58

$5. Tnh ton mng mm: 1. Khi nim :

+ Nn t yu mng n phi m rng ra n gn nhau, nn t bin dng nhiu, cn lm mng lin tc cng ln chu ln khng u.Ta i n gii php dng mng bng, bng giao nhau, hoc b. +Ti trng ngoi v phn lc nn mng b un mng b un li nh hng n s phn b phn lc nn. + Khng xt cng ca mng th tnh ton ch c ngha thc tin cho vic tnh ng xut cn tnh mng sai s s ln. + Tuy nhin n gin tnh ton ta ch xt khi bin dng un ln n mc no

1033

>=hl

EE

t o Eo - m uyn bin dng ca nn

E - m uyn n hi ca vt liu mng l - na chiu di mng h - chiu dy mng

Khi t s hai cnh: 7bl

mng dm; 7bl

mng bn

Bi ton: - Xc nh phn lc nn. - ln ca mng. - Kt hp vi ti trng ngoi tnh c kt cu mng. Cc cch tnh ton:

a, Theo s n gin: Lt ngc mng ln xem n nh 1 dm lin tc, chn ct nh gi ta, ti trng chnh l phn lc nn v xem phn lc nn phn b u vi

59

cng p. Tc l ta b qua bin dng ca bn thn Mng v b qua bin dng ca CT bn trn. Do n ch ng khi: CT bn trn l tuyt i cng phn lc nn phn b u (nh > 9 tng)

b, CT+M+N c gn vi nhau: v c tc ng qua li vi nhau nn ng n phi xem xt nh 1 th thng nht tnh ton tuy nhin kh khn v mt thut ton nn cha p dng rng ri

c, Hin nay ph bin dng cch ri rc kt cu: tc l tch ring mng v xt s lm vic ng thi ca mng v nn: Tc l xt mt KC c bin dng t trn nn cng c bin dng v xc nh ni lc trong mng Tnh ton kt cu trn nn n hi.

Do : Vic thit k mng bng mm bao gm:

+ Xc nh s b b rng mng tng t mng bng cng vi gi thit phn lc nn phn b u. + La chn s b KC mng ph hp b rng ni trn. + Tnh ton chuyn v mng, phn lc t v ni lc trong mng theo s : Dm trn nn n hi. + Kim tra cc trng thi gii hn ca nn. Nu khng tho mn tng kch thc mng, hoc tng cng ca mng ( sa i KC la chn) v xc nh li ni dung bc 3 + Thit k ct thp cho mng trn c s kt qu bc 3.

2. Tnh ton dm trn nn n hi: Xt mng dm:

Phng trnh trc vng ca dm:

60

[ ]bpqxd

dEJ xx

x=4

)(4

E.J - cng ca dm; b: b rng dm; : chuyn v ng Vi 2 n x v px ta tm thm 1 ph/trnh na. Quan h: ln ca nn vi p lc y mng

)(1)( xx pfS = )(2)( xx SfP =

Th hin c ch lm vic ( bin dng ) ca nn gi l m hnh nn. Cc m hnh nn t:

+ M hnh nn Winkler

xx sCp .= C: h s t l h s nn. Cn gi l m hnh nn bin dng cc b.

+ M hnh na khng gian bin dng tng th.

Theo But xi nt: RE

pSo

oyx pi

2),(

1=

+ M hnh lp khng gian bin dng tng th: ph trin M hnh trn nhng xt n chiu dy lp t nn chu nn Ha. Nn H>Ha th ly chiu dy Ha tnh H

61

theo m hnh Winkler (ng 4) theo m hnh na khng gian n hi (ng2) theo kt qu th nghim (ng 3)

Do thiu st ca 2 m hnh trn nn c rt nhiu m hnh khc c nu ra:

* M hnh nn mng. * M hnh nn tm. * M hnh nn n hi vi 2 h s nn. * M hnh lp n hi c chiu dy hu hn * M hnh nn na khng gian n hi c

oE bin i theo su... V vic la chon m hnh nn: T kt qu thc nghim cc tc gi a ra nhng kt lun:

1. ln ca t ngoi phm vi t ti tt rt nhanh: thng thng vng ln khong 0,3 0,5 ng knh tm nn. Nh vy t c tnh phn phi rt yu

2. khi m tng tnh phn phi gim, t bo ho nc tnh phn phi ca n khng ng k

3. mt bin dng ca nn khi xem l bn khng gian n hi tt qu chm so vi thc t quan st.

Do kt lun: t mm dng m hnh winkler hp hn, tuy nhin h s nn C khng c ngha r rng: n khng phi l hng s m n thay i ph thuc cng CT; khong tc dng ca ti trng. Nc ta ng bng sng Hng v Cu long s gp t mm, cha nhiu nc, nc ngm cao nn tnh phn phi ca t yu do nn chn m hnh Winkler 3. Tnh ton mng dm theo m hnh nn bin dng cc b: Phng trnh c bn:

xx CSp =

62

xx S= CbqS

dSd )(4)(4)(4

4

=+

Trong : xa.= 44EJCb

a = (1/m)

Dm khng ti: 0)(4)(44

=+ S

dSd

Gii tm S tm p a, Tnh dm di v hn trn nn n hi. Hai u dm xa ti trng khng b ln 2 u. 1. Lc P: Bi ton i xng ct i tnh

+++= eCCeCCS )sincos()sincos()( 4321 Xc nh C: H/s da vo /k bin: x () : C1=C2=0 nu khng S s khng bng khng.

Lc : += eCCS )sincos()( 43 (1) Ti x=0 ln s ln nht do :

0=dxdSx

(2)

0=

ddS

63

o hm (1) ( gc xoay =0) kt hp vi (2) ta c: ( ) ( )[ ] 0sincos 3434 =+ eCCCC 43 CC = (V cos 0 khi x=0)

Gi l Co ta c: += eCS 0)sin(cos)(

Co=? Ti x=0

20PQ =

Ti =a.x=0 0333

42

)( CEJaP

dSd

==

(SBVL)

30 8JaPC =

)sin(cos8

3 axaxeEJaPS axx +=

Sau xc nh: Mx; Qx: o hm lin tc

)cos(sin42

2

axaxea

Pdx

SxdEJM axx ==

axeP

dxSxdEJQ axx cos23

3

==

t )sin(cos1 axaxe ax +=

)cos(sin2 axaxe ax = Ta c: axe ax cos3

= 138

EJa

PSx = 138C

EJaPpx = 24

a

PM x = 32PQx =

: tra bng theo = a. x Smax ti x=0 ( ti P=0)

CbaP

EJaPS

28 3max==

baPp2max

= a

PM4max

= 2maxPQ =

64

Nhiu P: Dng phng php ng nh hng ca mng dm di v hn 1101 PSSM = 1MS ln ( vng) ti M do 1P gy ra. 10S tung ca ng nh hng ln. (do 1=P t ti im M) ly ti im 1xx =

ln ti M: =

=

n

iiioM PSS

1 tng t: MM QM ;

Cc trng hp khc tng t : M men tp trung ; Ti phn b cc b. Trong cc gio trnh nn mng c cc bng tra kt qu tnh ton cho cc trng hp c bn Lc tp trung, m men tp trung im gia dm di, bn vung hoc trn. C th gii bi ton mng dm, bn chu nhiu ti tp trung bng cch p dng cc bi ton c bn vi nguyn l cng tc dng. 4.Tnh ton mng dm theo m hnh nn l bn khng gian BDTT H p/trnh vi phn c bn i vi mng loi dm:

dxxxpE

bS ox

ox

o

ox )ln(

)1(2 2

1)(

2

)(

=

Thng l gii theo phng php gn ng, c th tm thy cc li gii in hnh: + Phng php M.I. Gorbunov Pasadov + Phng php I. A. Ximvulidi + Phng php ca Giemoskin + Phng php phn t hu hn. Trong cc gio trnh nn mng: Nguyn l ca cc phng php ny l chuyn s bi ton dm t trn nn n hi Dm trn cc gi hu hn ( dm lin tc) v dng phng php chuyn v trong c hc kt cu gii xc nh phn lc gi.

Phng php phn t hu hn: C th dng cc chng trnh phn mm dng cho vic tnh ton mng loi dm bn: V d chng trnh SAP 90; SAP 2000 vi m hnh Winkler

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* Phng php Phn t hu hn ng dng SAP 2000

Bng vic chia nh kt cu mng thnh cc phn t nh v h mng v nn c m hnh ho nh sau: + Kt cu mng: dm c thay th thnh cc phn t frame, bn l cc phn t shell. + Nn:

c thay th bng cc gi n hi spring l cc l so c cng: K=C. ax. bx ( ax; bx l din tch chu ti ca nt ang xt)

a, Mng dng dm n:

b, Bng giao nhau:

c, Mng b:

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Nh vy c th d dng k n s thay i ca EJ v K (trong mi phn t coi EJ, K l hng s khc nhau) *Cc phng php xc nh cng ca l xo:

+ Tra bng tm h s nn tng ng vi cc loi t( bng tra ph lc) + Cng thc thc nghim: Poulos; Terzaghi; Meyerhof + T kt qu th nghim bn nn.

* ng dng SAP 2000 gii ni lc mng: Bc1: to s hnh hc: frame (mng dm), shell (bn) Bc 2: khai bo s liu u vo: vt liu, tit din, ti trng, t hp nu c Bc 3: gn ti trng, iu kin bin. Bc 4: chia nh kt cu mng thnh cc phn t nh kch thc bng nhau ( kch thc phn t tu thuc vo mt bng mng v yu cu tnh ton cng nh cng chnh xc s lng phn t ln) Bc 5: tnh cng ca gi n hi thay th v gn cho tng nt ch din tch thay th. ttFcK .=

Bc 6: kim tra s liu nhp phn tch ni lc Bc 7: xem kt qu ni lc: + biu (M;Q) ca dm mng, phn lc ca cc gi n hi: Rz v Uz Bc 8: kim tra kt qu: Phn lc nn Rz lun dng (nn khng chu ko) nu Rz < 0 b lin kt ti cho Uz= 0 phn tch li ni lc. Bc 9: kim tra cng nn v tnh ton ct thp.

*Tnh lp h s nn:

+ Ti sao phi tnh lp: cng K tnh theo cc cng thc thc nghim khng chnh xc tnh lp tm K

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+ Cch tnh lp: Ln 1: tnh K1 theo cc cng thc thc nghim. phn tch kt qu c Rz1, Uz1 nu K ng th Uz1 = S; S - ln ca nn tnh cho mng c kch thc phn t theo phng php cng ln tng lp hc trong C hc t.