Giao Trinh Co Gioi Duong Day 2009

8/13/2019 Giao Trinh Co Gioi Duong Day 2009

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CHNG I : KHI NIM CHUNG

1.1. KT CU CA NG DY TRN KHNG

1.1.1. Kt cu chung- ng dy trn khng bao gm cc phn tnhct, mng ct, x

dy, dy dn in, cch in v cc phkin khc ca ng dy.

- Trong thnh phn ng dy trn khng (DK) cn c cc kt cu cn

thit m bo vic cp in cho cc h tiu thin khng bgin on v

ng dy lm vic bnh thng nhdy chng st, btiu st, thit bni t,

bkhrung v cc thit bph trkhc cho vn hnh. Cc phn t trn c

biu din trn hnh 1.1a v 1.1b

1 2 3 4 5

6

7

8

Hnh 1.1a : Bn vmt bng tuyn dy

Hnh 1.1b : Bn vmt ct dc tuyn dy

Khong no ct

Khong ct

Khong vt

f

h


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1.1.2. Nhim vv mt sthng scbn ca ng dy

1.1.2.1. Nhim v

- Ct in: dng gicho dy cch mt t v ni im ny n im

kia trn mt on cho trc.

- Dy dn in: c nhim vdn dng in i theo mt hng xc nh.

- Cch in: cch in gia cc pha v pha vi t, cnh dy dn mt

im xc nh.

- X ngang: nhim vdy dn v cch in ca ng dy.

- Dy chng st: nhim v tn dng in st xung t m bo an

ton cho ng dy vn hnh.

1.1.2.2. Thng scbn ca ng dy

- Khong ct: l khong cch nm gia tm ca hai ct in knhau c

treo dy dn.

- Khong no ct: bao gm mt skhong ct nm gia hai ct no dng

lm im ta cng dy ly vng.

- Khong vt: khong no hay khong ct i qua mt cng trnh no

hoc qua cc trng ngi tnhin nhknh, mng, sng, ngi, ao, h gi l

khong vt.

- Gc chuyn hng ( ): gc gia cc hng ng dy ti cc khong

ct knhau.

- cao ca dy dn (h): khong cch thng ng gia im thp nht(im vng) ca dy dn trong khong ct n hng mc cng trnh ct ngang

hay n mt t c gi l cao ca dy dn.

- im vng: im thp nht ca dy dn trong khong ct.

- vng ca dy dn (f): l khong cch thng ng gia im vng ca

dy dn trong khong ct vi ng nm ngang ni lin cc im gidy dn

trn ct.

- Lc cng ca dy dn: l lc dng cng v gi dy dn hay dy

chng st trn cc ct in. Lc cng dy thay i theo sc gi, nhit khngkh xung quanh, dy ca bng ng trn dy.

- H s an ton: T s gia ti trng nh nht lm ph hy mt thnh

phn no , trn ti trng thc tiu kin khc nghit nht gi l dtr

bn hay hsan ton ca cc thnh phn ng dy ti in trn khng.


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- ng sut chc ca vt liu: Tsti trng ln cc thnh phn ng

dy. Trn mt n vdin tch mt ct chu lc ca chng c gi l ng sut

chc ca vt liu.

- Gii hn bn: l ng sut chc ln nht cho php ca vt liu nu vt

qu gii hn no vt liu sbt u bph hy.

1.2. CC TRNG THI LM VIC CA NG DY

TRN KHNG

1.2.1. Trng thi lm vic bnh thng

- Khi quan st mi thit b, phkin trn ng dy u lm vic trng

thi bnh thng nhct in, dy dn, cch in

- Cc thng skthut u nm trng trsnh mc kcnhit .

V d:

5% 5%m m lv m mU U U U U +

lv mI 1.2.2. Trng thi lm vic khng bnh thng

-Khi quan st c mt hay nhiu thit b, bphn, phkin trn ng

dy lm vic tnh trng khng bnh thng. Nhct bnghing, dy dn b

qu trng, snh la nh, mi ni tip xc xu, dy tip t bt

-Quan st trn hthng ng ho ta thy thng skthut vn hnhvt qu trsnh mc, nhng nm trong phm vi cho php.

V d:

lv mI trong phm vi cho php

dmdmlvdmdmUUUUU %.10%.5 +


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10%

10%lv m m

lv m m

U U U

U U U

> +

<

1.3. CC YU CU KTHUT I VI NG DY

TRN KHNG

thc hin c cc ch tiu cp in ng dy ti in trn khng

phi m bo cc yu cu kthut trong qu trnh vn hnh.

1.3.1. m bo bn vng lu di

- Cc phn tca ng dy trn khng nhdy dn (DCS) v ct khng

c h hng lm cho ng dy phi ngng vn hnh trong trng thi vn

hnh bnh thng v sc.

-Khng thm bo tuyt i dy khng bao gihng, ct khng bao giv nu nhvy gi thnh ng dy srt t. Chm bo khnng xy

ra mc thp nht chp nhn c.

1.3.2. m bo cc chti in ca ng dy

Khng c xy ra cc tnh hung khc thng lm nh hng n ch

ti in ca ng dy. V ddy dn tin n gn nhau hoc chm nhau,

hoc chm vo dy chng st v cc vt ni t trong cc trng thi vn hnh

gy phng in hay ngn mch.

1.3.3. Khng lm nh hng n cc cng trnh ln cn- Khng nh hng n hot ng bnh thng ca cc cng trnh di

hoc ln cn ng dy trn khng, nhgiao thng di ng dy: ng st,

ng thy, ng dy thng tin.

- in p trn DK c thcm ng sang cc ng dy in v thng tin

nu chng gn nhau hoc giao nhau vi khong cch nh. in p cm ng ny

nu ln snh hng n cng tc ca cc ng dy. in p cm ng c th

gy nguy him cho cc ng dy.

1.3.4. m bo an ton- Khng c nh hng n an ton in vi ngi v gia sc hot ng

di ng dy hoc ln cn ng dy.

- in trng di ng dy 500kV nh hng n ngi v gia sc

di ng dy, phi c cc bin php hn chnh hng.


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1.4. CC TIU CHUN THIT KDK TRN 1000V

Cc yu cu kthut c thhin qua cc tiu chun thit km ngi

thit kphi p dng.

Di y l mt stiu chun kthut cbn, cc tiu chun khc c th

tham kho trong cc TCVN hoc quy phm trang bin QPTBVit Nam.

1.4.1. Khong cch an ton gia ng dy trn khng vi t v cc cng

trnh ln cn

1.4.1.1. Khong cch dc an ton nhnht ca dy dn vi mt t :

Khong cch (m)

U (kV) Khu vc

ng dn c

Khu vc

t dn c

Khu vc kh

qua li

Khu vc

ngi kh n 110

220

500

7

8

14

6

7

10

5

6

8

3

4

6

Bng 1.1 : Khong cch dc an ton

1.4.1.2. Khong cch an ton nhnht tdy dn ngoi vi phn nh ra ca

cng trnh, nh ca :

- Khu vc t dn c:

U 22kV : 2m ; U = 35kV : 3m ; U = 66 110kV : 4m

U = 220kV : 6m ; U = 500kV : 7m

- Khu vc ng dn c:

U 22kV : dy bc : 1m ; dy trn : 2m

U = 35kV : dy bc : 1,5m ; dy trn : 3,5m

U = 66 110kV : 4m

U = 220kV : 6m

U = 500kV : 7m- i vi ng dy 500kV yu cu in trng khng qu 5kV/m st

hnh lang tuyn dy.


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1.4.1.3. Khong cch gia 2DK i song song v i gn

Khong cch nm ngang gia cc dy dn ngoi cng khi dy khng b

lch khng c nhhn khong cch an ton ca hnh lang tuyn ca DK

in p cao hn.

1.4.2. Khong cch an ton nhnht gia cc dy pha vi nhau v vi dy

chng st

1.4.2.1. Khong cch gia cc dy pha :

V d:

U 35kV dng s treo, khong cch nh nht gia cc dy dn xc nh

theo cng thc sau :

0,65100

Uf + + (1.1)

Trong :

f -vng ln nht, (m)

U - in p danh nh, (kV)

-chiu di chui cch in, (m)

(Cc trng hp khc tham kho trong phn 2.3.1)

1.4.2.2. Khong cch thng ng gia dy dn v DCS:

gia khong ct, khng tnh n lch dy do gi, trong trng thi qu

in p kh quyn khng nhhn sliu trong bng sau :

Khong ct (m) 100 150 200 300 400 500 600 700 800 900Khong cch(m) 2 3,2 4 5,5 7 8,5 10 11,5 13 14.5

Bng 1.2

vng ca DCS khng c ln hn vng ca dy dn.

1.4.3. Tit kim dy dn ti thiu cho cc ng dy

1.4.3.1. Cho in p 110kV trlnhn chtn tht vng quang, phi chn

dy c tit din ti thiu 70mm2cho in p 110kV; 240 mm2cho 220kV.

1.4.3.2. Dy chng st :

Ud= 35kV dng dy 35mm2

Ud= 110kV dng dy 50mm2

Ud= 220 500kV dng dy 70mm2


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1.4.4. ng Sut Cho Php :

- Cho php hsan ton l tsgia gii hn bn ko ca dy dn

vi ng sut cho php cc ch:

nbat

cp

K

= (1.2)

trong :

nb -gii hn bn ko ca dy dn, daN/mm2

(tra bng phlc 1.1);

cp -ng sut cho php cc ch, daN/mm2.

- HsKatty thuc vo iu kin kh hu tnh ton :

Khi chnhit thp nht v chgi mnh nht ly Kat= 2,5

Khi chnhit trung bnh nm ly Kat= 4


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TTChkh

hu tnh ton

Nhit

(0C)

p sutgi

(daN/m2)

Ti trng tnghp

(daN/m.mm2)

ng sutcho php

(daN/mm2)

5- Min Bc

10- Min Trung1 Nhit thpnht (Tmin)

15- Min Nam

0 g1 cp

2Nhit caonht (Tmax)

40 0 g1 _

3Gi mnh

nht (qvmax)25 qvmax

2

vmaxggg += 212

cp

4Nhit

trung bnh(Ttb)

30 0 g1 cptb

5Qu in pkh quyn

(Up)20 0,1.qvmax

(*)2

vggg += 212

_

Bng 2.1

2.2. TNH TON DY DN

Mc ch tnh ton dy dn ng dy ti in l xc nh cc i lng vng (f) lc ko ca dy dn tc dng ln ct (Td). Cc i lng ny cn thit kim tra khong cch an ton t dy dn n mt t, la chn, kim tra ct,x, mng trong cc trng thi vn hnh.

2.2.1. Cc qui nh cn thit khi tnh ton dy dn

2.2.1.1. Qui nh v phn loi ng dy trn khng


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Theo bng 2.1

ng cp ng dyin p nh mc ca

ng dy (kV)Loi h dng in

I>35

35

Bt c loi no

Loi I v loi II

II35

1 20

Loi III

Bt c loi no

III 1 Bt c loi no

2.2.1.2. Qui nh v h s an ton

H s an ton:cp

gh

=n (2.1)

Trong : gh - ng sut gii hn ca dy dn v dy chng st N/mm2

cp - ng sut cho php ca vt liu lm dy dn N/mm2

Tr s ca h s an ton theo qui nh (bng 2.2)

Tnh cht khu vc v c tnh ca dy dn n

- Ni khng dn c

Dy nhiu si

Dy mt si

2

2,5

Ni ng dn c v khong vt quan trng

- Dy nhm nhiu si tit din ti 120 mm2

- Dy ng nhiu si tit din ti 70 mm2

- Dy thp tit din ti 25 mm2

- Cc dy trn nhng tit din ln hn

- Dy AC vi mi tit din

2,5

2,5

2,5

2


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2.2.1.4. Ti trng c hc tc ng ln dy dn

Dy dn v dy chng st chu nhng ti trng c gii ch yu sau y:

- Ti trng do trng lng bn thn dy;

- Ti trng do gi thi ln dy trong nhng khong ct;

- Ti trng do dn n nhit;

Trong tnh ton thng dng khi nim t ti, t ti l ph ti c gii tcng ln di 1m dy c tit din 1mm2. n v o t ti l: N/m. mm2.

1- T ti do trong lng bn thn dy dn

10001o

gg = (N/m.mm2) (1-1)

Trong :

go- Trng lng ring ca cht cu to dy dn (N/dm3)

* Vi dy vn xon chiu di thc t ln hn:

1000)1,0302,1(1

og

g = (N/m. mm2) (1-2)

F

FgFgg BBAA

1000)1,0302,1(1

+= (N/m. mm2) (1-3)

Trong :

Ag , Bg - Trng lng ring ca nhm v thp

AF , BF - Tit din ca phn nhm v phn thp

=F BA FF + - Tit din ca dy phc hp

2- T ti do p lc gi ln dy

* Sc p ca gi ln 1m dy

16.1000....81,9 2

1

vdCg

= (N/m) (1-4)

* T ti ca gi:

3

2

2 10.16

....81,9F

vdC

F

Pg

== (N/m. mm2) (1-5)


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Trong :

: H s biu th s phn b khng ng u ca gi trn khongct;

V = 20 m/s 1=

V = 25 m/s 85,0=

V = 30 m/s 75,0=

V = 40 m/s 7,0=

C: l h s ng lc ca khng kh ph thuc vo b mt chu gi vi

dy c d < 20mm C = 1,1;

d >20mm C = 1,2;

V: vn tc gi (m/s)

d: ng knh ca dy dn (mm)

3- T ti tng hp

22

213 ggg += (N/m.mm

2) (1-6)

2.2.1.5. Sc ko v vng ca dy trong khong ct

1- Khong ct ti hn

* Vi dy mt cht

21

2min )(.24

ggl cpth

=

bo

bo

* Vi dy phc hp

2

min

222

min )(.24

=

ACA

Ath

gg

l

min

bo

bo

bo

- Mc ch xc nh lth xem vi mt khong ct no , ng sut lnnht s xut hin khi no;

Nu l >lth th max xut hin khi bo


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Nu l < lth th max xut hin khi min

2- Phng trnh trng thi

- Vi dy mt cht

)(2424 2

22

2

22

mn

m

m

m

n

n

n

glgl

=

- Vi dy phc hp

)(2424

2222

mn

AC

AC

ACmAC

mACm

ACnAC

nACn

glgl

=

Phng trnh trng thi c s dng khi cn xc nh ng sut trng thim khi ht ng sut trng thi n thng gii phng trnh ny tm ngsut max sau xc nh vng ca dy dn.

3- vng

vng l khong cch theo phng thng ng t im thp nht ca dytrong khong ct n ng thng i qua 2 im treo dy dn trong mt khongct.

* Vi dy mt cht:

max

12

8

glf =

* Vi dy phc hp

max

12

8 AC

glf =

Trong :

cp - ng sut cho php ca vt liu lm dy

- h s dn n di ca vt liu lm dy

A - h s dn n di ca nhm

AC - h s dn n di ca dy phc hp AC

AFe

AAFeFe

ACaEE

EaE

+

+=

...


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Fe

A

F

Fa=

Trong :

Fea - h s dn n di ca thp

E- m un n hi ca vt liu lm dy

EA- m un n hi ca vt liu nhm

EFe- m un n hi ca vt liu thp

EAC- m un n hi ca vt liu lm dy phc hp

a

EEaE FeA

AC+

+=

1

.

- h s ko di n hi ca vt liu lm dy

AC - h s ko di n hi ca dy AC

AC

ACE

1=

,boAC minAC - ng sut dy AC lc bo v min

[ ]A

ACAoACAACAC

E

EE))(( min = Pmin

[ ]A

ACAoACAACAC

E

EE))(( boPbo =

o - nhit mi trng ch to dy, ly Co

o 15=

Bng thng s sau y cho php xc nh ng sut cho php, mun nhi v h s dn n di theo nhit ca cc loi dy dn.

c tnh c l ca dy bng 2.4

Vt liu dy dnTrng lng ring

go, (N/mm2)

,gh (N/mm2) E, (N/mm2)

ng ng 87,2 382 127.103 17.10-6

ng 87,2 529 127.103 18.10-6

Nhm 26,5 157 61,6.103 23,6.10-6


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Thp n 77 540 196.103 12.10-6

Thp bn 77 685 196.103 12.10-6

Thp trong AC

v dy chng st

77 1175 196.103 12.10-6

4- Sc ko ca dy

Vi dy mt cht: FT .= (N)

Vi dy AC:ACAC

FT .= (N)

Trong : FCAAC FFF += (mm2)

5- Xc nh khong cch an ton ti im bt k

Ngoi vic xc nh vng (f) kim tra khong cch an ton t imthp nht ca dy ti mt t (ho) cn cn kim tra khong cch an ton (h) tiim bt k, nu mt t khng bng phng, l ch ng dy ti in vt, ng t, xe la, ng dy thng tin...

=

l

X

l

fy 1

.4

Mt s khong cch an ton ti thiu theo bng 2.5

f

h

y

x


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2-3. Tnh ton ct

2.3.1. Ni dung tnh tonLa chn ng loi ct (chng loi, kiu dng, cao, bn) cho tng

tuyn ng dy v cho tng v tr ct trn tuyn.

Ct trung gian cn tnh ton kim tra trong trng hp lm vic bnh thng(dy khng t) vi iu kin bo ln.

Ct gc cn kim tra trong trng hp lm vic bnh thng, ct b ko vmt pha do sc ko ca dy dn hai na khong ct. Mc ch kim tra xem ctc cn t dy no hoc dng ct kp hay khng.

Ct cui cn kim tra 2 trng hp:

- Trng hp lm vic bnh thng, kim tra kh nng chu un ca ct khib cc dy ko v mt pha.

- Trng hp t mt dy ngoi cng, dy ngoi cng cn li s gymmen xon ln nht cho tit din t x.

2.3.2. Cc loi ct

Trn cc ng dy c cp in p t 35 kV tr xung thng dng ccloi ct b tng ct thp.

1- Ct trn (ct ly tm) k hiu: T hoc LT

Cc loi ct ny thng c ch to ti cc nh my, nh cc my ly tmvi ct thp ko trc (ng sut trc) hoc khng ko trc. c ch to vi cao ca ct 10m, 12m (LT10, LT12). Tu theo v tr s dng thc t ngi ta

Khong cch an ton ti thiu ti

mt t (m)

Khong cch an ton ti thiu khi

cc ng dy chung ct v giao

cho nhau (m)ng dy

Vng khng

dn

Vng khng dn

khong vtquan trng

Vi ng dy

in lc

Vi ng dy

thng tin

0,4 kV 5 6 2

(6 35) kV 6 7 2 2


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ch to ra cc ct c cao 16m, 18m, 20m (ct c chn di 6m, 8m chp nithnh cc ct 16m, 18m, 20m).

2- Ct vung hoc ch K (k hiu H, K)

Ct vung c ch to ti x nghip hoc ti ch, ct c chiu cao 7,5m,

8,5m, 9,5m, 10,5m s dng cc ng dy trung p v h p.2.3.3. S tnh ton ct

S tnh ton nh bng 2.6 dng kim tra kh nng chu un ca ccloi ct (trung gian, gc, cui) trong trng thi lm vic bnh thng.

Trong k hiu: Pd, Pc- p lc gi ln dy, ct

Td lc ko ca dy ln ct

Loi ct S u dy trn ct S tnh ton

1 2 3

Trung gian (6,10, 22,35)kV

Trung gian 0,4kV

Gc (6 35)kV

3Pd 1Pd

Pc2Pd

4Pd

Pc

4Pd

Pc4Pd

3Td

Pc

1Td

Pc2Td


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Gc 0,4kV

Cui (6 35)kV

Cui 0,4kV

Ch cn tnh ton kim tra xon vi ct ng dy trung p, v ct h ptreo 4 dy v 8 dy (i 2 tng). Nu t mt dy cng khng gy mmen xonln cho ct.

S tnh ton ct cui trung p c 2 cch nh hnh v:

Cc lc tc ng ln ct trong cc s tnh ton xc nh theo cng thcsau:

4Td

Pc

4Td

Pc4Td

3Td

Pc

1Td

Pc2Td

Pc

4Td

Pc4Td


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a- Lc tc ng l mt ct

FVCPc ....16

81,9 2= (N)

Trong :

- H s biu th s phn b khng ng u ca gi trn khongct

V = 20m/s = 1

V = 25m/s = 0,85

V = 30m/s = 0,75

V = 40m/s = 0,7

C- H s ng lc ca khng kh ph thuc b mt chu gi.

Vi ct phng: C = 1,5

Vi ct trn: C = 0,7

b- Lc gi tc dng ln dy

Lc gi tc ng ln dy dn trong mt khong ct di l tnh theo hnggi.

ldVCPd .....1681,9 2= (N)

c- Sc ko ca dy ln ct

Lc ko ca dy dn tit din F (mm2) ln ct xc nh theo cng thc:

T = .F (N)

Trong :

= minnu l lth

2.3.4. Tnh ton kh nng chu un ct b tng ct thp

p dng phng php trng thi gii hn ngha l tnh kh nng bn chcca vt liu.


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Cng thc tng qut:

MttMph

Trong :

Mtt mmen tnh ton ca tng cc lc tc ng ln tit din st mt

t ca ct.

== n

ptt MnnMM1

(Nm)

Vi n- h s qu ti (tra bng 2-7)

(Bng 2-7) H s qu ti trong ch bnh thng v s c

Ti trng ln ct v mng n

- Trng lng ct, mng, dy v cc ph kin

- p lc gi ln ct

- p lc gi ln dy

- Ti trng ngang do lc ko ca dy

1,1

1,2

1,2

1,3

M - Tng m men ngoi lc tc dng ln ct

Vi ct trung gian: pcpd MMM +=

Vi ct gc: Tdpcpd MMMM ++=

Vi ct cui: pcTd MMM +=

pdM - m men do lc gi tc dng ln dy gy ra

TdM - m men do lc gi tc dng ln ct gy ra

TdM - m men do lc ko ca dy gy ra


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cpM - m men cho php, chnh l sc bn gii hn chu un ca tit din

st mt t ca ct.

Ghi ch: Cc ti trng thng ng (trng lng ct, x, s, dy, ti trngxy lp) bnh thng khng gy m men un vi ct, nhng nu qu trnh lmvic, ct b un cong th cc ti trng ny cng gy nn mt m men un vi ct.

Trong trng hp cn thit, k n m men ny ngi ta tng tr s M tt ln10%.

* p dng thc t:

a. Ct ly tm:

Vi ct ly tm, nh ch to thng cho lc u ct (Pcp) v th iu kin anton ca ct khi chu un l:

cptt

tt Ph

M

P =

ttP -lc tnh ton qui v u ct (N)

h- chiu cao t nh ct ti mt t (m)

b. Ct vung:

an ton, ct vung phi c kim tra vi tit din rng, v b tngkhng chu ko, ch yu kim tra bn chu un vi ct thp.

m bo kh nng chu m men un ttM , din tch ct thp mt mt

ct phi tho mn iu kin:

)(.)(.. ahRm

M

ahRmm

MF

oatt

tt

oaa

tta

=

Trong :

m- h s iu kin lm vic vi kt cu

m = 1,1 vi ct c ti nh my

m = 1,0 vi ct c ti ch

attR - sc bn tnh ton ca thp (N/cm2), cho trong (bng 2 8)

(Bng 2 8) Sc bn tnh ton ca thp attR , N/cm2


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Dng ng sut CT3 CT5 25I2C

Ko

Nn

Ct

20600

20600

16500

23500

23500

16500

33400

33400

26700

Hoc )(.. ahFRmMM oaattcptt =

(Tit din rng ct vung)

2.3.5. Kim tra kh nng chu xon ca ct b tng ct thp

Ct chu m men xon ln nht khi t dy ngoi cng c khong cch nct ln nht. Khi m men xon tc ng ln ct l:

Mx= X.Td (N.m)

X- Khong cch t im treo dy ngoi cng n ct (m)

a. Kim tra xon ct ly tm:

(Kt cu thp trong ct ly tm)

Fa

aho

Td

aFd


24/95

24

Kim tra thp ai: Mx2.m.ma.Ra 1.Fa

Fa

Kim tra thp dc chu xon (thp dc ph)

1

1.2

u

FFRmmM aaa

x

Trong : Fd din tch mt si thp ai

F1 din tch vng trn thp ai

U1 chu vi vng trn thp ai

b. Kim tra xon ct vung:

Nu mua ct c sn cn kim tra:

)()( ahmR

M

ahRmm

MF

oa

tt

oaa

tt

a

=

Hoc MttMcp= m.Ratt(ho- a)

2 - 4. Tnh ton kim tra mng ct

Mng ct ng dy trong cc h thng cp in t 35kV tr xung thngdng 2 loi: mng chng lt v mng chng nh khi tnh ton mng cn ly trs an ton qui nh cho tng loi ct ng vi ch lm vic khc nhau.

(Bng 2 9) h s an ton cho mng chng lt v chng nh

Loi ctCh bnh

thngS c

Trung gian thng

Trung gian gc

No gc, no cui

Ct vt

1,5

1,8

2,0

2,5

1,3

1,5

1,8

2,0


25/95

25

2.4.1. Mng ngn

Ph bin l mng ngn khng cp, ch trong trng hp c bit nh ctvt sng, mi dng mng c cp.

Cng thc chung tnh ton kim tra:

)(1.

321 ott QFEFFSk

+

Trong :

5,015,1 21 +

++= tg

h

H

h

HF

( )

++= tg

h

dtgF 5,111 22

( ) tgh

dtgF ++= 23 1

( )[ ])1(5,0.. 2

++

+= Ch

tg

khbE ott

S- Tng lc ngang t ln ct

Qo- Tng trng lng t ln nn k c trng lng mng

,2

,ko, C tra bng

(Mng khng cp) (Mng c cp)

h

HS

b

d

h

H

S

b1

d

h

d1 b


26/95

26

Ring vi mng ngn c cp cn lu :

a, Qobao gm c trng lng t bao quanh mng

b, trnh s ph hoi ca t nm trn cp mng pha mt trc cn thomn iu kin:

'tttt EE

Trong :

ttE - Sc khng ca t pha mt sau mng

'tt

E - Sc khng ca t pha mt trc mng

+= )1(..

9

2 2221

' hChbkE ott

Tr s 2 tra bng

2.4.2. Mng chn su

Thng mng chn su khng c thanh ngang, chnh l phn ct c

chn su xung t (c khong 1,8 2,5m). Tuy nhin ch t xu hn hocmun gim bt chn su c th t thm thanh ngang mng chn su thngdng vng t tt nh trung du, min ni.

a. Mng chn su khng c thanh ngang:

Cng thc kim tra iu kin chng lt:

21. mbha

Sk

Trong :

n

Ha=

)2

45(. 2 += otgm

Vi l gc ma st ca t, tra bng

b- chiu rng tnh ton

- Vi ct trn c ng knh trung bnh phn chn su do


27/95

27

bo= do. kog

- Vi ct vung c mt chu lc b1th b = b1. kog

kog- Tra bng

S- Tng lc ngang tc ng ln ct

k- H s an ton

b. Mng chn su c thanh ngang:

di tnh ton ca thanh ngang m bo chng lt do

Theo cng thc:

'

11 )1.(..o

dfdhm

AL +

+=

Trong :

tgf = , tra bng

m- Tra bng

kSEA += )21( 2 vi

2

2mbh

E=

d1- ng knh hay b rng thanh ngang

'o

d - ng knh ct ti ch t thanh ngang

2 - c xc nh theo phng trnh

)(667,0)233,1( 1112 Hh

Eh

kS

h

h

h

h+=

h

H

S

do h

H

S

'o

d

L

h1


28/95

28

(mng chn su khng c thanh ngang v c thanh ngang)

2.4.3. Mng chng nh (mng no)

Kt cu nh hnh v:

Ghi ch:

1. Ct in

2. Dy no

3. Mng ct

4. Mng no

5. Mc buc dy no

Mng no thng c cu to hnh khi ch T hoc hnh ch nht

Kh nng chng nh ca mng no c xc nh nh sau:

* Khi gc < 75o

...2

1. 2 bhTk

Trong :

k- H s an ton

- Gc ma st trong ca t, tra bng

- Sc bn th ng ca t


29/95

29

)1(3

2)1( 222' BA

b

h +=

2

2'

)(

)(

SinCosCos

Cos

+=

Cc h s , , A, B tra bng* Khi gc 75o< < 90o

SCVTKodo + .

Trong :

o - Trng lng ring ca t p

Co- Lc dnh kt ca t p

S- Din tch xung quanh khi t b bt ln

Cc tr s o, Cov o theo

, kN/m3

15,2 16,7Loi t

o o Co o o Co

t ct 15,2 0,54 0,5C 16,7 0,84 0,8C

t st 15,5 0,44 0,4C 16,7 0,64 0,6C

2 -5. Cc bi ton ng dng

2.5.1. Xc nh chiu cao ct vt sng

Trn tuyn ng dy trn khng c in p 35kV v th x M c on phivt sng vi khong ct 200m. Ton tuyn dng ct b tng li tm LT 10,dy dn AC- 95. Vng dy i qua c tc gi v = 30m/s, yu cu la chnchiu cao ct loi ct thch hp cho 2 ct vt sng sao cho m bo khongcch an ton t dy dn n mt nc lc cao nht l 8m.


30/95

30

Bi gii:

Cn c vo hnh v khong ct vt sng ta nhn thy chiu cao ct vtcn tho mn:

321 hhhfhh o ++++

Trong :

f- vng ca dy dn

h1- Khong cch t im cao treo dy ti nh ct, bao gm chiu cao chuis v khong cch t x n nh ct.

Chui s gm 4 bt s 15 cm = 60cm

h1= 0,6 + 0,1 + 0,1 = 0,7

2h - Chiu cao t mt t chn ct ti mt 5,02 =h m

3h - chn su mng ct: 5,23h m

oh - Khong cch t im vng ca dy dn ti mt mc nc khi cao

nht cch mt l 1m.

Vy: 718 ==o

h m

Nh vy:

Cn phi xc nh vng ca dy dn:

Trc ht tnh cc t ti ca dy AC 95 theo cng thc:


31/95

31

F

FgFgg BBAA

10003,102,1(1

+=

(N/m.mm2)

Tra bng dy AC-95 c:

FA= 95,4 mm2

FFe= 15,9 mm2

EA= 61,800 N/ mm2

EFe= 196000 N/ mm2

Tra bng vng kh hu c g1= 36,5.10-3N/ m.mm2

g3= 70,2.10-3N/ m.mm2

CoA 110.23 6

= ;Co

Fe 110.12 6=

6==Fe

A

F

Fa

644

4646

10.18,1910.18,6.610.6,19

10.18,6.10.23.610.6,19.10.12

=+

+=

+

+=

AFe

AAFeFe

aEE

EaEa

Co1

NmmaEE

a

AFe

o /10.35,1210.18,6.610.6,19611 26

44=

+

+=

+

+=

Tnh ng sut dy AC-95 theo trng thi min v bo.

2157

==n

Agh

ACP N/mm2

[ ] 264

466min /76,9910.35,12.10.18,6

110.18,6).515)(10.18,1910.23(5,78 mmNAC ==

[ ] 264

466 /94,10510.35,12.10.18,6

110.18,6).2515)(10.18,1910.23(5,78 mmNAC ==

bo

Khong vt ti hn ca dy AC-95 l:

mgg

l

ACAC

Ath 25,186

76,99

5,36

94,105

2,70

)525(10.18,6.24)(2422

4

2

1

2

3

=

=

=

minBo

minBo


32/95

32

l > lth, vy ng sut ln nht trong dy xut hin khi c bo:

94,105== Bomax AC N/mm2

vng ln nht xut hin lc max v v = 0 m/s

Tnh ng sut tng ng:max

AC

)(24

.

24

.2

23

2

2

21

2

Bomax

Bo

Bo

max

max

=o

o

ACo

AC

ACo

AC

glgl

)2540(10.35,1210.18,19

94,105.10.35,12.2410.2,70.200

94,10510.35,12.24

10.5,36.2006

6

26

62

26

622

=

max

max

AC

AC

2

max

AC(

max

AC-23,24)=17,98.104

Gii ra cmaxAC = 65,1 N/mm

2

Tnh vng ca dy dn (f):

8,21,65.8

10.5,36.200

8

3221

2

===

maxAC

glf

Vy cao ca ct vt sng l:

h 0,7 + 2,8 + 7 + 0,5 + 2,5 = 13,5 m

Chn ct c cao 16m, ct ly tm ghp.

Mi b sng t mt ct hnh ghp t 2 ct ly tm 16m, c h thanhging tng bn vng. C th t thm dy no tng kh nng chng ltcho ct.

2.5.2. Tnh ton c gii ng dy h p

Mt tuyn ng dy trn khng c in p trn 0,4kV cp in cho mt

hp tc x nng nghip mt Tnh. Dng dy nhm 3A-50 + 1A-35.

1- Tuyn dy:Ly in t TBA ca x i dc theo ng lng di 300m,khong ct c thit k 40m, ton tuyn c 8 v tr ct. v tr 01 l ct xuttuyn, v tr 08 l ct cui, v tr 04 l ct gc, cc v tr cn li l ct trung gian,mt ng tuyn dy i qua hon ton bng phng khng c on no vt qua, ng giao thng, ng thng tin, ao h............

S bn v mt bng tuyn ng dy nh sau!


33/95

33

2- La chn ct, x, s

a- Chn ct ( theo thit k)

Ton tuyn chn ct vung (H7,5) c ch to ti nh my

- Kch thc ngn ct (15 15) cm

- Kch thc cho ct ( mt y) ( 27 36) cm

- Mc b tng: M 300

+ Ti cc v tr xut tuyn, ct gc, ct cui, dng ct kp.

+ Cc ct trung gian dng ct n.b- X: ( Theo thit k): S dng thp gc loi L 63636 v L 5050 5

- X ct xut tuyn, ct gc, ct cui dng x k hiu X2-8 c chiu di1,2m.

- Ct trung gian: dng x k hiu X1-4

- Tng s : X : X2-8= 03 b

X1-4= 05 b

c- Chn s cch in. ( theo thit k)

- Tt c cc v tr dng s cch in A- 102

- Tng s ton tuyn: 44 qu.

3- Tnh ton ng sut v vng ca dy A- 50 trong khong ct. Tra bngvi dy A 50, vng kh hu III. C g1= 27,2. 10

-3N/m.mm2

g2= 115. 10-3N/m.mm2, g3= 118. 10

-3N/m.mm2

cA 0

6

0

110.23 = EA= 6,18. 10

4N/ mm2

* Khong vt ti hn.

( ) ( )( ) 622

6

21

23

min

10.2,27118

52510.23.245,78

.24

=

=

gge baoA

Acpth

= 72,2m


34/95

34

Nhn thy khong ct chn l = 40m < 72,2m

ng sut max s xut hin trong dy khi min

Ly 2min /5,782157

2mmN

gh

AcpA ====

64

10.18,1610.18,6

1 ==

Gii phng trnh trng thi xc nh c ng sut max

( )6

6

26

622

2

21

2

max10.18,16

540.10.23

5,78.10.18,16.24

10.2,27.405,78

max..24

.

=

A

A

gl

( )6

6

26

622

26

622

max10.18,16

540.10.23

5,78.10.18,16.24

10.2,27.405,78

max10.18,16.29

10.2,27.40

=

A

A

( ) 37,304826,28. maxmax2 =

AA

Gii ra ta tm c 2max /44 mmNA =

Tnh vng ca dy A50 khong ct 40m.

mgl

fA

123,044.8

10.2,27.40

.8

. 322

max

12

===

4- Kim tra khong cch an ton t dy dn ti mt t.

h0= h - f- h1

Trong : h = 7,5 m

h1= 1m ( chn su ca ct)

h0= 7,5 - 0,123 - = 6,37m > hcp= 6m

5- Kim tra kh nng chu lc (un) ca ct trung gian.

- Cc ti trng tc ng ln ct.

+ Lc gi tc ng ln ct.

NFVCPC 10983,1.35.5,1.75,0.16

81,9..

16

81,9 22 ===

Trong : Din tch mt ct chu gi ( mt c) c cnh trn 15cm, cnh stmt t 25cm, cao 6,5m


35/95

35

23,15.6.2

025,05,1mF =

+=

- Lc gi tc ng ln dy.

P= g2.l.F = 11.5.10-3.40.50 = 230N

- Lc gi tc ng ln ct t trung tm mt ct, xc nh theo cng thc.

44,23

5,6.

25,015,0

15,015,0.2

3.

.2

21

21 =+

+=

+

+=

h

aa

aaH

Hnh v.

- Tng mmen un do ngoi lc tc ng ln ct.

Mtt= n im = 1,2(4.230 6,5 + 1098. 2,44) = 103909 Nm

- Kh nng chng un ca tit din ct st mt t.

MCP= mRatt.Fa.(h0- a) = 1,1.20.600.3.2,01 (40-3,5) = 48380 Nm

mtt < MCp Ct lm vic an ton.

6- Kim tra kh nng chu lc ca ct gc.

- C 2 kh nng xy ra tnh trng lm vic nng n vi ct gc

+ Lc ko dy ln nht ( V d lc min , V=0) lc ny ti trng t ln ct

ch l tng hp cc lc ko.

+ Lc gi bo ln nht: Lc ny ti trng t ln ct gm: lc gi tc dng

ln dy v lc ko dy 25

0

c ( khng phi ln nht)a- Xt trng hp lc ko dy ln nht.

- V dy to trc mt chu lc ca ct gc 600nn lc ko dy t vo ct lT1d= T, ng dy gm ( 3.50 + 1.35 ), nn tng lc ko dy t ln ct l:


36/95

36

Td= 3. max .50 + 1. max .35

Td= 3.78,5.50 + 1.78,5.35 = 14525,5 Nm

Mmen tnh ton tc ng ln tit din st mt t ca ct.

Mtt= 1,3.6,5.14525,5 = 122736 Nm

b- Xt trng hp gi bo ln nht.

- Lc gi tc ng ln ct. ( 2 ct)

PC= 2.1098 = 2196 N

- Lc gi tc ng ln dy ( 3 dy 50 + 1 dy 35)

Pd= ( 3.g2. F50+ 1.g2. F35) .l.sin 600

= ( 3. 115. 10-3.50 + 1. 138. 10-3.35).40. 23 = 764 N

- Lc ko dy cn tmbao

( = 250c ) bng cch gii phng trnh trng

thi theo A .min

=

5,78.10.18,16.24

10.115.4026

622

bao

bao

( )525.10.1618

10.23

5,78.10.18,16.24

10.2,27.406

6

6

622

Gii ra ta tm c bao = 61 N/m

- Lc ko dy ln ct l.

Td= 3.61.50 + 1.61.35 = 11285 N

Tng mn men tnh ton tc ng ln dy l.

mtt= 1,2.( 2196 . 2,44 + 764 . 65) + 1,3. 11285 = 136806 Nm

Kt lun vi 2 trng hp.

mtt> mcp= 2.48380 = 96760 Nm

* Bin php khc phc: ( c 2 bin php)

- Mua loi ct c sc bn cao ch to vi tho 2552C c Ratt= 33400N/cm2

- Khi mcp= 1,1. 33400.3.2,01.(40 - 3,5) = 78445 Nm

mtt= 136806 < 2. 7845 = 156890 Nm

- t dy no cho v tr ct gc nu a hnh cho php.


37/95

37

Hnh v:

7- Kim tra kh nng chu lc ca ct u, ct cui.

Ct u v ct cui lun chu lc dy ko v mt pha.

Hnh v.

Tng t nh ct gc, ct cui cng c tnh cho 2 trng hp.

a, Trng hp lc ko dy ln nht.

Td= 3. max .F50+ 1. max .F35

= 3.78,5.50 + 1.78,5.35 = 14525,5 N

Mtt= 1,3.6,5.14525,5 = 122736 Nm

b, Trng hp gi bo ln nht.

- Lc ny hng nguy him nht vi ct cui thi dc tuyn dy (Pd= 0)

- Lc gi tc ng ln ct:

PC= 2. 1098 = 2196 N

- Lc ko ca dy.

lc bo: c025= tnh c trn l mNbao /61=

Td= 3.61.50 + 1.61.35 = 11285 N

Tng mmen tnh ton ln mt A A ca ct l .

Mtt= 1,2 . 2196 . 2,44 . + 1,3. 11285 . 6,5 = 101787 Nm

Nh vy mtt > Mcp= 2.48380 = 96760 Nm

* Bin php khc phc.

- Mua ct c sc bn cao.

- t dy no nu a hnh cho php.


38/95

38

- Rt nhn khong ct.

8- Kim tra kh nng chng lt ca mng.

- Ton tuyn c 3 loi mng :

+ Mng ct trung gian: K hiu M1

+ Mng ct gc: K hiu M2

+ Mng ct cui: K hiu M2

Kch thc v chiu di t ct, mng so vi hng tuyn nh hnh v.

Hnh v.

Hnh a: mng ct trung gian ( M1)

Hnh b: mng ct gc (M2)

Hnh c: mng ct cui (M2)

a, Kim tra kh nng chng lt ca ct trung gian.

- Mng ngn c kim tra theo cng thc.

( )0321

..1

QFEFKn +

PCtra bng c h s an ton k = 1,5

Vi kch thc mng m1: (1 1,2 1 )m vng t st pha ct m t nhin tnhc cc tr s F1, F2, F3, En


39/95

39

CHNG III : GII PHP CT V X

3.1. MU

Thit kct l bi ton xy dng.Trong qu trnh lp n thit kng dy, thng thng ct c chn

theo c im ca ng dy, v sau c kim tra sc chu ti ca ct theo

cc iu kin tnh ton.

Loi ct v chiu cao ct c xc nh khi chia ct trn tuyn. Vn

cn li l la chn kt cu ct tha mn cc yu cu thit k.

ng dy c th sdng nhiu loi kt cu ct, trong , ba loi ph

bin nht l ct thp khung (ct sp st), ct b tng ly tm (BTLT) (bao gm c

loi dng lc) v ct thp n thn (TT). Ngoi ra, cn nhiu loi ct khc

nhct g, ct b tng ct thp (BTCT) vung, ct phi tiu chun

3.2. CT B TNG CT THP

Ct b tng ct thp (BTCT) gm hai loi l b tng ly tm (BTLT) v

ct b tng vung (BTV). Ngoi ra, ct BTLT c thchto di dng dng

lc gi l ct BTLT tin p, k hiu (BTTA).

3.2.1. Ct b tng ly tm

Ct BTLT c sn xut theo tiu chun TCVN 5847-1994 (hnh 3.1)

*) Kt cu ct

- Ct c mt ct trn vi cn 1,33 0,07

- Theo chiu di, ct BTLT c phn lm hai loi :

Loi c lin dng cho ct c chiu di di 14m.


40/95

40

Loi ni hai on bng bch hoc mng xng, dng cho ct c chiu

cao trn 14m

- Mi on ca ct coi nhmt, tun th theo cc quy nh, tiu chun

i vi mt ct.

*) Kch thc cbn

- ng knh ngoi u ct : 190mm

- Chiu dy lp b tng bo vct thp u ct : khng nhhn 50mm

- Chiu dy lp b tng bo vct thp y ct : khng nhhn 60mm

- lch tm ca ct v bch ni, % : 1,0

- cong ca ct theo chiu di, % : 1,0

- Xi mng cho b tng ph hp vi TCVN 2682-1992

- Ct cho b tng ph hp vi TCVN 1770-86

- cho b tng ph hp vi TCVN 1771-87

- Ct thp phi thit kv ph hp vi TCVN 1651-85

- Que hn dng loi c c tnh ph hp vi thp ct dc, ph hp TCVN

3223-89.

- Bch ni ct phi c lp bo vchng n mn.

- Mng xng ni ct khng phi m.

- Chiu dy lp b tng bo vct.

Mt trong : 5mm

Mt ngoi : 2mm- Lc u ct cho trong bng sau:

K hiu ctLc ko ngang u ct,

daN, khng nhhn

ng knh ngoi ca

ct, mm

10A

10B

10C

320

420

520

323

12A

12B12C

540

720900

350

14A

14B

14C

650

850

1100

387

16B 920 403


41/95

41

16C 1100

18B

18C

920

1100430

20B

20C

20D

920

1000

1300

456

Bng 3.1


42/95

42

Hnh 3.1


43/95

43

3.2.2. Ct b tng ly tm tin p (BTTA)

nng cao tui thv lc chu cn sdng ct BTLT dng lc (tin

p).

3.2.2.1. Vt liu sn xut ct b tng dng lc

*) B tng: i vi ct b tng dng lc mc b tng t M400 n

M500. Va dng lp kn cc khe thi cng, cc mi ni cu kin lp ghp v

lm lp bo v ct thp, bo vneo thp, phi c mc tM200 tr ln. Va

dng bm bo cc ng rnh ca thp phi c mc khng di M300 v phi d

chy, t bco ngt.

*) Ct thp:ct BTTA dng ct thp cng cao (Ra= 4500 kg/cm2).

3.2.2.2.c tnh ct b tng dng lc

- Ct b tng dng lc c khnng chng nt cao. T lm tng

cng, vn c nhp ln v chng thm tt. Dng ct BTCT dng lc s

khng sut hin cc khe nt trong b tng hoc hn chspht trin chiu rng

cc khe nt khi chu ti trng sdng.

- Ct b tng dng lc tit kim thp do vic sdng thp cng

cao. Trong ct BTLT cng ct BTCT thng khng sdng c thp cng

cao v nhng khe nt u tin b tng xut hin m ng sut trong ct thp

chu ko mi t t2100 3400kg/cm2

. Cn thp cng cao c tht ti10000 15000kg/cm2. Nu dng thp ny trong b tng thng sxut hin cc

khe nt rt ln vt qu gi trgii hn cho php nn phi dng b tng dng

lc.

- So snh vi cu kin BTCT thng, ct b tng dng lc dng t thp

hn trung bnh 45% (dng loi thp si cng cao). i vi ct b tng d

ng lc c chiu di ln, mc tit kim thp ko kh ln (t50 80% thp).

- Ct b tng dng lc nhc tnh chng nt v cng tt nn tnh

chng mi ca kt cu c nng cao khi lm vic vi ti trng lp li. Sdnhvy l do gim c schnh lch tng i gia ng sut cc i v cc

tiu ct thp. V thi vi nhng ct b tng dng lc qu di cn phi lp

ghp thnh hai on l cch ni rt tt gia cc on ring l.


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3.2.3. Ct b tng ct thp vung

- Ct b tng ct thp vung (hnh 3.2)c sn xut v sdng cc

a phng, cc xng nh, hoc l sn phm km cc nh my b tng ln

hoc sn xut trn cng trng xy dng ng dy. Phm vi sdng ct BTV

ngy cng thu hp v cc thng skthut, cht lng v mthut cng nghip

khng ph hp.

- Ch s dng cho xy dng mt s ng dy phn phi in 22kV

nhnh r, tuyn ngn, cung cp in tm hoc vn u thn chv ng dy

hp.

3.3. CT G

Ct ggm nhiu ct tiu chun v ct phi tiu chun.

3.3.1. Ct gtiu chun

Ct gtiu chun l ct gtm du v xly l ha m bo chng mt,

chu tc ng mi trng, cho php ct lm vic c 10-20 nm.

3.3.2. Ct gphi tiu chun

Ct gphi tiu chun l ct in lm bng nhiu loi gvi cht lng v

chiu cao ct khng c tiu chun ha. Ct gloi ny chyu xy dng

cc ng trc v nhnh h thnng thn, min ni. Cht lng cc loi gphi tiu chun rt thp, v khng nn khuyn khch pht trin.


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Hnh 3.2


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46

3.4. CT THP ST

3.4.1. Phm vi sdng

- Ct thp khung TK (hay ct thp st - TS) l ct thp c thit kch

to theo kt cu khung, lp ghp tcc thanh. Cc than c sn xut xng

gia cng ckh.

- Trc y, ct TS khng c m. Sau khi lp rp, ct c sn chng

r.Tngy cng nghmkm nhng nng pht trin, ct TS c tin hnh m

km nhng nng cc thanh sau khi sn xut v lp rp th.

- Cc thanh sau khi m, c vn chuyn n a im v lp ghp thnh

ct.

- Ct TS c u im chu lc ln, mng chu lc, tha mn i hi ng

dy ti nng (mng kp, phn pha, cdy ln, khong ct ln). Vic chto v

lp t n gin.

- Ct TS c nhc im l khi lng ct ln, lm tng gi xy dng

ng dy. Mt khc, ct chim khng gian ln, khng c li khi i vng

thng dn.

- Ct TS chyu p ng vic xy dng cc ng dy ti in 220kV,

500kV v cc ng dy 110kV cln, mng kp.

3.4.2. Cng nghchto ct thp st

3.4.2.1. Tiu chun vt liu:*) Vt liu chto ct TS l thp in hnh (u cnh) v thp tm.

- Lin kt gia cc thanh thp hnh v tm m (thp tm) ca ct TS bng

boulon.

- Thp hnh, thp tm v boulon u c mkm nhng nng.

- Thp hnh v thp tm chto ct TS ly theo tiu chun JIS G3101. C

hai loi vt liu thp cng thng v cng cao.

- Cng thng: loi thp theo tiu chun SS 400 (SS41) hoc CT3

theo GOCT 8509-72, k hiu l LGii hn chy :

C = 2300 (2500 daN/cm2)

Gii hn bn :D = 4000 (5200 daN/cm

2)

- Cng cao: loi thp SS 540 (SS55), SS 490, k hiu l L

Gii hn chy :C = 4100

Gii hn bn :D

= 5500


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- Vt liu dng chto ct TS l vt liu mi, khng ng g.

- Ct in c chto bi cc thanh st nguyn, khng hn ni

*) Bulng: boulon lp rp ct c bn cp 4.6 (SS400) i vi loi

16 v boulon thang leo.

- Bulng 20 v 24 c bn cp 5.6 (SS490)- Bulng ch to theo TCVN 1876-76 v TCVN 1897-76, i vi vng

m phng TCVN 204-77 v vng m vnh theo TCVN 130-77.

*) Cng nghgia cng st thanh v st tm

Ct TS c gia cng theo tiu chun 20 TCVN 170-89 do BXy dng

ban hnh. Cng nghgia cng ct TS gm cc cng on : ct st, khoan (hoc

t), un, hn, kim tra (kch thc, dung sai cho php, chiu cao, ngu

ng hn) v sa sai (nu c).

*) Cng nghct st gm ct ngui v ct nng

- Ct ngui: dng my ct - t lin hp hoc my ct t -dp (ci tin)

ct ngui cc loi st hnh (L45x45x4 n 200x200x14), c chiu dy sau

y :

Loi st SS 400 (thp cng thng) chiu dy di 16mm.

Loi st SS 540 (thp cng cao) chiu dy di 14mm.

- Ct nng: cc loi st chiu dy ln hn phi dng phng php ct

nng bng my hn hi.- Thp tm dng lm tm m (hoc c) c gc ln rt phc tp, c

ghp ct bng hn hi vi mi chiu dy.

- Nhng v tr st c ct bng hn hi phi gia cng li bng phng

php ct, gt.

*) t (hoc khoan):nhng vt liu c dy di y sc to l

bng phng php t (ckh hoc thy lc).

- t thy lc : loi st SS 400 : chiu dy 16mm v loi st SS 540:

chiu dy 10mm

- Khoan : nhng vt liu c chiu dy ln hn sc khoan ti ng

knh tinh hoc phi t mt lnh( 3mm) ri khoan ti ng knh chun.

- Khng c php t l, nu chiu dy vt liu ln hn ng knh l.

- Lt hoc khoan xong phi ng hnh dng theo thit kkhng c

cong, mo.


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- Cm hn vic tsa nhng lsai vtr, nu khng c sng ca c

quan thit k.

*) Un (chn): tt cnhng phn un ca thp cht lng cao u phi

un nng (nhit nung t8000C 9800C).

- Sdng que o nhit hoc ha kkim tra nhit un nng.

- Thp SS 540 c chiu dy di 10mm c thun ngui vi gc un l

50().

- Thp SS 400 c chiu dy di 10mm c th un ngui ti 600().

Vic un khp cnh v mcnh thp gc c ththc hin ngui vi gc un n

300().

- Nhng trng hp un ni trn, phi c nhng bin php thch hp

ngn nga sha gin khi mnhng km nng.

- Ti cc im un phi gic tit din phng v trn bmt khng cvt lm do dng ct vo.

*) Hn: hn v kim tra mi hn theo tiu chun Vit Nam : 20 TCVN

170-89.

- Khi hn kt cu thp, sdng phng php tng v bn tng

c nng sut cao, m bo kch thc chc v ctnh mi hn.

- Khi hn, phi chn ch cho m bo t l chiu cao, chiu di v

chiu su ngu ca mi hn t t1,3-1,5.

*) Kim tra

- Kim tra kch thc v dung sai cho php, i vi thanh st, tm m v

loi lst boulon.

- Kim tra bn ngoi v chc mi hn.

- Kim tra bn ngoi mi hn bng cch quan st.

- Mi hn tt l khng trn, chy, ngt qung, chuyn tip t ngt, khng

c vt nt v khng c khuyt tt vt qu qui nh.

- Kim tra chc ca mi hn bng cch siu m hoc tm du vo mtmt mi hn (khng t hn hai ln v thi gian mi ln 10 pht).

- Sau qut nc phn hoc nc cao lanh bmt mi hn ni khng

tm du v 4 gi.

- Nu mi hn kn th mt bn kia khng c vt du loang.


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*) Sa sai:qua kim tra, nu xut hin cc khim khuyt u phi nghim

tc sa sai.

- Sa sai mi hn, phi x l bng bin php hn p cc ch b t

qung, ming hn blm.

- Cc khuyt tt khc phi ty bvi chiu di bng chiu di kch thc

ca khuyt tt cng thm mi pha 15mm v sau hn p li.

- Kt cu hn c bin dng dc sa cha bng cc bin php nhit, c

hoc kt hp nhit v ccho ph hp quy nh.

*) Lp thct :ct TS sau khi chto xong nht thit phi c lp th

theo phng nm ngang trn mt chun.

Kim tra chi tit ch to v lp rp thct in ph hp vi tiu chun

TCVN 55.04-91 (kt cu thp tiu chun thit k).

3.5. CT THP N

3.5.1. Mu

3.5.1.1. Ct thp n tiu chun

- Ct thp n (vit tt T) gm loi tit din vnh khn dng c tit din

trn hoc a gic v ct thp hnh ch I. Cc loi ct Tc sdng trc

nm 1975 xy dng li in 66kV Si Gn (c) v ng dy 66kV a

Nhim Thp Chm Cam Ranh.

- Vit Nam hin nay c cssn xut ct Tdng cho pht trin liin chiu sng thv bt u dng nhiu mng phn phi v truyn ti.

- Ct thp n tiu chun c dng ng, tit din trn hay a gic, ng

dng l mt loi ct truyn thng c sdng phbin v ngy cng c a

chung.

3.5.1.2. Ct thp n phi tiu chun

Ct Tphi tiu chun do khch hng tto tngun thp ng hoc thp

gc xy dng tm nhnh hthcc hm, ph, vng ven . Vic sdng

cc loi ct T t to loi ny phi k thut v nguy him. N cng l nguyn

nhn gy scv tai nn vin.

3.5.2. K hiu m ct

*) nh m sct thun tin, chn m hiu ct gm mt chtheo sau

l ba nhm chs:

km xx/n yy


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k k t, chtit din ct ; C tit din trn ; P tit din a gic

- Cm sthnht:

m - mt con si sau k t, ngha nhsau

m = 0 - ct dng chyu cho mng phn phi.

m =1 v 2 - ct dng chyu cho mng cao p v siu cao p.

- Cm sthhai:

xx chchiu di tnh ton ct, ty n nh ct trng thi lp

rp.

- Cm sthba: gm hai phn n yy :

n - son

yy - lc ti cho php (lc u ct) tnh ra 1000 daN (tng ng

mt tn lc).

*) V d:

- P2-36/5-10 ct tit din a gic, dng chyu cho ng dy cao p,

di 36m, lc ct 10000 daN, nm on.

- C0-8/1-1.5 ct tit din trn, dng cho mng h p, di 8m, lc ct

1500 daN, mt on.

3.5.3. Lc u ct :

- Cn ciu kin thc t, cc ng dy phbin, xc nh lc ct 14

bc chia lm ba nhm nhsau :

Nhm nh: 500 daN tng ng 0,5 tn700 daN 0,7 tn

1000 daN 1,0 tn

1500 daN 1,5 tn

2500 daN 2,5 tn

Nhm trung : 5000 daN 5,0 tn

7000 daN 7,0 tn

10000 daN 10,0 tn

Nhm nng : 15000 daN 15,0 tn20000 daN 20,0 tn

35000 daN 35,0 tn

60000 daN 60,0 tn

100000 daN 100,0 tn

150000 daN 150,0 tn


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- Vcbn thang lc ct ny tha mn hu ht cc trng hp trong thc

t

- Tuy nhin, khi p dng, ngi thit k c th chn loi ct khc quy

chun thang lc ct ny v cn lp bng tnh thit kchn thng sct.

3.5.4. Chiu di ct

Chiu di ct gm cc thnh phn sau (hnh

3.3) :

- Chiu di chn trong mng Lm

- Chiu di tmt t n x di cng Lcs(chiu di cs)

- Chiu di lp x, tnh tx di cng ln

n nh ct L

Chiu di chn trong mng Lm= 0 khi ct

lp trn mng BTCT (ct c mt bch).

Chiu di nh ct Lph thuc cch b tr

dy trn ct.

Chiu di csgm khong cch an ton Lat, vng f, chiu di chui

cch in Lcv di dtrLdt:

Lcs= Lat+ f + Lc+ Ldt (3-1)

Chiu di ny c chn in hnh ha v cc s liu l chn sb

tnh ton thit kct.3.5.5. Kch thc cbn ca ct

Kch thc c bn ca ct gm c (khng k chiu di chn theo

chun ha) : Tit din ngn v gc, thun, chiu dy vt liu lm ct.

Ni dung tnh ton l xc nh quy cch kthut ct, gm :

- Kch thc tng on

- Khi lng tng on

- Khi lng ton ct

Bi ton 3.1 a ra kt qutnh ton xc nh cc kch thc cbn cact.


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3.5.6.thun, son v vt liu chto

3.5.6.1. thun

- thun ct l tl:

( )as

t

% 100b e top

c

D Dp

L

= (3-2)

trong :

Dbase- ng knh y, m;

Dtop - ng knh nh, m;

Lct - chiu di tnh ton, m.

- Cng c thtnh thun theo bn knh.

- thun cng ln, ct cng nhn. Tng thun scho php tit kim

vt liu (gim trng lng ct), gim lc tc ng ca gi ln ct. Tuy nhin,

sc chu lc phn trn ca ct sgim.- Chn thun ct trong gii hn :

p% max : 4,1

min : 1,0

u tin : 1,3 3,0

- Ct Dbaseln, cho php chn p% ln vn m bo sc chu lc pha

ngn v cho php gim trng lng ct.

3.5.6.2. Son

- Son thn ct phthuc vo khnng ca thit bgia cng ct (un,m).

- Thn ct cng di, i hi khnng gia cng ct ca thit bcng ln.

- Chiu di on thn cng ln, son thn cng t v cng tit kim

(gim trng lng ct, gim cng lp t). Tuy nhin, thn di se tng chi ph

vn chuyn trn n v.

- Hin ti, cc csgia cng u la chn chiu di gia cng ti a 12m.

Do , chiu di on ct cc i ly bng 12m. Trn cs, son thn ct

c xc nh.

3.5.7. Phng php ni ct

Ct c thai on trln cn thc hin ghp ni. C hai phng php c

bn ghp ni l ghp ni mt bch v ghp lng.


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3.5.7.1. Ghp ni mt bch

- Ghp ni mt bch thc hin tng tnhghp ni ct BTLT. u mi

on ct s hn mt bch. Cc thng s c bn ca

mt bch (Hnh 3.4)

ng knh ngoi Dbase

ng knh trong D0

ng knh vnh ai bulng Db

Chiu di mt bch tb

- Cc thng sny c xc nh qua cc c

tnh cl ca ct T.

- T im c bn ca phng php

ghp ni mt bch l gim c chiu di

thn ct, v nhvy gim c trng lng

chung ca ct.

- Phng php ghp ni mt bch

c nhiu hn ch, ng ch l :

Yu cu m bo trng ng tm cc on c ni.

Lp rp kh khn.

an ton chung ca ct gim do chi tit ghp ni tim n s c

(lng hoc mt bulng, bulng bst g)

Phi tin hnh sn phkn chni bo vchng g.- Chnh v l, phng php ghp ni mt bch t c sdng.

3.5.7.2. Ghp lng

Ghp lng l phng php ghp li dng tnh thun ca cc on ct,ngn on di lng vo gc on trn. Do tnh thun v trng lc ct, vicghp ni c m bo bn vng. Trong mt strng hp, c thsdng ccbu lng nh von ni.

Cc thng scbn ca on ni . t p% l thun.

Db1, Dt1 - ng knh gc v ngn on di

Db2,Dt2 - ng knh gc v ngn on trn

L1 - chiu di on di

L2 - chiu di on trn

L - chiu di ct gm hai on


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Lni- chiu di on ni

Ta c cc quan hsau :

Dt1= Db1 L1 p%

Dt2= Db2 L2 p%

Dt1= Db2 Lni p%Lct= (L1+ L2) Lni (3-3)

Thng s c bn l chiu di on n - Lni. Chiu di ny ph thuc

ung knh ct v lc u ct. Chiu di on ni cng ln, mi ni cng cht,

chu lc cng tt, nhng slm tng trng lng ct. V th, cn c di hp

l.

Lni= knix Dni.tb

( )2 2i.tb 2

t b

n

D DD

+= (3-4)

y, Dni.tb - ng knh trung bnh on ni

kni - hsni

kn= 2-3 : chn kn= 2,1 (3-5)

3.5.8.Gii php gc ct

Ct Tc tht trn mt bch (mng BTCT) hoc chn trong t (mng

t, mng ging chm, mng cn).

Vi ct chn trong t, khng c yu cu c bit g vgc ct. Chcnthit kcc lbulng cn thit lp cn theo yu cu thit k. Mt khc, lp

mcn c bdy tha mn yu cu chu n mn mi trng t.

Ct t trn mt bch l ct trn mng BTCT. Mng c thl mng tr,

mng bn ccao, mng cc.

Cc kch thc lin quan n mt bch gc ct nhsau :

- ng knh vnh bulng Dvnh

- ng knh mt bch Dbch- Qui cch v slng bulng.

- Cc chiu cao mng :

Cmng C0

Thn mng C2

y mng C3

Mt mng C1


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Ton mng L

Quy cch bulng v sbulng mng c chn theo thit kmng.

3.5.9. X

X l bphn lp trn ct, ty thuc vo cp in p, qui m ng dy,

cch btr dy trn ct.

C hai hnh thc lp x l x lin ct v x ri.

X lin ct chthch hp vi loi ct nhv thp, chyu cho mng chiu

sng v hth. Vi cc loi ct in lc, x ri thch hp cho vic chto.

X ri c hai loi thng dng l x cong (hnh 3.5) v x bt gic (hnh

3.6).

Hnh 3.5


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Hnh 3.6

3.5.10. Cng nghm

3.5.10.1. La chn phng php sn ph:

Ct Tc thp dng gii php sn hoc mkm nhng nng. Gii php

sn gm sn thng v phepoxy. Gii php sn thng hin khng cn thch

ng. Nn hu ht ct Tu sdng cng nghphsn epoxy hoc mkm

nhng nng.

Phng php mkm nhng nng c chn, v n thch hp vi yu cu

ct v cng nghtrong nc quen sdng.

3.5.10.2. Bdy lp m:Bdy lp mquan h tui th ca ct. Tui thct (vmt chng n

mn) l thi gian bt u dng ct cho n khi lp mbn mn ht tc dng

bo v.


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3.6. CC LOI CT IN KHC

Ngoi cc loi ct in trn, hin cn c ct BTCT c c, ct lm bng

cc loi vt liu sn c ti a phng (g, tre,...), chyu dng xy dng cc

ng trc v nhnh hthvng nng thn, min ni mi c in.

y l loi ct in c cht lng thp, tim n nhiu nguy csc. S

dng loi ct ny trong li in hthchl tpht, ngoi tm kim sot ca

cquan chc nng.

3.7. LA CHN CT V KCH THC CT

3.7.1. Chn loi ct

Hin ti, ba loi ct c xem l hp cch a vo n ng dy l

ct BTCT, ct thp khung v ct thp n. Cc loi ct khc chc dng

trong hon cnh c bit.

Trong cc loi ct BTCT, ct BTLT c coi l loi phthng. Cc loi

nhBTTA, BTCT vung c xem xt trong mt s trng hp c th. Tuy

nhin, theo xu hng tng tui th, tnh an ton v lc ct, ct BTTA sdn dn

thay thct BTLT.

*) Ct BTCT c ng dng chyu cho cc trng hp sau :- ng dy phn phi (n 35kV), tr trng hp ng dy tbn

mch trln.

- ng dy mch n n 220kV, cdy khng qu AC 185.

*) Ct thp n c dng chyu cho cc trng hp sau :

- ng dy i trong khu thpht trin, cn m bo chim t hnh lang

v phi hp cnh quan mi trng.

- ng dy i vng su, vng xa, vn chuyn kh khn.

- Cc trng hp ng dy qui m ln, nu chng minh c tnh kh thiso vi cc loi ct khc.

*) Ct thp khung c sdng cho cc trng hp cn li, khi vic dng

ct thp n v / hoc BTCT khng ph hp.


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3.7.2. Chn kch thc ct

Kch thc hnh hc ca ct gn lin vi vic btr dy trn ct, la chn

kch thc x v kt cu x, v quyt nh n chiu cao ct cng nhkhong

ct tnh ton.

Kch thc hnh hc ca ct quyt nh bi cc yu tsau y :

- Khong cch an ton gia cc dy dn vi nhau.

- Khong cch an ton gia cc dy mang in vi cc phn tcn li

ni t.

- Gc bo vca dy chng st (nu c).

- Sdy dn v dy chng st (nu c) cn btr trn ct.

- La chn kiu btr dy.

- Khong cch an ton ti mt t v cc cng trnh, vt thtrn mt t.


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Hnh 3.7


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CHNG IV : GII PHP MNG CT

4.1. CC HNH THC MNG

4.1.1. Mu

Ct thp n (T) c hai hnh thc nh vl chn trong t v t trn

mng (mt bch). Cc kiu mng hin sdng cho mng ct in u cho php

p dng cho ct T.

C bn loi mng phbin dng cho ct Tl mng t, mng cn,

mng ging chm v mng BTCT.

4.1.2. Mng t

Mng t, cn gi l mng chn trc

tip. Yu cu cbn l t c c tnh cl

tt n nh ct (Hnh 4.1).H mng c o hoc khoan theo

kch thc cn thit. y mng cn c tm

bng BTCT hoc thp.

Ct c t trong h mng, sau

ln bng t, c lp phbng t. Cng c

thsu dng b tng mc M50-100 chn

ct v lp phbmt tng tnh n nh

v chng n mn ct.4.1.3. Mng ging chm

Mng ging chm l loi mng t c bit. H mng c khoan su

theo kch cgc ct. Gc ct c th c lp bu lng nh v tng khnng

chng lt (Hnh 4.2).

Ct t vo h mng, sau c chn va b tng mc thp (M50-

M100). Lp va bo vc thcao hn mt t 0,25-0,35m bo vchng n

mn.

Mng ging chm c u im kt cu gn, cho php thc hin vng t

c cng trung bnh, gi thnh h. Thi cng mng ging chm i hi phi c

my khoan kch cph hp.

`

Lm

Ct T

t lp

Tm

Hnh 4.1 : Mng t

a)


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Hnh 4.2 Mng ging chm4.1.4. Mng cn

Mng cn l loi mng c lp thm thanh ngang vung gc vi tim

ng dy tng khnng chng lt ca ct.

4.1.5. Mng BTCT

Mng BTCT p dng cho ct TS, ct Tc mt bch v ccho ct BTCT

chu lc ln vng cng chu lc ca t khng ln.

Mng BTCT c thl mng bn, mng bn ccao (mng lmc), mng

cc ng, mng cc nhi, mng tr

Hnh 4.3 Mng b tng ct thp

Phn chn cakt cu ct

Phn ging chmbng thp c thvutthon ty cc yu cuvmng hocphng php xy lp

Cc bulng cdng gicu trc

ng chtrong khichn vo khongtrng gia ct vging chm

ai c sucnh hn voging chm

Tm chn va

Cc thanh

Vng gia phnchn ct v gingthp chm cln bng va

Mt bch

Bulng thp


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4.2. TNH MNG T V MNG C CHN B TNG

4.2.1. Mng t

Loi mng ny chyu l chng lt do lc ngang ca ti trng ngoi gy

ra. Ct chn su khng mng, dng cc loi t pha theo tlnht nh ri m

kxung quanh chn ct to thnh mng.

iu kin n nh ca ct chn khng mng :

21

k c d m gm b H n P (4.1)

trong :

- tlgia chiu cao ct (Hk) v chiu su chn ct (Hd);

- hsphthuc , tra phlc 4.2;

- tra bng phlc 4.2;

2 0452

= +

k d

m tg (4.2)

Bc - brng ct.

Ct trn: ng knh trung bnh phn chn ct

kdl hscn ca t (tra bng)

Hsan ton ca mng nm:

- Ct trung gian : nm= 1,5

- Ct gc, ct no : nm= 2

- Ct vt : nm= 2,5

Pg - tng cc lc ngang tc dng

ln ct, ln dy, kN, bao gm lc cng

dy, lc gi vo dy, lc gi vo ct.

Khi tnh ton, nu tha bt ngthc (4.1) ct s khng b lt. Mng t

c s dng cho loi ct chu ti trng

ngoi nh, vng t tt v khng c gi

bo xy ra.

P3

P2

P1P3

P2

P1

gi

Ct T

t nn

Bn chu lc

Hd

Hk

X

X

Hnh 4.4 Mng t


64/95

64

4.2.2. Mng chn b tng

Sdng b tng bao su chn ct Hd (hnh 4.5) nhm bo vchn ct,

chng ln, chng lt. y l loi mng ch yu chng ln, lt (do m rng

mng bng b tng)

Phng php tnh ton:

: tng ti trng thng ng.

Px : ti trng nm ngang (qui v lc u

ct)

Hk: chiu cao ct (phn trn mt t)

Hd: chiu su chn ct.

Dm, rm, hm : chiu di, rng, cao ca khi b

tng bao phmng.

ng sut cc i pha di mng :

mng n nh, chng ln :

(4.4)

Nu (9.4) tha mn, vic chn mng l ph

hp.

mng khng b lt cn phi tha mn

iu kin :

( )2 31

1+ k m gF E F G n P

F (4.5)

y,

F1 - h s nh hng ca chiu su chn

ct v loi t2

1 1,5 1 0,5

= + + +

k k

d

H HF tg

H H (4.6)

F2 v F3- hsphn khng ca mng c

nh bi cng thc :

Hd

Ct T

B tng

Hd (Thay i)

Hnh 4.5 : Mng chn b tng

Tm chu lc

Hd

Hk

Hd

x

Hm

Ct T

Px

dmMtbng mng

m

min

max

Hnh 4.6 : Mng chn b tng


65/95

65

( )22 1 1 1,5

= + +

m

m

dF tg tg

h (4.7)

( )23 1

= + +

m

m

dF tg tg

h (4.8)

Ek : L sc phn khng ca t

( ) ( )20,5 1

= + + +

m d ck d d

r H kE H C

tg

y,

kc - hscn phthuc vo loi t v kch thc ct (phlc 4.5)

C - lc dnh ca t

- hslin kt (phlc 4.4)

G - tng trng lng ct v b tng

t t c b t ng c b t ng m m mG G G G d r h= + = + (4.9)

nm - hsan ton ca mng

Pg - tng ti trng ngang tc dng vo ct + dy

dm, rm, hm- kch thc mng

4.3. MNG CN

tng momen chng lt cho ct, ngi ta dng thanh ngng bt vochn ct. Chiu su t thanh ngng t1/2 n 1/3 chiu su chn ct.

m bo an ton, i khi ngi ta t hai thanh ngng, mt y

mng, mt st mt t. Sbtnh chiu di thanh ngng bng cng thc :

( )( )

2

0

1 2

1

+= +

+

S m g

ng

k ng ng

E n Pl d

m h r tg (4.10)

y,

E - sc khng ca t

E = 0,5mk.bc.Hd (4.11)

- hstnh ti chn su ca ngng

lng - bdi thanh ngng

rng - brng thanh ngng

Hng - chn su thanh ngng

d0 - ng knh ct T ti cht ngng


66/95


67/95

67

2

6=

bhW (4.13)

b - brng mng

h - chiu dy bn mng

iu kin

Pmax< 1,2 Rtc (4.14)

Pmin> 0

Tnh lc ti mp cmng P0

ax 10 2

+=

tttt mP PP (4.15)

Chn chiu cao c ch ca bn mng :

00

0,4

tt tt

tr n

P bh L

b R

(4.16)

y, L - khong cch tmp mng n ni chiu cao mng thay i

btr- cnh trn ca mng

Rn - cng chu nn ca b tng

i.Tnh thp mng :

Theo phng I :

00,9=

Ia

a

MF

h R (4.17)

Theo phng II :

00,9=

IIa

a

MF

h R (4.18)

ii.Tnh theo cmng :

Dng lc ct Q, M ti chn cmng tnh thp cmng.

iii.Kim tra ln

= glN

h

F

(4.19)

Tnh ton theo phng php cng ln tng lp

1 10

= = iS S h

E (4.20)


68/95

68

4.5. MNG TR

Tnh bu lng theo vo cmng :

Trng thi chilc ca bu lng l va chu ko, va chu ct di tc dng

ca ti trng ngang.

ng lc nhcc i ca bu lng c xc nh bng cng thc :

axax 3,93 1

=

mm

b

G MP

N GD (4.21)

y, G - trng lng ct v phkin

Mmax - moment cc i chn ct

n - sbulng trn mt mng

Db - ng knh vng btr bu lng

ng lc ct trong mt bulng :

ax 0,8=m

QT

n (4.22)

y, Q - lc ct ti chn ct

Hs0,8 n sphn bkhng u ca lc ct cho mi bu lng

ng sut tc ng ln bulng :

axax 2

4

= mmP

d ; axax 2

4

= mmT

d (4.23)

y, d - ng knh hiu dng ca bulng :

2 2axax ax

14

2 2

+ + bm m m kmR (4.24)

m - hstn iu kin lm vic, dy m = 0,65

Rkb- cng thp tnh ton ca bulng

Ch :

i.Khi chn ng knh bulng theo tnh ton, nn tng thm 2 3 mm trong

mi trng n mn.ii.Bulng c ngm vo mng theo qui nh.

- i vi bulng cu to, chiu di phn bulng cm trong b tng bng

30d, d ng knh bulng.

- i vi bulng chu lc, chiu di phn cm trong mng b tng phthuc

vo cch cu to phn neo


69/95

69

Hnh 4.8 Mng bn ccao (mng lmc)

Tnh lc dnh ca bulng v b tng :

t nn

B tng lthm

Ct T

L

b

Hd

Hk

Pmax

PminP1tt

Mt bng mng

Cmng

Px

l


70/95

70

=N

dl (4.25)

y, N - lc ko nh

l - chiu di neo

Td

l u (4.26)y, - gii hn chy ca thp

Tnh mt bch y ct

ng sut nn ln mt cmng :

= b bd

N MmR

F W (4.27)

y, Fd- din tch mt bch y ct (vnh )

( )2 22 1

4 =

db bF ; 1 2 25+ D D cm (4.28)

W - moment khng mi ca vnh

( )4 42 1232

=

d

D DW

D (4.29)

m = 0,8 0,9

3= bb lt

d

FR R

F

(4.30)

y, Rb- cng p mt tnh ton ca b tng

Rlt- cng tnh ton lng trca b tng

iu kin :

n3 31,5 1,5chb b

d d

F F

F F> = (4.31)

Nu mt bch khng c gn tng cng, chiu dy mt bch chn theo:

1

3 b

a

S l R

= (4.32)

y, m = 0,65

Ra- cng tnh ton thp lm vnh

L - brng mp ngoi vnh

Nu vnh c gn tng cng, chiu dy vnh chn theo :

D1

D2


71/95

71

( )1

2 1 1

32

2 a

MS

D D d mR=

+ (4.33)

y, M - momen un ca on vnh gia hai gn tng cng c khong

cch b

Ra- cng thp chto vnh m = 0,65

D2D1 D1

D2

S1S1

l

Hnh 4.9. Mt bch (vnh ) Hnh 4.10. Mt bch (vnh )

y ct khng c gn y ct c gn

Gi trM cho bi bng sau (chn gi ln hn tnh ton) :

0,00

0,33

0,50

0,67

1,00

1,50

2,00

3,00

0,000,0078

0,0393

0,558

0,0792

0,123

0,131

0,133

0,50,428

0,319

0,227

0,119

0,124

0,125

0,125

Chn ct T

l

Vnh

Lbulng

ng knh ngoi vnh

Chn ct T

Sn tng cng

Vnh


72/95

72

Hnh 4.11. Bn ct v mng BTCT

Ct T

Px

Gi

t nn Bulng neo

Cmng

Mng hm

Hd

Btng lt mng

Mt bng mng

Hk


73/95

73

4.6. V DBI TP

V d4.1 - Tnh mng t, vi sliu u vo :

-Ct thp n di : 10m

-Lc u ct : P = 180daN = 1,8kN

-Trng lng ct : 167 kg

-ng knh nh : 140mm

-ng knh y : 310mm

-a cht : 3 0 '2,01 / , 15 11d g cm = =

Gii :

Bc 1 : Xc nh su chn mng :

Sbchn Hd= 2,5m

Bc 2 : n inh, ct chn su khng mng phi thoiu kin :21

K c d m gm b H n P

(a)

Trong ;

75003

2500K

d

H

H= = =

Tra phlc 4.2, t :

3= ;

1

0,0264 = 0 '

2 0 2 0 317 4145 20,1 45 34,14 / 2 2k d

tg tg kN m

= + = + =

bc= kd.dTb

kd- hscn ca t

dTb- ng knh chn ct

dTb= 310mm = 0,310m (ng knh y ct)

kd= 1,45 (tra bng phlc 3) bc= 1,45 x 0,310 = 0,45mnm= 1,3 - hsan ton

T(a) :

( )221 0,0264 34,14 0,45 2,5 2,535

2 k c dm b H

= =

nmPg=1,3 x 1,8 = 2,34 kN


74/95

74

So snh :

21 2,535 2,342 k c d m g

m b H n P kN

= < =

Sdng mng ct chn su l hp l.

V d4.2 - Tnh mng bao b tng chn ct, c sliu u vo :

Ct thp n di : 12m

Lc u ct : P = 700 daN = 7kN

Trng lng ct : 234kg

ng knh nh : 140mm

ng knh y : 310mm

a cht : 3 0 '2,01 / , 15 11d g cm = =

C = 0,209 kg/cm2= 20,9 kN/m3; Rd= 1,5kg/cm2

Gii :Bc 1: Chn sbsu chn ct, Hd= 1,8m

Bc 2: Ti thng ng : N = 1,1(Gc+Gm)

Gc trng lng ct, x s, dy dn, = 500kg

Gm trng lng mng b tng

Gm= {(0,75x0,75x1,3) (0,35x0,35x1,3)}2500 = 1427 kg

N = 1,1 (500+2535) = 3338,5 kg

Bc 3: Tnh ng sut cc i ti mng

max 1,2 dm m

NR

d

=

y, dm : Chiu di mng

: Chiu rng mng

Rd : Cng tiu chun ca t nn

T :

max

3338,5

0,95 0,95

=

= 3699 kg/m2< 1,2x15.000 kg/m2= 18.000 kg/m2

Mng khng bln.

Bc 4: Kim tra iu kin lt ca mng

( )2 31

1K m gF E F G n P

F+


75/95

75

21 1,5 1 0,5

k k

d d

H HF tg

H H

= + + +

( )210200 10200

1,5 1 15,11 0,51800 1800

9,70

tg

= + + +

=

( )22 1 1 1,5 mm

dF tg tg

h

= + +

( )( ) ( )20,75

1 15,11 1 1,5 15,111,5

1,28

tg tg

= + +

=

( )23 1 mm

dF tg tg

h = + +

( )( ) ( )20,75

1 15,11 15,111,5

0,806

tg tg= + +

=

( ) ( )20,5 1m d ck d d

H KE H C

tg

= + + +

( )( ) ( )

0,75 1,8 1,1580,5 2,01 1,8 20,9 1 0,577

0,760 0,760 15,11

135,31

tg

= + + +

=

Tra phlc 4.5 (vi1,8

2,4 1,1580,750

dc

m

HK

= = = )

( ) ( )2 31 1

1, 28 135,31 0,806 14,27 19,049,70k

F E F G kN F

+ = + =

NmPg= 1,5 7 10,5kN =

So snh : ( )2 31

19,04 10,5k m g

F E F G kN n P kN F

+ = > =

Kt lun :Mng lm vic n nh v khng bln, lt

V d4.3 - Tnh mng cn c sliu u vo :

Lc u ct : P = 700 daN = 7kN

Ct thp n thn di : 14m

Trng lng ct : 242,5kg

ng knh nh : 140mm


76/95

76

ng knh y : 340mm

a cht : 3 0 '2,01 / , 15 11d g cm = =

C = 0,209 kg/cm2= 20,9 kN/m3; Rd= 1,5kg/cm2

Gii :

Bc 1: Sbchn su chn ct, Hd=1,8m

Bc 2: Chiu su t thanh ngng, sb

1 1 11,80 0,90

2 3 2ng dh H m

= = =

Bc 3: mng lm vic an ton, chiu di thanh ngng c tnh bi :

( )( )

2

0

1 2

1S m g

ng

k ng ng

E n Pl d

m h r tg

+= +

+

Trong :

0,5 0,5 34,27 0,47 1,8 14,50k c dE m b H= = =

0 '2 0 2 0 315 1145 20,1 45 34,27 /

2 2k Wm tg tg kN m

= + = + =

1,46 0,327 0, 47c d tbb K d= = =

Bit :

0

1,85,50

0,327dH

d= = (d0= 131001,43% + 140 = 327,33)

Tra phlc 9.3 v 9.4 :1,46

dK = ; 0,58s =

Suy ra :

( )( )

2

0

14,50 1 2 0,58 1,5 70,327 1,19

34,27 0,9 0,45 1 15 11ngl m

tg

+ = + =

+

Chn lng= 1,20m

Dng 1 cn l = 1,20m , b = 0,45m

V d4.4 - Bng tnh mng lmc c sliu u vo :


Lc u ct : P = 1000 daN = 10kN

Trng lng ct : 430kg

ng knh nh : 271mm


77/95

77

ng knh y : 365mm

a cht : ( )0 2 202,01 ; 15 ; 100 / 1000 / W T E kg cm T m = = =

C = 0,03 kg/cm2 ; Rtc= 1,43kg/cm2 = 14,3T/m2

A = 0,32 ; B = 2,29 ; D = 4,85

Gii :Ti trng ng (trng lng ct + s+ x + dy dn) :

N1= 0,62T

Moment

M = 18,30Tm

Lc ct gy ra ti cmng :

Q = 1T

Chn mng c kch thc :

Chiu su chn mng h = 2mChiu di mng a = 2m

Chiu rng mng b = 2,4m

Chiu cao bn mng h2 = 0,3m

Chiu cao cmng h1 = 0,8m

Chiu rng cmng a= b = 1,10m

h0 = 0,25m

h0 = 0,30m

h0 = 27,5cmTi trng bn thn mng v t p :

N2= 19,21T

Ti trng ln y mng :

N = N1+ N2 = 19,83T

Kim tra ng sut gy ra trn nn :2

ax

1,9206

19,83 18,30

4,80 1,920m

abW

N MP

F W

= =

= + = +

= 13,662 T/m2< 1,2 Rtc= 17,203T/m2

2min

2,01 18,309,950 / 0

4,80 1,920

N MP T m

F W= = = >

Tha iu kin p lc nn.


78/95

78

Kim tra ng sut gy ra trn nn

2 2 24,13 / 4,13 / 1,434 / tcN

T m kg cm R kb cmF

= = = < =

Kim tra chiu dy bn mng

1 min

21

2ax 10

00

2,707

12,657 /

13160 / 2

0,65 0,20,4

tt

tt

tttt m

tt tt

tr n

P P x

x

P kg m

P PP kg m

P bh m

b R

= +

=

=

+= =

> =

Chn h = 30cm l ph hp khng bchc thng

Tnh thp bn mng v cmngTnh thp bn mng

Moment tng ng mt ngm I-I l :

2 ax 02 18576

tt tt

mI

P PM a L kgm

+= =

2ax min 11806 / 2

tt mtb

P PP kg m

+= =

0,450 0,4502,40 11806 717,2

8IIM kgm

= =

Din tch ct thp chu MI :

2

0

3,060,9

IaI

a

MF cm

h R= =

Din tch ct thp chu MII:

2

0

1,180,9

IIaII

a

MF cm

h R= =

Tnh thp cmng

20

0,24' '' n

MA

b h R= =

Tra bng 0,87=

2

0

27,317aa

MF cm

h R= =


79/95

79

V d4.5 -Bng tnh mng trsliu u vo :


Lc u ct : P = 1000 daN = 10kN

Trng lng ct : 420kg

ng knh nh : 271mm

ng knh y : 365mm

a cht : ( )0 2 202,07 ; 16 ; 100 / 1000 / W T E kg cm T m = = =

C = 0,308 kg/cm2 ; Rtc= 2,02kg/cm2 = 20,2477T/m2

A = 0,36 ; B = 2,43 ; D = 5

Gii :

Ti trng ng : N1= 0,87 T

Momen do thit bgy ra : M = 18,31 I.m

Lc ct gy ra ti chn ct : Q = 1 TChn mng c kch thc :

Chiu di mng a = 2m

B = 2m

h0= 0,5m

h = 2,2m

Chiu rng cmng a = b = 0,55m

Ti trng bn thn mng v t p

h0 = 0,45mN2= 21,63T

Ti trng ln y mng :

N = N1 + N2= 22,25T

Kim tra ng sut gy ra trn nn2

ax

1,3336

22,25 18,00

4,00 1,333m

bhW

N M

P F W

= =

= + = +

= 19,063 T/m2< 1,2 Rtc= 24T/m2

2min

2, 07 18,0014,018 / 0

4, 00 1,333

N MP T m

F W= = = >

Kim tra chiu dy bn mng


80/95

80

25,56 / N

T mF

= =

1 min

21

2ax 10

00

1387

15405 /

16540 / 2

0,375 0,20,4

tt

tt

tttt m

tt tt

tr n

P P x

x

P kg m

P PP kg m

P bh m

b R

= +

=

=

+= =

> =

Chn h = 50cm l ph hp khng bchc thng

Tnh thp mng

Moment tng ng mt ngm I-I l :

2 ax 12 281372

tt tt m

IP PM b L kgm += =

2ax min 16540 / 2

tt mtb

P PP kg m

+= =

1,450 1,4502,00 16540 8693,8

8IIM kgm

= =

Din tch ct thp chu MI :

2

0

24,810,9

IaI

a

MF cm

h R= =

Din tch ct thp chu MII:

2

0

7,950,9

IIaII

a

MF cm

h R= =

Tnh thp cmng

20

0,13n

MA

b h R= =

Tra bng 0,995=

2

0

12,20aa

MF cm

h R= =


81/95

81

Phlc 4.1 -Cc thng sc trng ca t tnh mng chng lt

Loi tGc

lF tg= 2

345

2kkN

m tgm

= +

t st v ct ngm nc

t st v ct m tnhin

t st mm ngm nc

t st mn

t st rt mn

nhln ct ngm nc

Si ct ln ngm nc

Ct nhsng nc

Ct nhsch v t

Ct nhsch v khCt ln ln si kh

t st ln

Gch vn ct t

20

40

2

40

45

25

25

15

25

3540

35

30

0,364

0,839

0,364

0,839

1,000

0,466

0,266

0,468

0,466

0,7000,839

0,700

0,577

38,0

67,7

36,0

12,2

104,5

48,3

45,8

31,7

48,3

57,881,0

50,7

52,9

18,6

14,7

17,6

15,7

17,6

23,5

18,6

18,6

19,6

15,717,6

13,7

17,6

Phlc 4.2- Hs v1

tnh mng chng lt, ct chn su khng mng

k

d

H

H=

1

1

2

3

4

5

6

7

8

9

10

17,68

14,06

12,61

12,13

11,81

11,55

11,28

11,15

11,03

10,91

17,68

28,12

37,83

48,52

59,05

69,30

79,96

89,20

99,27

109,1

0,0566

0,0356

0,0264

0,0206

0,0169

0,0144

0,0127

0,0112

0,0109

0,0009


82/95

82

Phlc 4.3 Hscn kdtnh mng chng lt, ct chn su khng mng

Tl

kch

thc

chn

ct

0

dd

H

b = hoc

0

dd

H

b =

Gc

l10 9 8 7 6 5 4 3 2 1

15

20

25

30

3540

45

1,450

1,667

1,920

2,210

2,9803,020

3,550

1,405

1,600

1,828

2,089

2,4222,818

3,293

1,380

1,583

1,736

1,968

2,2642,616

3,040

1,315

1,466

1,644

1,847

2,1062,411

2,785

1,700

1,400

1,552

1,785

1,9482,212

2,530

1,225

1,333

1,480

1,605

1,7902,010

2,775

1,180

1,287

1,368

1,484

1,6321,809

2,020

1,135

1,200

1,276

1,363

1,4741,505

1,705

1,090

1,133

1,184

1,242

1,3461,404

1,516

1,045

1,067

1,092

1,121

1,1501,202

1,255

Ch thch : Hk - l chiu cao tmt n nh ct

Hd - l chiu su ct

b0, d0 - l brng hay ng knh trung bnh ca phn ct chn

su di t

Phlc -Hs , v tnh mng ngn chng lt

15

20

21

22

23

2425

26

27

28

29

0,760

0,700

0,687

0,675

0,663

0,6500,637

0,625

0,616

0,600

0,589

0,577

0,490

0,472

0,455

0,440

0,4220,406

0,390

0,379

0,360

0,347

2,30

3,30

3,55

3,88

4,30

4,655,20

5,60

6,30

6,96

7,70

30

31

32

33

34

3536

37

38

39

40

0,577

0,565

0,555

0,543

0,531

0,5210,510

0,498

0,488

0,478

0,467

0,333

0,320

0,308

0,295

0,282

0,2710,260

0,248

0,238

0,228

0,218

8,75

10,10

11,50

13,22

15,50

18,4024,00

30,50

37,05

52,00

70,85


83/95

83

Phlc 4.5 -Hscn kctnh mng ngn chng lt

dc

m

H

r =

Loi t

0,6 0,8 1 2 3 4

Ct nhno nc

t st pha, ct pha no nc

t ln ct dm no nc

Ct mn t

t c mn rc m t

Ct kh mn sch

Ct ln mn rc kh

t st pha, ct pha m tnhin

t rng khCt to ln dm kh

t ct kh rt cht

1,03

1,04

1,06

1,06

1,07

1,10

1,10

1,12

1,121,12

1,15

1,04

1,05

1,08

1,08

1,10

1,13

1,13

1,16

1,161,16

1,21

1,05

1,07

1,10

1,10

1,12

1,16

1,16

1,20

1,201,20

1,21

1,09

1,13

1,18

1,18

1,24

1,32

1,32

1,40

1,401,40

1,51

1,14

1,20

1,28

1,28

1,36

1,47

1,47

1,61

1,611,61

1,77

1,18

1,27

1,37

1,37

1,48

1,63

1,63

1,87

1,871,87

2,02

Ch thch : Hd - l chiu su chn ct

rm - l chiu rng ca mng


84/95

84

Phlc 4.6 -Qui cch v moment chng lt mng bn c c

Kch thc, mm Vt liu Moment chng lt, tn x mt

t kh, 0 Ngm nc, 0 T

T

Lo

i E A B C D

Th

p

kg

B

tn

g35

0

300 250 300 250 200

1MT

1

80

0

100

0

100

0

100

0

100

0

12,5

80,9

1,8

10,2 1,6

6,8

1,5

4,2

1,07

3,5

1,00

0,92

2MT

2

80

0

100

0

160

0

100

0

100

0

13,5

41,1

4,8

15,2

4,0

12,0

3,35

8,3

2,1

6,2

1,85

4,25

1,7

3,0

3

MT

3

80

0

120

0

160

0

120

0

100

0

13,5

4

1,3

5

5,5

18,85

4,8

13,6

4,1

9,55

2,8

7,5

2,3

5,5

2,1

3,7

4MT

4

80

0

140

0

180

0

120

0

120

0

13,5

4

1,5

96,4

20

0,0

15,1

5,0

10,8

3,25

8,4

2,9

6,2

2,6

4,3

5MT

5

80

0

160

0

200

0

120

0

140

0

13,5

4

1,8

2

7,8

22,2

6,9

16,6

6,6

12,4

4,25

9,4

4,1

2,3 3,9

5,3

6

MT

6

80

0

160

0

200

0

120

0

140

0

13,5

4

1,9

8

8,25

23,6

8

18

7,7

13,3

4,9

10,2

4,7

7,9

4,5

6,0

7MT

7

80

0

160

0

220

0

120

0

160

0

13,5

4

2,2

3 25,3

9,5

20,1

9, 2

15

5,8

11

5,5

8,9

5,25

6

8MT

8

80

0

160

0

220

0

120

0

180

0

13,5

4

2,4

828,6

11,4

22

11

17,1

6,9

12,6

6,6

10,5

6,3

9,1

9MT

-T1

80

06,5 5,0 3,5 - - -

Ghi ch :

Tsng vi H = 1,3m ; mu sng vi H = 2,3m

Mng MT-T1 l mng thanh ngng, ct 12m chn 2m,

ct 18-20m chn su 3m.


85/95

85

Phlc 4.7 -Mmen chng lt ca mng bn kiu khi

Moment chng lt, tn x mt

t kh t ngm ncTTKch thc

a x b, mm

Chn

su,

m 00 25 = 0

0 30 = 0

0 35 = 0

0 25 = 0

0 20 =

1 800 x 1300 1600 3,6 2,7 2,0 1,3 -2 1000 x 1600 1600 4,8 3,9 3,0 2,4 1,7

3 1300 x 1800 1600 6,0 5,0 3,7 3,1 2,4

4 1400 x 2000 1600 7,2 6,4 4,7 4,3 3,9

5 1600 x 2200 1600 11,3 9,9 6,9 6,4 6,0

Phlc 4.8 -Cc chtiu cl t

t kh t ngm nc

TTGc

0 t

nhin

H

skt

dnh,

C

Hs

trt

Dungtrng t

nhin0 ,tn/m3

Lc khngR,

daN/m20 m R

1 200 0,0670,39

8- - - 1 2,04 0,4

2 250 0,0920,46

61,6 3,94 0,8 1 2,46 0,6

3 300 0,1210,57

71,7 5,1 1,2 1 3,0 1,0

4 350 0,1580,70

01,75 6,46 1,5 - - -

5 400 0,2020,83

71,80 8,37 2,0 - - -


86/95

86

Phlc 4.9 -Chtiu cl t mt svng c trng

TTc tnh t t

nhin

t

yu

Rung

nc

i ct

gn bin

Trung

du

Min

ni

Min

ni

Cao

nguyn

1 M tcht t (1) (2) (3) (4) (5) (6) (7)

2

m tnhin

ww ,% 4,7 30 22-24 31 22-24 22-24 22-25

3Dung trng w ,

g/cm31,45 1,67 1,65-1,7 1,97

1,68-

1,7

1,8-

2,21,37

4Ttrng ,

g/cm32,61 2,67 2,71 2,69 2,72 2,72 2,7

5 st 1,92 0,67 - 0,17 - - -

6Hsnn a,

daN/cm20,191 0,037 0,02 - - -

0,025-

0,029

7Lc dnh kt, C

daN/cm20,82 0,12

0,25-

0,280,37 - -

0,3-

0,45

8Gc ma st

trong, 0


87/95

87

Phlc 4.10 -Mng no

Thp

(kg)

B

tng

M20

0

(kg)

Khi

lng

(kg)

Lc chng nh, tn, ng vi t,

/ (T/m3)

t kht ngm

ncTT

Loi

mng

neo

Kch

thc

(m)

6 27

50

60

400

1,6

350

1,6

300

1,53

25

0

1,40

300

0,9

250

0,9

200

0,9

1MN12-4

0,2

0,419,14

0,750,05

8150

13,8

11,5

9,50

6,90

6,404,90

4,10

2MN

15-5

1,5

0,5

33,

910,75

0,09

1230

18,

8

15,

0

11,

5

9,1

08,20

6,4

0

5,4

0

3MN

18-6

1,8

0,6

46,

001,04

0,13

0325

21,

0

20,

1

17,

3

14,

49,70

7,9

0

6,6

0


88/95

88

CHNG V

NG DY TRN KHNG TRONG VN HNH

5.1. NHNG VN CHUNGC ba chlm vic ca ng dy trn khng.

1. Chlm vic bnh thng : Trong khng c dy bt, chbnh

thng xy ra trong mi thi tit.

2. Chsc : Trong chny c dy dn hoc dy chng st bt.

Ngi ta chxt trng hp t mt dy chng st hoc t mt dy pha nguy

him nht. Chtrong trng hp cn thit mi xt t ng thi hai dy.

3. Ch thi cng : L chtrong ng dy c thi cng, cc dy

c ko v treo ln ct. Chthi cng xy ra trong trng thi bo,

5.2. CHLM VIC BNH THNG

5.2.1. Slch i ca cc chui s- khong ct i biu

- Khi thi cng cc chui sc treo thng ng, c ngha l ng sut (

im treo dy thp nht) trong dy trong mi khong ct ca mt khong no

khi bng nhau.

- Khi iu kin thi tit thay i, ng sut trong mi khong ct trnn khc

nhau, to ra lc tc ng ln s theo hng dc ng dy. Lc ny lm cho

chui sng blch i cho n trng thi cn bng, tc l lc ko bng 0, khi

ng sut trong cc khong ct li bng nhau.

-Nh vy, trong vn hnh ng sut trong cc khong ct ca mt khong

no lun bng nhau. Trong mi iu kin thi tit, nhng cc chui s b

lch khi vtr thng ng theo chiu ca dy.

- Trong cc khong no khc nhau th ng sut khc nhau, schnh lch ny

gy ra lc ko tc ng ln ct no.- ng sut ca mt khong no c xc nh qua phng trng thi ca

khong no nhsau :

( )

= 2 2 2 2

0 0 02 2

024 24b bg El g El E T T


89/95

89

- Trong : lbl khong ct i biu ca cc khong no v c xc nh

nhsau :

=

=

=

3

1

1

n

ii

b n

ii

l

l

l

(5.1)

Trong l1, l2 lnl khong ct trong no.

5.2.2. Slch i ca dy dn v chui sdo gc

5.2.2.1 Xt lch vo trong ca chui s

+ Trng thi chui sbthi lch vpha ct, khong cch tchui sn

ct c th nguy him trong trng thi bo v trong trng thi qu in p kh

quyn.

Hnh 5.1

- G = P.lTL, vi lLTl khong ct trng lng, P l trng lng dy (daN/m)

- GS= n.GOS + Gpk, vi n l sbt s; GOSl trng lng mt bt s, Gpk l

trng lng ca cc phkin.

-QV= PV.lG, vi PVl p lc gi tnh trong trng thi bo v qu in p kh

quyn; lGl khong ct gi.

-Thnh vta c :

=

+2

qv V

S

k Qtg

GG

(5.2)

kqvl hshiu chnh theo p sut gi (Qv), tra bng.

- lch ca chui sl :

= .sinr (5.3)

Dx

a

QV

PT

rT


90/95

90

-Tnh a = Dxr cho tng trng thi ri so snh vi khong cch an ton yu

cu.

+ Nu khong cch a khng c m bo khi ta phi dng tcn bng

Cch xc nh trng lng ca tcn bng CCBnhsau :

tacp rcp sin cp cptg

li c =+ +

2

qv V

cpS

CB

k Qtg

GG G

(5.4)

=

2qv V S

CB

cp

k Q GG G

tg (5.5)

5.2.2.2. Trng hp slch ra ngoi

- Chui sv dy dn bgi thi ra ngoi, khong cch gia dy dn gia

khong ct v vt cn xung quanh c tht n gi trnguy him.-Xt trng hp nguy him nht l slch ra ngoi ca chui spha ngoi

cng nhhnh sau :

Hnh 5.2- lch ca dy im bt ktrn khong ct :

(

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Giao Trinh Co Gioi Duong Day 2009