Thiet lap he phUorng tnnh giai bai toan dam tren nen dan hoi chiu tai trong tinh bang phUorng phap phan tilT bien Establishment of system of equations for solving beam on elastic foundation under static load problem using boundary element method
Ngay nhan bai: 23/02/2017 Ngay siiS bai: X4/3/2017 Ngay chap nhan d i n g : 5/4/2017
TOM TAT
Bai bao trinh bay du6ng loi thiet lap he phUdng trinh dai sd giai bai toan dam tiin nen dan hoi b^ng phUong phap phSn tii hiin.
Cac ham nghiem chuyen vi va npi ldc cua dam dUcfc xay ddng tren ccf sd ldi giai Cochy ciia phUcfng trinh vi ph5n cO b^n dam trin nen dan hoi theo mo hinh Winkle. Td do, thiet lap he phddng trinh dai sd xac dinh cac an s6 npi li^c va chuyen vi tai bien, xay ddng phdong trinh giSi cho cac ph§n td dSm m i u Td khoa: DSm tren nen dan hdi, phdcfng phap phan tii bi6n
ABSTRACT
This paper presents how to formulate the system of algebraical equations in solving the problem of beam on elastic foundation using boundary element method. The functions of internal forces and displacement were established based on Cochy solving results of fimdamental differential equations of beam on elastic foundation in accordance with Winkle model. From there, the system of algebraical equations for determining internal forces and displacement on boundary of element was established and the solving equations for form beam element were formulated.
Key words: Beam on elastic foundation, boundary element method
TS. VU THI BICH QUYfiN
Giang vien. Khoa Xay ddng. Tracing DH Kien triic Ha Npi
Vu Thj Bich Quyen
1 . Dat van de
Dam trgn n^n dan hoi la mot ket cau piid bien trong xSy dUng. Viee tim nghiem giai tieh tucjng minh trong bai toan cJam tren nin dan hoi tL/ong doi philc tap va chi thUe hiSn duoe eho mot so trUcJng hop dan giSn.Vdi cac bai toan phiJc tap thong thucfng can sCf dung phuong phap s6. C^c phuang phap so pho bien tai Viet Nam n h u phan tCf hCiu han (FEM), sai phan hitu han (FDIVl) chl cho ket qua gan dung tai nut cCia phan tCl, Phi/ong phap phan til bJSn (BEIW) [1 ] la phUOng phap giai tfch s6 duoc xay difng tren ca sd ldi giai phuong trinh tich phan bien, cho ph^p xSc dinh nghiem ehinh xac la cac ham chuyen vi va noi lye doc theo true thanh, nhung con I't ddoc silr dung tai Viet Nam. Trong bai bao, tae gia xin trinh bay each thi^t lap he phUOng trinh giai bai toan ciam tren nen dan hoi bang phUOng phap ph^n t d bien, t U d o xay dung phuong trinh g\i\
cho dc phan tUdam vdi cac dieu kiSn bien khac nhau (cac phan tilr mau).
2 . 1 . PhiTcfng trinh v i phan C0 ban dam tren nen dan hSi theo m o hinh Winkle
Cac mo hinh dam tren n^n dan hoi dUoc xay dung tren ca sd giS thiet ve tuong tac giua dam va dat nin. Bai toan dam tren nen dan hoi 1^ m6t bai toan si&u tinh dac biet, trong do phSn ldc nen la mot he lUc phan bo lien tuc phu thuoc vao bien dang cua dam va quan mem ve mo hinh nen.
Do do, trong phUOng trinh vi phan do vong ciJa dam tren n^n dan hoi ngoai yeu to tai trong tac dung eon co them dac trUng cua dat nen.
Cac mo hinh nen duoc ehia thanh mo hinh mot, hai va ba thdng s d Trong bai bao SLf dung mo hinh nen mpt thong so Vi/inkle[2](hinh1) vdi giii thiet moi quan he gifla phan lue va chuyen vi la tuyen linh hay cudng do phan ldc cOa nen tai mot diem ty le v6i d6 Idn eiia nen tai diem do.
R(x) Hinh 1 Dam tr^n nen dan hai theo mo hinh Winkler
Xet dam tren nen dan hoi co chieu dai I ch[u tac dung tai trong q{x}
bao gom cae tai trong va mo men udn nhd hinh 1.
Phdong trinh vi phan lien he do vong y(x) dgc theo true x va t^i trong q(x) CO dang:
. 5 W
v " ( x ) + 4 a ' y ( x ) - (2.1) EI
V6i: El la do CLfng chiu uon cOa mat cat ngang, k la he so nen: Cac moi quan he giUa noi lUc va chuyen v r
04.2017i!IBIim|103
^ = <p(x)A= M(x),dV^ Q(x)
dx ^ dx- EI 'dx' EI
(2.2)Nhan ddoc nghiem cua phdOng trinh vi phan thuan nhSt.
trong do: cp(x), M{x), Q(x) - ham gde xoay, mo men xoSn vi lUc e^t ngang.
2.2. Xac d i n h nghiem t o n g quat cilia phtTorng trinh vi phan Theo toan hoe giSi tieh phdong trinh (2,1 )c6 nghiem tong quat.
Y = y ( x ) + y-(x) (2-3) Trong do:
y { x ) -nghiem cua phdong trinh vi phan thuan nhat, y { x ) - n g h i & m rieng eda phdong trinh viphSn khong thuan nhat.
Xac ^ n h nghiem eda phiTcmg trinh vi phan thuan nhat Nghiem cua phuong trinh vi phan thuan nhat (tuong dng vdi trddng hap thanh chda ke den tai trong tac dung) va cac dao ham ed dang.
y ( x ) = chax[C|Cosax + C , s i n a x ] + shax[C3C0sax + CjSinax] (2.4) Trong do Ci.is^ la eae hang sd tieh phan.
Tinh dao hSm y (x), sil dung (2.3) thay x=0 vao cac ham, nhan ddoc:
y ( 0 ) = y „ = C, y'(0} = <Po = a C , + a C , y'(0)=
' EI = 2a-C, (2.5)
Td (2.5) cd the xac dinh cac hSng sd tich phan theo c^c thdng so chuyen vi va dao ham chuyen vi ten bien x=0.
^
c,-y.
' 2a 4a'EI
' 2a 4a'EI
2a-EI
Thay (2.6) vao (2.4) nhSn ddoe nghiem cua phdOng trinh vi phan thuan nhlt:
, chax.smax+shcK.cosc«cpn Yofx)=chax.cosax.yQ +
2 a shcocsinax M,, chax.sinax-shax.cosc(x Q,
2 a ' e 4 o^EI Dat
A|{ax) = chax.cosax , , , chax sinax +shax.cosax A, (ax) =
2a . , , shax.sinax A, ( a x ) - —^
2a
, , , chax.sinax - shax.cosax
A , (ax) = - ^ (2.8) 4a
A j (ax) = -a (chax.sinax - shax.cosax) A^(ax) = —2a^shax.sinax
A , ( ax) = - 2 a ' (chax.sinax +shax.cosax)
y(x) = A,(ax).yo + A,(ax)9<, - A,(ax)- • A > x ) ^ • (2.9) Bieu thde (2.9) b i l u dien nghiem cCia phdong trinh vi phan thuSn nhat phu thugc cac thong sd hinh hgc va tinh hoe tai biin x=Q cua phan td. DUdi goc do co hoc, nghiem ciJa phddng trinh vi phan khdng thuan nhat la dng xd ve chuyen vi eda m6i diem tr€n thanh do dc tSc dgng cCia chuyen vj, dao hdm chuyen vj (goc xoay), mo men udn va ldc cat t^i bien x=0.
Cae hSm gde xoay, md men uon va luc cSt ngang duocx^c djnh bSng each lay dao ham y(x}, sddung (2.3), nhan hai ve vdi El nhan duoc:
Ely(x) EI<,(x) I
M(x)
A,(x) A.(x) -A.(x) -A.(x)- A,(x) A.(x) -A,(K) -A,(X) -A.(x) -A,(x) A,(x) A . ( X ) -A,(x) A ( x ) A,(x) A,(x)
Ely(O)' E1<P(0) M(0) Q(0).
Xac dinh nghiem rieng cuia phu'cmg trinh v i phan khdng thuan nhat
Nghiem rieng cua phuang trinh vi phan khdng thuan nhat y ( x j IS dng x d v 4 chuyin VI eda thanh t d cae nguyen nhan tai trong tac dung tren thanh Do 3d nghiem neng ciia phdang trinh eo the xic dinh theo dang ham nghiem cua phUOng trinh vi phan thuan nhat. Ham chuyin in do tac ddng eda md men va tai trgng tren thanh tcSl diem cd tga dg x Idn hon vi tri diem dat lue co dang nhd ham chuyen vi do md men va ldc tai bien x=0 tdi cac diem tr^n thanh De b i l u dien tinh gian doan cOa h^nn tSi trong sddung ham dOnvlHeviside [4] va quy dde d a u " + " nhu(2.11).
i^-^l
Jo, if x < a x-a, if x > aH ( x - a ) . 0, if
(2.111
l * i i i - i ^ l
t ^ -^
F^^^x„J
Hinh 2 Tai trong tac dung len thanh
Nghiem neng eda phuong trinh vi phan do teii trong tac dung nhii trong hinh 2 bao gom cic thanh phan:
- Thanh phan ham nghiem do tac dgng md men M (2.12) y l i ( x ) = — A 3 [ a ( x - X M ) J
- ThJnh phan ham nghiem do tac dong luc t | p trung P
y ; ( x ) - - ^ A , W x - X , ) . ] (213) - Thdnh phan ham nghiem do tac d6ng t i l trong phan bd q
Ji\ trong phdn bd dUoe bieu dien thdng qua hhm gian doan Heviside
q,(x) = cl[H(x-x.,)-H(x-x.,)]
Tdi trgng phan bo duac md hinh hda thanh tdp hop ede tai trgng tap trung tiin ehilu dai phan b d Hop ldc eda tdi trong phan bd ddae xac dinh bdng tich phan tr^n doan c6 tae dung eua tdi trgng phdn bo
i n i l H D i i i ! [ I B » 04.2017
y ; ( x ) - j ^ A , K x . a . ] q [ H ( x - x „ ) - H ( x - x „ ) ] d 5
g H(a(x-x^,)^ l-cha{x-Xq,),cosa(x-x^,)^
4 E I a ' | . [ H ( a ( x - x , , ) . ) - Q h a ( x - x „ ) . c o s a ( x - x „ ) . ]
TH (ax) - chax.cosaxl Bat A, - ^ ' ' ; J
4a'
y',(x)--i[A,(x-x„).-A,(x-x„).] (2.14)
Khi do nghiem rieng ehuyin vi goe cua phuong trinh vt phan ddoe viet dddi dang mgt b i l u thde
y'(x)-—A,[a(x-x^,).] + —A,[a(x-Xp)]
^^ ^^ (2.15) + | - [ A , ( X - X „ ) , - A , ( X - X ^ , ) J
EI'- ^ ^ -'
Nhdn y ' ( x ) vdi El, sd dung ii&n h§ (2.3) nhan dUOc
dung ldi gidi so, thay cac gia tri x=0 va x=l vao cac ma tran Y, A, B trong h f (2,18), he (2.18) trd thanh he phdang trinh dai so (2.19).
Y ( 1 ) = A ( 1 ) X ( 0 ) + B ( 1 ) ^ A ( 1 ) X ( 0 ) - Y ( 1 ) = - B ( ] ) j2^gj Cdc ^n so nam trong hai ma tran X(0) vd Y(l) Khi chda gan cac dieu bi^n tinh hoe va hinh hoc tai nut, sd an sd Idn han so phdang trinh. Do dd, can gdn edc d i l u kien bi^n tinh hoe va hinh hoe tai eac nut de khd bdt an. Cdc di4u kien bien tinh hgc ddoc xac djnh dua trgn cac phuong trinh cdn bSng li/c va md men tai nut, Cae dieu kien bien hinh hoe ditac xac djnh dda vdo d i l u kien ehuyin vi tai bien eda edc cdc p h i n td.
Sau khi da khd bdt an can thde hi§n vide di chuyen thong sd an trong Vic t d Y(l) tdi vecta X(0) de va thiet lap he phdOng trinh dai sd giai bai toan xac dmh cae thong sd ndi ldc va chuyen vi tai bien
A ' ( 1 ) X * ( 0 , 1 ) = - B ( 1 ) (2.20) 2.4. Xay diTng phi/dng t r i n h xac d j n h t h d n g s6 bi£n cho cac phan
tiJTmdu
Phan tdrdSm hai dau tiTdo (hinh 3) M '^ ' P
fEIy«(x)l Elip«(x) M - ( x )
I Q - W J -B,(x)l -BM
-B.,(x)
-B,(x)J
I • - - I
Hlnh3 Dam hai dau tudo
Dam hat tfau t y do dattren nen dan hoi (h)nh 3) co cac dieu kien bien sau M(0)=Q{0)-M(l)=Q(l)=0
GSn cac dieu kien bien vSo phuong trinh (2.20) n h i n duoc
B,(x)-M.A,[a(x-x„).] + P.A,[o(x.x,)]+q[A,(x-x„).-A.(x-x„).]
B,(x)-M.A,[a(x-x„)J + P.A,(a(x-x,)J+q[A,(x-i„)_-A,(x-x^).]
B,(x)-M.A,[o(x-^„).l^•P.AJa(x-x,)l + q[A,(x-x„).-A,(x-x,,),]
B,(x)-M.{.4n".A,[c.(x-x„).]) + P.A,[c.(x-x,)]
+ q [ A , ( x - x „ ) . - A , ( x - x , 4 , ]
(2.17) Nghiem t o n g quat eda phuong t r i n h co dang
(EI.Y(x)] r A , { x ) A , ( x ) - A , ( x ) - A . ( x ) ] f E ] y ( 0 ) l ( B , ( X ) EI.a.(x) I A , ( x ) A , ( x ) - A , ( x ) - A , { x ) Ei,>{0) B , ( x ) M ( x ) - A . ( x ) - A , ( x ) A , ( x ) A , ( x ) M ( 0 ) - B , { X ) I Q(x) ) [-A,(,) A ( x ) A,(x) A , ( X ) J [ Q ( 0 ) J [-B,(X) Viet (2.17) dddi dang rut gon
Y(x)=A{x)X(0) + B(x) (2.18) Trong do:
Y{X) - ma trdn cot ham chuyen vi va noi luc dgc theo true thanh (vecta trang thai cua thanh);
A{x) - ma tran vudng nghiem ca bdn eua phuang trinh vi phdn thuan nhat;
X(0) - ma tran cot ehuyin vl vd n$i luc tai d i l m cd tga do x=0 [vecto thong so ban dau),
B[x) - ma tran cdt ham chuyen vi vd ndi ldc do tdi trgng tac dung dgc theo true thanh (vecta tdi trong);
2.3. Thiet lap h f phiTOng trinh dai sd xac d i n h cac t h d n g s6 bien d i m tr€n nen dan hoi
He (2.18) la nghiem hdm gidi tich ngi ldc vd chuyin vi dam tren nen dan hoi chiu tai trong tinh theo md hinh Winkle Trong dd, cae an so can tim Id cdc thdng sd noi lUc va chuyen vi tai nut bien eda thanh Oe xay
A,(l) A,(l) -A,(l) -A.(l) A,(i) A,(l) -A.(l) -A.(l) -A,.(i) -A,(l) A,(l) A,(l)
•A,(l) A ( i ) A,(l) A,(I)J Ely{0) El(p(0) M ( 0 ) - 0
| Q ( O ) - O [ E!.Y{i)
EI.O{i)
| M ( I ) = O
| Q W - O [B,(l)
B.(l) B,(l)
M
[Ely(O) JEI(P(0)
I 0 I 0
EI.Y(1) EI.<1>(1) M ( l ) - 0 Q ( l ) - 0
B,(l)l
8,(1)1
B . ( l ) |
B,(1)J
Thde hien viec loai bo cac thong sd ed gid tri bdng khong, di chuyen eac an sd tdY(l) sang X(0), thiet lap vec to A" vd X" trong (2.20) theo cdc bude:
BUdc 1. Do cae thdng sd md men va lUc cSt (tai hang 3 va 4) cua A(0) c6 gid tri bang khdng, gan gia tri bang khdng vao cae sd hang trong cdt tuang dng [cot 3 va 4) cua A(l)
' A , ( I ) A , ( 1 ) 0 o' A J ( I ) A , ( 1 ) 0 0 - A „ ( 1 ) - A 5 ( 1 ) 0 0 - A , ( [ ) - A , { 1 ) 0 0
BUdc 2. Chuyen edc thdng sd khac khong, EI.Y(I) vd El.*(l), t d hang 1 va 2 cua Yd) sang hang 3 va 4 eda X{0). Bi ddm bdo EI.Y(I) vd EI.<B(I} v l n ndm trong phdang trinh hang t h d 1 va 2 eda he (2.19), can them cae thong sd bu vdo A(l,3) va A(2,4) de ehuyin vi tri vdo A
(1) A , ( i ) C 5 ) Ol(EIy(0) j,EI.Y(l)l [B,(I)
A, (1) A, (I) 0 O El<p(0) ^ ^ EI.<I.(I) 1 _ B, (1)
A ( l ) -As(l) 0 0 EI.Y(I) / ' M ( 1 ) = 0 63(1) -A,(l) -A,(l) 0 0 [El.<t>(l))^ [Q(1) = OJ [B,(I) Nhan dddc he phdang trinh (2.21) xac dinh ndi ldc va chuyen vi tai bien p h d n t d d a m t r e n n l n dan hdi h a l d d u t i / d o
" A , ( I ) A , (I) - 1 A , ( I ) A , (I) 0 A ( l ) - A , { i ) 0
.-A,(l) A ( 0 0
Thay n g h i f m eda (2.21) vao (2.18) nhdn ddoe hdm gidi tich n chuyen vj eda dam.
0 ' - 1 0 0
[Eiy(o)
Elq)(0) E].Y(1) EI.O(I)
[B,(l)l
B.(l) B.(l)
[B.(I)J
04.2017 ssiisnon i m i ;
TUang tu, xdy dung phan t d m a u cho cae dam cd dieu kifn bien khac.
Phan tiSrdIm lien k£t hai d&u gdi tifa [hinh 4]
ii'inh4.DSmh3tdlugoiti/a A,(l) 0 -A,(l) A,(l) 0 -A,(l) -A,(l) 0 A,(l) -A.(l) 0 A,(i)_
(Ely(O).Ol El9(0) K
1«(0)
= 0 fI Q(») H
0 A,(l) 0 -A,(l) 1 A,(l) 0 -A,(l) 0 -A,(l) 0 A,(l) 0 -A,(l) - 1 A
(i)J
E).Y(1) = 0 k EI-*(I) M(l) = 0
U Q C )
EI(D(I)]
EI(|)(0) 1 QC) I Q(o) J
B,(l) B,(I) B,(l) B,(l)
B,(l) B.(l) B,(l) B.(l)
Phan t d dam hai dSu ngam (hinh 5)
Hinh 5. D^m hai dau ngam 0 -A, (I) -A,(l)- 0 -A, (I) -A, (I) 0 A, (I) A . (I) 0 A, (I) A, (I)
0 -A,(l) -A,(i)- 0 -A, (I) -A, (I) 0 A,(l) A, (I) 1 A , ( l ) A,(l)
Ely(O) = 0 EI(p(0) = 0 M(0) Q(o) M(l)"
Q ( 0 M(0) Q(o)
EI.Y{i) = 0 EI.<I>(l) = 0
^ M ( l )
^ Q ( 0
"B.(01
B , ( I ) B.(l) B . ( l ) |
1
1
B,(l)l
B.(l) B,(l)
B.W
0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 1
-A,(l) -A,(l) A,(l) A , ( l ) -A,(l) -A=(l) A,(l) A . ( l )
-A.W"
-A3(l) A,(1) A,(l)
-A.(0
-A,(l) A , ( l )
A,(I)J
E l y ( O ) ^ EI(p(0) = 0 '
1V1(0)\
Q(») . EI<I)(1)1
QW M(0)
II Q W J [
El Y(l) - 0 EHI)(1) M ( 1 ) = 0 Q(i) E I . Y ( l ) - O l
EI.<1>(1) I M ( I ) - 0 I
Q(i) J
B,(l) . B.(l)
B.(l) .B.(l] B,(1}1 B.(l) B.W B.(l)|
5. Nhan xet
He phuong trinh dai so t h i l t lap cho cdc phan til mau dam tren nen dan hoi Id co sdcho viec sd dung phddng phdp sd (vdi sUtra giCip eda cae phan mem lap trinh toan hgc) giai bai toan phde tap. Hdm nghifm noi ldc vd c h u y i n VI xde dmh theo (218) hoan todn trdng khdp vdi nghiem gidi tich. Id du d i l m ro ret cua phddng phdp phan t d bien.
TAIUEUTHAMKHAO
f l l P K. Banerjee and R. Butterfield, Boundary Element MeOiods in Engineenng Sdaice, McGraw-Hill Book Company [UK) LitDited,1981.
[2] Edward Tsudic,^nfl/w/j of structures an Elastic foundation. J. Hoss Publishing, USA, 2013 [li'^iiTM&iCwjen.PhMigphdppbantdbiengidibiiitadnfinbhitbanhbieRdangddnlm, Tap 2-Tuyen tap H6i nghi Khoa hoc toan quoc Cohoc vat ran bien dang lan thiJ12,flan3ng, 2015.
[41 flajapfia B A., KonaiiieHso CH, QMuteHme ipyuKUun a ladaiia Memmm. - KKEB.:
HavK0BaflyMKa,1974.-191c
D i m m o t d a u tUdo, m 6 t d a u n g a m ( h i n h 6}
. - 1 - Hinh e. Dam mot dau tudo, mot dau ngam
0 -A,(l) -A.(l) 0 -A,(l) -A,(l) 0 A,(l) A.(l) 0 A,(l) A,(l)
[Ely(O) = 0 1 EI(|)(0) = 0 M(0)
I Q(»)
-I 0 -A,(l) -Aj(l) 0 -1 -A,(l) -A,(l) 0 0 A,(l) A , ( l ) 0 0 A , ( l ) A
WJ
Ely(l) El(p(l) M(0)
IQC)
< - f E I . Y ( I )
<-4 EI.<D(I) M ( 1 ) . C [ Q ( I ) . O
B,(l)l
B.(l) B,(l)
B.(l)]
B,(0
B.(l) B,(l)
B.O)
Dam m d t dau n g a m , m o t dau goi t u a (hinh 7)
Hinh? Dammot dau ngam, motdautJdo
