Chuong 2 - Dap BT va BTCT tren nen mem(mo)-Nguyen Phuong Mau.pdf

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Chng 2. p b tng v b tng ct thp

trn nn mm

Bin son: PGS.TS Nguyn Phng Mu

p trn b tng trn nn mm khc vi p b tng trn nn nhng im sau: p trn nn mm thng c y rng hn, do sc khng trt ca nn nh v ti trng n v cho php b. Do , vic xy dng cc p cao trn nn mm thng tn km v nhiu khi khng th thc hin c (chiu cao p khng vt qu 4050 m). V vy, khi thit k loi p ny, cn xem xt k cc c trng a k thut ca vt liu nn.

2.1. c im a cht nn v cng tc chun b

Theo thnh phn ht, t c chia ra thnh cc loi ph thuc vo kch thc ht:

+ tng: c kch thc ln hn 300 mm ;

+ cui v dm: c kch thc 300 150 mm ;

+ Si v sn: c kch thc 150 2 mm ;

+ Ht ct: c kch thc 20,06 mm ;

+ Ht bi : 0,06 0,002mm ; + Ht st: c kch thc nh hn 0,002 mm ;

+ Ht mn: tp hp ca ht bi v ht st ;

+ Ht th : cc ht c kch thc ln hn ht bi ;

+ t hu c : t c di tch thc vt v ng vt ;

+ t ht mn : t, gm hn 50% trng lng l nhng ht c kch thc nh hn 0,08 mm ;

+ t ht th: t, gm hn 50% trng lng l nhng ht c kch thc ln hn 0,08mm ;

+ t cui si: t ht th, trong thnh phn ch yu l cc cui si ;

+ t ct : t ht th, trong thnh phn ch yu l cc ht ct ;

+ t bi : t ht mn, trong hm lng st chim t hn 20% trng lng ca thnh phn ht mn ;

+ t st: t ht mn, trong hm lng st chim hn 20% trng lng ca thnh phn ht mn ;

+ t ri : t, trong bn chng ct ch yu ph thuc vo lc ma st gia cc ht ;

+ t dnh: t, trong bn chng ct gm lc ma st gia cc ht v lc dnh gia cc ht :

+ Tnh do: tnh cht ca vt liu c kh nng chu c bin dng tc thi khng n hi, c bin dng th tch khng ng k v khng b rn nt ;


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+ Tnh nn: kh nng bin dng ca t di tc ng ca lc nn ;

+ Gii hn chy: hm lng nc ranh gii quy c gia trng thi do v trng thi chy ca t ;

+ Gii hn do: hm lng nc ranh gii quy c gia trng thi do v trng thi cng ca t.

I. c tnh ca t dnh i vi st, hm lng ht c ng knh d


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II. c tnh ca t khng dnh t khng dnh nhn chung c chia thnh ct v t c kch thc ln. Nh c

cp trc y, t ny l t cha cc mnh v tn tch, v, mnh v, dm v cui si.

ruaceous ca thi k k t v tin k t lun lun b xi mng ho bi cc lp khc. v ximng ho c gi l dm kt, cui kt. Ht cui si b xi mng ho gi l gravelite. ruaceous thng l v, tn tch ca cc loi khong vt khc nhau. Chng cha hn 50% cc mnh v c d>10mm. Cc thng s khc ca chng l: gc ni ma st, 33350; lc dnh, 0,40,5T/m2, K = 1,75.10-11,75.10-2cm/s; rng 0,350,37.

Ngi ta cng hay s dng ct c rng 3540%, gc ma st trong 30350. H s thm ca ct l K = 1,75.10-21,75.10-5cm/s. Trng lng ca 1m3 ct t 1519 KN.

Cc loi t khng dnh c cc t trng khc nhau tu thuc vo iu kin hnh thnh ca chng. Tuy nhin chng c mt im chung in hnh l khng c lc dnh.

Nhm ct v ct mn hay c gi l ct chy cha cc mnh vn c kch thc t 0,25 n 0,5mm ti 8096% trong trng thi bo ho c cc c tnh nh sau: c th c gc ngh nh 370 ng vi m 1314% v gim ti 00 ti m 1718% v kh nng chu ti nh. Ct mn khng bo ho c th c rng ti 4250% v trng lng ring 1315 KN/m3. V vy cc loi ct ny di tc dng ca ti trng ng c th b st ti ln.

Trn nn cui si v t ri, ta c th xy dng p c ct nc cao ti 3040m , cn trn nn ct ta c th xy dng p c ct nc cao ti 2030m.

Ct chy c s dng trong nn p c ct nc thp khi cc gii php sau c tin hnh: mt bn c c b tr ton b dc theo ng vin thm, nn phi c tng cng v gia c bng phun pht va ho hc...

2.2 Thit k p b tng trn nn mm, ng vin thm ca p

Trong thit k p, gii php c coi l hp l nht l gii php c th tho mn cc yu cu v cng , n nh ca p v nn, a ra c phng php thi cng c li nht trong iu kin c th, p ng cc yu cu v mt vn hnh, c gi thnh r nht.

p b tng trn nn t c mt ct ln nhm m bo s phn b ng sut ng u trn ton b mt tip xc ca nn vi cng trnh.

I. Cc b phn ca p: cc b phn ca p c th chia lm hai phn: 1. Phn di: t su trong nn t, chng hn nh bn y mng, bn y sn

thng, h lu, b tiu nng v tng chng thm ca b tiu nng, l thot nc b tiu nng, c chng thm ..v.v...

2. Phn trn: b tr pha trn bn y, chng hn nh phn trn nc, cc tr pin v cu giao thng,.v.v...


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V nguyn tc p trn nn mm thng c ng vin thm pht trin theo phng ngang v cc b phn c thit k nhm trit tiu nng lng tha x v h lu cng trnh v m bo cho y lng sng khng b xi l v bo mn nh hng n s n nh ca p.

Vic xy dng cc p khi ln tng i d dng (trng lng ca p c quyt nh ph thuc vo cc iu kin n nh v sc khng ct). Cn cc p rng (p c trng lng nh hn) cn t khi lng b tng nhng li cn hm lng thp cao hn. m bo an ton n nh cho p rng, ta cn tin hnh thm mt s bin php thi cng, iu ny lm cho vic xy dng p tr nn phc tp hn. V vy, khi xy dng p, ngi ta phi so snh cc gii php thay th khc nhau tm ra gii php ti u (xy p khi ln hay xy p rng).

a)

b)

Hnh 2-1. p trn trn nn mm v cc b phn ca n.

(a) mt ct dc; (b) mt bng; 1. lp bo v sn ph bng st; 2. cc tm gia c; 3. tr; 4. ca van sa cha v cu giao thng; 5. ca van chnh; 6. thn p; 7. hnh lang thot nc trong thn

p; 8. bn y tiu nng; 9. sn sau; 10. h xi sn sau; 11. hc xp sn sau; 12. tng lc ngc; 13. l thot nc; 14. c thp.

Hnh 2-2 m t cc mt ct ca p trn trn nn t c xy dng Nga trong cc giai on khc nhau. Vi cng mt chnh lch ct nc thng h lu (25m), cc p khi ln xy dng t nm 1929 n 1935 cn 15001600m3 b tng trn 1m chiu di p, ln hn t 4550% so vi khi lng b tng trn 1m chiu di p ca cc p c xy dng nm 1954 (10001200 m3 b tng trn m). Lng thp dng xy p nm 1929-1935 khong 35 kg/1m3 b tng, nm 1951-1954 l khong (55 66)kg/1m3 b tng. R rng l vic la chn p khi ln hay p rng v mc gim trng lng ca p ph thuc vo cc yu t kinh t v k thut in hnh ca khu vc xy dng.

4

1 2

14 12 1312

8 109

11

7

6

5

3

h3

13


76

a)

e)

(b)

(f)

(c)

(g)

(d)

(h)

Hnh 2-2. Mt s p trn trn nn mm c xy dng Lin x

t nm 1929 n 1954.

Hnh 2-3. p y c xy dng tnh H Ty Vit Nam.

1929

1935

1948

1950

1951

1954

19491929

14.8014.00

7.00

3.007.00

12.30

-2.00

Tm b tng Dm tiu nng

t st chng thmC btng ct thp C thp C btng ct thp

80.60 19.70 13.00 15.00 15.00

2.60

143.30

5.00


77

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110

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7.5

0

6.0

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9.0

5

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80

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- Nn mng cng trnh thng c thit k di dng tm: tm phng hoc tm c tng chn khay c hai pha thng hu v h lu.

- Tng chn khay c xy dng nhm mc ch:

(1) To lin kt tt hn gia y mng v nn cng trnh;

(2) Ngn nga thm tip xc;

(3) Tng tnh n nh chng ct ca p. Chn khay thng su 23m, chiu rng ca bn chn khay nhn chung 3m.

Cc phn trn ca p c b tr sao cho ti trng ca kt cu phn trn cng vi cc

lc khc s phn b mt cch hp l. ng sut trn mng ca p, c c trng bi h s K=smax/smin; trong : smax, smin l ng sut ln nht v nh nht ti y p. H s ny cng xem xt n s phn b khng u ca ng sut. i vi nn st, K 1.21.5; i vi nn ct K < 1.53.

II. S hnh thnh ng vin thm ca p. S ca ng vin thm di y cng trnh ph thuc vo kt cu a cht, loi nn

v cc yu cu i vi ng vin thm. Cc yu cu i vi ng vin thm l phi m bo n nh thm cho nn, gim nh lc thm ln y p.

Khi thit k ng vin thm, thng xut pht t nhng quan im sau:

(1) Bn cnh cc thnh phn theo phng ngang nn c cc thnh phn theo phng ng. Theo quan im ca thu lc th on ng vin thm theo phng ng c hiu qu tiu hao ct nc thm tt hn on ng vin nm ngang. su ca bn c thng ly t (0,51,5)H, vi H l ct nc ln nht ca p.

(2) Bn c i c s dng nhiu hn, v vic tng thm mt bn c u sn ph thng lu l phng n tit kim hn so vi vic tng chiu di ca bn c th nht.

(3) Khi tng thm dy th khng nn ng c n tng khng thm. Trong trng hp ny ta dng bn c v chn khay treo.

(4) Bn c theo phng ng trn t st vi h s thm nh l khng hiu qu.

(5) Mt kt cu chng thm theo phng thng ng (tng chng thm, thng chm) c h thp ti tng khng thm s ngn chn c hon ton dng thm. Trong trng hp ny, sn thng lu khng cn na (v trng hp ny gi l ng vin su).

(6) Chn khay h lu c dng h gradient ct nc thm ca ra. iu ny s lm tng tnh n nh thm ca t nn, nhng s lm tng p lc y ni di y p.

Khi b tr thit b thot nc (TBTN) cn ch ti cc im sau y:

(1) TBTN l mt gii php hiu qu nhm gim p lc thm tc ng ln y p. V vy TBTN c thit k trn nn ca p v lp bo v h lu (s I, IV, V).

(2) C th sn thng lu khng cn n TBTN hoc ch mt phn ca sn cn TBTN. (S II).

(3) Trong trng hp khng cn h thp p lc y ngc ln y p hoc khi trong nn c t bi tch thng lm tc nghn TBTN th khng cn thit k TBTN na.


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(4) TBTN theo phng dc v phng ngang c thit k khi trong nn c lp t thm nhiu nm di t thm t. Ti trng tc ng ln tng kp ca t khng thm s to ra p lc y ngc, lc tc ng ny hng ln trn lm gim s n nh ca p v ca nn.

Hnh 2-5. ng vin thm di t ca p trn trn nn mm

1,2. cc phn khng thot nc v thot nc ca ng vin; 3. tng khng thm nc;

4. thot nc theo phng ng; 5. t c tnh thm tng.

III. La chn lu lng x ca p trn nn mm.

Lng nc c x ra qua p trn hoc ca ra ca p c quyt nh qua vic tnh ton thu lc c tnh n s bin dng ca dng chy, s x nc qua nh my thu in, cng v.v... Lu lng nc x trn mt n v chiu rng ca c quyt nh bng cch so snh cc ch s kinh t, k thut ca cc phng n khc nhau ca p. Khi so snh cc phng n, ngi ta khng ch xem xt tnh kinh t m cn xem xt c iu kin hot ng v phng php thi cng ca cng trnh.

Lu tc bnh qun cho php sn h lu c xc nh ph thuc vo loi t nm di sn: i vi t ct, vra = 2,53,0 m/s; i vi t st, vra = 3,03,5 m/s ; i vi t na , vra = 3,54,0 m/s; i vi t , vra = 5 m/s.

Sau khi xc nh c cao trnh ca sn h lu v mc nc sng trong iu kin t nhin, tc nh c hh , lu lng x ti sn sau c xc nh nh sau:

qra = Vra hh (2 - 1)

Vi lu lng x qua p Qs v lu lng n v x qra, chiu rng ca sn sau B = Qs/qra. Cn ch rng tng chiu rng ca trn phi nh hn tng chiu rng ca sn h lu. Lu lng n v x ca trn xp x bng:

qx= (1,201,25).qra (2 - 2) i vi qx nht nh, cng thc tnh kh nng x ca trn cho php xc nh c b

rng ngng trn. Sau khi chia p thnh cc khoang, gi thit hnh dng ca u tr, xc nh nh hng ca mc ngp ca trn th ta s xc nh c cao trnh ngng trn.

1 2

3

I

12 2

II

1 2

3 5

4 4

III

21

IV

1 2

V

1 2

VI


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2.3 Cc b phn ca ng vin thm

I. Sn trc: Tu theo mc thm, sn trc c chia thnh cc loi nh sau: sn trc khng

thm hoc thm t vi h s thm c khong 10-6 cm/s. Loi sn trc khng thm c dng cho nn t st hoc cha st, cn loi sn trc t thm c dng cho nn ct hoc ct.

Chiu di sn trc phi c xc nh trn c s tnh ton bn thm ca nn. Tt c cc loi sn trc tr sn trc bng btng th phi c ph bng mt lp t bo v khng mng hn 0,3m; Lp t bo v ny phi c ph ln trn bng mt lp gia c bo v chng xi l do dng nc mt.

Chiu di sn trc thng t (11,5)H, nhiu khi ln n 2,5H nhng khng c ln hn :

l = ...2 0 TK

K ; (2-3)

trong :

d - Chiu dy sn trc ; T - su tng t thm nc pha di y p ;

K - H s thm ca t nn ;

K0 - H s thm ca sn trc.

V mt kt cu, sn ph c chia lm hai loi: sn ph n hi v sn ph cng.

Sn ph n hi l sn ph c kh nng thch ng s bin dng ca nn. Sn ph loi ny c lm bng cc vt liu c bin dng: st, st, atphan, t st nn v cc loi vt liu tng hp. Tnh thm ca vt liu phi nh hn 50 ln so vi nn t. Chiu di ca sn ph c xc nh thng qua vic so snh cc gii php thay th khc nhau ca ng vin thm.

i vi ct nc ln n 15m, ta thng s dng loi sn ph st, st, than bn. i vi nhng ct nc cao hn, ta thng s dng sn ph lm bng t st nn, b tng ct thp c qut sn ph, atphan hoc vt liu tng hp. dy ca sn ph bng st, st v than bn trong trng hp ny c ly d DH/Jcp, trong : DH - chnh lch p lc tc ng t pha di v pha trn sn trc ; Jcp - gradient chng thm cho php ca vt liu (i vi t st 610, st 45). Chiu dy ti thiu ca pha u sn trc khong 0,5m; cui sn trc khong 1 2m.

Trong khu vc s b tr sn trc, t c p ln nn c m cht theo cc lp, dy ca cc lp ph thuc vo loi m . i vi cc loi m nh th chiu dy ca lp l khong 1015cm, cn i vi loi m nng hn th chiu dy ca lp l 25cm.

Trong qu trnh m nn, t c lm m ti m ti u. mt s p, ngi ta s dng thnh cng phng php p t st trong nc m khng cn m. Phng php ny tit kim hn so vi phng php m kh.


81

Sn ph bng st m nn cha 2025% st, 3040% ct v 3540% cui si. trn cng ca sn ph, ngi ta thng thit k mt lp b tng.

Gia sn ph v p khng c php hnh thnh khe nt, v mt khe nt rng ch 1cm cng c th lm sn ph mt tc dng. Do gia sn ph v p b tng cn c nghing.Ngoi ra ngi ta cng dng n cc loi st, st n hi.

Hnh 2-6. S sn trc neo

1. lp bo v b tng; 2. lp khng thm nc;

3. tm cch nc; 4. bitum; 5. st; 6. b tng ca p;

7. neo sn trc; 8. tm b tng ct thp; 9. c; 10. mattic asphal ; 11. dm trn c.

Sn ph bng vt liu cng thng lm bng b tng hoc b tng ct thp. Sn ph loi ny c chia ra thnh nhiu mng c s lin kt vi nhau, c cc khi vt liu khng thm nc trong lin kt . Khe nt c th xut hin trong cc mng, do lp lt bitum vi cc khi vt liu c gia c bng si thu tinh hoc cc vt liu khng thm nc c b tr trong cc lp chng ln nhau nhm ngn nc thm qua. i vi ct nc 10m, ngi ta s dng cc tm b tng khng cn sn chng thm. dy ca tm trong trng hp ny thng da trn gradient ct nc ca dng thm cho php (Jcp < 2030)

Sn ph neo c s dng khng ch lm tng ng vin thm m cn gim cc lc gy trt cho p. Sn ph neo c dng bn b tng ct thp dy t 0,40,7m.

II. Cc bn c. C l vt liu tiu hao ct nc: khi c c tr s ct nc trn cc on ca ng vin

di t sau c gim i v cc dc o p dc ng vin di t cng cng gim i. Ngoi ra cc bn c cn c tc dng:

- Ngn cn s pht trin xi ngm trong vng t nn;

- Bo v nn p khi b mi xi do dng chy mt gy ra (c h lu);

- Ngn cn hin tng tri t pha di p di tc dng ca trng lng p (iu ny ch c th xy ra trong trng hp nn t yu v khng ng nht) ;

- Cho php thc hin vic ni tip thn p vi tng khng thm nc v do to thnh s su ca ng vin di t.

Vic b tr c h lu s gy ra s tng p lc y ngc ln y p. trnh nhc im trn, c h lu trong cc trng hp ny phi c c l. Khi tnh ton thm, cc hng c c c l khng c tnh n.

Khi b tr cc hng c, khng c dng cc hng c qu ngn (th d nh hn 23m). T chc thi cng ng cc vn c qu ngn s khng kinh t. Phi nh chiu di vn c

758

2111 3 4 5 6

910


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thp trong cc n thit k p ph hp vi chiu di vn c sn c. Phi tnh n trong mt s trng hp c th hn c thp (theo chiu di) tng chiu su vn c, (c th ti 30 40m).

Trong trng hp nn khng ng nht c cc lp kp thm nc nm ngang th tu theo kh nng m hng c phi ct qua cc lp kp .

Khng cho php gia mi c v mt ca lp khng thm c mt khong cch tng i nh (th d, nh hn 0,5 0,1T). Trong trng hp ny trnh xy ra tc thm ln gia mi c v tng khng thm nc, hng c phi c ng su vo tng khng thm.

Khi tng khng thm khng phi l (loi t st) phi ng su hng c vo tng khng thm nc vi su d. Trong trng hp ny xut pht t cc tr s ct nc trc v sau vn c, bng tnh ton c th xc nh c tr s d. Khi tnh ton phi xt hng c n thun c chiu su d chu tc dng ca ct nc Z.

Khi tng khng thm l , vic ni tip gia vn c vi nn s rt kh khn. Trong trng hp ny nn thm nc c ngn trn ton b chiu su xung n tn tng khng thm bng cc vt ngn dng tng rng su bng b tng.

Khi di p l hng c treo th khong cch gia chng khng nh hn 2s, trong :

s- chiu su c ng trong t. y cn ch vn sau: Nu ct nc tn tht mt hng c c chiu di bng s l hf th hai hng c cng c chiu di nh vy b tr hng n cch hng kia vi mt khong cch ln hn 2s, ct nc tn tht s bng 2hf (vi cng mt lu lng q). Nu khong cch gia hai hng c trn nh hn (1,5 2,0)s, th tn tht tng cng v ct nc trn hai hng c trn s nh hn 2hf tc l trong trng hp ny v mt thm hng c s c s dng khng hon ton.

Khi b tr c nn p, cn phi xt n tnh thm nc ca chng do s lin kt khng kn ca cc vn c. Khi thi cng ng c vo trong t phi nht t dnh vo cc ngm c khe h cc lin kt gia cc vn c l nh nht.

Khi thit k ni tip u c vi phn b tng ca p, phi d kin hnh thc kt cu ca phn ni tip sao cho cc lc thng ng t thn p khng truyn xung c. Khi xem xt kh nng truyn lc ngang ln u c t pha cng trnh cn ch cc iu kin sau y:

- Lc ngang hng v pha h lu c th y nghing u c v pha h lu v phn trn ca u c v pha mt thng lu ng thm c th ngn i ;

- Lc ngang truyn ln u c trong thi gian khai thc cng trnh c th c gi tr thay i tu theo ct nc tc dng ln cng trnh ;

- Khi cc hng c c chiu di kh ln (c su) v ngm ni tip gia cc vn c c gii quyt kn nc tt, vic truyn lc ngang ln u c khng nguy him nh trng hp c ngn ;

- Trong mt s trng hp khng truyn lc ngang ln u c thng lu di p, khng nn ni trc tip hng c ny vi chn khay thng lu p m nn ni vi phn cui ca sn ph ni tip vi chn khay ni trn.

Vic s dng c kim loi mi trng n mn phi c lun chng ring.


83

Chiu di (chiu su ng c) ca c di sn ph v c thng lu di p khi chng l c treo, phi c xc nh trn c s tnh ton bn thm ca nn. Khi tnh ton phi so snh cc phng n ng vin c kh nng chng thm tng ng nhau nh c cc chiu di ca sn ph v c khc nhau (th d cc phng n c sn ph tng i di v hng c ngn v cc phng n sn ph tng i ngn v hng c di).

Bn c thng c lm bng thp, b tng ct thp, b tng ct thp d ng lc cn bn c bng g t c s dng.

Chiu di ca c chng thm bng thp phng khong 1225m, cn chiu di ca c hnh lm sng c th ti 50m (bng cch s dng thit b rung hoc bng tc ng ca vic xi t).

Tng chng thm bng b tng ct thp v b tng ct thp d ng lc thng c dng nhiu hn so vi tng chng thm bng thp v chng c sn xut, thi cng ngay ti cng trng.

dy ca tng chng thm t 1050 cm.

Hnh 2-7. Lin kt bn c vi bn y s dng n cht

1. c; 2. giy du; 3. vn nh hng; 4, 6. vn chn;

5. np ng t cch nhau 1 m; 7. hn hp bitum ;

8. bao ti; 9. h to sn cch nhau 4 m; 10. mc

va trong h.

i vi tng chng thm chn khay cc b (c treo), khong cch gia cc tng chng thm khng c nh hn tng chiu di ca bn c, v trong mt s trng hp ngoi l khng c nh hn 0,75 gi tr ny. Trong cc trng hp khc, tc ng ca bn c gim mt cch ng k. Cc bn c c lin kt vi bn y theo cc cch sau: phn trn ca bn c c gn vo cc nm (cht) khng thm c bit, cc cht ny c nh v trong b tng bn y. S hnh thnh ca cc khe nt trong matt ph thuc vo thi gian v s phn tch ca cc khp ni. iu ny c th gy ra cc l hng m nc c th chy qua lm cho chn khay khng cn tc dng.V vy ta cn phng bng cch un nng matt cho chy vo cc cht thng qua cc h d phng.

III. Tng, mng chng thm v chn khay su. Ngi ta thng s dng cc kt cu ny khi cc iu kin v a cht khng cho php

xy dng mt tng tm. Chng hn trong trng hp c mt hm lng ln tng, v.v... trong nn. Chng c to ra: (a) bng cch ct cc rnh (tng) trong cc h o, y nhanh nc ngm ra ngoi, (b) bng cch ng cc rung hoc cm cc tr trn c ng knh ln ni vi cc lin kt c bit hoc cm cc khun dn nhm hnh thnh mt tng chn khay lin tc; (c) bng cch b tng di nc vo cc rnh c o bng gu xc, my o...; (d) bng cch s dng cc thng chm, y l mt phng php him khi

10

9

87

6

5

4

3 12

4045


84

c s dung trong k thut thu li; (e) bng cch bm ximng v xi mng st vo ct v cui ct vi h s thm khng ln hn 10-1 cm/s.

Cc chn khay b tng dng ni tip tt hn gia p v nn (nhm mc ch ngn nga thm tip xc nguy him). Cc tng rng b tng su phi c b tr thay cho hng c trong trng hp khng th ng c c vo t nn hoc trong trng hp cng trnh c bit quan trng.Thng b tr chn khay hoc tng rng thng lu di p.

Chn khay h lu di p c b tr tch thit b tiu nc di p khi h lu v c th bm nc t thit b tiu nc di p v h lu bng my bm t trong hnh lang kim tra b tr trong thn p. Vic bm nc khi thit b tiu nc l cn thit, th d kim tra s lm vic ca thit b tiu nc.

Khi s dng tng rng thng lu, c th b tr tng rng h lu cm su xung di tn tng khng thm nc nhng phi b tr cc l thot nc tng ny nhm m bo ct nc di p ng vi mc nc h lu.

Cc tng rng su chng thm bng b tng thng thng cn c tch khi phn mng p bng khp ni bin dng c vt chn nc tng ng.

Khi b tr c h lu do c khe h gia cc vn c, chiu su chn khay h lu d, phi tho mn iu kin d 2b, trong l chiu rng vn c.

Mng pht chng thm: cc mng chng thm ny c thc hin vi nn khng phi l bng cch pht vo khong rng ca t nn va xi mng, va t st c ph gia ho do, va xi mng-plime, va plime, v.v...Chiu dy ca mng chng thm k t trn xung di phi gim dn.

C th s b coi nh i vi mng chng thm, gradien chng thm ln nht cho php (khi nc thm qua mng chng thm theo hng ngang) bng 2 3. Trong t bi tch cc mng chng thm c th b tr n su bt k.

Hnh 2-8. Chn khay thng lu di p

I- b tng st; II- thit b tiu nc ; III- tng lc ngc ; A - t c m nn cht

IV. mng p su tm y p trong nn c xc nh bng tnh ton tnh hc v tnh thm. V

mt n nh ca p, nu c th phi b tr mng p ln tng t tt, c tr s h s ma st trong ln.

Trong trng hp s p khng c thit b tiu nc (hnh 2-9) xut pht t tr s d=Sra xc nh theo cng thc :

Sra = (0,05 0,10)T , (2-4) nhng khng ln hn :

II III

I dA


85

Sra = (0,05 0,10)lo , (2-5) ng vin ca p c th thit k theo mt trong hai phng n sau y:

- p c cc chn khay (hnh 2-9a)

- p khng c chn khay (hnh 2-9b)

Hnh 2-9. Thit k mng p.

I-tng lc ngc

V l do kinh t, nn p dng phng n th nht (hnh 2-9a), n nh tr s d vi tnh ton lm sao on MN ca p nm trn t tt v t thm nc. Ch trong trng hp gp loi t kh o ho cho chn khay th mi loi b phng n (hnh 2-9a) v chuyn sang phng n p khng c chn khay (hnh 2-9b).

Cu to ch i ra ca dng thm h lu. Trong vng mt ct t ch dng thm i ra bao gi cng phi b tr thit b tiu nc lc ngc bo v.

Lc ngc cn phi c thit k theo tiu chun thit k tng lc ngc cng trnh thu cng.

Thit b tiu nc nm ngang b tr di sn tiu nng, di p v di sn ph phi c lm bng vt liu ht ln. Chiu dy nh nht ca thit b tiu nc theo yu cu v cu to v thi cng quy nh bng 0,2m. Vic dn nc t thit b tiu nc v h lu cng nh kh nng tiu nc ca n (c xt n kh nng cho nc qua lc ngc) thng thng phi c thit k sao cho tn tht ct nc khi chuyn ng dc thit b tiu nc l khng ng k. Vi iu kin trn, ct nc dc theo ton on ng vin di t thc di cc b phn thm nc ca cng trnh, thc t s ng vi mc nc h lu.

Thit b tiu nc cng vi lc ngc phi c p cht xung nn bi trng lng cc b phn bn trn ca cng trnh. iu ny c bit quan trng trong trng hp nn l loi t st c kh nng mt dn bn b mt khi khng c ti trng.

Nn chn cu to ch ra ca dng thm nh gii thiu (hnh 2-10a) vi tr s Sra ln. Thng thng khng c php s dng s b tr thit b tiu nc nh (hnh 2-10b) vi Sra=0 ; s b tr thit b tiu nc nh (hnh 2-10c) vi tr s d ln c th chp nhn c.

Mc nc TL

Mc nc HL

M Nd'

lo

d

Mc nc TL

Mc nc HL

d

b)a)

IT

I

lo T


86

Hnh 2-10. Cu to ch ra ca dng thm h lu

2.4 Tnh ton Thm vng quanh, thm vai p b tng ni tip vi b

I. M t dng thm vng quanh tr bin.

Trong trng hp chung, khi mng tr bin khng t trn tng khng thm (tng khng thm nm kh su) v c thm vng quanh tr bin, s chuyn ng ca nc ngm s c dng khng gian. Khi , cng vi dng thm c p di p, cn c c dng thm khng p vng quanh tr bin.

Hnh 2-11. Tr bin c tng cnh thng gc

I- p trn ; II- Tng dc ca tr bin; III- Tng khng thm;

IV- Cc ng dng; V- ng ng p

hT

MNTL

h

MNHL

h

OO Z

C

D

1 2

3 4m

III

h

T

A

B

C

D

AB

3

2

4

6

5

1

IV V

V

III

Mt bng

A - B C - Db)a)

c)

1

Sra

raS =0

A

a)

b)

c) ra = -Sd


87

hnh 2-11 nhng ng dng l ca phn dng chy khng p v cc ng ng p ca dng thm. Tit din t ca dng vo l mi dc v y thng lu; ca dng ra l mi dc v y h lu.

Nu nh mng ca tr bin khng tip gip vi tng khng thm nm su, th c th xut hin thm dng thm bn p di cc tng ca tr bin.

Trong mt s trng hp, c th xy ra dng thm b sung, t b ra h lu.

C th coi nh ng bo ho bao quanh mt trong ca tr bin (ng vin di t 1-2-3-4-5-6, hnh 2-11c); phn bo ho chy theo tng dc ca tr bin c biu th trn hnh 2-11a (ng 3-4). R rng l phn ng bo ho ny quyt nh tr s p lc ca nc ngm ln tng dc ca tr bin.

Nu nh v ng dng thm theo ng 1-2-3-4-5-6 (hnh 2-11c) ri trin khai n ra trn mt mt phng, th ta nhn c hnh nh nh hnh 2-12. Hnh nh ny tng t nh hnh nh dng thm qua p t trn nn thm nc.

Hnh 2-12. ng bo ho quanh tr bin-I

Nh vy, khi tnh ton thm vng quanh tr bin, ta c th p dng phng php ging nh khi tnh ton thm qua p t trn nn thm nc.

II. Tnh ton dng thm qua p trn nn thm nc. (theo phng php ca S.N. Numrp)

1) Trng hp p t ng cht:

Khi trnh by phng php tnh ton ny, ta s dng hnh 2-13 biu th mt ct ngang p t khng c thit b tiu nc, trn nn thm nc.

Hnh 2-13. S tnh ton thm ca p t

hT h

MNTL

MNTLZ

1

2 3 4 5

6

I

12

h

T T hh

1

2

x

A' A

4' 1'

0,4h L0,4h

L

MNTL

ab

B

m2

m 1

M

3' 2'

K

q

2" 3"

4"1"

B' MNHL

N

0

1 2


88

Ta k hiu v tr mp nc tng ng thng lu v h lu A v B. t v pha phi ca A v B nhng on tng ng bng 0,4h1 v 0,4h2, y h1 v h2 l chiu cao mc nc thng lu v h lu so vi mt tng khng thm MN. Kt qu l ta nhn c mt khi t hnh ch nht 4' - 4" - 3" - 3' nm trn tng khng thm MN. Bit chiu su nc thng lu, h lu khi t ny (h1, h2), ta tm c lu lng n v ca dng thm qua p t ang xt theo cng thc ca Duy-puy:

KL2

hhq

o

22

21

-= (2-6)

trong : Lo = Lyp + 0,4h1 + 0,4h2 , (2-7)

Lyp - khong cch theo mt nm ngang gia cc im mp nc A v B ;

Lo - chiu rng ca khi t p ch nht m ta thay th cho p t.

Nh bit, khi thay th nh trn, ta coi tn tht nc nm thng lu p v nn ca n bng tn tht nc trong khi t hnh ch nht 1" - 2" - 3" - 4", rng 0,4h2.

Bit lu lng q - xc nh theo cng thc (2 - 6) ta thit lp ng bo ho A"-B" i vi khi t hnh ch nht quy c 4' - 4" - 3" - 3' bng cch dng cng thc ca Duy - puy:

( )2221o

21 hhL

xhh --= (2-8)

trong : x v h - cc kch thc nh biu th (hnh 2-13).

Cui cng, ta la bng mt un thm cc on cong cha bit A - a v B - b sao cho A - a vung gc vi mi dc thng lu ti A, b - B tip tuyn vi mi dc h lu ti B ( y b qua on dng thm i ra mi dc h lu).

Kt qu ta s c ng bo ho A - a, b - B i vi p trn nn thm nc.

Khi phn nm h lu ca p c b tr thit b tiu nc th theo quan im thu lc ta s c p t c mi dc h lu thng ng a - b t theo trc thit b tiu nc. Khi quy i p t loi ny thnh khi t ch nht, ta s c hnh dng ng bo ho nh hnh 2-14.

Ghi ch: Trong trng hp nm h lu khng c thit b tiu nc, tr s 0,4h2 khng phi tnh t ng thng ng 1" - 2" i qua mp nc h lu nh (hnh 2-13), m l t ng thng ng k qua im gia on dng thm i qua mi dc h lu.

Chiu cao So ca on dng chy i qua mi dc h lu trong trng hp mi dc h lu kh (khi h2 = T) c th xc nh theo cng thc:

( )Kq

m7,0S o2o +=

trong : ( )1'o22

1o

h4,0L2Th

Kq

+-

=

y, 'oL - khong cch nm ngang t mp nc mi thng lu ti chn dc h lu.


89

Hnh 2-14. S tnh ton thm ca p t c mi dc h lu thng ng

2) Trng hp p c li gia:

y cng nh trng hp trn, nm thng lu v h lu p (cng vi nn ca n) c thay th bng cc khi t hnh ch nht (hnh 2-15, ch gch cho). Kt qu nhn c mt khi t hnh ch nht 4' - 4" - 3" - 3' c li gia.

Hnh 2-15. S tnh ton thm ca p t c li gia

Ta xt khi t hnh ch nht ny theo phng php quy c m nhiu ngi bit ca N.N.Pavlpski. Sau ta hiu chnh ng bo ho nhn c t khi t hnh ch nht ny v tim ng bo ho cn bit.

Ghi ch: y ta khng xt trng hp m li gia khng t ti tng khng thm. 3) Vng hot ng thm nn p:

p dng phng php nu trn, c th dng ng bo ho i vi p t theo phng trnh Duy - puy trong iu kin khng thm mt su hu hn. Tuy nhin, tng khng thm trn thc t c th nm su v hn. Trong trng hp ny, dng ng bo ho, phi s dng khi nim vng hot ng thm nn p.

Nu tng khng thm nm su th chiu dy vng hot ng thm ly bng:

Thng = 0,5L", (2-9)

trong : L' - chiu rng p mt nn ;

Thng - chiu su vng hot ng thm di mt y h lu.

Sau khi tnh Thng theo (2-9), ta xc nh v tr tnh ton ca tng khng thm ( v ng bo ho) nh sau:

hT

MNTL

m 1

1

T h

0,4h1

a

0,4h2

2

b

MNTL

0,4h1 0,4h2

h

T

1

MNTL

hT

2

Li gia

A

B

3' 3"

4' 4"


90

a) Nu: Tthc Thng (2-10) th tr s Ttt ly bng:

Ttt = T ; (2-11)

b) Nu nh Tthc > Thng (2-12)

th tr s Ttt ly bng:

Ttt = Thng = 0,5L' (2-13)

trong : Tthc - chiu su thc ca tng khng thm ;

Ttt - chiu su tnh ton ca tng khng thm.

Cc tr s Tthc v Ttt o t mt y h lu.

Cn xt n cc trng hp sau:

a) Khi Tthc > Thng th v tr ng bo ho thc t khng ph thuc vo v tr ca tng khng thm;

b) Khi 0 < Tthc < Thng (2-14)

th khi Tthc tng, ng bo ho ca p t s gim mt cht.

c) Khi tho mn iu kin (2-14) th v tr ng bo ho du sao cng t ph thuc vo v tr ca tng khng thm. Do , khi tnh ton s b, thin v an ton, tr s T i khi ly bng 0, tc l s v ng bo ho i vi p xy dng trn nn thm nc vi gi thit rng nn khng thm nc (lc ny tt nhin phi p dng phng php trnh by trn).

III. n gin ho vic lp ng bo ho khi thm vng quanh tr bin.

chuyn dng thm, nh m t im 1, sang dng gi l "dng thm phng", (Khi tnh ton c th p dng phng php gii bi ton thm phng ca F.Forkhgymer), ta thc hin gi thit n gin ho nh sau:

1. Ph hp vi im 2, 3 nu, ta coi rng:

khi: Tthc 0,5% (2-15) th tng khng thm tnh ton trng vi tng khng thm thc ;

nu nh: Tthc > 0,5% (2-16)

v tng khng thm tnh ton nm su di y h lu bng:

Ttt = 0,5'ol (2-17)

y: 'ol - chiu di hnh chiu ca tr bin theo hng trc tng dc.

2. Ph hp vi 2 im bn trn ta thay cc mi dc thng lu, h lu ca khi t tip gip vi tr bin bng cc mi dc thng ng chm tng khng thm.

Ta hy v nhng mi dc thng ng tnh ton ny cch mp nc 1 khong nh sau (hnh 2-16)

a) i vi mi dc thng lu:

a1 = 0,4h1 ; (2-18)

b) i vi h lu:


91

a2 = 0,4h2 ; (2-19)

trong : h1 v h2 - vt cao ca mc nc thng lu v h lu trn tng khng thm tnh ton.

Bng kt qu ca s n gin ho ny, tu theo hnh dng kt cu ca tr bin, ta c th nhn c cc s khc nhau ca dng thm trn mt bng hnh 2-16 gii thiu mt vi v d v cc s nh vy.

Hnh 2-16. n gin ho cc dng hnh hc ca tr bin

3. Ta hy quy c rng tt cc cc tng ca tr bin c chn su ti mt bng khng thm tnh ton.

4. t p sau lng tr bin c coi l ng nht v ng hng.

5. B qua dng thm ngm chy t b ra, ch xt nc thm t thng lu v h lu (hnh 2-11).

6. Cui cng, b qua on nc chy ra t mi dc thng ng (tnh ton) h lu; tr s ny trong trng hp ny s rt nh.

Khi s dng nhng gi thit nu ta nhn c dng thm tnh ton c c trng mt cch gn ng bi cc tit din t hnh tr vi cc ng sinh thng ng; cc ng dn hng ca nhng tit din t ny s l cc ng ng p ca mt gim p (mt bo ho).

IV. Lp ng bo ho quanh tr bin theo phng php ca F.Forkhgymer. Tm y tng tng.

Theo phng php ca F.Forkhgymer, xc nh chiu su ct nc h (tnh t ng bo ho n tng khng thm tnh ton) mt im m no y trn ng bo ho (hnh 2-11), c th vit phng trnh sau:

l '

aa2

1

Mp nc

o

Mp nc Mp nc

l 'o

A1A2

4III

1

B1 B2

2 3I IIa1 2a

zv

z

zc 12

Mp nca1

2a

12

Mp nca1Tiu nc

III

aa1 21 4

2

a) b)

c) d)

e) f)

3


92

( ) 22r22212 hhhhh +-= (2-20) trong :

hr - ct nc tnh i im tng ng ca tm y tng tng c ng vin di t ging nh ng vin di t ca tr bin (hnh 2-16) khi tng khng thm su v hn T = v khi mt chun O - O nm ngang vi mc nc h lu. Tm y tng tng dng cho cc s tr bin (hnh 2-11) v (hnh 2-16a), c trnh by trn (hnh 2-17). Hnh 2-17 ch ra im m tng ng cng nh mt chun O - O.

Hnh 2-17. Tm y tng tng (c bn thm tng ng vi tr bin ang xt)

ng vi mt chun nu, tr s hr i vi im m ca s (hnh 2-17) (c c trng bi dng thm p lc), bng:

Zhh fr = (2-21)

trong :

Z - ct nc trc tm y tng tng, ly bng ct nc Z tr bin (bng chnh lch mc nc thng h lu, hnh 2-11) ;

hf - ct nc im m ca tm y tng tng hoc bng chnh tn tht ct nc t im m n h lu (hnh 2-17).

Xt rng (hnh 2-11):

h1 - h2 = Z (2-22)

Th t (2-20) v (2-21), ta c cng thc tnh ton dng v ng bo ho quanh tr bin nh sau:

( ) 22f21 hhhhh ++= (2-23) y, tr s hf i vi im m cho phi c xc nh t s xem xt y tng tng ng vi ct nc Z, bng ct nc tc dng ln tr bin.

Tr s hf i vi im m bt k ca mt s s tr bin no nu trong (hnh 2-17), c tm theo phng php h s sc khng. Bit hf i vi cc im khc nhau ca ng vin di t ca tr bin, theo cng thc (2-23), c th tm chiu su h cc im y v theo v c ng bo ho quanh tr bin.

Mc nc TLP

PP

Zh0 0

a' 1

2

3 m 4 6

5

b'

T =

fMc nc HL


93

V. Cc nhn xt b xung v cch lp ng bo ho quanh tr bin

1. Gi thit ch yu nht trong s nhng gi thit nu im 3 l gi thit thay cc phn mng "treo" ca tr bin (nu c) bng cc phn mng quy c cm xung ti tng khng thm tnh ton (im III.3).

Gi thit ny khng thin v tnh ton.

V l do trn, cng nh xt nh hng ca tng khng thm tnh ton i vi v tr ng bo ho (im 2) khi c cc phn mng ca tr bin khng cm xung ti tng khng thm, nn bo m an ton, ta n nh tng khng thm tnh ton nm ngang mc y h lu (gi thit T = 0). Lm nh vy vic tnh ton s n gin hn phn no.

2. Khi tnh tr bin, c th gp trng hp tng dc AB ca tr bin tip xc vi li gia bng t loi st t thm nc hoc vi hng c (hnh 2-18) khi , tnh ton, tr bin thc cn c thay th bng "tr bin quy c".

Hnh 2-18. Trng hp tr bin ni tip vi li gia bng t st hoc

mng ngn di dng hng c

Khi chuyn t tr bin thc sang tr bin quy c cn b li gia (hoc bng c thng ng) ri ko di tng dc ca tr bin thm 1 on lB nh sau:

a) Trng hp khi b li gia:

( )1

B KKtx1ll +-= (2-24)

b) Trng hp khi b tng c

f+= llB (2-25)

trong :

l - chiu di thc ca tng dc tr bin ;

t - chiu dy trung bnh ca li gia ;

K1 - h s thm ca t lm li gia ;

K - h s thm ca t cn li ;

f - chiu dy quy i ca hng c .

3. giai on tnh ton nht nh, cc s tr bin (hnh 2-16), cn xem nh cc s tm y tng tng vi T = , vi ct nc trn chng l Z bng ct nc tc dng vo

Li gia

A B

C

A B

a) b)l l

t


94

tr bin. chnh t vic xem xt cc s nh vy m ta xc nh c cc tr s hf trong cng thc (2-23).

Trn hnh 2-17 trnh by mt tm y tng tng tng ng vi s tr bin trn (hnh 2-16a) s tm y tng tng ny d dng gii c theo phng php h s sc khng.

Ta hy gii thch thm cch tin hnh gii mt s s c th trnh by trong hnh 2-16b, c, d, e, g theo phng php h s sc khng:

a) S hnh 2-16b:

Khi xt s ny, ta cn bit h s sc khng i vi on nn ca tm y tng tng I v II b gii hn thng v h lu khng phi bng cc ng nm ngang m bng cc ng cong A1B1 v A2B2.

V i vi nhng on nh vy, ta khng c cc tr s h s sc khng z, nn c th tin hnh nh sau i vi trng hp ny:

Vch tit din thng ng I - II (hnh 2-16b). Trong tit din ny ta nhn c khi t hnh ch nht. Tnh ton ng bo ho cho khi t ny theo phng trnh ca Duy-puy (vi gi thit l bi ton phng), ta d dng c th tm c tn tht ct nc Dh1 trn chiu di 1-2 ca khi t v tn tht ct nc Dh2 trn on 3-4.

Sau , khi tnh ton tr s hf ta xt tm y tng tng (hnh 2-19a), coi ct nc tc dng trn n bng (Z = Dh1 - Dh2).

Ghi ch: nh r thy trn (hnh 2-19a) nn ca tm y ny c gii hn bi cc ng thng, ng 1-1 v 2-2.

T vic xem xt tm y tng tng nh vy, ta nh c tr s hf i vi cc im khc nhau ca ng vin di t theo phng php h s sc khng.

Hnh 2-19. Cc s b sung ca tr bin

Hin nhin l h s sc khng i vi cc on I v II s bng 0,5zc, y zc l h s sc khng i vi hng c thng thng bn trong.

S

Z= h - h

MNTL

MNHLD D1 2

1'

1

2'

2

I' II'

l

Tiu nc

S

A C

B

E D

S

l

Tiu ncAB F

E

C

D

a) b)

c)


95

b) S hnh 2-16 c,e:

Cc s ny c th tnh ton theo phng php h s h s sc khng bng cch gii nu im trn.

i vi s hnh 2-16c, cng c th lm nh sau: n nh tit din t quy c ch vo, tit din nm ngang v thng ng (ng nt t trong hnh v).

i vi tit in vo nm ngang, ta tm tr s zvo ; i vi tit din t vo thng ng, ta tm h s sc khng bng 0,5zc.

Tr s cha bit z i vi tit din t vo thc (tit din nghing) tm c bng cch ni suy gia tr s zvo v 0,5zc'

c) S hnh 2-16d:

y, ta cn xc nh tr s z i vi cc b phn hi nghing (khng phi nm ngang) ca ng vin, v d nh b phn 1-2 ca ng vin.

R rng l khi xc nh z trong trng hp ny cn phi s dng cng thc i vi h s sc khng ca b phn ng vin nm ngang zng, sau khi thay l bng hnh chiu ca ng 1-2 trn ng nm ngang v thay T bng gi tr trung bnh no ca tr s ny.

Ghi ch: trng hp ny cng thc nu, cc i lng S = 0.

d) S hnh 2-16f:

Trong trng hp ny cn v thm ng thng, ng 1-2-3. i vi on nn nm bn tri ca ng 1-2-3, h s sc khng phi ly bng na tr s h s sc khng i vi hng c n (0,5zc), coi chiu su ca hng c ny bng chiu di ca on 1-2. B phn nghing 2-4 ca ng vin di t cn xt nh nu im trc.

e) S hnh 2-19b:

S ny ca tr bin c bin i thnh tm y tng tng, c trng bng ng vin di t BED; tit din t ch vo l ng thng ng AB, ch ra l ng thng ng vch theo trc thit b tiu nc ca tr bin.

Tng h s sc khng ca s ny bng:

ngc21

z+z=zS (2-26)

trong : zc - h s sc khng i vi hng c thng thng bn trong ;

zng- h s sc khng ca on nm ngang, tnh theo cng thc:

nghdng T

2/Sl -=z (2-27)

Cc kch thc l v S ch ra trong hnh v:

Thng - chiu su vng hot ng thm.


96

g) S hnh 2-19c:

S ca tr bin c bin i thnh tm y tng tng c trng bng ng vin di t AEFC; tit din t ch vo l AB, ch ra l CD (vch theo trc thit b tiu nc ca tr bin).

Tr s Sz i vi s ny bng:

44,0T

2/Sl

dc +

-+z=zS

ngh

(2-28)

y, cc k hiu l v S ch ra trong hnh v; cc k hiu cn li trnh by trn.

2.5 Cu to p v b tr ni tip h lu

I. Cu to p. p c chia thnh nhiu phn bi cc khe ln trnh nt ngang p do hin tng

ln khng u v cc bin dng do nhit.

p thng c chia cng vi ton b cc lin kt nm ngang theo dc tr pin trnh hin tng ln khng u gia cc tr (hnh 2-20e) dn n kt ca van.

Bn cnh cc khe ln c nh, trong thi gian thi cng ngi ta cng thng chia p thnh cc khi khc nhau bi cc khe thi cng.

Chiu rng ca cc khe ln c nh c la chn sao cho khng mt b phn no b y chng ln nhau do s ln khng u v s bin dng ca cc b phn.

Thng thng, khe lin kt c thit k nh sau: gn bn y, chiu rng ca khe l 12cm pha h lu v 34cm pha thng lu, phn trn ca bn y mng l 1015cm. Chiu rng ca khe lin kt ph thuc vo c trng a cht ca nn mng v s thay i nhit ca mi trng (hnh 2-20).

(a) (b) (c)

(d) (e)

Hnh 2- 20. Phn chia p thnh cc n nguyn

(a) mt khoang; (b) hai khoang; (c)3 khoang; (d) tr pin ct ri trn;

(e) phn chia dc theo tr pin; 1- khe rnh; 2- ngng trn;3- tr; 4-bn y;

5-khe ni gia cc n nguyn

1

2 3

4

5


97

Tnh khng thm nc ca cc khe c m bo bng cc thit b chn nc dc, ngang, n v kp (hnh 2-21). Cc thit b chn nc c b tr di p di dng bao

ti nha ng. phn trn, cc khp ni c lm bng thp khng g, cao su, cc tm b tng ct thp c t trong cc h ng y asphan.

Hnh 2-21. V tr ca vt chn nc trong cc khe gia cc b phn ca p

1-lp lt y; 2- cht ng vin thm trong; 3- ng phn cch gia phn rng v hp ca khe; 4- cht ti b phn hp nht ca lin kt; 5- cht ti b phn rng nht ca khe;

6- vt chn nc pha bn ngoi; 7- ging nha ng

II. B tr ni tip h lu. Khi thit k p x nc trn nn mm, phi dng ch chy y lm ni tip thng

h lu ch yu, khi vng dng chy b co hp, trn on tiu nng phi d kin cc kt cu tiu nng v phn dng.

Trong b tiu nng nn s dng cc kiu vt tiu nng chnh sau y:

- Tng tiu nng lin, t cch mt ct co hp mt on bng 0,8 chiu di ca nc nhy (chiu di ca nc nhy xc nh bng tnh ton vi b tiu nng) hoc cch mt

khong 3h khi c tr s K

00 h

T=e bin i trong phm vi 0,2 12 (trong h l chiu su

dng chy cui on nc nhy).

T0 - t nng ca dng chy trc cng trnh chu p bng hiu ca mc nc thng lu c xt lu tc n gn v mt b tiu nng ;

hK - chiu su phn gii ca dng chy.

- Tng tiu nng t qung, b tr cch mt ct co hp mt khong 3h khi tr s e0 = 2 6.

- Tng tiu nng phn dng, gm hai tng to thnh mt gc hng ngc dng chy, gc t cc tng phn dng c th bin i trn chiu rng ca b tiu nng, cn bn thn tng phn dng c th c chiu cao thay i.

- Vt tiu nng c dng kt hp hai m hnh thang v tng tiu nng pha h lu.

Chiu di sn sau (nm ngang, nm ngang c mt on nghing hoc nm nghing) phi c xc nh t iu kin lm cho cc biu lu tc ca dng chy c san bng dn trn ton b chiu di ca sn sau hoc trn mt phn ca n (sn sau rt ngn).

i vi nhng p cp I, II v III thng thng phi thit k sn sau dng cc tm btng hoc btng ct thp ti ch.

i vi nhng p cp IV v V cho php d kin sn sau dng hoc r , tm btng hoc btng ct thp lp ghp.

21

3

4

5

6

7


98

Trng hp dng cc tm btng v btng ct thp c sn lm sn sau, phi d kin lin kt chng bng ct thp m bo tnh n nh ca chng chng li tc dng thu ng ca dng chy.

Chiu dy ca cc tm btng b tiu nng v sn sau phi c xc nh bng tnh ton, xut pht t iu kin m bo cng v n nh ca chng. Phi xem xt kh nng gim chiu dy cc tm b tiu nng v sn sau bng cch phn nh nh cc khp ni nhit-ln v b tr cc ging tiu nc.

Kch thc trn mt bng ca cc tm phi c xc nh t iu kin m bo s n nh chng trt v y ni, cng nh m bo kh nng btng mi tm thnh mt khi.

Ging tiu nc phi c tit din trn mt bng t 0,25x0,25 n 1x1m tu theo b dy ca tm b tiu nng v sn sau, cng nh iu kin thi cng.

Trn mt bng cn b tr cc ging theo kiu hoa th trong mt hng c cch 5 10m lm mt ging (tu theo kch thc ca cc tm) v cc hng ging khng nh hn 5m, ng thi din tch cc ging tiu nc khng c nh hn 1,5% din tch ton b cc tm gia c.

Khi dng cc tm btng hoc btng ct thp lm sn sau, cho php khng lm ging nc.

cui sn sau phi d kin b tr mt kt cu c dng tng thng ng, hoc rnh phng xi, hoc phn gia c chuyn tip c th bin dng c, hoc t hp cc kt cu bo v cho sn sau, cc m bin v tng phn cch khi b xi l.

Tng thng ng cui sn sau (c dng tng btng hoc btng ct thp, tng c kt cu phng hoc t ong, ci g trong b v.v...) phi c thit k cm xung ht chiu su ca lp t c kh nng b xi l. Khi chiu su xi l qu ln, c th lm tng ng khng cm xung ht xung ht chiu su xi l, nhng phi lm thm mt on gia c chuyn tip mm c kh nng bin dng sau tng .

Khi dng chy c t lu ln v t nn l loi t d b xi l, phi d kin b tr rng phng xi cui sn sau, cng vi phn gia c chuyn tip mm mi dc pha thng lu v y rnh phng xi.

Vic xc nh mi dc pha h lu ca rnh phng xi phi xut pht t iu kin n nh ca n trong thi gian thi cng.

Mi dc pha thng lu ca rnh phng xi phi c quyt nh c xt n iu kin thu lc ca s tn dng, n s b tr mt on sn sau nm nghing hoc on gia c chuyn tip mm c kh nng bin dng.

on gia c chuyn tip mm c kh nng bin dng phi c thit k di dng cc tm btng v btng ct thp ring r lin kt bn l (khp) vi nhau, di dng si hoc , r , rng , rng cy hoc m cnh cy trn c hoc si, hoc di dng t hp cc kiu gia c trn.

Cn phi la chn kiu gia c trn c s so snh cc ch tiu kinh t k thut ca cc phng n nu ra, c xt n cc iu kin thu lc, chiu su xi cho php v cc yu t khc.


99

2.6 Tnh ton n nh ca p trn nn mm theo s trt phng

Khi tnh n nh p b tng trn nn mm, ta thng tnh ton theo 3 s : trt phng, trt su v trt hn hp. Trong mc ny ta xt s trt phng.

Mt trt c th l mt nm ngang hoc mt nghing. Trt theo mt nm nghing xy ra khi y p nm nghing hoc khi c tng chng thm (hnh 2-22b), n nh c tnh ton da trn mt trt . Nu mt lp c cc c trng trt nng nm khng qu su trong nn th n nh c tnh ton c i vi trt trn mt phng ca y p v trt trn mt tip xc ca t, c xem xt tho ng n trng lng ca khi t gia mt ct 1-1 v 2-2.

a) c)

b) d) Hnh 2-22. S tnh ton n nh ca p theo s trt phng.

(a,c,d)- mt phng trt nm ngang ;

(b)- mt phng trt nm nghing ; (0-0) - mt trt.

Hnh 2-22 m t cc trng hp khc nhau, c xem xt n trng lng ca khi t.

Khi mt trt xy ra trn mt phng nm ngang, h s an ton n nh Kt c biu din ph hp vi iu kin cn bng gii hn:

Kt= ( )[ ]

Q.nm.F.cEtg.WWV

N.nR.m

c

idnth

c

++j--= , (2 - 29)

trong : V- tng cc lc tc dng theo phng ng (trng lng bn thn, nc) ;

Wth, Wn - p lc thm v y ngc ;

c - lc dnh n v, F - din tch y mng ;

Ei - p lc t ch ng trc tip ln thng lu ;

Q- tng cc lc theo phng ngang ;

j - gc ma st trong, nc- h s t hp ti trng, m - h s iu kin lm vic (i vi kt cu b tng v b tng ct thp trn nn t thng ly m=1).

Khi mt trt nghing v thng lu, h s n nh ca p tng. Trong trng hp ny y p c thit k nghing hoc y ca chn khay h lu cn phi nng ln (hnh 2-23). Khi , h s m bo n nh trt c xc nh nh sau:

0 0Ea

0 0E

1

E

1

a b

0

0E

2

2

Ea

b0 0

E

1

1

0 0

2

2

Eba


100

Ks=( )

( ) cdnth

nVQ

FctgWWQV

.sin.cos...sin.cos.

-+j--+

, (2 - 30)

trong : F - din tch mt trt.

Hnh 2-23: S tnh ton n nh ca p, trt phng trn mt nghing

Sn trc neo c tc dng lm gim ti trng y ngang nh lc ma st v lc dnh gia sn v nn. Cc lc tc dng gm : p lc nc W1 bng trng lng nc tng ng vi s a, b, c, d (hnh 2-24); trng lng G ca sn trc v phn t pha di tm y thng lu; p lc y ngc Wn v p lc thm Wth tc ng hng ln trn. Ma st sinh ra di y ca mt on p l:

Ts = (W1t+G-Wth-Wn).m.tgj trong : m - h s xt n iu kin lm vic ca sn trc (m = 0,8).

Cng thc ny c s dng xc nh s n nh ca p c sn trc neo vo p v ch khc vi cng thc (2 - 29) bi c thm lc chng trt Ts.

Hnh 2-24. S tnh ton ti trng ln sn trc neo

A. sn trc neo ; S. c ; D. thot nc

Ks = ( )[ ]

Qn

TmFcEtgWWV

c

Sidnth

.... +++j--

. (2 - 31)

Trong cng thc trn, ly m = 1.

E

W

W

W + W

T

0

W

W

0

2

43

1

a

th dn

W

GSDA

b

a

c

H

L

W

W

h

H

12

34

567

h

d

1

th

n


101

2.7 n nh ca p vi s trt su v trt

hn hp I. Sc chu ti ca nn. Hnh 2-26 minh ha cc ti trng theo phng ng v phng ngang tc ng ln bn

y p. Cc ti trng ny bao gm ng sut php s v ng sut tip t ti y mng p. i vi s ct phng, mi quan h gia s v t trng thi gii hn c biu din di dng: t = s.tgj + c.

T suy ra gii hn ca ng sut tip t ng bin vi ng sut php s (hnh 2-25, ng OD).

a) b)

Hnh 2-25. S tnh ton trt su v trt hn hp

(a) cung trt ; (b) s thay i gii hn ng sut ph thuc vo ng sut php

Do cc ng sut tng, quan st mt trt phng ta thy n mt lc no ng sut t ti gi tr ti hn. Khi s > sgh s xy ra trt do mt n nh nn. Trong trng hp ny, khi s tng, thot u sc chng trt tip tc tng (phn AB ca ng ABC trong hnh v 2-25b) v sau gim. V tr ca im A trn ng thng t = s.tgj + c ph thuc vo loi t v lch tm ca hp lc tc dng ln y mng. Khi hp lc ny gn nh st mt thng lu ca y p th ng trt nm v tr su nht (hnh 2-25b).

ng cong trt 1, 2 v 3 tng ng vi cc im a, a1, a2 ca hp lc. Cc im A, A1, A2 trn th biu din quan h t = f(s) xc nh sgh v ph hp vi kh nng chu ti ca nn v sgh tng.

Trt hn hp (hnh 2-26a), trong trng hp ny, trt phng xy ra ti mt mng AB v phn trt cn li s xy ra cng vi s y tri ca t dc theo ng cong BCD - l trt su. Trong hnh 2-26, nn cng trnh c phn ra 3 vng r rt: vng I - vng bin dng do, vng II - vng dch chuyn c bn, vng III - vng trng thi ng sut b ng.

a) b)

Hnh 2-26. S s dch chuyn cung trt su v trt hn hp

Trt su v trt hn hp c quan st trong nhng trng hp khi cc iu kin sau y khng tho mn:

- i vi nn cha ct ht th, rn v lp t st na rn.

a a a3

12

1 2

AA

A 1

23

B

C0

gh

D

2

1

tang

B

245-

DIIIA B

CC

III

245-

1

2

IIIIII

A,B

C

D


102

N.B

1

max , (2 - 33)

( )

4..

.1.2 >

+

o

o

h

tK , (2 - 34)

trong : smax, stb - cc gi tr ng sut php ln nht v trung bnh ti y p ; B- chiu rng y p ;

g1, g - dung trng ca t nn (y ni) v trng lng nc ; N - Ch s m hnh ho, khi khng c ti liu th nghim ly N = 3 ; K - h s thm ca t nn ;

j - gc ma st trong ; c- lc dnh n v ca t ; t0 - thi gian thi cng ; e - rng ban u ca t;

a = 12

21

--

- h s nn ln ;

(e1, e2 - rng tng ng vi s1 v s2) ; h0 - chiu dy ca lp t c kt.

h0 = 2hh 21 + (h1 - chiu dy lp t cha st, h2 - chiu dy lp t gia y p v lp

cha st, h0 khng ln hn B, khi h2 =0, h1 ly khng ln hn B). Hin nay c mt s phng php tnh ton n nh cng trnh theo s trt su

v s trt hn hp.

II. Trt su vi mt trt tr trn trn nn ng nht. Khi tnh ton n nh ca p b tng theo s mt trt tr trn, mt trt gi thit

l mt cung trn (hnh 2-27) i qua im bin pha thng lu ca y p hoc chn khay thng lu, n nh ca p cng vi nn c quyt nh bi h s an ton Kt - t s gia tng m men cc lc chng trt v tng m men cc lc gy trt.

Hnh 2-27. S tnh ton n nh i vi mt trt trn gii hn, nn ng nht

T

G

VR

R

A

Q

V

B Q

ra

O

1

N1 T2

O1S1

Wth T3 N2

S2


103

Kt =

gt

ct

M

M (cc gi tr momen ly vi tm O ca cung trt).

Hp lc R theo phng ng (trng lng p, sn, nc pha trn v di khoang p, trng lng ca t pha trn mt phng i qua bin ca chn khay) v phng ngang (p lc nc tc dng t pha trn v pha di khoang trn) c di tng ng v mt phng nn AB v c phn tch thnh cc lc theo phng ng V v lc theo phng ngang Q. Cc lc V v Q c p dng cho cung trt v phn tch thnh cc thnh phn php tuyn v tip tuyn:

N1=V.cosb, S1=V.sinb; N2=Q.sina, S2= Q.cosa. Trn cung trt, trng lng ca phn t trt b y ni trong nc bng:

G = gn.2r.cos.sin

180.

a- ,

trong : gn=gbh - (1 - n).gn - dung trng y ni ca t. Gc a, b v bn knh r c ch ra trong hnh 2-27. Cc lc S1 v S2 c hng v pha di khoang p v c xu hng quay mt phn ca nn v kt cu xung quanh im O. Cc lc ma st T1 = V.cosb.tgj ; T2 = G.tgj v T3 = Q.sina.tgj s c tc dng chng trt. Trong trng hp t dnh, lc dnh tc dng ln tng phn ca y cung trt ly bng c = 2arc cng c tc dng chng trt. Trong biu thc ny, c l lc dnh n v. Ngoi ra lc thm tc ng ln mt phn ca nn. Lc ny c xc nh thng qua li thm (hnh 2-28). Phng ca lc th tch g.J.w tc dng ln mi mt li trng vi phng ca ng dng trung bnh trong . Lc thm tng cng s bng tng cc vect biu din lc thm thnh phn trong tng li.

Hnh 2-28. S tnh ton lc thm Wth

Khi nc = 1, m = 1, h s Kt bng:

Kt = ( )

r

aWQV

crtgGQV

th .cos.sin.

...2.sin.cos.

++

+j++b , (2 - 35)

trong : a - cnh tay n ca lc thm.

III. Trng hp nn khng ng nht . Nu nn bao gm cc lp c cc ch tiu c l khc nhau (gi, ci, ji) (hnh 2 -29) th tnh

ton n nh c th hin vi cc tiu chun khc nhau ca cc ch tiu c l t nn.

A

a

b

cd

B


104

W

0

A

V

B

T Q sin

Q

V

0'

N

a =n.bn

3

2

1

noV

III

III

-I-II

-III

l i

r

b bQ

q

n

1

1 2

2

aM P

n

n1Vn2Vn3V

nn

sn

1 1 1C

2 2 2C

3 3 3C

Vng t nn, s th hin cc ng sut sinh ra do trng lng ca kt cu v ti trng theo phng ng ln y p, sn v sn sau c chia thnh cc thi t theo phng ng c chiu rng b. Cc lc theo phng ng tc ng trong phm vi ca mt thi c

cng li v di n cung trt. Chng hn, Vn=Vno+Vn1+Vn2+Vn3 tc dng ln thi th n. y, Vno l ti trng do trng lng ca kt cu v nc tng ng vi s ca cc ng sut (do cc lc theo phng ng) tc dng theo chiu dc ca thi t, Vn1, Vn2, Vn3 l trng lng ca cc lp t bo ho 1, 2 v 3 trong mi thi t .

Thi t c ng tm OO k hiu l thi s 0, cc thi pha bn tri ca thi s 0 c k hiu du + gi l thi +I, +III, +III, ..., +n,...v cc thi pha bn phi k hiu l - gi l -I, -II, -III.,... , -n,...

Lc Vn gy nn m men quanh tm O: M (Vn) = an.Vn = n.b.Vn

Lc Vn nm y cung trt ca thi vi du + gy ra m men trt, vi du gy nn m men chng trt.

Cc lc ma st v lc dnh c tc dng chng trt. Thnh phn php tuyn ca Vn l Nn = Vn.cosbn.

Lc thm tc dng ln t, lm gim thnh phn php tuyn Nn bi gi tr :

Wtn =n

nn cosb.h. ,

trong : hn l ct nc p lc thm ti tm ca y thi.

Hnh 2-29. S tnh ton n nh ca p theo mt trt tr trn gii hn,

nn khng ng nht

M men ca lc ma st tc dng ln cung trt v m men ca lc dnh chng li s

trt: j-= rtgWNTM ntnnn .).()( ; = rlcCM nnn ..)( ;

trong : l1 , l2 ,.... ln l chiu di ca y cung trt ca thi ang xt.


105

Chng ta thay th cc thnh phn theo phng ngang W1 v W2 ca p lc ct nc thng lu v h lu bi Q. Bng vic p dng cc lc trc i Q ti im B, chng ta c m men ca cc cp lc Q.(r.cosa - q) = Qh v lc Q ti im B hng v pha thng lu. M men ca lc ma st r.tg.sin.Q)T(M tbQ ja= s xut hin do kt qu ca s phn tch trn. Gc ma st trong ca tng lp t nh hng ti cc gi tr ng sut tip trn phn y cung trt OB. V vy, ta thng biu din gi tr trung bnh ca gc ma st:

tgjtb = n

nn

llll

tgltgltgltgl

++++j++j+j+j

..........

321

332211 (2 - 36)

trong : l1, l2, l3...,ln l chiu di y cung trt ca cc thi trong on OB c cha cc gi tr ji khc nhau. H s an ton chng trt ca p m bo chng li s trt cng vi t nn c xc nh theo cng thc sau:

Kt = ( )[ ]

+

j++j-b

h.Qsin.V.rtg.sin.Ql.CtgWcos.V.r

nn

tbnnntnnn (2 - 37)

Nu mt lp t c gc ni ma st v h s trt nh (v d, lp 2 vi b mt MP trn hnh 2-29) nm v tr nng, th s mt n nh ca nn c th xy ra khng ch khi trt dc theo cung trt m cn trt dc theo mt lp t yu .

S tnh ton nh trn hnh 2-22d. Khi trt c gii hn t pha trc v pha sau bi cc mt thng ng 1-1 v 2-2.

2.8 Tnh ton ng sut y p

Vic tnh ton ng sut ti mt tip xc gia y p v nn p l nhm xc nh kh nng chu ti ca nn v tnh ton n nh ca cng trnh.

Hnh dng ca s ng sut php y p (hnh yn nga, parabol) ph thuc vo cng, chiu su ca nn v cc vng truyn ti. i vi nn c tnh n hi thay i, s ng sut php y c ch ra trong hnh 2-30 cho trng hp nn bin dng v ti trng tp trung. i vi nn c tnh n hi cao, s ng sut php c li ln th gi tr ng sut cc bin y nh. Tnh n hi ca nn c xc nh qua cng thc: (M.I. Gorbunov-Posadov)

31

30

hElE

10t , (2 - 38)

vi: E0 v E1 - m un n hi ca nn v ca vt liu thn p; l = 1/2 chiu rng hoc 1/2 chiu di ca nn ; h = chiu cao ca dm .

Khi t th nn cng tuyt i; khi t 0 th nn l n hi hon ton.

ng sut y p tng ng vi cc trng hp bin dng phng c tnh ton nh sau:


106

(a) Nu nn gm cc loi t khng dnh c cht tng i JD 0,5, th ng sut php c tnh t cng thc nn lch tm.

(b) Trong trng hp t dnh v khng dnh c JD > 0,5; ng sut y p c tnh t cng thc nn lch tm v s dng phng php l thuyt n hi, chiu su ca tng t nn b gii hn l: 0,3B i vi t ct v 0,7B i vi t st.

y, B l kch thc y p (t thng lu v h lu). i vi cng trnh cp III v cp IV trn t khng dnh v i vi cng trnh cp IV trn t dnh, ng sut y p ch c xc nh t cng

thc nn lch tm. ng sut php cc im gc ca mt mt ct p hoc vai p sa, sb, sc, sd, c xc nh bng cng thc nn lch tm:

si = y

y

x

x

W

M

W

M

F

V , (2-39)

vi: V- lc php tuyn (c xt n p lc ngc); F- din tch y; Mx, My, Wx, Wy- mmen un v m men chng un vi cc trc qun tnh ch yu ca y p.

Trong trng hp ti trng tc ng i xng i vi mt trong cc trc th:

Mx = 0 hoc My = 0.

2.9 Tnh ton bn ca thn p I. Lun im chung p b tng c chiu di ln thng c phn thnh nhiu on (n nguyn) lm

vic c lp. Cc on ny tip gip vi nhau ti v tr khp ni. Khp ni thng c t ti tr pin, n c chia thnh hai na gi l bn tr. Mi bn tr gn vi mt on ca p.

Mi on ca p c th c mt hay nhiu khoang (ca) trn. u mi on c th l tr bin (khi tip gip vi b) hay bn tr (khi tip gip vi on khc). Ngn cch gia cc khoang (ca) trong mi on l cc tr pin. Khi tnh ton bn ca p, ngi ta xt ring cho tng on.

Ni dung ca tnh ton bn l xc nh cc tr s ng sut ti cc v tr khc nhau trong thn p, trn c s tin hnh kim tra bn v cng v b tr ct thp khi cn thit.

Do c im cu to ca p, khi tnh ton bn ca mi on p cn phi xt theo kt cu khng gian. Vic gii bi ton khng gian tm gi tr ng sut ti cc im khc nhau trong mt on p thng gp nhiu kh khn v thng phi p dng mt s gi thit lm n gin ho bi ton v hnh dng kt cu, phn b ti trng, quan h vi nn...

Hnh 2-30. Cc s ng sut tip xc php tuyn i vi cc loi nn c tnh

do khc nhau.

P

t=0

t=1

t=51.0

0.8

0.6

0.4

0.2

P/l

2l

h


107

Do tnh gn ng ca cc kt qu tnh ton, vic kim tra iu kin bn ca kt cu cn phi m bo mt mc d tr cn thit.

Mc chnh xc ca s bi ton cng thng c phn bit theo tm quan trng (cp) cng trnh v giai on thit k.

1- Tnh ton bn chung ca cc on p cc cp ni chung v c bit khi cng trnh l cp I v II c tin hnh nh i vi kt cu khng gian trn nn n hi v c tnh theo cc phng php c hc kt cu hay l thuyt n hi. c ng dng ph bin nht hin nay l cc phng php s - phng php sai phn hu hn hay phn t hu hn.

2- Tnh ton bn ca p khi cng trnh l cp III, IVv V cng nh khi thit k s b cc p cp I, II c th tin hnh bng cc phng php c hc kt cu. Khi ngi ta xt ring cc bi ton kt cu theo phng dc (dc theo dng chy) v phng ngang (vung gc vi dng chy).

II. Tnh ton bn ca p theo phng php c hc kt cu. 1. Tnh ton on p c ngng trn thc dng.

Trng hp ny cc on p c xem nh kt cu c sn chng l tr hay bn tr (xem hnh 2-26a, b). Cc p hai tng v p c l x su c tnh ton nh kt cu hp. Khi trong mt ct tnh ton ch a vo mt phn theo chiu cao ca cc tr v bn tr gii hn bi mt phng nghing 450 so vi phng ngang v i qua im u v cui ca bn mng (hnh 2-31a).

a) b) c)

d)

Tu theo c im kt cu ca on p m c th xut cc s tnh ton khc nhau v phng php tnh ton gn ng tng ng. Chng hn ta xt vic tnh ton tm mng ca mt on p c hai khoang. on p (hnh 2-31b) c xem nh vt cng tuyt i so vi t nn. Ngoi s un chung th trong tm mng cn pht sinh s un cc b trn cc din tch S1 v S2. Vic tnh ton tm mng v un cc b c th thc hin nh i vi tm gi ln ba cnh, v c mt cnh t do thng hay h lu. Cc ti trng cn

Hnh 2-31. S tnh ton mt on p trn c

ngng thc dng

a. mt ct dc dng chy; b. mt ct ngang; c. biu ng sut trn mt ct B - B khi xt un chung(I), un cc b(II), v un tng cng(III); d. mt ct bng; p. phn lc nn tnh theo phng php nn lch tm.

B

B

p

A - A

45 45

A

A

P

I II III

B - B

2211

2211

S2

S2S1

S1


108

tnh n l: trng lng bn thn, phn lc nn, ti trng nc, p lc thm v y ni t pha di...

Mt trong nhng phng php gii bi ton ny l phng php dm trc giao. Cc tm c chia thnh nhng dm dc c ngm cng l tr gia v gi t do l bn tr (hnh 2-31d). Khi cn phi thc hin iu kin cn bng v un ca dm ngang v un ca dm cng xon trung tm I-I ti cc im trn mt ct ngang ca chng. Theo kt qu tnh ton s xy dng c cc biu ng sut tng cng bao gm ng sut khi tnh un chung v un cc b.

2. Tnh ton p c ngng trn nh rng

y l trng hp thng gp i vi cc p dng ngng thp v cc cng l thin trn h thng ti, tiu, phn l... Tm y ca p thng c chiu dy khng ln v khng c chnh lch nhiu v chiu dy cc phn khc nhau ca mt on p. Vic tnh ton kim tra iu kin bn v b tr ct thp cho bn y thng c tin hnh theo phng ngang, cn ct thp theo phng dc th b tr theo cu to. C th p dng cc phng php tnh ton sau:

a - Phng php dm o ngc: phng php ny p dng vi cc cng trnh nh, nn t cng. Khi , coi rng phn lc nn tc dng ln y p phn b tuyn tnh theo phng dc dng chy v phn b u theo phng ngang.

Khi xt chung cho c p, dng cng thc nn lch tm xc nh phn lc nn. Sau xt ring tng di ngang ca on p c chiu rng n v (hnh 2-32).

a) b)

c)

d) Hnh 2-32. S tnh ton bn y p theo phng php dm o ngc

a. ct dc p; b. ct ngang on p (theo B-B); c. s dm o ngc;

d. biu momen un (o ngc).

Xem bn y l mt dm lin tc gi ln cc ph v bn tr, c ti trng tc dng l phn lc nn p ti v tr ca di trn mt ct dc. Dng phng php c hc kt cu xc nh ni lc (M, Q), t tnh ton c ct thp theo phng ngang ca p.

Phng php ny c nhc im l cha xt c quan h gia vng ca bn y v tr s phn lc nn; khi tnh mi ch xt ring tng di m cha k n tnh ton khi ca on p.

B

B

y"y'

1m

p

B - B

p

p

+ +

-M


109

b - Phng php dm trn nn n hi: phng php ny cng xt n ni lc trn tng di theo phng ngang, nhng c k n tnh ton khi ca on p v quan h gia vng ca di vi cng phn lc nn ti cc im tng ng.

u tin cng xt ton b on p, dng cng thc nn lch tm xc nh s b phn lc nn (tc l coi phn lc nn phn b tuyn tnh theo phng dc v u theo phng ngang).

Ngoi lc tc dng ln mt di ngang bao gm:

- Lc t cc tr truyn xung a v lc tp trung t tm y tr Pi' ;

- Cc lc phn b u trn di: trng lng nc pha trn y q0, trng lng tm y q1, p lc nc y ngc q2 ;

- S b xem phn lc nn trn di l phn b u (q3).

- Lc ct khng cn bng Q t cc di bn cnh. Tr s ca Q c xc nh t phng trnh cn bng tnh theo phng thng ng:

Q + SPi' + 2l.Sqj = 0 , (2-40)

trong : SPi'-tng cc lc tp trung t m truyn xung ;

Sqj- tng (i s) cc lc phn b u :

Sqj = q0 + q1 + q2 +q3 ; 2l- chiu di ca di.

Sau khi xc nh c Q t phng trnh (2-40), ta cn phn phi Q cho cc phn (tr v bn y) ca mt bn thuc di ang xt. rng, quy lut phn b ng sut ct trn mt bn nh sau(cng thc Jurapxki):

t.bc = cS.JQ

, (2-41)

trong : t - cng ng sut ct ; b - chiu rng lt ct theo mt phng nm ngang ;

J - momen qun tnh ca mt bn ;

Sc - momen tnh ca phn mt bn b ct ly i vi trc trung ho ca ton mt bn.

Tr s Q v J ca mt bn c xc nh nn biu phn b ca t.bc ng dng vi biu phn b Sc. Tin hnh v biu Sc (hnh 2-33), xc nh phn din tch A1, A2 tng ng vi phn tr v bn y, t tnh c cc phn lc ct Q phn cho tr v bn y:

Hnh 2-33. S mt bn ca di tnh ton v biu Sc.

Trc trung ho

A2

A1

Sc


110

- Cho tr: 21

11 . AA

AQQ

+= ;

- Cho bn y: Q2 = Q - Q1.

Phn lc ct Q1 c phn cho cc tr theo t l din tch:

=i

i1

"i F

F.QP , (2-42)

trong :

"iP l phn lc ct phn cho tr th i c din tch Fi.

Phn lc ct Q2 c phn u cho bn y:

l2

Qq 24 = (2-43)

Ngoi ra khi phn tch lc cn phi xt n nh hng ca ti trng bn nh trng lng t p sau lng tr bin, hay p lc y mng bnh qun ca on p bn cnh, ti trng t mt ng giao thng truyn ti, momen do cc lc ngang gy ra...

S ti trng cui cng ln di nh hnh 2-34

Hnh 2-34. S ngoi lc cui cng tc dng ln di bn y.

Trong s :

- Lc tp trung truyn t m Pi = Pi' + Pi" ;

- Lc phn b trn di: q = q0 + q1 + q2 + q4 ;

- Cng ti trng bn, pha gip t: S ; pha gip vi on khc: q3 ;

- Mo men do lc ngang ti tr bin: Mb ;

- Lc truyn t xe chy trn ng: q5.

Ni lc trong di ca bn y c xc nh theo phng php dm trn nn n hi. C th gii theo cc phng php ca Winkler, Jenmonskin hay phng php tra bng ca Gorbunop - Poxadop.

Khi xt nh hng ca ti trng bn cn lu :

- Nu ti trng bn lm tng momen un bn y th xt nh hng hon ton.

Tr bin Bn tr Khp ni

Mi h mng

P3P2P1S

q5

2lMb

q3q


111

- Nu ti trng bn lm gim momen un bn y vi t p bn l st th khng xt n nh hng ny, nu l t ct th xt 30-50% nh hng ca ti trng bn.

- Ch xt nh hng ca ti trng bn phn b trong phm vi chiu di 2l tnh t mp bin ca on p ang tnh.

3. Tnh ton tr v bn tr

Tr hay bn tr chu tc dng ca ngoi lc nh p lc nc truyn t ca van, trng lng bn thn tr v cc b phn t ln n (cc cu v ti trng trn cu)... Tr bin cn chu tc dng ca p lc t. Cch thc truyn p lc nc t ca van ph thuc vo loi ca van (van phng, van cung, van trc ng...)

Khi tnh ton thng xt cc trng hp sau:

- Trng hp thi cng: tr chu tc dng ca trng lng bn thn v cc my mc, thit b thi cng truyn xung. Tr lm vic nh cu kin chu nn lch tm.

- Trng hp lm vic, ca van ng: tr chu tc dng ca p lc nc rt ln t ca van truyn n. V th cn kim tra n nh trt ca tr (khi tr lm tch ri bn y), hoc kim tra kh nng b ct mt lin kt tr v bn y. i vi ca van phng cn kim tra kh nng chu lc ca tr ti mt ct c khe van. Nu l van cung th cn phn tch ng sut ca tr di tc dng ca p lc nc truyn tp trung b phn b t cng van, v kim tra bn cc b ca khu vc xung quanh b t.

- Trng hp sa cha, khi dng phai chn nc thng v h lu khoang p v bm ht nc trong khoang ra kim tra sa cha, v khoang bn cnh vn m bnh

thng (hnh 2-35). Lc ny tr lm vic nh mt kt cu chu nn v un hai phng. ng sut ln nht v nh nht pht sinh ti mt mt ct ngang ca tr c xc nh theo cng thc:

y

y

x

x

W

M

W

M

F

P =sminmax , (2-44)

trong :

SP - tng cc lc thng ng tc dng ln m ;

SMx, SMy - tng momen i vi trc x v y ti mt ct xt (hnh 2-21) ;

Wx, Wy- mun chng un i vi trc x v trc y ;

F - din tch mt ct tnh ton.

Hnh 2-35. S lc tc dng ln tr, trng hp kim tra, sa cha.

Ngoi ra, tu theo c im cng trnh v tnh hnh lm vic, c th xt thm mt s trng hp tnh ton khc. Kt qu tnh ton s xc nh c cc trng hp lm vic bt li nht b tr ct thp, hoc la chn kch thc tr cho thch hp.

i vi p c ca van phng, mt ct xung yu ca tr l mt ng i qua khe van. Khi ca van ng, p lc nc truyn vo tr c th ko t tr theo mt ct xung yu ny.

xy


112

Trong tnh ton, thng xem b tng khng tham gia chu ko; khi din tch ct thp Fa t hai bn khe van theo phng ca p lc nc c tnh theo cng thc:

a

a R.mWF = , (2-45)

trong :

W-tr s tnh ton ca p lc ngang ca nc truyn qua ca van tc dng vo mt tr ;

m - h s iu kin lm vic ;

Ra - cng chu nn tnh ton ca ct thp ;

i vi tr p c ca van cung, cn xt bi ton tr chu lc phc tp l nn, un theo hai phng v chu lc y tp trung vo tai tr.

Bi ton phn tch ng sut tr khi chu lc y t cng van c th gii theo m hnh nm v hn chu lc tp trung ti nh, c th dn n cc bng tra (xem, ph lc I1 I6, Thit k cng ca tc gi Trnh Bn-L Ho Xng). Tuy nhin ngy nay, cc bi ton loi ny thng c gii bng phng php phn t hu hn.

III.Tnh ton bn ca p hay cc b phn ca n bng phng php PTHH.

Nh trn ni, khi tnh ton cc on p thuc cng trnh cc cp, nht l cp I, II ni ring, hay khi phn tch ng sut tr van cung ca cc p ni chung th phng php tin cy hin nay l phng php PTHH. Nguyn l c bn ca phng php ny c trnh by trong chng 1-p b tng trng lc. Vic tnh ton p b tng trn nn mm c tin hnh c xt n lc tng tc gia cng trnh v nn. Cn bi ton phn tch ng sut tr van cung thng c xt theo s tm ngm vo bn y.


113

Ti liu tham kho chng 2

1- TCXDVN 285-2002. Cng trnh thu li. Cc quy nh ch yu v thit k, NXB Xy dng H Ni, nm 2002.

2- ng vin di t ca p trn nn khng phi l - Quy trnh thit k 14TCN 58-88, B Thu li, nm 1988.

3- TCVN 5747-1993 t xy dng v phn loi . NXB xy dng, 1993

4- TCVN 4235-86. Nn cc cng trnh thu cng. Tiu chun thit k. NXB Xy dng, H Ni, 1988.

5- 14TCN 123-2002. Tiu chun nghnh. t xy dng cng trnh thu li - phn loi, B Nng Nghip v PTNN, 2002.

6- 14TCN 56-88. Thit k p b tng v b tng ct thp. Tiu chun thit k, B thu li 1988.

7- Trnh Bn, L Ho Xng. Thit k cng - NXB Nng nghip, H Ni, 1988.

8- Gio trnh thu cng- B mn thu cng - i hc Thu Li. Thu cng tp II, NXB Nng nghip, H Ni, 1989.

9- L.N.Raxxcarop v Nak. Cng trnh thu li phn I (Bn ting Nga), NXB xy dng, Matxcova, 1996.

10- Hydraulic Structures Vol1 Edited by N.M.Grishin Mir PublishersMoscow (Bn ting Anh)

11- Jezy Hydrostatick A Jezy Automatick S Vyvenm (Bn ting Tip Khc)

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