bai giang mo-tru cau-gs trung

7/31/2019 Bai Giang Mo-Tru Cau-GS Trung

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GS.TS.Nguyn vit Trung M tr v gi cu dm

1 3/26/2008

M tr v gi cu dmM tr v gi cu dmM tr v gi cu dmM tr v gi cu dmChng I:

M, tr cu dmM, tr cu dmM, tr cu dmM, tr cu dm

1. khi nim chung v m tr, cu1. khi nim chung v m tr, cu1. khi nim chung v m tr, cu1. khi nim chung v m tr, cu (1 Tit)M tr cu l mt b phn quan trng trong cng trnh cu, c chc nng kt cu nhp, truyn

cc ti trng thng ng v ngang xung t nn.

M cu l b phn tip gip gia cu v ng, m bo xe chy m thun. M cu cn c tcdng nh tng chn t nn ng u cu nn ng khng b ln st, xi l. M cu chnh dng khng i xng v chu p lc mt pha.

P

M m

Tng thn

B m

Tng nhTng cnh

Nn m

Hnh 1.1. Cu to chung m

Tng nh l b phn chn t sau dm ch hoc dm mt cu, c chiu cao tnh t mt cun mt k gi

M m l b phn k gi cu, chu p lc trc tip t kt cu nhp truyn xung.

Tng thn l b phn tng nh v m mTng cnh l cc tng chn t chng st l ca nn ng theo phng ngang cu

Mng m l b phn tng trc hoc tng thn v tng cnh

Nn m l cng trnh chng si l, ln st ta luy nn ng ta v tr u cu ng thi c tc dngnh mt cng trnh dn dng chy, tu theo dc taluy, vn tc nc, nn m c th p t gia cc, gia c hc hoc lm di dng tng chn.

Tr cu c tc dng phn chia nhp, truyn phn lc gi t hai u kt cu nhp, hnh dng trcu i xng theo dc v ngang cu v phi m bo cc yu cu v:

+ M quan

+ Thng truyn

+ Va x tu thuyn

+ Tc ng ca dng chy


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2 3/26/2008

M

Tr

Kt cu nhp

M

Hnh 1.2. B tr chung

V mt knh t, m tr cu chim 1 t l ng k, i khi n 50% vn u t xy dng cngtrnh.

M tr l kt cu phn di, nm trong vng m t, d b xm thc, xi l, bo mn vic xydng, thay i, sa cha rt kh khn nn khi thit k cn ch sao cho ph hp vi a hnh, a cht,cc iu kin k thut khc v d on trc s pht trin ca ti trng.

V vy, m tr cu phi m bo nhng yu cu v kinh t, k thut, xy dng v khai thc. mbo yu cu v kinh t, k thut ngha l m tr s dng vt liu mt cch hp l, c kch thc c bnc chn sao cho c tr s nh nht m vn m bo v cng , cng, n nh khng b xil, ln, st. m bo v yu cu xy dng ngha l s dng nhng kt cu lp ghp, ch to sn trong

cng xng, c gii ho thi cng. m bo yu cu v khai thc: cho php thot nc m thun dicu, bo m m quan ca cu, khng cn tr s i li di cu trong cu vt, chng bo mn b mtm tr.

Phn loi m tr cu

- Theo s tnh hc

+ M tr cu dm ( cu bn, dm gin n, lin tc, mt tha): Di tc dng ca ti trng thngng, ch c phn lc gi thng ng V

Tr

MM

Hnh 1.3. M tr cu dm

+ M tr cu khung: M vn ging cu dm nhng tr lin kt ngm vi kt cu nhp. Nh vy trchu mmen rt ln B tr c ct thp thng v ct thp d ng lc.

Hnh 1.4. M tr cu khung+ M tr cu treo: M phi c kch thc ln chu lc V,H cu to phc tp

H

V

LC


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3 3/26/2008

Hnh 1.5. M tr cu treo

+ M tr cu dy vng: M chu lc nh ti m b tr gi chu lc nh v m phi nng chu lc c nh. M khng chu lc y ngang do dy neo c neo vo u dm cng. Tr thp cuchu lc ch yu, cc dy neo truyn ti trng vo tr thp truyn xung mng tr thp phi cng chu c lc tc dng ca cc ti trng

Hnh 1.6. M tr cu dy vng

- Theo cng dc cu

+ M tr cng: Kch thc ln, trong lng ln. Khi chu lc bin dng ca m tr tng i nhc th b qua. Mi tr c kh nng chu ton b ti trng ngang theo phng dc cu t kt cu nhp

truyn n v ti trng ngang do p lc t gy ra.Loi m tr ny p dng cho cu nh, cu trung v cu ln

+ M tr do: Kch thc thanh mnh, cng nh gm: X m, cc (ct). Trn m tr ch c gic nh hoc khng cn gi. Khi chu lc ngang theo phng dc cu ton b kt cu nhp v tr s lmvic nh 1 khung v khi lc tc dng ngang s truyn cho cho cc tr theo t l cng ca chng.

Lc ny cu lm vic nh 1 khung nhiu nhp gim lc ngang tc dng ln tr.

Tuy nhin m tr do chu va x km cc sng c thng thuyn, cy tri khng lm c.Nhng vi loi m tr ny cho php s dng vt liu hp l hn nn gim c kch thc m tr.

p dng cho cu nhp nh v chiu cao khng ln lm

- Theo vt liu

+ B tng, xy

+ BTCT

- Theo phng php xy dng

+ Ton khi ( ti ch)

+ Lp ghp

+ Bn lp ghp

2. cu to m, tr do. cu to m, tr do. cu to m, tr do. cu to m, tr do (4 Tit)

I. Cu to tr doI. Cu to tr doI. Cu to tr doI. Cu to tr doTr do thng c hai dng: cc, ct.

1. Tr do dng cc

y l dng chnh ca tr do trong cc cu nhp nh c chiu di nhp L 20m, H 6m. Tuynhin dng ny khng p dng c trong trng hp kh ng cc: t rt rn, t ln nhiu tng, m ci hoc chiu di cc di qu khng thch hp lm tr do.


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Tr cc thng c p dng cc thung lng kh cn v n l phng n n gin nht.

Thng s dng s 1lin vi s lng nhp t 15 hoc s 2, 3 lin.

Tr neo Tr nhit Tr neo Tr do

L in b in L in g ia

Hnh 1.7. B tr tr do

Cc lin c phn cch bi nhng tr c bit gi l tr nhit . Tr nhit c 2 cc ringbit, c 2 x m ring. C 3 phng n phn chia nh sau:

- ltc (4045)m lm s 1 lin- Cu lm 2 lin khi chiu di 1 lin (3540)m- Cu gm nhiu lin khi chiu di lin bin (3545)m v chiu di lin gia (4045)mKhi tr c chiu cao ln H = 78m, tng cng cng theo phng dc v ton cu cng

nh gim bt ni lc i vi tr do trong lin ngi ta b tr 1 tr c cng ln hn cc tr khc gi ltr neo.

V tr tr neo:

- Lin bin t tr s 2 chu lc ngang cho tr b.- Lin gia, ti tr gia lin, chuyn v do nhit u c 2 pha.

Cu to:

Cc:Tit din cc thng c dng ch nht, c cnh ln song song phng dc cu, ct thp ch btr trn 2 cnh ngn, nh vy s tng mmen qun tnh tr theo phng chu lc bt li. Tuy nhin m bo tnh mm ca tr, chnh lch gia hai kch thc tit din cc khng nn ly ln qu. Tit

din cc thng c kch thc: 2535, 3035, 3540.Chiu di cc c chn theo chiu cao tr v chiu su ng cc. Theo QT 79, cc ng su

trong tng t chu lc ti thiu 4m. Ct thp s dng trong cc l ct thp thng v ct thp d nglc. Cc quy nh v ct thp nh hnh v sau:

d=18-22

35

30

d=6

Hnh 1.8. Ct thp cc

X m:X m lm vic nh dm lin tc, kch thc nh hnh v 2.3.


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5 3/26/2008

=(15-20)cm >=40cm

>=40cm

(120-150)cm

>=4m

70-80

Hnh 1.7. Kch thc x m

Thng thng x m gm hai loi:

- X m lp ghp: trong cc khi lp ghp c cha cc l hnh chp ct ng vi v tr u cc.- X m ti ch: ct thp ch d=(2024)mm, ct thp ai d=6mm

5cm>=40cm

BT M300-400

X m

5cm

Cc

(16-20)cm

Hnh 1.9. Cu to x m lp ghp

30cm 30cm 35cm

d=6mm

d=12-14mm

d=18-32cmI

I

I - I

Hnh 1.10. Ct thp x m ti ch

2. Tr do dng ctS dng: Khi tr cao n 6m, chiu di nhp l = 3040m, vn tc nc Vnc > 1m/s.

Cu to:

Tr c th c 1, 2 hay nhiu ct thu thuc vo kh cu v kch thc ct. Ct c th c tit dinvung, ch nht hoc hnh vnh khn. Ct vnh khn hay c s dng do gim c trng lng khilp rp.


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5m

10-12cm

120-140cm

1m

70cm

(0.8-1)m

B tng >30cm

Hnh 1.11. Tr do dng ct

Ct c th t trc tip ln mng chung hay ring hoc trc tip ln t nn nu tng t khnng chu lc.

II. Cu to m doII. Cu to m doII. Cu to m doII. Cu to m do

Thng c dng cc, ct, tng mng.

1. M do dng cc

S dng:Khi chiu cao t p H 6m, chiu di nhp l < 40m l loi dng ph bin v n gin nht.

Cu to:Cng ging nh tr do n gm cc, x m nhng khc tr l trn x m c tng nh vtng cnh c nhim v chn t.

Khi chiu cao t p H 2m, l 20m c th ch dng 1 hng cc.

Khi H, l ln b tr thm 1 hng cc xin.

X m c chiu cao h 40 cm, ton b thn cc nm trong t p nn m v c kch thc nhcc ca tr do.

H

40cm

Hnh 1.12. Cu to m do dng cc

2. M do dng ctGing nh m do dng cc nhng do ng knh ct ln hn nn c th m ch cn 2 ct ng.

Khi l (1215)m dng 2 cc ng 0.8m ng su 8m.

Khi l (1824)m dng 2 cc ng 1.0m ng su 12m.


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7 3/26/2008

H

40cm

Hnh 1.13. Cu to m do dng ct

3. M c dng tng chn

Khi chiu cao t p v chiu di nhp khng ln lm, nht l cu vt ng, cu trong thnhph c th dng m dng tng mng bng BTCT c sn tng cng tam gic.

u im ca loi ny l t p pha trc khng ln vo phn khng gian di gm cu cccu vt ng gim c chiu di nhp.

Khi H, l ln, m c cu to cc tng song song ring r gim p lc t ln tng, nhgnhc im l nu nn p khng tt t s chui ra pha trc.

b)

Tng chn

Sn tng cng

Tng cnh

(35-40)cm

Tng dc

Tng cnha)

Tng nh

Hnh 1.14. Cu to m dng tng chn

a) Tng chn dc

b) Tng chn ngang

3. cu to m, tr cng. cu to m, tr cng. cu to m, tr cng. cu to m, tr cng (4 Tit)

I. Cu to tr cngI. Cu to tr cngI. Cu to tr cngI. Cu to tr cng

Tr cng gm 3 b phn chnh: M, thn v mng tr. Trn nhng sng c dng nc chy xithoc c kh nng va p ca tu b, cy tri c th t b phn chng va x cho tr.

1. M tr

M tr chu ti trng trc tip t kt cu nhp v truyn xung thn tr. Kt cu nhp ta trn mtr thng qua gi cu. Ti ch t gi cu, m tr thng b tr li ct thp chu ng sut cc b c


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bc (5 5) cm. Mt trn ca m tr phi to dc t nht 1:10 thot nc. B tng m tr thng sdng M250 hoc M300.

Cu to:

Cu to m trnh hnh v 2.10.

Hnh 1.15. Mt s dngtr cng thng gp

a) Tr c thn hp

b) Tr c thn rng

c) Tr thn ct

>40cm

(1-3)m

>40cm

10-15cm

>1m

0.6-1m

a)

b)

c)

Ct thp ca m trc b tr ph thuc vo cu to thn tr

+ Tr c thn rng: ct thp m tr t theo cu to

N1 N2 d=10@200-250

N1

@200-250d=10 N2

Hnh 1.16. Ct thp m tr c thn rng

+ Tr c thn hp: ct thp m tr phn hng phi c t theo tnh ton.

S : Dm ngm mt u

Ti trng: Trng lng bn thn m tr

Trng lng k gi


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9 3/26/2008

Phn lc gi do tnh ti: R t

Phn lc gi do hot ti: Rh (c xt n h s phn b ngang)

d=22-30

d=8-10

d=12-14

d=14-22

N1 ( CT chu ko)

N2 ( CT ai)

N3 ( CT dc ph)

N4 ( CT chu nn)

@120-200

N2 11 - 1

1

Hnh 1.17. Ct thp m tr c thn hp

+ Tr thn ct: p dng trong cu dn thp c ng xe chy di, cu dm nhp l = 2030m.Ct thp chu lc ca x m thng c ng knh d=20mm, c b tr nh sau:

d=18-32cmI

d=6mmI - I

d=12-14mm

I

Ct

d=80-200(300)cm

Hnh 1.18. Ct thp m tr thn ct

k gibng BTCT M300, c li ct thp theo tnh ton. Li ct thp thng c cc kch thcsau:

d = (812)mm c khi n 14mm

@ = (8080 120120)mm

Khong cch cc li phi tho mn yu cu cu to tc khong cch cc li (50 70)mm.

V d: Cu BTCT d ng lc L = 33m k cc gi di ng v c nh cn cu to thm b kgi k cao gi c nh.

20

Gi di ng Gi c nh

Hnh 1.19. B k gi

Kch thc c bn ca m tr:

- Chiu cao m:Hm 40cm m bo cho kt cu nhp truyn phn lc qua m vo thn tr.


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- Chiu rng m B (dc cu):

b0 b'0

15-20cm

b1 b'1

B

Hnh 1.20. B tr gi trn m tr (dc cu)

Gi: - khe h gia 2 u kt cu nhp

+ Nu trn tr t 2 gi c nh th ly min = 5cm

+ Nu trn tr t 1 gi c nh + 1 gi di ng =5cm + to l

Trong : - to chnh lch nhit ( gia nhit khi t dm ln gi vi nhit nng hoclnh nht

- - h s bin dng do nhit ca kt cu nhp

- l chiu di nhp tnh ton

+ Nu trn tr t hai gi di ng:

=5cm +1 to l1 + 2 t

o l2

Gi: b1

, b1

l khong cch t tim gi n u mt kt cu nhp, nhp tri v phi

bo, bo l kch thc tht di ca gi theo dc cu

ao, ao l kch thc tht di ca gi theo ngang cu

(1520)cm l khong cch t mp tht gi n mp k gi

a l khong cch mp k n mp m tr

a xc nh theo Quy trnh, amin = 1520cm

b0

a 15-20cm

k gi

b0

a0

Hnh 1.21. B tr gi v k gi


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Bmin = +b1 +b1 +2

bo +2

'bo +2(1520) + 2a

V ddm d ng lc l = 33m c B = 2m

- Chiu di m A (ngang cu)Gi: a1 khong cch t mp k n mp m tr

a1 = (3050)cm tu loi gi cua2 khong cch tim cc dm ch theo ngang cu

n s dm ch theo ngang cu

Amin = (n - 1) a2 + ao + 2( 1520) +2a1

a0 k gi

a1

a2

Hnh 1.22. B tr gi k v kt cu nhp gi theo ngang cu

2. Thn tr

Thn tr lm nhim v truyn p lc t m tr xung mng v chu cc lc ngang theo phngdc cu v ngang cu. Mt ct ngang ca tr trong phm vi lng sng phi c dng r nc tt. Thntr phi chu c va p do cy tri, cc nhp c tu thuyn qua li cn phi chu c va ca tu.

Hnh dng mt ct ngang thn tr ph thuc vo iu kin dng chy di cu.

- Gim xi l lng sng v h chiu cao nc dng thng lu cu

p dng cho cu cn, tr hai ct

Rb

- Trnh to thnh cc dng soy ngm gn tr

- Dng chy mnh

R = b / 2

D thi cng

Thn tr ch nht p dng cho cu cn

Hnh 1.23. Mt s mt ct ngang thn tr

Sn bn c th nghing


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1

40(20)

Hnh 1.24. Sn nghing

Tr cu hin i c sn bn thng. Tit din tr c chn theo tit din trn nh mng

Mt s loi thn tr khc cng c s dng:

+ Tr thn c rng ( B tng, xy hoc BTCT)

300cm

(75-100)cm

Hnh 1.25. Tr rng lng

+ Tr thn hp BTCT

+ Tr thn ct BTCT c d = 80 200 ( 300)cm

3. Mng tr

- Mng tr c nhim v truyn ti trng t thn tr m xung t nn bn di v xung quanh.Ngoi ra mng tr cn c nhim v phn b lc t thn tr xung 1 din tch rng hn mbo chu lc cho t nn v n nh cho tr. su t mng cn phi m bo cho trkhng b mt n nh, nghing lch hoc b ph hoi do xi l gy ra. u trn ca cc phi cngm vo trong b hay x m BTCT mt tr s theo tnh ton ng thi phi ngp su vo trong

b mt on khng nh hn 2 ln chiu dy thn cc, vi cc cc ng knh d 60cm thkhng c nh hn 1.2m. Vi cc cc cho ct thp chn vo trong b th cc phi ngm vo b

(1015)cm v ct thp nm trong b t nht l 20 ln ng knh ct thp g v 40 ln ngknh ct thp trn trn.

- Kch thc:quy nh nh hnh 1.25. m bo s truyn ti trng ng u xung cc cc thchiu dy b phi 2m.


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>25cm >25cm2m

Hnh 1.26. Cu to mng tr

a) Mng cc ngb) Mng cc ng knh ln

- Cao nh mng:Ph thuc vo iu kin a cht, a hnh, kinh nghim ca ngi thit k.+ Nu mng nng: Cao nh mng phi nm ngang hoc di mt t t nhin khong

(0.51)m.

+ Nu l mng cc: B thp: y mng n ng xi l phi tho mn h hmin (hnh 1.26a)( t xung quanh mng chu c lc ngang)

B cao: Cao y b, cao nh mng nm v tr bt k (hnh1.26b)

MNTN MNTN

0.5m

hmin

a)

b)

Hnh 1.27. Cao nh mng

- Cao y mng:+ Nu mng nng: y mong phi nmg di ng xi l 2.5m.

+ Nu l mng cc: Cc phi cm vo tng t chu lc 4m.

4. Mt s dng tr cng

- Tr ton khi:Nh trnh by trn- Tr lp ghp:


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14 3/26/2008

L 40X40

(0.5-1)m

Cho lng ct thpvo l sau khi lp

xong v btng bt kn

0.6-1m

ti ch

ti ch

Lp ghp 30d

>1.1 D

coi lngm

a) b)

Hnh 1.28. Tr lp ghp

D - ng knh ct

d - ng knh ct thp ct

II. Cu to m cngII. Cu to m cngII. Cu to m cngII. Cu to m cng

Trong cng trnh cu, m thuc kt cu phn di c chn trong t, nm trong vng m t

chu xm thc ca xi l. M c cc chc nng c bn:+ kt cu nhp

+ Chu ti trng thng ng v nm ngang t kt cu nhp truyn xung

+ Chu p lc t y ngang

+ B phn chuyn tip v bo m xe chy m thun t ng vo cu

+ m bo chng xi l b sng

Cu to chung m:

P

M m

Tng thn

B m

Tng nhTng cnh

Nn m

Hnh 1.29. Cu to chung m

1. Cu to m ch U.

c im: + Nn t ch gii hn trong tng trc thot nc tt hn m vi

+ Chiu cao t p H =46m (c khi n 810m)

+ p dng cho c cu t v cu ng st

+ n nh chng lt, chng trt tt.

a) Cu to m U b tng, xy.

Tng nh:Chiu dy: Trn: 60cm

Di: b1 = (0.50.6) h1


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Chiu cao: h1 = hd + hgi + h k

h k =20cm

Chiu di ( ngang cu) = Bcu

1b1

Bcu

h1

20-50

10

>60cm

Hnh 1.30. Cu to tng nh

M m: BTCThm 40cm

hm

(10-15)cm(10-20)cmH

b

10

1

Bcu

(10-15)cm (10-15)cm

Hnh 1.31. Mt ct dc, ngang m

Tng thn:+ Chiu dy ti mt ct nh mng: b = (0.350.4)H. Vi m c chiu cao t p H > 8m tng

trc c th nghing 10:1.+ Chiu di (ngang cu) = [ Bcu 2(1015)cm ].

Tng cnh:-Theo phng dc cu:+ Xc nh chiu di tng cnh cn c vo: dc taluy nn m 1:n

ngp su ca tng cnh m vo nn ng (s)

hm

1:n

sLc >60cm

80-10

0cm

(0.35-0.4)H

a) b)

Hnh 1.32. Tng cnh dc cu

Theo Quy trnh: H 6m dc 1:1 ( cu t) ; 1:1.25 ( cu ng st)


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H >(612)m dc 1:1.25 ( cu t) ; 1:1.5 ( cu ng st)

(s) H 6m s =0.65mH > 6m s = (0.751)m

C th xc nh chiu di tng cnh theo cng thc sau:

Lc = n H + s

+ on thng tng cnh: Tu ngi thit k. C th ly = h 1 hoc = (80100)cm.

+ on xin: (6:1) (4:1)

-Theo phng ngang cu:Kch thc tng cnh nh hnh 1.31b

B mng:Ging trb) Cu to m U BTCT

Cc kch thc c xc nh tng t nh trn.

Tng mng hn do c b tr ct thp: Tng nh: b1 =(3050)cm

Tng cnh: bc = (4050)cm

Tng cnh c hng ln: (1:1) (1:1.5)

h1

1:1.5

(3-4)m

1:1

(30-50)cm

P

Hnh 1.33. Cu to m U BTCT

2. Cu to m vi

a) M vi b tng, xy

S dng khi chiu cao t p H = (520)m

GiA l giao im tng trc v m m. QT 79 quy nh: Taluy nn m phi cch A 1 khong 30cm

B l giao im tng trc v nn m: B phi cao hn MNCN 25cm


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70-100

A

1

70-100

10

1

(0.35-0.4)H

B3-6

75-100 >60

Hnh 1.34. Cu to m vi b tng, xy

b) M vi BTCT

M vi BTCT thng c 2 loi: M vi tng dc

M vi tng ngang

+ Tng nh, tng cnh dy 30cm.

+ Tng thn: . Nhiu tng: bt = (3540)cm. 2 tng: bt = (70100)cm.

Tng dc

a)

bt2.5

1

30

bt

30b)

a 0.3a

Hnh 1.35. Cu to m vi BTCT

a) Nhiu tng dc

b) Hai tng dc

4. khi nim tnh ton m, tr cu. khi nim tnh ton m, tr cu. khi nim tnh ton m, tr cu. khi nim tnh ton m, tr cu (3 Tit)

1. Khi nim chung

Khi thit k m tr cu trc ht chn chn loi m tr cu, s b xc nh kch thc cc titdin. Sau tin hnh theo cc trnh t sau:

- Chn s tnh ton.- Xc nh cc loi ti trng i vi tit din cn tnh ton ca cc b phn m tr.- Lp cc t hp ti trng nhm xc nh cc tr s ni lc bt li rt c kh nng xut hin

trong qu trnh xy dng v khai thc cng trnh.

- Kim tra li cc tit din theo cc trng thi gii hn.


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2. Cc ti trng tc dng ln m tr cu

a) Trng lng bn thn

Xc nh theo kch thc hnh bao ca cc bn v k thut. Khi tnh ton nn chia m tr thnhcc khi hnh hc n gin tnh th tch, trng lng v cnh tay n t trng tm ca cc khi nyn 1 trc no cn tnh mmen.

Cng thc tnh ton:Q = V

Trong : - trng lng ring ca vt liu

V - th tch m tr

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

= - 1 (T/m3)

b) Phn lc gi di tc dng ca trng lng bn thn kt cu nhp

Xc nh da vo thit k c th. V d nhp dm gin n:

Rt = g L /2

Trong : g trng lng bn thn kt cu nhp trn 1 n v chiu di nhp.

L chiu di nhp tnh ton.

c) Trng lng t p

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

P = H (T/m2)

Trong : - trng lng ring ca t, = 1.8 T/m3.

H - chiu cao t p.

d)p lc ngang ca t

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

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

ep = tc H

Trong : H chiu cao tng t tnh ton.

=

245tg o2 - h s p lc ngang ca t.

, tc gc ma st trong, dung trng th tch ca t.

Khi y mng t cch mt t t nhin 3m coi p lc y ngang ca t phn b theo quylut ng thngHp lc y ngang tnh theo cng thc:

HBe21

E p=

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

B - chiu rng tnh i ca m


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B xc nh nh sau:

b1 2b2 B = b.

b1 > 2b2 B =2 b2.

Vi m cc (ct) nu chiu rng tng cng cc cc (ct) < 1/2 chiu rng mtr th B =2 b ( b chiu rng cc hoc ct)

Vi m cc (ct) nu chiu rng tng cng cc cc (ct) 1/2 chiu rng mtr th B ly bng khong cch mp ngoi ca cc (ct).

E

H/3

H

15mmt u kt cu nhp b tr gi di ng theo 2 phng (Hnh 2.1b).

- Trong cu dm lin tc thng thng t gi c nh trn cc tr gia nhp m bo chuynv theo phng dc cu khng qu ln (Hnh 2.1c).

- Cc cu gin n nhiu nhp: Vi cu gin n nhiu nhp c nhiu cch b tr gi cu. Thngthng trn 1 tr thng b tr 1 gi c nh v 1 gi di ng m b khe co dn khng qu ln. Tuynhin, nu tr yu v a cht ti ni t tr yu c th t 2 gi di ng (Hnh 2.1.d).

1 2 3a) b)

c)

d)

Hnh.2.1. B tr gi cu

1. Gi c nh: 2. Gi di ng theo 1 phng; 3. Gi di ng theo 2 phng


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2. cu to cc loi. cu to cc loi. cu to cc loi. cu to cc loi gi cugi cugi cugi cu (1 Tit)

Cu to gi cu ph thuc vo tr s p lc truyn ln gi, i vi gi di ng cn ph thuc vo chuyn dch ca u kt cu nhp. Chiu di nhp cng ln, cu to gi cu cng phi hon chnh m bo chuyn v v xoay t do ca u kt cu nhp.

1. Cu to gi cu dm BTCT

a) Gi tip tuyn- p dng: Dng cho cu bn, dm c sn c L=(1220)m.

- Cu to: Gm 2 bn thp c chiu dy (3050)mm gi l tht gi. Tht trn l 1 tm thp phng chn vo cc thanh thp neo chn sn trong dm BTCT. Tht di c mt tip xc hnh tr c lin ktvi thanh thp chn sn trong k gi. Cu to gi c nh v gi di ng ch khc nhau ch: Gi cnh c cht hoc vu ngn cn chuyn v ca tht trn i vi tht di. Gi di ng c t bn np sn bn ngn khng cho tht trn chuyn v theo phng ngang so vi tht di.

Gi c th chu c phn lc gi n 300T

1

2 43

5

1

Hnh.2.2. Gi tip tuyn

1. Thp neo; 2. Tht trn; 3. Tht di; 4. Thp bn; 5. Cht thp (gi c nh)

b) Gi con ln BTCT

- p dng: Dng cho nhp L 20m.- Cu to: Gm 2 tm thp b mt hnh tr, gia l khi BTCT c M300400, chiu cao gi khong

(6070)cm.

Gi c th chu c phn lc n 80T

BTCThg

Hnh 2.3. Gi con ln BTCT

c) Gi con ln thp ct vt

- p dng: Dng cho nhp L = (3033)m.

- Cu to: Ging nh gi con ln BTCT nhng khi BTCT c thay bng con ln thp vt gc.


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Con ln thpct vt

hg

Hnh 2.4.Gi con ln thp ct vt

d) Gi cao su phng

- p dng: Dng cho nhp L = (3033)m.

- Cu to: C cc bn thp dy 5mm nm gia cc lp cao su. Cc bnthp c tc dng nh cc ctthp ngn cn v tng cng ca gi khi chu phn lc thng ng. Nh tnh cht n hi ca cao su,tit din u dm c th chuyn v xoay v chuyn v trt. Gi c dng hnh trn hoc hnh ch nht.

Hin nay loi gi cao su phng ch nht c s dng trong d n ci to nng cp cc cu trn Qucl 1. Gi c th chu ti trng n 200T.

910

1110

147

20

1472

0

thp

cao su

20

23

20

2

2 20

2

2

93

20

2

Hnh 2.5. Gi cao su phng dng trong Cu Non Nc (Hng OVM)

e) Gi bn thp phng

-- p dng: Dng cho cu bn hay cu dm c sn L 12m.

- Cu to: Gi c nh l bn thp phng t (1020)mm c nh bng cht thp 25. Gi di ng gm2 bn thp, gia m bng Aming hay 1 lp than ch chng r v gim ma st m bo cho chngc th trt ln nhau. Bn trn hn vo ct thp chn trong dm, bn di hn vo ct thp chn trongm, tr.

f) Gi cao su hnh chu- p dng: Dng cho nhp L =(40130)m.

- Cu to: Gm 1 tm cao su hnh trn (1) t trong 1 b phn bng thp hnh chu (2). Trong gi ding chuyn v trt ca gi do tm teflon PTFE (polytetra fluoroethylene). Tm trt teflon PTFE ct trong khc lm ca bn thp. Trn mt tm trt PTFE l 1 l thp hp kim c chiu dy 1mm. TmPTFE c chiu dy t (48)mm. gi di ng c theo 1phng ngi ta t thm 1 bn np dhng. Gi c nh c np y di.


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Chu c phn lc gi (1002600)T

Giong cao su

Np y

Chu thp

L thp hp kim

Tm cao su

a PTFE

Bn trt thp

Hnh 2.6. Gi cao su hnh chu

2. Gi ca cu thp

a) Gi cu nh, chung

Dng cc gi tip tuyn, gi con ln thp vt gc dng cho gi di ng

b)Cu nhp trung v ln- p dng: Dng cho nhp L 50m.

- Cu to: (hnh v)

S con ln khng nn qu 4.

a)

Bu lng neo

Tht trn

b)

Hnh 2.7. Gi cu thp nhp ln

a) Gi c nhb) Gi di ng

3. ni dung tnh gi bn phng v gi tip tuyn. ni dung tnh gi bn phng v gi tip tuyn. ni dung tnh gi bn phng v gi tip tuyn. ni dung tnh gi bn phng v gi tip tuyn (0.5 Tit)

1. Tnh gi bn thp phng

- Ni dung tnh:Xc nh kch thc a, b v chiu dy ca thp gi


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28 3/26/2008

a

b

I

A

bo

b

bo

I

I

L

I

Hnh 2.8. S tnh gi bn thp phng

- A l p lc tc dng ln gi ang xt do ti trng tnh ton

- Din tch F ca bn thp phi tho mn iu kin

F = a b bR

A

Rb l cng chu nn cc b ca b tng b k gi ( Thng dng M250)

T sba

thng chn t 11.5 T xc nh c a, b

tnh chiu dy , coi ng sut di y tht di phn b u:

=ba

A

Tht di coi nh ngm tit din I-I lm vic nh congson di tc dng ca ti trng phn bu q theo chiu rng a:

q =bA

Chiu di congson l: L =2

bb o

Tit din tnh ton l tit din ch nht (a ) do ta c s tnh nh hnh 2.8.

T ta c:

MI-I = ( )2

obbb8A

WI-I = 6a 2

Bn thp phi tho mn: tuRWM=

tu

o

Rba

A32

bb

2. Tnh gi tip tuyn


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Gi cu c tnh vi phn lc gi A do ti trng tnh ton (c tnh n n, 1+)

Chn kch thc gi:

a1b

R

aA

Hnh 2.9. Kch thc gi

+ Theo phng ngang cu (a) gn bng chiu rng bu dm hoc cnh di dm I

+ Theo phng dc cu (b): b = cm)2018(RaA

em

Bn knh con ln c xc nh theo iu kin nn (quy c) theo tit din i qua ng knh conln.

A = Rc 2 R a1

Trong : Rc cng tnh ton chu ct ca thp lm gi, Rc = 0.04 Ro

Ro cng chu lc dc trc ca thp lm con ln, R o = 200kG/cm2.

R bn knh con ln

a1 chiu di ng tip xc

( 2 R a1 l din tch quy c)Kim ton tht gi thp theo iu kin chu un: Coi phn lc A phn b u trn din tch,

mmen ti tim tht gi (coi nh ti ngm ca congson).

8b.A

4b

.2b

.bA

M ==

Chn trc chiu dy tht gi, = (3050)mm.

uRWM=

Trong : 6.aW2

=

bai giang mo-tru cau-gs trung

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