Đồ án kĩ thuật thi công 1

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


t

Q

=0.5 T)

5,250,55,75

yctait

QQQ

=+=+=

(T)


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


t

Q

=0.3 T )

3

.

1

3

.

0

1

=

+

=

+

=

t

thap

yc

Q

Q

Q

(T)


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


t

Q

=0.3 T )

0,880,31,18

yccoct

QQQ

=+=+=

(T)


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:



6194

15,49

400

n

==

tam Chon 16 tam

+ Ho mong truc 9 D; 9 C co kch thc vach:



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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