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.
+ Qu trnh thi cng ph thuc nhiu vo iu kin thi tit.
+ 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.
+ Qu trnh thi cng ph thuc nhiu vo iu kin thi tit.
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
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
Trng lng h dm mt cu:
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:
Do ti trng HL93 + ngi (LL + PL)
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
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 =
0.30 T/m
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:
Ni lcNguyn nhnTrng thi gii hn
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
Trng lng lan can, gii phn cch, b bo:
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
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 = 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
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 A1
Din tch ng nh hng p lc m: w = 65.5
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:
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 + 0.976x14.5 + 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 )
Vy: LL= max(LL(Xe tii) , LL(Xe 2 trc) ) = 301.717 ( T )
T hp ta trng tc dng ln m A1 Phng n 2
Vi 4 ln:
Ni lcNguyn nhnTrng thi gii hn
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
Xe 2 trc thit k v ti trng ln thit k + ngi:LL(Xe 2 trc)= 6 x 0.65
x 1.25 x ( 1 x 11 + 0.991 x 11 ) + 6x0.65 x 1.25 x 0.93 x 65.5
= 403.73 ( T ) LL = 452.57 T.
Ni lcNguyn nhnTrng thi gii hn
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
Vy s cc dng ti m A1 l:
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:
Ti trng thng xuyn (DC , DW): gm trng lng bn thn tr v trng lng kt
cu nhp:
I.8. Trng lng bn thn tr:
Pthp = 2.4 x Vthp = 2.4 x 6504.61= 15611.064( T )
I.9. Trng lng kt cu nhp(H dm mt cu, kt cu bn mt cu, lp ph, lan
can, g chn):
I.10. V ng nh hng p lc gi:
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:
Do ti trng HL93 + ngi (LL + PL)
I.12. 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
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 =
0.30 T/m
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 )
I.13. 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 + 0.995 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:
Ni lcNguyn nhnTrng thi gii hn
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
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 = 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
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 = 16.25
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:
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 + 0.868x14.5 + 0.735x3.5 ) +
4x0.65x0.93x16.25
= 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 )
Vy: LL= max(LL(Xe tii) , LL(Xe 2 trc) ) = 116.405 ( T )
T hp ti trng m A1 PA2 vi 4 ln xe chy:
Ni lcNguyn nhnTrng thi gii hn
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:
LL(Xe tii)= 6x0.65x1.25x(1x14.5 + 0.868x14.5 + 0.735x3.5 ) +
6x0.65x0.93x16.25
= 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
Vy s cc dng ti m A1 l:
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:
I.15. Trng lng bn thn tr:
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):
Trng lng h dm mt cu:
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
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:
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)
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 : H s xung kch ( lc ng ) ca xe, 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 = 0.93 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 )
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 1.25 x 0.93 x 32.5
= 143.727( T )
Vy:LL = max ( LL(Xe tii) , LL(Xe 2 trc) ) = 156.907 ( T )
T hp ti trng tr P2 PA3:
Ni lcNguyn nhnTrng thi gii hn
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:
Ti trng thng xuyn (DC , DW): gm trng lng bn thn tr v trng lng kt
cu nhp:
I.17. Trng lng bn thn tr:
Pthp = 2.4 x Vthp = 2.4 x 3160= 7584( T )
I.18. 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
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
Trng lng h dm mt cu:
gdm = 22.29x 2.4 = 53.496 T/m
I.19. V ng nh hng p lc gi:
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:
Do ti trng HL93 + ngi (LL + PL)
I.21. 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
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 =
0.30 T/m
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 )
I.22. 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
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
Ni lcNguyn nhnTrng thi gii hn
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
Lp t khung nh v ng vch, h ng vch gi thnh h khoan.
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 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
= 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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