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I
O AN KY THUAT THI CONG 1 GVHD : Th.S Ks: TRAN THOAI CHAU
GII THIEU CONG TRNH
Dien tch xay dng : 9m x 23,2m
Mong cong trnh la mong coc.
Cong trnh s dung giai phap mong coc ep BTCT.
Coc dai: Lo = 24 m gom 3 oan C1, C2, C3 moi oan 8 m
Tiet dien coc : 200x200
Sc chu tai cua coc la Pc = 30 T
Lc ep toi a au coc la Pmax = 60 T
Lc p nh nht: Pmin = 45 T
I. THI CONG COC EP BETONG COT THEP
1. Chon may ep coc :
Theo TCXDVN 286-2003 : ong va ep coc Tieu chuan thi cong va nghiem thu.
a. Chon kch ep :
Pkch
R = 12000
9000
2200
1.4 Pmax = 84 T
b. Chon oi trong : oi trong la cac cuc BTCT co kch thc (0,7x1x3)m
Ptai = 5,25 T
Poi trong = 1.1 Pmax = 1.1x60 = 66 T
So lng cuc tai can :
66
12,57
5,255,25
doitrong
P
n
===
(cc)
Chn 14 cc, b tr 7 cc cho mi ben dan ep.
2. Chon may cu :
a. Cu dm chnh :
+ Sc truc :
ycdct
QQQ
=+
g
Q
: Trong lng dm chnh (thep hnh ch I, 9mx2,2mx0,7m, chon
g
Q
2 T)
t
Q
:Trong lng day treo ( chon
t
Q
=0,3 T)
20,32,3
ycgt
QQQ
=+=+=
(T)
+ o cao nang can thiet :
puli
t
ck
at
ke
yc
h
h
h
h
h
H
+
+
+
+
=
0,310,721,55,5
=++++=
(m)
Vay :
2,3
yc
Q
=
(T);
5,5
yc
H
=
(m);
10,5
yc
R
=
(m);
9,9
L
=
(m)
b. Cu dm ph: Vi dm ph c kch thc 2,2mx1mx0,4m => kch thc v ti trng nh hn dm chnh nn sc cu s tha.
THANH KE
DAM CHNH
DAM PHU
300
700
400
400
2200
400
c. Cu tai :
+ Sc truc :
t
tai
yc
Q
Q
Q
+
=
tai
Q
: Trong lng 1 cuc tai( kch thc: 0,7 m 1m 3m => Qtai = 5,25 T)
t
Q
:Trong lng day treo ( chon
t
Q
=0.5 T)
5,250,55,75
yctait
QQQ
=+=+=
(T)
+ o cao nang can thiet :
puli
t
ck
at
ct
yc
h
h
h
h
h
H
+
+
+
+
=
R = 12000
9000
2200
2,4210,711.56,62
=++++=
(m)
Vay :
5,75
yc
Q
=
(T);
6,62
yc
H
=
(m);
10,2
yc
R
=
(m);
10,1
L
=
(m)
c. Cu thap :
+ Sc truc :
t
thap
yc
Q
Q
Q
+
=
thap
Q
: Trong lng thap ( chon
thap
Q
= 1 T )
t
Q
:Trong lng day treo ( chon
t
Q
=0.3 T )
3
.
1
3
.
0
1
=
+
=
+
=
t
thap
yc
Q
Q
Q
(T)
+ o cao nang can thiet :
puli
t
ck
at
ct
yc
h
h
h
h
h
H
+
+
+
+
=
1,519.50.51.514
=++++=
(m)
KCH THUY LC
KHUNG CO NH
600
X
600
KHUNG DI ONG
350
X
350
1500
1500
1000
8500
500
1500
500
1000
1000
1000
350
1300
350
1000
1000
1000
500
9000
11859
15581
Vay :
3
.
1
=
yc
Q
(T);
14
yc
H
=
(m);
12
yc
R
=
(m);
16,5
L
=
(m)
d. Cu coc :
+ Sc truc :
t
coc
yc
Q
Q
Q
+
=
coc
Q
: Trong lng coc,
coc
Q
= 0.2x0.2x8x2.5x1.1 = 0,88 T )
t
Q
:Trong lng day treo ( chon
t
Q
=0.3 T )
0,880,31,18
yccoct
QQQ
=+=+=
(T)
+ o cao nang can thiet :
puli
t
ck
at
ct
yc
h
h
h
h
h
H
+
+
+
+
=
1,518.50.51.513
=++++=
(m)
1500
1000
9500
500
1500
1500
500
1000
1000
1000
350
1300
350
1000
1000
1000
500
9000
11968
16304
KCH THUY LC
KHUNG CO NH
600
X
600
KHUNG DI ONG
350
X
350
1500
1500
300
1000
700
2000
1500
9849
700
1300
1300
2400
1300
1300
700
9000
10500
Vay :
1,18
yc
Q
=
(T);
13
yc
H
=
(m);
12
yc
R
=
(m);
15,6
L
=
(m)
Bang tong hp :
Cau kien
yc
Q
(T)
yc
H
(m)
yc
R
(m)
L (m)
Dm chnh
2,3
5,5
10,5
9,9
Tai
5,75
6,62
10,2
10,1
Thap
1.3
14
12
16,5
Coc
1.18
13
12
15,6
MAX
5,75
14
12
16,5
Ta chon can truc t hanh banh xch KX 8161, tra bng trang 214 (so tay chon may xay dng nm 2008)
Can truc ma hieu KX 8161 co cac thong so sau:
Chieu dai tay can L = 20 m
Sc nang
16.5100
QT
=-
Tam vi ln nhat
max
18
Rm
=
Tam vi nho nhat
m
R
6
min
=
cao nng hn = 12 19.6 m
3. Thi cong ep coc : (xem ban ve)
Cc bc thi cng, kim tra v nghim thu qu trnh p cc p dng theo tiu chun xy dng Vit Nam TCXD VN 286: 2003 ng v p cc.
Cc cao - so vi m chun - 0.400.
a) Giai on chun b
Chun b mt bng :
Mt bng thi cng phi c dn dp thng thong m bo cho qu trnh vn chuyn v thi cng.
Tp kt my mc :
My mc v thit b phc v thi cng p cc c la chn phn chc my p.
Cn tp kt my mc, kim tra v m bo s lng, kh nng vn hnh lm vic ca my.
Tp kt v chun b cc :
Phi tp kt cc trc ngy p t 1 n 2 ngy.
Khu xp cc phi t ngoi khu vc p cc, ng i vn chuyn cc phi bng phng, khng g gh li lm.
Cc phi vch sn trc thun tin cho vic s dng my kinh v cn chnh khi p cc, m bo trc cc khi ni p khng lch qu gii hn cho php gy nh hng n s lm vic chu ti cc.
Trc khi tin hnh p cc phi kim tra li ti hin trng bng sng bt my hoc bng siu m.
Cc on cc c vt nt rng hn 0.3mm th khng c p.
Cn loi b nhng cc khng cht lng, khng m bo yu cu k thut.
Chun b ti liu :
Phi c y cc bo co kho st a cht cng trnh, kt qu xuyn tnh...
H s k thut v sn xut cc.
H s k thut v thit b p cc.
Vn bn v cc thng s k thut ca cng vic p cc do thit k a ra
Lc p gii hn ti thiu yu cu tc ng ln nh cc Pepmin cc t sc chu ti d tnh.
Lc p ln nht cho php tc ng ln nh cc Pepmax.
nghing cho php khi ni cc.
Phi c nht k p cc.
b) Giai on p cc th
Trc khi p cc i tr, phi tin hnh p lm th nghim nn tnh cc ti nhng im c iu kin a cht tiu biu nhm la chn ng n loi cc, thit b thi cng v iu chnh phng n thit k nu c s c.
S lng cn kim tra vi th nghim nn tnh l 1% tng s cc p nhng khng t hn 2cc..
Quy trnh p cc th phi m bo thc hin hon ton ging nh quy trnh p cc i tr.
c) Giai on p cc i tr
Trnh t p cc
Thi im p cc do thit k qui nh.
S p cc chn theo s p theo nhm cc v p t trong ra ngoi m bo lp nn ct khng b nn gy chi gi kh khn cho vic p cc cc cui cng v m bo SCT thc ca cc.
Trnh t p cc :
Vn chuyn v lp rp thit b p cc vo v tr p m bo bo an ton.
Chnh my ng trc ca khung my, ng trc kch v ng trc ca cc thng ng v nm trong mt phng, mt phng ny vung gc vi mt phng chun nm ngang.
nghing khng qu 0,5%
Chy th my kim tra tnh n nh ca thit b (khng ti v c ti).
Kim tra cc v vn chuyn cc vo v tr p.
Lp on cc u tin C1.
on cc u tin phi c dng lp cn thn, cn chnh trc C1 trng vi trc ca kch i qua im nh v cc.
lch tm khng qu 1cm.
u trn C1 gn cht vo thanh nh hng ca khung my. Nu khng c thanh nh hng th y kch (hoc u pittong) phi c thanh nh hng.
Tin hnh p on cc C1.
iu chnh van tng dn p lc. Nhng giy u tin p lc u nn tng chm, u, on C1 xuyn vo vo t mt cch nh nhng vi vn tc xuyn khng qu 1cm/s.
Khi pht hin thy nghing phi dng li, cn chnh ngay.
on C2 l cc trung gian.
Lp ni v p on cc tip theo (on C2)
Kim tra b mt hai u C2, m bo 2 u cc c b mt tht phng.
Kim tra cc chi tit ni cc v chun b my hn.
Lp C2 vo v tr p. Cn chnh ng trc ca C2 trng vi trc kch v ng trc C1.
nghing C2 khng qu 1%.
Tc dng ln cc mt lc to tip xc khong 3 4 kG/cm ri mi tin hnh hn ni cc theo quy nh ca thit k.
Tin hnh p C2. Tng dn p lc nn my p c thi gian cn thit to lc p thng ma st v lc khng ca t mi cc cc chuyn ng. Thi im u C2 xuyn vo t vi vn tc khng qu 1cm/s.
Khi C2 chuyn ng u th mi cho cc chuyn ng vi vn tc khng qu 2cm/s.
on C3 l cc cui cng.
Cc bc tin hnh ni cc v p cc C3 c thc hin tng t cc C2 bn trn.
Kt thc vic p cc t yu cu khi tho cc iu kin sau :
Chiu di on cc c p khng nh hn chiu di ngn nht do thit k qui nh tc mi cc t su xp x su tnh ton thit k, m bo ph hp vi SCT d tnh Ptk.
Lc p ti thi im cui cng phi t tr s Pep =60 T trn sut chiu su xuyn ln hn ba ln cnh cc. Trong khong vn tc xuyn khng qu 1cm/s.
Cc c ngm vo lp t tt chu lc 1 on t nht 3-5 ln cnh cc (khong 1.5m) k t lc p lc kch thay i ng k.
Trng hp khng t 1 trong cc iu kin trn th bo cho ch cng trnh v thit k x l.
Nht k p cc
Nht k ghi chp theo mu quy nh bi TCXD 190 : 1996.
Ghi nhn ch s nn u tin khi cc cm su vo t 30-50 cm.
Sau , mi ln cc xuyn thm 1m th ghi nhn li gi tr lc p ti thi im .
Nhng thi im lc p tng ng k 1 cch t ngt th phi c ghi chp r rng.
giai on cui : l giai on k t lc lc p t 80% gi tr lc p ti thiu theo quy nh n khi kt thc p cc. Ghi nhn thi im u ca giai on ny. Bt u t thi im ny ghi nhn gi tr lc p vi mi on xuyn 20 cm cho n khi kt thc.
Lu khi p cc
Mt s s c thng gp khi p cc :
Khi lc nn tng t ngt tc l mi cc gp lp t cng hn (hoc gp di vt cc b), cn gim tc nn cc c kh nng vo t cng hn (hoc kim tra di vt ch x l) v gi lc p khng vt qu gi tr ti a cho php.
Cc nghing qu qui nh, cc p d dang, cc b v x l bng cch nh ln hoc thay th.
Gi tr lc p ti thi im cui cng khng nh hn Pepmin m bo Ptk cho cc v khng ln hn Pepmax gy ph hy cc.
Kim tra SCT cc sau khi p
Sau khi p cc xong, tin hnh kim tra sc chu ti ca cc p bng th nghim nn tnh cc theo tiu chun hin hnh.
4/ Thi cong van chong vach ng:
Chon c thep:
Loai c van thep Larsen SP II khong neo co:
Be rong: B = 400 mm
Cao: H = 100 mm
Day: t = 10,5 mm
Cheu dai mot oan: L = 10 m
So lng van can dung cho ca cong trnh:
+ Co 3 ho cha mong M2, M4 s dung vach ng co kch thc vach:
Cao 1600 mm; Rong 4344 mm
So lng van can dung:
4344
10,75
400
n
==
tam chon 12 tam
Vay ba ho se la: 36 tam
+ Ho mong truc 7 D, 7 C, 8 D, 8 C co kch thc vach la:
Cao 1600 mm; Rong 6194 mm
So lng van can dung:
6194
15,49
400
n
==
tam Chon 16 tam
+ Ho mong truc 9 D; 9 C co kch thc vach:
Cao 1600 mm; Rong 4264 mm
So lng van can dung:
4264
10,66
400
n
==
tam Chon 12 tam
Vay tong so lng van can dung cho ca cong trnh: N = 12 +16 + 12 = 40 tam.
Chon may thi cong:
- Chon phng phap thi cong c bang bua rung nen c
- Chon s bo may thi cong c thep theo So tay chon may thi cong xay dng cua thay Nguyen Tien Thu, trang 55. Chon may ep c ma hieu VPP 2A, co cac thong so sau:
+ Cong suat: 40 Kw.
+ Lc rung ln nhat: 250 kN.
+ Tan so rung: 1500 vong/ phut.
+ Trong lng: 2,2 T.
I. THI CONG AO T
700
700
2420
1000
700
1000
1500
1500
1000
1000
1000
500
9000
10199
10094
1. Phng an ao ho mong:
o h c mi dc kt hp vi o vch thng ng c s dng dng c chng vch h.
2. Tnh khoi lng at ao :
Cao o mat at t nhien: -0.400
Cao o ay mong: -1.900
Chieu day cua lp be tong lot: 100 mm
dc t nhin ca t:
1
0,67
i
=
Ta c kch thc h mng:
+ Chiu su h o:
1,6
d
Hm
=
+ B rng ca mi dc:
1,6
1,072
1
0,67
d
H
Bm
i
===
+ Khong m rng thao tc y h mng xo = 0,5 m.
Vi cc h o c ming h m rng chng ln nhau ta o h lin tc.
a/ Khi lng t o trc H, G, F:
Mng M3:
Kch thc y h:
a = am + naxo = 1,12 + 20,5 = 2,12m
b = bm + nbxo = 1,1 + 20,5 = 2,1m
Kch thc ming h:
c = a + naB = 2,12 + 21,072 = 4,264 m
d = b + nbB = 2,1 + 21,072 = 4,244 m
Khi lng t o ca mng M3 trc H, G, F(s lng: 3 h):
(
)
(
)
(
)
(
)
1
3
3
6
1,6
32,122,14,2644,2442,124,2642,14,244
6
50,44
H
Vabcdacbd
m
=++++
=++++
=
Mng M2 v M4 chung mt h o nn ta c:
Kch thc y h:
a = 0,6 + 1,35 + 1,1 + 0,5 = 3,55 m
b = 0,52 + 1,2 = 2,2 m
Kch thc ming h:
c = 3,55 + 1,072 = 4,622 m
d = 2,2 + 1,0722 = 4,344 m
Khi lng t o ca mng M2, M4 trc H, G, F(s lng : 3 h):
(
)
(
)
(
)
(
)
2
3
3
6
1,6
33,552,24,6224,3443,554,6222,24,344
6
65,1
H
Vabcdacbd
m
=++++
=++++
=
b/ Khi lng t o trc E:
Mng M3:
3
1
3
50,44
16,813
33
V
Vm
===
Mng M2 v M4:
Kch thc y h:
a = 4,05 m
b = 2,2 m
Kch thc ming h:
c = 4,05 + 21,072 = 6,194 m
d = 2,2 + 21,072 = 4,344 m
Khi lng t o ca mngM2, M4 trc E :
EMBED Equation.DSMT4
(
)
(
)
(
)
(
)
4
3
6
1,6
4,052,26,1944,3444,056,1942,24,344
6
27,43
H
Vabcdacbd
m
=++++
=++++
=
c/ Khi lng t o trc D, C:
Hai mng M3 trc 9:
Kch thc y h:
a = 5,05 m
b = 0,52 + 1,1 + 0,5 = 2,12 m
Kch thc ming h:
c = 5,05 + 1,072 = 6,122 m
d = 2,12 + 21,072 = 4,264 m
Khi lng t o ca mng M3 trc 9 D, C:
(
)
(
)
(
)
(
)
5
3
6
1,6
5,052,126,1224,2645,056,1222,124,264
6
27,49
H
Vabcdacbd
m
=++++
=++++
=
d/ Khi lng t o trc D, C:
Mng M1, M2, M3, M4 trc 7, 8:
Kch thc y h:
a = 5,105 m
b = 4,05 m
Kch thc ming h:
c = 5,105 + 1,072 = 6,177 m
d = 4,05 + 21,072 = 6,194 m
Khi lng t o:
(
)
(
)
(
)
(
)
6
3
6
1,6
5,1054,056,1776,1945,1056,1774,056,194
6
46,54
H
Vabcdacbd
m
=++++
=++++
=
Vy tng th tch t cn o:
123456
3
50,4465,116,81327,4327,4946,54
233,813
VVVVVVV
m
=+++++
=+++++
=
e/ Khi lng t o bng c gii:
Ta c: H = H 0,2 = 1,6 0,2 = 1,4 m
Mng M3 trc H, G, F:
(
)
(
)
(
)
(
)
1
3
3
6
1,4
32,122,14,2644,2442,124,2642,14,244
6
44,13
H
Vabcdacbd
m
=++++
=++++
=
Mng M4 v M2 trc H, G, F:
(
)
(
)
(
)
(
)
2
3
3
6
1,4
33,552,24,6224,3443,554,6222,24,344
6
57
H
Vabcdacbd
m
=++++
=++++
=
Mng M3 trc E:
3
1
3
44,13
14,71
33
V
Vm
===
Mng M2 v M4 trc E:
(
)
(
)
(
)
(
)
4
3
6
1,4
4,052,26,1944,3444,056,1942,24,344
6
24
H
Vabcdacbd
m
=++++
=++++
=
Mng M3 trc 9 D, 9 C:
(
)
(
)
(
)
(
)
5
3
6
1,4
5,052,126,1224,2645,056,1222,124,264
6
25,23
H
Vabcdacbd
m
=++++
=++++
=
Mng M1, M2, M3, M4 trc 7 D, 7 C, 8 D; 8 C:
(
)
(
)
(
)
(
)
6
3
6
1,4
5,1054,056,1776,1945,1056,1774,056,194
6
40,72
H
Vabcdacbd
m
=++++
=++++
=
Tng khi lng t o bng c gii:
Vcg = V1 + V2 + V3 + V4 +V5 + V6
= 44,13 + 57 +14,71 + 24 + 25,23 + 40,72 = 205,79 m3
f/ Khi lng o t bng th cng:
Vtc = V Vcg = 233,813 205,79 = 28,023 m3
g/ Tng khi lng t p:
V mng ca cng trnh l mng nng nn th tch t p ly bng 2/3 th tch t o:
3
22
233,813155,9
33
dap
VVm
===
3. Chon may ao :
- Ho ao nong, khoi lng ao at bang may la 205,79 m3 < 20000 m3 nen ta chon may ao 1 gau nghch co dung tch gau
3
65
,
0
4
,
0
m
q
=
.
- Chon may ao EO-3322B1 (dan ong thuy lc) co cac thong so ky thuat :
MA HIEU
q (m3)
may
R
max
(m)
ho (m)
Hao (m)
a (m)
b (m)
c (m)
EO-3322B1
0.5
7.5
4.8
4.2
2.81
2.7
3.84
)
(
75
.
6
25
.
5
5
.
7
)
9
.
0
7
.
0
(
)
9
.
0
7
.
0
(
max
max
m
R
R
may
dao
=
=
=
=> chon
)
(
7
.
6
max
m
R
dao
=
min
1,07211,54,572()
dao
talyatmay
Rblrm
=++=++=
=> chon
min
4,6()
dao
Rm
=
=> khoang lui =
maxmin
6,74,62,1()
daodao
RRm
-=-=
6,1222
15,061()
2222
do
at
BC
Rlm
=++=++=
=> chon Ro = 5,1 (m) <
dao
R
max
Nang suat cua may :
3
(/)
d
cktg
t
K
NqnKmh
K
=
q : dung tch gau.
K : he so ay gau, phu thuoc vao loai gau, cap va o am cua at.
K = 1,4
Kt : he so ti cua at, Kt = 1,2
Ktg : he so s dung thi gian, Ktg = 0,75
Nck : so chu ky xuc trong 1 gi
ck
ck
T
N
3600
=
Tck = tck.Kvt.Kquay : thi gian cua 1 chu ky.
tck = 17(s) : thi gian cua 1 chu ky.
Kvt = 1,1 : he so phu thuoc vao ieu kien o at.
(o at len thung xe)
Kquay = 1 : he so phu thuoc vao
quay
j
can vi.
Tck = tck.Kvt.Kquay = 17.1,1.1 = 18,7 (s)
193
7
,
18
3600
3600
=
=
=
ck
ck
T
N
)
/
(
4
,
84
75
,
0
.
193
.
2
,
1
4
,
1
.
5
,
0
3
h
m
N
=
=
Nang suat may trong 1 ca (8h) :
)
/
(
2
,
675
8
.
4
,
84
3
ca
m
N
ca
=
=
So ca may ao :
233,813
0,35
675,2
ca
V
nca
N
===
4/ Chn my vn chuyn t:
i vi cng trnh ny khi o bng my cn vn chuyn t i ngay nhm trnh ng lm chm tin thi cng cng trnh. Do lng xe vn chuyn c xc nh theo nng sut my o.
- Ta chn t vn chuyn FK117-FD
+ Dung tch thng l 5,3m3
+ Di : 6.785 m
+ Rng : 2.475 m
+ Cao : 2,94 m
- Khi o bng th cng v khi lng t nh nn ta s ti ch nhm rt ngn thi gian vn chuyn t nhm m bo tin thi cng.
III/ THI CONG COP PHA
- Dung go nhom VI vi
3
500/
go
daNm
g
=
- ng suat uon
[
]
2
5
/
10
.
8
,
9
m
daN
=
s
- ng suat nen
[
]
2
6
/
10
.
04
,
3
m
daN
n
=
s
- ng suat keo
[
]
2
6
/
10
.
74
,
6
m
daN
k
=
s
- Moun an hoi
2
10
/
10
.
2
,
1
m
daN
E
=
Tai trong thang ng:
Khoi lng the tch cua cop pha a giao
Chon van day
mm
van
20
=
d
2
.500.0,0210(/)
tc
copphagovan
qdaNm
gd
===
2
.10.1,111(/)
tttc
copphacoppha
qqndaNm
===
Khoi lng n v the tch cua betong cot thep:
3
/
2600
m
daN
btct
=
g
- San (hs = 100 mm):
)
/
(
260
1
,
0
.
2600
.
2
m
daN
h
q
btct
btct
tc
btct
=
=
=
g
)
/
(
312
2
,
1
.
260
.
2
m
daN
n
q
q
tc
btct
tt
btct
=
=
=
- Dam: (hd= 350 mm):
)
/
(
910
35
,
0
.
2600
.
2
m
daN
h
q
btct
btct
tc
btct
=
=
=
g
)
/
(
1092
2
,
1
.
910
.
2
m
daN
n
q
q
tc
btct
tt
btct
=
=
=
Tai trong do ngi va dung cu thi cong:
2
/
250
m
daN
q
tc
nguoi
=
)
/
(
325
3
,
1
.
250
.
2
m
daN
n
q
q
tc
nguoi
tt
nguoi
=
=
=
Tai trong do am rung:
2
/
200
m
daN
q
tc
dam
=
)
/
(
260
3
,
1
.
200
.
2
m
daN
n
q
q
tc
dam
tt
dam
=
=
=
Tong tai trong ng :
- San:
2
10260250200720(/)
tc
qdaNm
=+++=
2
11312325260908(/)
tt
qdaNm
=+++=
- Dam:
2
109102502001370(/)
tc
qdaNm
=+++=
2
1110923252601688(/)
tt
qdaNm
=+++=
Tai trong ngang:
Ap lc ngang cua betong mi o vao cop pha
- Dam : H = 0,35 m
)
/
(
875
35
,
0
.
2500
.
2
m
daN
H
p
bt
tc
=
=
=
g
)
/
(
5
,
1137
3
,
1
.
875
.
2
m
daN
n
p
p
tc
tt
=
=
=
Tai trong do chan ong phat sinh khi o betong vao cop pha
)
/
(
400
2
m
daN
p
tc
=
)
/
(
520
3
,
1
.
400
.
2
m
daN
n
p
p
tc
tt
=
=
=
Tong tai trong ngang :
- Dam:
)
/
(
1275
400
875
2
m
daN
p
tc
=
+
=
)
/
(
5
,
1657
520
5
,
1137
2
m
daN
p
tt
=
+
=
1. COP PHA COT C3 (250 X 400): (tnh toan vi tai trong ngang)
Tnh toan van khuon :
Dung van co be rong b = 250 mm va 440 mm,
mm
van
20
=
d
Lc phan bo tren 1 m dai cop pha:
.11275.11275(/)
tctc
van
ppdaNm
===
.11657,511657,5(/)
tttt
van
ppdaNm
===
+ Kiem tra ben:
2
2
max
.
1657,50,44
40,11()
88
tt
van
pl
MdaNm
===
)
(
10
.
67
,
6
6
02
,
0
.
1
6
.
3
5
2
2
m
b
W
-
=
=
=
d
[
]
5252
max
5
40,11
6,01.10(/)9,8.10/
6,67.10
M
daNmdaNm
W
ss
-
===
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