Tinh Toan Mong.ok

Kin trc cng trnh

TI : Trung tm thng mi v vn phng cao cp nn mng cng trnh

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Mc lc phn nn mng:

I. nh gi c im cng trnh.

II. nh gi iu kin a cht cng trnh.

II.1. a tng.

II.2. Bng ch tiu c l ca cc lp t.

II.3. nh gi tnh cht xy dng ca cc lp nn.

III. Phn tch v la chn gii php nn mng.

III.1. Khi nim v cc chng v cc ma st.

III.2. Nghin cu v cc ma st.

III.2.1. Phn tch cc ma st n chu ti trng thng ng.

III.2.2. Xc nh chuyn v ca cc n.

III.2.3. Bn cht ca cc ln thnh phn trong cc ma st.

III.2.4. Tng quan v ma st m v nhng nh hng ca n n

sc chu ti ca cc.

III.3. Nghin cu v cc khoan nhi.

III.3.1. Tng quan.

III.3.2. Gii php mt bng mng.

III.3.3. Cc gi thit tnh ton.

III.3.4. Ti trng.

III.3.5. Thit k cc khoan nhi.

III.3.6. Tnh ton s lng cc khoan nhi cho cng trnh.

III.3.7. Tnh ton i mng.

IV. Kim tra nn mng cng trnh theo phng php phn t hu hn.

IV.1. Xy dng m hnh khng gian.

IV.2. Ti trng tc dng ln mng cng trnh.

IV.3. nh gi sc lm vic hiu qu ca cc khoan nhi.

IV.4. Phn tch ng sut trong i mng b.

V. Thit k tng Diaphragm.

V.1. Tng quan v la chn phng php tnh ton tng Diaphragm.

V.2. Xc nh s tnh tng Diaphragm.

V.3. Xc nh su chn tng, lc dc thanh chng v mmen thn

tng theo tng giai on thi cng.

V.4. Xc nh s b kch thc cu kin.

V.5. Kim tra n nh kt cu chn gi tng Diaphragm.

V.6. Tnh ton v b tr ct thp cho tng Diaphragm.

I. nh gi c Im cng trnh.

Cng trnh Trung tm thng mi v vn phng cao cp gm c 41 tng ni v 3 tng hm.

Cng trnh c ct sn tng 1 l , ct t nhin .

Tng hm -1 nm ct 4.2 m ;

Tng hm -2 nm ct 7.8m ;

Tng hm -3 nm ct 11.4m ;

Chiu cao t ct ( 0.00 n nh to nh l +150,3m

Khi tnh ton nn mng theo TTGH II, cn khng ch ln gii hn v ln lch gii hn ca cng trnh c th s dng cng trnh mt cch bnh thng, v ni lc b sung do s ln khng u ca nn gy ra trong kt cu siu tnh khng qu ln kt cu khi h hng v m bo m quan ca cng trnh :

Trong :

St: ln tuyt i, ln nht ca mt mng (cm).

(S: ln lch tng i gia hai mng.

Do c im cng trnh l kt cu nh cao tng bng thp vi h ging, do theo TCXD 45-78: ln gii hn tuyt i cho php Sgh = 8(cm), ln lch cho php l (S = (Smax-Smin)/L = 0,002. II. nh gi iu kin a cht cng trnh :

II.1. a tng :

Theo kt qu kho st a cht cng trnh, a cht di l khoan su 50m gm cc lp t nh sau:

+ Lp 1 t 0(1,2 m l lp t lp , = 18KN/m3.

+ Lp 2 t 1,2( 5,8 m l lp st do cng , N = 18.

+ Lp 3 t 5,8 ( 9,2 m l lp st pha do mm , IL = 0,5 , N = 12.

+ Lp 4 t 9,2 ( 12,8 m l lp st pha do chy , N = 4.

+ Lp 5 t 12,8 ( 19,5 m l lp ct pha do , N = 21.

+ Lp 6 t 19,5 ( 26,3 m l lp ct bi cht va , N = 30.

+ Lp 7 t 26,3 ( 38,4 m l lp ct ht trung cht , N = 55.

+ Lp 8 t 38,4 ( 50 m l lp ct th cui si , N = 80.

Mc nc ngm xut hin su - 6,5 m k t mt t thin nhin.

a tng c phn chia theo th t t trn xung di nh sau:

II.2. Bng ch tiu c l ca cc lp tTheo kt qu kho st a cht cng trnh v kt qu th nghim trc tip vi

t cho kt qu c l ca cc lp t trong bng ch tiu c l nh sau:

LpTn tChiu

dy (m)(tn(KN/m3)(h(KN/m3)W

(%)WL(%)WP(%)k

(m/s)((()CII(KPa)m(MPa-1)E

(MPa)

1t

lp1.218---------

2St

Do cng4.618,226,93950303,1.10-1013190,117,5

3St pha

do mm3.417,526,63845311,0.10-81150,207

4St pha

do chy3.618,526,833,236222,5.10-816100,1210

5Ct pha

do6.719,226,52024182,1.10-718250,0914

6Ct bi

cht va6.81926,526--3,1.10-630-0,1310

7Ct ht trung cht12.120,126,416--2,0.10-438-0,0340

8Ct th cui si> 11.620,126,416--2,0.10-343-0,0350

II.3. nh gi tnh cht xy dng ca cc lp t nn:

a. Lp t 1: Lp t lp c chiu dy 1,2m.

Phn b mt trn ton b khu vc kho st c chiu dy 1,2m . L lp t yu v kh phc tp, nn cht cha n nh. V vy khng th lm nn cho cng trnh nn lp ny phi c bc b ht.

b. Lp t 2: Lp St do cng c chiu dy 4,6m. - Kt qu th nghim xuyn tiu chun:

N = 18 ba / 30cm.

- st:IL=== 0,45

0,25 < IL = 0,45 < 0,5 ( t trng thi do cng.

- H s rng: e = -1 = -1 = 1,054 < 1,1

- H s nn ln: 0,1MPa-1 < m = 0,11MPa-1 < 0,5MPa-1

( kh nng chu nn trung bnh.

- Mun bin dng: 5MPa < E = 7,5 Mpa < 10MPa ( trung bnh.

( Kt lun: Lp 2 l St do cng c kh nng chu ti trung bnh, mt khc do vi cng trnh nhiu tng th chiu dy lp t kh mng khng thch hp lm nn mng.

c. Lp t 3: Lp St pha do mm, chiu dy 3,4 m.

- Kt qu th nghim xuyn tiu chun :

N = 12 ba / 30 cm.- st:IL == = 0,5 < 0,75

( t trng thi do mm

- T trng:( = =

- H s rng: e = -1 = -1 = 1,098 >1 ( yu

- Trng lng ring y ni: (n = = = 7,914 KN/m3- H s nn ln: 0,1MPa-1 < m = 0,2 MPa-1 < 0,5MPa-1

( kh nng chu nn trung bnh.

- Mun bin dng: E = 5MPa < E = 7Mpa < 10MPa ( trung bnh.

( Kt lun: Lp 3 l St pha Do mm c kh nng chu ti trung bnh, tnh nng xy dng yu, bin dng ln ln. Do khng th lm nn cho cng trnh c.d. Lp t 4: Lp St pha do chy, chiu dy 3,6 m.


N = 4 ba / 30 cm.- st:IL == = 0,8 > 0,75

( t trng thi do chy ( yu

- T trng:( = =

- H s rng: e = -1 = -1 = 0,93 10MPa ( tt.

( Kt lun: Lp 5 l Ct pha Do c kh nng chu ti tng i tt, tnh nng xy dng kh tt, bin dng ln nh, chiu dy lp t bng 6,7m. V vy c th lm nn cho cng trnh c.f. Lp t 6: Lp Ct bi cht va c chiu dy 6,8m .


N = 30 ba / 30 cm.

- T trng:(= = = 2,65

- H s rng: e = -1 = -1 = 0,757

0,6 < e = 0,757 < 0,8 ( ct trng thi cht va.

- Trng lng ring y ni: (n = = = 9,389 KN/m3- H s nn ln: 0,1 MPa-1 < m = 0,13 MPa-1 < 0,5 MPa-1

( ct bi c kh nng chu nn ln trung bnh.

- Mun bin dng: E = 10 MPa ( tt

( Kt lun: Lp 6 l Ct bi Cht va c kh nng chu ti tng i tt, tnh nng xy dng kh tt, kh nng chu nn ln trung bnh, chiu dy lp t bng 6,8m. C th lm nn cho cng trnh c.

g. Lp t 7: Lp Ct ht trung cht, chiu dy 12,1m.

- Kt qu th nghim xuyn tiu chun:

N = 55 ba / 30 cm.

- T trng:

(= = = 2,64

- H s rng: e = -1 = -1 = 0,524

e = 0,524 < 0,6 ( ct trng thi cht.


( ct ht trung cht c kh nng chu nn ln tt

- Mun bin dng: E = 40 MPa > 30MPa ( rt tt

- Gc ma st trong: ( rt tt.

( Kt lun: Lp 7 l Ct ht trung Cht, c kh nng chu ti ln, tnh nng xy dng tt, bin dng ln nh, chiu dy lp t ln (12,1 m). Do dng lm nn cho cng trnh cao tng s tt.

h. Lp t 8: Lp Ct th cui si, chiu dy 11,6m.

- Kt qu th nghim xuyn tiu chun:

N = 80 ba /30 cm ( t rt cht.

- T trng:

( = = = 2,64

- H s rng: e = -1 = -1 = 0,524

e = 0,524 < 0,6 ( t trng thi cht.


( ct th cui si c kh nng chu nn ln tt

- Mun bin dng: E = 50 MPa > 30MPa ( rt tt

( Kt lun: Lp 8 l Ct th cui si, c kh nng chu ti ln, tnh nng xy dng tt, bin dng ln nh, chiu dy lp t ln (11,6 m) v cha kt thc trong phm vi l khoan 50m.

III. phn tch v La chn gii php nn mng.

Vi cc c im a cht cng trnh nh gii thiu, cc lp t trn l t yu xen kp khng th t mng cao tng ln c, ch c lp cui cng l ct th si cui c chiu dy khng kt thc ti y h khoan l c kh nng t c mng cao tng.

Vi quy m v ti trng cng trnh nh vy, gii php mng su (mng cc) l hp l hn c.

Cn c vo phng thc truyn ti trng ngoi thng qua cc vo t, thng chia cc lm hai loi: Cc chng v cc ma st. III.1. Khi nim v cc chng v cc ma st

III.1.1. Cc chng.

Ni chung cho rng, ti trng ngoi c truyn vo trong t ch thng qua mi cc th gi l cc chng.

Trn thc t, mt bn ca cc vn c lc ma st nhng khi thit k ch xt n kh nng chu ti trng ngoi do phn lc u mi cc (sc khng mi cc) m khng xt n kh nng chu ti do ma st mt bn cc. Cc chng ch s dng khi mi cc i vo tng hay tng t rn chc. ln ca cc chng rt b, s chuyn dch ca cc mt ct cc ch yu l do nn p n hi ca bn thn cc.

Trong thc t xy dng c 3 loi cc sau: Cc ng, Cc khoan nhi v cc Barets.

III.1.2. Cc ma st.

Mt b phn hay ton b ti trng ngoi tc dng ln u cc c truyn vo trong t xung quanh cc thng qua ma st mt bn cc, cc gi l cc ma st.

Khi tnh ton cc ny s xt n ma st gia t v mt bn cc, ng thi c xt n sc khng u mi cc. Nu ton b ti trng ngoi u truyn vo trong t thng qua ma st mt bn cc th gi l cc ma st thun ty. Trong thc t xy dng, nu t xung quanh cc v nn di mi cc u l t yu th cc c coi l cc ma st thun ty.

Iii. 2. Nghin cu v cc ma st:

III.2.1. Phn tch cc ma st n chu ti trng tnh thng ng

III.2.1.1 Sc chu ti dc trc ca cc n

Sc chu ti cc hn ca cc c chia thnh sc khng bn v sc khng mi nh sau:

Trong :

+ Sc khng bn:

y: - l ma st bn n v ca cc

- chu vi thn cc;

- chiu di on phn t cc m trn c coi l hng s;

- din tch xung quanh ca on phn t cc.

+ Sc khng mi:

y: - sc khng mi n v ca cc;

- tit din ngang mi cc;

Sc chu ti nn ca cc l:

trong : - h s an ton, thng ly t 2 n 4;

w - 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;

Nhiu nghin cu thy rng: Di ti trng cho php, chuyn v ca cc kh nh. Mc d chuyn v nh, sc khng bn ca cc vn c huy ng kh ln. Tuy nhin, ti chuyn v , sc khng mi mi ch huy ng mt phn nh . Do , ta nn dng hai h s an ton nh sau:

H s an ton cho sc khng bn thng ly bng , cn h s an ton cho sc khng mi trong khong .

Ta cn thit k sao cho ti trng tc dng nh hn sc chu ti, tc l:

Cc gi tr: ma st n v thn cc (), lc dc thn cc (), v chuyn v mt ct cc (s) s thay i khi ti trng tc dng ln u cc Q tng dn. Khi u Q tng i b, ch yu do lc ma st bn on thn cc pha trn chu. Khi Q tng n mt gi tr no th mi cc chuyn dch, phn lc u cc mi bt u huy ng mt cch r rng.

Cn c ti liu th nghim khi chuyn dch tng i ga cc v t khong (vi t st) hoc (vi t ct) th ma st n v mt bn cc t trng thi gii hn. V khi chuyn dch u mi cc t ln ng knh cc, th phn lc u mi cc mi t trng thi gii hn.

Chuyn v tng i

ca cc v t S huy ng sc khng

III.2.1.2. Xc nh lc nn dc trc ca cc n

Da theo s lm vic ca cc trong t khi

c ti trng tc dng u cc Q.

Ti su z (z < l, l - l chiu di cc) th ta

xc nh ni lc thn cc theo cng thc ca

Sc bn vt liu:

Cng thc tnh lc dc thn cc:

;

Trong : l lc dc thn cc su z;

l lc tp trung t ln u cc;

l chiu di on cc;

l ma st n v mt bn cc, c

gi tr khng i trong on li III.2.2. Xc nh chuyn v ca cc n

Vi mc ch thit k v tm hiu r qu trnh lm vic ca cc, ln ca cc c th tch thnh 3 phn:

Trong :

l ln tng cng u cc n

l ln do chuyn v ca thn cc, gy ra do s bin dng cong ca t

xung quanh cc (t b bin dng cong do c s dnh bm gia cc v t);

l ln do bin dng thn cc, gy ra do s nn p n hi vt liu lm cc;

l ln do ti trng mi cc (gi tr ni lc thn cc ti su mi cc)

truyn vo nn.

III.2.3. Bn cht ca cc ln thnh phn v cch xc nh

a. Trng thi ban u

Sau khi ng cc vo t, cha c ti trng tc dng u cc. t xung quanh cc v t nn di mi cc s phc hi tr li nhng bin dng do qu trnh ng cc (b qua nhng bin dng d).

Lc ny cc mt ct lp t mi su c dng nm ngang v vung gc vi thn cc. ln u cc trng thi ny (trng thi ban u) bng khng. Cc hnh vung A, B biu th cc phn t quy chiu trong lp t, ab l mt ng nm ngang quy chiu trong t .

b. Trng thi th 2.

Sau tc dng ln u cc mt ti trng tp trung Q = Q1 cha ln, th cc lp gn pha trn vng u cc t bt u b bin dng cong. ng quy chiu ab ban u s chuyn dch thnh ab, cc phn t quy chiu b mo i thnh A, B. Gia b mt cc v phn t t A xut hin ng sut tip , gia hai phn t A v B c ng sut tip , gia phn t B vi cc phn t bn ngoi l ng sut.

Hin tng xon trn chng t rng: ti trng u cc Q = Q1 truyn vo trong t xung quanh cc thng qua cc ng sut tip. Cc ng sut tip s gim dn khi cc phn t xa dn cc () do s phn tn ng sut vo khng gian.

ln ca cc trng thi ny gi l sbd.

Tip tc tng ti trng tc dng u cc, cc lp t pha di su vng mi cc cng b cc ng sut tip gy nn bin dng un cong, gi tr sbd tng dn.

Nhng cha c s chuyn dch tng i gia cc v t.

c. Trng thi th 3.

Tip tc tng ti trng u cc t gi tr Q = Q2 < [Q], n khi cc lp t khng b un cong theo ln ca cc na, lc ny cc huy ng sc khng mi r rt. Ni lc ti mi cc s truyn xung nn (t di mi cc) v ln tng thm ca c cc lc ny .

Cc gi tr sbd v c xc nh nh th nghim nn tnh cc n ti hin trng. Ring gi tr cn c xc nh bng cch sau:

n gin, ta gi thit nn l nn Winkler, v do phn lc nn mi cc v chuyn v ca mi cc t l bc nht, tc l:

hoc

Suy ra:

Trong :

lc dc thn cc ti v tr mi cc,

l din tch tit din mt ct mi cc

h s phn lc nn

nn p n hi thn cc do lc dc trc cc () gy ra c xc nh theo cng thc ca Sc bn vt liu:

Trong : l bin dng n hi ca on cc c chiu di li ;

l lc dc thn cc trung bnh trong on li ;

l din tch tit din mt ct thn cc

l mun n hi ca vt liu lm cc

V ln u cc do nn p n hi thn cc l:

, ( chia cc thnh n on li )

Cc lm vic di tt trng Q2 th biu din ln th biu din lc dc

d. Trng thi th 4.Tip tc tng ti trng n Q = Q3 > [Q], th lc dc ti mi cc s ln hn sc khng mi cc, t s b ph hoi ct. Lm xut hin s chuyn dch tng i gia cc v t xung quanh cc, chuyn dch ny tng nhanh mt cch ng k, v hiu qu tr thnh mt cc khc vi cc ban u, c chiu su ln hn v trong cc iu kin t c th l khc trc.

Kt lun: Vic huy ng sc khng trong t i lin vi s bin dng; lng dch chuyn ca cc cn thit lm bin dng t ti im ph hoi th ph thuc vo tnh cht ca t, ti trng u cc, kch thc cc, vt liu lm cc v cch h cc vo t.

III.2.4. Tng quan v ma st m v nhng nh hng ca n n s lm vic ca cc

III.2.4.1. Ma st m (negative skin friction) v nguyn nhn hnh thnh

Thng thng khi ng cc vo tng t c kt n nh, nu u cc chu ti trng tc dng th cc s b ln xung. Lc mt bn cc s chu lc ma st do t gy ra theo chiu hng ln, c tc dng cn tr s ln xung ca cc. Lc ma st ny gi l ma st dng.

Trong thc t xy dng, c nhiu trng hp ngc li l do mt s nguyn nhn no lm cho tng t xung quanh cc b ln, ln ca tng t vt qu ln ca cc. Nh vy trong trng hp ny ln ca t lm gia tng ln ca cc, c ngha l t nn khng to c ma st dng m to ma st m.

R rng ma st m va c tc dng gia tng ln ca cc, va lm tng lc dc thn cc, gy nh hng ca chng n sc chu ti v s lm vic ca cc.

III.2.4.2. Nguyn nhn hnh thnh ma st m

Tnh hnh sn sinh ra ma st m c nhiu nguyn nhn, v d:

Xung quanh cc l t st mm c kt cha n nh.

Xung quanh cc l t mi p, t s b c kt do trng lng bn thn

Trn mt t b cht ti vi din rng, khin t xung quanh cc b nn cht.

Lp t nn quy hoch c chiu dy trn mt mt

Ph ti trn nn kho ln hn 20 Kpa

S gim th tch t do cht hu c trong t b phn hy

S tng ng sut hu hiu trong t do mc nc ngm b h thp

S tng cht ca t ri di tc dng ca ng lc

S ln t ca t khi b tng m do ngp nc

Khi ng cc gy ra s tri ln i vi nhng cc ln cn c h trc ...

III.2.4.3. S hnh thnh ma st m i vi cc n v cch xc nh

I.1. im trung tnh v phm vi xut hin ca ma st m trn cc n

V mt nguyn nhn no , lm cho t xung quanh cc b nn p hoc b dch chuyn i xung. S dch chuyn ny s tc ng n cc trong hai trng hp sau y:

+ Trng hp 1:

Cc cha chu ti trng tc dng ln u cc, th chuyn dch ca t s lm cc chuyn dch ko theo hoc gy nn ti trng ban u cho cc do ma st m gia cc v t. n khi tng t n nh khng cn ln na, lc ny cc trong iu kin t l hon ton khc trc.

+ Trng hp 2:

Cc chu ti trng tc dng ln u cc, c ngha cc b ln vo t, ln ny c phn tch trong mc III.2.3 (gi l ln ban u). Khi tng t b ln xung, hin tng ny s nh hng n s lm vic ca cc nh th no, ta cn phn tch hai trng hp sau c th xy ra nh sau:

a) Nu dch chuyn tng i ca tng t c chiu dy no nh hn ln ca cc ( ln ban u), dn n lc ma st dng ban u gia tng t v cc s b gim i.

b) Ngc li, dch chuyn tng i ca tng t c chiu dy no ln hn ln ca cc ( ln ban u), dn n khng nhng lm trit tiu hon ton lc ma st dng m cn gy ko cc xung bng ma st m () xut hin trong chiu dy tng t . Dn n cc b ln thm v lc dc thn cc tng ln

xc nh c khi tng t b ln xung gy nh hng mc no ti s lm vic ca cc th ta cn thc hin cc bc sau:

Bc 1. Xc nh ln ca ton cc Ta cn xc nh c cc thnh phn ln, cch xc nh trnh by trong mc II.2. Sau v c biu nn ln ca cc im trn cc theo chiu di thn cc c dng hnh 2.4.d.

Bc 2. Xc nh ln ca tng t xung quanh cc

V phm vi ca ti mi ch nghin cu nguyn nhn gy ln t do ti trng t p, C ngha sau khi xy dng cng trnh trn nn mng cc (cc ma st), sau ta phi p t tn nn, hoc p t sau tng chn. Di ti trng ca t p, lm cho t nn b ln xung. ln ny s c xc nh bng cng thc:

Trong : - ln ca lp t i-j ;

; l h s n hng ca t, thng ly ;

l ng sut gy ln trung bnh gia lp t i-j ;

l dy ca lp t i-j ;

l mun n hi ca lp t i-j.

V ln tng cng (tnh theo phng php cng ln tng lp) l:

( Chia tng t thnh n lp nh )

Sau khi tnh c ln ca tng lp t, ta s v c biu nn ln ca tng t xung quanh cc, th c dng nh sau:

Biu ln ca t

Bc 3: Xc nh im trung tnh im trung tnh l im m ti su khng c s chuyn dch tng i ca cc v t. c th hin bng s giao thoa biu ln ca t v biu ln ca cc.

Sau khi ta tm c chuyn dch ca cc v ln ca t c biu th bng biu (trong mc III.2.3). em chp hai biu trn cng mt h trc ta , hai biu ny giao nhau u th l im trung tnh.

C hai trng hp xy ra khi xc nh im trung tnh:

Trng hp 1:

im trung tnh nm trong phm vi chiu di cc.

Kh nng ny thng xy ra khi lp t m mi cc cm vo l lp t c cng tng i ln (nh ct ht to cht hoc t cng). Nguyn nhn, tng t cng vng mi cc lm gim hoc trit tiu ton b chuyn dch ca nhng lp t pha trn, v ln ca t lc ny ch yu do s nn p ca nhng lp t pha trn.

Mt khc, chnh lp t cng vng mi cc cng lm gim ln ca cc, nn trong trng hp ny im trung tnh s gn st b mt lp t tt vng mi cc.

Biu th bng th sau:

a) b) c) d)

S lm vic ca cc khi xy ra ma st m

Trong :

+ Hnh a. S chuyn dch tng i gia cc v t xung quanh cc;

+ Hnh b. th biu din s chuyn dch tng i gia cc v t

xung quanh cc. Trong , ng 1 biu th chuyn dch

ca cc, ng 2 biu th chuyn dch ca t; o1 l im

trung tnh; + Hnh c. th biu din phm vi v tnh cht phn b lc ma

st trn b mt thn cc;

+ Hnh d. th biu din lc dc thn cc khi chu nh hng ca

ma st m.

Trong trng hp ny, lc dc thn cc s c tnh nh sau:

Ti v tr im trung tnh th lc dc thn cc l ln nht. Bi ti , lc dc thn cc b nn bi ma st m t trn xung v ma st dng t di ln.

Lc dc thn cc ti v tr im trung tnh o1:

vi l tng lc ma st m tc dng ln mt cc n

; ( )

Lc dc thn cc ti v tr mi cc:

vi l tng lc ma st dng cn li tc dng ln mt cc n

; ()

Trng hp 2:

im trung tnh nm di phm vi chiu di cc (nm di mi cc). Kh nng ny thng xy ra khi lp t m mi cc cm vo l lp t c cng trung bnh (nh t ct ht mn, ht trung hoc t st cng). Nguyn nhn, lp t vng mi cc cng b ln khi xy ra hin tng ma st m, ln ny tng i ln, ln hn chuyn v ca mi cc. Dn n trn ton b chiu di cc, th chuyn dch ca t ln hn chuyn dch ca cc, do im trung tnh (im trung tnh o) s nm di mi cc.

Biu th bng th sau:

a) b) c) d)

S lm vic ca cc khi xy ra ma st m

Trong :

+ Hnh a. S chuyn dch tng i gia cc v t xung quanh cc;

+ Hnh b. th biu din s chuyn dch tng i gia cc v t xung quanh cc. Trong , ng 1 biu th chuyn dch

ca cc, ng 2 biu th chuyn dch ca t; im trung

tnh o o1 nm di su mi cc; + Hnh c. th biu din phm vi v tnh cht phn b lc ma st

trn b mt thn cc;

+ Hnh d. th biu din lc dc thn cc khi chu nh hng ca

ma st m.

Trong trng hp ny, lc dc thn cc s c tnh nh sau:

Cc b ko xung bi ton b ma st m trn b mt thn cc, nn lc dc thn cc ti u cc l ln nht.

Lc dc thn cc ti u cc:

; ()

im trung tnh cho ta kt lun rng:

+ Pha trn im trung tnh th chuyn dch tng i ca t ln hn ca cc. Nn trong phm vi , cc b t ko xung, c ngha on cc chu nh hng ca ma st m.

+ Ngc li, pha di im trung tnh th chuyn dch tng i ca cc ln hn ca t, nn cc vn huy ng ma st dng mt bn cc.

Kt lun chung : Do mc c kt ca t tng theo thi gian, cho nn chuyn dch ca t v chuyn dch mt ct cc u l hm s ca thi gian. Do vy, v tr ca im trung tnh, lc ma st v lc dc thn cc u bin i tng ng. Di tc dng ca ti trng ln u cc, nu chuyn dch cc mt ct cc n nh, sau t xung quanh cc mi xy ra c kt th tc v mc c kt s l nhn t ch yu nh hng n tr s v quy lut phn b ca ma st m. Mc c kt cao, ln mt t ln th im trung tnh s dch xung lm lc ma st m tng nhanh. Trong qu trnh , ln mi cc cng tng cao, d nhin dn n s chuyn dch tng i gia cc v t s gim nh v lc ma st m s gim dn v t ti trng thi n nh.

III.2.4.4. Gii thiu mt s phng php tnh ton ma st m i vi cc n .

a) Vi t st mm hay c cng trung bnh, ly tr s ma st m , vi cu l cng chng ct khng thot nc.

b) Nu mt t p hoc cht ti gy ra ma st m th ly tr s ma st m nh sau:

+ Khi H < 2m th khng xt ma st m.

vi H l cao t p, nu cht ti th H s l tr s tnh i ; l dung trng ca t, q l ti trng phn b .

+ Khi . Xt ma st m cho lp t yu v cc lp t p ln trn:

, vi l lc ma st dng gii hn

+ Khi H > 5m, ly

c) Cn c p lc hu hiu thng ng ) ti im trung tnh tnh ton lc ma st m theo cng thc sau:

k0 - h s p lc ngang

- gc ni ma st hu hiu

- h s, ly nh sau:

vi cc di 15 m

vi cc di 40 m

vi cc di 60 m.

Sau khi phn tch s lm vic ca cc ma st phn trn, ta quyt nh khng la chn loi cc ny bi c cc l do nh sau:

1. V cng trnh ny xy dng qun Cu Giy thuc thnh ph H Ni, do khi thi cng s gp mt s bt li sau:

Gy n trong thnh ph nu h cc bng cch ng.

Gy gy rung chuyn a cht (nu ng) v gy tri t i vi cng trnh ln cn.

Kh khn v chun b kho bi c cc hoc vn chuyn cc

Khng thch hp cho nhng gii php thi cng tng ngm (chng hn nh thi cng tng ngm bng phng php Top Down)

2. V quy m cng trnh rt ln, nn kt cu s gp nhng bt li sau:

Khi cng trnh b dao ng di tc dng ca ti trng gi hay ng t, th chnh s rung ng s lm gim dn lc ma st bn gia cc v t, dn n s lm gim dn sc chu ti ca cc.

Mt bng tng hm rt ln, do nu xy ra hin tng ln lch tm s to nn nhng mmen bt li cho kt cu.

3. Khng thch hp vi c im a cht xy dng cng trnh

Rt d xy ra hin tng ma st m i vi cc, v hai l do sau:

a. S tng ng sut hu hiu trong t do mc nc ngm b h thp:

Vic bm ht nc ngm phc v cho sinh hot v cng nghip h thp mc nc ca cc tng cha nc di t, lm gim p lc nc l rng, tng ng sut hiu qu trong t. iu ny s dn n bin dng ln b mt.

Theo bo co ca Hi ngh quc t v ln t ln th 4 c t chc ti thnh ph Houston (M) t 12 n 18 thng 5 nm 1991, th trn th gii hin ti c trn 50 thnh ph ln nh chu nh hng ca qu trnh bin dng ln, trong c 17 thnh ph chu hu qu nng n hn c. Tiu biu l thnh ph Mexico City, Houston, Bngcc,Tokyo, Osaca, Thng Hi, Qung Chu, i Bc, Venice...

Tc ln thnh ph Bng Cc o c l hn 100 mm/nm lan rng trn mt din tch 800 km2.

Houston l mt thnh ph ng dn ng nam bang Texas. Nc phc v sinh hot, cng nghip Houston u c bm t lng t. Mc ln Houston c ni vt qu 2 m. n nm 1981 din tch b ln ln n 80 km2. T nm 1969 n nm 1974 , mi nm thit hi do bin dng ln b mt gy ra cho Houston c chng 31,7 triu USD. Tokyo c khong 3000 km2 b bin dng ln b mt do bm ht nc, mc ln ln nht ti 5 m.

Osaca din tch b ln t nm 1935 n nm 1970 l 500 km2, mc ln l 3 m. V th chi ph cho vic sa cha cc cng trnh xy dng l rt tn km.

Vi tc khai thc nc ngm thnh ph H Ni nh hin nay, th bin dng ln b mt c th xy ra.

b. Cc ma st nm trong nn c lp t yu cha c kt n nh.

y cng l mt vn thi s nc ta, v cu to a tng ng bng Bc b thng l t yu, c chiu dy ln, c kt cha n nh.

Iii. 3. Nghin cu v cc khoan nhi III.3.1. Tng quan:

Cc nhi c s dng trong vic xy dng nh cao tng. Nh cao tng c nhng c im ng ch :

*Ti trng tp trung thng ng chn ct ln ng k. Ngoi ra di chn cu thang v thang my, chn nhng vch cng cng c nhng ti trng kh ln. Ti trng ngang cng nh vn n nh ca nh cao tng l nhng bi ton cn c xem xt mt cch nghim tc.

* Nh cao tng rt nhy vi ln, c bit l ln lch. Ln kiu g cng gy ra nhng tc ng mnh m n s lm vic tng th ca cc kt cu nh.

*Trong tnh trng th ca ta hin nay, nh cao tng s c xy dng nhiu trong khu ng dn c, mt nh c sn kh dy c. Vn bo m an ton cho cc cng trnh c l mt c im xy dng nh cao tng nc ta.

T nhng c im nu khi qut m gii php chn cho mng nh cao tng hay thy l mng cc nhi.

Nhng u im ca mng cc nhi c th tm tt :

@ Khi thi cng cc khoan nhi cng nh s dng cc khoan nhi m bo an ton cho cc cng trnh hin c chung quanh. Loi cc khoan nhi t su khng gy ln nh hng ng k cho cc cng trnh ln cn.

@ Qu trnh thc hin mng cc, d dng thay i cc thng s ca cc (chiu su, ng knh) p ng vi iu kin c th ca a cht di nh.

@ Cc khoan nhi tn dng ht kh nng chu lc ca b tng mng cc do iu kin tnh ton theo lc tp trung.

@ u cc c th chn cao tu cho ph hp vi kt cu cng trnh v quy hoch kin trc mt bng.

@ Rt d dng lm tng hm cho nh cao tng.

Do cc khoan nhi m bo c:

- ln cho php

- Sc chu ti ca cc

- Cng ngh thi cng hp l khng lm h hi n cng trnh xy dng.

- t hiu qu - kinh t - k thut

Kt lun:

Vy s dng gii php mng cc khoan nhi cho cng trnh

ny l hp l.

III.3.2. Gii php mt bng mng:

V cng trnh c 3 tng hm, nn ta chn gii php mt bng mng nh sau:

Mng c kt cu dng b ton khi bng btng ct thp.

Ton b kt cu b mng s c phn tch bng phng php Phn t hu hn, nhm xc nh s lm vic khng gian ca n.

III.3.2. Cc gi thit tnh ton:

Vic tnh ton mng cc i thp da vo cc gi thit ch yu sau:

- Sc chu ti ca cc trong mng c xc nh nh i vi cc n ng ring r, khng k n nh hng ca nhm cc.

- Khi kim tra cng ca nn t v khi xc nh ln ca mng cc th ngi ta coi mng cc nh mt mng khi quy c bao gm cc v cc phn t gia cc cc.

- V vic tnh ton mng khi quy c ging nh tnh ton mng nng trn nn thin nhin (b qua ma st mt bn mng), cho nn tr s mmen ca ti trng ngoi ti y mng khi quy c c ly gim i mt cch gn ng bng tr s mmen ca ti trng ngoi so vi cao trnh y i.

- i cc xem nh tuyt i cng.

III.3.4. Ti trng:

- Ti trng tc dng ln mng cng trnh gm c:

+ Tnh ti

+ Hot ti

+ Ti trng gi (gi tnh v gi ng).

+ Ti trng ng t.

Mng cng trnh c tnh da theo gi tr ni lc nguy him nht truyn xung mng. S liu ni lc chn ct ta c khi chy chng trnh Sap2000_8.32.

III.3.5. Thit k cc khoan nhi.a. Chn vt liu lm cc.

- B tng mc 300#, Rn = 130 (kG/cm2) = 1300T/m2

- Ct thp chu lc nhm AII c Ra = 2800 (kG/cm2) = 28000T/m2- Ct ai nhm AI c Ra = 2300 (kG/cm2) = 23000 T/m2 Cc chu ti trng ngang, hm lng ct thp ( ( (0,4 1,0)%. chn c kch thc v chiu su h cc thch hp nht cho iu kin a cht v ti trng ca cng trnh, cn phi a ra cc phng n khc nhau so snh la chn. Do hn ch v thi gian, chn phng n tnh cc c kch thc :

- D = 1,0m ( Fb = m2.

- Ct thp dc chu lc gi thit gm 1422 c Fa = 53,2cm2 = 53,2(10-4 m2

(

- S b chn cc kch thc:

+ Chiu cao i mng l h = 2,5 m .

( y i c t su 13,3 m so vi ct thin nhin (hay su 14,3 m so vi ct ( 0.00 ).

+ Chn cc cm su vo lp ct th cui si (lp t 8) on 5,0 m. Cht lng b tng cc khoan nhi phn u cc thng km do p v btng u cc cho cha ct thp ra mt on 1m v ngm vo i. Phn cc ngm vo i 40 (cm).

( Tng chiu di cc l : 1,0 ( 0,4 ( (43,4 13,3) = 31,5m.

b. Tnh ton sc chu ti ca cc khoan nhi:

1/ Theo bn ca vt liu lm cc:

- Sc chu ti ca cc theo vt liu lm cc c xc nh theo cng thc:

Pv = (.(m1m2RbFb + RaFa)

Trong :

( : h s un dc, vi mng cc i thp khng xuyn qua bn,

than bn ( =1.

Rb : Cng chu nn tnh ton ca b tng, vi b tng mc

M300# c Rn = 130 (KG/cm2) = 1300(T/m2).

Ra : Cng chu nn tnh ton ca ct thp, vi ct thp nhm

AII c Rn = 2800 (KG/cm2) = 28000 (T/m2).

Fb: Din tch tit din ca b tng Fb = 0,785(m2).

Fa: Din tch tit din ca ct thp dc Fa = 53,2(10-4 (m2).

m1 : h s iu kin lm vic khi b tng qua ng dch chuyn

thng ng th m1 = 0,85.

m2 : h s iu kin lm vic k n phng php thi cng cc,

y khi thi cng cn dng ng vch v b tng di huyn

ph st, nn m2 = 0,7.

Vy ta c :

Pv = 1((0,85(0,7(1300(0,785 + 28000(53,2(10-4) = 756,16 (T).

2/ Theo kt qu xuyn tiu chun SPT

- S dng s liu th nghim xuyn tiu chun SPT tnh ton sc chu ti gii hn ca cc theo cng thc ca Nht Bn cho trong TCXD 205:1998

- Sc chu ti cho php ca cc:

PSPT = (T)Trong :

+ Na: ch s SPT ca t di mi cc, mi cc nm trong lp ct th cui si c N = 80.

+ Ns: ch s SPT ca lp ct bn thn cc; bn thn cc c cc lp ct: lp ct pha do 5 (N5 = 21), lp ct bi cht va 6 (N6 = 30), lp ct ht trung cht 7 (N7 = 55), lp ct th cui si 8 (N8 = 80).

+ Ls (m): chiu di on cc nm trong t ct, cc xuyn qua cc lp t ct: lp ct pha do 5 (L5 = 6,7m), lp ct bi cht va 6 (L6 = 6,8m), lp ct ht trung cht 7 (L7 = 12,1m), lp ct th cui si 8 (L8 = 5,0m).

+ Lc (m): chiu di on cc nm trong t dnh (t st). Lc = 0 v u cc nm di cc lp st.

+ CU: lc dnh khng thot nc ca t theo SPT, CU = 0+ (: h s ph thuc phng php thi cng cc, cc khoan nhi ly ( = 15

+ F: Din tch tit din ngang di chn cc, F = Fb = 0,785m2.

- Vy:

- Sc khng mi: = (T).

- Sc khng thnh: =

Vy :PSPT = 314 + 296 = 610 (T).Da vo kt qu tnh sc chu ti ca nn theo iu kin bn vt liu lm cc P v v theo kt qu xuyn tiu chun SPT l PSPT ta c:

P v = 756,16(T) > PSPT = 610(T). Mt khc hai gi tr ny chnh nhau khng qu ln, nn m bo iu kin kinh t cng nh an ton. Sc chu ti ca cc:

P = min{PV, PSPT}

Do vy ta chn : P = P SPT = 610 (T) tnh ton cc.

III.3.6. Tnh ton s lng cc khoan nhi cho cng trnh:

III.3.6.1. Tnh ton s lng cc khoan nhi cho mng C - 7.Ti trng tnh ton

Ti trng tnh ton c s dng tnh ton nn mng theo trng thi gii hn th nht.

- Ti trng ct tng ngm 3 (phn t 1220 c ni lc ln nht) truyn xung mng theo c 2 phng :

= -90,66T/m;

= 88,72 T/m;

= -28,05T;

= -26,32T ; = - 2020T.

Ti trng tiu chun

Ti trng tiu chun c s dng tnh ton nn mng theo trng thi gii hn th hai.

Ti trng tnh c l ti trng tnh ton, mun c t hp cc ti trng tiu chun phi lm bng t hp khc bng cch nhp ti trng tiu chun tc dng ln cng trnh. Tuy nhin, n gin quy phm cho php dng h s vt ti trung bnh n =1,15. Nh vy, ti trng tiu chun nhn c bng cch ly t hp cc ti trng tnh ton chia cho h s vt ti trung bnh.

= -90,66/1,15 = - 78,83 T/m ;

= 88,72/1,15 = 77,15 T/m ;

= -28,05/1,15 = - 24,4 T ;

= -26,32/1,15 = -22,88T ;

= - 2020/1,15 = - 1756,5 T ;

c. Xc nh s lng cc khoan nhi:

- S lng cc s b:

(cc)

- Do mng chu ti lch tm nn ta c s cc tnh ton l :

nc = 1,3(3,31 = 4,63 (cc).

( Chn thc t n'c = 5 cc b tr cho mng di ct C7.

- Khong cch gia cc tim cc ( 3.D = 3.100 = 300cm . Theo kt qu ni lc t trng trnh Sap 2000 - 8.3.2. Ta thy ni lc ct C7 v cc ct C5, C8, C10, ct K7, K5, K8, K10, ct 3E, 3H, 3I, ct 12E, 12H, 12I c ni lc gn bng nhau, do di chn cc ct ny ta u b tr 5 cc khoan nhi.

III.3.6.2.Tnh ton s lng cc khoan nhi cho mng di li cng. T kt qu phn tch bng Sap 2000_8.3.2 th tng cng lc dc ca 34 cy ct trong qun th li cng l: .

- S lng cc s b:

(cc) - Do mng chu ti lch tm nn ta c s cc tnh ton l :

nc = 1,4(25,86 = 36,2 (cc). ( Chn thc t n'c = 44 cc b tr cho mng di qun th li cng. III.3.6.3. Tnh ton s lng cc khoan nhi cho mng di vch cng gc cng trnh. T kt qu phn tch bng Sap 2000_8.3.2 th tng cng lc dc ca 3 cy ct chnh v mt cy ct c phm vi chiu cao trong 3 tng hm trong vng vch cng l: .

- S lng cc s b:

(cc) - Do mng chu ti lch tm nn ta c s cc tnh ton l :

nc = 1,4(6 = 8,4 (cc). ( Chn thc t n'c = 9 cc b tr cho mng di vch cng. III.3.6.4. Tnh ton s lng cc khoan nhi cho mng di ct ca phn m rng phn ngm cng trnh. Phn m rng tng ngm cng trnh c tt c 66 ct. Lc dc ln nht trong ct ny l 367,3 (T).

- S lng cc s b:

(cc)

- Do mng chu ti lch tm nn ta c s cc tnh ton l :

nc = 1,4(0,6 = 0,84 (cc). Do di mi ct ta ch cn b tr mt cc khoan nhi.

III.3.6.5. Tnh ton s lng cc khoan nhi cho mng ngoi vng chu lc do ct gy ra: trnh gy mmen ln cho b mng, nn ta cn b tr thm cc cc khoan nhi trong cc vng trng ca b mng.

V tng s cc khoan nhi cn b tr thm l: 36 cc.

Nh vy di i mng b ca cng trnh c tt c 252 cc khoan nhi.

III.3.7. Tnh ton i mng.

kim tra cc iu kin lm vic ca i mng b v b tr ct thp b mng, ta ch c th tnh c i mng ring bit di ct theo phng php thng thng.

Do ta gi vng i mng cc b gi thit trong khi mng b l mt khi i coi nh lm vic ring r di tc dng ca ti trng chn ct.

By gi ta s tnh ton khi i cc gi thit ny di ct C7.

- Tnh p lc trung bnh y mng :

V khong cch trnh nh hng ng sut l 3d, cho nn gi thit P s c truyn trong hnh vung c cnh l 3d, ng thi gi thit l p lc y mng phn b u v bng vi sc chu ti ca cc, do gi thit ny l khng ng, cho nn khi tnh ton c s cc s b cn phi tng s cc ln ph hp vi mmen lch tm, ph hp vi p lc tiu chun Rtc tc dng xung y mng tnh ton ln khi kim tra TTGH v ln. Ta c :

Trong :

+ P: sc chu ti ca cc = min{PV, PSPT}= 610(T).

+ D: ng knh cc D = 1,0m.

- Din tch s b y i gi thit :

Trong :

+ Ntt0 : lc dc tnh ton xc nh ti nh i, ly gi tr ln nht

khi t hp ti trng tc dng theo hai phng:

= 2020(T).

+ h : chiu su t y i

h = 2,5 (m).

+ n : h s vt ti, n = 1,1.

+ (tb : l tr trung bnh trng lng ring ca i, y i nm di Mc nc ngm nn ly (tb= 1,5 (T/m3).

Nh tnh phn trn, ta cn 5 cc khoan nhi cho ct ny.

- Khong cch gia cc tim cc ( 3.D = 3.100 = 300cm .

- Khong cch t tim cc n mp i ( 0,7.D = 0,7.100 = 70cm, ly bng 100cm. V mmen lch tm ca ct C7 khng ln, nn ta chn hnh dng ca i l hnh vung.

- Mt bng b tr cc cho mng nh hnh v sau :

T mt bng b tr cc ta c din tch y i thc t l:

Ftt = 6,0(6,0 = 36,0 (m2).

Trng lng ca i sau khi b tr cc:

Ntt = n.Ftt. h.(tb= 1,1(36,0(2,5(1,5 = 148,5 (T)

Lc dc tnh ton xc nh n ct i:

Ntt = N0tt + Ntt = 2020 + 148,5 = 2168,5(T).

V mng chu ti lch tm theo hai phng (phng trc x v y), lc truyn xung cc c xc nh theo cng thc sau:

Pttmax,min =

EMBED Equation.3Trong :+ nc = 5 l s lng cc trong mng.

+ Mxtt v Mytt : mmen un tnh ton tng ng quanh trc x v trc y

MXtt = M0Xtt + Q0Ytt(h = 90,66 + 28,05(2,5 = 160,785 (Tm).

MYtt = M0Ytt + Q0Xtt(h = 88,72 + 26,32(2,5 = 154,52 (Tm).

Vi:

M0xtt, M0ytt l mmen un tnh ton nh i quanh trc X v Y;

Q0xtt, Q0ytt l lc ct tnh ton nh i theo trc X v Y;

h = 2,5m l chiu cao i.

+ xmax, ymax (m): khong cch t tim cc n trc Y, X.

+ xi, yi (m): khong cch t trc cc th i n cc trc i qua trng tm din tch tit din cc cc ti mt phng y i (xem s b tr cc). Thay s vo ta c :

Pttmax,min =

EMBED Equation.3=

Vy: Pttmax = 473,11 (T);

Pttmin = 394,28 (T);

Ptttb = 433,7 (T);

Trng lng tnh ton ca cc (k n y ni):

Pc = 31,5(0, 785(1,5(1,1 = 40,8(T).

Trng lng tnh ton ca t (k n y ni):

P = 0,785((6,2(0,996 + 6,8(0,94 + 12,1(1,08 + 5(1,08) = 24,36(T).

Kim tra lc truyn xung cc:

+ Pc - P = 473,11 + 40,8 24,36 = 489,55 (T) < PSPT = 610(T). Tho mn iu kin lc truyn xung cc. Chnh lch lc truyn xung cc v sc chu ti ca cc khng ln lm v thin v an ton, ta chn cc c ng knh v chiu su chn cc nh trn l t yu cu.

Mt khc Pttmin = 394,28 (T) > 0 nn ta khng phi tnh ton kim tra theo iu kin chng nh .

a. Kim tra nn mng cc theo iu kin bin dng :

Vi quan nim nh ma st gia mt xung quanh cc v t bao quanh, ti trng ca mng c truyn trn din rng hn, xut pht t mp ngoi cc ti y i v nghing mt gc ( = (tb/4, gi l gc m ca mng. Gc m ca mng: do nn t di mng c ln cht v coi nh lin kt li thnh mt, cho nn coi nh nn t trong phm vi cc v gc m rng l mt mng khi quy c (tnh n nh hng ca ma st ti s lm vic ca mng su - khc vi mng nng).Trong : (tb = : gc ma st quy i ca khi mng quy c (gc ma st tng ng ca ton b nn t trong khi mng quy c).

Vy : ( = (tb/4 = 32,81o/4 = 8,2o

- Chiu di, rng y khi mng quy c:

LM = BM = L + 2.H.tg(() = 5,0 + 2(30,3(tg(8,2o) = 13,73(m).- Xc nh trng lng tiu chun ca khi mng quy c :

+ Trng lng trong phm vi i tr ln n ct sn tng hm xc nh theo cng thc: Ntc1= LM ( BM ( h ( (tb = 13,73 ( 13,73 ( 2,5 ( 1,5 =706,9 (T).

+ Trng lng lp ct pha do 5 trong phm vi chiu dy 6,4 m ca lp ct c tr i phn cc chim ch (c k n (n):

Ntc4 = (13,732 - 4(0,785)(6,4(0,9962 = 1182 (T).

+ Trng lng lp ct bi cht va 6 trong phm vi chiu dy ca lp ct c tr i phn cc chim ch (c k n (n):

Ntc5 = (12,892 - 4(0,785)(6,8(0,9389 = 1040,75 (T).

+ Trng lng lp ct ht trung cht 7 trong phm vi chiu dy ca lp ct c tr i phn cc chim ch (c k n (n):

Ntc6 = (13,732 - 4(0,785)(12,1(1,0764 = 2414,4 (T).+ Trng lng lp ct th cui si 8 trong phm vi chiu dy ca lp ct c tr i phn cc chim ch (c k n (n):

Ntc7 = (13,732 - 4(0,785)(5(1,0764 = 997,7 (T).

+ Trng lng tiu chun cc trong phm vi khi mng quy c :

Ntcc = 5(0,785(30,3(2,5 = 237,86 (T).

( Trng lng khi mng quy c : Ntcq = + Ntcc = 6579,4 tn- Tr tiu chun lc dc xc nh n y khi quy c:

NZtc = N0Ztc + Ntcq = 1756 + 6663,07 = 8336 (T).

- Mmen tiu chun tng ng trng tm y khi quy c:

Mtc = M0tc + Q0tc.(h+ L)

h': chiu cao t im t lc n y mng = 2,5m.

L: chiu di cc = 30,3m.

+ Mmen quanh trc X:

MXtc = = 90,66 +28,05((2,5+30,3) = 1011 (Tm).+ Mmen quanh trc Y:

MYtc = = 88,72 +26,32((2,5+30,3) =952 (Tm).

- lch tm :

+ Theo trc X :

eX = =

+ Theo trc Y:

eY = =

- p lc tiu chun y mng khi quy c do ti trng tiu chun :

=

Vy:

(tcmax = 48,74(T/m2).

(tcmin = 39,69 (T/m2).

(tctb = 44,22 (T/m2).

- Cng tnh ton ca t y khi mng quy c:

RM =(A.BM.(II + B.HM.(II + D.CII)

Trong : Ktc, m1, m2: tra bng.

+ Ktc = 1: cc ch tiu c l ca t ly theo s liu th nghim

trc tip vi t.

Tra bng 3-1(Sch Hng dn n Nn v Mng), ta c :

+ m1 = 1,4 : loi t ct cui si.

+ m2= 1,1 : v cng trnh c s kt cu cng.

+ Tr tnh ton th hai ca gc ma st trong lp ct th cui si l

(II = 43o, tra bng 3-2 Sch Hng dn n nn v mng ta c:

A = 3,12 ; B = 13,5 ; D = 13,31 ;

+ Trng lng ring t di y khi quy c:

(II = (n8 = 1,076 (T/m3).

+ Trng lng ring t t y khi quy c tr ln :

- Lc dnh n v di y khi quy c: CII = 2 Kpa = 0,2 T/m2.

( Vy ta c c:

RM=(3,12(13,73(1,076+13,5(35,85(1,03+13,31(0,2) = 842,8(T/m2).

Kim tra iu kin p lc tiu chun y mng :

tho mn iu kin ng sut ln nht khng vt qu nhiu p lc tiu chun (20%), ng sut trung bnh khng vt qu p lc tiu chun tho mn iu kin khng ch bin dng do di y mng, do tnh ton vi TTGH II nn cho php vt qua p lc tiu chun bi v mng chu ti trng lch tm. Nn t di mng c bin dng do nh, c th tnh ton ln theo l thuyt nn bin dng tuyn tnh. Tnh ton ln theo phng php cng ln tng lp.

So snh : (tcmax = 48,74 (T/m2) < 1,2.RM = 1011,4(T/m2).

(tctb = 44,22 (T/m2) < RM = 842,8 (T/m2).

Vy ta c th tnh ton c ln ca nn theo quan nim nn bin dng tuyn tnh. Trng hp ny nn t chn cc tr xung c chiu dy tng i ln, y ca khi quy c c din tch b nn ta dng m hnh nn l na khng gian bin dng tuyn tnh tnh ton.

- ng sut bn thn :

+ Ti y lp ct pha do 5:

(btz = 6,4 = 6,4 ( 0,996 = 6,37 (T/m2).

+ Ti y lp ct bi cht va 6:

(btz = 6,4+6,8 = 6,37 + 6,8 ( 0,939 = 15,76 (T/m2).

+ Ti y lp ct ht trung cht 7:

(btz = 6,4+6,8+12,1 = 12,76 + 12,1 ( 1,076 = 29,78 (T/m2).

+ p lc bn thn ti y khi quy c:

(bt = 29,78 + 5,0 ( 1,076 = 38,2 (T/m2).

- ng sut gy ln :

+ ng sut gy ln y khi quy c :

(glz = 0 = (tctb - (bt = 44,22 38,2 = 6,02 (T/m2).

+ ng sut gy ln su Z di y khi quy c :

(glzi = Koi.(glz=0 (T/m2).

Vi K0 tra bng theo 2z/BM v LM/BM = 1,0

+ ln ca lp phn t th i:

Si =

- Nhn xt :

Ti im Z = 0 ta c :

(glz=0 = 6,02 (T/m2) < 0,2.(btz=0 = 0,2(38,2 = 7,64 (T/m2).

( Khng cn tnh ln theo quy phm. su chn cc m bo mng gn nh khng b ln.

( iu kin ln tuyt i tho mn.

b. Tnh ton bn v cu to i cc:

B tng i s dng b tng mc M300#

- Xc nh chiu cao i cc theo iu kin m thng: v thp m thng th y thp nm trm ra ngoi trc cc cc. Nh vy i khng b m thng.

- Lp Btng lt y i dng va Ximng, ct, gch v hoc 4x6, M75 dy 100mm, lp btng lt ging cng dy 200mm.

S thp chc thng xc nh chiu cao i cc

c. Tnh ton mmen v thp t cho i cc:

+ Mmen tng ng vi mt ngm I-I :

MI = r1.(P1 + P2)

y : P1 = P2 = Pmax = 473,11 (T); r1 = 1,47 (m).

MI = 1,470(473,11 + 473,11) = 1391 (Tm).

+ Mmen tng ng vi mt ngm II-II :

MII = r2.(P1 + P3)

y : P3 = Pmin = 433,7(T); P1 = Pmax = 473,11(T); r2 = 1,55 (m).

MII = 1,55(433,7 + 473,11) = 1405(T.m).

- Ct thp b tr theo hai phng, mt lp trn v mt lp di.

+ Ct thp theo phng Y-Y t di c tnh nh sau :

Chn lp bo v ct thp l a = 3,5cm. D kin dng thp (22 nn chiu cao lm vic: h0 = 250 - 40 - 3,5 - = 205,4cm = 2,054(m).

Chn thp AIII c Ra = 3600 Kg/cm2 = 3,6(T/cm2).

Ta c:

Chn 34(28 c Fa = 209,4(cm2), khong cch gia tim 2 ct thp cnh nhau l: a = 180 (mm); Chiu di mi thanh bng 6 (m).

+ Ct thp theo phng X t trn c tnh theo cng thc :

Chn lp bo v ct thp l a = 3,5cm. D kin dng thp (22 nn chiu cao lm vic: h0 = 250 - 40 - 3,5 - = 203,2cm = 2,032(m).

Chn thp AIII, c Ra = 3600 Kg/cm2 = 3,6( T/cm2).

Ta c:

Chn 34(28 c Fa = 209,4(cm2), khong cch gia tim 2 ct thp cnh nhau l: a = 180 (mm); Chiu di mi thanh bng 6 (m). Sau y l s b tr thp cho cc vng i gi thit di cc ct ct C7 v cc ct C5, C8, C10, K7, K5, K8, K10, 3E, 3H, 3I, 12E, 12H, 12I.

IV. kim tra nn mng cng trnh theo phng php phn t hu hn.

IV.1. Xy dng m hnh khng gian:

Sau khi xc nh c s b tng s cc khoan nhi dng chu ti cng trnh, v d kin chiu dy ca b mng l 2,5m, ta dng c m hnh khng gian ca ton th mng cng trnh bng phn mn Sap 2000_8.3.2 nh sau:

Chn phn t b mng l phn t Solid phn t lm vic 3 chiu.

Chn cc khoan nhi l dng Frame ct cao 3,6m, ng knh 1m. Chn cc c coi nh dng ngm cng.

( chiu cao cc 3,6m l mt thng s khng mang ngha, nhng cng khng c ly qu ln, bi nu ly qu ln th cc s mnh).

V i mng rt dy v rt cng, cng vi s ln trong cc khoan nhi l rt b (v cc c cm su vo lp si 5m) cho nn ta c th b qua ch s n hi ca t. M coi m hnh mng nh dng sn nm.

Khng k n s tham gia chng ca tng Diaphragm xung quanh b mng.

Khng k n thnh phn lc nn ngang do p lc t v nc ngm thng qua tng Diaphragm.

Mc nc ngm c th dao ng ln theo cc ma, nn ta s tnh ton mng vi hai trng hp :

+ Khng c p lc y ni ca nc ngm;

+ C c p lc y ni ca nc ngm;

IV.2. Ti trng tc dng ln mng ca cng trnh:

+ Tnh ti (TT): l ti trng do lc ca cc ct truyn xung v trng lng bn thn i mng.

+ Hot ti (HT): l ti trng gy ra do p lc ca nc ngm. Trng hp nc ngm b h thp hn so vi cao trnh y i mng th HT = 0.

T hp ti trng: TH = 1,1*TT + HT.

S tnh ca h mng cng trnh nh sau:

M hnh 3d:

IV.3. nh gi sc lm vic hiu qu ca cc khoan nhi:

Xt thy vic xc nh sc lm vic hiu qu ca cc khoan nhi s tt nht khi cc lm vic trong trng hp mc nc ngm b h thp hn so vi cao trnh y mng. Do ta s phn tch s kt cu mng theo phng php phn t hu hn m khng xt ti p lc nc ngm. Cch phn tch ny s thin v an ton v s lm vic bn vng ca cng trnh.

Sau khi xy dng m hnh khng gian v nhp cc s liu tnh ton trong chng trnh Sap 2000_8.3.2, kt qu phn tch cho ra sc chu ti ca mi cc khi lm vic thc t. iu c th hin bng di y:

TABLE: Element Forces - Frames

FrameStationOutputCaseCaseTypeP

TextmTextTextTon

10THCombination-342.68

11.8THCombination-337.08













































































































































































































































































































































































































































































































































































































































































































































































Da vo kt qu phn tch trn cho ta thy :

+ Ti trng dc trc trong cc ln nht l Nmax = - 521,9 tn;

+ Ti trng dc trc trong cc nh nht l Nmax = - 218,5 tn;

+ Ti trng dc trc trung bnh trong 252 cc Nmax = - 356 tn;

T kt qu trn ta kt lun rng:

V lc dc trc ln nht trong cc l Nmax = - 521,9 tn < P = 610 tn (sc chu ti ln nht ca cc) . Do cc tho mn iu kin an ton v cng .

Ch s lm vic hiu dng ln nht ca cc l :

Ch s lm vic hiu dng nh nht ca cc l :

Ch s lm vic hiu dng trung bnh ca cc l :

IV.4. Phn tch ng sut trong i mng b

Theo kt qu phn tch ta thy i mng b ca cng trnh lm vic trong vng ng sut an ton.

V Thit k tng diaphragm

V.1. Tng quan v la chn phng php tnh tng Diaphragm

Hin nay c rt nhiu phng php a ra tnh ton tng trong t c cc thanh neo nh:

Phng php Sachipana (Nht bn);

Phng php n hi c xt n ma st gia t v tng chn (Quy phm thit k mng cng trnh xy dng ca Nht bn);

Phng php c xt n nh hng ca lc trc thanh chng cc tng thanh chng theo tin trin ca vic o t;

Phng php S gia;

Phng php phn t hu hn (trn nn n hi, PTHH trn bn mng n hi, PTHH c xt quan h ca t vi tng chn);

Phng php B.N.Gimoskin cho php tnh ton c xt n lin kt mt pha ca kt cu vi nn;

Phng php ca Blium-Lomeier xem p lc t pha trc tng l phn b u, cng cng ca n phn b pha di

thun li v n gin trong vic xc nh p lc ln tng chn ta dng phng php Sachipana gn ng ca Nht Bn, vi phng php ny tnh Lc trc thanh chng v Mmen thn tng khng bin i theo qu trnh o t.

Mt s gi nh a ra tnh ton nh sau:

+ Trong t c tnh dnh, thn tng c xem l n hi di v hn.

+ Phn lc chng hng ngang ca t bn di mt o chia lm hai vng: Vng do t ti p lc t b ng c cao l; Vng n hi c quan h ng thng vi bin dng ca thn tng.

+ Sau khi khi lp t chng th xem chng nh bt ng.

+ Sau khi lp t tng chng di th xem tr s lc trc ca tng chng trn khng thay i, cn thn tng t di ln vn duy tr v tr nh c.

Phng php gii gn ng ch dng hai phng trnh cn bng tnh:

(Y = 0

(MA = 0

(Y = -

Vi ( = ( - (Suy ra:

Nk = (.h0.xm + (.h0k2/2 - W.xm2/2 v.xm - - (.h0k.xm/2 + (.xm2/2 (1)

(2)Thay (1) vo (2) ta c:

(W - ()xm3/3 ((.h0k/2 v/2 W.hkk/2 + x.hkk/2- (.h0k/3) xm2 -

- ((.h0k v (.h0k/2).hkk.xm [+(.h0k2(hkk h0k/3)/2] = 0 (2a)

Vi: ( - p lc tnh ch ng tc dng ln trn mi mt di tng (theo chiu cao tng).

( - p lc tnh ca t tc dng ln trn mi mt di theo chiu cao tng.

w - p lc nc.

wx + v tr s p lc t pha b ng

xm chiu su tng trong t ti thiu trong tng giai on o t.

Tng c cm vo cc lp t nh sau:

Lp t lp: c chiu dy 1,2 m.

Lp St do cng: c chiu dy 4,6 m v ( = 18,2 kN/m3, c = 19, ( = 13o Lp St pha do mm: c chiu dy 3,4 m v ( = 17,5 kN/m3, c = 5, ( = 11o Lp St pha do chy: c chiu dy 3,6 m v ( = 18,5 kN/m3, c=10, ( = 16o Lp Ct pha do: c chiu dy 6,7 m v ( = 19,2 kN/m3, c = 25, ( = 18o Lp Ct bi cht va: c ( = 19 kN/m3, ( = 30oV.2. Xc nh s tnh ca tng Diaphragm:

S tnh ton tng trong t c th hin t qu trnh thi cng sau: o t n su - 3,4 m (k t mt t) th ta t mt t cy chng th nht (tng qut th cy chng y c th l neo, sn tng hm btng ct thp hoc h cc thanh chng bng thp hnh).

o t n su - 7m th t tip t cy chng th hai.

o t n su -10,6m th t tip t cy chng th ba.

V tip tc o t n su y i 13m kt thc qu trnh o t.

Ta thy rng: thc t cng trnh trn mt t xung quanh tng u c cht cc vt liu hoc ln tri xy dng, v c th c cc phng tin nh i li ln trn mt t h mng cng trnh, v th c cht mt phn ti trng phn b u q, gi thit q = 1 T/m2 = 10 kN/m2.

Tng tng ko di lng tng chn n chiu cao h. T c th xc nh cc tr s p lc t theo l thuyt p lc t Raikine, cn c vo mc nc ngm tnh ton p lc nc, ly 1m theo chiu di thn tng tnh.

V.2.1. Tnh p lc ca t v nc ln tng:

1. Tnh p lc t ch ng tc dng ln tng:

su Z = 1,2 m:

= (q + (.h) tan2(450 -) 2c.tan(450 -)

= (10 + 17 x 1,2) tan2(450) = 30,2 (KN/m2).

su z = 3,4 m:

su z = 5,8 m:

su z = 7,0 m:

su z = 9,2 m:

su z =10,6 m:

su z = 13,3 m:

2. Tnh p lc nc ngm:

Trong qu trnh thi cng, phi lun m bo c mc nc ngm thp hn mt t o l 1m. Do s chnh p lc nc s c tnh nh sau:

Giai on 1: o n su - 3,4 m (so vi mt t) th mc nc ngm t nhin ang su - 6,5m, m bo iu kin thi cng t ra, do giai on ny, p lc nc ngm hai bn tng l cn bng nhau.

Giai on 2: o t n su 7m (so vi mt t), h thp mc nc ngm bn trong lng tng xung 1m so vi mt t thi cng, do chnh lch mc nc su ny l . p lc nc c tnh nh sau:

Giai on 3: o t n su 13,3 m (so vi mt t), h thp mc nc ngm bn trong lng tng xung 1m so vi mt t thi cng, do chnh lch mc nc su ny l . p lc nc c tnh nh sau:

3. Tnh p lc t b ng

su 7m. PP = (.x.tan2(450 + ) + 2c.tan(450 +)

= 17,5*x*tan2(450 + ) + 2*5.tan(450 +)

= 25,75.x + 12,131.

Vy c, w7 = 25,75 v v7 = 12,131

V.3. Xc nh su chn tng, lc dc thanh chng v mmen thn tng theo tng giai on thi cng.

Khi xc nh chiu su tng trong t ngoi vic tng lm vic mt cch bnh thng cn ch n vic h su thm ngn nc. Tr s phn cm su thm c ly nh sau:

Trong cht: 0,5 - 1m, trong bin cht v trong st cht: 0,75 1,5m, trong st do v st do: 1,5 - 2m. Vi cng trnh ny, o t n ct 13,3m (lp ct pha do) nn ta chn cm thm l xm.

Giai on I. o t n su -3,4m (tnh t mt t) v t hng thanh chng th nht.

Sau khi o t n su 3,4m (so vi mt t), ta tin hnh t hng chng th nht. Gi l lc dc trong thanh chng th nht. Ti thi im ny, p lc t ch ng tc dng ln tng cn b, nn coi .

Mmen un thn tng ti su ny l: .

Tip tc o t th p lc t ch ng tng dn v p lc t b ng gim dn, do xut hin lc nn trong hng thanh chng ny.

Giai on II. o t n su -7m (tnh t mt t) v t hng thanh chng th hai. Ti giai on ny, gi lc dc trong hng chng th nht l

S thanh chng k = 1, hok = 7m, hkk = h1k = 3,6m, Nk = . Dng cng thc (2a) tm xm:

(W - ()xm3 ((.h0k v W.hkk +x.hkk - (.h0k) xm2 -

- ((.h0k v (.h0k).hkk.xm [+(.h0k2(hkk h0k)] = 0

Xc nh cc h s :

( = = = 6,25 ; ( = = = 4,1 ; ( = ( - ( = 6,25 4,1 = 2,15

( (25,75 4,1)xm3 (6,25*7 12,131 25,75*3,6 - 2,15*7) xm2 -

- (6,25*7 12,131- .2,15*7).3,6.xm [6,25*72*(3,6 7)] = 0 ( 7,22xm3 + 35,55*xm2 86,74xm 194 = 0

( xm3 + 4,93xm2 12xm 26,87 = 0

Gii bng phng php th dn, ta c: xm 1 = 2,8 m

Dng cng thc (1) tm lc trc thanh chng th hai:

Nk = (.h0k.xm + (.h0k2 - W.xm2 v.xm - - (.h0k.xm + (.xm2 (1)

= 6,25*7*2,8 + 6,25*72 - 25,75*2,82 12,131*2,8

- *2,15*7*2,8+ 4,1*2,82 = 135,7 kN = 13,57 tn

=> Mmen un thn tng to su 7m l : .

Biu ni lc thanh chng v thn tng ti giai on thi cng II

Giai on III. o t n su 10,6m (tnh t mt t) v t hng chng th ba.

Vi k = 2; Nk = = 135,7 kN; hok = 10,6 m, h1k = 7,2 m;

hkk = h2k = 3,6 m; Nk = ; w; v; (; (; (. Dng cng thc (2a) tm xm:



( (25,75- 4,1)xm3 (6,25*10,6 12,13125,75*3,6- 2,15*10,6)xm2

- (6,25*10,6 12,131- .2,15*10,6).3,6.xm

- [135,7*7,2 135,7*3,6 + 6,25*10,62*(3,6 10,6)] = 0 ( 7,22xm3 + 26,89xm2 153,8xm 513,2 = 0

( xm3 + 3,72xm2 21,3xm 71,1 = 0


Dng cng thc (1) tm lc trc thanh chng :

Nk = (.hok.xm + (.h0k2 - W.xm2 v.xm - - (.h0k.xm + (.xm2 (1) => = 6,25*10,6*4,5 + 6,25*10,62 - 25,75*4,52 12,131*4,5 -

- 135,7 - *2,15*10,6*4,5 + 4,1*4,52 = 188,8 kN = 18,85 tn

=> Mmen un thn tng ti su 10,6m l: .

Biu ni lc thanh chng v thn tng ti giai on thi cng III

Giai on IV. o t n su 13,3m (tnh t mt t) - kt thc qu trnh o t.

Vi k = 3; Nk = = 135,7 kN; Nk = = 188,8 kN; hok = 13,3 m,

h1k = 9,9 m; h2k = 6,3m ; hkk = 2,7 m ; Nk = ; w; v; (; (; (.

Dng cng thc (2a) tm xm:



( (25,75- 4,1)xm3 (6,25*13,3 12,13125,75*2,7- 2,15*13,3)xm2

- (6,25*13,3 12,131- .2,15*13,3)*2,7*xm

- [(135,7*9,9 + 188,8*6,3) (135,7 + 188,8)*2,7 +

+ 6,25*13,32*(2,7 13,3)] = 0 ( 7,22xm3 + 8,79xm2 153,1xm 701 = 0

( xm3 + 1,22xm2 21,2xm 97,1 = 0


Dng cng thc (1) tm lc trc thanh chng :

Nk = (.hok.xm + (.h0k2 - W.xm2 v.xm - - (.h0k.xm + (.xm2 (1) => = 6,25*13,3*5,64 + 6,25*13,32 - 25,75*5,642 12,131*5,64 -

- 135,7 - *2,15*13,3*5,64 + 4,1*5,642 = 204,5 kN = 20,45 tn

=> Mmen un thn tng ti su 13,3m l: => .

Biu ni lc thanh chng v thn tng ti giai on thi cng IV Xt thy, nu x = 5,64m th chn tng Diaphragm s nm trong lp Ct pha do, m lp ny l lp yu, c N = 21. Nn ta phi ko di cho chn tng nm trong lp Ct bi cht va (c N = 30) mt on l 2,5 m. Cc l do khc cho s la chn ny l:

Bn thn tng trong t khng chu ti trng ca cng trnh, m ch chu trng lng bn thn v mt phn ti trng t cc sn tng hm.

Tng trong t khng c php ln, bi nu xy ra ln th s ph v lin kt gia n vi sn tng hm di cng, iu ny dn n nc ngm s thm thu vo hm

Vy tng di ca tng Diaphragm l: H = 22cm.

V.4. Xc nh s b kch thc tit din tng Diaphragm:

Cc kch thc s b ca tng nh sau:

- Chiu dy tng chn theo c s sau:

+ Theo yu cu chng thm.

+ T Mmen trong tng, ta tnh c chiu cao lm vic ca tng (h0) thng qua cng thc: h0 = . Vi b chiu rng ca di tng cn tnh ton. Trong bi ton ny, chn b = 0,8m (theo chiu cao tng).

+ Cn c vo cng ngh v phng tin thi cng thc t. Thng thi cng cp tng bng gu ngom, c cc kch thc gu: 600, 800, 1000mm.

+ Chn theo kinh nghim.

V.5. Kim tra n nh kt cu chn gi:

Kim tra chng tri y h mng c xt ng thi c c v ( theo cng thc sau:

(*)(Cng thc kin ngh ca nh nghin cu Ung Bnh Gim - i hc ng T - Trung Quc c tham kho t cng thc xt kh nng chu lc ca nn t ca Prandtl v Terzaghi).

Trong :

KL ( 1,2 1,3, do b qua tc dng chng tri ln ca cng chu ct pha sau tng. Vi iu kin t nn ca cng trnh nh nu trong phn I, ta chn KL = 1,3 (thin v an ton).

D- chn su ca thn tng xc nh theo cng thc chng lt (So snh vi tiu chun hng dn thc hnh v nn mng - Tiu chun BS 8004:1986 ca Anh, Nh xut bn Xy Dng 2002). Nh tnh ton trn, c D = 8,7m

q = 1T/m2: Siu ti t trn mt t.

(1: Tr bnh qun trng lng t nhin ca cc lp t pha ngoi h o (t mt t n y tng).

(1 = (kN/m3).

(2- Tr bnh qun trng lng t nhin ca cc lp t pha trong h o (K t mt o n y tng).

(kN/m3).

Nc , Nq h s tnh ton kh nng chu lc gii hn ca t (Xc nh theo Terzaghi). Ti y h o su 13,3m (thuc lp ct pha do), c c = 25, ( = 180. Vy cc gi tr v Nc c xc nh nh sau:

= 5,64

= 14,27 Thay vo cng thc (*) ta c:

> 1,3.

Vy h mng tho mn iu kin chng tri.

V.6. Tnh ton ct thp chu lc cho tng:

- Xc nh ct thp chu lc:

Ct mt di tng Diaphragm c chiu rng 1m theo chiu di tng, coi cc v tr neo nh cc gi ta. T , coi di nh dm n gin lin tc gi ln cc gi ta l cc v tr b tr cc hng chng.

Cng thc tnh dm n gin, tnh vi gi tr m men ln nht:

Chn b tng mc 300, c Rn = 130 kg/cm2.

Chn thp chu lc AII, c Ra = 2800KG/cm2.

+ M = 50 (T.m)

EMBED Equation.3 < Ao = 0.428.

= 0,5(1+) = 0,9636Fa = =

Chn 32a200, c Fa = 40,2 cm2 , thp c b tr i xng.

=100% = 0,54 %

Ct thp ngang chn theo cu to (20a300.

- Pha trong lng thp c b tr cc thanh ngang (20 c khong cch:

> ng knh ng Tremie.

( 200mm.

A = ( 10D chu lc.

( 1,5B Vi B l chiu dy tng

Vy chn khong cch cc thanh thp ngang A = 800mm.

Ti v tr cao ca sn v v tr i mng, ta phi t thm ct thp chu lc ct, c th:

Trong chiu cao 1m vng cao mc sn v 3,7m ti v tr i mng, ta t xen k thm thp (20 vo gia hai thanh (32a200 .

Trong vng ny, ct thp ngang c t dy hn, (20a200.

Hnh v b tr thp trong tng Diaphragm nh sau:

VII. Tnh chuyn v ca tng bng phn mm Sap 2000 _8.3.2:

1. S tnh ton:

Tit din cu kin nh sau:

Kt qu ca chuyn v thn tng ti cao cc mc sn t chng trnh Sap_2000. 8.3.2:

TABLE: Joint Displacements

JointOutputCaseCaseTypeU1U2U3R1R2R3

TextTextTextmmmRadiansRadiansRadians

17APLUCLinStatic4.86E-03-2.64E-06-1.02E-041.10E-054.34E-051.03E-04

19APLUCLinStatic4.77E-032.31E-06-8.57E-055.34E-06-7.93E-062.07E-04

21APLUCLinStatic3.90E-033.07E-06-5.38E-05-2.14E-06-1.66E-042.55E-04

23APLUCLinStatic2.16E-032.22E-06-2.10E-05-7.88E-06-3.45E-041.53E-04

Ta thy rng chuyn v ln nht ca tng l ti nh tng, c tr s 4,86mm, tr s ny rt nh, nn phng n b tr h thng chng tng nh vy l hp l.

PAGE 1gvhd : gs tskh nguyn trm . SVTH : L Xun Tng .

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