/ . 1. -^
PHUONG PHAP PHAN TU HUU HAN TINH TAM CHU NHAT CHjU UON CO XET OEN ANH H U A N G C O A BIEN OANG TRUOT NGANG
GS.TSKH. HA HUY Cl/GfNG ThS. NGUYEN THUY ANH
Triicfng Dai hoc Cong nghe Giao thong van tai
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1. Cac lien h# cd bin Dga tren cae gia thie't
Mat trung binh cua tam khdng bj bien dgng.
Tiet dien phang thing gdc vdi mat trung binh trUde khi bien dgng thi sau khi bien dgng vin phang nhUng khdng thang gdc vdi mat trung binh (do cd xet de'n inh hudng cua lgc cat Q nhung khdng xet de'n bien dgng do Oz gay ra) nhgn dUdc cac lien hp sau:
Bien dgng trupt ngang a a Y = — 0 Y = — 0 Gdc xoay do mdmen
8w 8w
^^ ax ^^^^ ay ^'
Bien dgng udn
X x = - acp^
ax A*y
ay
(1)KHOA HOC - CONG NGHE Bien dgng xoan
A/XV
d<^^d%
^ dy 5x _ Cac mdmen ndi lgc
M , = D ( x , + | i X y ) My=D(Xy + ^Xx)
M.,=D(l-^)x.,
2. PhUdng phap phan tii hufu han. Cac ham noi suy
PhUdng phap phan tCf hOu hgn (PTHH) chia cdng trinh thanh nhOng phan nhd dupe gpi la phan tCf. Viec tfnh toan dupe thgc hien ddi vdi moi phan tCf, sau dd ket ndi chung Igi vdi nhau ed dupe toan bd cdng trinh.
Cac ham ndi suy dupe viet theo tpa dp tg nhien dupe dung vUa de md ta trgng thai (vf dg ehuyen vj cua dam, tam v.v...) va cd the vUa de md ta dgng hinh hpc (vf dg, dam cong, vd ...) cua cdng trinh cho phep de dang lap trinh va tgo dieu kien tg ddng hda qua trinh tfnh toin (phan tCf hufu hgn dung ham ndi suy nhu vay dupe gpi la phan \is dang thdng so. Isoparametric finite element, [2]). Cae ham ndi suy viet theo tpa dp tg nhien do B.
Irons va O. Zienkiewicz dUa ra nam 1968 [3].
Do kfch thude phan \Cs nhd, trgng thai (vf dg ehuyen vj cua dam, tam...) cua cae diem trong moi phan tCf khac nhau ft eho nen cac ham ndi suy dupc dung la cae da thUe bae thap, vf dg ddi vdi dp vdng cua dam ham ndi suy thudng dung la eac da thUe bae ba theo tpa dp x, ddi vdi dp vdng cua tam la cae da thUc bac ba theo tpa dp x va bae ba theo tpa dp y v.v.. Vi dung eac da thUe bac tha'p cho nen cac luc tac dgng trong moi phan tCf eung nhu lgc quan tfnh (bai toan ddng luc hpc) deu phai qui ve eac nut. Ham ndi suy dupe chpn sao cho ket qua tfnh la on djnh: ke't qua la duy nhat, thay doi be
eua dieu kien bien hoge dieu kien ban dau khdng lam thay doi ket qua tfnh.
Ham noi suy do vong (chuyen vj ) cua phan tii
Phan ti^f ehuf nhat ddn gian va cd ban dupe dung rdng rai khi tfnh tam dupc trinh bay tren hinh 1.
i(-i,-i)
•—
Ax 2(1,-1)
•
c
Ay3(-1,1) 4(1,1)
Hinh 1 Phan tCl chO nhat trong he toa dd tu nhien
Dp dai that eua cac egnh phan tCf la Ax, Ay, trong he tpa dp tg nhien lay bang 2, cd 4 nut, mdi nut cd 3 thdng sd la cac ehuyen vj wi, gdc xoay cua dp vdng theo x,
G^i = 9 w / 9 x , gdc xoay eua dp vdng theo y, e^. = aw / Sy, i=1,2,3,4 la sd thU tg eua 4 nut cua phan \is vdi cae tpa dp tu nhien trinh biy tren hinh 1. Tong cdng ed 12 thdng sd can xac djnh.
Gpi w la veetd cdt chUa 12 thdng sd nut cua phan ti^f (vie't theo ngdn nguf lap trinh MatLab):
| w } = [w, W2 W3 W4
6x1 e , , 9,3 e , , (2) 6yl 6y2 6y3 ^yA
Gpi N la veetd ddng ehUa cae ham ndi suy N j = N j ( x , y ) tUdng Ung (viet theo ngdn nguf lap trinh MatLab):
[N] = [N^„N^„N^„N^„
N.,,N^„N^„N^„ (3)
thi ehuyen vj We tgi moi diem cua phan tCf dupe xac djnh theo cae thdng sd nut bang
w'=[N]{w} (4)
TAP CHI CAU D U O N G VIET NAM
Cac ham ndi suy eua [N] dupe xae djnh N^, = ( - x + l ) ( - y + l ) ( 2 - x - y - x ^ - y ^ ) / 8 nhu sau: Vi ph'an \is ed 12 thdng sd eho nen N,3 = ( - x + l ) ^ ( - x - l ) ( y + l)Ax/16
ed the chpn da thUe bae ba ddi vdi x va bae >.T . s 2 2 ba ddi vdi y nhung thieu (tfch eua da thUe bae ^ ^ ^^^'^ ^^^"^ ^ ^^^^ + x - y - x - y ) / 8
3 theo X vdi da thUe bae 3 theo y ed 16 sd N,4 = ( x + l ) ' ( x - l ) ( y + l)Ax/16
hgng, ehl can lay 12 sd hgng) de xa'p xl ham N^j = ( - x + l)(y + l ) ( 2 - x + y - x ^ - y ^ ) / 8 chuyin vj We N^, = ( - y + l ) ^ ( - y - l ) ( - x + l)Ay/16
w'=a,+a,x + a,y-^a,x' + a,xy + a,/ N^, =(x + l)(y + l)(2 + x + y-x^-y^)/8 ^^^
+ a,x'+a,x'y + a,xy'+ay N , =(-y + l)^(-y-l)(x + l)Ay/16
3 3
+ a,,x y + a^,xy N,, =(-x + l)'(-x-l)(-y + l)Ax/16
hay dUdi dgng ma tran N^j = ( y + l ) ^ ( y - l ) ( - x + l)Ay/16
w^=[l X y x' xy y' N,^ =(x + l)'(x-l)(-y + l)Ax/16 x' x'y xy' y' (5) N^^ =(y + l)2(y_i)(x + l)Ay/16
{^/-[^O' ^1' ^2' ^3'
a^; a^; a^; a^; (6) a
xV xy']{a}
{a} l i veetd cdt: ^ ^ ^ *^^ 9'^ ^^^ '^ ""^^"^ " 9 ^ ^ ' ^^^ *'^"
dUa ra cae ham ndi suy tren.
Cd the neu nhgn xet rang, gdc xoay SwVSx theo (9) l i ham bac 2 theo x va bac
^g', &,„; a,J m5t thgo y, tUdng tu, gdc xoay dw'/Sy l i Thay tpa dp 4 nut vao (5) ta ed 4 ehuyen vj ^^^ ^gc 2 theo y v i bac mdt theo x.
nut va dd la 4 sd hgng dau eua (2). Lay dgo ^ ^
^ , \ ,, . Ham noi suy li/c cat Qx ham (5) theo x, thay tpa dp 4 nut vao bieu thue
nhan dupe, ta cd 4 so hgng tiep theo eua (2). ^ie'n dgng trupt ngang do luc cat Qx gay La'y dgo ham (5) theo y, thay tpa dd 4 nut vao '^ '^"^ ^^^V ^ ° ' 9°^ ^o^V e . - a w / 5 x , cho bilu thUe nhan dupe, ta ed 4 sd hgng eud'i cua " ^ " ^gng h i m Ige eat dupe chpn se la dgng (2). Bang each lam nhu vay, viet dupe ^^"^ 9°^ ^°^y- '^ ^^ *^^^^ ^go hai dd'i vdi x v i bac nha't dd'i vdi y. Oa thUe niy ed 6 sd hgng vi vay phan \is lgc eat Qx phai ed 6 nut nhu
{w}=[B].{a} (7)
6 day [B] la ma tran cd kfch thude tren hinh 2.
(12,12). TU (7) xac djnh {a} theo cae thdng Gpi { Q j l i veetd cdt chUa c i c thdng sd sd nut {w} bang each nghjeh dao [B]: lgc cat tgi eie nut phan tCf
{a} = [ B ] - ' { w } = [ C ] { w } (8) { Q j K Q x , Qx2 Qx3 ^^^^
DUa {a} xac djnh theo (8) vao (5) ta se Qx4 Qx5 Qxef
cd bilu thUc xac djnh ehuyin vj moi diim eua Gpi [NQx] la veetd ddng chUa cae ham phan tCf theo cac thdng sd nut eua nd (bieu ndi suy
thUe 3), tu dd tfnh dUdc cac ham npi suy [N].
Vdi each lam tren, ta nhan dupc c i c bieu [ ^ e . ] = [^exi ^Q.2 ^0.3 thUesau Ng^, NQ^, TV^^J
KHOA HOC - CONG NGHE
1(-1,-1)
N^,=x(,x-\)(\-y)IA
N^,=x{x-\){\ + y)IA N^,={\-x'){\-y)l2
N^,={\-x')(\ + y)ll
(17)4(-1,1) 5(0,1) 6(1,1)
Hinh 2- Phan tCl lUc cat Qx
Lgc cat tgi moi diim trong phan Xil Q^
dupe xac dinh bang
Q : = [ N < , . ] { Q . } (12) Q^ dUde vie't dudi dgng da thUe bae hai
theo X va bac nha't theo y :
Qx=ao+aiX + a2y
+ a3xy + a^x^ + ajX^y hay dudi dgng ma tran
Q : = [ 1 X y xy X^ x V ] { a } (13) {a} la veetd cdt ehUa cac he sd eua da thUe
{a} = [ao;. a,; a^; a,; a^; a J (14)
Dua tpa dp 6 diim eua phan Xil lUe cat v i o (13) se xae djnh 6 lgc eat tgi nut cua veetd (10) ta cd phUdng trinh xac djnh lgc cat tgi nut theo cac hp sd eua da thUe
{Qx} = [B]{a} (15)
[B] la ma tran ed kfch thudc (6,6). NgUpe Igi, bang each nghjeh dao ma tran [B] ta cd cac cdng thUe xae djnh cae he sd cua da thUe (veetd a ) theo cac lgc cat tgi nut:
{a} = [Br{Q.} = [C]{Q.} (16)
Dua {a} xae djnh theo (16) vao (13) ta se ed eac cdng thUe xae djnh lgc cat tgi moi diem eua phan tCf theo lgc cat tgi nut {Qx}- nhu vay se tim dupe eac ham ndi suy lgc cat [NQx]. Cich.tim cac ham ndi suy hai chieu dga tren ham ndi suy mdt chieu da bie't se nhanh hdn.
Vdi cae each lam tren tac gia nhan dupc c i c ham ndi suy cua lgc cat (12) sau
NQ^,=x{x + \\\-y)IA NQ^,=x{x + \)(\ + y)IA
Ham n9i suy li/c cat Qy
Ta chpn ham ndi suy lUc eat Qy theo dgng ham gdc xoay 9^; = 5 w / 5 y , la da thUc bae hai theo y va bae nha't theo x. Nhu vay phan tCf lgc c i t Qy ed 6 nut nhu tren hinh 3
TUdng tg, tU cac ham ndi suy [NQx] vie't cac ham ndi suy [NQy] nhu sau
NQ^,=y{y-\){\-x)IA No,2 = {\-y'){l-x)l2 N^,=y{y + \){\-x)IA
N^,=y{y-\){\ + x)IA (18) N^,={\-/)(\ + x)l2
Noy, = yiy+m+x)iA
i(-i,-i) A x 4(1,-1)
2(-1,0)<i
r
Ayo5(1,0)3(-1,1) 6(1,1)
Hinh 3- Phan tCf lUc cat Qy
Cac phan Xil lgc cat (hinh 2 va 3) va eac ham ndi suy (17) va (18) la cua t i e gia b i i bao nay.
3. IVIa tran do cufng phan tii
Phan Xil ehuyen vj (hinh 1) ed 12 an, phan Xil lgc cat Qx (hinh 2) ed 6 an, phan tCf lgc cat Qy (hinh 3) cd 6 an, tong cdng ed 24 an. Gpi
[U] la veetd cdt chUa 24 an cua phan Xil
{U}=[w Q, Q J (19)
TAP CHI CAU D U O N G VIET NAM
Oe trinh bay sau nay dupe de dang, ta md rpng veetd ddng ham ndi suy ehuyen vj (bilu thUe 2) thanh veetd ddng cd 24 thanh phan (them 12 sd khdng vao cudi):
M = [iV, 0,0,0,0,0,0,0,0,0,0,0,0] (20)
Bay gid ehuyin vj moi diim trong phan tCf, cdng thUc (3), dupe viet Igi bang
w'=[N]{U} (21)
TUdng tg, md rdng veetd ddng eie ham ndi suy Ige cat (11) bang c i c h them 12 sd khdng phfa trUde v i 6 sd khdng phfa sau de cd veetd ddng 24 thinh ph'an :
[^e^] = [0.0,0,0,0,0,0,0,0,0,0,0,
iVg,, 0,0,0,0,0,0] ^^^^
Bay gid lgc cat Qx tgi moi diim phan Xis vie't Igi nhu sau
Q : - K X ] { U } (23) TUdng tg, md rpng veetd ddng eie h i m
ndi suy lgc eat NQy bang cich them 18 sd khdng phfa trudc d l cd veetd ddng 24 thinh phan :
iVg^= [0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,iVe,] ^^"^^
phan Xis ve chieu dai that Ax, Ay eua nd khi lay tfch phan.
C i c dgi lUpng bien phan trong (26) l i cic"
bien dgng Xx^Zy^Zxy'/x'/y ddc lap dd'i vdi cae ndi lgc tUdng Ung. Tam cd lien ket hai chieu (lien ke't giO) eho nen dieu kien cgc trj eua (26) l i
dZ I
= 1
dy dU; dU;
-' dxdy(^^) = 0
* • *? | _ ^ xfj
Bay gid lgc cat Qy tgi mdi diem phan tU viet Igi nhu sau
Q;=[NQy]{U} (25)
Biet cac bilu thUe xac djnh chuyin vj (21), lgc cat Qx (23) va lgc cat Qy (25) ta tfnh c i c lien he (1).
Xem phan XiS nhU mpt tam, ta viet lUpng CUdng bijfe (ehuyin ddng theo phudng phap nguyen ly egc trj Gauss) nhu sau
, , [M,XX + MyX, + 2M,,x,, + Q:Y, + Q;y, ] Z= J j _,..j..,Ax Ay,
i=1,2,3,...24 (27) Lfng vdi mdt i cho trudc tfnh tfch phan (27)
ta dupe 1 ddng ed 24 cdt eua ma trin dd eUng phan Xis Ae. Thgc hien eie phep tfnh vdi i = 1,2 24 ta dupe ma tran dp cUng phan XiS Ae (24x24)
4.Vi dg tfnh toan:
SCf dgng phUdng phip tren d l giii b i i toin tam ngim bdn egnh chju t i c dgng cua t i i trpng phan bd deu thda man hai dieu kien bien tren mdi egnh tam (ly thuye't tam Kirchhoff).
Xet tam chO nhat ed kfch thddc axb. Dieu kien bien tren moi egnh tam nhu sau (hinh 4)
w = 0 < p , = 0
0///////////////? ^
H' = 0 (p^ = 0
^ \ ' v \ \ \ \ \ \ \ \ \ \ \ \ \
(26) -',-•', dxdy(—-y) ^ mm
He sd nam trong ngoac ddn eCia bilu thUe (26) la d l dua chieu d i i 2 ddn vj cua
Xl
Hinh 4- Dieu kien bien cua tam ngim bon canh
Chia tam thanh nxn phan Xis ed dp dai c i c egnh Ax, Ay.
Trong trudng hpp n = 2 tam dupe chia thinh 4 phan Xil nhu sau (hinh 5)
KHOA HOC - CONG NGHE
4f • hf I • • •
m-
Hlnh 5. Vf du chia tam thanh 2x2 phan tQ Gpi nw la sd an chuyin vj, ngx la sd an gdc xoay cua dp vdng theo x , ngy sd an gdc xoay cua dp vdng theo y, nqx la sd an luc cat theo X, nqy l i sd an luc cat ttieo y tgi cac nut eua tUng phan Xil trong toan tam. Ta ed cac mang an sau:
O O P 0 0 0 O O P P 0 0
p I 0 I i 11 1 I 0 I n p I 0 I 1 11 1 I 0 r~o
P O P p p P P O P p p P 2
0 4
0 0 p
p p 5
0 0 6
P P P
3 0 7 4
0 8
p p p
5 P P
S 0 0
P 0 0
7 P 9
IP P P
p p p
12 0 13
12 P 13
0 0 0
16 P P P
P 11
p p p
14 0 15
14 P 15
0 0 0
P P 17
18 0 2P
0 0 21
P 0 22
0 0 23
0 0 24
111 P
2!f
20 P 26
21 0 P
22 0 0
23 0 0
24 0 0
25 P 27
28 P P
P P P
30 31 32
30 31 32
0 P P
36 0 0 P
P 29
P P P
33 34 35
33 34 35
P P P
0 0 37
Trong trudng hdp nay ta cd sd an sd (thdng sd' nut) k = 37. Biet dUde cac an cua moi phan Xil, dung ma tran dp eUng phan Xil ta se xay dgng dupc ma tran tong the eua tam theo c i c dieu kien bien va dieu kien lien tge ve gdc xoay theo chieu x va y.
Vdi cac mang an nhu tren ta tha'y dieu kien ve ehuyin vj tgi cae bien eCia tam dUde tg ddng thda man. Ta cd dieu kien eua gdc xoay theo cae bien nhu sau:
Tgi egnh x = 0:
g.=cpi-^Q':=^g.=cp:-^Q':-o
Gh a
Gh
Gh Tgi egnh x = a:
^ 6
-cpi-^Q^:=o
a Gh Tgi egnh y = 0:gl=<Py
"" Qf=0 g.=<pf-^Qf=^
Gh Gh
^1 10
e!'=o
Tgi egnh y = b:
Dieu kien lien tuc theo chieu x ve gdc xoay:
a
~Gh
f
gx. 5 ^ /o22 ] I „ 6 <^ /n23
.^^ Gh^' ) r^ Gh^'
0Dieu kien lien tge theo chieu y ve gdc xoay:
g 14
a
^ 1 3 _ _ ^ g 3 2 _
^y Q^'^y I
r
.14(p. Gh Q\ = 0 Khi dd ma tran toan tam A la ma tran vudng 51x51 vdi 37 thdng sd nut va 14 an phg dd ehinh la eac thUa sd Lagrange.
Xet tam chju t i e dgng eua tai trpng phan bd deu. Vi tai trpng chi t i e dgng tgi nut cd ehuyen vj nen veetd tai trpng B la veetd cdt ed kfch thude (51x1 )va trong trUdng hpp niy ch? cd tai trpng dat tgi hang 1 vdi gia trj P = q*Ax*Ay/4.
Cudi cung ta nhgn dupc phUdng trinh ma tran sau:
A.X = B
TAP CHI CAU D U O N G VIET NAM
Vdi X la ma tran cdt (51x1) gdm 51 an sd ndi tren.
Giai he phUdng trinh tren nhan dupe c i c an cua bai t o i n (eac*thdng sd nut va c i c an sd phg)
Trinh bay tren ehl dung vdi trudng hpp chia tam thinh 2x2 phan Xis. De tfnh toan cho ke't qua chfnh xae thi phai chia tam thanh nhieu phan Xis hdn nOa.
T i c gia vie't ehUdng trinh trong mdi trudng Matlab d l giai b i i t o i n nay. Khi dd ke't qua thu dupe la bieu dd dp vdng, sg bien thien Ige cat va md men xoan tren cac egnh tam, dp vdng va md men tgi diim bat ky tren tam vdi cic trudng hpp chieu day tam thay doi (tUc h/a thay doi).
C i c bilu dd 1, 2, 3 va 4 dudi day la bilu dd dp vdng, sg bien thien Ige eat v i md men xoan tUdng Ung vdi trUdng hpp h/a = 1/100 v i tam dupe chia thanh 20x20 phan Xis.
0 0
Bieu do 2: SU phan bd md men xoan tren canh x=0
Bieu do 3: SU phan bd lUc cat Qx tren canh x=0
Bleu dd 1- DUdng dd vdng
Bieu do 4: SU phan bd lUc cit Qy tren canh x=0
Ke't q u i tfnh dp vdng v i md men udn tgi diim giOa tam vdi c i c trudng hpp chieu d i y tam thay doi dupe trinh biy d bang 1
BSng 1. Tam vuong (a=b) 4 egnh ngam co xet bien dang tri/pt, he so Poisson cua vat ii#u ^l = 0.3
h/a (1) 1/1000
1/100 1/10
w (qa*/D) x=a/2; y=b/2
(2) 0.00127 0.00127 0.00127
w (qaVO) x=a/2; y=b/2
(3) 0.00127 0.00127 0.00151
Mx (qa^) x=a/2; y=b/2
(4) 0.02308 0.02308 0.02336
My (qa^) x=a/2; y=b/2
(5) 0.02308 0.02308 0.02336
KHOA HOC - CONG NGHE
h/a (1) 1/5 1/3
w (qa*/D) x=a/2; y=b/2
(2) 0.00127 0.00127
w (qaVO) x=a/2; y=b/2
(3) 0.00217 0.00370
Mx (qa^) x=a/2; y=b/2
(4) 0.02373 0.02400
My (qa^) x=a/2; y=b/2
(5) 0.02373 0.02400
Op vdng d cdt (2) tUdng Ung vdi trudng hpp khdng xet anh hudng cua bien dgng trupt ngang (G->oo) [1], dd vdng d cdt (3) tUdng Ung vdi trudng hdp ed xet de'n anh hudng eua bie'n dgng trupt ngang.Theo ke't qua tfnh toan eua vf dg tren ta tha'y: Khi xet anh hudng cua bie'n dgng trupt ngang, dp vdng cua tam thay doi khi ty le h/a thay doi. Vdi tam ed cung chieu day thi dp vdng eua tam cd xet de'n anh hudng cua bien dgng trupt ngang tang len ding k l .
4. Ket luan:
PhUdng phip phan Xis hUu hgn vdi viec Iga chpn ham ndi suy dd vdng dgng bac ba theo X va y va xay dung them c i c ham ndi suy lgc cat Qx la da thUc bac hai dd'i vdi x, bae nha't dd'i vdi y, ham ndi suy lgc cat Qy l i da thUe bae nha't theo x, bae hai theo y giup viec giai bai toin tam chju udn ed xet bien dgng trUpt ngang d i dang hdn, bao dam dp chfnh xac tdt. C i c ke't qua tfnh toan ed tfnh hdi tg v i khdng xay ra hien tupng
"sheariocking" khi md dun trupt G^oo hoge khi ty le h/a ri't nhd (h/a =1/100-4-1/1000) •
TAI LIEU THAM KHAO:
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Journal of Computational and Applied Mathematics 200 (2007) 606 - 627, Elsevier press. Available online at
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[8] B.A.KHcejiCB (1973), PACHET nJlACTHH, MOCKBA, CTP0HH3MT
[9] Nguyen Thuy Anh, Ly thuyet tam cd xet bien dgng trupt, Tgp chf Cau dUdng thing 12 nam 2010.
TAP CHI CAU D U O N G VIET NAM