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TRNG I HC XY DNG N TT NGHIP KHOA CNG NGH THNG TIN

TRNG I HC XY DNG N TT NGHIP KHOA CNG NGH THNG TIN

THIT K S B

CHNG I: GII THIU CHUNG

1.iu kin thy vn.

Qua cc s liu thy vn c cung cp ta nhn thy iu kin tnh hnh thy vn ti khu vc kh n nh.

Cc s liu tnh ton thu vn dng trong thit k :

Mc nc cao nht

:Hmax = +9.1.0 m

Mc nc thp nht

:Hmin = -4.20 m

Mc nc thng thuyn

:Htt = +8.10 m

2.c im a tng.

Dc theo tim cu,ti cc v tr tr d kin khoan thm d 1 l khoan (LK3 ~ HTV), tnh cht a tng t trn xung c th nh sau:

Lp 1: st.

Lp 2: Ct ht nh.

Lp 3: Ct pha st.

Lp 4: Dm cui.Lp 5: dip thch.

I.1. CC PHNG N KT CU CU.

Phng n 1: Cu chnh dm lin tc BTCT DL 5 nhp + cu dn nhp n gin bn lp ghp tit din ch I.S nhp: 2x33+75+3x120+75+2x33 m

Cu chnh dm lin tc BTCT DL tit din hp thi cng bng phng php c hng cn bng, chiu cao dm thay i t 7,0m cho ti 3,0m.

- Cu dn l cu nhp n gin dm I, chiu cao dm l 1,7m. Chiu di ton cu l 642800 mm.

Phng n 2: Cu chnh l cu dy vng 3 nhp + cu dn nhp n gin bn lp ghp tit din ch I.S nhp: 33+(144 + 288 + 144)+33 m-Cu chnh l cu dy vng 2 mt phng dy 2 nhp i xng (144+144), dm cng BTCT vi chiu cao khng i 2.0m.. -Cu dn l cu nhp n gin dm I, chiu cao dm l 1,7m. -Chiu di ton cu l 642400 mm.

Phng n 3: Cu chnh l cu Extradosed + cu dn nhp n gin bn lp ghp tit din ch I.- S nhp: 33 +(80+3x140+80)+ 33 m

- Cu chnh l cu Extradosed 3 nhp lin tc (90+140+90), tit din hp thi cng bng phng php c hng cn bng, chiu cao dm thay i t 4,5m cho ti 2,5m.

- Cu dn l cu nhp n gin dm I, chiu cao dm l 1,7m.

- Chiu di ton cu l 646400 mm.Cc phng n b tr chung cu so snh, thc hin trn bng sau:

PAThng thuynKh cuS SL(mm)Nhp chnhNhp dn

180x108.0 +2x0,752x33+75+3x120+75+2x33642800Lin tcDm I

280x108.0 +2x0,7533+(144 + 288 + 144)+33642400Dy vngDm I

380x108.0 +2x0,7533 +(80+3x140+80)+ 33646400ExtradosedDm I

Sau khi thit k s b cho 3 phng n trn, tin hnh phn tch, so snh cc hiu qu kinh t x hi, ca tng phng n, la chn phng n thit k k thut.

I.2-u v nhc im ca cc phng n.

Phng n 1.

u im:+ Kt cu nhp chnh thi cng theo phng php c hng cn bng l phng php thi cng quen thuc vi cc nh thu trong nc.

+ ng n hi lin tc, xe chy m thun.

+ Vt c nhp tng i ln.

+ Cu bng BTCT nn chi ph cho cng tc duy tu bo dng trong giai on khai thc thp.

Nhc im:

+ Qu trnh thi cng ph thuc nhiu vo iu kin thi tit.

+ S lng tr nhiu nh hng n lu thng dng chy.

+ Kt cu nng n.

Phng n 2.

u im:

+ S tr trn dng ch t do t nh hng n dng chy, thun li cho giao thng.

+ Kt cu c chiu cao kin trc nh, thanh mnh.

+ Kt cu cu v cng ngh thi cng hin i ph hp vi khuynh hng pht trin ca ngnh cu ng Vit Nam.

+ Hnh dng kin trc p, m quan ph hp vi cnh quan thin nhin.

+ ng n hi lin tc, t khe co gin, xe chy m thun.

+ Vt c nhp ln.

Nhc im:

+ Cng ngh thi cng phc tp i hi c trnh k thut cao, thit b tin tin.


+ Chi ph cho cng tc duy tu bo dng trong giai on khai thc ln.

Phng n 3.

u im:

+ u im ni bt ca cu Extradosed l v mt kt cu. Kch thc dm nh hn phng n cu lin tc do kt cu nng n hn v s lng dy t hn phng n cu dy vng nn chi ph cho bo dng cp vng trong giai on khai thc tn km.

+ Chiu cao thp thp hn phng n cu dy vng nn thi cng n gin hn.

+ S tr trn dng ch t do t nh hng n dng chy, thun li cho giao thng

+ Hnh dng kin trc p, m quan ph hp vi cnh quan thin nhin.

+ ng n hi lin tc, xe chy m thun.

Nhc im:

+ Cng ngh thi cng phc tp i hi c trnh k thut cao, thit b tin tin.


CHNG II: PHNG N CU LIN TC

II.1. GII THIU PHNG N.

S nhp : (33x2 + 75 + 3x120 + 75 + 33x2) m

Kh cu : 8 + 2 x 0.75 m.

Trc dc cu: Mt phn cu nm trn ng cong trn c R = 6000m, Kt cu phn trn:

Cu chnh dm lin tc 5 nhp (75 +3x120 + 75). Dm lin tc tit din hp. Chiu cao dm trn tr l H = 7.0 m, gia nhp l h = 3.0 m, phn dm c chiu cao khng i l h = 3.0 m.

Kt cu phn di:

M: Hai m i xng, loi m BTCT ch U, BTCT tng thng, t trn mng cc khoan nhi ng knh D = 1.5m.

Tr:

Tr nhp chnh: dng ch c BTCT, t trn mng cc khoan nhi ng knh D = 1.5m

Kt cu khc:

Khe co gin bng cao su.

Gi cu bng cao su.

Lan can cu bng b tng v thp ng

Lp ph mt cu 9cm:

Btng nha ht va 5cm

Lp bo v (b tng li thp) 3cm

Lp phng nc 1cm.

II.2. CHN TIT DIN.

Dm hp phn cu chnh:

Trn gi : H = (1/15 1/20)Lnhp Gia nhp : h = (1/30 1/ 45)Lnhp ; khng nh hn 2m.

Vi Lnhp = 120m, chn H = 7 m, h = 3.0 m.

Phn dm c chiu cao khng i h = 3.0m,

y dm bin thin theo quy lut ng cong c phng trnh l:

Y = X2 + h ,m

Vi L l chiu di cnh hng cong, L = 58 m.Vy ta c phng trnh ng cong bin di y dm hp l:

Y = X2 + 3.0 ,m

Chiu dy bn y ti nhp chnh thay i theo ng parabol t chiu dy ti gi l 100 cm n chiu dy gia nhp l 30 cm.

Y = X2 + 30 ,cm

Chiu dy bn y ti nhp bin thay i theo ng parabol t chiu dy ti gi l 120 cm n chiu dy gia nhp l 30 cm.

Y =

Chiu dy sn dm thay l 60cm.

Hnh II.2: Mt ct ngang dm hp c chiu cao thay i nhp chnh.

Mt cu c dc ngang 2% v dc dc thay i t 0% n 4 % trn ng cong trn R = 6000m.

Tnh ti phn b ca lp ph mt cu:

glp = 0.05x2.3 + 0.03x2.4 + 0.01x1.5 = 0.202 T/m2

Hnh II.3: Mt ct dc dm hp vi gc ta ti y t hp long.

Tnh chiu cao trong t y dm hp bin ngoi theo ng cong c phng trnh l:

Y1 = a1X2 + b1a1 = = 1.1891 x 10-3 ; b1 = 3.0 m

Chiu cao phn t hp long v phn t trn tr l khng i. Bng s liu chi tit xem ti Bng II.1Bn mt cu ti ch dy 20 cm.

Bng II: Tnh khi lng cnh hng.

STTTn tChiu di t (m)X (m)Chiu cao hp (m)Chiu dy bn y (m)Chiu dy sn (m)Chiu rng bn y (m)Din tch mt ct (m2)Th tch V(m3)Khi lng V(tn)

Cnh hng nhp c chiu cao dm thay i

1K02607.001.000.5005.5013.30026.60063.840

2K05587.001.000.5005.5013.30066.500159.600

3K13536.340.880.5005.5012.33537.00588.811

4K23505.970.820.5005.5011.79835.39384.942

5K33475.630.760.5005.5011.29133.87481.298

6K43445.300.700.5005.5010.81732.45077.880

7K53415.000.650.5005.5010.37331.12074.687

8K63.5384.720.600.5005.509.96134.86483.673

9K73.534.54.420.550.5005.509.52033.31979.967

10K83.5314.140.500.5005.509.12131.92476.618

11K93.527.53.900.460.5005.508.76530.67873.627

12K103.5243.680.420.5005.508.45229.58170.994

13K113.520.53.500.390.5005.508.18128.63368.719

14K124173.340.360.5005.507.95331.81076.345

15K134133.200.340.5005.507.74430.97674.341

16K14493.100.320.5005.507.59130.36372.872

17K15453.030.310.5005.507.49329.97471.937

180.5HL113.000.300.5005.507.4527.45217.884

Tng582.5151398.037

on c trn gio c dnh

1DG153.00.3000.5005.507.452111.178268.272

BngIII:. Khi lng cng tc m: MCao (m)Vtngcnh

(m3)Vthn m

(m3)Vi cc (m3)Vm (m3)

A19.0910.69152.28190.00352.97

A29.0910.69152.28190.00352.97

Tng cng705.94

Bng IV: Khi lng cng tc tr:

TrCao (m)Vthn tr (m3)Vicc (m3)Vmu tru (m3)VTr (m3)

P16.34235.01472.50707.51

P29.00333.61472.50806.11

P312.45461.5075601217.50

P417.32642.0275601398.02

P523.23861.0975601617.09

P616.43609.0375601365.03

P79.00333.61472.50806.11

P86.34235.01472.50707.51

Tng cng8624.88

II.3. CU TO M TR CU.

M: Hai m i xng, loi m nng ch U, BTCT tng thng, t trn mng cc khoan nhi ng knh D = 1.5m.

Bn qu : Hay bn gim ti c tc dng lm tng dn cng nn ng khi vo cu, to iu kin cho xe chy m thun, gim ti cho m khi hot ti ng trn lng th ph hoi. Bn qu c t nghing 10%, mt u gi ln vai k, mt u gi ln dm k bng BTCT, c thi cng lp ghp.

Tr: Tr c, BTCT, t trn mng cc khoan nhi ng knh D = 1.5 m

Hnh 5: Cu to m A1.

II.4. VT LIU.

Kt cu nhp:

B tng mc : #500 dng cho kt cu nhp.

#300 dng cho kt cu m , tr, cc khoan nhi.

Thp thng : AI c Ra = 1900 kG/cm2 AII c Ra = 2400 kG/cm2 Es = 2x106 kG/cm2 (5.4.3.2 22 TCN 272 - 01)

Thp cng cao :

Cng chu ko : fpu = 18600 kG/cm2

Gii hn chy : fpy = 16000 kG/cm2 (= 0.85fpu)

S cng ko : RH1 = 14000 kG/cm2 (= 0.76fpu)

Cng tnh ton : RH = 12800 kG/cm2 (=0.8fpy)

Es = 1.95x106 kG/cm2 (5.4.3.2 22 TCN 272 - 01).

II.5. TNH TON KHI LNG KT CU PHN TRN.

II.5.1. Khi lng phn kt cu nhp.

a. Phn cu chnh.

Ta lp bng tnh xc nh th tch cc khi c hng:

(Xem bng II)

Th tch phn dm hp c chiu cao thay i:

Vh thay i = 8 x 582.515 = 4660.12 m3 Th tch on c trn gio c nh:

Vh DG = 2 x 111.178 = 222.36 m3

Th tch bn tng nhp chnh l:

Vnhip chinh = Vh thay i + Vh DG = 4660.12+222.36= 4882.48 m3

II.5.2. Khi lng cng tc phn kt cu m tr:a) Khi lng m cu: Tng khi lng cng tc b tng m: Vm = 633.44 (m3)

(Xem bng III)

b) Khi lng tr cu: Tng khi lng cng tc b tng tr: Vtr = 4974.17 ( m3 )

(Xem bng IV)

Khi lng cng tc b tng m tr:

Vtr m = VTr+ VM = 4974.17+ 633.44 = 5680.11 ( m3 )

II.5.3. Khi lng cng tc lan can, g chn v lp ph mt cu:

Lan can:VLan can = 2x ALan can x Llan can = 2x0.22 x 530 = 233.2 (m3)

G chn+ b bo:VG chn = 2xA G chnxL G chn = 2x0.16 x 530=169.6 (m3)

Tng khi lng cng tc b tng lan can, g chn:

VLan can+g = 232.2+169.6= 401.8 (m3)

Th tch b tng li thp lp bo v (thuc lp ph mt cu):

VLp bo v = 0.03 x A = 0.03 x 17 x 530 = 270.3(m3)

Din tch lp phng nc dy 1cm:

APhng nc = 17 x 530 = 9010 m2 Th tch b tng nha:

VB tng nha = 0.05 x A = 0.05 x 17x530 = 450.5( m3 )

II.6. XC NH SC CHU TI CA CC KHOAN NHI:

II.6.1. Phn tch iu kin a cht:

Cn c vo cc s liu thm d a cht ta nhn thy tnh hnh a cht ca khu vc l kh n nh .

Dc theo tim cu,ti cc v tr tr d kin khoan thm d 1 l khoan (LK3 ~ HTV), tnh cht a tng t trn xung c th nh sau:

Lp 1: st. Lp 2: Ct ht nh.

Lp 3: ct pha st.

Lp 4: Dm cui.

Lp 5: dip thchSc chu ti ca cc theo vt liu

Theo iu [A5.7.4.4] sc chu ti ca cc theo vt liu lm cc tnh theo cng thc sau: Pr = Pn Trong :

Pr - sc khng lc dc trc tnh ton c hoc khng c un (N),

Pn - sc khng lc dc trc danh nh c hoc khng c un (N),

i vi cu kin c ct ai xon tnh theo cng thc

Pn = 0.85 x ( 0.85fc x (Ag-Ast) + fyAst )

fc - cng quy nh ca b tng tui 28 ngy, tr khi c quy nh cc tui khc, fc = 300kg/cm2 Ag - din tch nguyn ca mt ct (mm2),

Ast - din tch ca ct thp thng trong mt ct(mm2)

Ly Ast = (1,5-3)%Ag

fy - gii hn chy quy nh ca ct thp (MPa): fy = 3450kg/cm2 - h s sc khng quy nh [A 5.5.4.2] c = 0,75.

II.6.2.1. Xc nh sc chu ti theo vt liu vi D=1m:

Chn cc khoan nhi bng BTCT ng knh D = 1,0m, khoan xuyn qua cc lp t khc nhau.

B tng cc cp 30MPa.

Ct thp chu lc 16(28 c cng 345MPa. ai trn (10 a200.

Ct thp chu lc v ct thp cu to cc khoan nhi c b tr nh trong bn v ct thp cc khoan nhi.

Theo 5.7.4.4 - TCTK : i vi cu kin c ct ai xon th cng chu lc dc trc tnh ton xc nh theo cng thc:

Pr = (.Pn.

Vi Pn = Cng chu lc dc trc danh nh c hoc khng c un tnh theo cng thc:

Pn = (.{m1.m2.fc.(Ag - Ast) + fy.Ast}

= 0.80x{0.85x0.85x fcx(Ag - Ast) + fyxAst}.

Trong :

+ ( = H s un dc, ( = 0,80

+m1,m2 : Cc h s iu kin lm vic.

+fc , fy : Cng chu nn nh nht ca btng v cng chy do quy nh ca thp (MPa).

+Ac,Ast : Din tch tit din nguyn ca mt ct , ca ct thp dc (mm2).

Vi vt liu v kch thc ni trn ta c:

Pr = 1x0.80x[0.85x0.85x30x + 345x16x ]

= 16160.103(N).

Hay Pr = 16160 (KN).II.6.3.1. Xc nh sc chu ti theo vt liu vi D=1,5m:

1- Chn cc khoan nhi bng BTCT ng knh D = 1,5m, khoan xuyn qua cc lp t khc nhau.

2- B tng cc cp 30MPa.

3- Ct thp chu lc 24(28 c cng 345MPa. ai trn (10 a200.

Pr = 1x0.80x[0.85x0.85x30x + 345x24x ]

= 34447.103(N).

Hay Pr = 34447 (KN).2.3.1.2Sc chu ti ca cc theo t nn:

Theo [A10.7.3.2] sc khng ca cc c tnh theo cng thc sau:

Qr = Qn = qpQp + qsQs - Qpile (N)

Vi: Qp = qpAp ; Qs = qsAs

Trong :

Qp = Sc khng ca mi cc (N)

Qs = Sc khng ca thn cc (N)

Qpile : trng lng cc.

qp = 0.5 h s sc khng ca mi cc

qs = 0.65 h s sc khng ca thn cc

qp = Sc khng n v ca mi cc (Mpa)

qs = Sc khng n v ca thn cc (Mpa)

Ap = Din tch ca mi cc (mm2)

As = Din tch ca b mt thn cc (mm2)

Xc nh sc khng n v ca mi cc qp (Mpa), v sc khng mi cc Qp:

Cc ta trn nn theo iu[10.7.3.5]

V cc cc c khoan ngm vo lp dip thch nn c th b qua sc khng ct mt bn .Tnh nh cc chng .

Trong quy trnh khng cp n cng thc tnh ton sc khng ca cc khoan trong cng ,do ly cng thc tnh sc khng n v danh nh ca cc ng trong :

Sc khng n v danh nh ca mi cc qp ca cc cc ng n bng MPa c th tnh nh sau:

qp = 3 qu Ksp d(10.7.3.5-1)

trong :

(10.7.3.5-1)

y:

qu = cng nn dc trc trung bnh ca li (MPa) = 10Mpa

d = h s chiu su khng th nguyn (DIM)

Kps = h s kh nng chu ti khng th nguyn, t Hnh 1 (DiM)

sd = khong cch cc ng nt (mm) =200mm

td = chiu rng cc ng nt (mm)=3mm

D = chiu rng cc (mm) = 1200mm

Hs = chiu su chn cc vo trong h tnh =7.35m

Ds = ng knh ca h (mm)=1300mm

qp = 3 qu Ksp d=3x10x0.135x3.3 = 13.464 Mpa

Sc chu ti ca cc tnh ton l :

QR = x qp xAs

Vi l h s sc khng ca mi cc trong . 0.5 .

As din tch ngang ca mi cc .

QR l sc khng tnh ton ca mi cc.

QR = 0.5x13.464x103x3.1415x1.22/4=7609KN =760.9 T

T cc kt qu tnh trn chn sc chu ti ca cc l:[N] = Min (Pr, Qr) = 760.9 T

[N] = Min (Pr, Qr) = 760.9 T

II.7. XC NH S LNG CC M A1.

Ti trng thng xuyn (DC , DW): Gm trng lng bn thn m v trng lng kt cu nhp:

Trng lng bn thn m:

PM = 2.4xVM = 2.4 x 352.97 = 847.128 T

Trng lng kt cu nhp ( H dm mt cu, kt cu bn mt cu, lp ph, lan can, g chn):

Trng lng h dm mt cu:

gdm = 76.56 T/m

Trng lng bn mt cu:

gbn= 0.2x19x2.5= 9.12 T/m

Trng lng lp ph:

glp ph = ( 0.05x2.25 + 0.03x2.4 + 0.01x1.5 ) x 17 = 3.392 T/m

Trng lng lan can, gii phn cch, b bo:

glan can = (2 x 0.22+2x0.16+0.25) x 2.4 = 2.424 T/m

V ng nh hng p lc gi ti m M0

Din tch ng nh hng p lc m: w = 16.5

DC= PM+ (gdm + gbn + glan can ) x w

= 847.128 + (76.56+ 9.12 + 2.424 ) x 16.5

= 2300.84 T

DW = glp ph x w= 3.392 x16.5= 55.97 T

Hot ti:

Do ti trng HL93 + ngi (LL + PL)

Xe ti thit k v ti trng ln thit k + ngi:

LL = n.m.(1+ IM/100).(Pi .yi )+ n.m.Wln.w

PL = 3.pngi.w

Trong :

LL: hot ti xe

PL: ti trng ngi i

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : Lc xung kch, khi tnh m tr c th (1+ IM/100) = 1.25

Pi , yi :Ti trng trc xe, tung ng nh hng.

w: Din tch ng nh hng.

Wln , pngi : Ti trng ln v ti trng ngi, Wln = 0.93T/m , pngi = 0.3 T/m

LL(Xe tii)= 4x0.65x1.25x(1x14.5 + 0.95x14.5 + 0.899x3.5 ) + 4x0.65x0.93x42.5

= 296.92( T )

PL = 3 x 0.3 x 42.5 = 38.25T

Xe 2 trc thit k v ti trng ln thit k + ngi:

LL(Xe 2 trc)= 4 x 0.65 x 1.25 x ( 1 x 11 + 0.985 x 11 ) + 4x0.65 x 0.93 x 42.5

= 173.73 ( T )

Vy: LL= max(LL(Xe tii) , LL(Xe 2 trc) ) = 123.83 ( T )

T hp ti trng m A1 PA1:

Ni lcNguyn nhnTrng thi gii hn

cng I

(T)

DCDWLLPL

(gD = 1.25)(gW = 1.5)(gLL = 1.75)(gPL= 1.75)

P(T)4591.5144.16296.9238.256542.16

Py i = 6542.16 ( T )

Ti m A1 ta dng loi cc 1.5m, L=70m. a cht tnh ton l a cht ti l khoan LK3

Vy s cc dng ti m A1 l:

nc = = 13.33 (cc). Chn 15 cc

II.8. XC NH S LNG CC TR P3.

Ti trng thng xuyn (DC , DW): gm trng lng bn thn tr v trng lng kt cu nhp:

I.1. Trng lng bn thn tr:

Ptr = 2.4 x Vtr = 2.4 x 1268.66= 3044.784( T )

I.2. Trng lng kt cu nhp(H dm mt cu, kt cu bn mt cu, lp ph, lan can, g chn):

Trng lng lp ph:

glp ph = ( 0.05x2.25 + 0.03x2.4 + 0.01x1.5 ) x 17 = 3.392 T/m


glan can = (2 x 0.22+2x0.16+0.25) x 2.4 = 2.424 T/m


gdm lin tc = 18.463x 2.4 = 44.311 T/m

I.3. V ng nh hng p lc gi:

I.4. Din tch ng nh hng p lc tr: w = 120m2DC= PTr + (g dm lintc + glancan)xw

= 3044.784 + (44.311 + 2.424) x 120

= 8652.98( T )

DW = glp ph x w = 3.392 x 120= 407.04( T )

Hot ti:


I.5. Xe ti thit k v ti trng ln thit k + ngi:

LL = n.m.(1+).(Pi .yi )+ n.m.Wln.w

PL = 3.pngi.w

Trong :

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : Lc xung kch (lc ng ) ca xe, khi tnh m tr c th (1+IM/100) = 1..25




LL(Xe ti)= 4x0.65x1.25 x (1x14.5 + 0.964x14.5+ 0.964x3.5) +4x0,65x1.25x0.93x120

= 466.219 (T )

PL = 3 x 0.30 x 120

= 108( T )

I.6. Xe 2 trc thit k v ti trng ln thit k + ngi:

LL(Xe 2 trc)= 4 x 0.65 x 1.25 x (1 x 11 + 0.99 x 11) + 4 x 0.65x1.25 x 0.93 x 120

= 433.84(T)

I.7. 90% 2 xe ti thit k v ti trng ln thit k + ngi:

LL(2Xe ti)=0.9[4x0.65x1.25x(1x14.5+0.964x14.5+0.964x3.5+0.839x14.5+ 0.803x14.5+ 0.768x3.5) + 4x0,65x1.25x0.93x120]

= 497.10 (T )

Vy: LL= max (LL(Xe tii) , LL(Xe 2 trc) ,LL2xe ti) = 497.10( T )

T hp ti trng tr P3 PA1:


cng I

DCDWLLPL

(gD = 1.25)(gW = 1.5)(gLL = 1.75)(gPL= 1.75)

P(T)8652.98407.04497.10108.0012485.7

Py i = 12485.7 ( T ).

Ti tr P3 ta dng loi cc 1.5m, L = 80m. a cht tnh ton l a cht ti l khoan LK3~HTV.

Vy s cc dng ti m P3 l:

nc = = 19.303 (cc). Chn 20 cc

TrS ccD (m)Chiu di (m)

P1201.580

P2201.580

P3201.580

P4201.580

MS ccD (m)Chiu di (m)

M A1151.570

M A2151.570

TNH TON GI THNH V TNG MC U T PHNG N 1TTHng mcn vKhi lngn gi ()Thnh tin ()

Tng mc u t(A+B+C)159,581,036,400

AGI TR D TON XY LPAI+AII124,045,380,200

AIGi tr d ton xy lp chnhI+II119,274,404,000

IKT CU PHN TRN87,569,156,000

1Btng nhp lin tcm37550.3586,000,00045,302,148,000

2B tng c trn giom31482.6246,000,0008,092,656,000

3Ct thp thng dm lin tc(0.21T/m3)T1585.612,000,00019,027,200,000

4Ct thp UST dm lin tc(0.05T/m3)T400.518,000,0007,209,000,000

5Ct thp thng dm dn(0.17T/m3)T211.6512,000,0002,539,800,000

6Ct thp g chn lan can (0.1T/m3)T546,500,000351,000,000

7B tng g chn lan canm3537.92800,000430,336,000

8B tng atphanm3452.711,300,000588,536,000

9Gi cu ( loi ln)ci6913,000,000897,000,000

10Gi cu ( loi nh)ci148,000,0001,024,000,000

11Khe co dn 10cmm808,000,000640,000,000

12Khe co dn 5cmm302,500,00075,000,000

13Lp phng ncm29054.2120,0001,086,480,000

14in chiu sngCt368,500,000306,000,000

IIKT CU PHN DI31,705,248,000

1Cc khoan nhi D=1.5 mm5603,500,0001,960,000,000

3Ct thp m , tr (0.15 T/m3)T887.3047,000,0006,211,128,000

4B tng trm35209,421,300,0006,781,700,000

5Cng trnh ph tr%11963.47%15,552,420,000

AIIGa tr xy lp khc%AI4%4,770,976,160

BCHI PH KHC8,938,816,801

Chi ph khc%A6%7,442,722,810

CD PHNG, TRT GI%A+B20%26,596,839,390

CHNG III: PHNG N CU DY VNG

III.1. PHNG N THIT K.

Ta nhn thy tuyn i qua sng rng c thng thuyn, kh thng thuyn cp I (10x80m). ng thi do iu kin a hnh lng sng ti lch ch kh su. Trn c s kin ngh phng n cu dy vng 3 nhp .

Trc dc cu:mt phn cu nm trn ng cong trn c R = 10000m,

III.2 KCH THC C BN.

III.2.1. Kt cu nhp chnh.

S cu dy vng: 131+262+131 m

III.2.1.1. Chiu di khoang:

Chiu di khoang dm nh hng ti:

Tr s m men un cc b ca dm ch trong phm vi khoang

Ni lc, ln ca dy vng v h neo c

Cng ngh thi cng dm v dy.

an ton ca cng trnh khi dy c s c v to thun li khi sa cha ,thay dy

Hin nay cu dy vng thng c thi cng theo cng ngh hng, khoang dm ngn c nhiu thun li.Cn c vo cc iu kin trn chn chiu di khoang nh sau:

Nhp gm 72 khoang trong :

4 khoang p tr thp mi khoang di 12m

68 khoang cn li u nhau mi khoang 7m

III.2.1.2. S lng dy v tit din dy.

Theo s lng khoang v chiu di khoang chn th s lng dy nhp gia l 68 dy.

Hin nay cc b cp cng cao trong cu dy vng thng c t hp t cc tao cp n v cc tao cp n d vn chuyn, lp t v thch hp vi cc h thng neo. Do s dng cc tao cp n loi 15.2mm gm 7 si thp f5.

Cng gii hn ca cp lm dy vng fu=1860 Mpa

Din tch tit din ngang danh nh A=1.44 cm2Trng lng trn 1m di g=1.099 kg/m

Thit k mt ct ngang dm chnh:

Dm cng ng vai tr c bit quan trng trong cu dy vng, nh hng n kh nng chu ti trng, cng, n nh, cng ngh thi cng v gi thnh cng trnh.

Do dm cng ch yu chu nn nn dng BTCT. V vy quyt nh chn mt ct ngang dm cng l loi dm BTCT gm 2 dm ch tit din hnh thang to vt that gi,dm cng lin kt vi nhau bng dm ngang v bn mt cu.

Chiu cao dm ch: vi h 3 nhp ,2 mt phng dy:

Vi L = 262m , h = 2m( h/l=2/262 = 1/131

Chiu cao bn mt cu hb = 25cm.

Chiu cao dm ngang:hdn = 1.5m, dy 30cm, b tr cch nhau 3,5 m ti cc im neo dy v gia khoang

Hnh 6.Kch thc mt ct ngang cu dy vng.

III.2.1.3. Thp cu.

Chiu cao thp cu trc tip nh hng n gc nghing ca cc dy vng,gc nghing ca dy li quyt nh cng v cc ch tiu kinh t k thut ca cu.

III.2.1.4.1. Chiu cao thp cu:

Theo yu cu v vng ca nt dy treo l nhnht:

Trong :

Si , li:lc dc v hnh chiu ca dy vng th i ln phng dc cu

E, Ai: cng chu ko ca dy vng th i

ai :gc nghing ca dy vng th i

Ta thy yi nh nht khi sin2ai = 1 2ai = 90o ai = 45o

Theo yu cu v chuyn v ca nh thp cu l nh nht:

Trong :

So :lc dc trong dy neo

H :chiu cao thp cu

E(Ao: cng chu ko ca dy neo

a0 :gc nghing ca dy neo so vi phng ngang

( ( nh nht khi sin2ao = 1 2ao = 90o ao = 45oNh vy khi gc nghing tt nht trn quan im cng ca h:

Gc nghing ca dy neo (o=450Gc nghing ca dy vng (i=450Tuy nhin gc nghing ca dy neo ln th thp cu s rt cao, lm tng kch thc v khi lng vt liu.V mt chu lc thp cu lm vic nh mt thanh chu nn un ,thp cao gy bt li khi chu un dc,lm tng lc nh, c bit cng ngh thi cng gp nhiu kh khn.Do chiu cao tt nht ca thp cu cn c xc nh xut pht t tng gi thnh ca thp,dm v dy nh nht m bo bn v cng ca h.Thc t cho thy gc nghing hp l v chu lc v kinh t ca dy vng nghing nht l 22o ( 26o.T xc nh c chiu cao hp l ca thp cu.

Gc nghing ca cc dy vng cn li c la chn trn c s m bo cng tt nht ca h v trnh m men un ln trong thp. Do kin ngh dng s dy hnh r qut l hp l nht, n khc phc c nhc im ca s dy ng quy v song song.

Gc nghing tt nht ca dy vng thoi nht t 220(250 s m bo gi thnh chung ton cu nh nht.

Chn gc nghing ca dy vng xa nht l :(min= 230([220(250]

III.2.1.4.2. Tit din thp cu.

S dng thp c dng hnh ch H, din tch ti thiu ca thp c th xc nh theo cng thc:

=

Trong :

+ At : Din tch ct thp (thp 2 ct)

g , p:Tnh ti v hot ti tnh ton phn b u tc dng ln mt dn dy

l1 , l2 : Chiu di nhp bin v chiu di nhp chnh l1 = 131m, l2 = 262m

Rt :Cng vt liu lm thp ,b tng Mc M#300 cp R28= 30 Mpa

a :Gc nghing ca chn thp so vi phng ngang, a = 90o ( :H s phn phi ngang ca xe thit k i vi AH gi dm ti thp

Pht :Ti trng xe thit k, coi gn ng xe l lc tp trung ng ti v tr thp cu.

Tnh ti:

+ Trng lng bn thn ca h dm mt cu:

Trng lng bn thn dm ch:

Vi Fdc= 9.791m2 , g = 24 KN/m3 (9.791(24 = 234.984 KN/m

Trng lng dm ngang:

Dm ngang tit din 30(150 cm,chiu di 20.52m b tr cch u nhau: 3.5 m

g1dn= 0.3((1.5 - 0.25)(9.3 (24 = 97.09 2KN

+ Trng lng dm ngang trn mt mt di cu:

gdn= 27.794 KN/mVi :

n : s dm ngang,n= 150 L : chiu di nhp cu chnh,L= 524m

Vy trng lng bn thn ca h dm mt cu l:

g1= 234.984 + 27.794= 262.778KN/m

Trng lng lp ph mt cu

glp ph = ( 0.05x225 + 0.03x24 + 0.01x15 ) x 17 = 33.92 kN/m


glan can = (2 x 0.22+4x0.16+0.25) x 24 = 31.92 T/m

Vy tnh ti tnh ton phn b u trn mt dn dy:

g = 0.5((1.25(g1+1.5(glp ph+1.25(glan can)

=0.5((1.25(262.778+1.5(33.92 +1.25(31.92)= 209.626KN/m

Hot ti:

Xc nh h s phn phi ngang:

Hnh 7.ng nh hng p lc gi dy vng

H s phn phi ngang ca ti trng ln:

wln = 0.5((0.956+0.579+0.485+0.107)(6 = 6.378

Vy p = (w (m (WW+vWbohanh)

=1.75((0.65(6.378(3.1+0.650x3)= 25.903KN/m

Ti trng xe:

H s phn phi ngang ca ti trng xe ti thit k HL93.Xe ti thit k coi l lc tp trung ng ti thp cu.

Hnh 8.ng nh hng p lc gi dm cng ti thp

0.5(yi = 0.5((0.989+0.857+0.768+0.636+0.348+0.305+0.217+0.085)

=2.103

Ti trng hot ti tnh ton tc dng ln thp:

Pht = m((((I+IM/100)((xe(P=0.65(1.75((1+0.25)(2.103((145(2+35)

= 971.816KN

Vy din tch thp cu ti thiu:

=

= 4.23 m2

Chn tit din ca thp cu c dng hnh hp thay i t nh thp n chn thp:

Ta c din tch nh nht ca 1 thp l Amin= 6.438 m2 > 4.23 m2 -> t

Tit din dnh thp Tit din neo dy

Tit din thn thp

Hnh 9.Tit din chn thp

III.2. TNH TON S B DY VNG.

Trong cu dy vng, dy lm vic nh gi n hi chu ko, ni lc trong dy t tr s ln nht khi hot ti ng trn ton cu.Vy lc dc trong dy thoi nht gia nhp di tc dng ca tnh ti v hot ti l ln nht xc nh theo cng thc gn ng:

Trong :

g,p : tnh ti v hot ti tnh ton (ti trng ln) phn b u trn ton cu,(g+q)/2 tnh cho mt mt phng dy.

g = 0.5((1.25(g1+1.5(glp ph+1.25(glan can)

=0.5((1.25(262.778+1.5(33.92 +1.25(31.92)= 209.626KN/m

p = (w (m (WW+vWbohanh)

=1.75((0.65(6.378(3.1+0.650x3)= 25.903KN/m

Pi :Ti trng trc xe thit k

yi :Tung AH

:H s phn phi ngang ca xe thit k

d,dg:Chiu di 2 khoang dm nm k nt dy thoi nht

ag:Gc nghing ca dy vng thoi nht gia nhp

Hnh 10.ng nh hng gi dy vng- xc nh

= 0.5((0.918+0.805+0.73+0.616+0.447+0.333+0.258+0.145) = 2.216

Xp xe ln ng nh hng p lc xc nh ni lc trong dy vng thoi nht

+Do xe ti thit k:

Hnh 11.ng nh hng p lc gi dy vng-xe ti

+Do xe hai trc thit k:

Hnh 12.ng nh hng p lc gi dy vng-xe hai trc

Vy Smax = 5010.2 KNNi lc trong cc dy vng cn li trong phm vi nhp c xc nh theo cng thc:

ai: gc nghing ca dy vng th i,

i = 2(34Ring dy neo lm vic bt li nht khi hot ti ng kn nhp khi ni lc trong dy neo xc nh theo cng thc:

Trong : St 1:Ni lc trong dy neo do tnh ti

V cc dy neo b tr i xng qua thp cu nn:

+ Sh1:Ni lc trong dy neo do hot ti:

Trong : -

Vi : =5010.2 3755.48= 1254.72 KN Sh i:Ni lc trong dy vng th i do hot ti

a1: gc nghing ca dy neo

Tit din dy vng c xc nh theo cng thc:

Trong :

Si :Ni lc do tnh ti v hot ti trong dy vng th i

R :Cng tnh ton ca vt liu lm dy: R = 0.45(Rg (Tnh theo t hp chnh)

Rg :Cng gii hn ca vt liu lm dy, Rg = 1860 Mpa

R = 0.45(1860 = 837 Mpa

Tit din cc dy vng c t hp t cc tao cp ng knh 15.2mm c din tch f = 1.44cm2(gm 7 si f5)

S tao cp trong tng dy vng l:

Kt qu tnh ton c th hin trong bng

DyGc

()L

(m)Si

(kN)Sti

(kN)Sih

Aiyc

(m2)S taoChn

Aithc(m2)

123134.715010.20.0059941.6420.00605

224127.744813.10.0057539.9400.00576

324120.784813.10.0057539.9400.00576

425113.864632.20.0055338.4390.00562

526106.964465.70.0053437.1380.00547

627100.954311.90.0051535.8360.00518

72893.2734169.90.0049834.6350.00504

83086.5043915.30.0046832.5330.00475

93279.83694.20.0044130.7310.00446

103473.183500.80.0041829.1300.00432

113666.673330.50.0039827.6280.00403

123960.2933110.720.0037225.8260.00374

134254.1132925.60.0034924.3250.0036

144748.1972676.70.0031922.2230.00331

155242.6552484.30.0029720.6210.00302

166037.6532260.50.0027018.8190.00274

176933.4332096.90.0025117.4180.00259

186932.9762096.91462.1525.1204.430.0025117.4180.00259

195937.2242283.82101.3571.9319.980.0027318.9190.00274

205242.1122484.32511.8622.1416.070.0029720.6210.00302

214647.5642721.42834.1681.53514.320.0032522.6230.00331

224253.5832925.63031.9732.67591.50.0034924.3250.0036

233859.533179.73214.9796.3681.680.003826.4270.0039

243565.853413.03341.9854.7760.60.004128.3290.0042

253272.33694.23459.8925.1852.280.0044130.6310.00446

263078.893915.33533.2980.5922.450.0046832.5330.00475

272985.5744037.93568.31011.2960.80.0048233.5340.0049

282792.1243123635.11079.71045.10.0051535.8360.00518

292699.0494465.73666.91118.41092.020.0053437.1380.00547

3025105.864632.23697.51160.01142.00.0055338.4390.00562

3124112.624813.03727.11205.31196.20.0057539.9400.00576

3223119.625010.23755.51254.71254.70.0059941.6420.00605

3322126.675010.23755.51254.71254.70.0059941.6420.00605

3422133.475010.23755.51254.71254.70.0059941.6420.00605

III.3. TNH TON KHI LNG.III.5.1. Tnh ton khi lng phn kt cu nhp.

III.5.1.1. Phn cu chnh. Th tch khi c phn dm ch c chiu cao khng i h =2 m

V dm ch= A((l = 9.791(524 = 5130.48 m3 Th tch btng dm ngang:

Vdm ngang = n(V1dn= 150(0.3((1.7-0.25)(20.52 = 1338.93 m3

(Th tch b tng phn nhp cu chnh:

V btngcu chnh =5130.48 +1338.93 = 6459.41 m3

III.5.2. Khi lng m cu.

(Tng khi lng cng tc b tng m: Vm = 633.44 m3III.5.3. Th tch b tng thp cu.

Th tch thp cu:

Phn nh thp: chiu di 30.2 m ,din tch mt ct ngang trung bnh :A=7.22m2, th tch:

V= 7.22(30.2 = 218.044 m3Phn thn thp :chiu di 29 m,din tch mt ct ngang trung bnh Ath= 6.92m2, th tch:

Vth= 6.92(29 = 200.68 m3Phn chn thp :chiu di 18 m,din tch mt ct ngang trung bnh Ath= 13.89 m2, th tch:

Vch= 13.89(18 = 250.02 m3

Phn dm ngang pha trn :chiu di 13.4 m,din tch mt ct ngang trung bnh Ath= 2.5 m2, th tch:

Vnt=13.4(2.5 = 33.5 m3

Phn dm ngang pha di :chiu di 23 m,din tch mt ct ngang trung bnh Ath= 4.5 m2, th tch:

Vnd=23(4.5 = 103.5 m3

Phn b thp th tch:

Vb= 5(48(18 = 4320 m3(Tng th tch b tng 2 thp:

Vthp = 4((218.044 +200.68 +250.02)+2((33.5 +103.5 +4320) = 13009.22 m3III.5.4. Tnh khi lng lan can v lp ph mt cu.

Th tch b tng lan can: VLan can = ALan can(Lcu= 1.12( 524 = 586.88 m3Din tch lp phng nc:

APhng nc = 20.52 x524 = 10752.48 m2Din tch b tng nha:

VB tng nha = 17 x524= 8908 (m2)

III.4. TNH S B S LNG CC CA M.

III.6.1. S lng cc m A1.Ti trng thng xuyn (DC , DW): Gm trng lng bn thn m v trng lng kt cu nhp:

Trng lng bn thn m:

PM = 2.4xVM = 2.4 x 352.97 = 847.128 T



gdm = 5130.48x2.4/524 = 23.5 T/m

Trng lng bn mt cu:

gbn= 0.2x20.5x2.5= 10.25 T/m

Trng lng lp ph:

glp ph = ( 0.05x2.25 + 0.03x2.4 + 0.01x1.5 ) x 17 = 3.392 T/m


glan can = (2 x 0.22+2x0.16+0.25) x 2.4 = 2.424 T/m

V ng nh hng p lc gi ti m A1


DC= PM+ (gdm + gbn + glan can ) x w = 847.128 + (23.5+ 10.25 + 2.424 ) x 65.5

= 3216.525T

DW = glp ph x w= 3.392 x65.5= 222.176 T

Hot ti:



LL = n.m.(1+IM/100).(Pi .yi )+ n.m.Wln.w

PL = 3.pngi.w

Trong :

LL: hot ti xe

PL: ti trng ngi i

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : Lc xung kch, khi tnh m tr c th (1+IM/100) = 1.25




LL(Xe tii)= 4x0.65x1.25x(1x14.5 + 0.976x14.5 + 0.934x3.5 ) + 4x0.65x1.25x0.93x65.5

= 301.717( T )

PL = 3 x 0.3 x 65.5 = 58.95T

Xe 2 trc thit k v ti trng ln thit k + ngi:LL(Xe 2 trc)= 4 x 0.65 x 1.25 x ( 1 x 11 + 0.991 x 11 ) + 4x0.65 x 1.25 x 0.93 x 65.5

= 269.152 ( T )


T hp ta trng tc dng ln m A1 Phng n 2

Vi 4 ln:


cng I

(T)

DC

(gD = 1.25)DW

(gW = 1.5)LL

(gLL = 1.75)

P(T)3216.525222.176269.1524824.94

Py i = 4824.94 ( T ). Vi 6 ln: xe 3 trc

LL(Xe 3 trc)= 6x0.65x1.25x(1x14.5 + 0.976x14.5 + 0.934x3.5 ) + 6x0.65x1.25x0.93x65.5 = 452.57 T


= 403.73 ( T ) LL = 452.57 T.


cng I

(T)

DC

(gD = 1.25)DW

(gW = 1.5)LL

(gLL = 1.75)PL

(gll=1.75)

P(T)3216.525222.176452.5758.955249.08

Py i = 5249.08 TXc nh s cc

Ti m A1 ta dng loi cc 1.5 m. L=60m a cht tnh ton l a cht ti l khoan LK3


nc = 1,5=1,5. = 15.05 ccChn 15 cc D =1.5m L=60m.

III.6.2. Xc nh s cc thp T1.

Xc nh ti trng tc dng ln thp T1:



Pthp = 2.4 x Vthp = 2.4 x 6504.61= 15611.064( T )



I.11. Din tch ng nh hng p lc tr: w = 196.5

DC = Pthp + (g dm + glancan)xw

= 15611.064+ (23.5 + 2.424) x 196.5

= 20705.13( T )

DW = glp ph x w = 3.392 x 196.5 = 666.528( T )

Hot ti:




PL = 3.pngi.w

Trong :

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : Lc xung kch (lc ng ) ca xe, khi tnh m tr c th (1+IM/100) = 1.25




LL(Xe ti)= 4x0.65x1.25 x (1x14.5 + 0.983x14.5+ 0.967x3.5) + 4x0.65x0.93x196.5

= 579.586 (T )

PL = 3 x 0.30 x 196.5

= 176.85( T )


LL(Xe 2 trc)= 4 x 0.65 x 1.25 x (1 x 11 + 0.995 x 11) + 4x0.65 x 0.93 x 196.5

= 546.46 (T)

I.14. 90% Xe 2 xe ti thit k v ti trng ln thit k + ngi:

LL(2Xe ti) = 0.9[4x0.65x1.25x(1x14.5+0.983x14.5+0.967x3.5+0.926x14.5+ 0.909 x14.5 +0.877x3.5) + 4x0,65x0.93x196.5] = 607.34 (T ) Vy: LL= max (LL(Xe tii) , LL(Xe 2 trc) ,LL2xe ti) = 607.34( T )

T hp ti trng tc dng ln thp T1 phng n 3:


cng I

DCDWLLPL

(gD =1.25)(gW = 1.5)(gLL =1.75)(gPL=1.75)

P(T)20705.13666.582607.34176.8528253.62

Py i = 28253.62 ( T ).

Ti Thp T1 ta dng loi cc D = 2 m L =70m. a cht tnh ton l a cht ti l khoan LK3

Vy s cc dng ti thp T1 l:

nc = 1,5=1,5. =22.34 (cc). Chn 24 cc

III.5. TNH TON GI THNH V TNG MC U T PHNG N 2.

TTHng mcn vKhi lngn gi ()Thnh tin ()

Tng mc u t(A+B+C)172.216.839.400

AGI TR D TON XY LPAI+AII135.390.597.000

AIGi tr d ton xy lp chnhI+II130.183.266.300

IKT CU PHN TRN79.775.105.200

1Btng nhp dy vngm36459.416,000,00038.756.460.000

2Ct thp thng dm chnh(0.23T/m3)T1485.6612,000,00017.827.971.600

3Dy vng

( bao gm ph kin i km)T68925,000,00017,225,000,000

4Ct thp g chn lan can (0.1T/m3)T58.6886,500,000381.472.000

5B tng g chn lan canm3586.88800,000469.504.000

6B tng atphanm3890.81,300,0001.580.400.000

7Gi cu ( loi ln)ci8130,000,0001.040.000.000

8Khe co dn 10cmm408,000,000320,000,000

9Lp phng ncm210752.48120,0001.290.297.600

10in chiu sngCt1048,500,000884.000.000

IIKT CU PHN DI50.408.161.130

1Cc khoan nhi D=1.5 mm18002,000,0003.600.000.000

2Cc khoan nhi D=2.0 mm33604,500,00015.120.000.000

3Ct thp m (0.15 T/m3)T105.8917,000,000741.237.000

4B tng tr, thpm313009.221,300,00016.911.986.000

5Cng trnh ph tr%I7%14,034,938,138

AIIGa tr xy lp khc%AI4%5.207.330.653

BCHI PH KHC

Chi ph khc%A6%8.123.435.817

CD PHNG, TRT GI%A+B20%28.702.806.560

CHNG IV: PHNG N CU EXTRADOSED

IV.1. GII THIU PHNG N.

S nhp : 3x33 + (90 +140+90) + 3x33 m

Kh cu : 4 x 3.5 + 2 x1.5 m.

Trc dc cu: Mt phn cu nm trn ng cong trn c R = 7000m, phn cn li nm trn ng thng c dc dc id = 2.0 %.

Kt cu phn trn:

Cu chnh EXTRADOSE 2 mt phng dy 3 nhp (90 +140 + 90). Dm lin tc tit din hp. Chiu cao dm trn gi l H = 4.5 m v gia nhp l h= 2.5m.

Cu dn dm n gin bn lp ghp tit din ch I mi bn. Chiu cao dm khng i h = 1.7m, mt ct ngang gm 8 dm I.

Kt cu phn di:

M: Hai m i xng, loi m nng ch U, BTCT tng thng, t trn mng cc khoan nhi ng knh D = 1.5m.

Tr: Tr c, BTCT, t trn mng cc khoan nhi ng knh D = 1.5 m.

Thp: Thp cng BTCT tit din ch nht, t trn h mng cc khoan nhi ng knh D = 2 m.

Kt cu khc:

Khe co gin bng cao su.

Gi cu bng cao su.

Lan can cu bng b tng v thp ng

Lp ph mt cu:

Btng nha ht va 5cm

Lp bo v (b tng li thp) 3cm

Lp phng nc 1cm.

Vt liu:

B tng mc : #500 dng cho kt cu nhp, thp.

#300 dng cho kt cu m , tr, cc khoan nhi.

Thp thng : AI c Ra = 1900 kG/cm2 AII c Ra = 2400 kG/cm2 Es = 2x106 kG/cm2 (5.4.3.2 22 TCN 272 - 01)

Thp cng cao :

Cng chu ko : fpu = 18600 kG/cm2

Gii hn chy : fpy = 16000 kG/cm2 (= 0.85fpu)

S cng ko : RH1 = 14000 kG/cm2 (= 0.76fpu)

Cng tnh ton : RH = 12800 kG/cm2 (=0.8fpy)

Es = 1.95x106 kG/cm2 (5.4.3.2 22 TCN 272 - 05).

IV.2. LA CHN CC THNG S HNH HC C BN.

H dm mt cu:

S dng dm hp BTCTDL, bn mt cu dy 300mm, chiu dy sn dm 600mm. Neo c t trong neo c b tr hai bn cnh dm.

Chiu cao ca hp ti thp l 4.5m v chiu cao gia nhp l 2.5m.

Tng b rng pha trn ca hp l: 21.3m, chiu rng pha di l 16m.

Ti thp b tr hai li i 1.7x1.0 c vt gc 0.15x0.15m.

Mt ct ngang dm.

Cu to thp:

Thp cu c dng ch H, tit din dm ti thp cu to nh dm ngang tng n nh cho hai ct thp(Hnh v).

Thp cu c xy dng trn mt b mng cc. Chiu cao thp cu tnh t nh b cc l 43.55m. Mi ct thp c tit din khng i 3.0m x 2m t nh n cao mt cu v thay i t 3.0mx3.25m n 5.0mx3.25m t y dm n chn ct thp. Do cc dy thoi, gc to bi hai dy i xng nh. Do , s dng kt cu yn nga, nhm gim thiu chiu cao ct thp.

Cu to thp cu.

Mng thp: dng mng cc BTCT, khoan nhi ng knh 2m.

Dy cp ngoi (Extradosed cables):

S b tr dy theo hnh r qut vi hai mt phng dy thng ng cch nhau 20.5m (bng khong cch tim hai thp). Cc cp dy vng c b tr i xng qua mt phng thng ng i qua tim dc ca cu. Gc nghing ca dy so vi phng nm ngang nh nht (dy ngoi cng) l 120, v ln nht (dy gn thp cu nht ca nhp bin) l 180. Ton cu c 56 dy.

H cp dng cc tao song song ng knh 15.2mm. Cp c m v bc nha chng g. Cc tao cp c t trong ng bo v. Khong cch gia cc im neo dy ti mt cu bng chiu di cc t c dm ti v tr neo dy 5.0m

Theo ASTM A416-80: 1 tao ng knh danh nh l 15.2mm c:

Cng gii hn ca cp lm dy vng Rtc = 1860Mpa

Din tch tit din ngang danh nh A = 1.44 cm2

Trng lng trn 1m di g = 1.099 kg/m

Cc dy vng c cng lun trong qu trnh thi cng ln dy vng cng vi dm chu ti trng thi cng. ( Xem bng IV.1)

Bng IV.1. Tnh ton s tao cp vng Phng n 3:

Dy sGc

()

Chiu di dy

(m)

S tao chnDin tch cp

hu hiu

(cm2)Khi lng

(kg)

11430.914253633.97

21436.025253639.591

31441.138253645.211

413.546.252253650.831

513.551.368253656.453

61356.465253662.055

71361.578253667.674

Tng cng323.74175355.79

Tng s tao cp ton cu l: 175(2(2= 700 (tao)

Tng chiu di dy ton cu: 323.74(2(2= 1294.96 m

Tng khi lng dy : 355.79(2(2= 1423.16 kg

IV.3. TNH TON KHI LNG CNG TC KT CU PHN TRN.

IV.3.1. Khi lng cng tc phn kt cu nhp

IV.3.1.1. Phn cu chnh.

Din tch trung bnh mt ct dm chnh on thay i l :

Adm chnh= (16.14+28.24)/2 = 22.29 m2Th tch dm chnh l:

Vdm chnh = 120x 22.29+200x16.14 = 7132.8 m3

IV.3.1.2. Phn cu dn.

4- Din tch trung bnh ca mt ct ca 1 dm ch I l:

(0.6545x30+1.198x3)/33=0.704 m2.

5- Th tch ca 1 dm ch I l :

Vchu I =33x0.704=23.229 m36- Th tch ca ton b phn dm cu dn l:

Vdam dan = 6x8x23.229=1114.99 m3.

7- Th tch bn mt cu dn l:

Vban dan = 19x0.2 x 198 =752.4 m3.

8- Th tch phn nhp dn l:

Vnhip dan = Vdam dan +Vban dan = 1114.99 + 752.4 = 1867.39m3.

Th tch b tng kt cu nhp ton cu

Vkt cu nhp ton cu = Vdm chnh + Vnhp dn = 9000.19 (m3)

IV.3.2. Khi lng m cu.

Bng IV.2. Khi lng m cu phng n 3:

M Cao (m)Vtngcnh

(m3)Vthn m

(m3)Vi cc (m3)Vm (m3)

A1 8.02710.69152.28153.75316.72

A28.02710.69152.28153.75316.72

Tng cng633.44

Tng khi lng cng tc b tng m: Vm = 633.44 (m3)

IV.3.3. Khi lng tr cu.

Bng IV.3: Khi lng cng tc tr phng n 3:TrCao (m)Vthn tr (m3)Vicc (m3)Vmu tru (m3)VTr (m3)

P17.589.12309.8360.06459.01

P210.44124.06309.8360.06499.95

P312.284145.97309.8360.06515.86

P411.565137.429309.8360.06507.319

P58.563101.755309.8360.06471.645

P67.589.12309.8360.06459.01

Tng2912.794

Tng khi lng cng tc b tng tr: Vtr = 2912.794 ( m3 )

IV.3.4. Tnh khi lng thp.

Phn trn ca thp c chiu cao l 12 m c tit din trung bnh l 6 m2Vtrn=12x6= 72 m3 Phn di thp c chiu cao l 18.5 m c din tch trung bnh l 8.0 m2Vdi= 18.5x8.0= 148 m3 Th tch phn y i l:

Vi= 5 x34 x16= 2720 m3 Tng th tch thp cu l:

Vthp= 2x2x(72+148 )+2x2720= 6320 m3 Khi lng cng tc b tng m tr:

Vtr m = VTr+ VM = 6320+ 2912.794 + 633.44 = 9938.734 ( m3 )

IV.3.5. Khi lng lan can, g chn, lp ph mt cu.

Lan can:

VLn can = 2x ALan can x Llan can = 2x0.22 x 518.6 = 228.184 (m3)

Phn cch gia:

Vphn cch = Aphn cch xLphan cch =0.25x518.6= 129.65(m3)

G chn+ b bo:

VG chn = 2xA G chnxL G chn = 2x0.16 x 518.6 = 165.952(m3)

Tng khi lng cng tc b tng lan can, g chn:

VLan can+g = 228.184+129.65+165.952 = 523.786(m3)

Th tch b tng li thp lp bo v v (thuc lp ph mt cu):

VLp bo v = 0.03 x A = 0.03 x 17 x 518.6 = 264.486(m3)

Din tch lp phng nc dy 1cm:

APhng nc = 17 x 518.6 = 8816.2 m2 Th tch b tng nha:

VB tng nha = 0.05 x A = 0.05 x 17x518.6 = 440.81( m3 )IV.4. TNH TON S B S CC.

IV.4.1. Sc chu ti ca nn t.

Do ging nhau v a cht phng n mt nn s liu a cht v sc chu ti ca nn t v vt liu ly theo tnh ton phng n I

IV.4.2-Tnh s cc m A1.

Ti trng thng xuyn (DC , DW): Gm trng lng bn thn m v trng lng kt cu nhp:

Trng lng bn thn m:

PM = 2.4xVM = 2.4 x 352.97 = 847.128 T



gdm = 0.704 x 2.4 x8 = 13.517T/m

Trng lng bn mt cu:

gbn= 0.2x19x2.5= 9.12 T/m

Trng lng lp ph:

glp ph = ( 0.05x2.25 + 0.03x2.4 + 0.01x1.5 ) x 17 = 3.392 T/m


glan can = (2 x 0.22+2x0.16+0.25) x 2.4 = 2.424 T/m

V ng nh hng p lc gi ti m M0


DC= PM+ (gdm + gbn + glan can ) x w

= 847.128 + (13.517+ 9.12 + 2.424 ) x 16.25

= 1254.369 T

DW = glp ph x w= 3.392 x16.25= 55.12 T

Hot ti:



LL=n.m(1+IM/100)(Pi.yi)+n.m.Wln.w

PL = 3.pngi.w

Trong :

LL: hot ti xe

PL: ti trng ngi i

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : Lc xung kch, khi tnh m tr c th (1+IM/100)= 1.25





= 116.405( T )

PL = 3 x 0.3 x 16.25 = 14.625T

Xe 2 trc thit k v ti trng ln thit k + ngi:

LL(Xe 2 trc)= 4 x 0.65 x 1.25 x ( 1 x 11 + 0.963 x 11 ) + 4x0.65 x 0.93 x 16.25

= 95.434 ( T )


T hp ti trng m A1 PA2 vi 4 ln xe chy:


cng I

(T)

DCDWLL

(gD = 1.25)(gW = 1.5)(gLL = 1.75)

P(T)1254.36955.12116.4051854.35

Py i = 1854.35 ( T )T hp ti trng m A1 PA2 vi 6 ln xe chy:


= 203.52( T )LL(Xe 2 trc)= 6x 0.65 x 1.25 x ( 1 x 11 + 0.963 x 11 ) + 6x0.65 x 0.93 x 16.25

= 115.45 ( T )Ni lcNguyn nhnTrng thi gii hn

cng I

(T)

DCDWLLPL

(gD = 1.25)(gW = 1.5)(gLL = 1.75)(gPL= 1.75)

P(T)1254.36955.12203.5214.6252023.4

Py i = 2023.4 ( T )Ti m A1 ta dng loi cc 1m. a cht tnh ton l a cht ti l khoan LK3


nc = 1,5=1,5. =5.61 (cc). Chn 6 cc

Vy dng 6 cc D=1.5 ,L=60m

VI.4.3.Tnh s cc tr P2

Ti trng thng xuyn: (DC , DW) gm:


Ptr = 2.4 x Vtr = 2.4 x 499.95= 1199.88(T)

Trng lng kt cu nhp ( h dm mt cu, kt cu bn mt cu, lp ph, lan can, g chn):


gdm = 0.704 x 2.4 x8 = 13.517T/m

Trng lng kt cu bn mt cu:

gbn= 0.2x19x2.5= 9.12 T/m

Trng lng lp ph:

glp ph = ( 0.05x2.25 + 0.03x2.4 + 0.01x1.5 ) x 17 = 3.392 T/m


glan can = (2 x 0.22+2x0.16+0.25) x 2.4 = 2.424 T/m

V ng nh hng p lc gi:

Din tch ng nh hng p lc tr: w = 32.5

DC= PTr+ (gdm + gbn + glan can) x w= 1199.88+ (13.517+ 9.12 + 2.424 ) x 32.5

=2014.36 ( T )

DW= glp ph x w = 3.392 x 32.5 = 110.24 ( T )

Hot ti:

I.16. Do ti trng HL93 + ngi (LL + PL)



PL = 3.pngi.w

Trong :

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : H s xung kch ( lc ng ) ca xe, khi tnh m tr c th 1+IM/100= 1.25



Wln , pngi : Ti trng ln v ti trng ngi, Wln = 0.93 T/m , pngi = 0.3 T/m

LL(Xe tii)= 4 x 0.65 x 1.25 x (1 x 14.5 + 0.868 x 14.5 + 0.868 x 3.5) + 4x0.65x 1.25 x 0.93 x 32.5

= 156.907( T )

PL = 3 x 0.3 x 32.5 = 29.25 ( T )


= 143.727( T )

Vy:LL = max ( LL(Xe tii) , LL(Xe 2 trc) ) = 156.907 ( T )

T hp ti trng tr P2 PA3:


cng I

DCDWLLPL

(gD = 1.25)(gW = 1.5)(gLL = 1.75)(gPL= 1.75)

P(T)2014.36110.24156.90729.253009.08

Py i = 3009.08 ( T ).

Ti tr P2 ta dng loi cc 1.5m. a cht tnh ton l a cht ti l khoan LK3

Vy s cc dng ti tr P3 l:

nc = 1,5=1,5. =8.34 (cc). Chn 10 cc

Vy chn 10 cc D=1,5m L=60m.

IV.4.4 Xc nh s cc thp T1.

Xc nh ti trng tc dng ln thp T1:



Pthp = 2.4 x Vthp = 2.4 x 3160= 7584( T )


Trng lng lp ph:

glp ph = ( 0.05x2.25 + 0.03x2.4 + 0.01x1.5 ) x 17 = 3.392 T/m


glan can = (2 x 0.22+2x0.16+0.25) x 2.4 = 2.424 T/m


gdm = 22.29x 2.4 = 53.496 T/m


I.20. Din tch ng nh hng p lc tr: w = 115

DC = PTr + (g dm + glancan)xw

= 7584+ (53.496 + 2.424) x 115

= 14014.8( T )

DW = glp ph x w = 3.392 x 115 = 390.08( T )

Hot ti:




PL = 3.pngi.w

Trong :

n : S ln xe , n = 4.

m : H s ln xe, m = 0.65

IM : Lc xung kch (lc ng ) ca xe, khi tnh m tr c th (1+IM/100) = 1.25




LL(Xe ti)= 4x0.65x1.25 x (1x14.5 + 0.969x14.5+ 0.952x3.5) + 4x0.65x0.93x115

= 381.688 (T )

PL = 3 x 0.30 x 115

= 103.5( T )


LL(Xe 2 trc)= 4 x 0.65 x 1.25 x (1 x 11 + 0.991 x 11) + 4x0.65 x 0.93 x 115

= 349.25 (T)

I.23. 90% Xe 2 xe ti thit k v ti trng ln thit k + ngi:

LL(2Xe ti) = 0.9[4x0.65x1.25x(1x14.5+0.969x14.5+0.952x3.5+0.862x14.5+ 0.831x14.5 +0.800x3.5) + 4x0,65x0.93x115] = 423.51 (T ) Vy: LL= max (LL(Xe tii) , LL(Xe 2 trc) ,LL2xe ti) = 423.51 ( T )

T hp ti trng tc dng ln thp T1 phng n 3


cng I

DCDWLLPL

(gD =1.25)(gW = 1.5)(gLL =1.75)(gPL=1.75)

P(T)14014.8390.08423.51103.519025.88

Py i = 19025.88 ( T ).

Ti Thp T1 ta dng loi cc 2 m. a cht tnh ton l a cht ti l khoan LK3

Vy s cc dng ti thp T1 l:

nc =1,5 =1,5= 15.044 (cc).

Chn 18 cc D=2m L=70m.

IV.5. XC NH GI THNH V TNG MC U T PHNG N 3.

TTHng mcn vKhi lngn giThnh tin ()

Tng mc u t(A+B+C)197,985,583,100

AGi tr d ton xy lpAI+AII163,354,441,500

AIGi tr d ton xy lp chnhI+II157,071,578,400

IKt cu phn trn88,276,973,300

1B tng nhp chnhm37132.86,000,00042,796,800,000

2B tng nhp dnm31867.396,500,00012,138,035,000

3Ct thp thng dm chnh (0.21T/m3)T1497.8912,000,00017,974,656,000

4Ct thp ST + dy vng (0.05T/m3)T356.6418,000,0006,419,520,000

5Ct thp thng dm dn (0.17T/m3)T317.45612,000,0003,809,475,600

6Ct thp g chn lan can (0.1T/m3)T52.3796,500,000340,460,900

7B tng g chn lan canm3523.786800,000419,028,800

8B tng atphanm3440,811,300,000573,053,000

9Gi cu ( loi ln)ci6130,000,000780,000,000

10Gi cu ( loi nh)ci968,000,000768,000,000

11Khe co dn 10cmm208,000,000160,000,000

12Khe co dn 5cmm1102,500,000275,000,000

13Lp phng ncm28816.2120,0001,057,944,000

14n chiu sngCt908,500,000765,000,000

IIKt cu phn di68,794,605,080

1Cc khoan nhi D=1.5 mm49203,500,00017,220,000,000

2Cc khoan nhi D=2 mm25204,500,00011,340,000,000

4Ct thp m tr thp (0.15 T/m3)T1490,817,000,00010,435,670,700

5B tng m tr + thpm39938.7341,300,00012,920,354,200

6Cng trnh ph tr%I7%16,878,580,184

AIIGi tr xy lp khc%AI4%6,282,863,135

BChi ph khc

Chi ph khc%A6%9,801,266,492

CD phng trt gi%A+B20%34,631,141,600

CHNG V: BIN PHP THI CNG CH YU CA 3 PHNG NV.1. THI CNG M TR TRN CN.

* iu kin thi cng:

Vi mc nc thi cng nh trn th 2 m thi cng trn cn.

Mt bng thi cng tng i bng phng.

* Cc bc thi cng ch yu:

San i mt bng bng my i, nh v tim m, tim cc.

Lp t khung nh v ng vch, h ng vch gi thnh h khoan.

Lp dng my khoan chuyn dng, tin hnh thi cng cc khoan nhi ng knh D = 1.5m.

V sinh l khoan, h lng thp, b tng cc.

o h mng bng my xc kt hp th cng, p u cc, v sinh h mng, hon thin h mng.

lp b tng m dy 10cm, cp 15 Mpa ti cao y i.

b tng ti ch b m, thn m, tng cnh.

Hon thin m: tho d vn khun, p t nn.

V.2. THI CNG TR DI NC.

* iu kin thi cng:

Thi cng tr di nc nhng kh su

Mt bng thi cng tng i bng phng.

* Cc bc thi cng:

nh v tim tr bng my trc c. kt hp vi h sn o


Dng my khoan t trn h ni khoan to l cc ng thi gi thnh l khoan bng ng vch v dung dch bentonite.

V sinh h khoan, th lng thp, b tng cc.

Dng gi ba t trn h ni kt hp vi cn cu ng cc nh v, lp t vnh ai khung dn, rung h cc vn thp n cao thit k.

Xi ht t trong vng vy cc vn thp, p u cc, v sinh h mng.

Lp t vn khun, ct thp b tng b, thn m, tr.

V.3. THI CNG KT CU NHP CHNH 3 PHNG N.

V.3.1.Phng n 1: Cu lin tc.

* Trnh t thi cng dm lin tc.

Thi cng khi K0:

M rng tr, lm gi tm, lp t dn gio, vn khun, ct thp..tin hnh BT khi K0 trn nh tr.

Thi cng cc t tip theo bng cch c hng cn bng ra hai bn:

Lp t cc thanh DL trn nh tr, cng ko cc thanh DL gi n nh cho cnh hng trong qu trnh thi cng.

Lp xe c trn khi K0,nh v xe c,tin hnh c hng ra hai bn tr.

Lp t vn khun, ct thp, ng gen, b tng cc t.

Sau khi b tng t cng tin hnh cng ko cp DL.

Di chuyn xe c c t tip theo.

Thi cng on trn gio.

Lp t gio.

Lp t vn khun, ct thp, ng gen, b tng.

Sau khi b tng t cng tin hnh cng ko cp DL.

Hp long nhp bin:

Sau khi c hng n t cui cng, gng u cnh hng vi khi c trn gio.

Di chuyn xe c hp long nhp bin.

Lp t vn khun, ct thp, b tng t hp long nhp bin

Hp long nhp 1 v 4:

Gng 2 u cnh hng nhp 1 v 4 li vi nhau.

Di chuyn xe c, lp t vn khun, ct thp, BT khi hp long.

Sau khi b tng t cng tin hnh cng ko cp DL chu mmen dng, di chuyn xe c, gii phng lin kt tm trn cc nh tr.

Hp long nhp gia:

Qu trnh hp long tng t nh cc nhp trn.

V.4.1. Phng n 2: Thi cng cu dy vng.

V.4.2.1. Thi cng thp.

* iu kin thi cng

- Thp c thi cng di nc.

- Khi lng ln

- Mng o su

* Bin php thi cng chnh ca thn thp.

Lp t vng vy cc vn.

San i mt bng bng my i, nh v tim m, tim cc.


Lp dng my khoan chuyn dng, tin hnh thi cng cc khoan nhi ng knh D = 2m.

V sinh l khoan, h lng thp, b tng cc.

Dng ba kt hp vi cn cu ng cc nh v, lp t vnh ai khung dn, rung h cc vn thp n cao thit k.

o h mng bng myo gu ngm hp si ht, p u cc, v sinh h mng, hon thin h mng.

lp b tng bt y dy 2m, cp 15 Mpa ti cao y i.

+ b tng 8m phn chn thp bng dn gio.

+ Dng h vn khun leo, lp t ct thp v cc chi tit chn sn phc v thi cng, b tng thp.

+ b tng tng phn thn thp bng cn cu kt hp thng v vi bm b tng.

+ Khi xong t u tin tin hnh di chuyn vn khun leo ln cc t cn li cho n ht ton b thp cu.

+ Hon thin thp: tho d vn khun, hon thin thp.

V.4.2.2. Thi cng kt cu nhp phn cu dy vng.

* Cc bc thi cng:

Phn thi cng trn gio m rng thp:c khoang K0

+ Lp dng h thng gio thi cng khoang dm K0 u tin i xng qua thp

+ Lp dng vn khun, ct thp

+ b tng khoang K0 bng cn cu thp kt hp thng v vi bm

+ Bo dng b tng

+ Lp t dy vng v cng s chnh dy vng cho khoang K0

c cc khoang tip theo :c hng cn bng

+ Lp dng ng trt cho xe c

+ Lp dng xe c hng chuyn dng i xng 2 bn thp

+ Lp t vn khun, ct thp khoang K1 trn dn gio treo ca xe c

+ b tng khoang K1 bng cn cu thp kt hp thng v vi bm

+ Bo dng b tng.

+ Lp t dy vng v cng s chnh dy vng cho khoang K1 theo thit k

+ Tip tc di chuyn xe c thi cng cc khoang tip theo

Sau khi c xong mi khoang phi tin hnh lp ngay dy vng ca khoang v cng s chnh trc khi chuyn sang khoang mi

Hp long nhp chnh:

+ Di chuyn xe c hp long nhp chnh

+ Tin hnh nh v 2 u dm cng bng cc my trc a

+ Lp dng vn khun, ct thp, b tng cho t hp long

+ Bo dng b tng

Cn c biu ni lc v bin dng thc t tin hnh iu chnh dy vng ln cui nhm t c trng thi ti u trc khi a cng trnh vo khai thc.

Hon thin cu:

+ b tng cc lp mt cu, lp lan can, thit b chiu sng, thot nc.

V.4.2. Phng n 3: Thi cng cu Extradosed.V.4.3.1. Thi cng thp. Thi cng thp ging nh thi cng tr di nc.

V.4.3.2. Thi cng kt cu nhp phn cu chnh.

Thi cng khi K0:

Lp t dn gio m rng tr, lp vn khun, ct thp... tin hnh b tng khi K0 trn nh tr.

c hng cn bng ra hai bn:

Lp xe c hng cn bng trn khi K0, tin hnh c hng ra hai bn tr.

Lp t vn khun, ct thp, ng gen, b tng cc t.

Sau khi b tng t cng tin hnh cng ko cp DL

Di chuyn xe c c t tip theo.

Bt u t t K6 n K12 sau khi b tng t cng tin hnh cng dy vng, ri cng ct thp DL.

Thi cng on trn gio:

Lp dng gio, vn khun, ct thp b tng.

Sau khi tin hnh c hng cn bng t cui cng, di chuyn xe c c khi hp long ca hai nhp bin:

Trc khi b tng phi ging gi hai u.

Sau khi b tng t cng , tin hnh tho xe c, cng cp chu m men dng.

Hp long nhp nhp gia .

Qu trnh hp long tng t nh cc nhp trn.

CHNG 1: TNG QUAN1.1. t vn .

M tr cu l b phn quan trng trong cng trnh cu, c chc nng k kt cu nhp, tip nhn v truyn cc ti trng xung nn t.

Tr cu c xy dng gia hai nhp k nhau v chu p lc truyn ti trng t kt cu nhp. Tr nm phn long song c th chu tc dng ca dng chy, lc va p ca thuyn b

M cu c xy dng v tr tip gip gia ng v cu, ngoi nhim v k kt cu nhp n cn c vai tr ca mt tng chn m bo n nh cho nn ng u cu. Do ngoi cc phn lc truyn t kt cu nhp, m cn chu tc dng ca nn t. tnh ton v thit k thng phi p dng theo nhiu phng php v cc trng thi gii hn khc nhau, cng vic tnh ton thit k phc tp ny li ch c th p dng c vi cng trnh hin thi, i vi cc cng trnh gn tng t th li phi tnh li t u. Do c th gim ti cng vic lp i lp li ny cho ngi k s vic pht trin v xy dng phn mm tnh ton m tr cu l rt cn thit, t vic c th gii quyt c cho cng trnh hin thi n cn c p dng cho cc cng trnh tng t sau ny. Trn thc t c rt nhiu phn mm ra i phc v cho vic tnh ton M Cu nh: MO-2K5, SAP 2000, Midas/Civil

1.2. Cc phn mm tnh ton m cu

1.2.1. Phn mm tnh ton m cu MO-2K5 theo TCVN 272 2005 (Mo2K5 - Ch nghim ti KS Lm Hu Quang).a. u im

Xc nh nhanh c ni lc tnh ti ca m cu

Xc nh nhanh cc ti trng tc dng ln m cu

Th hin Tng Qut v hnh dng v kich thc m cu

b. Nhc im

Nhp qua Excel, kh mt thi gian

Cha t hp ni lc

1.2.2. Phn mm SAP 2000

SAP2000 l mt cng c mnh, c tin cy cao v c th s dng trong hu ht cc bi ton tnh ton kt cu. Chng trnh s dng phng php phn t hu hn kt hp vi cc thut ton x l s lm nn tng.

a. u im :

V kh nng tnh ton: Kh nng tnh ton mnh, h tr nhiu loi kt cu lm vic nhiu trng thi khc nhau chu tc ng ca nhiu loi ti trng.

V vt liu: Phn mm c th m t vt liu ng hng, trc hng, d hng hay vt liu c cc tnh cht phi tuyn.

V mt ti trng tc dng: Sap2000 h tr rt tt s a dng v th loi l: Tnh ti vi cc loi lc, nhit , gi ln. Hot ti vi nhiu loi xe tiu chun, xe do ngi dng t nh ngha tc dng ln nhiu ln phc tp ph hp vi nhiu quy trnh. Ti trng ng vi nhiu phng php tnh ton tin tin : ti trng thay i theo thi gian, ph phn ng

V kt qu tnh ton, thit k y , tin cy. Xut kt qu ra mn hnh ha, vn bn hay my in. Hn na, xut c cc kt qu dng tp tin cho cc chng trnh thit k sau tnh ton. Sap2000 hon thin hn cc phin bn trc, tch hp phn thit k mt ct thp v b tng ct thp vo chng trnh. Do vy, kt qu tnh ton kt cu s c s dng ngay trong phn thit k mt ct.

b. Nhc im:

Tnh tng thch v h tr cc bi ton chuyn bit.

Nhng quy nh tnh ton s dng trong chng trnh ch yu thch hp vi cc nc u M.

Tnh ton theo quy nh ca ta cng c th thc hin c nhng i hi ngi dng phi mt thi gian.

Do SAP2000 h tr tnh ton gn nh tt c cc loi kt cu do vy so vi cc chng trnh chuyn dng v tnh ton thit k cu n khng th thun tin bng.

1.2.3. Phn mm Midas/Civil

MIDAS/Civil l phn mm tch hp hon chnh h tr thit k kt cu cu. Bng s kt hp cc tnh nng phn tch kt cu vi cc tnh nng c bit phn tch c bit dnh cho k thut cu. Midas Civil h tr phn tch v thit k cu b tng, b tng d ng lc, cu thp, cu dy vng, cu dy vng, cu lin m, v.v.

a. u im:

Kh nng m hnh ha: chng trnh h tr nhiu m hnh kt cu, c bit l kt cu cu, cung cp nhiu loi mt ct khc nhau. Kh nng m t c vt liu ng hng, trc hng, d hng hay vt liu phi tuyn. Chng trnh c nhiu cng c trc quan h tr vic m hnh ha trc tip, ngoi ra ngi dng cn c th m hnh kt cu hoc mt ct thng qua Auto CAD.

V ti trng: chng trnh h tr y v a dng v cc th loi nh: tnh ti (vi cc loi lc, nhit , gi ln, d ng lc), hot ti (vi nhiu loi xe tiu chun k thut, xe do ngi dng nh ngha) m hnh ti trng ng vi cc gii php tin tin.

Giao din v tc tnh ton: chng trnh hot ng nh mi trng Windows, giao din thn thin, kh nng tnh ton mnh. Kh nng nhp v xut d liu: d liu u vo c th c nhp trc tip hoc nhp t cc file ca chng trnh khc, kt qu tnh ton c th xut ra ngoi mn hnh ha, vn bn hay my in, hn na c th kt xut kt qu dng tp tin cho cc chng trnh thit k sau s dng.

Kh nng phn tch bi ton cu: y l mt tnh nng mnh ca chng trnh, MIDAS cung cp nhiu phng php phn tch kt cu hin i, c bit l phn tch phi tuyn tnh v cc giai on thi cng.

b. Nhc im:

L mt phn mm thit k theo tiu chun nc ngoi, khng ph hp vi cc tiu chun, quy phm ca Vit Nam.

+Ngn ng nc ngoi.1.3. ti.

Trn c s phn tch u v nhc im t cc phn mm trn v khc phc mt s nhc im ca cc phn mm trc nh: ngn ng s dng, tiu chun thit k v cc quy phm Hng n mt phn mm s c ngn ng giao din bng ting Vit v c p dng theo cc tiu chun ca Vit Nam hin hnh. T ti s a ra nghin cu trong n tt nghip ny l: THIT K V KIM TON M CH U.

CHNG II: C S L THUYT

2.1. S liu chung.

Thit k cu qua sng

Loi cu

Tn m tnh ton

Quy trnh tnh ton : Tiu chun thit k cu 272 -05 TCVN

2.2. Kt cu phn trn.

Chiu di nhp tnh ton

Loi dm

S lng dm ch

S lng dm ngang

Chiu cao dm ch

Chiu cao g lan can

Chiu cao lan can

Kh cu

B rng mt cu

Chiu cao bn dm ch

Din tch dm ch

Din tch dm ngang

2.3. S liu m.

Loi m : m ch U..

Nn t nhin.

2.4. Tnh ti.

Trng lng bn thn kt cu phn bn trn,

Trng lng bn thn m

Trong :

- Trng lng ring ca vt liu,

V- Th tch m cu,

Khi b phn m tr nm di nc khi tnh n nh phi xt n tc dng ca p lc thu tnh. Khi trng lng ring l:

=-1 (T/m) Trng lng t p,

Trng lng t p trn cc b mng v cc thnh nghing ca m cu:

P= * H

Trong : trng lng ring ca t, = 1,8 T/m.

H Chiu cao t p.

p lc ngang ca t

Rt quan trng khi tnh m. i vi tr th tu loi, c th tnh hoc khng tnh tu theo mc nh hng.

Theo QT 79 p lc y ngang tnh theo cng thc:

Ep =

Trong : H chiu cao tng t tnh ton.

=tg2 - H s p lc ngang ca t,

- Gc ma st trong v dung trng th tch ca t

Khi y mng t cch mt t t nhin 3m coi p lc y ngang ca t phn b theo quy lut ng thng.

(Hp lc y ngang tnh theo cng thc:

E=

Trong : ep v H - p lc nm ngang ca t v chiu cao tng t.

B - chiu rng tnh i ca m.

b1 2b2 ( B = b. b1 > 2b2 ( B =2

Vi m cc (ct) nu chiu rng tng cng cc cc (ct) < 1/2 chiu

rng m tr th B =2 ( b chiu rng cc hoc ct)

Vi m cc (ct) nu chiu rng tng cng cc cc (ct) 1/2 chiu

rng m tr th B ly bng khong cch mp ngoi ca cc (ct).

Cnh tay n ca hp lc cch y mng 1 khong :

E= H/3

2.5. Hot ti. Hot ti 3 trc,

Hot ti 2 trc,

Ti trng ln,

2.6. Ti trng gi

2.6.1. Ti trng tc dng ln cng trnh.

Ti trng gi ngang :

Pd = 0,0006V2*A*Cd > 1,8A (KN)

Trong :

V : Tc gi thit k,

A : Din tch kt cu chn gi khng c hot ti tc dng,

Cd: H s sc cn =1.

Tnh cho gi tc dng ln kt cu phn trn,

Ti trng gi tc dng ln m,

2.6.2. Ti trng gi tc dng ln xe c.

2.6.2.1, Ti trng gi ngang cu,

Ti trng gi ngang tc dng ln xe c : 1,5 kN/m

Chiu di tc dng : 30 m,

im t lc cch mt ng 1,8 m

2.6.2.2, Ti trng gi dc cu,

- Ly bng ti trng gi ngang,

2.7. T hp ti trng

TTGH DCDWLL, BR

CE, PLWSWLWA

S dng1110,311

Cng 11,251,51,75001

Cng 31,251,51,350,411

2.8. Kim ton.

2.8.1.Kim tra cu kin chu un.

2.8.1.1. Sc khng un:

Mr = 0.9*As*fy*(h0-a/2)

Momen tnh ton do ngoi lc:

Mu = Max(M1,M2,M3)

Kim tra Mr < Mu2.8.1.2.Hm lng c thp ti thiu.

Hm lng thp ( = Kim tra:

( 2.8.1.2.Kim tra sc khng un. Mrc = 0.623(1000(Kim tra:

Mr min(1.2Mrc,Mu)

2.8.2.Kim tra nt.

*

Trong :

dc : chiu dy phn b tng t th chu ko ngoi cng cho n tm ca thanh, dc khng c ln hn 50.

Z : Thng s v b rng vt nt khng ly qu 30000,

A : Din tch phn b tng c cng trng tm vi ct thp ch chu ko.

2.8.2.Kim tra cu kin chu ct.

Sc khng ct danh nh do ng sut ko trong:

Vc = 0.083( (bw(d((0.001

Sc khng ct ca ct thp chu ct.

Vs = Av*fy*d/(s*tan(())*1000)

Vn = 0.25(fc(bw(d

Sc khng ct tnh ton:

Vr = min(fv(Vn, fv((Vc+Vs))

Lc ct tnh ton:

Vu = H (lc ct trng thi cng I)

Kim tra:Vu VrCHNG III: THUT TON

3.1. S phn r chc nng.

Trn c s thit k chng trnh bo gm 2 chc nng ln l tnh ton v thit k m cu, ta xy dng s phn r chc nng ca chng trnh nh sau:

3.2. Thut ton tng quan.

3.2.1. Thut ton con tnh ton.

a) Thut ton con: Tnh ti m

b) Thut ton con: Hot ti

c) Thut ton con: p lc t tnh

d) Thut ton con: p lc t do hot ti EL

e) Thut ton con: Ti trng gi ngang

f) Thut ton con: Ti trng gi ng

3.2.2. Thut ton con nhp d liu u vo

a) Thut ton con:Nhp thng s vt liu: trng lng b tng, cng b tng, trng lng thp, cng thp

b) Thut ton con:Nhp kt cu bn trn : DC, DW, Ln, Bc, Ln

c) Thut ton con: Hot ti do 1 ln

d) Thut ton con: Nhp kch thc

CHNG IV: NGN NG LP TRNH.

IV.1. Gii thiu mt s ngn ng lp trnh

PASCAL : l ngn ng lp trnh c bn.u im:

Pascal dng ngn ng st vi ngn ng t nhin hn do n thn thin vi ngi lp trnh hn. Pascal kt hp c tnh gn, d nh, kh nng truy cp thp cc cu trc d liu a dng. Pascal l ngn ng lp trnh c nh kiu r: cc i lng ( bin v hng) c khai bo s dng vi kiu d liu ny th khng th em dng ln vi kiu khc. Pascal l ngn ng lp trnh c cu trc. Tnh cu trc ca Pascal c th hin qua 3 yu t : cu trc trong d liu, cu trc trong cc ton t v cu trc trong cng c th tc. c tnh sng sa, d hiu, d c ca n gip ngi mi dng c th d dng hc vit mt chng trnh my tnh. chng trnh vit gn , dch nhanh, khng ngng c ci tin p ng yu cu ngi s dng.

Nhc im:

Bn cnh Pascal cng cn khng t hn ch, giao din tng tc km vi ngi s dng , h tr ho khng mnh m.

Vi nhng chng trnh ln th dng ngn ng lp trnh c cu trc qun l s l rt kh khn. C: l ngn ng tt cho lp trnh h thng v pht trin ng dng.

u im:

- Chng trnh vit bng C l tp hp cc hm ring bit, gip cho vic che giu m v d liu tr nn d dng. Hm c vit vi nhng ngi lp trnh khc nhau, khng nh hng n nhau v c th c bin dch ring bit trc khi gip ni thnh chng trnh.- C l mt ngn ng rt mnh m mm do, linh hot, c mt th vin gm rt nhiu cc hm ( FUNCTION ) c to sn. Ngi lp trnh c th tn dng cc hm ny m khng phi to mi. Hn na ngn ng C c h tr rt nhiu cc php ton nn ph hp gii quyt cc bi ton k thut c nhiu cng thc phc tp. Ngoi ra C cng cho php nh ngha thm cc kiu d liu tru tng khc- Mt c im ni bt ca C l C c tnh tng thch cao. Chng trnh vit bng C cho mt loi my hoc h iu hnh khc. Hin nay hu ht cc loi my tnh u c trnh bin dch C. Nhc im:

-Tuy vy, C ch thch hp vi nhng chng trnh h thng hoc nhng chng trnh i hi tc. Cn vi nhng bi ton ln v phc tp th cng nh Pascal , C rt kh kim sot chng trnh.

C++ : l ngn ng lp trnh hng i tng u tin.

u im:- Ngn ng C++ c pht trin t ngn ng C. N mang y cc c tnh ca C.- C++ l ngn ng lp trnh hng i tng, do vy n c y cc tnh cht ca mt ngn ng lp trnh hng i tng. - Cc c tnh ca C++ cho php ngi lp trnh xy dng nhng th vin phn mm c cht lng cao phc v nhng n ln, chng trnh ln nh cc h son tho, chng trnh dch, cc h qun tr c s d liu, cc h s truyn thngNhc im:

C++ khng hng i tng hon ton m l a hng. V C++ h tr c lp trnh hng hnh ng. Visual Basic 6

- VB6 c s dng to giao din ho ngi dng ( Graphical User Interface hay vit tt l GUI ). C sn nhng b phn hnh nh gi l Controls.

- VB6 cha n hng trm cu lnh (Commands), hm (Function), v t kho (Keywords), rt nhiu commands, fuctions v t kho lin h trc tip n MSWindows GUI. Nhng ngi mi bt u c th vit chng trnh bng cch ch hc vi commands, functions v keywords.

Visual Basic . Net (VB.net)

VB.net l ngn ng lp trnh hng i tng ( Object Oriented Language OOL) do microsoft thit k li t con s khng. VB.net khng k tha VB6 hay b sung t VB6 m l mt ngn ng lp trnh hon ton mi trn nn Microsofts.Net Framework. Do n cng khng phi l VB phin bn 7. Tht s y l ngn ng mi v rt li hi, khng nhng lp nn tng vng chc theo kiu mu i tng nh cc ngn ng lp trnh hng mnh khc vang danh nh C++, Java m cn d hc, d pht trin v cn to c hi hon ho gip ta gii quyt nhng vn khc mc khi lp trnh. Hn na, d khng c kh khn g khi cn tham kho, hc hi hay o su nhng g xy ra bn trong.VB.net gip ta i ph vi cc phc tp khi lp trnh trn nn Windows. IV.2 La chn ngn ng lp trnh

Da vo nhng kin thc tch ly c trong qu trnh hc tp v nghin cu cc ngn ng trn em thy: VB.net l ngn ng m em thnh tho nht, ngoi ra n cn c rt nhiu im mnh nh:

VB.net l ngn ng c dng kh ph bin hin nay.

VB.net l ngn ng n gin c nhiu h tr, c bit h tr lp trnh hng i tng kh tt.

VB.net c th kt hp vi cc cng c m rng lm p thm giao din chng trnh (VD: B Dotnetbar)

VB.net cho php thit k chng trnh bng cc modul c lp gip ngi lp trnh d dng pht trin phn mm m khng nh hng ti cc modul khc.

VB.net l ngn ng thng dch, c tnh mm do cao. Cho php kt hp vi cc c s d liu bn ngoi x l.

T rt nhiu nhng u im trn nn ngn ng VB.net c dng vit chng trnh cho n tt nghip.

IV.3Cc Form trong chng trnh.

Form chng trnh chnh gii thiu v cc menu chc nng.

Giao din chng trnh.

Menu chn tp:

M tp: chn m n tp c trc.

Lu tp: chn lu tp ang m.

Thot : kt thc chng trnh.

Menu nhp d liu:

Thng s vt liu: m form vt liu.

Nhp kt cu bn trn: m form kt cu bn trn.

Hot ti: m form hot ti.

Nhp kch thc m: m form kch thc m.

Menu tnh ton:

Tnh ti m: m form tnh ti m.

Hot ti m: m form hot ti m.

Ti trng gi: m form ti trng gi.

p lc t: m form p lc t.

Menu t hp ti trng:

T hp ti trng: m form t hp ti trng.

Menu kim ton:

Kim ton : m form kim ton.

Form vt liu:

Chn v d xem cc thng s mu c sn, chn nhp mi nhp li cc thng s, chn combobox mc b tng v loi thp thay i s liu cn tnh, sau nhn lu d liu ri tip tc.

Form kt cu bn trn:

Lm tng t nh form vt liu , c th nhn vo nt trc quay li form vt liu.

Form hot ti:

Sau khi nhp xong cc thng s form kt cu bn trn , nhp cc gi tr ti trng xe thit k cho form hot ti ri tip tc.

Form nhp kch thc m:

Lm tng t nh cc form trc ,nhp y cc s liu ri tip tc.

Form tnh ti m:

Sau khi nhp y cc thng s chuyn sang phn tnh ton tnh ti ,nhn kt qu hin th kt qu ri nhn tip tc tip tc tnh ton.

Form hot ti m:

Lm tng t nh tnh ton tnh ti m, form ny c them phn v ng nh hng, v ri tip tc.

Form p lc t:

Form ny c 2 tab ,phi nhp p lc t tnh trc ri chuyn sang nhp s liu p lc t do hot ti ri tip tc.

Form ti trng gi:

Form ny tng t nh form p lc t ,nhp y ri tip tc.

Form t hp ti trng: mt ct tng nh.

T hp ti trng cho 4 mt ct khc nhau, mi bng th hin gi tr ti trng tc dng trn tng mt ct

Mt ct tng thn.

Mt ct tng cnh.

Mt ct y mng.

Form kim ton:

Kim tra kh nng chu lc ca m ti cc mt ct nh: kim tra chu un, kim tra chu ct , kim tra nt xem m c t kh nng chu lc hay khng.

CHNG CHNH TRNH

TR GIP

XUT KT QU

NHP S LIU

KIM TON

TNH TON

Kim ton tng nh

Tnh ti

c trng vt liu

Xut kt qu ra Excel

Gii thiu

Kim ton tng thn

Hot ti

Ti trng xe

Xut kt qu ra Word

Hng dn

S liu kt cu

Kim ton tng cnh

Ti trng gi

p lc t, ti trng cht thm

Kim ton y mng

Nhp kch thc m

T hp ti trng

Kt thc

Xut kt qu

Tnh ton

Nhp d liu u vo

Bt u

145KN

145KN

35KN

9.3KN/m

1

0.87

0.899

33000

4.3

4.3

110KN

110KN

9.3KN/m

1

0.985

33000

1.2

k0

K1

K2

K3

K4

K5

K6

K7

K8

K9

K10

K11

K12

K13

K14

K15

5@3000=15000

6@3500=21000

4@4000=16000

14000/2=7000

60000 (1/2 Nh?p )

hl

2000

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

NG THANH KHU_9330.55_55TH1 10

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