90

chng V mng cc

5-1.Khi nim chung: Mng cc c s dng rng ri trong xy dng:

Khi ti trng cng trnh tng i ln, Khi lp t tt xut hin tng i su, Khi c yu cu cao v hn ch bin dng cng trnh. Mng cc gm 2 phn: Cc: l b phn chnh nhim v truyn ti trng CT vo t thng qua ma st bn thnh cc v phn lc u mi cc. i cc: l cu kin dng bn c nhim v: Nhn ti trng t CT v Phn phi ln cc cc. T mt t n y i gi l chiu su chn i h

Nu h su cho t t y i tr ln tip nhn c ti trng ngang th: mng cc gi l MC i thp, ngc li gi l MC i cao. Cc trong MC i thp chu ti trng ngang khng ng k m ch yu chu ti ng. Cc trong MC i cao chu ti trng ngang ca CT kh ln nn i hi phi c cng chng un ln. Cc cng trnh giao thng thng p dng loi mng ny gim bt kh khn cho thi cng.

Cc lm vic chia thnh 2 loi: Cc chng: Nu mi cc nm trong lp t c tnh bin dng ln rt thp hoc thc t coi l khng bin dng. Ton b ti trng truyn xung mi cc; ma st xung quanh thn cc khng ng k thng l nn . Tnh ton cc chng ch yu da vo kh nng lm vic VL

91

Cc treo ( cc ma st): Ti trng truyn xung mt phn mi cc, mt phn xung quanh thn cc. Tnh ton cc treo ch yu da vo kh nng lm vic t nn.

(Phn ln cc cng trnh ca ta s dng cc ma st) 5.2 Cu to cc: Phn loi cc: Theo vt liu: G, BTCT, thp Theo cng ngh thi cng cc:

Cc c sn: ng; p; rung, xon.. Cc ti ch: khoan nhi, Ba rt

Phn loi theo mc thay th t: Khng thay th: cc c hoc rng c bt u ng. Thay c: khoan nhi

Thay mt phn: khoan mi + ng hoc p 5.2.1. Loi cc c sn h bng ba thng (cc ng) 1, Cc g: Dng trong CT nh hoc tm thi.

u im: cc g nh vn chuyn d dng, ch to n gin, khng cn thit b ng ln, nu cc g lun nm di mc nc ngm th thi gian s dng rt lu Nhc: Sc chu ti nh, hn ch v chiu di D b xm thc & d b ph hoi khi ng - cn ngm tm Creozot p lc cao) Khi dng cc g y/c ch nhng im sau: - /k thng chn 20 30 cm, L= 5 8m, g dng cc nn chn loi g cng thng, cong khng nn qu 1/100 so vi chiu di. - nh cc c bo v bng ai thp dy 8mm rng 710cm, mi vt nhn, bt thp d ng, khng b dp nt. Do g him nn hin nay t dng. 2.Cc BTCT: Dng ph bin u im: - iu kin p dng rng ri hn ( lu nc mn) tit din, chiu di c th theo mun, cng vt liu cc tng i ln. Nhc im: - Trng lng qu ln, gy kh khn cho vn chuyn v h cc. Cn khi lng thp ln cho vn chuyn, cu lp. Thi cng cc hay b nt ( khc phc bng cch dng BTCT ng sut trc) Tit din: trn, tam gic, I, T, rng.. Vi cc BTCT thng thng c cu to nh sau: L = 425m c khi ti 100m. Cc on thng 68 m ( c khi ln hn - ty thuc a im v thit b thi cng) ni li vi nhau trong khi h. T s gia chiu di L ca cy cc v b rng b ca tit din cc thng khng ln qu 100. nh cc tng cng li thp a = 50mm

92

Mi tng cng thm thp dc v ai thp ai hai u dy hn a= 510cm thp 68mm Thp dc 4-8 thanh CT5; thp ai CT3, c th dng ai xon. Thng thng ngi ta b tr ct thp nh sau: Cc 2020 dng 144 ; Cc 2525 dng

164 ; Cc 3030 dng 184 ; Mi ni cn t c kh nng chu ti t nht l tng t nh cc tit din khc ca cc, tt nht l dng lin kt bu lng chu tt ti trng ng khi thi cng. B tng cc M 250 # ( thng dng 300#); B tng cho cc khoan nhi ngoi /k v cng b ng phi c st ln m bo tnh lin tc ca cc ( st khong 7,5cm n >15 cm - ph thc /kin s dng; tham kho TCXD) Thng thng BTcc khoan nhi c hm lng xi mng khng < 350kg/m3

C mc treo cc v vn chuyn. Lp b tng bo v: a 3cm.

T l 3,12,1 DA

V d cc:

tit din 20 20 cm ( A=25cm) tit din 25 25cm ( A=32cm) tit din 30 30cm ( A=34cm) tit din 35 35cm ( A=40cm)...

Lu : Vi cc c yu cu k thut cao( a cht phc tp, ti ln, iu kin thi cng kh khn...) cn ch to cc c l rng vi /knh khng < 30mm kim tra thng ng ca cc sau khi thi cng.

ngng cccc CCcc tittit didinn nhnh

93

Mi ni cc

on ni on mi cc

L1

on ni

2L L3

Lc

b

a a

L

Mc cu Thp dc

u cc Mi cc

(thp AI)

350

30

290

30

30 290 30

350

350

A

72

31

40

D

100100100

50

50

50

50

A-a

86

70

3020 20

8 8

= D

D

B-B

50

b b

a a

1 1

94

3, Cc thp: Thng dng nhng CT vnh cu hoc nhng CT chu ti trng ngang ln cng c trng hp dng lm CT tm thi nh cc vng vy thi cng mng. Trong trng hp c bit mi dng cc thp. *Tm li: Loi cc ch to trc a n v tr ri h cc bng phng php ng

u im: Cht lng bn thn cc c th kim sot c v dng loi b khi pht hin thy khng m bo, tc thi cng nhanh v thun li. Nhc im: Vi cc ng l ting n v rung ng. 5.2.2. Loi cc h bng phng php xon ( cc xon) Phm vi: cng trnh:

Cu, cng, thu li, ct in cao th, cc neo khi th nghim nn tnh. Cu tao: Thn: BTCT, Thp : c dng mi khoan; thp, gang, hoc trong BT ngoi bc thp. Cch h: nh thit b to m men xon.

u im: h cc m thun. ph hp xy chen Chu nh tt do mi m rng 5.2.3.Loi cc h bng phng php xi nc. Cu to: tng t nh cc ng

Khc: gia hoc xung quanh c gn thm cc ng dn nc. Cch h: ch xi nc trng lng bn thn cc lm cho n h gn n chiu su thit k khi cn cch su thit k 1 1,5m dng ba ng Va xi nc, va dng ba ng. 5.2.4.Loi cc h bng phng php rung ( mng cc ng) Cu to: Cc ng l kt cu v mng bng thp hoc BTCT c dng hnh ng. Cch h: Do tc dng rung ca my chn ng.

u im: C th cng nghip ho, dng hiu qu trong loi t khng dnh hoc t dng ht Tn t vt liu Tn dng cao kh nng lm vic ca vt liu mng C th a mng xung su m khng cn n mng ging chm (l loi mng c hi cho sc kho)

95

C th thi cng quanh nm v ton b cng tc trn mt nc ci thin /kin lao ng Vit nam: cc cu Hm rng /k = 1,55m Nc ngoi: /k n 6m. 5.2.5.Loi cc ti ch ( Cc nhi) Cc s c ch to ngay ti ch m cc s lm vic sau khi xy dng CT Nguyn tc: Khoan to l, c ng vch hoc khng ng vch, hnh dng c th trn hoc ch nht (Cc Baret) BT; BT c st cao, qua ng dn. Cu M thun: cc /k = 2,5m; L 100m C/trnh 27 Lng h: cc Bart = 1m 2,8m n 1,5m 2,8m. u im: Sc chu ti ln Khng phi vn chuyn nn lng ct thp trong cc t hn so vi cc ng, Thi cng m thun nn t gy chn ng cho CT ln cn nh phng php ng Nhc im: Vic kim tra hnh dng cng nh cht lng cc rt b hn ch, c bit l khi thi cng khng c ng vch.

Thit b thi cng

cng knh. Nu cc b hng

kh thay th.

1 1

1 1

D

Chi

u

di

to

n c

c

2000 - 3000

2000 - 3000

1-1

D

2-2

D

96

5.2.6.Loi cc h bng phng php p tnh:

C tn cc Mga thi k u dng sa cha CT nghing ln l chnh Nguyn l:

Cc ch to thnh on ngn. Dng kch thu lc p tng on xung t. Dng ngay trng lng bn thn CT lm i trng

p xong on ny t on khc ln Lin kt: bng hn hoc cht

Sau khi p xung su th/k lin kt cc vi mng c. Hin nay cng ngh ny rt pht trin Vit nam, dng trong c sa cha v xy mi. Thit b p ln n trn 300 tn Ti trng lm vic ca cc trn 100 tn Chia thnh 2 loi: Phng php p trc: cc c p trc khi xy dng CT lc i trng neo kch l i trng hoc cc neo Phng php p sau: l h cc c cha sn i Khi xy xong 1 phn CT th p cc Sau lin kt cc vi i.

97

pp ""trtrcc" " pp ""sausau""

5.3. Cu to i cc: Vt liu: BT; BTCT i c th: ton khi hay lp ghp. Dng: nh i ph thuc y CT

y i ph thuc b tr cc C th: n, Bng , B Chiu dy i tnh ton.

i c th: lin tc hoc thnh bc Cu to chung i cc nh sau:

h

h

o

>=100

>=100

Thp tng cng

>1,2 m

>2D

v

>=

98

Lin kt cc vi i ph bin hn c l lin kt ngm.

Nu u cc khng th p ly thp ch lin kt vo i (trng hp thi cng di nc nh tr cu chng hn) chiu di cc trong i phi khng t hn 2D hoc 1,2m nu D > 600

Trng hp p u cc neo thp vo i: 20 i vi thp gai. 40 i vi thp trn.

on cc trong i ch yu cu n 100 l .

Trng hp c bit cc c th khng lin kt trc tip vi i m thng qua tng gim chn c p dng trong cc vng c ng t, lp m trung gian bng vt liu ri ( dm, si, ct th v ct trung)

Khong cch t mp cc ngoi cng n mp i ph thuc vo sai s cho php khi ng cc, tc l ph thuc vo ng knh cc, ng thi khng < 100mm

Vi MC i cao: Nn tng cng ct thp cho i bng cc li 20 25 t cch nhau 10 20 cm nh cc nn t cc li ct thp 12 cch nhau 10 20cm Cc mp phi c tng cng bng cc thanh ct thp un mc cu.

BT yu cu M 200# 5.4.S lm vic gia cc n v nhm cc: + S lm vic ca 1 cc n v ca 1 cc trong mng cc l khc nhau, tuy nhin trong cc phng php tnh hin nay u coi Sc chu ti (SCT) trong 2 trng hp l nh nhau cn phi nghin cu thm hon chnh nht l vi cc ma st. + Cc cc thng c ng thnh nhm khong cch gia cc cc thng t 3 n 4 ln ng knh hoc cnh ca cc. Vi cc chng th c th khng thy c du hiu khc nhau nh th. L do: cc chng ti ch yu truyn qua mi cc m mi cc li nh nn t gy nh hng n cc cc xung quanh. Vi cc ma st din tch phn b p lc ca ti CT xung nn t mi cc ln hn nhiu so vi cc chng nn khong cch gia cc cc b tr c nh hng ln n /k lm vic ca cc cc xung quanh. Nu khong cch ln biu p lc do ti truyn xung mt phng mi cc gia cc cc khng chm nhau th lc SCT ca MC c th xem nh bng tng SCT ca mi cc n. K/q th cc cho thy: khi khong cch gia trc cc cc a 6D khng nh hng c th coi nh cc n. + Khi cng tr s ti trng ln cc n v ln mi cc trong nhm. p dng li gii ca Mindlin ta thy ng sut z ti im trn trc cc do c nhm gy ra ln gp vi ln so vi ng sut di cc n, chiu rng hiu qu ca nhm cng gp mt s ln so vi cc n. do ln ca nhm s ln hn cc n.

99

Ngi ta gi l hiu ng nhm: c th vit: P u nhm = n P u n

Trong : = h s nhm. n = s lng cc trong nhm. P u n = sc chu ti ca mt cc n. (cn nghin cu cch thi cng v cc loi t khc nhau rt ra tr s ) SCT ca cc gm hai thnh phn:

Sc khng bn, to nn bi ma st v lc dnh ca t v cc mt bn cc. Sc khng mi.

Nhiu nghin cu thy rng: Sc khng bn t cc hn rt nhanh ( chuyn v khong 3 5mm. Nu cc nhi c thnh bn rt nhm, th sc khng bn c th t cc hn chuyn v ln hn, khong 10 15mm. Ngc li sc khng mi t cc hn rt chm. Di ti trng cho php, chuyn v ca cc [ ]s kh nh, do sc khng mi mi ch huy ng mt phn nh (trong khi , sc khng bn ca cc huy ng kh ln) 5.5. Xc nh sc chu ti ca cc: Cc trong mng c th b ph hoi do 1 trong 2 nguyn nhn sau: Vt liu cc b ph hoi Pvl Pmin (Pvl; Pn) [P] thit k Do t nn b ph hoi Pn

Sc khng tng

Sc khng bn

Sc khng mi

Sc khng

Chuyn v tung igia cc v ts

Qp/Fp

QpQf/Ff

Qf

{Q}

Qu

0

100

Mt s nh ngha:

Sc chu ti cc hn (Pu): l ti trng m ti vt liu hoc t nn b ph hoi. Sc chu ti cho php [ ]P : l ti trng m ti cc lm vic an ton (vi mt h s an ton sF thng ln hn 2) A. Xc nh SCT theo phng dc trc: 5.5.1.Theo Vt liu: 1. Cc b tng ct thp ch to sn tit din c, chu nn. '( )vl b n a aP m R F R F= +

nR cng chu nn tnh ton ca b tng. F din tch tit din ngang ca cc.

'

aR cng chu nn tnh ton ca ct thp.

aF din tch ct thp dc.

bm h s /k lm vic ca BT. i vi cc c tit din nh hn 30cm; mb=0,85

h s un dc. Khi cc xuyn qua lp bn, st yu v mng cc i cao, s un dc c k n trong phm vi chiu di t do ca cc. Chiu di t do ca cc c tnh t y i n y lp t yu ( tra bng). Cn li ly =1 2. Cc ng BTCT c lp y phn rng bng b tng: ' ' '( 2,5 )vl b n a a ax axP m R F R F R F= + +

lF din tch tit din ngang ca li BT

'

ctF din tch tit din ngang tnh i ca ct thp xon ' l x

ax

x

D FFB

pi=

lD ng knh li BT

xF din tch tit din ngang ca thanh thp xon

xB bc ca ct thp xon.

axR cng chu nn tnh ton ca ct thp xon. 5.5.2. Theo t nn: 1. Phng php th nghim: 1. Phng php nn tnh:

101

102

Vit nam Ta c cc quy phm nh sau: 88-82; 189-190; 269:2002 TCVN: 269: 2002 Cc Phng php th nghim bng ti trng tnh p dc trc: Thi gian ngh ca cc: Thi gian ti thiu: 21 ngy cc khoan nhi 7 ngy vi cc khc Quy trnh gia ti tiu chun: (C th gia ti theo chu k) Gia ti tng cp n ti trng th nghim ln nht theo d kin. Mi cp 25% ti thit k. Cp mi ch c tng khi tc ln u cc t n nh quy c nhng khng qu 2h. Gi cp ti ln nht cho n khi ln u cc t n nh quy c hoc 24 gi, ly thi gian no ln hn. Sau khi kt thc gia ti, nu cc khng b ph hoi th gim ti v 0, mi cp gim ti bng 2 ln cp gia ti, thi gian mi cp 30 pht, ring cp 0 c th lu hn nhng khng qu 6 gi.

103

n nh ln quy c: Khng qu 0,25mm/h i vi cc chng vo lp t hn ln, t ct, t st t do cng n cng

Khng qu 0,1mm/h i vi cc ma st trong t st do mm n do chy. i vi cc thm d: th nghim n 250% 300% ti thit k i vi cc kim tra: th nghim n 150% n 200% ti thit k. Trnh by kt qu: V Biu quan h ti trng chuyn v( P-S) V Biu quan h chuyn v thi gian( S t) Da vo xc nh Sc chu ti gii hn Da vo th P S

Trng hp ng cong bin i nhanh, th hin r im c dc thay i t ngt ( im un), sc chu ti gii hn bng ti trng tng ng vi im ng cong bt u thay i dc.

Nu ng cong bin i chm, kh hoc khng th xc nh chnh xc im un th cn c vo cch gia ti v quy trnh th nghim chn phng php xc nh sc chu ti gii hn ( xem ph lc sau)

Vi quy trnh 88-82: Sc chu ti ca cc c ly nh sau:

Pgh c xc nh da trn trn biu ti trng - ln ng vi tr s ln bng: S* = .Sgh

Trong : Sgh = ln gii hn cho php ca cng trnh = h s chuyn i k n s khc nhau gia thi gian tc dng ca ti trng khi

th nghim v ca ti trng thc t. Quy phm Vit nam cho php ly = 0,2

25

25

40

45

35

30

20151050

S*

10

15

20

5

45403530

Pgh

S(mm)

P(t)

Ti trng cho php c ly: [ ]s

ghFP

P = ; (Fs = 1,25)

104

Ghi ch: Gi tr sc chu ti gii hn xc nh theo phng php khc nhau c th khc nhau * Sc chu ti cho php:

[ ]s

gh

FP

P = ; Fs thng thng ly = 2. Tuy nhin h s an ton c th cao hn hoc thp

hn tu thuc mc quan trng ca CT, iu kin t nn, c im cc v phng php th nghim. (Tham kho Ph lc E TCXDVN 269: 2002) Theo quy phm ASTM D-1443 th c 7 cch gia ti khc nhau. Tuy nhin quy trnh

gia ti nhanh ( QL Quick Load) thng c s dng nht. Quy trnh ny nh sau: D tnh ti nn ti a Pmax. Thng thng Pmax= 2 3 [P]; [P] l sc chu ti thit k.

Mi cp s l 1/20 Pmax ( nh vy s c 20 cp ti) Mi cp ch gi 2,5 n 5 pht. Sau gim ti.

Nhn xt:

- c coi l phng php tin cy nht. - lu, cng knh, tn km.

Bi 1

Kt qu th nghim nn tnh cc BTCT tit din (30 30) cm2 cho trong bng sau. Hy xc nh ti trng gii hn ln cc v ti trng cho php ca cc theo TCVN. Bit rng ln cho php ca cng trnh [ S] = 6cm P(tn) 5 10 15 20 25 30 35 40 42 44 S(mm) 2 3,5 5,1 6,9 8,8 12,6 17,2 24,3 30,4 35,2

Bi lm:

105

S(mm)

0

P(t)5

10

15

20

25

30

35

40

45

5 10 15 20 25 30 35 40 45

12mm

28T

Trc ht da vo kt qu th nghim ta xy dng th quan h S=f(P) nh hnh trn, ng vi ln quy c S* = 0,2[S] = 12mm ta xc nh ti trng tng ng. T th ta c: Pgh = 28T

v [ ] TFP

Ps

gh 4,2225,1

28===

chn [P] = 22 T 2.Phng php ng: * Nguyn l: Sau khi h cc n su thit k, cc ngh, tin hnh ng th vi ba trng lng Q, cao ri ba H, o ln 1 nht ba gy ra l e - chi ca cc - xuyn su bnh qun ca cc do 1 nht ba. (ng cc th phi tin hnh t 3 n 5 nht ba. cao ri ba phi ng u cho tt c cc nht, v ly gi tr trung bnh.)

Pgh 1/e Thit b: Ba ng cc: Q; H Dng c o ln.

Din dch kt qu:

Theo Gerxvanov:

QHe

nFqQqkQnFnFP

TNgh

+

++

+=

21

2

22

F = din tch tit din ngang ca cc. Q = trng lng phn ba ri q = trng lng cc bao gm c m cc, m cc v cc dn nu c.

HL

106

H= chiu cao ri ba tnh ton. k1= h s phc hi khi va chm; khi thp vi gang va chm vi g th ly k1 = 0,2 n = h s kinh nghim, ph thuc vt liu lm cc v cch ng cc.

theo bng. H s n

Loi cc v m cc n(kG/cm2) Cc g: - c m - khng m Cc b tng ct thp c cc dn Cc thp: - c m g - c m thp v cc dn - c cc dn, khng m

8 10 15 20 30 50

Ti trng tnh ton c xc nh nh sau:

tc

gh

kP

P = ktc = ly theo ph lc A TCXD 205:1998 ( =1,25 hoc 1,4)

TCXD 205: 1998 ( Quy phm mi ca VN) c kin ngh 2 cng thc tnh Pgh ng vi trng hp e 0,002m v e< 0,002m.

Khi chi o c meTN 002,0

( )

++

+++= 141

2 11

2

www

www

nFeEnFMP

cn

cn

gh

Trong : n = h s ly bng 150t/m2 i vi cc BTCT c m cc. F= din tch cc. M = h s kiu ba, ly bng 1 khi ng cc bng ba tc dng va p cn khi h

cc bng ba rung th ly theo bng. E = nng lng tnh ton ca 1 va p ba theo bng. ef = chi thc t, bng ln ca cc do mt va p ca ba cn khi dng

my rung l ln ca cc do cng ca my trong thi gian 1 pht,m. e = chi n hi ca cc xc nh bng my o chi. w = trng lng ca phn va p ca ba, T. wc = trng lng ca cc v m cc, T. w1 = trng lng ca cc dn ( khi h cc bng rung w1=0),T. wn = trng lng ca ba hoc ca my rung,T. = h s phc hi va p, khi ng cc ng b tng ct thp bng ba tc ng va

p c dng m m g ly 2 = 0,2 h bng rung ly 2 =0

Khi chi o c meTN 002,0< nn chn ba sao cho tho mn /kin trn nu khng c th i c thit b ng th tham kho cng thc trong TCXD 205:1998.

107

Hin tng chi gi khi ng cc: chi o c ngay sau khi va ng cc xong khc vi chi thc. Vi t ct chi o c ngay sau khi ng cc < chi thc. Vi t st th ngc li.

L do: Khi ng cc do rung t dn cht li lm ma st tng ln chi nh i. Sau

mt thi gian cc tnh cht ca t c phc hi chi o c ln hn.

Vi t st khi ng t b p cht nn nc b thot ra, nhng mi ch thot nhiu ra pha xung quanh cc lm ma st gim i nn chi tng ln, mt thi gian t hi phc chi gim i.

Vi: t ct sau: 23 ngy th nghim t st 1530 ngy hoc lu hn.

Theo tiu chun ca H lan:

( )eqQHQPgh

+=

2

75

=ghPP ; H s an ton thng ly bng = 6 ( B tc k s)

Bi 2:

Kt qu ng th cc bng ba dizen kiu ng c: Trng lng qu ba: Q= 12,5kN. Trng lng ton phn ca ba: Qn = 26kN Chiu cao ri ti a ca qu ba: H= 3m Cc b tng ct thp c tit din (30 30) cm2, trng lng cc: q= 20,5kN Cc c m lt bng g. Trng lng m g v tht thp ca my trn u cc: q1= 2KN Kt qu th cho chi ca cc l: e = 0,008m

Hy xc nh sc chu ti ca cc.

Bi lm:

Khi th ng cc ng, nu chi thc t o c ef 0,002m; Qu xc nh theo cng thc:

]1)(.41[2 1

12

++

+++=

WWWWWW

nFenFMQ

cn

cn

f

pu

Nu chi o c < 0,002m th nn i thit b c nng lng ln hn h cc. Nu khng i c th dng cng thc trong 205: 1998 n = h s kinh nghim ph thuc vt liu lm cc v cch ng cc tra bng:

Cc BTCT c m lt bng g: n =1500kpa

108

F = din tch cc M = h s ly bng 1 khi ng, cn khi rung tra bng loi t di mi cc. p = nng lng tnh ton ca mt va p ca ba tra bng loi ba. ef = chi thc t bng ln ca cc do mt va p ca ba. W = trng lng phn va p ca ba. Wc = trng lng ca cc v m cc. W1 = trng lng ca cc dn. Wn = trng lng ca ba. = h s phc hi va p khi ng cc v cc ng BTCT bng ba c dng m m g ly 2 = 0,2 Thay s:

=

++

+++= ]1

25,2026)25,20.(2,026

.

008.0.09,0.15003.5,12.9,0.41[

21.09,0.1500

uQ

KNQu 7744,534)19226,8.(5,67]16289,0.1251.[5,67]15,485,30

.

08,11351.[5,67 ==+=+=

Sc chu ti cho php ca cc:

tc

tca k

QQ = ; ktc = 1,4 h s an ton; du

tc kQQ = ; k = 1 h s an ton theo t

Vy TKNKNQa 383824,17744,534

===

3, Phng php th ng bin dng ln: ( PDA- Pile Driving Analyzer) L phng php ca gio s G.C. Goble, Hc vin cng trnh Case ca M o ng xc nh gn ng sc chu ti ca cc n v phn on cht lng thn cc, gi l phng php Case. Phng php Case l phng php o ng v phn tch dng l thuyt truyn sng

lm c s. ng thi nghin cu ch to thit b phn tch ng cc PDA c th ngay hin trng thu c cc tham s nh sc chu ti ca cc, cht lng thn cc, ng sut ng cc, nng lng ng ba v tnh nng lp m.

Thit b th nghim gm 2 b phn: + Thit b ng: Phi lm cho gia cc v t c chuyn v tng i, nng lng tc ng ln u cc phi tng i ln, v vy phi dng ba nng ng ln u cc. Trng lng ba thng ly bng 1% sc chu ti ca cc. + My o: C nhiu: M, H lan, Thu in, Trung quc. Hin nay c phn mm CAPWAP vn s dng kt qu o c trong th nghim PDA qu trnh phn tch ca CAPWAP cht ch hn, chnh xc hn v a li nhiu kt qu hn so vi phng php Case trn.

109

4.Xc nh SCT theo phng php thng k: Da vo k/qu chnh l nhiu s liu thc t v th nghim th tnh cc h trong cc loi t khc nhau tm tng quan gia ma st v phn lc t mi cc vi mt s ch tiu c l.

Vi cc chu nn:

+=

=

n

iiitt

tc

FRlumk

P i1

211

Hnh 2. B tr th nghim PDA

Hnh 3. S thit b th nghim PDA

110

Vi cc chu ko: in

iitt

tc

lumk

P =

=

11

1

P = sc chu ti tnh ton. m = h s /k lm vic ca cc trong t theo TCXD 205: Cc ng tit din

111

Bng: Cng tnh ton ca t di mi cc, nR (kPa)

t loi ri cht va si Ct to Ct va Ct nh Ct bt

t st c ch s st Il bng

Chiu su h cc (m)

0,0 0,1 0,2 0,3 0,4 0,5 0,6 7000 3400 2200 5 8800 6200

4000 2800 2000

1300 800

7300 3700 2400 7 9700 6900

4300 3300 2200

1400 850

7700 4000 2600 10 10500 7300

5000 3500 2400

1500 900

8200 4400 15 11700 7500

5600 4000

2900 1650 1000

4800 20 12600 8500 6200 4500

3200 1800 1100

23 13400 9000 6800 5200 3500 1950 1200 35 15000 10000 8000 6000 4100 2250 1400

S trn cho t ct, s di cho t st. Bng: Cc h s 1 v 2 dng trong cng thc thng k

H s iu kin lm vic ca t c k n mt cch c lp vi nhau khi tnh

ton sc chu ti ca cc

Phng php h cc

Di mi cc, 2 mt bn cc, 1 1.H cc c, cc rng bt mi bng ba 1 1 2.Cc ng vo l khoan mi, vo trong t cha khoan ti thiu 1m, vi ng knh l khoan Bng cnh cc vung Nh hn cnh cc vung 5cm Nh hn cnh cc vung hoc ng knh cc

trn 15cm

1 1 1

0,5 0,6 1

3.H cc trong t ct c xi nc nhng ng tip m cui cng khng xi nc

1

0,9

Bi 3

Hy d bo sc chu ti ca cc BTCT tit din vung (25 25) cm2, di 12m c ng vo nn t c a tng gm 3 lp (theo th t trn xung) nh sau:

Ct pha do dy 6m c st Il = 0,6 Ct bt cht va dy 4m St do cng Il = 0,3

(cha kho st ht chiu dy trong su h khoan 18m) Bit rng nh cc cch mt t 0,5m; y i cch mt t 1,0m

112

Bi lm:

1, i vi cc chu nn: Chu vi tit din cc u = 0,25m 4 = 1,0m Tit din ngang cc Fc = 0,25 0,25 = 0,0625m

2 Chiu di lm vic ca cc lc = 11,5m; Mi cc su 12,5m k t mt t. S lm vic ca cc nh trn hnh v. Da vo s lm vic ta thy:

Cc phn lp 1;2;3 thuc lp t ct pha do; Cc phn lp 4;5 thuc lp t ct bt cht va; Phn lp 6 thuc lp t st do cng.

Sc chu ti ca cc d bo theo cng thc sau:

66

1RFluP ci

iigh +=

=

Ct pha do

Ct bt cht va

6000

4000

2500

0.000

500 800

200

1000

300

800

-1.500

-5.000

-7.000

-9.000

-11.250

-12.500

-3.000

St do cng

-1.000

-6.000

-10.000

-12.500

0.000

113

Tra bng: Trong : hi su trung bnh ca mi lp t chia.

Lp t Loi t li (m)

hi (m)

i Kpa

1 Ct pha do Il=0,6

1 2 2

1,5 3,0 5,0

10 14 17

2 Ct bt cht va 2 2

7,0 9,0

32 33,5

3 St do cng Il=0,3

2,5 11,25 47

R6=f(Il=0,3; L6=12,5m) tra bng = 3750 Kpa

Sc chu ti gii hn ca cc:

1,0.{[1.10 2.14 2.17 2.32 2.33,5 2,5.47]

0.0625.3750} 320,5 234, 4 554,9 55,5ghP

Kpa T

= + + + + + +

+ = + = =

[ ] 55,5 401, 4

gh

tc

PP T

k= = ; i n v: 1 kPa = 0,1T/m2

2, i vi cc chu ko:

cii

igh GluP +==

6

1

TKpalu ii 325,320]47.5,25,33.232.217.214.210.1.{[0,1..6

1=+++++=

TmTmGc 88,1/5,2.12.25,0.25,0 3 == ; [ ] TTP 1568,1488,18,1288,15,232

=+=+=

3.Mt s cng thc thc nghim xc nh SCT ca cc: Cng thc tnh SCT ca cc gm 2 thnh phn + Tng lc khng bn quanh cc Qs + Lc khng ti mi cc Qc 1.Da vo kt qu xuyn tnh: CPT

cc

n

i

ciigh qFkqluP +=

1

ciq = sc khng xuyn trung bnh ca lp t th i cq = sc khng xuyn trung bnh ca t trong phm vi 3d pha trn chn cc v 3d pha di chn cc.(d l ng knh hay cnh ca cc)

kc = h s tra bng.

hd

l1l2

114

s

ghd F

pP = theo TCVN 205:1998: Fs = 23

Bng: H s chuyn i ki v i t kt qu CPT i vi cc BTCT

H s ki H s i Gi tr cc i ca i Loi t qc (kPa) Cc

khoan Cc ng Cc

khoan Cc ng Cc

khoan Cc ng

t loi st chy,bn

0-2000 0,4 0,5 30 30 15 15

t loi st cng va

2000-5000

0,35 0,45 40 40 (80) 35

(80) 35

t loi st, cng n rt cng

>5000 0,45 0,55 60 60 (80) 35

(80) 35

Ct chy 0-2500 0,4 0,5 960) 120

(60) 80

35 35

Ct cht va

2500-10000

0,4 0,5 (100) 180

100 (120) 80

(120) 80

Ct cht n

rt cht

>10000 0,3 0,4 150 150 (150) 120

(150) 120

Cc ga tr trong ngoc c th s dng khi:

i vi cc nhi, thnh h c gi tt, khi thi cng khng gy ph hoi thnh h v b tng cc t cht lng cao. i vi cc ng c tc dng lm cht t khi ng.

Bi 4:

Hy d bo sc chu ti theo t nn ca cc BTCT tit din (25 25) cm2, di 12m c thi cng theo phng php p trc vo nn t c a tng gm 3 lp nh sau (k t mt t xung)

ct pha do dy 6m: sc khng xuyn qc = 15 kG/cm2

bn st dy 4m: sc khng xuyn qc = 4 kG/cm2

ct ht trung cht va: sc khng xuyn qc = 45 kG/cm2

(cha kho st ht chiu dy trong su h khoan 18m) Bit rng nh cc cch mt t 0,5m; y i cch mt t 1,0m

115

Bi lm S lm vic ca cc v a tng nh hnh v:

Theo kt qu xuyn tnh, ta xc nh c chiu di cc qua cc lp t v cc gi tr i

v Rn theo cc bng sau: ccn

i

ciigh qFK

qluP += 1

; Kc ph thuc loi cc v phng php h cc: tra bng l1 = 5m; 1 = 30

l2 = 4m; 2 = 30 l3 = 2,5m; 3 = 100 k3 = 0,5 Sc chu ti gii hn ca cc:

=+++= 45.5,0.625]10045

.250304

.4003015

.500.[100ghP

32 667,0+14 062,0 = 46729kg = 46,7T

[ ] TTpP gh 197,185,27,46

5,2===

2.Da vo kt qu xuyn tiu chun ( SPT) Theo Mayerhof:

s

tbp

tbgh NulKFNKP += 21 s

tbp

tb NN ; = tr s SPT trung bnh trong khong 1d di mi cc v 4d trn mi cc, v ca cc lp t dc theo thn cc.

K1 (kN/m2) = 400 cc ng; =120 cc nhi

K2(KN/m2) = 2 cc ng ; = 1 cc nhi

[ ]s

gh

FP

P = 35,2 =sF ( TCVN 205:1998)

5000

4000

0.000

-1.000

-6.000

-12.500

1000

2500

116

Theo cng thc ca Nht bn:

[ ]

++= ucLLNFNP

n n

cisisia1 1

2,031

aN - ch s SPT trung bnh trong khong 1d di mi cc v 4 d trn mi cc (d l ng knh hay cnh ca cc)

siN - ch s SPT trung bnh dc thn cc ca lp ct th i m cc i qua.

siL - chiu di on cc nm trong lp t ct th i m cc i qua, m;

siL - chiu di on cc nm trong lp t st th i m cc i qua, m; - H s, ph thuc vo phng php thi cng cc; cc BTCT thi cng bng phng php ng hoc p: =30; cc khoan nhi =15. Bi 5: Cng nn t nh trn, chiu di cc nh trn. Kt qu th nghim SPT cho kt qu sau: su th nghim (m)

2 4 6 8 10 12 14 16

Gi tr N 3 3 1 1 20 21 22 22

Lc dnh ca lp ct pha c=1,1T/m2 ; lp bn st c=0. Hy d bo sc chu ti ca cc.

Bi lm:

* Theo cng thc ca Nht bn: [ ]

++= ucLLNFNP

n n

cisisia1 1

2,031

; 21aN

[ ] ( ){ }( )[ ] ( ) T

P

183,1815,154,3931155,104,39

31

25,0451,15,2212,00625,0213031

=+=++=

++=

Lu : Theo TCXD205: 1998: * Cn k n trng lng cc nh ti trng tc dng trong tr/ hp cc chu ma st m.

Sc chu ti cho php ca cc l: [ ] wFP

wwFP

Ps

ghdatcoc

s

gh== )(

w = l hiu s gia khi lng bn thn cc v khi lng bn thn t do n chim ch c xt n lc y acsimet ca phn cc di mc nc ngm. Sc chu nh cc hn ca cc theo t nn ly bng tng ma st bn cc c k thm trng lng cc.

* SCT trng ngang cc hn ca cc c tnh ton khi cc chu tc dng ng thi ca m men un, lc ngang, lc dc trc v phn lc ca t nn - Tham kho ti liu.

SCT theo phng ngang trc xc nh theo th nghim hin trng hoc theo kinh nghim nh bng trong gio trnh NM.

117

5.6.Tnh ton mng cc i thp: Ni dung: 1.Chn loi, kch thc i v cc. 2.Xc nh SCT ca cc. 3.S b xc nh s lng cc. 4.B tr cc trn mt bng v ng. 5.Kim tra theo cc trng thi gii hn. nu khng tho mn thay i li.