NEN MONG - BKDN.pdf

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1/ Coc th nghim2/ Cac khi b tng lam i trong3/ Kch thuy lc4/ Thin phn k5/ Dm gn TP k6/ H dm tai

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AI HOC A NNG TRNG AI HOC BACH KHOA KHOA XY DNG CU NG

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B MN C S KY THUT XY DNG

BAI GIANG

NN MONG

A NNG 2006

I HC BCH KHOA

Trng I HC BCH KHOA NNG Nhm chuyn mn CH-Nn Mng B mn C s k thut Xy dng Bi ging Nn v Mng

CHNG I. MT S VN C BN TRONG THIT K NN MNG 1. CC KHI NIM C BN 1.1. Mng Mng l b phn chu lc t thp nht, l kt cu cui cng ca nh hoc cng trnh. N tip thu ti trng cng trnh v truyn ti trng ln nn t di y mng.

Mt mong

G mong

ay mong

Mong

C mong

Nn

b

a

Cu kin bn trn

hm

1.2. Mt mng B mt mng tip xc vi cng trnh bn trn (chn ct, chn tng) gi l mt mng. Mt mng thng rng hn kt cu bn trn mt cht to iu kin cho vic thi cng cu kin bn trn mt cch d dng. 1.3. G mng Phn nh ra ca mng gi l g mng, g mng c cu to phng sai lch v tr c th xy ra khi thi cng cc cu kin bn trn, lc ny c th x dch cho ng thit k.

Hnh 1.1 Nn v mng1.4. y mng B mt mng tip xc vi nn t gi l y mng. y mng thng rng hn nhiu so vi kt cu bn trn. S d nh vy bi v chnh lch bn ti mt tip xc mng - t rt ln (t 100 - 150 ln), nn m rng y mng phn b li ng sut y mng trn din rng, gim c ng sut tc dng ln nn t. * Khi nim v p lc y mng: p lc do ton b ti trng cng trnh (bao gm c trng lng bn thn mng v phn t trn mng), thng qua mng truyn xung t nn gi l p lc y mng.

N

G

hm

pdtb

Cng thc: axb

GNtb

+= (1.1) Trong :

N - Tng ti trng thng ng tnh n mt nh mng. G - Trng lng ca vt liu mng v phn t nm trn mng. * Khi nim v phn lc nn: Khi chu tc dng ca p lc y mng, nn t di y mng cng xut hin phn lc nn, c cng tr s nhng ngc chiu vi p lc y mng.

Hnh 1.2: p lc y mng v phn lc nn

Cng thc: axb

GNp tb+== (1.2)

Vic tnh ton phn lc nn c ngha rt ln cho vic tnh ton bn, n nh ca mng sau ny.

nng 9/2006 CHNG I TRANG1


1.5. Nn Nn l phn t nm di y mng, tip thu ti trng t mng truyn xung. Ngi ta phn nn lm hai loi: + Nn thin nhin: L nn khi xy dng cng trnh, khng cn bin php no x l v mt vt l v c hc ca t. + Nn nhn to: L loi nn khi xy dng cn dng cc bin php no ci thin, lm tng cng kh nng chu ti ca t nn. 1.6. ngha ca cng tc thit k nn mng Khi tnh ton thit k v xy dng cng trnh, cn ch v c gng lm sao m bo tho mn ba yu cu sau: 1- Bo m s lm vic bnh thng ca cng trnh trong qu trnh s dng. 2- Bo m cng ca tng b phn v ton b cng trnh. 3- Bo m thi gian xy dng ngn nht v gi thnh r nht. Vi yu cu th nht th nu cng trnh c ln, hoc ln lch, hoc chuyn v ngang qu ln th cng trnh khng th lm vic bnh thng, ngay c khi n cha b ph hu. Vi yu cu th hai: Cng cng trnh ngoi vic ph thuc vo cng bn thn kt cu, mng, cn ph thuc rt ln vo cng ca t nn di y cng trnh. Do vy cng tc kho st, thit k v tnh ton nn phi cht ch v chnh xc m bo an ton cho cng trnh. Vi yu cu th ba: th vic tnh ton, thit k v chn bin php thi cng hp l c nh hng rt ln n thi gian thi cng cng trnh. Thng thng vic thi cng nn mng thng mt nhiu thi gian, do vy yu cu ny cn c th hin tnh hp l v cht ch. Gi thnh xy dng nn mng thng chim 20-30% gi thnh cng trnh ( i vi cng trnh dn dng). Vi cng trnh cu, thu li t l c th n 40-50%. Kinh nghim thc tin cho thy hu ht cc cng trnh b s c u do gii quyt cha tt cc vn v thit k nn mng Do vy, vic nghin cu, tnh ton, thit k nn v mng mt cch ton din c ngha rt quan trng i vi ngi k s thit k nn mng. 2. PHN LOI MNG V PHM VI S DNG 2.1. Phn loi theo vt liu:

Thng thng s dng cc loi vt liu lm mng nh sau: Gch, hc, , b tng, b tng ct thp

+ Mng gch: S dng cho cc loi mng m cng trnh c ti trng nh, nn t tt, s dng ni c mc nc ngm su.

+ Mng hc: Loi lng ny c cng ln, s dng nhng vng c sn vt liu.

+ Mng g: Cng nh, tui th t, t c s dng, thng s dng cho cc cng trnh tm thi, hoc dng x l nn t yu.

+ Mng thp: t c s dng lm mng v thp d b g do nc trong t v nc ngm xm thc.



+ Mng b tng v b tng ct thp: Cng cao, tui th lu, c s dng rng ri trong xy dng cng trnh. Vi loi mng ny yu cu b tng Mc200. 2.2. Phn loi theo cch ch to mng:

Theo cch ch to mng ngi ta phn ra hai loi: mng ton khi v mng lp ghp.

+ Mng ton khi: Thng s dng vt liu l b tng hc, b tng v b tng ct thp, loi mng ny c s dng nhiu.

+ Mng lp ghp: Cc cu kin mng c ch to sn, sau mang n cng trng lp ghp. Loi mng ny c c gii ho, cht lng tt tuy nhin t c s dng v vic vn chuyn kh khn. 2.3. Phn loi theo c tnh tc dng ca ti trng:

Theo c tnh tc dng ca ti trng ngi ta phn thnh mng chu ti trng tnh v mng chu ti trng ng:

+ Mng chu ti trng tnh: Mng nh, cng trnh chu ti trng tnh. + Mng chu ti trng ng: Mng cng trnh cu, mng my, mng cu trc

2.4. Phn loi theo phng php thi cng: Theo phng php thi cng ngi ta phn thnh mng nng v mng su: * Mng nng: L mng xy trn h mng o trn, sau lp li, su chn

mng t 1.23.5m. Mng nng s dng cho cc cng trnh chu ti trng nh v trung bnh, t trn

nn t tng i tt (nn t yu th c th x l nn). Thuc loi mng nng ngi ta phn ra cc loi sau:

+ Mng n: S dng di chn ct nh, ct in, m tr cu + Mng bng: S dng di cc tng chu lc, tng ph hoc cc hng ct,

mng cc cng trnh tng chn. + Mng bn (mng b): Thng s dng khi nn t yu, ti trng cng trnh

ln, hoc cng trnh c tng hm. * Mng su: L loi mng khi thi cng khng cn o h mng hoc ch o

mt phn ri dng phng php no h, a mng xung su thit k. Thng s dng cho cc cng trnh c ti trng ln m lp t tt nm tng su.

Mng su gm c cc loi sau: + Mng ging chm: l kt cu rng bn trong, v ngoi c nhim v chng

p lc t v p lc nc trong qa trnh h v to trng lng thng ma st. Sau khi h n su thit k th ngi ta lp y (hoc mt phn) b tng v phn rng. S thi cng mng ging chm t trng nh hnh v (1.3).

Vic ly t di y ging c th bng nhn cng o t v a ln trn, ngoi ra c th dng vi xi p lc ln xi t v ht c t v nc ra ngoi, h ging xung cao thit k.

* u im: - Mng c kch thc ln, kh nng chu ti rt ln. - Thi cng thit b n gin. * Nhc im: - Khng ph hp khi nc ngm ln hoc c nc mt. - Nng sut khng cao.



-Thi gian thi cng lu. Nhn xt: Mng ging chm ph hp khi xy dng mng cu ln v iu kin

thi cng ph hp. Tuy nhin cn cn nhc gia cc phng n mng su p ng yu cu v tin thi cng v nng sut lao ng.

Thng ng vt liu

1. c t u tin 2. o h ging 3. c t th 2

Hnh 1.3. S h ging chm + Mng ging chm hi p: Khi gp iu kin a cht thu vn phc tp ngi ta thay mng ging chm

bng mng ging chm hi p. Nguyn tc lm vic ca n l dng kh nn vo bung kn ca ging nh sc p ca kh m nc b y ra ngoi tao iu kin kh ro cng nhn o t. S thi cng Ging chm hi p nh trn hnh (1.4).

Sau khi hon thnh cng tc to mt bng thi cng, li ct bng thp c lp trc tip trn nn v ng v tr. Phn trong ca li ct c y ct v cng tc b tng khoang lm vic c thc hin. Vic lp t cc thit b v b tng tng cho Ging cng vi cng tc o t c thc hin ng thi. Sau khi hon thnh cng vic thi cng tng ging, np Ging (sn trn) c xy dng v pha trong khoang lm vic c bm y b tng. Kh nng chu ti ca t trc tip di y ca Ging c khng nh bng th nghim kim tra kh nng chu ti bng tm nn, thc hin trong lng khoang thc hin.

1. Chun b mt bng thi cng

Thep li ct

2. Lp t li ct bng thep

3. b tng ln th nht 4. Cng tac ao t va lp t kt cu May ao chuyn dung

May nen kh

Li ngi ln xungKhoang ngi

Khoang vt liu

Thung cha t

Cu banh xchLi ngi ln xung



May bm b tng

Chen b tng khoang lam vic B tng khoang lam vic

Thep ng cho tru

B tng san trn

6. B tng san trn5. b tng khoang lam vic Hnh 1.4 Trnh t h mng Ging chm hi p

nh gi u nhc im: * u im: - Vng chc, chu ti ln - t nh hng n mi trng. - Hiu qu kinh t cao. - Thi gian thi cng ngn. - tin cy cao. * Nhc im: Vic thi cng mng nh hng nhiu n sc kho ca cng nhn khi o ging trong iu kin p sut cao. Cn nghin cu pht huy nhng u nhc im v hn ch thp nht nh hng n sc khe ngi lao ng, c th ch to robot o trong ging l hp l nht, va hiu qu va khng nh hng n sc khe con ngi. Nhn xt: Vi nhng u khuyt im nh trn, mng ging chm hi p ph hp khi lm mng cho cc cng trnh cu ln, cc tr thp cu dy vng, cu treo dy vng nhp ln, ng cc m neo cu treo chu lc nh ln Tuy nhin cn khc phc nh hng n sc khe ngi lao ng nh nu. + Mng cc: Gm cc cc ring r, h xung t v ni vi nhau bng i cc.

Mng cc s dng cc loi vt liu nh: G, thp, b tng v b tng ct thp.

Thng s dng cho cc cng trnh chu ti trng ln, cng trnh trn nn t yu nh m tr cu, cu cng, b k

Thuc loi mng cc c nhiu loi, y da vo phng php thi cng ta chia thnh cc loi sau: (i vi cc b tng ct thp)

Hnh 1.5: Mng cc trong tr cu

Cc b tng ct thp c sn: Loi cc ny c ch to nng 9/2006 CHNG I TRANG5


sn trn cc bi c, tit din t 20x20cm n 40x40cm,sau h cc bng phng php ng hoc p.

Cc b tng ct thp ti ch (cc khoan nhi): Dng my khoan to l sau a lng thp vo v nhi b tng vo l. Cc c ng knh nh nht d=60cm, ln nht c th t d=2.5m.Chiu su h cc n hn 100m.

3. KHI NIM V TNH TON NN MNG THEO TRNG THI GII HN

3.1. Khi nim v trng thi gii hn: Trng thi gii hn l trng thi ng vi khi cng trnh khng iu kin s dng bnh thng (vng qu ln, bin dng ln, nt qu phm vi cho php, mt n nh) hoc b ph hon ton. Theo quy phm mi, vic tnh ton nn mng theo 3 trng thi gii hn (TTGH)

+ Trng thi gii hn1: Tnh ton v cng n nh ca nn v mng. + Trng tha gii hn 2: Tnh ton v bin dng,ln ca nn mng. + Trng thi gii hn 3: Tnh ton v s hnh thnh v pht trin khe nt (ch s

dng cho tnh ton kt cu mng). 3.2. Khi nim v tnh ton mng theo TTGH: Nh mi kt cu chu lc khc, kt cu mng c th phi tnh ton thit k theo ba trng thi gii hn: trng thi gii hn th nht, th hai v th ba.

Ngoi ra, v mng lm vic chung vi nn cho nn c th xy ra mt dng ph hng khc l mng b lt hoc trt trn nn. Khi b mt n nh nh th, mng khng cn lm vic c na,cng trnh b b hng mt d bn thn mng khng t ti TTGH no trong 3 TTGH k trn. Do vy khc vi kt cu chu lc khc, ngoi 3 TTGH thng thng, mng cn c th tnh theo TTGH v n nh (lt v trt) trn nn.

- Nhng mng chu ti trng ngang ln m lc thng ng nh (Nh cc tng chn t, mng neo) th phi tnh theo TTGH v n nh trn nn.

- Mng bn y ca cc b cha vt liu lng, mng t trong mi trng c tnh n mn mnh phi tnh theo TTGH3.

- Nhng mng dng tm mng, bin dng ln th phi tnh theo TTGH2. - Tt c cc loi mng u phi tnh ton theo TTGH1. i vi mng ca hu

ht cc nh Dn dng v Cng nghip th ch cn thit k v tnh ton theo TTGH1 m thi. 3.3. Khi nim v tnh ton nn theo TTGH

Khng nh nhng kt cu chu lc lm bng nhng vt liu khc, nn t ch c hai TTGH: Trng thi gii hn th nht (v cng ) v TTGH th hai (v bin dng). TTGH th ba v s hnh thnh v pht trin khe nt) khng c ngha i vi nn t. 3.3.1. Tnh ton nn theo TTGH1: Theo TCXD 45-70, i vi cc loi nn sau:

- Cc nn t st rt cng, ct rt cht, t na v .(1)



- Cc nn t mng thng xuyn chu ti trng ngang vi tr s ln (Tng chn, chn)

- Cc nn trong phm vi mi dc ( trn hay ngay di mi dc) hoc lp t mm phn b rt dc th phi tnh ton thit k theo TTGH1.

- Cc nn t thuc loi st yu bo ha nc v than bn. Cc nn t (1) ch bin dng rt nh di tc dng ca ti trng cng trnh,

ngay c khi ti trng t n ti trng cc hn ph hng nn t th bin dng vn cn b. Do vy nhng loi nn ny khi chu tc dng ca ti trng, s dn ti TTGH1 trc khi xut hin TTGH2. Cng thc kim tra:

atK

N (1.3) Trong : N - Ti trng ngoi tc dng ln nn trong trng hp bt li nht. - Sc chu ti ca nn theo phng ca lc tc dng. Kat H s an ton, ph thuc loi nn v tnh cht ca ti trng, cng trnh, do c quan thit k quy nh. 3.3.2. Tnh ton nn theo TTGH2

Vic tnh ton nn theo TTGH2 c p dng cho tt c cc loi nn tr cc loi nn nu (1). Mc ch ca vic tnh ton l khng ch bin dng tuyt i v chuyn v ngang ca nn khng vt qu gii hn cho php, m bo iu kin lm vic bnh thng ca cng trnh. Cc iu kin: S < [S] S < [S] (1.4) U < [U] Trong : S, S, U - chuyn v ln, ln lch v chuyn v ngang do ti trng gy ra.

[S], [S],[U] - chuyn v ln, ln lch v chuyn v ngang gii hn. 3.4. Cc loi ti trng v t hp ti trng 3.4.1. Cc loi ti trng 3.4.1.1. Ti trng thng xuyn v ti trng tm thi Ti trng thng xuyn: L ti trng tc dng trong sut thi gian thi cng v s dng cng trnh: Trng lng bn thn kt cu, p lc t, p lc nc Ti trng tm thi: Ch xut hin trong mt thi k no trong thi cng hoc s dng cng trnh, sau gim dn hoc mt hn. Tu theo thi gian tn ti, ngi ta phn ti trng tm thi thnh: + Ti trng tm thi tc dng lu di (di hn): Trng lng thit b, vt liu cha + Ti trng tm thi tc dng ngn hn: Trng lng ngi, xe my thi cng, ti trng gi, p lc sng + Ti trng tm thi c bit: Xut hin trong trng hp rt c bit khi thi cng hoc khi s dng cng trnh (ng t, s c cng trnh) 3.4.1.2. Ti trng tiu chun v ti trng tnh ton Ti trng tc dng ln cng trnh c phn thnh ti trng tiu chun v ti trng tnh ton:


Trng I HC BCH KHOA NNG Nhm chuyn mn CH-Nn Mng B mn C s k thut Xy dng Bi ging Nn v Mng + Ti trng tiu chun: L ti trng ln nht, khng gy tr ngi, lm h hng v khng lm nh hng n s lm vic bnh thng khi s dng cng nh khi sa cha cng trnh. + Ti trng tnh ton: Ti trng xt n kh nng c th xy ra s khc nhau gia ti trng thc v ti trng tiu chun v pha khng c li cho s lm vic bnh thng ca cng trnh. Ti trng tnh ton c xc nh bng cch nhn ti trng tiu chun vi h s vt ti tng ng: Ntt = n. Ntc (1.5) Vi n l h s vt ti, ly nh sau:

Trng lng bn thn cc loi vt liu: n=1,1. Trng lng cc lp t p, lp cch m cch nhit n=1,2. Trng lng cc thit b k thut (k c trng lng vt liu cha trong thit b

khi n hot ng) ly n=1,2. Trng lng thit b vn chuyn:n=1,3.

3.4.2. Cc t hp ti trng Khi tnh ton cn xt cc t hp ti trng sau:

+ T hp ti trng chnh: (t hp c bn): Bao gm cc ti trng thng xuyn, cc ti trng tm thi di hn v mt trong cc ti trng tm thi ngn hn.

+ T hp ti trng ph: (T hp b sung): Bao gm cc ti trng thng xuyn, cc ti trng tm thi di hn v hai hoc nhiu hn hai ti trng tm thi ngn hn.

+ T hp ti trng c bit: Bao gm cc ti trng thng xuyn, cc ti trng tm thi di hn, mt s ti trng tm thi ngn hn v ti trng c bit. * Vic tnh ton nn mng theo bin dng tin hnh vi t hp chnh (t hp c bn) ca cc ti trng tiu chun. * Vic tnh ton nn mng theo cng v n nh tin hnh vi t hp chnh, t hp ph hoc t hp c bit ca cc ti trng tnh ton. 3.5. Cc h s tnh ton Khi tnh ton nn mng theo trng thi gii hn, ngi ta thng dng cc h s sau y: + H s vt ti n: Dng xt ti s sai khc c th xy ra ca ti trng trong qu trnh thi cng v s dng cng trnh. Tu loi cng trnh m ngi ta quy nh h s vt ti l bao nhiu. Tu theo tnh cht tc dng ca ti trng tc ng ln cng trnh m n c th ln hn hoc b hn 1. + H s ng nht K: Dng xt ti kh nng phn tn cng ca t ti cc im khc nhau trong nn do tnh cht phn tn v cc ch tiu c hc gy ra. V t c tnh ng nht km nn K thng b hn 1. + H s iu kin lm vic m: Dng xt ti iu kin lm vic thc t ca nn t. Tu iu kin c th m m c th ln hn hoc b hn 1. H s iu kin lm vic xc nh theo cc s liu thc nghim.



4. CC TI LIU CN THIT THIT K NN MNG Trc khi thit k nn mng ca cng trnh no , ngi thit k phi c cc ti liu c bn sau y: 4.1. Cc ti liu v a cht cng trnh v a cht thu vn Ni dung ca cc ti liu ny bao gm: - Bn a hnh, a mo ni xy dng cng trnh, quy m, v tr cc cng trnh xy trc lm c s chn phng n mng hoc x l nu c. - Cc ti liu khoan a cht, hnh tr l khoan, mt ct a cht, cu trc a tng, ngun gc, chiu cao mc nc ngm, kt qu kho st bin ng ca nc ngm - Kt qu th nghim nh gi cc tnh cht ca nc ngm, trnh tc ng xu n nn mng sau ny. Kt qu th nghim cc ch tiu c hc, vt l ca cc lp t: Thnh phn ht, dung trng, t trng, m gii hn chy, m gii hn do, h s thm, gc ni ma st, lc dnh, cc kt qu th nghim ct, nn, kt qu th nghim xuyn ng SPT, kt qu th nghim xuyn tnh CPT, ct cnh, CBR .v.v. lm c s, nn tng quyt nh phng n mng. 4.2. Cc s liu v cng trnh v ti trng - Hnh dng, kch thc y cng trnh. - c im cu to ca cng trnh (cng trnh c tng hm hay khng, c b tr h thng ng nc, ng cp, ng hm ni gia cc cng trnh ln cn hay khng). - Cc ti liu v chi tit cc cng trnh bn trn v cc ti trng tc dng, c th nh sau: + Trng lng bn thn: Tnh t kch thc hnh hc ca cc kt cu truyn xung. + Trng lng cc thit b cha hoc thit b thi cng. + p lc t, p lc nc. + p lc gi, cng , hng gi. + p lc sng. + p lc thm. + Lc va ca tu b. + Ti trng chn ng v cp ng t ca tng vng nu c. 5 XUT SO SNH V CHN PH5.1. Chn chiu su ch

NG N MNG n mng

ng l khu c

ti y ng

chn mng sao

. iu kin a cht v a cht thu vn n

Q

NM

hm

Vic chn chiu su chn mbn nht trong cng tc thit k nn mng. su hm k t mt t thin nhin m gi l su chn mng Vic la chn chiu sucho hp l n ph thuc vo cc yu t c bn sau: 5.1.1

Hnh 1.6: Chiu su chn mng

y l yu t nh hng nhiu nht


Trng I HC BCH KHOA NNG Nhm chuyn mn CH-Nn Mng B mn C s k thut Xy dng Bi ging Nn v Mng v n chiu su chn mng, trong xc nh v tr lp t chu lc l quan trng nht. Lp t chu lc l lp t tt tip xc trc tip vi y mng. Theo Gs Berezantex,

ic ch

nhng lp t sau y khng nn dng lm lp t chu c:

iu kin a cht ni xy dng, ta xt mt vi s in

l t ct ri, t st nho, st cha nhiu hu c hoc st c h s rng e> 1,1; st c e>1,0; hoc ct c e>0,7. xt nh hng ca hnh nh sau:

(a)

t tt t yu t yu

t tt

t tt

t yut tt

(b) (c) (d)

S n quyt nh, t

n mng ph thuc ch yu vo phng

: Nu lp t yu mng th t mng vo lp t tt 25 - 30 cm cn

h c th t mng, nhng phi m bo chiu

ht thu vn hi tu

nn ph thuc vo v tr cc lp t, trng thi vt

2- Phi t y mng vo lp t tt chu lc t 15-20cm. tnh nn ln ca lp

t yn kt cu ca t v

thp hn mc nc ngm (c k n s ln xung a n

nh hng ca tr s v c tnh ca ti trng u chn mng mng ta ln

Hnh 1.7: Cc s in hnh ca nn t khi chn su chn mng - a: Trng hp ny chiu su chn mng ch yu do tnh touy nhin khng t mng trong lp t trng trt v nn t nh mng thp hn

mt t t nhin 25 - 30cm trnh va chm. - S b: Trng hp ny su chphp x l nn. - S cnu lp t yu dy th tr li s b. - S d: Nu lp t tt dy tsu t tt di y mng, nu lp t tt mng th tr li s b hoc c. * Ch : Khi chn chiu su chn mng theo cc iu kin a cp n theo cc quy tc sau y: 1- Chn lp t chu lc cal ca chng, phng php xy dng mng, tr s ln gii hn v s n nh ca nn. 3- Khng nn di y mng c mt lp t mng nu ln hn nhiu so vi tnh nn ln ca lp t nm di. 4- Nn t mng cao hn mc nc ngm gi ngukhng phi tho nc khi thi cng. 5- Khi chiu su chn mng c ) th phi gii quyt gi nguyn kt cu t trong nn khi o h mng v xy mng. 5.1.2. Nu ti trng cng trnh ln th nn tng chiu scc lp t cht hn nm di v gim ln.


Trng I HC BCH KHOA NNG Nhm chuyn mn CH-Nn Mng B mn C s k thut Xy dng Bi ging Nn v Mng Khi mng chu ti trng nh (hng ln) hoc ti trng ngang, momen ln (lch tm ln) th yu cu phi ngm su mng n sau thch hp m bo n nh cho mng. 5.1.3. nh hng ca c im cu to cng trnh Khi chn chiu su chn mng, cn phi k n c im ca nh v cng trnh (nh c tng hm, c ho, h, c ng lin lc ngm ) cng cn ch n vic t ng dn nc bn trong cng nh gn nh v cng trnh. 5.1.4. nh hng ca mng cc cng trnh ln cn Thng thng ngi ta chn chiu su chn mng ngang vi cao trnh y ca cc mng chnh ca nh v cng trnh ln cn. Ch c php t cao hn khi m bo gi c kt cu ca t nm trn chiu su chn mng ca nh hoc cng trnh ln cn. Nguyn tc chung ca cc gii php k thut nhm khc phc nhng tc ng xu ca mng mi tc ng ln mng nh hoc cng trnh c l hn ch n mc thp nht cc p lc t mng nh mi tc dng ln mng nh c k bn. Mt s gii php t mng:

hi=0

,5-0

,6m

Mong cu Mong mi Ging mong gi cn bng

Mong cu

Mong miMong cu Mong mi

Tng c

ly=1,0-1,2m

Hnh 1.8: Mt s gii php t mng khi c mng cng trnh ln cn 5.1.6. nh hng ca bin php thi cng mng Tu theo phng php thi cng m kt cu ca t nn c th b ph hoi. Nu bin php thi cng khng m bo gi nguyn c kt cu t nn khi o h mng di mc nc ngm th phi ly chiu su chn mng ti thiu cho php v din tch y mng tng n tr s ln nht. Khi bin php thi cng m bo gi nguyn c kt cu t nn (ht nc tng su, dng ging chm hi p) th cho php mng c din tch y mng b nht, t su tng i ln.



5.2. xut, so snh v chn phng n mng. Cng nh i vi nhiu cng trnh khc, khi thit k nn mng, nhim v ca ngi thit k phi chn phng n tt nht c v kinh t v k thut. Thng thng vi nhim v thit k cho, vi cc ti liu v a cht cng trnh, a cht thu vn, ti trng, ... ngi thit k c th ra nhiu phng n nn mng khc nhau nh : - Phng n lm nng trn nn thin nhin. - Phng n mng nng trn nn nhn to. - Phng n mng cc. - Phng n mng ging chm, ... Mi phng n ln c th xut nhiu phng n nh v d phng n mng nng c th l: mng n, mng bng hay mng b; phng n mng cc c th l : cc di, ngn, cc ng, cc p, cc nhi, ... v mi phng n nh cng c th c nhiu phng n nh hn, khc nhau v hnh dng, kch thc v cch b tr. Tuy nhin tu loi cng trnh, c im, qui m v tnh cht v do kinh nghim ca ngi thit k m ngi ta c th xut ra mt vi phng n hp l so snh v la chn phng n ph hp nht. Khi thit k s b so snh phng n ngi ta da vo ch tiu kinh t quyt nh (dng tng gi thnh xy dng nn mng ). Khi thit k k thut th ngi ta kt hp c hai ch tiu kinh t v k thut ng thi vi iu kin v thi gian thi cng quyt nh phng n. Vic so snh la chn phng n nn mng l mt cng vic kh khn v quan trng. Mun gii quyt tt cng vic ny, ngi thit k phi nm vng nhng l thuyt tnh ton trong C hc t v Nn mng kt hp vi kinh nghim tch lu trong qu trnh thit k v thi cng xut v la chn phng n ti u nht v nn mng ca cng trnh xy dng.



CHNG II: MNG NNG TRN NN THIN NHIN 1. KHI NIM CHUNG 1.1. nh ngha Mng nng l nhng mng xy trn h o trn, sau lp li, chiu su chn mng khong di 23m, trong trng hp c bit c th su n 5m. So vi cc loi mng su, mng nng c nhng u im: + Thi cng n gin, khng i hi cc thit b thi cng phc tp. Vic thi cng mng nng c th dng nhn cng o mng, mt s trng hp vi s lng mng nhiu, hoc chiu su kh ln c th dng cc my mc tng nng sut v gim thi gian xy dng nn mng. + Mng nng c s dng rng ri trong cc cng trnh xy dng va v nh, gi thnh xy dng nn mng t hn mng su. + Trong qu trnh tnh ton b qua s lm vic ca t t y mng tr ln. 1.2. Phn loi mng nng 1.2.1. Da vo c im ca ti trng Da vo tnh hnh tc dng ca ti trng ngi ta phn thnh : + Mng chu ti trng ng tm. + Mng chu ti trng lch tm. + Mng cc cng trnh cao (thp nc, ng khi,...). + Mng thng chu lc ngang ln (tng chn, p nc, ...). + Mng ch yu chu ti trng thng ng, m men nh. 1.2.2. Da vo cng ca mng + Mng tuyt i cng: Mng c cng rt ln (xem nh bng v cng) v bin dng rt b (xem nh gn bng 0), thuc loi ny c mng gch, , b tng. + Mng mm: Mng c kh nng bin dng cng cp vi t nn (bin dng ln, chu un nhiu), mng BTCT c t l cnh di/ngn > 8 ln thuc loi mng mm. + Mng cng hu hn: Mng B tng ct thp c t l cnh di/cnh ngn < 8 ln. Vic tnh ton mi loi mng khc nhau, vi mng mm th tnh ton phc tp hn. 1.2.3. Da vo cch ch to Da vo cch ch to, ngi ta phn thnh mng ton khi v mng lp ghp. + Mng ton khi: Mng c lm bng cc vt liu khc nhau, ch to ngay ti v tr xy dng (mng ti ch). + Mng lp ghp: Mng do nhiu khi lp ghp ch to sn ghp li vi nhau khi thi cng mng cng trnh. 1.2.4. Da vo c im lm vic Theo c im lm vic, c cc loi mng nng c bn sau : + Mng n: di dng ct hoc dng bn, c dng di ct hoc tng kt hp vi dm mng. + Mng bng di ct chu p lc t hng ct truyn xung, khi hng ct phn b theo hai hng th dng my ng bng giao thoa. + Mng bng di tng: l phn ko di xung t ca tng chu lc v tng khng chu lc.

nng 9/2006 CHNG II TRANG 13


+ Mng bn, mng b : mng dng bn BTCT nm di mt phn hay ton b cng trnh. + Mng khi: l cc mng cng dng khi n nm di ton b cng trnh. Theo cch phn loi ny ta s nghin cu cu to chi tit ca mt s loi thng gp. 2. CU TO CC LOI MNG NNG THNG GP 2.1. Mng n. Mng n c ch to, kin thit di chn ct nh dn dng nh cng nghip, di tr dm tng, mng m tr cu, mng tr in, thp ng ten, ... Mng n c kch thc khng ln lm, mng thng c y hnh vung, ch nht, trn, ... trong dng ch nht c s dng rng ri nht.

(a) (b) (c) (d)

Hnh 2.1: Mt s loi mng n a. Mng n di ct nh: gch, xy, b tng, ... b. Mng n di ct: b tng hoc b tng ct thp. c. Mng n di tr cu. d. Mng n di chn tr in, thp ng ten. Thuc loi mng n, ta xt cu to chi tit cc loi sau 2.1.1. Mng n di tng

1

2

4

3

5 Mng n di tng c p dng hp l khi p lc do tng truyn xung c tr s nh hoc khi nn t tt v c tnh nn ln b. Cc mng ny t cch nhau t 36m dc theo tng v t di cc tng gc nh, ti cc tng ngn chu lc v ti cc ch c ti trng tp trung trn cc mng n, ngi ta t cc dm mng (dm ging).

Hnh 2.2: Cu to mng n di tng 1. Bn mng, m mng;2. Ct truyn lc bng b tng; 3. Dm mng; 4. Lp lt tng; 5. Tng nh.

2.1.2. Mng n di ct v di tr Mng n di ct lm bng hc nh hnh (2.3a). Mng b tng v b tng hc cng c dng tng t. Nu trn mng b tng hoc mng hc l ct thp hoc b tng ct thp th cn phi cu to b phn t ct, b phn ny c tnh ton theo cng ca vt liu xy mng.



Cc mng n lm bng gch xy loi ny, khi chu tc dng ca ti trng (Hnh 2.3b) ti y mng xut hin phn lc nn, phn lc ny tc dng ln y mng, v phn mng cha ra khi chn ct hoc bc b un nh dm cng xn, ng thi mng c th b ct theo mt phng qua mp ct.

1 2 h

l

r(a) (b)

g

ng vt liu nh. Mt bin ca mtruyn

Do v s h/l (gia chiu cao v rn

Hnh 2.3a: Cu to mng n bng hc Hnh 2.3b: S lm vic ca mng 1. ng truyn ng sut; 2. Gc m

y t ca bc mng) phi ln khi phn lc ng phi nm ngoi h thng ng nn r ln v c

ng sut trong khi mng. Do vy quy nh mng cng hay mng mm, ngi ta da vo gc . i vi mng cng phi b hn max no , ngha l t s h/l khng c nh hn cc tr s sau :

p lc trung bnh di y mng P 1,5kG/cm2 P > 1,5kG/cm2

Mc B tng Loi mng 100 < 100 100 < 100

Mng bng 1,5 1,35 1,75 1,5 Mng n 1,65 1,5 2,0 1,65

p lc trung bnh y mng di Mng hc & BT hc khi P mc 2,5k 2G/cm

51

1 ( 2,5kG /cm1,5 1,75 2,00

h t s h/l ca cc bc pha trn phi

hc 30h b , mng gch xy th

ng cn kh g ch t s h/l m cn c nthc hp l ca mng v ct thp.

TRANG 15


Thuc loi mng n b tng ct thp c th ngi ta dng mng n BTCT ti ch khi m dng kt cu lp ghp khng hp l hoc khi ct truyn ti trng ln. Mng b tng ct thp ti ch c th c cu to nhiu bc vt mng.

50

H

50

100b200

50 50 Lp va ximng Mac 50

The

p c

u l

p75

200 va2

00 100

b50 5050 5050 b

>200 5

0

bc75

The

p c

u l

p

Lp va ximng Mac 505050

>200

2 86a200

200

Hnh 2.4 Cu to mt s mng n BTCT ti ch

Di cc mng b tng ct thp, thng ngi ta lm mt lp m si c ti cc cht dnh kt en hoc va xi mng, hoc bng b tng mc thp hoc b tng gch v. Lp m ny c cc tc dng sau: + Trnh h xi mng thm vo t khi b tng. + Gi ct thp v ct pha v tr xc nh, to mt bng thi cng. + Trnh kh nng b tng ln vi t khi thi cng b tng.

- Mng n b tng ct thp lp ghp di ct c cu to bng mt hoc nhiu khi, gim trng lng, ngi ta lm cc khi rng hoc khi c sn vic cu lp thi cng d dng.

I1. Ban2. Sn3. Ngam b tng4. Ct

I

I

I-I

a b

I

I - I

1

2

3

1

2

3 2.2. Mng bng v mng bng giao thoa

Hnh 2.5: Cu to mng lp ghp



Mng bng l loi mng c chiu di rt ln so vi chiu rng, mng bng cn c gi l mng dm, c kin thit di tng nh, mng tng chn, di dy ct. 2.2.1. Mng bng di tng Mng bng di tng c ch to ti ch bng khi xy hc, b tng hc hoc b tng hoc bng cch lp ghp cc khi ln v cc panen b tng ct thp. Mng ti ch ti dng nhng ni m vic lp ghp cc khi l khng hp l.

Hnh 2.6: Cu to mng bng di tng bng xy hoc BTCT Mng bng di tng lp ghp: Cu to gm hai phn chnh: m v tng. m mng bao gm cc khi m, cc khi ny thng khng lm rng v

c thit k nh hnh sn. Cc khi m c t lin nhau hoc vi nhau gi l m khng lin tc. Khi dng cc khi m khng lin tc s lm gim c s lng cc khi nh hnh nhng s lm tr s p lc tiu chun tc dng ln nn t tng ln mt t.

Tng mng c cu to bng cc khi tng rng hoc khng rng v c thit k nh hnh sn.

h

b

a

b

ah

bh

l

I I

II II

II

II - II

I - I

a) b)

c) d)

b.

Tng

m mong

Khi tng mong

a,b - m mongc,d - Tng mong

Hnh 2.7: Cu to mng bng lp ghp 2.2.2. Mng bng di ct



Mng bng di ct c dng khi ti trng ln, cc ct t gn nhau nu dng mng n th t nn khng kh nng chu lc hoc bin dng vt qu tr s cho php.

Dng mng bng b tng ct thp t di hng ct nhm mc ch cn bng ln lch c th xy ra ca cc ct dc theo hng ct .

Khi dng mng bng di ct khng m bo iu kin bin dng hoc sc chu ti ca nn khng th ngi ta dng mng bng giao thoa nhau cn bng ln theo hai hng v tng din chu ti ca mng, gim p lc xung nn t.

Trong cc vng c ng t nn dng mng bng di ct tng s n nh v cng chung c tng ln. Mng bng di ct c ti ch. Vic tnh ton mng bng di ct tin hnh nh tnh ton dm trn nn n hi.

a. Mong bng di ct b. Mong bng giao thoa

Hnh 2.8: Mng bng di ct v mng bng giao thoa

b

a

a

b

LC

LC

L

C=40

0-800

t m chtNhi va Ximng

Hnh 2.9: Mng bng lp ghp



I

I

II

II

I-I II-II

Hnh 2.10: Cu to chi tit mng bng BTCT 2.3. Mng b

L mng b tng ct thp lin khi, c kch thc ln, di ton b cng trnh hoc di n nguyn c ct ra bng khe ln.

Mng b c dng cho nh khung, nh tng chu lc khi ti trng ln hoc trn t yu nu dng phng n mng bng hoc mng bng giao thoa vn khng m bo yu cu k thut. Mng b hay c dng cho mng nh, thp nc, xil, bunke b nc, b bi...

Khi mc nc ngm cao, chng thm cho tng hm ta c th dng phng n mng b,lc mng b lm theo nhim v ngn nc v chng li p lc nc ngm. Mng b c th lm dng bn phng hoc bn sn.

A A B

C D

B

C D

A-A B-B

D-DC-C

a) b)

d)c)

Hnh 2.11: a) Mng b bn phng; b) Mng b bn phng c gia cng

m ct; c) Mng b bn sn di ; d) Mng b bn sn trn



Loi mng bn c th dng khi bc ct khng qu 9m, ti trng tc dng xung mi ct khng qu 100T, b dy bng ly khong 1/6 bc ct.

Khi ti trng ln v bc ct ln hn 9m th dng bn c sn tng cng ca mng, b dy ly khong 1/8-1/10 bc ct, sn ch nn lm theo trc cc dy ct .

Mng b s dng c kh nng gim ln v ln khng u, phn phi li ng sut u trn nn t, thng dng khi nn t yu v ti trng ln. Vic tnh ton mng bn (mng b) c tnh nh bn trn nn n hi. Cc mng B tng ct thp dng hp dng di nh nhiu tng cng thuc loi mng ny.

E E

E-E

Cc mng ny gm hai bn (trn v di) v cc sn tng giao nhau ni cc bn li thnh mt kt cu thng nht 2.4. Mng v: Mng v c nghin cu v p dng cho cc cng trnh nh b cha cc loi cht lng (du, ho cht...), nh tng chu lc.. Mng v l loi mng kinh t vi chi ph vt liu ti thiu, c th chu c ti trng ln, tuy nhin vic tnh ton kh phc tp. Hnh 2.12: Mng hp

3 XC NH KCH THC Y MNG THEO IU KIN P LC

TIU CHUN CA NN T 3.1. Xc nh p lc tiu chun ca nn t Nh ta bit trong l thuyt C hc t: Nu ti trng tc dng trn nn nh hn mt gii hn xc nh ( ) th bin dng ca nn t ch l bin dng nn cht, tc l s gim th tch l rng khi b nn cht, tt dn theo thi gian v nhng kt qu thc nghim cho thy gia ng sut v bin dng c quan h bc nht vi nhau.

1ghP

Nu ti trng tc dng ln nn tip tc tng vt qua tr s th trong nn t hnh thnh cc vng bin dng do do cc ht t trt ln nhau, th tch t khng i v khng nn cht thm. Lc ny quan h gia ng sut v bin dng chuyn sang quan h phi tuyn.

1ghP

z=b/

4

Q

NM

Giai oan

2p(kG/cm)

S(mm)

Pgh1

nen cht

bin dang deoGiai oan

S(mm)

Thi gian T

p

Vung bin dang deo

Hnh 2.13



thit k nn theo trng thi gii hn v bin dng th trc ht phi khng ch ti trng t ln nn khng c ln qu mt tr s quy nh m bo mi quan h bc nht gia ng sut v bin dng, t mi xc nh c bin dng ca nn v tt c cc phng php tnh ln u da vo gi thit nn bin dng tuyn tnh.

1ghP

Ti trng quy nh gii hn ( Pgh1) gi l ti trng tiu chun, hay p lc tiu chun ca nn hay cn gi l p lc tnh ton quy c ca nn. Khi thit k nn mng hay c th l xc nh kch thc y mng th ngi thit k phi chn din tch y mng rng v sao cho ng sut di y mng bng hoc nh hn tr s p lc tiu chun. Vic xc nh p lc tiu chun ca nn t l cng vic u tin khi thit k nn mng, c th xc nh p lc tiu chun theo hai cch sau y. 3.1.1. Xc nh p lc tiu chun theo kinh nghim Tu theo tng loi t v trng thi ca n, theo kinh nghim ngi ta cho sn tr s p lc tiu chun Rtc ca nn nh trong bng sau: Bng 2.1:Tr s p lc tiu chun Rtc ca nn theo kinh nghim Tn t Rtc(kG/cm2) Tn t Rtc(kG/cm2) t mnh ln trng thi 1. t d to c ct nhi trong k h 6,0 t loi st H s st B 2.Cui si l mnh v kt tinh 5,0 (dnh) rng e B=0 B=13. Dm, mnh v trm tch 3,0 8. ct 0,5 3,0 3,0 t ct Rtc(kG/cm2) 0,7 2,5 2,0 trng thi 9. st 0,5 3,0 2,0 t mnh ln Cht Cht va 0,7 2,5 1,8 4. Ct th khng ph thuc m 4,5 3,5 1,0 2,0 1,0 5. Ct va, khng ph thuc m 3,5 2,5 10. St 0,5 6,0 4,0 6. Ct nh: 0,6 5,0 3,0 a. t m 3,0 2,0 0,8 3,0 2,0 b. Rt m 2,5 1,5 1,1 2,5 1,0 7. Ct bi a. t m 2,5 2,0 b. Rt m 2,0 1,5 c. Bo ha nc 1,5 1,0 * Ghi ch: vi cc tr s e, B trung gian, xc nh Rtc bng cch ni suy. Cc tr s trong bng ng vi b rng mng b=1m, hm=1,5 - 2m. Nu b # 1m v hm b # 1,5m th phi hiu chnh: Rtc = R.m.n (2.1) Trong : R - Tr s p lc tiu chun tra theo bng trn. m - H s hiu chnh b rng mng. Khi b 5m th m = 1,5 cho t ct, m = 1,2 cho t loi st. Khi 1 < b < 5m th:



1.4

)1( += bm (2.2) = 0,5 cho t ct. = 0,2 cho t st. n - H s iu chnh su t mng. n = 0,5 + 0,0033.h (khi h < 1,5m)

)200h(k.R.m

1n += (khi h > 2m) (2.3) - Dung trng ca t (tnh ra kG/cm3), h - Chiu su chn mng (cm), k = 1,5 cho t st, k = 2,5 cho t ct, v k = 2,0 cho t st v ct. * Ngoi ra, i vi cc loi t p dng lm nn cng trnh, loi t ny tuy c nhc im l bin dng ln v tnh khng ng nht cao, nhng mt iu kin thch hp n vn dng lm nn cng trnh tt. Theo quy phm, i vi nn t p n nh, tr s p lc tiu chun ca mt s loi t nh sau: Bng 2.2 p lc tiu chun trn nn t p n nh

Rtc (kG/cm2) X ht to, ct va v ct nh

Ct x, x nh, t loi nh

bo ho nc G

t p

G 0,5 G 0,8 G 0,5 G 0,8 1. t san lp theo quy hoch, c m cht 2. t thi b cng nghip c m cht 3. t thi b cng nghip khng m cht 4. t , b thi cng nghip c m cht. 4. t , b thi cng nghip khng m cht.

2,5 2,5 1,8 1,5 1,2

2,0 2,0 1,5 1,2 1,0

1,8 1,8 1,2 1,2 1,0

1,5 1,5 1,0 1,0 0,8

* Ghi ch : Tr s Rtc trong bng dng cho mng c chiu su chn mng h1 > 2m, khi h1 < 2m th tr s Rtc phi gim xung bng cch nhn vi h s K:

1

1

2hhhK += (2.4)

i vi t , b thi cng nghip cha n nh th Rtc nhn vi h s 0,8 Tr s Rtc trung gian ca bo ho G th ni suy. 2.1.2. Xc nh bng cch tnh theo quy phm Theo TCXD 45 - 70 v 45 - 78 cho php tnh ton tr s p lc tiu chun ca nn t khi vng bin dng do pht sinh n su bng 1/4 b rng mng b. Biu thc tnh ton Rtc theo TCXD 45 - 70: Rtc = m[(Ab + Bh) + D.ctc] (2.5) Biu thc tnh ton Rtc theo TCXD 45 - 78:

]c.DBhAb[K

m.mR tc'tc

21tc ++= (2.6) Trong : Rtc - Cng tiu chun ca nn t m - H s iu kin lm vic ca t nn



m=0,8 Khi nn t l t ct nh, bo ho nc m=0,6 - Khi nn t l ct bi, bo ho nc m=1 trong cc trng hp khc ctc - Lc dnh tiu chun ca t di y mng. - Dung trong trung bnh ca t di y mng '- Dung trong trung bnh ca t trn y mng Ktc - H s tin cy, nu cc ch tiu c l c xc nh bng th nghim trc tip i vi t th Ktc ly bng 1,0. Nu cc ch tiu ly theo bng quy phm th Ktc ly bng 1,1. m1,m2 ln lt l h s iu kin lm vic ca nn v h s iu kin lm vic ca cng trnh tc dng qua li vi nn, ly theo bng sau:

Bng 2- 3: Tr s ca m1, m2

Loi t H s

H s m2 i vi nh v cng trnh c s kt cu cng vi t s gia chiu di

4 1,5 t hn ln c cht nht l ct v t st, khng k t phn v bi

1,4 1,2 1,4

Ct mn : - Kh v t m

- No nc

1,3

1,2

1,1

1,1

1,3

1,3Ct bi : - Kh v t m

- No nc

1,2

1,1

1,0

1,0

1,2

1,2t hn ln c cht nht l st v t st c st B 0,5 1,2 1,0 1,1

Nh trn c st B > 0,5 1,1 1,0 1,0 - A,B,D cc h s ph thuc vo tr s gc ni ma st tc tra bng: Bng 2.4 : Tr s A, B v D

Tr s tiu chun ca gc (gc ma

st trong tc (o)A B D

0 0,00 1,00 3,14 2 0,03 1,12 3,32 4 0,06 1,25 3,51 6 0,10 1,39 3,71 8 0,14 1,55 3,93 10 0,18 1,73 4,17 12 0,23 1,94 4,42



14 0,29 2,17 4,69 16 0,36 2,43 5,00 18 0,43 2,72 5,31 20 0,51 3,05 5,66 22 0,61 3,44 6,04 24 0,72 3,87 6,45 26 0,84 4,37 6,90 28 0,98 4,93 7,40 30 1,15 5,59 7,95 32 1,34 6,35 8,55 34 1,55 7,21 9,21 36 1,81 8,25 9,98 38 2,11 9,44 10,80 40 2,46 10,84 11,73 42 2,87 12,50 12,77 44 3,37 14,48 13,96 45 3,66 15,64 14,64

* Nhn xt: Vic xc nh p lc tiu chun theo kinh nghim (tra bng) thng thin v an ton, cc tr s nu ra trong bng i din cho mt dy cc tr s dao ng trong din rng. Trong thc t th cc loi t rt phong ph v loi v trng thi nn xc nh Rtc t cch tra bng thng t chnh xc v khng cht ch v l thuyt. C th s dng tr s ny trong thit k s b, hoc cho cc cng trnh nh t trn nn t tng i ng nht, cng trnh loi IV v loi V. Xc nh Rtc theo TCXD 45 - 70 v 45 - 78 cng cha cht ch lm v mt l thuyt v s pht trin ca vng bin dng do ca t cng khc vi vt th n hi. Tuy nhin khi vng bin dng do cn nh th sai khc cng khng ln, hin nay trong thit k ngi ta hay s dng tr s ny. Trong mt s nghin cu gn y cho thy c th s dng cng tnh ton ca t nn trong tnh ton kch thc mng bng cch tnh ton cng chu ti ca t nn theo cng thc ca Terzaghi hoc Berezantev ri chia cho h s an ton (Fs = 2 - 2,5). Theo quan im ny cho rng ly cng tnh ton nh vy va m bo iu kin bin dng, va m bo iu kin chiu ti. Tuy nhin vic ly tr s Fs chnh xc l bao nhiu th cng cha thng nht. Do vy vic tnh cng tnh ton ca nn t theo phng php no sao cho ph hp vi thc t ca nn t v tnh cht cng trnh m bo ti u trong thit k xy dng cng trnh. 3.2. Xc nh din tch y mng trong trng hp mng chu ti trong ng tm Xt mt mng n chu ti trong ng tm nh hnh v (2.14): Trong iu kin lm vic, mng chu tc dng ca cc lc sau: - Ti trng cng trnh truyn xung mng qua ct mt nh mng: tcON - Trng lng bn thn mng: tcmN



- Trng lng t p trn mng trong phm vi kch thc mng ; tcN

F=axb

tcNdNdtc

tcNo

ptc

b

a

hm

- Phn lc nn t tc dng ln y mng ptc. Biu ng sut tip xc di y mng l ng cong, nhng i vi cu kin mng cng, ta ly gn ng theo dng hnh ch nht. iu kin cn bng tnh hc: (2.7) FpNNN tctctcmtcO .=++Trong : F - Din tch y mng Trng lng ca mng v t p trn mng c th ly bng trng lng ca khi nm trong mt ct t y mng: (2.8) tbmtctcm hFNN ..=+ Hnh 2.14 Trong : tb - Dung trng trung bnh ca vt liu mng v t p trn mng, ly bng 2 - 2,2 (g/cm3) hoc 2 - 2,2 (T/m3). hm - su chn mng. T (2.7) v (2.8) ta c: F.p.h.FN tctbmtc0 =+ Suy ra:

mtb

tc

tco

hpNF

.= (2.9) m bo iu kin nn bin dng tuyn tnh th p lc do ti trong tiu chun ca cng trnh gy ra phi tho iu kin: ptc Rtc (2.10)

Do : mtb

tc

tco

hRNF

. (2.11) Cng thc (2.11) cho php xc nh c din tch y mng F khi bit ti trng ngoi tc dng , p lc tiu chun RtcoN

tc v chiu su chn mng hm. y cn ch rng tr s Rtc ly theo kinh nghim th xc nh c s b din tch y mng F, cn nu Rtc xc nh theo cng thc (2.6) v (2.7) th tham s b rng mng b phi

gi thit trc, sau khi c c din tch y mng F, chn t s ba= tm c

cnh a v kim tra li iu kin F* = axb F. * Xc nh kch thc hp l ca mng n Vic chn kch thc hp l ca mng n y ta cn tm b rng b ca mng v t t s a = .b tm c cnh di a v so snh vi din tch yu cu. Phng php ny xut pht t iu kin: (2.12) tctctb Rp = Vi : Rtc - Cng tiu chun ca nn t



- Cng p lc trung bnh tiu chun do ti trng cng trnh gy ra ti y mng.

tctbp

axb

GNptcotc

tb+= (2.13)

G - Trng lng ca mng v t p trn mng

hoc mtbtcotc

tb hbNp .. 2

+= (2.14) Trong :

ba=

Thay (2.6) v (2.14) vo (2.12) bin i ta c phng trnh bc ba xc nh b rng mng nh sau: (2.15) 0. 2213 =+ KbKbTrong :

mtbtcm

hMcMhMK .... 3211 += (2.16)

...32 mN

MKtco= (2.17)

Vi : M1, M2, M3 - Cc h s ph thuc vo gc ni ma st tc ca t nn, tra bng (2.5). m - H s iu kin lm vic, ly bng 1. - Dung trng ca t nn di y mng. Gii phng trnh (2.15) tm c tr s b - b rng ca mng, t xc nh a da vo iu kin a = b v c c din tch y mng. * Xc nh kch thc hp l ca mng bng i vi mng bng c chiu di ln hn nhiu ln so vi b rng, khi tnh ton ngi ta ct ra 1m di tnh ton, do vy tr s p lc trung bnh tiu chun ti y mng s l:

mtbtcotc

tb hbNp .+= (2.18)

Thay (2.6) v (2.18) vo (2.12) bin i ta c phng trnh bc hai xc nh b rng mng bng nh sau: (2.19) 0. 212 =+ LbLbTrong :

.

..

.. 3

211 m

hMcMhML mtbtcm +=

..32 mN

MLtco=

M1, M2, M3 - Cc h s ph thuc vo gc ni ma st tc ca t nn, tra bng (2.5). Gii phng trnh (2.19) tm c b rng hp l ca mng bng theo iu kin p lc tiu chun. Vic xc nh kch thc mng t vic gii phng trnh (2.15) v (2.19) th khng cn kim tra li iu kin (2.12).



Bng 2.5: Cc Tr s M1, M2, M3tc() M1 M2 M3 tc() M1 M2 M3

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22

74,97 38,51 26,36 20,30 16,66 14,25 12,52 11,24 10,24 9,44 8,80 8,26 7,8 7,42 7,08 6,08 6,54 6,32 6,12 5,91 5,78 5,64

229,2 114,6 76,3 57,2 45,7 38,1 32,6 28,5 25,3 22,7 20,6

18,82 17,32 16,04 14,93 13,95 13,08 12,31 11,62 10,99 10,42 9,90

70,79 34,51 22,36 16,30 12,66 10,25 8,52 7,24 6,24 5,44 4,80 4,26 3,80 3,42 3,08 2,80 2,54 2,32 2,12

1,942 1,783 1,640

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

5,51 5,39 5,29 5,19 5,10 5,02 4,94 4,87 4,82 4,75 4,69 4,64 4,60 4,55 4,52 4,47 4,44 4,41 4,38 4,35 4,32 4,30 4,27

9,12 8,88 8,58 8,20 7,85 7,52 7,21 6,93 6,66 6,40 6,16 5,93 5,71 5,51 5,31 5,12 4,94 4,77 4,60 4,44 4,29 4,14 4,00

1,511 1,393 1,287 1,188 1,099 1,017 0,944 0,872 0,808 0,749 0,694 0,643 0,596 0,552 0,512 0,474 0,439 0,406 0,376 0,347 0,321 0,296 0,273

* Mt s cch gn ng xc nh din tch y mng F + Xc nh Rtc theo cc bng (2.2) hoc (2.3) tu thuc vo tnh hnh c th ca t nn hoc theo gi tr Rtc do th nghim cung cp. Thay tr s Rtc vo cng thc (2.11) xc nh c din tch y mng F, t chn cc kch thc chi tit cho ph hp. Vi mng vng hoc ch nht: F* = axb Vi mng hnh trn : F* = .R2 + Xc nh kch thc mng theo kinh nghim: Chn trc mt tr s kch thc y mng axb no , t kt hp vi iu kin t nn tnh ra Rtc v sau kim tra li iu kin: , nu cha tho mn th chn li v kim tra cho n khi t yu cu, thng thng chn kim tra n ln th hai hoc ba l t.

tctctb Rp

3.3. Trng hp ti trng lch tm



Mng chu ti lch tm l mng c im t ca tng hp lc khng i qua trng tm din tch y mng. Thng l mng cc cng trnh chu momen v ti trng ngang. lch tm e c tnh nh sau:

tctc

NMe = (2.22)

min>0 max>0

Oeb

a

b

ea

Trong : Mtc - Gi tr momen tiu chun ng vi trng tm din tch y mng. Ntc - Tng ti trng thng ng tiu chun tc dng ln mng.

Hnh 2.15: Mng chu ti lch tm 3.3.1. Trng hp lch tm b Trng hp ny lch tm e < a/6, biu ng sut y mng nh hnh v (Hnh 2.15). Vic xc nh kch thc y mng trong trng hp ny ging nh i vi mng chu ti trong ng tm, sau tng din tch tnh ln chu mo men v lc ngang bng cch nhn vi h s K (K= 1,0 - 1,5), khi momen v lc ngang b th ly K b v ngc li. Flch tm = K.Fng tm (2.23) Sau khi chn c kch thc y mng cn kim tra li iu kin p lc: tctc R2,1max (2.24) tctctb R Vi: mtb

y

tco

x

tco

tcotc

minmax, h.WM

WM

FN += (2.25)

hay: mtbbatcotc

minmax, h.)be6

ae6

1(axbN += (2.26)

mtbtco

tcmin

tcmaxtc

tb h.axbN

2+=+=

3.3.2. Trng hp mng chu ti trng lch tm ln Dng biu ng sut trong trng hp ny nh hnh v v 0,0 minmax , trng hp ny sau khi chn din tch y mng cn kim tra li theo iu kin lch tm.

L>0,25aNtcmin0 Lu : Tng ti trng tiu chun t cch mp mng mt on L 0,25a phn cnh mng khng b tch khi mt nn qu 25%.

Hnh 2.16: Mng chu ti lch tm ln



3.4. Mt s bin php lm gim hoc trit tiu phn biu ng sut m di y mng + Thay i kch thc, hnh dng mng

2N1 2N2

max>0min>N1

2N22N1

max>0min>0

Hnh 2.17 . + Thay i trng tm mng

b

a

max>0min0 max>0

Tm ct trung tm mong

Dch tm ct v pha minmaxHoc m rng ay mong v pha

Hnh 2.18 + Cu to h thng dm, ging mng chu momen.

Dm ging doc

Dm ging ngang

Hnh 2.19: Dm v ging mng trit tiu ng sut do lch tm gy ra



3.5. V d mu: V d II-1: Xc nh s b kch thc y mng di ct hnh ch nht kch thc 30x40cm vi t hp ti trng chnh ti mt mng l: N=90,0T, M=2,40Tm, Q=1,20T. Nn t gm hai lp c cc ch tiu c l c bn nh sau: Lp trn: st do cng c: =1.95T/m3, = 200, c=1,8 T/m2Lp di: ct do c: =1.95T/m3, = 220, c=2,0 T/m2Gii :

+ Xc nh ti trng tiu chun ca t hp ti trng chnh : TQQTMMTNN tcotcotco 0,12,1/,0,22,1/,0,752,1/ ======

+ Vt liu lm mng c chn l Btng ct thp. + Chn chiu su chn mng l hm = 2m. + Xc nh kch thc y mng : Do mng chu ti trng lch tm nn kch thc y mng phi tha mn hai

iu kin sau y: - ng sut trung bnh ti y mng phi nh hn hoc bng cng p lc

tiu chun ca nn t. - Tr s ng sut ln nht ti y mng phi nh hn hoc bng 1,2 ln cng

p lc tiu chun ca nn t.

)2(R2,1

)1(Rtcd

max

tcdTB

Kch thc hp l nht ca y mng c xc nh t iu kin (1) trong trng hp xy ra phng trnh. T ta c phng trnh xc nh b rng mng nh sau:

b3 + K1.b2 - K2 = 0

Trong : K1 = M1.h + M2.

h..Mc tb3

K2 = M3 . ..mN tco

- Vi tc = 200 tra bng (2.5) ta c: M1 = 5,91; M2 = 10,99; M3 = 1,942

- H s iu kin lm vic, m = 1 - Chiu su chn mng hm = 2m

- c = 0,18 kG/cm2 = 1,8 T/m2

- tb l dung trng trung bnh ca t ngay y mng v vt liu lm mng, ly tb = 2,2 (T/m3) - : l dung trng ca lp 1, =1,95 (T/m3) - Chn = 1,4 =

ba

K1 = 5,91 . 2 + 10,99. 95,18,1 - 1,942.

95,12.2,2 = 17,58



K2 = 1,942. 4,1.95,1.10,75 = 53,35

Thay vo phng trnh trn ta c phng trnh sau: b3 + 17,58.b2 - 53,35 = 0

Gii phng trnh ny bng phng php th dn nghim b 1,663 (m), chn b = 1,7 (m) Do t s = 1,4 =

ba a = 1,4.1,7 = 2,38, chn a = 2,4 (m)

Vy kch thc s b y mng c chn l : b = 1,7m, a = 2,4 m + Tnh Cng tiu chun Rtc ca nn t Cng tiu chun Rtc ca nn t c xc nh theo cng thc sau:

Rtc = m. [(A.b + B.hm). + D.c ] Vi: m=1 ; b=1,7m ; hm=2m ; =1,95(T/m3) ; c = 0,18 KG/cm2= 1,8 T/m2, = 200 Tra bng (2.4) ta c: A = 0,51, B = 3,06, D = 5,66. Suy ra:

Rtc = (0,51.1,7 + 3,06.2 ) 1,95 + 5,66.1,8 = 23,8 (T/m2) + Xc nh ng sut di y mng :

=be6

ae6

1b.a

N batcdtc

Min,Max

Trong : )(952,922.4,2.7,1.2,20,75.. ThFNN tbtcotcd =+=+= eb = 0; 0533,00,75

2.0,10,2. =+=+= tc mtco

tco

a NhQMe m

)/(75,19)/(82,25

4,20533,0.61

4,2.7,1952,92

2

2

, mTmTtc

MinMax =

=

)/(78,224,2.7,1

0,752.2,2. 2mTF

Nhtco

tbtctb =+=+=

Kim tra iu kin

==


+ Chn kch thc ban u: b rng mng b=1,6m + Xc nh Rtc theo TCXD 45-78:

]c.DBhAb[K

m.mR tc'tc

21tc ++= Vi = 220 tra bng (2.4) ta c : A=0,61 ; B=3,44 ; D=6,04 m1 = 1,0 ; m1 = 1,4; Ktc = 1,1 ; 3m/T87,12/)94,18,1(' =+=Thay vo c: 2tc m/T3,29]9,1.04,687,1.5,1.44,394,1.6,1.61,0[

1,14,1.1R =++=

+ Din tch y mng yu cu:

254,25,1.23,29

2,1/15,80.

mhR

NFmtb

tc

tco ==

Mng chu ti trng lch tm, ta tng kch thc mng ln bng cch nhn vi h s K=1,2 2* 05,354,2.2,1. mFKF ===Vy cnh di ca mng l: a=F*/b=3,05/1,6=1,905m; ta chn a=2m + Xc nh ng sut di y mng :

mtbbatcotc

minmax, h.)be6

ae6

1(axbN +=

Vi : mN

hQMe tco

mtco

tco

a 054,02,1/15,802,1/4,1.5,12,1/25,2. =+=+= , eb =0

Vy : 2max /27,275,1.2)2054,0.61(

6,1.22,1/15,80.)61( mTh

ae

axbN

mtba

tcotc =++=++=

2min /47,205,1.2)2054,0.61(

6,1.22,1/15,80.)61( mTh

ae

axbN

mtba

tcotc =+=+=

2/87,235,1.26,1.2

2,1/15,80. mThaxbN

mtb

tcotc

tb =+=+= Kim tra iu kin: 22max /16,353,29.2,12,1/27,27 mTRmT tctc === 22 /3,29/87,23 mTRmT tctctb ==t yu cu, vy kch thc mng chn F=axb = 2x1,6m l hp l. V d II-3: Xc nh s b kch thc mng bng di tng dy 20cm vi t hp ti trng chnh ti mt mng l: N=30T/m, M=2,5Tm/m. Nn t gm hai lp c cc ch tiu c l nh v d 2. Gii:

+ Chn vt liu : mng b tng ct thp + Chn su chn mng : hm = 1,2m + Chn kch thc ban u: b rng mng b=1,5m + Xc nh ng sut di y mng (vi mng bng ta ct ra 1m di tnh ton):



mtbbtcotc

minmax, h.)be6

1(xb1

N +=

Vi : mNMe tc

o

tco

b 083,02,1/302,1/5,2 === , ea =0

Vy : 2max /22,255,1.2)5.1083,0.61(

5,1.12,1/30.)61(

1mTh

be

xbN

mtbb

tcotc =++=++=

2min /13,145,1.2)5,1083,0.61(

5,1.12,1/30.)61(

1mTh

be

xbN

mtbb

tcotc =+=+=

2/67,195,1.25,1.12,1/30. mTh

axbN

mtb

tcotc

tb =+=+= + Xc nh Rtc theo TCXD 45-78 nh v d 2 ta c: Rtc = 29,15T/m2 + Kim tra iu kin:

22max /98,3415,29.2,12,1/22,25 mTRmT tctc === 22 /15,29/2,67,19 mTRmT tctctb ==t yu cu, vy b rng mng bng chn b = 1,5m l hp l.

4 TNH TON NN THEO TRNG THI GII HN V BIN DNG (TTGH II)

4.1. Khi nim: Sau khi xc nh c kch thc y mng theo iu kin p lc tiu chun, ta phi kim tra li nn theo trng thi gii hn v bin dng, hay cn gi l TTGH II. Ni dung ca phn tnh ton ny nhm khng ch bin dng ca nn, khng cho bin dng ca nn ln ti mc lm nt n, h hng cng trnh bn trn hoc lm cho cng trnh bn trn nghing lch ln, khng tho mn iu kin s dng. m bo yu cu trn th ln ca nn phi tho iu kin: Stt [Sgh] (2.27) Trong : Stt - ln tnh ton ca cng trnh thit k [Sgh] - Tr s gii hn v bin dng ca cng trnh, tr s ny ph thuc vo: + c tnh ca cng trnh bn trn: Vt liu, hnh thc kt cu, cng khng gian v tnh nhy cm vi bin dng ca nn... + Ph thuc vo c tnh ca nn: Loi t, trng thi v tnh bin dng ca t, phn b cc lp t trong nn... + Ph thuc vo phng php thi cng. Tr s ln gii hn [Sgh] theo TCXD quy nh tu thuc vo tnh hnh c th ca cng trnh, ly theo bng sau: Bng 2.6 Tr s gii hn v ln ca mng

Tr s [Sgh] (cm) Kt cu nh v kiu mng Trung bnh Tuyt i



1. Nh Panen ln, nh Blc khng c khung 2. Nh bng tng gch, tng Blc ln, mng n c: L:H 2,5 ( L chiu di tng, H chiu cao L:H 1,5 3. Nh tng gch, tng Blc ln c ging BTCT hoc gch, ct thp. 5. Nh khung trn ton b s 5. Mng BTCT kn khp ca l nung, ng khi, thp nc. 6. Mng nh cng nghip mt tng v nh c kt cu tng t khi bc ct l: 6m 12m

8 8 10 15 10 30 - -

- - - - - - 8 10

Ngoi ra ta cn c bit ch n chnh lch ln hay ln khng u ca cc mng trong cng mt cng trnh. Nu tr s ny ln s gy ra s phn b li ni lc trong kt cu bn trn, lm nt gy kt cu. chnh lch ln c nh gi qua cc i lng: - ln lch tuyt i: S = S2 S1 [Sgh] (2.28) - nghing ca mng hoc cng trnh: L t s gia ln ca cc im bn ngoi ca mng ( hoc cng trnh) vi kch thc (chiu di, chiu rng) qua im y:

L

SStg 12 = (2.29) LS1 S2Gc nghing:

LSSarctg 12 = (2.30)

Tr s gc nghing ny phi b hn tr s gc nghing gii hn, quy nh theo quy trnh. Hnh 2.20 4.2. Tnh ton ln ca mng Hin nay c nhiu phng php khc nhau tnh ton ln ca nn mng, mt s phng php c trnh by k trong gio trnh C hc t. Trong ni dung ny ch gii thiu nhng bc c bn ca phng php cng ln tng lp. y l mt trong nhng phng php c ch nht v cho kt qu gn st vi thc t nht. Ni dung ca phng php cng ln tng lp: 1. Chia nn t di y mng thnh nhiu lp c chiu dy hi (0,2 - 0,4)b hoc hi 1/10 Ha, vi b l b rng mng, Ha l chiu su vng nn p. 2. Tnh v v biu ng sut do trng lng bn thn t: (2.31) iibtzi h = 3. Xc nh p lc gy ln: gl (2.32) mtb hgl . = Trong : - p lc trung bnh ti y mng do ti trng cng trnh v trng lng mng, t p trn mng gy ra:

tb



axb

GNtcotb

+= (2.33) - Dung trng ca lp t t mng hm - Chiu su chn mng 4. Tnh v v biu ng sut do ng sut gy ln gy ra: (3.34) gloizi K .= Vi Koi =f(a/b,2z/b) tra bng trong sch C hc t. 5. Xc nh chiu su vng nh hng Ha, theo TCXD 45-70, Xc nh Ha da vo iu kin ni c : glzibtzi .2,0= 6.Tnh ton ln ca cc lp t phn t Si theo cc cng thc:

iioi

i

iioii

iii

ii

ii

iii

hpE

S

hpaS

hpe

aS

heeeS

..

..

..1

.1

1

1

21

==

+=+=

(2.35)

7. Tnh ton ln cui cng ca mng:

(2.36) =

=n

iiSS

1

Xc nh e1i v e2i tng ng vi cc tr s p1i v p2i vi

2

1

1

zibt

zibt

ip +=

(2.37)

2

1

12

zigl

zigl

ii pp ++=

(2.38)

Ha

p1i

btzp(kG/cm)

zO p2ip1i 2

p2i

b

hm

ing cong nen lun

e1i

hi

e2i

p(kG/cm)2

e 5. TNH TON NN THEO TRNG THI GII HN V CNG Hnh 2.21: S chia lp t v ng cong nn ln

(TTGH I)



5.1. Khi nim Khi ti trng ngoi vt qu kh nng chu lc ca nn t, nn b ph hng v mt cng , n nh, lc ny nn c xem l t n trng thi gii hn th nht. i vi nn , khi t n TTGH1 th nn khng cn kh nng chu ti na v nn b ph hoi. i vi nn t, khi t n TTGH1 th xy ra hin tng ln t ngt, lm ph hng cng trnh bn trn. Phm vi s dng tnh ton nn theo TTGH1: + Nn . + Nn st rt cng, ct rt cht, t na . + Nn st yu, bo ha nc v t than bn. + Nn t mng thng xuyn chu ti trng ngang. + Nn ca cng trnh trn mi dc. Ti trng tnh ton: Dng ti trng tnh ton v t hp b sung. iu kin kim tra: Mun cho nn t khng b ph hng, mt n nh (trt, tri) th ti trng truyn ln mng cng trnh tc dng ln nn t phi c cng nh hn cng gii hn ca nn t y.

atK

N (2.39) Pgh

S(mm)

p(kG/cm)2Pgh

Nn t mm yu

Nn t cng

Trong : N - Ti trong cng trnh tc dng ln mng - Kh nng chu ti ca nn theo phng tc dng ca ti trng Kat - H s an ton, do c quan thit k quy nh, h s ny ph thuc vo cp nh, cp cng trnh, ngha v hu qu ca vic nn mt kh nng chu ti, mc nghin cu cc iu kin ca nn t, thng chn >1. Khi tnh ton nn theo TTGH1, lc ny ti trng khi gn t n TTGH1 l rt ln, ti nhiu vng trong nn t quan h ng sut bin dng khng cn bc nht na, lc ny khng th gii quyt bi ton theo kt qu ca l thuyt n hi na m phi gii quyt theo hai hng s trnh by cc mc sau. 5.2. Sc chu ti ca nn i vi nn , tnh nn ln ca n rt b, khng ng k, moun bin dng ca c th ln hn moun bin dng ca t hng ngn ln. C khi ng sut tc dng ln nn gn t n tr s ph hoi m bin dng ca n cn rt b. V vy ngi ta khng cn kim tra bin dng ca nn m ch cn tnh ton v kim tra nn theo TTGH1 v cng .

Hnh 2.22: Quan h P-S

Nn aNn t

2p(kG/cm)

S(mm)

Hnh 2.23



Sc chu ti tnh ton R ca nn c xc nh theo biu thc: R = k.m.Rn (2.40) Trong : Rn - Cng ph hoi ca mu khi b nn mt trc trng thi bo ho nc. k - H s ng nht m - H s iu kin lm vic i vi cc trng hp c th, cn tin hnh th nghim xc nh cc tr s cn thit. Khi khng c s liu, ngi ta thng ly k.m = 0,5. 5.3. Sc chu ti ca nn t 5.3.1. Phng php gii tch Vic tnh ton sc chu ti ca nn t c gii thiu k trong C hc t . y ch gii thiu li mt s biu thc tnh ton sc chu ti c bn:

q=.h

p

pgh

z

b Y

5.3.1.1. Phng php ca Xocolovski a: nn t chu ti trng thng ng, lch tm (Hnh 2 - 24) Ti trng gii hn trong trng hp ny c tnh theo cng thc sau: pgh = pT .(c + q.tg)+ q (2.41) Trong : pT : h s khng th nguyn ph thuc vo YT v , tra bng (2-7)

Hnh 2.24

YT = ctgq +. . y Vi: 0 y b (2.42)

T cng thc (2-41), ta suy ra cc trng hp c bit sau: + Khi mng t trn mt t dnh ( h=0, c0) th: pgh= pt .c (2.43)

Trong : pT ph thuc vo YT = yc.

Khi mng t trn t ct ( c=0, q 0, h/b< 0.5) pgh =q(pT .tg+ 1) (2.44) Trong : pT = tgq .

. y Bng 2- 7: Tr s ca pT.



() YT

5 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40

0 6,49 8,34 11,0 14,8 20,7 30,1 46,1 75,3 0,5 7,73 0,02 12,5 17,9 27,0 43,0 73,8 139 1,0 6,95 9,64 13,8 20,6 32,3 53,9 97,1 193 1,5 7,17 10,20 15,1 20,1 37,3 64,0 119 243 2,0 7,38 10,80 16,2 25,4 41,9 73,6 140 292 2,5 7,56 11,30 17,3 27,7 46,4 82,9 160 339 9,0 7,77 11,80 18,4 29,8 50,8 91,8 179 386 3,5 7,96 12,30 19,4 31,9 55,0 101,0 199 432 4,0 8,15 12,80 20,5 34,0 59,2 109 218 478 4,5 8,33 13,20 21,4 36,0 63,8 118 237 523 5,0 8,50 13,70 22,4 38,0 67,3 127 256 568 5,5 8,67 14,10 23,3 39,9 71,3 135 275 613 6,0 8,84 14,50 24,3 41,8 75,3 143 293 658

b. Nn t chu ti trng nghing, lch tm (hnh 2 - 25):

Thnh phn thng ng ca ti trng gii hn (pgh) trong trng hp ny c xc nh nh sau: pgh = N..y + Nq..h + Nc.c (2.45)

Trong : N, Nq, Nc- cc h s sc chu ti ca t ph thuc vo gc ma st trong ca t v gc nghing ca ti trng, ly theo bng (2.8). Thnh phn nm ngang gh ca ti trng gii hn xc nh theo cng thc: gh = pgh . tg (2 .46) Biu ti trng tnh theo cng thc (2.45) c dng hnh thang, cc tr s ca pgh ti im y = 0 v y = b c tnh nh sau (b: chiu rng ca mng hnh bng)

Pgh

gh

q=.h

b

Z

Y

P

Hnh 2-25

p = N( )0=ygh q..h + Nc.c (2.47)

p = p + N( )bygh = ( )0=ygh ..b Hai thnh phn thng ng v nm ngang ca tng hp lc ti trng gii hn xc nh theo cc cng thc sau y:



pgh = 2

1 .(pgh(y=0) + pgh(y=b)).b

gh = pgh.tg i vi trng hp ti trng lch tm nh trn (c hai trng hp a v b) thc ra nu mun tnh ton sc chu ti ca nn cho cht ch th khng nhng ch kim tra tr s pgh v p, m cn phi kim tra c im t ca ti trng na (im t ca pgh phi trng vi im t ca p do ti trng ngoi tc dng. Nhng theo li gii ca V.V.Xcolovxki th ti trng gii hn pgh ch c mt im t nht nh vi lch tm egh:

egh = 3

b . ( ) ( )( ) ( )

+

+==

==23

pppp.2

0yghbygh

0yghbygh (2.49)

Bng 2-8: Tr s ca Nq, Nc v N 5 10 15 20 25 30 35 40 45

0 NqNcN

1,57 6,49 0,17

2,47 8,34 0,56

3,94 11,0 1,4

6,40 14,90 3,16

10,70 20,7 6,92

18,4 30,2 15,32

33,30 46,20 35,16

64,20 75,30 86,46

134,50 133,50 236,30

5 NqNcN

1,24 2,72 0,09

2,46 6,56 0,38

3,44 9,12 0,99

5,56 12,52 2,31

9,17 17,50 5,02

15,60 25,40 11,10

27,90 38,40 24,38

52,70 61,60 61,38

96,40 95,40

163,30

10 NqNcN

1,50 2,84 0,17

2,84 6,88 0,62

4,65 10,00 1,51

7,65 14,30 3,42

12,90 20,60 7,64

22,80 31,10 17,40

42,40 49,30 41,78

85,10 84,10 109,50

15 NqNcN

1,77 2,94 0,25

3,64 7,27 0,89

6,13 11,00 2,15

10,40 16,20 4,93

18,10 24,50 11,34

33,30 38,50 27,61

65,40 64,40 70,58

20 NqNcN

2,09 3,00 0,32

4,58 7,68 1,19

7,97 21,10 2,92

13,90 18,50 6,91

25,40 29,10 16,41

49,20 48,20 43,00

25 NqNcN

2,41 3,03 0,38

5,67 8,09 1,50

10,20 13,20 3,84

18,70 21,10 9,58

26,75 35,75 24,86

30 NqNcN

2,75 3,02 0,43

8,94 8,49 1,84

13,10 14,40 4,96

25,40 24,40 13,31

35 NqNcN

3,08 2,97 0,47

8,43 8,86 2,21

16,72 15,72 6,41

40 NqNcN

3,42 2,88 0,49

10,15 9,15 2,60

45 Nq 3,78

(2.48)



NcN

2,70 0,50

5.3.1.2. Phng php ca Terzaghi

K.Terzaghi a ra cng thc tnh ti trng gii hn trng hp bi ton phng nh sau:

c.Nh..N2b..Np cqgh ++=

(2.50) Z

b

p q=.h

/4/2 /4/2

q=.h

/4/2/4/2

Trong : N, Nq v Nc - Cc h s sc chu ti, ph thuc vo gc ma st v tnh theo biu (hnh 2.27). N

Hnh 2-26: S tnh ton i vi bi ton phng ca K.Terzaghi

goi ra K.Terzaghi cn a ra cc h s kinh nghim vo cng thc (2-50)

(2.51)

(2.52)

4 0

o3 0

o2 0

Nq

cNN

Nc , qN N

tnh ti trng gii hn trong trng hp bi ton khng gian. - i vi mng vung c cnh l b:

pgh = 0,4.N..b + Nq..h + 1,3Nc.c - i vi mng trn c bn knh R:

pgh = 0,6.N..R + Nq.h + 1,3.Nc.c o o

6 0 5 0 4 0 3 0 2 0 1 0 0 2 0 4 0 6 0 8 00o

o1 0

.3.2. Phng php gi xc nh kh nng chu ti ca t c nh kh nng chu

ti

n t khng ng nht, gm hai hoc ba lp t c ch tiu c l khc nhau

ng t trn mt tng t phn b rt dc.

Hnh 2-27: Biu tra N, Nq v Nc 5

Trong trng hp khng th dng phng php gii tch xca nn c, lc ny ta phi s dng phng php gii. Mt s trng hp hay

gp l: - N- Ph ti hai bn mng chnh lch nhau qu 25% - Mng t trn mi dc, mt di mi dc hoc m



Trong nhng trng hp trn ta phng phdng p gii vi gi

thi

(2.53) Tro

t mt trt l mt tr trn. Theo phng php ny, ngi ta tnh theo s bi ton phng: Ct ra mt on di 1 n v tnh ton. Vi nhng mng bng, tng chn t, nn ng c chiu di ln mi ph hp vi bi ton phng. Nhng vi mng hnh ch nht ngi ta vn tnh theo s bi ton phng thun tin v thin v an ton. Ni dungca phng php nh sau: + Gi thit mt trt l mt cung trn i qua mp mng tm O1, bnknh R1. Chia lng th trt thnh nhi+ Ta xt mnh th i:

- Tng cc lc tc dng ln mnh i:

u mnh bng cc mt ct thng ng (Hnh2.28)

Gi = pi + qi ng : .ii Fq =

- Din tch mnh th i

ti y mng. i

i v khng c pi

ii bp = . p Vi iF - Dun ag trng c t p C ng ng sut tnh ton bi b rng mnh th i nhng mnh nm ngoi phm vi y mng th Lc gy trt mnh th i: iig sin. Lc gi mnh th i: Lc ma st: iii tgg .cos.

Lc dnh: ii lc . Trong : gc nghin bn knh vi g ca I tm trt i

on cung trt th i :

li Chiu di cung trt ci,i tr s lc dnh v gc ni ma st trong H s n nh Ki cho mnh trt i

ii

iiiii

gtrii gRM

K sin.== gi lctggRM )cos( + (2.54)

Xt ton b lng th trt gm n m nh, ta c h s n nh:

=

+== i

iiiiig

lctggRMK 1

)cos( (

=

n

iii

n

gtri

gRM1

sin. 2.55)

Trong : Mg, Mgtr l mo men gi v momen gy trt ca lng th trt

i

i

O

gi.cosigi

biCB

AD pgh

gi.sini

li

c.li

R

b

p

Hnh 2.28



Sau khi xc nh c cc tr s K i vi mi cung trt, ta chn tr s nh nht Kmin xt n nh ca nn. Mun nn n nh phi tha mn iu kin sau: KK min > (2.56)

rong T K - h

c, nu di y mng

nh lt c tin hnh so vi

u kin kim tra:

s n nh cho php, ly t 1,2-1,5. 5.4. Kim tra lt v trt cho mng 5.4.1. Kim tra n nh lt

Trong qu trnh chu lxut hin biu ng sut m, tc min < 0

th mng c kh nng b lt, do vy cn phi kim tra n nh lt ca mng.

Vic kim tra n trc i qua mp ngoi ca y mng (im

O) di tc dng ca t hp ti trng tnh ton bt li nht. i

[ ]gi KMK = lglM

(2.56) Trong : - Tng momen gi mng

y vi en gy lt cho mng, ly vi mp mng.

khng nm

ng ngang Q s lm cho m

m bo mng khng b trt th phi tha m

rong : - T th

h ti y

hF.

hm

No

max>0min1). Q tng ti trng ngang tc dng ln mng.

f H s ma sat gia t v nn ph thuc v s f ca hoc b tng vi cc loi t khc nhau ly theo bng sau: Bng 2.9: Tr s ca f



Loi t di y mng Tr s f 1. t st v nham thch c b mt b bo mn 0,25 2. t st trng thi cng 0,3 3. t st trng thi do 0,2 4. Ct m t 0,55 5. Ct m 0,45 6. st trng thi cng 0,45 7. st trng thi do 0,25 8. ct trng thi cng 0,5 9. ct trng thi do 0,35 10. t 0,75

Trong thc t i vi cc mng ca cc cng trnh xy dng dn dng v cng

nghip

6. TNH TON MNG THEO TRNG THI GII HN I tn

hu lc ca mt mng n nh hnh v. B qua lc ngang v ma st

ti trng cng trnh tc dng trn

din tch y m

c nh vy, mng c kh

t ct trc

2. ca ng thng

3. ca momen un. Trong phm vi chn ct hoc

Tnh ton mng theo trng thi gii hn I, hay ni cch khc l tnh ton bn ca m

o ca mng, bc mng.

No

MoQo

maxmin

1

, cc iu kin lt v trt u tha mn. iu kin ny cn c kim tra cht ch i vi cc cng trnh c din tch y mng hp, chiu cao ln, chu ti trng ngang, ti trng nh ln nh thp ngten, thp nc, tr in 6.1. S h ton

Ta xt trng thi c trn mt bn ca mng. Vt th mng chu tc

dng ca cc lc sau: - Lc tc dng do

2

Vt nt

tttt

Hnh 2.31: Cc hnh thc ph hoi ca mng khi chu ti

ton din tch y mng trn mt din tch hp (chn ct hoc chn tng chu lc).

- Phn lc nn tc dng trn ton ng, c chiu ngc li. Trong iu kin chu lnng b ph hng theo cc kiu sau: 1. Mng b chc thng bi ng su

tip trn tit din xung quanh chn ct hoc chn tng (ng 1 trn hnh v). Mng b chc thng do tc dng sut ko chnh, lc ny mt ph hng l mtng. (ng 2 trn hnh v). Mng b nt gy do tc dng

nghing 45o so vi phng

chn tng, cng ca kt cu mng rt ln, nn c th xem mng b ngm ti , phn mng cha ra ngoi chn ct (hoc chn tng) b un nh dm cng xn.

ng. Ni dung chnh l xc nh kch thc ca mng v cu to cho hp l, m bo cho mng khng b ph hng theo nhng kiu nu trn. Vic tnh ton gm hai ni dung chnh sau y:

- Tnh ton chiu ca



- Tnh ton v b tr ct thp i vi mng b tng ct thp Kh n sung.

o iu kin ct trc tip

i t h ton mng theo TTGH I dng ti trng tnh ton, t hp b6.2. Xc nh chiu cao ca mng cng 6.2.1. Xc nh chiu cao mng cng theXt s mng cng chu tc dng ca ti trng nh hnh v: iu kin bn ca mng:

cc

tto RN = hu.

(2.58)

Trong : - ng sut ct do ti trng cng trnh gy ra

cng trnh t

ln m chiu cao ca mng tnh theo iu kin

bn chn bn ta c:

ttoN -tng ti trng thng ng tnh ton ca tc dng ln mng ti mt nh mng.

u- chu vi tit din ngang ca ct hay tngng hc ng ct T iu ki

c

ttoNh c Ru. (2.59)

Theo kinh nghim cho thy nu mng c cu to p l

hiu cao mng theo iu kin

ng chu un nh hnh v (2.33). Khi

iu kin bn:

h th iu kin ph hoi ny lun tha mn. Trong thit k mng c th tnh ton chiu cao mng t cng thc (2.59) hoc chn mt gi tr ri kim tra li theo cng thc (2.58). 6.2.2. Xc nh cbn chng un Xt mt mchu tc dng ca ti trng ngoi (N,M,Q), di y mng pht sinh phn lc nn, phn lc ny gy ra momen un phn cha ra ca mng (phn ny lm vic nh dm cng xn) nn c th gy ra nt gy mng.

kuRM W

(2.60)

men un do phn lc nn gy ra ti tit din tnh ton (I-I) v (II-II) Trong : M mo

2maxmax ).(.125,0... ccII aabaaaabM =42 ctttt=

2maxmax ).(.125,04

.2

.. cttcc

ttIIII bba

bbbbaM ==

(Lu : y thin v an ton ta s dng tnh ton momnen ti tit din) tn

No

Qo Mo

Mt pha hoai

hc

Hnh 2.32: Dng ph hoi th nht

hu

No

MoQo

min maxtt

a

b

I

I

II IIbc

ac

Hnh 2.33

maxtt

W momen chng un ca tit din h ton:



6. 2uII hbW =

6. 2uIIII haW =

Rku cng chu ko khi un ca vt liu mng T thay vo (2.60) ta tnh c chiu cao ca mng theo iu in bn chu momen un nh sau:

ku

ttc

Iu R

aahmax

)(87,0 (2.61)

ku

ttc

Iu R

bbhmax

)(87,0 (2.62)

Chiu cao mng chn: ),max( IIuIuu hhh =Lu : khi tnh ton mng b tng chu un, dng iu kin (2.60) khi xc nh momnen chng un W, c k n tnh khng n hi ca vt liu. Theo TCXD 41-70 cho php tnh gn ng nh sau:

5,3. 2uhbW =

T kt hp vi iu kin (2.60) ta rt ra:

ku

ttc

Iu R

aahmax

)(66,0 (2.63)

ku

ttc

Iu R

bbhmax

)(66,0 (2.64)

6.2.3. Xc nh chiu cao mng theo iu kin chng chc thng trn mt phng nghing Theo iu kin ny ngi ta cho rng nu mng b chc thng th s chc thng xy ra theo b mt hnh chp ct c cc mt bn xut pht t chn ct, v nghing mt gc 45o so vi phng thng ng.

mng khng b chc thng th sc chng chc thng ca thn mng phi ln hn lc gy ra chc thng. iu kin bn: (2.65) ntbkttct hURP ...75,0 Trong : tt - Lc chc thng tnh ton, c tnh bng hiu s gia lc dc tnh ton v phn lc nn trong phm vi y thp chc thng.

ctPttoN

cttttb

tto

ttct FNP .=

vi: b.a

N2

tto

ttmin

ttmaxtt

tb

F din tch

=+= ct y thp chc th ng


ac

bc b

a

ttmaxmin

Qo Mo

o

hn =45o

bct

act

Hnh 2. 34


Fct = act.bct , vi : tghaa ncct 2+= , tghbb ncct 2+= ; ac, bc cnh di v rng ca ct. 0,75 h s thc nghim, k n s gim cng chc thng ca b tng so vi cng chu ko. Utb Chu vi trung bnh ca thp chc thng

2

dttb

UUU

+= Vi Ut = 2(ac + bc), Ud = 2(act + bct)= 2(ac + bc+4hntg) Utb = 2(ac + bc+2hn) (vi =45o, tg = 1) hn chiu cao mng tnh theo iu kin chng chc thng Rk cng chu ko tnh ton ca b tng Thay cc gi tr tm c vo (2.65) v gii phng trnh bc hai, tm c gi tr ca hn, hoc c th chn trc gi tr ca hn ri thay vo (2.65) kim tra cho tha iu kin bn.

Chiu cao mng chn cui cng h=max(hc, hu, hn) V d II-4: Xc nh chiu cao mng ca mng la chn kch thc trong v d 1 Gii: Cc thng s s b: axb= 2,4x1,7m, hm=2m, Ntt = 120,3T, Mtt=3,2Tm

Vt liu mng: B tng ti ch mac 200, cng tnh ton: Rn=900T/m2, Rku=65T/m2

ng sut tnh ton ti y mng: , , .

2ttmax m/T68,33= 2ttmin m/T28,25=

2tttb m/T48,29r ==

+ Chiu cao mng xc nh theo iu kin bn chng un:

)m(77,065

48,29)65,04,2(66,0R

r)aa.(66,0hK

cII

m ==

)m(62,065

48,29)30,07,1(66,0R

r)bb.(66,0hK

cIIII

m == V mng thit k l mng B tng ct thp, ton b ng sut ko do momen un gy ra do ct thp tip thu nn ta chn chiu cao mng hm = 0,7m. + Chiu cao mng bo m bn chng chc thng iu kin bn: ntbkttct hURP ...75,0Vi: - Pct = NTT - r (ac + 2hn) . (bc + 2hn) Pct = 120,3 - 29,48. (0,65 + 2hn) . (0,30 + 2hn)

- Ut = 2(ac + bc), Ud = 2(act + bct)= 2(ac + bc+4hntg) Utb = 2(ac + bc+2hn )=2(0,95+2hn) hn = ho Chiu cao lm vic ca mng

Thay vo iu kin bn, ta c bt phng trnh sau: 120,3 - 29,48 (0,65 + 2ho) . (0,30 + 2ho) 0,75.65. 2 . (0,95 + 2ho).ho

Gi s chn ho = 0,65 (m), thay vo bt phng trinh ta c: 120,3 - 29,48 (0,65 + 2.0,65) . (0,30 + 2. 0,65) 0,75. 65. 2 . (0,95 + 2.0,65).0,65 28,32 < 142,6 tha mn

Vy ta chn chiu cao mng hm = ho + 0,05=0,65+0,05 = 0,7 (m)



S tnh ton chu un v chc thng nh hnh v:

act

bct

o=45 ho

No

MoQo

min maxtt

2400

1700bc

acac

bcIIII

I

I

1700

2400

ttmaxmin

Qo Mo

No

hu Hnh 2.35

6.3. Tnh bn ca mng b tng ct thp 6.3.1. Xc nh chiu cao ca mng Btng ct thp

Chiu cao ca mng btng ct thp phi c tnh ton v kim tra theo iu kin chc thng (2.65) v ch thay chiu cao hn bng chiu cao ho. S d vy l v mc d l mng btng ct thp nhng ngi ta vn t ra yu cu l mng bn chng chc thng m khng c ct thp. o=45 ho

No

MoQo

min maxtt

6.3.2. Tnh bn chu un ca mng BTCT Tnh bn chu un ca mng BTCT tc l tnh

ton xc nh hm lng ct thp cn t trong mng chu momen un. Khi tnh ton ct thp trong mng ngi ta da vo hai gi thit sau: Hnh 2.36

- Ton b ng sut ko do ct thp tip thu. - Cnh tay n ngu lc ly bng 0,9ho vi ho l chiu cao lm vic ca mng:

ho= h-c, vi c l chiu dy lp btng bo v. Din tch ct thp trong mng tnh theo biu thc:

aao

tttd

a RmhMF

...9,0= (2.66)

Trong : Ra Cng chu ko tnh ton ca ct thp ma H s iu kin lm vic ca ct thp trong mng ly t 0,85-0,95.

tttdM - Momen ti cc tit din tnh ton (M

I-I, MII-II). Sau khi xc nh c hm lng ct thp, chn ng knh ct thp, tnh ton s thanh v b tr ct thp cho mng.



IIac

bc ba

100 100

I

I

II

2

1

1

2

3

4

100

30d

Ct thep s 1: Chu lc do mo men tai mt ngam I-ICt thep s 2: Chu lc do mo men tai mt ngam II-IIY cu: Ct thep co >10 mm, khoang cach a = 10-25cmCt thep 3- Ct thep chu lc cua ct, b tr oan ch trn cot mt oan L = 30d (d - ng knh ct thep)Ct thep 4 - Ct thep ai, 6 8 , a = 20cm

V d II-5: Tnh ton v b tr ct thp cho mng xc nh kch thc nh v d II-4:

Hnh 2.37: B tr ct thp cho mng

Gii:

Chn s dng thp mng loi AII c Ra=26000T/m2

Tnh mmen un ln nht - Theo phng cnh di

)(19,19)65,04,2(.7,1.48,29.125,0)(...125,0 22 TmaabrM c

IIMax ===

- Theo phng cnh ngn

)(33,17)3,07,1(.4,2.48,29.125,0)(...125,0 22 TmbbarM c

IIIIMax ===

Tnh v b tr ct thp Theo phng cnh di

25

ct0

IIMaxI

a cm6,1226000.65,0.9,010.19,19

R.h.9,0M

F ===

Chn 13 12 c Fa = 14,69 cm2 Bc ct thp theo phng cnh di l:

cm58,1312

5,3.2170a ==


540

100

4

3

2

1

1

2

II

I

I

100100

2400

1700

300

400II

418

8a200

1212a210

1312a135

1212a210

1312a135

350

Hnh 2.38


chn a = 13,5cm=135mm Theo phng cnh ngn

2

ct0

IIIIMaxII

a cm39,1126000.65,0.9,033,17

R.h.9,0M

F ===

Chn 12 12 c Fa = 13,56 cm2 Bc ct thp theo phng cnh ngn l:

cm18,2111

5,3.2240a == chn a = 21cm=210mm B tr ct thp nh hnh v bn. 7. TNH TON MNG MM 7.1. Khi nim v mng mm v m hnh nn 7.1.1. Khi nim Tnh ton mng mm thuc phn Tnh ton dm trn nn n hi mt b phn ca c hc cng trnh. B phn c hc ny xt n vic tnh ton cc loai kt cu nh: mng bng, mng bng giao thoa, mng bn, mng hp, mng p thy in, tm trn ng t, tm sn bay Hin nay, cc cng trnh nh cao tng, ti trng ln c xy dng ngy cng nhiu, nhiu khi phi xy dng trn nn t yu. Do vy cc loi mng bng, mng bng giao thoa, mng b, mmg hp c s dng nhiu. Do vy vic nghin cu tnh ton lai mng ny l cng vic ht sc cn thit phc v cng tc thit k nn mng. m bo nn mng cng trnh iu kin chu lc v bin dng. Khc vi mng cng, mng mm c kh nng b un ng k di tc dng ca ti trng cng trnh, Bin dng un ny c nh hng nhiu n s phn b li ng sut tip xc (phn lc nn) di y mng. Do vy khi tnh ton ta khng th b qua bin dng un ca bn thn kt cu mng, hay ni cch khc l cn phi xt n cng ca mng. Tuy nhin n gin trong tnh ton, ngi ta ch xt n cng ca mng trong nhng trng hp mng c bin dng un ln n mt mc no . Theo QP 20-64 nhng mng tha iu kin sau:

10.10 33

0 >=hl

EEt (2.67)

th cn xt ti cng ca mng. Trong : Eo Moun bin dng ca t nn, E Moun n hi ca vt liu lm mng, h chiu dy ca mng, mng c t 10 c xem l mng mm, mng c t s hai cnh l/b 7 coi nh mng dm, l/b


p(x)

q(x)

x

w

xw(x

)

Hnh 2.39: S tnh dm trn nn n hi Di tc dng ca ngoi lc q(x) v phn lc nn p(x), mng dm b un, trc vng ca dm c xc nh theo phng trnh vi phn sau:

bxpxqdx

xwdEJ )].()([)(44

= (2.68) Trong : b b rng dm W(x) chuyn v ng ( vng) ca mng EJ cng chu un ca mng Di tc dng ca p lc y mng (bng nhng ngc chiu vi phn lc nn p(x)) mt nn b ln xung. Gi S(x) l ln ca nn th iu kin tip xc gia mng v nn sau khi ln l: W(x) = S(x) (2.69) Nh vy ta c hai i lng cha bit l W(x) hay S(x) v p(x) m ch c mt phng trnh (2.68) gi th cha . Do vy d gi c bi ton cn phi thit lp thm mt phng trnh th hai th hin quan h gia ln ca nn v p lc y mng, ngha l: S(x) = F1[p(x)] (2.70) Hoc p(x) = F2[S(x)] (2.71) Mi quan h ny th hin c ch lm vic ca nn di tc dng ca ngoi lc m ngi ta cn gi l m hnh nn. Ngha l nn t c m hnh sao cho gn st vi thc t nht m bo s lm vic ca mng trong nn t gn ging vi m hnh. 7.1.2. Cc loi m hnh nn 7.1.2.1. M hnh nn bin dng cc b (Winkler) M hnh ny cho rng ln ca nn, mng ch xy ra trong phm vi gia ti. Gi thit ca loi m hnh nn ny l mi quan h bc nht gia p lc v ln (m hnh ny do gio s ngi c Winkler xut nm 1867) C ch ca m hnh ny c biu din bng quan h: P(x) = C.S(x) (2.72) Trong : C l h s t l, cn gi l h s nn, th nguyn l lc/th tch (T/m3, kN/m3, N/cm3) v c coi l khng thay i cho tng loi t, c th tra bng theo cc ti liu tham kho hoc tnh ton t kt qu th nghim. S(x) ln ca t trong phm vi gia ti Quan h (2.72) ngha l cng phn lc ca t nn ti mi im t l bc nht vi ln n hi ti im .



M hnh nn Winkler c biu din bng mt h thng l xo t thng ng, di bng nhau v lm vic c lp vi nhau (Hnh 2.40). Bin dng ca l xo (c trng cho ln ca nn) t l bc nht vi p lc tc dng ln l xo. Theo m hnh ny ch nhng l xo nm trong phm vi phn b ca ti trng mi c bin dng. Do vy m hnh ny cn gi l m hnh nn bin dng cc b.

PLo xo

Lo xo chu nen

Hnh 2.40: C ch m hnh nn Winkler

M hnh ny c nhc im nh sau: Quan nim cho rng ln ch xy ra

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