chuong-3-nen móng
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$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
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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:
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[ ]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
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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 =
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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
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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 =
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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.
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