hn ln n nh theo yu cu th cn phi c bin php rt ngn thi gian nn trc C th lm cch sau: lm h thng thot nc theo chiu thng ng. Ging ct (SW- Sandy Well)

hoc bc thm (PVD - Prefabridated Vertical Drainage)

84

b, Rt ngn thi gian nn trc: Nn t st no nc c nn di ti trng, ng thi trong nn b tr cc SW hoc PVD theo chiu thng ng to /k cho nc l hng thot ra ngoi lm cho qu trnh c kt ca t tin hnh c nhanh chng v tng bn cho t yu no nc. + Ging ct: Thi cng tng t nh cc ct tuy nhin tha hn, v nhim v chnh y l tng nhanh tc c kt ca nn. + Bc thm: V nguyn l thot nc tng t ging ct. Cu to bao gm 2 phn: - Li: bng cht do, dy 23,5mm, c rnh, ng dn nc. - V lc: bng vt liu tng hp, ch cho nc trong t thot qua nhng ngn cn c cc ht t chui vo bn trong. Bc thm c ch to di dng cun lin tc vi chiu di tng i ln. thi cng nh sau: mt my cy chuyn dng kp mt u bc n su xung t n su thit k, ct bc u trn sau rt cn ln, li bc trong t. Dng bc thm din tch tit din bc i v hnh trn tng ng:

theo Hansbo : 2

bdw+

=

;

- chiu dy ca PVD thng thng t 3-5mm (c bit mm10= ) b - chiu rng bng PVD, hin nay b=100mm. ln theo thi gian trong trng hp ny c th tnh c da trn ph/tr vi phn c kt i xng trc ca Rendulic. c kt c tnh nh sau:

)1)(1(1 zrt UUU = tU - c kt ton phn ca t.

zr UU ; - c kt theo chiu xuyn tm v theo chiu thng ng.

( )r r

U f T= v ( )z zU f T= Trong : t

a

cT rr 24

= ; thc

T zz 2=

a- khong cch gia cc ging ct h- chiu dy lp t

V ln theo thi gian : SUS tt .= Tham kho ti liu chuyn ngnh

85

c, Chiu dy m ct trn nh vt thot nc ng (VTN)

m ct trn nh VTN tip nhn nc t cc VTN v tiu thot ra ngoi phm vi nh hng ca ti trng do phi m bo / kin thot nc.

Chiu dy ca m phi c xc nh trn c s ln d kin c th xy ra di tc dng ca ti trng cng trnh S vi chiu cao d phng khng t hn 30cm : H = S + 30 (cm)

Trong : H = chiu dy lp m, cm.

S = ln cui cng ca nn do ti trng thit k gy ra, cm.

5. Phng php phn p: Mc ch tng sc chu ti ca nn t yu. bng cch tng p lc bn hq .= trong cng thc tnh Sc chu ti : (C t) B phn p s dng nhiu trong xy dng nn ng p cao qua vng t yu nhm tng cng s n nh cho cng trnh. B phn p cn c tc dng phng l, chng sng, chng thm nc..v.v. Bin php ny ch thch hp khi din tch t ginh cho ng khng b hn ch. Ni dung: Dng vt liu nh: si, , t, ct p ln nn hai bn cng trnh to ra p lc bn q.

Xc nh kch thc b phn p l vn mu cht trong vic tnh ton thit k b phn p.

C 3 phng php gn ng: - Da vo s hnh thnh ca vng bin dng do pht trin 2 bn cng trnh. - Da vo gi thit mt trt ca nn t c dng l hnh tr trn (xem c t) - Da vo l lun CBGH xc nh mt trt v suy ra trng thi gii hn ca

t nn (do phc tp nn thit k t dng)

Phng php da vo s pht trin ca vng bin dng do : T /k CBGH ca 1 in trong nn C t khi ti trng cng trnh l hnh

thang th vng bin dng do di cng trnh c dng hnh tri xoan lm gia. Phm vi pht trin ca vng bin dng do cho php bng mt na khong cch hai mp ngoi ca b phn p. T /k CBGH ca 1 im trong nn (c t) ta thy im c mt trt i qua l im m ti gc =

86

gc lch gia ng sut php v ng sut tng o. gc ma st trong ca t Biu din bng biu thc:

Vi t ri:

sin = 31

31

+

; sin2 = ( )( )222 4

yz

yzyz

+

+;

Vi t dnh: tng t ta c:

sin =

gccot23131

++

; sin2 = ( )( )222

cot24

gcyzyzyz

++

+;

By gi ta tm nhng im c cng ri ni li ta c ng ng tr s gc lch. Khu vc ng vi ng cong = l khu vc bin dng do. Cch v vng bin dng do nh sau : Chia nn t thnh mt h li vung sau xc nh ri ni li. Cch lm: Khi c kch thc ng t p ta gi thit b rng v chiu cao b phn p sau kim tra ri iu chnh cho ph hp. Khi mi hay mi ng cao th nn dng phn p nhiu cp gim chnh lch ca tng phn p ln mt nn.

6. Gia c xi mng - t:

Nguyn l: - Trn t di su ln trn cng vi xi mng nh nhng phn ng ho l xy ra lm cho t mm yu ng cng li thnh th cc trong t c cng nht nh.

- Phn ng ho l gia t + xi mng v qu trnh ng cng bao gm. - Phn ng thu ho + nc to thnh cc hp cht ca Canxi. - Cc ht st tc dng vi cc cht thu ho ca xi mng bn thn cc cht thu

ho ng cng to thnh b khung xng trong t gia c. - Phn ng ccbont ho to thnh ccbont canxi khng tan.

Thi cng: - My trn di su ti ch, hin nay Vit nam dng my d 40cm, su 20m Vt liu:

+ Xi mng silicat 400# hoc xi mng x qung. + T l nc/ xi mng = 0,4 0,5.

+ Lng xi mng/ t = 7% 15%. - B tr trn, khong cch, su trn ph thuc vo:

+ H mng + a cht + CT xy dng

L

l

P

q

87

+ CT ln cn

Cng dng: - t trn xi mng c cc c trng c ci thin tt, trng lng ring tng t, cng

chu nn 1 trc Ru = 0,5 40 kg/cm2 ( ln hn nhiu so vi t t nhin)

- V vy c tc dng gim ln, tng sc chu ti, chng thm, gim nh hng ti CT ln cn v hn ch tnh bin long, chy ca nn t

- Phng php ny c s dng rt ph bin to tng vy, tng chn, gia c h mng, mi dc v nn CT.

Phm vi s dng: - Thch hp vi cc loi t yu khc nhau t bn n st do bo ho cha hu c

khng nhiu, pH tng i cao. - CT va v nh. - Hiu qu ph thuc rt ln vo cng ngh (ct gt, nho trn, bm va xi mng) 7. Vi a k thut:

1. Cu to: Vi a k thut (geotextile), l nhng tm hoc di bng di c cun li thnh cun di (t vi chc m n 300m/cun). y l loi vt liu c kh nng thm nc. Vi a k thut (VKT) c ch to t Polymer tng hp hoc cc sn phm ca n, dng tm hoc bng chia thnh hai loi dt v khng dt. Vi a k thut c tnh dai, bn, thot nc hoc cch nc. 2. Cc chc nng ca VKT: + Lm lp ngn cch gia cc lp vt liu vi nhau. + Gia cng t yu. + Lm tng chn t (tng chn ct mm). + Lm tng lc nc sau lng tng chn. + Vt liu thm h mc nc ngm. + Chng si mn. + Lm ng a k thut phc v cc cng trnh tiu thot nc. 1. phn cch 2. tiu thot nc 3. thm 4. bo v 5. gia c t

6. gia c nn ng 7. gia c mi dc.

88

3.Thit k s dng vi a k thut theo chc nng. 3.1. Cc c tnh ca vi a k thut. Bao gm cc c tnh v khi lng ring, kch c l, kch c tm vi, vn tc thm, cng chu ko, x rch m thng, dn di, bn, kh nng chu nhit, nh sng, ma st gia t v vi... 3.2.Thit k lp vi a k thut ngn cch cc lp t khc nhau: - Trng hp ny hay gp trong lnh vc ng t, my bay, sn gn, nn nh. Vi a k thut c dng ngn cch lp dm, ct vng vi lp t nn. - Tnh ton chng phng ca vi do tc ng ca cc vin dm, si cui. - Tnh ton chc thng. - Tnh ton chu ko. 3.3.Thit k Vi a k thut tiu nc, lc. Ni dung bao gm:

- Tnh thm - Gi t li. - Tnh bn lu, khi vi b bt kn.

3.4. Thit k vi a k thut gia cng. Ni dung tnh ton cc lp vi trong nn gia c, tng chn, mi dc bao gm hai ni dung c bn sau:

- Tnh ton n nh (trt, lt) - Tnh ton ln, ln lch.

C ch: t l vt liu c bn nn tng i cao, khng chu ko. bn ca t ch b gii hn bi sc khng ng sut ct. Mc ch kt hp t vi ct m y l vi a k thut (c th l li a k thut, hoc ct cng hoc ct thp m trn hoc c g tng ma st gia t ct.) tip nhn lc ko hoc ng sut ct do gim c ti trng ph hng t v ng sut ln hoc v bin dng qu mc. iu ny gn tng t nh nguyn l b tng ct thp. Khi c ct c coi l mt khi vt liu hn hp c cc c tnh c ci thin, c bit kh nng chu ko, chu ct so vi ring t. C th tham kho trnh t thit k trong ch dn ca BS 8006: 1995. Hin nay nhng tnh ton ny s dng phn mm Plaxis, Geoslope ang ph bin v c tin cy cao.

89

AAA

A

AA

Tnh ton nn p gia c vi a k thut bng chng trnh Plaxis