L Thuyt n Hi 1 M U Tiliucbinsonnhmmcchcungcpchohcvinnhngkhinimcscal thuyt phn tch cc kt cu cng trnh ni chung v kt cu tu thy ni ring. Trong ti liu, cc nguyn tc c bn sau y c gng th hin: Nguyn tc u tin l bo m tnh nht qun, tnh logic trong ton b ti liu c trnh by, trnh dng bt c mt gi nh m h no lm c s cho mt phng php tnh no . Mi gi thit c tnh gn ng lun c i km bi nhng li gii thch v tnh ph hp ca chng. Nguyn tc th hai l chp nhn vic gii thch thm khi cn thit. Nguyn tc ny xut pht t c im ca qu trnh nhn tcn c s lp li. Tnh cht lp li l mt trong cc cng c quan trng ca qu trnhnhnthc.Vicgiithchthmcmcchtrnhhiuvnmtcchsailch,trnhngnhn. Nguyn tc th 3 l cc ch chnh ca cc ni dung c trnh by lun c tm lc, nh ngi c c c nhn thc tng th vi y nhng ni dung chnh yu v nhng vn c qua.LThuyt n Hi gii cc bi ton lin quan n vic xc nh ng sut v bin dng, xut hin trong vt th n hi, di tc dng ca lc ngoi. y cng chnh l vn gii quyt trong mn hc sc bn vt liu. Tuy nhin, trong gio trnh sc bn vt liu, nhiu gi thit tnh ton c th khc nhau c s dng, nhm thu c nhng li gii gn ng cho cc bi ton ring bit v do , ch p dng c cho chnh cc bi ton ny thi. L Thuyt n Hi t ra mc tiu l tm nhng li gii chnh xc, da trn cc gi thit chung v tnh cht ca vt th kho st m khng ph thuc g vo hnh dng vt th cng nh tnh ring bit ca ti trng tc dng ln vt th. . . Vt th kho st trong L Thuyt n Hi c gi thit l c tnh lin tc, tc, vt th kho st lun in y khng gian m n chim ch, trc cng nh sau khi b bin dng. Ta coi l trong mi th tch bt k, d nh n u, cng cha v s cc phn t v tc dng ca phn vt th b ct b ln phn kho st c th nh gi bng tr s trung bnh ca s thay i lc tng tc gia cc phn vt th nm v haiphacamtct.Ccchuynvlnhnghmlintccatoccim.Tnhchtlintccho php ng dng gii tch cc i lng v cng b vo vic nghin cu bin dng ca vt th n hi. Sai s lin quan n vic s dng tnh cht ni trn l c th b qua trong cc bi ton thc t, v n chng k khi xc nh ng sut trn cc din tch vi kch thc c ca khong cch phn t v khi xc nh cc chuyn v ca cc im m khong cch gia chng cng vo c khong cch gia cc phn t. Ngoi ra, cng cn phi gi thit rng, c th p dng cc nh lut ca ca tnh hc v ng lc hc cho cc phn t nh tu , t vt th kho st. Cc vt th n hi, l i tng nghin cu ca mn hc, cn c nhiutnh cht khc m ta s cp n sau ny khi thit lp cc phng trnh c bn ca L thuyt n hi. L Thuyt n Hi 2 Chng I TON HC C S Cc ng dng ca L Thuyt n Hi i hi s hiu bit v nhiu lnh vc ton hc khc nhau. Bn thn L Thuyt n Hi c xy dng trn c s ng dng nhiu i lng bin i (cc bin) khc nhau, trong c cc bin v hng, cc vector, cc trng tensor vphi s dng nhiu n cc php tnh tensor. Vn dng cc nguyn l ca C Hc Cc Mi Trng Lin Tc, L Thuyt n Hi c xy dng di dng mt tp hp cc phng trnh o hm ring m li gii ca chng c tm trong cc min trng vi khng gian m vt th kho st chim ch. gii cc phng trnh ny, trong nhiu k thut khc nhau, thng phi dng n phng php Fourier, cc k thut bin phn, cc php bin i tch phn, cc bin s phc, l thuyt th nng, phng php sai phn hu hn, phng php phn t hu hn, phn t bin,... V th cho nn, nm c cc c s ca L Thuyt n Hi, cn c mt cn bn ton hc tng xng. Mc ch ca chng ny l cung cp cho ngi c mt s chuyn ton hc thit yu, phc v cho vic lnh hi cc c s ca L Thuyt n Hi. Cc chuyn ton hc khc s c trnh by tm lc cc chng, ni m chng c s dng. 1.1 Cc nh ngha v v hng, vector, ma trn v tensor. Trong L Thuyt n Hi thng s dng nhiu loi bin khc nhau, trong c cc loi bin c cc bin v hng. Cc bin v hng ch biu th v ln (ca mt i tng no ) ti mi mt im ring bit trong khng gian (m vt th kho st chim ch). Cc v hng thng gp l: t trng vt liu , module n hi E, h s Poisson , module trt G. Mt loi bin khc cng c s dng nhiu, l cc i lng vector. Cc bin vector c th biu din c theo cc thnh phn ca n trong cc h ta 2 hoc 3 chiu. V d v cc bin vector l: Chuyn v ca mt im vt cht trong vt th n hi, gc xoay ca mt vt th Cc cng thc ca L thuyt n Hi cn c din t vi vic s dng cc bin ma trn. Cc bin ma trn thng c dng khi cn biu th mt i lng vi nhiu hn 3 thnh phn. Ma trn ng sut v ma trn bin dng l nhng v d v bin ma trn. Nh s thy trong cc chng di y, biu th ng sut hoc bin dng ti mt im thuc khng gian 3 chiu, i hi ng thi 9 thnh phn (trong c 6 thnh phn c lp nhau). Trong trng hp ny, ch cn s dng mt bin di dng ma trn vi 3 hng v 3 ct l . Tm li, trong h ta Cc 3 chiu, cc bin v hng, bin vector, bin ma trn c th c biu din (v d) nh sau: T trng (v hng)=; Vector chuyn v = 3 2 1e e e u + + = v ; Ma trn ng sut=[ ]((((

= =z zy zxyz y yxxz xy x ; trong ,3 2 1, , e e el 3 vector n v c s trn 3 trc ta . TrongvicbiutccvncaLThuytnHi,vimtbinnhiukhicnihis thnh phn nhiu hn th na. Khi phi vn dung n cc php tnh tensor vi vic s dng cc cck hiuchs(IndexNotation).iunychophpbiuthttcccbinvccphngtrnhdn1theo mt s tiu chuNn ha duy nht. C th nimt cch nginrng: cc bin v hng, bin vector, bin ma trn v cc bin khc cp cao hnu c th biu din bi cc tensor vi cc cp khc nhau. 1.2 K hiu ch s (Tensor) v cc php tnh K hiu ch s l s gn nht m nh mt tp hp cc s (l cc phn t hay cc thnh phn) c th c biu din bi ch mt k hiu c km cc ch s. V d nh c 3 sa1, a2 , a3 c th biu din bi mt k hiu ch s (tensor) ai trong , ch s i bin i (chy) trong phm vi 1, 2, 3. Cng tng t, 1 L cc phng trnh ch o m vic gii chng cho php tm c li gii ca mt vn (mt bi ton) no .L Thuyt n Hi 3 k hiu aijbiu th cho c tp hp 9 s: a11, a12,a13,a21,a22,a23,a31,a32,a33. Phm vi ca cc ch s i v j u l 1, 2, 3.C nhiu cch biu din khc nhau i vi tp hp cc s ni trn, tuy nhin, cch thng dng nht l s dng s c lin h n cc dng thc vector v dng thc ma trn nh sau: ((((

=321aaaia ; ((((

=33 32 3123 22 2113 12 11a a aa a aa a aija (1.1) Trong dng thc ma trn, a1i biu din hng u tin, ai1 biu din ct u tin. Vi cc hng v ct khc cng c th biu din theo cch tng t, tc ch s u tin th hin s th t hng, ch s th 2- s th t ct.Mt cch tng qut, aij...k vi N ch s phn bit biu th c n 3N s phn bit. Cn thy r rng ai v aj cng biu th 3 s nh nhau, cng nh aij v akl cng biu th mt ma trn.Cc php tnh cng, tr, nhn (tch) hay bng nhau ca hai khiu ch s cng c nh ngha theo cch thng thng. Chnghn nh, php cng v php tr c nh ngha bi: ( (( (( (( (( (( ( ( (( (

= == = ( (( (( (( (( (( ( ( (( (

= == = 33 33 32 32 31 3123 23 22 22 21 2113 13 12 12 11 113 32 21 1;b a b a b ab a b a b ab a b a b ab ab ab akl ij i ia a b a . (1.2) Nhn vi mt v hng c xc nh nh sau: ((((

=((((

=33 32 3123 22 2113 12 11321;a a aa a aa a aaaaij i a a(1.3) Ngoi tch gia hai k hiu vi cc ch s phn bit trong trng hp n gin c xc nh theo khun mu sau: ((((

=3 3 3 3 1 33 3 2 3 1 23 1 2 1 1 1b a b a b ab a b a b ab a b a b aj ib a (1.4) Cc ton t trn y tun th lut kt hp, lut giao hon v lut phn phi. Cc nh lut ny c minh ha qua cc v d sau: ( ) ( )( ) ( )( ) .;;;;k ij k ij k k ijl jk i l jk ii i i i i iij kc a b a c b ac b a c b ac b a c b aa b b aa b b a+ = +=+ + = + +=+ = +k iji i i i(1.5) Ton t bng (biu th bi du "=") ch c th t gia hai k hiu c cc ch s phn bit ph hp (dng nht) vi nhau v biu th quan h bng nhau ca cc thnh phn tng ng ca hai k hiu.Chng hn:ai = bi c ngha l: a1 = b1; a2 = b2; a3 = b3 ; cn quan h aij = bij c ngha tng t, l: a11 = b11;a12 = b12; a13 = b13....;a33 = b33.iu ng ch y l bo m s ging nhau gia cc ch s tng ng v hai pha ca du "=".Chnh v th, cc quan h c dng nh:ai = bj hay aij = bkl c ngha m h (khng r rng) vccchstngngkhnggingnhau.Tmlilccchsphnbittngnghaiv trong quan h "bng" phi ng nht. L Thuyt n Hi 4 Qui c ch s cm: cho tin li trong cch din t, ta qui c rng: nu nh mt ch s di no xut hin hai ln trong mt s hng th n c ngha l tng ly theo ch s ny khi n chy t 1 n 3, v d: + + = + + = ===313 3 2 2 1 13133 22 11ji i iiii iib a b a b aa a a aj ijb aa(1.6) Cn nhc li rng, theo qui c trn,.. = = =mm jj iia a a . nn cc ch s lp (tc xut hin hai ln) c tn l ch s cm. Cc ch s khng c n nh (tc khng phi l mt con s c th) v khng lp li c gi l ch s t do hoc ch s phn bit. Qui c trn y c gi l qui c ch s cm hoc qui ctng.Quictngskhngpdngnunhxuthindugchdimttronghaichslp hoc, n gin hn, nu c ch thch "khng tng" bn cnh. Vic mt ch s xut hin qu 2 ln trong cng mt s hng (chng hn: aiii, aijkl bkml clkj) s c ngha m h v cn trnh s dng. Trong mt k hiu,tcngcaquicchscmivihaichsgingnhaugilphpco.Chnghnnh:aii chnh l kt qu ca aij khi co hai ch s i v j . Php co c s dng khi thc hin ngoi tch gia hai k hiu ch s m mt ch s trong mi k hiu (tensor) trng vi ch s ca k hiu kia, lm ny sinh ton t ni tch; chng hn, aij bjk l ni tch thu c t ngoi tch aij bmknh php co thc hin i vi hai ch s j v m. Mt tensoraij...nm...kl c gi l i xng i vi n v m nu nh tha mn aij...m..n..kl=aij..n..m..kl(1.7) cn nu nh tha mnaij...m..n..kl = -aij..n..m..kl(1.8) thcgilixnglchhocphnxng.Cththyrng,vihaitensormmttronghaili xng i vi hai ch s no cn tensor kia li l phn xng cng i vi hai ch s ny ngoi tch ca hai tensor ny bng 0. Tc: nu aij..m..n..kl i xng i vi m v ncn bpq...m..n..l l phn xng i vi m v n thaij..m..n..kbpq..m..n..l = 0.(1.9) Ta c th vit ng nht thc thng dng sau ( ) ( )( ) [ ] ija a a a a a aij ji ij ji ij ij+ = + + =2121,(1.10) trong , s hng th nht ( )( )ji ij ija a a + =21 l phn i xng cn s hng th hai [ ]( )ji ij ija a a =21 l phn phn xng. Vy l: mt tensor aij bt k c th biu din bng tng ca phn i xng v phn phn xng. Mt tensor i xng ija ch c 6 thnh phn c lp nhau cn nu ija l phn xng, cc thnh phn trn ng cho aii (khng tng) ca n bng 0 v nh vy, ch c 3 thnh phn c lp. V rng [ ] ija ch c 3 thnh phn c lp nn c th biu th n nh k hiu vi ch s n, ai , chng hn. 1.3 Delta Kronecker ij v K hiu hon v ijk Tronglthuyttensor,cmtkhiuchscbitthngdng,lDeltaKronecker,c nh ngha bi ((((

===1 0 00 1 00 0 1khi 01j ij i khiij

(1.11)L Thuyt n Hi 5 Trong l thuyt ma trn, k hiu trn y chnh l ma trn n v. Delta Kronecker l mt tensor i xng. Cc tnh cht lm cho Delta Kronecker c ng dng rng ri bao gm: 3;;; 1 3;= == = = == == == = = == == == = = == == == = = == == == =ij ij ii ij ijij ik jk ik jk iji ij i j iji i iiji ij ; a a a a ; a a a a ; a a ; j (1.12) BncnhDeltaKronecker(cngilkhiuKronecker),cmtkhiukhc,cgilK Hiu Hon V (ijk ), cng c s dng rt rng ri. K hiu hon v c nh ngha nh sau:

03 2 1 , , 13 2 1 , , 1=lai con hop truong cua nghich vi hoan la khi -cua thuan vi hoan la khi, , k j i, , k j iijk(1.13) v th cho nn,0; 1 ; 1322 221 132 312 321 312 231 123= = = = = = = = .Nhnxt: Trong tng s 27 k hiu hon vkhc nhau c 3 k hiu nhn gi tr 1, ba k hiu nhn gitr -1, cn li l 0; K hiu hon v l phn xng i vi mt cp 2 ch s bt k ca n. K hiu c bit ny rt tin li trong vicnh lng cc nh thc v cc tch vector. nh thc ca mng aij c th c vit di hai dng tng ng: [ ]333 32 3123 22 2113 12 11detk ij ija a aa a aa a aa a a a a a a aj2 i1 ijk 3k 2j 1i ijk= = = = (1.14) trong , biu thc u tin theo k hiu ch s ca (1.14) th hin khai trin theo hng trong khi biu thc th hai lkhai trin theo ct. Vn dng tnh cht kr kq kpjr jq jpir iq ippqr ijk =(1.15), c th biu din nh thc di dng khc [ ]kr jq ip pqr ijk ija a a a61det = (1.14) 1.4Php bin i ta

Trong thc t, thun tin li cho cc ng dng khc nhau, ngi ta thng biu th cc bin n hi, nh chuyn v, ng sut, bin dng, cng nh phi vit cc phng trnh theo nhiu h ta khc nhau. iu ny i hi phi c cc qui tc bini c bit i vi cc bin v hng, vector,ma trn, cng cc bin cp cao hn khc, tng ng vi vic chuyn i h ta ni trn. tng ny gn lin vi nh ngha c s ca bin tensor v qui lut bin i cc ta c lin quan i vi cc tensor. Ta hn ch ch tho lun v php chuyn h gia cc h ta -Cc (De Cartre) .Hy kho st hai h ta -Cc, nh biu din trn hnh H1.1: L Thuyt n Hi 6 H 1 (ngun): (x1, x2, x3) v h 2 (ch): (x'1, x'2 , x'3), ch khc nhau v hng ca cc trc ta . Cc vector n v c s ca h ta ngun l{ } { }3 2 1, , e e e e =i v ca h ta ch: { } { },3,2,1,, , e e e e =i. a vo k hiu( )j ix x , ' cos =ijQ . (1.16) S dng k hiu ny, c th biu din cc vector c s trong h ta c du phNy (ch) theo cc vectorc s trong h khng c du phNy (ngun):3 2,33 2,23 2,1e e e ee e e ee e e e33 32 1 3123 22 1 2113 12 1 11Q Q QQ Q QQ Q Q+ + =+ + =+ + = (1.17) Hay, nu nh dng k hiu ch s, c th vit li (1.17) di dng: je Q eij,i=(1.18) Ma trn[ ]ijQgi l ma trn bin i ta , hay gn hn, l ma trn xoay (h ta ). Tng t, vic chuyn h ta ngc li cng c th vit theo cng mt kiu nh trn: j ji ie' Q e = (1.19). By gi, vi mt vectorvbt k, c th biu din trong c hai h ta nh sau: jv v vv v ve' v' e' e' e' e v e e e vj 3 3 2 2 1 1i i 3 3 2 2 1 1= == = + ++ + + ++ + = == == == = + ++ + + ++ + = == =' ' ' (1.20) Thay (1.19) vo phng trnh u ca (1.20) ta c jv v e' Qji i= == = , nhng t phng trnh th 2 ca (1.20), jv' e' vj= == =nn ta li cjv vij iQ ' = == =(1.21) Tng t, thay (1.18) vo phng trnh th hai ca (1.20) ta c jv' vji iQ = == = (1.22) Lurngtrongphpbinitrn,bnthnvectorlkhngi(tcginguyndiv hng).Cc quan h (1.21) v (1.22) to nn qui tc bin i cc thnh phn -Cc ca mt vetor khi h ta -Cc vung gc thay i hng. V th cho nn, khi bit cc thnh phn ca mt vector trong L Thuyt n Hi 7 mthta,nhdngquanh(1.21)hoc(1.22),tacthtnhcccthnhphncavectorny trong h ta th 2, c hng xc nh so vi h ta th nht. V php bin i ni trn ch thc hin i vi cc h ta vung gc nhau nn n phi chu mt s rng buc nht nh, l nhng rng buc i vi ma trn cosine ch phng Qij. Ta hy xc lp cc rng buc ny.Trn c s ca cc quan h (1.21) v (1.22), c th vit kv v vjk ji j ji iQ Q ' Q = == = = == =(1.23). T tnh cht (1.12) ca Delta Kronecker biu thc trn c th c vit nh sau: ( (( ( ) )) ) . 0 = == = = == =k ik jk ji k jk ji k ik Q QQ Q v hay v vV phng trnh trn ng vi mi kvnn biu thc trong du ngoc trn trong v bn tri phi bng 0, tc .ik Q Qjk ji=(1.24) Mt cch tng t, dng (1.21) v (1.22) loi tr iv(ch khng phi iv' ), ta c.ik Q Qkj ij= (1.25) Cc quan h (1.24) v (1.25) chnh l cc iu kin trc giao mcc cosine ch phng (1.16) phi tha mn v cng l cc rng buc cn tm.Xc nh nh thc ca mt trong hai ma trn trn, ta c [ ] . 1 det =ijQCc ma trn tha mn quan h trn y c gi l ma trn trc giao, cn cc php bin i (1.21) v (1.22) thuc v cc php bin i trc giao. 1.5 Tensor -Cc Cc v hng, vector, ma trn cng nh cc bin cp cao hn khc u c th c biu din theo mt phng thc tng qut, l theo cc k hiu ch s. Theo quan nim ny, cc bin, theo mi kiu,uccoinhcctensorvicpkhcnhau.Cctnhchtcaphpbinitrnhbytrnycho vector cng chnh l tnh cht ca php chuyn i tng qut i vi cc tensor ny. Ta ch hn ch trong cc php chuyn i gia cc h ta -cc. Khi , cc quan h ca php bin i i vi tensor cc cp khc nhau (gia hai h ta -Cc) c th c vit nh sau: quat. tong cap:4; cap; 3 captran); (ma hai cap(vector); mot caphuong); (vo = == = = == = = == = = == = = == = = == =pqr..s ms kr jq ip ijk...mls kr jq ip ijklkr jq ip ijkjq ip ijp ip ia ...Q Q Q Q a'a Q Q Q Q a'a Q Q Q a'a Q Q a'a Q a'pqrspqrpqzero cap a a'(1.26) Theo cc nh ngha trn, v hng l tensor cp zero, vector l tensor cp 1, ma trn l tensor cp 2.. . Nh vy l quan h (1.26) xc lp qui tc bin i i vi cc thnh phn ca tensor -Cc cp bt k di tc ng ca php xoay Qij.L thuyt chuyn i trn y t rart hu ch trong vic xc nh cc ng sut, bin dng v chuyn v theo cc h ta khc hng nhau. Mt s tensor c c im l cc thnh phn ca n lun khng i di tc ng ca bt k php chuyn(xoay)tano.Cctensornycghpchungvomtloi,giltensornghng.D thy rng, Delta Cronecker ij c tnh cht ni trn v v vy, n l tensor ng hng cp 2. K hiu hon v ijkcng l tensor ng hng, cp 3.L Thuyt n Hi 8 Cn phn bit r gia cc thnh phn ca mt tensor vi bn thn cc tensor ny. Tr li vi cng thc biu din vectorv (tensor cp 1): 3 1e' e' e' e e e v3 2 2 13 3 2 2 1 1v v vv v v+ ++ + + ++ + = == =+ ++ + + ++ + = == = (1.27) Cng tng t nh trn, tensor cp 2 - A c th biu din bi 'j'i'ij j i ij3 1 33 2 1 32 1 1 313 1 23 2 1 22 1 1 213 1 13 2 1 12 1 1 11e e A e ee e e e e ee e e e e ee e e e e e A= == = = == =+ ++ + + ++ + + ++ ++ ++ + + ++ + + ++ ++ ++ + + ++ + = == =Ar r r r r rr r r r r rr r r r r rA A AA A AA A A (1.28) Dng tng t cng c th dng biu din tensor cp cao hn. Dng biu din (1.28) ca tensor c tn l dng dyadic (dyadic notation). Nhiu khi dng dyadic cn c biu din thng qua tch tensor: je ei .Cc quan h (1.27) v (1.28) cho thy, mt tensor bt k c th biu din theo cc thnh phn ca n trong mt h ta bt k. Cc thnh phn ca tensor thay i theo tng h ta (cn bn thn tensor th khng i). C th chn tensor ng sut, m chng tip theo ta s nghin cu chi tit, lm v d. Cc thnh phn ng sut l cc thnh phn ca mt tensor cp 2, biu th trng thi ng sut ti mt im. Cc thnhphnnythayitheo(nhhngca)htacchn thhintrngthingsutcn chnh bn thn tensor ng sut th khng i (vn l tensor c trng cho trng thi ng sut kho st). V d 1.1:Cho bit cc thnh phn ca cc tensor cp mt v cp hai nh sau: ((((

=(((

=4 2 33 2 03 0 1;241ij ia a Hy xc nh cc thnh phn ca cc tensor ny i vi h ta ch, xc nh bng cch xoay h ta ngun mt gc 600 quanh trc x3 theo hng ngc chiu quay ca kim ng h khi nhn ngc vi trc x3 (tc khi nhn t trn xung).Hnh v H1.2 biu th h ta ngun v h ch cng vi gc gia cc trc ca hai h ta ny.Ma trn xoay trong trng hp ang xt s lL Thuyt n Hi 9 ((((

=((((

=1 0 00 2 / 1 2 30 2 3 2 / 10 cos 90 cos 90 cos90 cos 60 cos 150 cos90 cos 30 cos 60 cos0 0 00 0 00 0 0ijQ Tc ng ca php bin i h i vi vector xc nh nh phng trnh th 2 ca(1.26) ((((

+=((((

((((

=22 3 23 2 2 / 1241.1 0 00 2 / 1 2 30 2 3 2 / 1'iai vi tensor cp 2, cn dng phng trnh th 3 ca (1.26) v thu c kt qu: ,4 2 / 3 3 1 3 2 / 32 / 3 3 1 4 / 5 4 / 33 2 / 3 4 / 3 4 / 7

1 0 00 2 / 1 2 30 2 3 2 / 14 2 33 2 03 0 11 0 00 2 / 1 2 30 2 3 2 / 1( (( (( (( (( (( ( ( (( (

+ ++ + + ++ += == =( (( (( (( (( (( ( ( (( (

( (( (( (( (( (( ( ( (( (

( (( (( (( (( (( ( ( (( (

= == = = == =Tpq jq ip'ija Q Q a trong , ch s trn "T" ch php chuyn v i vi cc thnh phn ca ma trn. 1.6 Tr chnh v hng chnh ca tensor i xng cp 2 Khi xoay h ta (-Cc), cc thnh phn ca tensor thay i theo v tn ti mt h ta xc nh no m, vi n, cc thnh phn ca tensor c gi tr cc tr (cc tiu hoc cc i). iu ny c th nhn thy bng trc gic khi kho st s thay i gi tr cc thnh phn ca mt vectorkhi xoay h ta . Nu chn h ta ch sao cho trc 3' xtrng vi vectorvth, khi , vector ny s c cc thnh phn trong h ta ch l:{ } v v 0 0 = . Trong trng hp ny, 2 trong 3 thnh phn c gi tr bng 0 trong khi thnh phn cn li c gi tr ln nht, bng di ca vector. Thuc tnh nu ra trn y t ra c bit hu dng i vi tensor i xng cp 2, l loi tensor c s dng biu din ng sut/bin dng ti mt im trong vt th n hi. Nu tn ti sv vector n v) , , (3 2 1n n n n sao choi n aj ij=(1.29), thhng ca vector) , , (3 2 1n n n nc gi l hng chnh hay vector ring ija cn s c gi l tr chnh hay tr ring ca ca tensor i xng, cp 2, ija . Quan h (1.29) c th vit di dng ( ) 0 = j n aij ij Quan h trn y chnh l mt h gm 3 phng trnh i s tuyn tnh thun nht vi 3 Nn s 3 2 1, , n n n . H phng trnh ny s c cc nghim khng tm thng (khng ng thi bng 0) nu nh tha mn: [ ] 03 2213= + + = a a aijI I I det aij(1.30)Trong ,L Thuyt n Hi 10 ( (( ( ) )) )[ [[ [ ] ]] ]ijaaiiaIa aa aa aa aa aa aIa a a Iaa a a aaij ij jj iidet21333 3113 1133 3223 2222 2112 11233 22 11 1= == =+ ++ + + ++ + = == = = == =+ ++ + + ++ + = == = = == =(1.31) Cc v hng a a aI , I , I3 2 1 c gi, tng ng, l cc bt bin c bn th nht, th hai v th 3 ca tensor ija , cn quan h (1.30) c gi l phng trnh c trng. ng nh tn gi ca chng, cc bt bin trn y khng thay i gi tr theo php bin i h ta . Cc nghim ca phng trnh c trng (1.30)chnhlcctrring vvivicthayngcmimttrongccnghimnyvophngtrnh (1.29), ta tm c hng chnh) n , n , n (3 2 1n . Nu cc phn t ca ija l thc th c th chng t rng cc nghim 3 2 1, , cng l thc. Ngoi ra, nu nh cc nghim l phn bit nhau th cc hng tng ng vi cc tr ring s vunggc nhau.Nh vy, c th kt lun rng, mi mt tensor i xng cp 2 c t nht3hngchnhvunggcnhauvcnhiunht3trchnhlnghimcaphngtrnhctrng. Vi k hiu cc hng chnh l () ( ) ( ) 3 2 1n n n , ,tng ng vi cc tr ring 3 2 1, , , c th xy ra cc kh nng sau y: 1.C 3 tr chnh phn bit nhau: khi , ba hng chnh tng ng l duy nht; 2.C 2 tr chnh bng nhau( )3 2 1 = : hng chnh () 1nl duy nht; mi hng vung gc vi () 1n u c th coi l hng chnh tng ng vi 3 2, ; 3.C ba tr chnh bng nhau: mi hng khi u l hng chnh. Tensor l ng hng, nh cp trn y.Nhvylvimttensorixngcp2btk,baogicngcthccmthta -Ccthunsaochocchngchnhcatensornmdctheotrccan.Htrcnhtrngilh trcchnhcatensor.Trongtrnghpny,ccvectornvcschnhlcchngchnhnv() ( ) ( ) 3 2 1n n n , , , v nh vy, trong h ta chnh, tensor ni trn thu gn v dng ng cho ((((

=3210 00 00 0ija (1.32) C th thy rng cc bt bin c s nh ngha bi (1.31) c th c biu din theo cc tr chnh nh sau . I; I; Iaaa3 2 1 31 3 3 2 2 1 22 2 1 1 =+ + =+ + =(1.33) Cc tr ring c tnh cht cc tr. Nu ta sp xp cc tr ring theo trt t 3 2 1 > > , th 1s l gi tr ln nht trong cc phn t trn ng cho trong khi 3l gi tr nh nht trong s cc phn t trn ng cho (trong mi h ta ). Tnh cht ny c vn dng trong L Thuyt n Hi khi cn tm gi tr ln nht trong cc thnh phn ng sut/bin dng ti cc im trong vt th n hi. V d 1.2:Hy xc nh cc bt bin, cc tr chnh v cc hng chnh ca tensor i xng cp 2 sau y: ((((

=3 4 04 3 00 0 2ijaL Thuyt n Hi 11 Cc bt bin xc nh theo cng thc (1.31): ( ) . I; I; a Iaaiia50 16 9 23 4 04 3 00 0 225 6 25 63 00 23 44 33 00 22 3 3 2321 = == = =+((

+ == + = = Phng trnh c trng: [ ]( )( ), 5 ; 2 ; 5; 0 25 2; 0 50 25 2 det3 2 122 3 = = = = = + + = ij aij trong trng hp kho st, c 3 tr ring l phn bit nhau. Vi nghim51 = , phng trnh (1.29) tr thnh h phng trnh: ()() ()() () = = + = 0 8 4; 0 4 2; 0 31312131211n nn nn H ny cho nghim chuNn ha sau:( )( ) . 5 / 23 21e e n + =Mt cch tng t cho hai hng chnh cn li: ( ) ( )( ) 5 2 /3 2312e e n ; e n = =D dng chng t c rng, cc hng ny vung gc nhau. Trn hnh H1.3 biu din cc hng chnh trong h ta cho ban u cng vi h ta thun mi ('3'2'1, , x x x ). Vi h ta mi ny, ma trn chuyn h, nh ngha bi (1.16), s l ((((

=5 2 5 1 00 0 15 1 5 2 0/ // /Qij L Thuyt n Hi 12 im cn ch y l cc hng ca ma trn ijQ cng chnh l cc vector ring ca tensorija .S dng ma trn bin i ta ny, ta tnh c cc thnh phn ca tensor cho trong h ta mi: ((((

=5 0 00 2 00 0 5ij' aKtqunyphhpvilthuytbiuthbiquanh(1.32)ratrnyrng,tensorni trntronghtachnhphicdngngcho,vi3thnhphntrnngchochnhlcctr ring. 1.7 Cc php tnh i vi vector v ma trn TrongLThuytnHisdngrngriccphptnhiviccvector,matrnvtensor. Ccphptnhnybaogm:Nitch(dotproduct)vngoitch(crossproduct)giaccvectorcng nhiu tch ma trn, tensor khc. Tt c cc ton t trn u c th din t mt cch hiu qu nh s dng k hiu ch s tensor hon chnh.1.Trctintahyxtcctchthchiniviccvector.Choccvectora vb ,vicc thnh phn -cc l i ib a , . Tch v hng (scalar product) hay cn gi l ni tch, c nh ngha biib a b a b a b aia.b = + + =3 3 2 2 1 1 (1.33) V trong biu thc trn, cc ch s lp li nn kt qa phi l mt v hng, hay l mt tensor cp 0. Khi , di ca vector c th c biu din bi( ) ( )2121i ia.a a a a = =(1.34) Tch vector, hay cn gi l tch cho (cross product) c vit di dng ib a b b ba a a X ee e eb ak j ijk3 2 13 2 13 2 1= == = = == =(1.35) trong , iel cc vector n v ca h ta . D nhn thy rng, kt qu ca tch vector l mt vector vi cc thnh phn l kb a j ijk(vi i=1, 2, 3). Mt loi tch thng dng khc l tch tam bi v hng (scalar triple product) c nh ngha bi kc b a c c cb b ba a aXj i ijkc ab = == = = == =3 2 13 2 13 2 1(1.36) 2.Tip n l cc tch ma trn thng dng. Vi cch vit thng thng ca ma trn v vector, cc tch gia ma trn[ ] A A =v vector a c vit nh sau [ ]{ }{ }[ ]ia A A aA a a Aij ij iT Tij j j ijA a A aa A Aa= = == = =(1.37) trong ch s trn "T" mt vector biu th php chuyn v i vi vector ny. Php chuyn v ny ch n gin bin vector ct (3x1) thnh vector hng (1x3) v ngc li. Ch rng cc tch, xc nh theo (1.37) trn y, u cho kt qu l mt vector. Cc biu thc dng ny cha ng nhiu ni tch trong cc s khiuchs.Lurng,trnhtcxuthinccthnhphncatchkhnglmthayiktqu cui cng.L Thuyt n Hi 13 Gia hai ma trn A v B tn ti cc tch sau y: [ ][ ]( ), B A Tr TrB A TrB AB AB Aij ijT Tji ijjk jiTkj ijTjk ij) (AB ) (AB; AB; B A; AB; B A AB= ===== =(1.38) trong , TA l chuyn v cn Tr(A) l vt caA, c nh ngha bi . A A A Tr33 22 11+ + = ==iijiTijA (A); A A(1.39) Cngnhtrnghptchvevtor,ivitchmatrn,khiquitctngtheochscmchiu lc,ktquskhngphthucvothtlitkcaccshng.iunykhngthchiulBA AB = . Ta bit biu thc ny ni chung l khng ng. 1.8 Cc php tnh i vi tensor -cc a s cc bin dng trongLThuyt n Hi c xc nh trong khnggian ca vt th kho st.Ccbinnylhmcataccimthuckhnggiannitrn.Trongccbitoncsph thuc vo thi gian, cc bin ny cn c th bin i theo thi gian. V th cho nn, cc v hng, ma trn vbintensortngqutmtanghincucnglcchmcaccbinkhnggian(3 2 1, , x x x ).Vcc phng trnh ca L Thuyt n Hi cha ng c cc ton t vi phn v tch phn nn ngi c cn c nhng hiu bit nht nh v cc php tnh lin quan i vi trng tensor -cc.Khi nim trng i vi cc thnh phn ca tensor c th biu th nh sau ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ).3 2 13 2 13 2 1x a x a a a; x a x a a a; x xij i ij ij iji i i i ii= = == = == = =,x ,x x,x ,x xa a ,x ,x x a a cho tin li trong biu din, ta a thm k hiu du phy (comma notation) dng biu th cc o hm ring: ,...ijaxa, axa a,xak,k ij ijj , ii,i = == = = == = = == =C th thy rng, nu nh cc ch s ca o hm (ring) l phn bit th cp ca tensor (sau tc ng ca ton t o hm ring) tng ln 1. Chng hn nh, sau tc ng ton t o hm ln vector ia s thu c tensor cp 2, j i ,a , c dng ma trn sau: ( (( (( (( (( (( (( (( (( (( (( (( (( (( ( ( (( (

= == =332313322212312111,xaxaxaxaxaxaxaxaxaaj i S dng cc ta -Cc ( z y x , , ), ta kho st cc o hm nh hng ca mt hm v hngf , theo hng s: dsdzzfdsdyyfdsdxxfdsdf++=L Thuyt n Hi 14 Vector n v ca hng s c biu th bi: 3 2 1e e e ndsdzdsdydsdx+ + =V th cho nn, o hm nh hng c th biu din qua tch v hng nh sau: fdsdf = == = n.(1.40)trong ,f gi l gradien ca hm v hng f , c nh ngha bi zfyfxfgradf f++= =3 2 1e e e (1.41) cn ton t vector c gi l ton t del. z y x ++=3 2 1e e e (1.42) (nhc li: iel cc vector n v ca h ta ). Cctonttrnycngnhiutonthudngkhcscvndngtrongtrnghpivicc tensor -Cc. Chol hm v hng cnu l vector, cc php tnh ton t vi phn quen thuc u c th vit di dng k hiu ch s nh sau : Gradient ca mt v hng : ;, i ie = Gradient ca mt vector : ;, j i j iu e e u = Laplacian ca mt v hng :ii ,2. = = (1.43) Divergence ca mt vector: i iu,. = u Curl ca mt vector :i j k ijku X e u, = Laplacian ca mt vector :i kk iu e u,2= Nu nhv l cc trng v hng cnu vvl cc trng vector th tn ti mt lot cc ng nht thc nh sau: ( (( ( ) )) ) ( (( ( ) )) ) ( (( ( ) )) )( (( ( ) )) ) ( (( ( ) )) ) ( (( ( ) )) )( (( ( ) )) ) ( (( ( ) )) )( (( ( ) )) ) ( (( ( ) )) )( (( ( ) )) ) ( (( ( ) )) ) ( (( ( ) )) )( (( ( ) )) ) ( (( ( ) )) )( (( ( ) )) ) ( (( ( ) )) ) . . .21; .; 0 .; .; 0; . . .;; . . .; . 2;222 2 2u u u u u uu u uuu u u uu u uu u u = == = = == = = == = = == = = == = = == = + ++ + = == = + ++ + = == = + ++ + + ++ + = == = + ++ + = == = X XX XXXX u X v XX X X v rr (1.44)C th chng minh tnh xc thc ca cc ng nht thc trn y bng cch vn dng cc k hiu ch s theo cc nh ngha (1.43).Phn tip theo trnhbymt s nh l c s dng trongLThuyt n Hi v l kt qu ca cc tnh ton tch phn i vi cc vector v tensor. 1.8.1 nh l GaussL Thuyt n Hi 15 GisSlmtbmtlintctngphn,baoquanhkhnggianV.Nunhulmttrng vector lin tc v c o hm cp 1 lin tc trong V, th dV dSV S=u n u . . (1.45) trong ,n l vector php tuyn ngoi trn mt S. Kt qu ny cng ng cho cc tensor cp bt k, tc: =S Vk k ij kdV dS, ...a n aij...k(1.46) 1.8.2 nh l Stocks Gi s S l mt mt cong h (two-sided surface), c bao bi mt ng cong kn, lin tc tng on C. Nu nhn lin tc cng vi o hm cp 1 ca n trn Sth: ( (( ( ) )) ) dS XS Cn u dr u . . = == = (1.47) trong tch phn ng l dng khi min S nm bn tri hng di chuyn dc theo ng cong C theo hng can - vector n v php tuyn ngoi ca mt S. V mt ln na, kt qu trn cng ng cho cc tensor cp bt k, tc =C Sr s k ij t k ijdS dx n a arst, ... ... (1.48) 1.8.3 nh l Green phng p dng nh l Stockes cho min S phng vi trng vector chn2 1e e u g f + = , ta thu c kt qu ( ). + =|||

\|S Cgdy fdx dxdyyfxg (1.49) Tip tc, chn mt trong hai hm f= 0hoc g = 0 , c kt qu ==C CySxSdS fn dxdyyfdS gn dxdyxg ;(1.50) 1.8.4 nh l gi tr zero Cho k ij...f l trng tensor cp bt k, vi cc thnh phn khng m, xc nh trong min V. Nu nh tch phn ca k ij...f trong min V trit tiu th k ij...fphi trit tiu trong V, tc: = =Vk ij k ijV dV 0 0... ...f f (1.51) 1.9 H ta cong trc giao 1.9.1 Cc cng thc tng qut trong h ta cong trc giao t bi ton v pht trin li gii ca L thuyt n hi trong trng hp min kho st c gii hn bi cc b mt cong, cn thit phi s dng h ta cong. Ta xt trng hp tngqut, trong ,cc ta congca h ta trc giao c k hiu bi 3 2, ,1 cn trong h ta -Cc, vn nh trc y, l 3 2 1, , x x x .Trn hnh H1.6 biu din cc vector n v ca h ta cong trc giao tng qut. Ta gi thit l tn ti php bin i thun nghch ( (( ( ) )) ) ( (( ( ) )) )3 2 1 3 2 1, , ; , , m m m mx x x x x = == = = == = . (11.52)Chiu di phn t trong khng gian c th c biu din trong h ta cong nh sau ( ) ( ) ( ) ( )23 322 221 12 d h d h d h ds + + = , (1.53)trong , 3 2 1, , h h hl cc h s t l. Ni chung, h s t l l cc hm khng m ca v tr. Hy xc nh quan h gia cc vector n v c s trong h ta -cc, 3e , e , e2 1, v cc vector n v c s trong L Thuyt n Hi 16 h ta cong, 3e , e , e2 1) ) ). S dng phng php tng t nh thc hin trong mc 1.5, kt hp vi (1.53) ta c ;1;1;13 3 332 2 221 1 11kkkkkkkkkkkkxh dsdxxh dsdxxh dsdxe e ee e ee e e= == == =))) (1.54)Trn c s ca iu kin trc giao, ij j i e e =) ). , t (1.54) thu c ( )( )( ) .;;3 3232 2221 121 ===k kk kk kx xhx xhx xh (1.55) Quan h (1.54) cho thy, i lngrkrkrxh =1Q(khng tng theo r). (1.56) biu th tensor bin i, cho php xc nh vector n v c s ca h cong theo vector n v c s ca h -Cc. i lng ny ng vai tr tng t nh tensor bin i ijQxc nh bi (1.16) i vi h ta -Cc. Ccthnhphntnhincavectorhaytensortrongtrnghpkhostcngnginlcc thnhphntrongh-Ccccb,tip xcvihtacongtitngimtrongkhnggian.Vth cho nn, mt tensor, a, trong h ta cong tng qut c th c biu din di dng . ...... ... s pqskqjpi k ija Q Q Q a => < (1.57)L Thuyt n Hi 17 trong , s pq...al cc thnh phn trong h ta -Cc xc nh cn > < k ij...a l cc thnh phn trong h ta cong. Lu : n s c mt ca du ch s l dnh cho phn t trong h ta cong.Bn thn mt tensor bt k c th biu din ctrong c hai h ta : .... k k ije ... e e a ...e e e a aj i ij...k k j i) ) )> >> > < >> > < >> > < < ) ;1 1 e e u e e uue eue euu eueu ||

\|+ ||

\|++= |||

\|+=> E (4.54) 2. Trng thi ct thun ty: Trong trng thi ct phng thun ty, cc thnh phn ng sut c xc nh bi((((

=0 0 00 00 0ij (4.55) Cng thc th nng bin dng s l: ( ) ( ) + =+= 1 22122E EW . (4,56) S dng tnh cht th nng dng v kt qu0 > E bn trn, t (4.56) ta c 1 0 1 > > + .(4.57) L Thuyt n Hi 68 3. Trng thi nn thy tnh: Trong trng thi nn u theo mi phng (gi l nn thy tnh), cc thnh phn ng sut cho bi ((((

=pppij0 00 00 0 .Th nng n v trong trng trhi ny c xc nh bi ( ) ( ) 2 1233232122 2 = +=EppEpEW . (4.58) S dng mt ln na tnh cht th nng dng v0 > E , t (4.58) c: 210 2 1 < > . (4.59) Kt hp (4.59) vi (4.57) ta xc nh c gii hn ca h s Poisson: 211 < < . (4.60)

4.5.3 Th nng bin i th tch v Th nng bin i hnh dng. Ta kho st th nng bin i n v th tch (gi tt: th nng th tch) ca vt th ng hng. Nh ch ra trn y, vt th ng hng c th chu c p sut rt cao, m vn khng c bin dng d, do , trong l thuyt do hin i, trong biu thc th nng tng qut ca vt th, ngi ta tch ring phn th nnglin quan n s bin i hnh dng.Vi vt th n hi l tng, c th biu th th nng di dng tng ca th nng bin i th tch v th nng bin i hnh dng (gi tt: th nng hnh dng): ,0 hdW W W + = (4.61) trong,W0vWhd,tngng,lcng(thnng)tiutnchovicthayithtchvchothayi hnh dng. V s bin i th tch lin quan n (hay, c quyt nh bi) ng sut trung bnh , cho nn, khi gn0 ; = = = = = =zx yz xy z y x vo cng thc (4.39), ta c ( ).22 1 320EW=(4.62) Nh cc cng thc (4.39) v (4.62) c th vit cng thc th nng hnh dng ca vt th nh sau ( ) ( ) ( ) ( ) [ ] . 2412 2 2 2 2 20 zx yz xy z y x hdGW W W + + + + + = = (4.63) hay, khi thay( )x x x + + =31, ta c ( ) ( ) ( ) ( ) [ ]. 61212 2 2 2 2 2zx yz xy x z z y y x hdGW + + + + + =(4.64) Nh vy, th nng n v c thvit di dng ( ) ( ) ( ) ( ) ( ) [ ] . 66162 12 2 2 2 2 2 2zx yz xy x z z y y x z y xE EW + + + + + ++ + +=(4.65) Cng thc (4.65) cho php ta xc nhgii hnca s bin i cacchng s n hi,m trn y ch ra. Tht vy, bo m hm W lun dng i hi phi tho mn iu kin. 01, 02 1>+>E E (4.66) L Thuyt n Hi 69 VE>0 , ta c : . 0 1 , 0 2 1 > + > T suy ra, h s Poisson ch c th bin i trong gii hn 211 < < . TM LC CHNG IV nhlutHookexcnhtnhchtvtliucamhnhtnhtontheo,trongiukinbin dng b, cc thnh phn ng sut c quan h tuyn tnh vi cc thnh phn bin dng.Tnh cht c hc ca vt th n hi tuyn tnh c xc nh bi cc hng s h ca vt liu m hnh. Gia cc hsh c cc quan h xc nh Cc hsh c th biu th qua cc hng s k thut. Cc hng s k thut c c qua o c thc nghim tnh cht c hc ca vt liu. Cc hng s k thut bin i trong nhng gii hn xc nh. Vt liu h trc hng c 3 MPH X vung gc nhau ti mi im. Vi vt th h ng hng, bt k mt phng no i qua im kho st cng l MPHX. Th nng bin dng h l cng cc lc ngoi tch ly trong qu trnh lm vt th h bin dng.Th nng lun dng khi vt th c bin dng Th nng bin dng tng cng c phn tch thnh tng ca th nng bin i hnh dang v th nng bin i th tch L Thuyt n Hi 70 Chng V THIT LP BI TON L THUYT N HI 5.1 Cc phng trnh c s tng qut ca l thuyt n hi Trng thi vt l ti im vt cht trn vt th n hi c xc nh bi cc i lng: 6 thnh phn ng sut, 6 thnh phn bin dng v 3 thnh phn chuyn v. Cc i lng ni trn c gi l cc bin c bn ca bi ton l thuyt n hi. Cc bin c bn ti mt im bn trong vt thphi chu s rng buc bi cc quan h c thit lp c t cc chng II, chng III v chng IV, bao gm:h cc phng trnh cn bng:. 0; 0; 0= +++= +++= +++zzyzxzyzy y xyxzxyxxFz y zFz y xFz y x (5.1) cc quan h ng sut-bin dng :;;;zwyvxuzyx=== .21;21;21||

\|+=|||

\|+=|||

\|+=xwzuzvywyuxvzxyzxy(5.2) h phng trnh cc tng thch: ; 2; 2; 2222222222222222x zezexez yeyezey xexeyezx x zyzzyxy yx =+ =+ =+ .;;222|||

\|++= |||

\|++= |||

\|++= yexezex y xexezeyey x zezeyexex z yezxyz xyzyz xyzxyxyzxyzx(5.3) nh lut Hooke tng qut: .;;;;;66 65 64 63 62 6156 55 54 53 52 5146 45 44 43 42 4136 35 34 33 32 3126 25 24 23 22 2116 15 14 13 12 11zx yz xy z y x xzx yz xy z y x yzzx yz xy z y x xyzx yz xy z y x zzx yz xy z y x yzx yz xy z y x xc c c c c cc c c c c cc c c c c cc c c c c cc c c c c cc c c c c c + + + + + =+ + + + + =+ + + + + =+ + + + + =+ + + + + =+ + + + + = (5.4). Trong 6 phng trnh tng thch (5.3) trn y, ch c 3 phng trnh c lp nhau v l cc iu kin cn v cho cc thnh phn chuyn v l n tr v lin tc (cho vt th n lin), khi cc thnh phn bin dng xc nh trc. Cc phng trnh trn y c gi l cc phng trnh c s chung ca l thuyt n hi. V thc cht,cc phng trnh c s chung bao gm 15 phng trnh vi phn v cc quan h i s, c lp nhau, gia cc thnh phn ng sut, cc tp bin dng v cc tpchuyn v. hon tt vic thit lp bi ton cn phi thit lp cc iu kin bin. Cc iu kin bin c xc nh ty theo iu kin vt l, hnh hc ti tng im trn mt bao vt th kho st. Trong khi cc phng trnh c s l nh nhau th cc iu kin L Thuyt n Hi 71 binlikhcnhau,tythucvotngbiton.ccligiicaccbiton,cnbitcchxc nh ng cc iu kin bin. Cc phng trnh c s kt hp vi cc iu kin bin s hnh thnh cc bi tongitrbincbn(gittlbitonbin)calthuytnhi.Chaihnhthcbitonbin thng gp, l: bi ton theo chuyn v v bi ton theo ng sut. V, ni chung, rt kh tm li gii ca bi ton bin nn c kh nhiu cc chin thut gii quyt vn c ra phc v cho mc ch ny.Trc khi tho lun v cc bi ton bin c bn, ta tho lun thm v cc iu kin bin v cch xc nh ng cc iu kin bin. Phn cui chng nu ra cc chin thut thng s dng v gii thiu cc nguyn l tng qut c ng dng rng ri khi gii quyt cc bi ton l thuyt n hi. Cc quan h cp n trong chng ny v v sau ch yu c m t di dng v hng hoc ma trn. 5.2 iu kin bin Nu nh cc gi tr cc bin bn trong vt th ch chu s rng buc ca cc phng trnh c s th cc bin c bn ti cc im trn mt bin cn phi tun th thm cc cc iu kin bin. V mt vt l, cc iu kin bin phn nh tng tc trn bin gia vt th vi mi trng cn v mt ton hc, cc iu kin bin kt hp vi cc phng trnh cn bng to nn cc bi ton bin hon chnh. Thng thng, iu kin bin cho bit vt th c gi cn bng trong khnggian bng cch no vchu tc dng th notccitngbnngoi.Cciukinnycxcnh,vmttonhc,thngquavicch nh (cho bit trc) cc chuyn v hoc lc mt ti cc im trn mt bao vt th. Nu k hiu b mt cho trc chuyn v l Su (gi tt l mt bin chuyn v ) cn b mt cho trc lc mt, (mt bin lc), l ST th bmttonbcavtth,S,bngtngcamtbinchuynvvmtbinlc:S=Su +ST.Nu chuyn v bit trc l bng 0, th mt bin c gi l bin c nh. Nu trn mt bin khng c lc tc dng v chuyn v cng khng bit trc, ta c iu kin bin t do. Nichung,trnmimtbintntimtiukinbin:hoclbinlchoclbinchuynv. Tuy nhin,theo hai hng khc nhau trn cng mt mt bin c th tn ti iu kin binchuyn v theo hng ny v iu kin bin lc cn theo hng kia. Trong v d v tm ch nht chu ko i xng trn hnh H5.2, hai cnh song song vi trc y l cc bin lc, cn hai cnh kia l cc bin khng chu lc, cng vn thuc loi bin lc. Tuy nhin, nu s dng tnh cht i xng ca bi ton, ch cn kho st mt na tm, nh trn hnh, v trc i xng tr thnh mt "bin". Cc im trn trc i xng khng cchuyn v ng theo phng x (l iu kin binchuyn v) v khng chu lc mt (l iu kin bin khng chu lc)theophngy,do,trcixngvalbinlcvalbinchuynv.Ttnhinccyutchuyn v v lc ca loi iu kin bin ny phi tng ng vi nhau. Thit lp cc iu kin bin l bc rt quan trng trong qu trnh lp bi ton v gii quyt bi ton l thuyt n hi. Vic xc nh sai cc iu kin bin c th bin bi ton ny thnh bi ton khc hoc lm cho li gii sai lch v nhiu khi l khng gii c. K nng thit lp ng cc iu kin bin cn c rn luyn k thng qua thc hnh gii cc bi tp. L Thuyt n Hi 72 Vic xc lp iu kin bin chuyn v c th thc hin mt cch trc tip v, ni chung, l n gin hn so vi iu kin bin lc. Vic p t iu kin bin lc ti mt im trn bin c ngha l p t iu kin cn bng ca t din phn t ti im ni trn (xem 2.2). Da trn kt qu thu c t chng II, iu kin cn bng ca t din phn t c biu din (di dng v hng) bi phng trnh :,;;z z y yz x xznxz zy y y x xynyz zx y yx x xnxn n n Tn n n Tn n n T + + =+ + =+ + =(5.5) trong , z y xn n n , , l cc cosine ch phng php tuyn ngoi ca im trn bin trong h ta kho st;

nznynxT T T , ,l cc thnh phn ca vector lc mt n v ti im kho st;

zx yz xy z y x , , , , ,l cc thnh phn ng sut ti im kho st. Cng thc (5.5) cho thy, tha mn iu kin cn bng trn bin, lc mt n v phi c quan h xc nh vi cc thnh phn ng sut ti cc im trn bin. Vic xc lp iu kin bin lc cho mt bi toncthchnhlvicxcnhcccosinechphng z y xn n n , , cngviccthnhphnlcmt nznynxT T T , , ticcimtrnbintxclpcccquanhdng(5.5),mccthnhphnng sut ti im trn bin phi tha mn.H5.3 l cc v d n gin, c th v cc iu kin bin. L Thuyt n Hi 73 5.3 Cc bi ton bin Giimtbitonlthuytnhicthlvicxcnhphnbngsut,bindngv\hocchuyn v tha mn cc phng trnh c s (5.1) cng cc iu kin bin c th ca bi ton. Nh trn y cp n, cc phng trnh c s kt hp vi mt loi iu kin bin xc nh s hnh thnh mt bi ton gi tr bin tng ng. Ta gi tt bi ton gi tr bin l bi ton bin. C 3 bi ton bin bin c bn, c xc nh nh sau: 1. Bi ton bin lc: Xc nh s phn b ca cc ng sut, chuyn v v bin dng ti cc im trn vt th n hi cn bng khi cho trc cc lc khi v s phn b ca cc lc tc dng ln b mt c m t bi: ( ) ( )Si jSinjx f x T = (5.6) trong , Six biu th im trn mt bin (chu lc) cn( )Si jx fbiu th gi tr ca lc mt ti im trn bin ni trn. 2. Bitonbinchuynv:Xcnhsphnbcaccngsut,chuynvvbindngticc im trn vt thn hi cn bng khi cho trc cc lc khi v s phn b ca chuyn v trn b mt c m tbi: ( ) ( )Si jSi jx g x u =(5.7) trong , Six biu th im trn mt bin (chu lc) cn( )Si jx gbiu th gi tr cachuyn v ti im trn bin ni trn. 3.Bi ton bin hn hp: Xc nh s phn b ca cc ng sut, chuyn v v bin dng ti cc im trn vt th n hi cn bng khi cho trc cc lc khi v s phn b ca cc lc tc dng ln b mt chu lc STc m t bi (5.6) v s phn b ca chuyn v trn b mt Su c m t bi (5.7). 5.4 Tnh duy nht nghim ca bi ton bin Taxtbitonviiukinbinhnhp,tctntingthicmtbincv,Sulnmtbin chulcST.Tnhchtduynhtnghimbitongatrbincchngminhbngphngphpphn chng.Gistntingthinghimphnbitnhauhai: () ( ) (){ }1 1 1i ij iju e v ( ) ( ) ( ){ }2 2 2i ij iju e ,cho L Thuyt n Hi 74 cng mt bi ton vi lc mt, lc khi cng nh cc iu kin bin ht nh nhau. Hiu ca hai nghim trn c biu din bi: () ( )() ( )( ) ( ).;'2 12 12 1i i iij ij ijij ij iju u ue e e = = = (5.8) V hai nghim (1) v (2) tha mn iu kin cn bng vi cng mt lc khi nh nhau nn hiu ca chng s phi tha mn iu kin cn bng khng lc khi, tc: 0,=j ij(5.9) Cng nh vy, hiu hai nghim s phi tha mn iu kin bin 0, tc iu kin khng c lc mt trn mt chu lc ST cng nh khng c chuyn v trn mt bin cv, hay: 0 = =j ijnin T trn ST (5.10) v c iu kin khng chuyn v trn bin chuyn v Su :0 =iutrn Su(5.11) Ta c th bin i biu thc th nng vt th ng trng thi hiu ca hai nghim nh sau: ( )( ) = = = = =Vi j ijSi j ijVi j ijV Vi ij j i ijVij j i ijVij ijVdV u dS u ndV u dV u dV udV u dV e WdV,, ,,2 (5.12) trong,( )i j j i iju u, ,21 = ltensorxoay[xemcngthc(3.8)mc3.1].cktqutrn,vn dng nh l divergent chuyn tch phn trn th tch v tch phn mt tng ng. Ngoi ra, cn nh rng ngoi tch ca mt tensor i xng vi mt tensor phn xng lun bng 0, nn c0 =ij ij . Vi lu rng mt bin bao gm haiphn: bin chu lc v bin cv, tc u TS S S + = , trn c s (5.9)(5.12)ta c 0 2 =VWdV .(5.13) V th nng W l hm xc nh dng nn t (5.13) ta c0 = Wtrn ton min V v t ,suy ra rngcc thnh phn ng sut v cc thnh phn bindng u trit tiu, tc0 = =ij ije , trn ton min.V 0 =ijetrn ton min nn vt kho st khng b bin dng m ch c th chuyn v nh vt rn. Nhng v trn bin chuyn v phi tha mn iu kin0 =iunn suy ra rng, chuyn v cng trit tiu, tc,0 =iutrn ton min V. Ni cch khc, hai nghim cho trc phi ng nht, hay, nghim bi ton l duy nht. Lu rng, nu nh khng tn ti mt bin chuyn v th () 1iuv ( ) 2iusai khc nhau bi chuyn v ca vt rn. 5.4 Dng chuyn v v dng ng sut ca cc phng trnh c s chung Vic tm li gii bt k mt trong 3 bi ton bin trn y u ht sc phc tp. phc v cho vic tm cc li gii ny, ta tip tc rt gn, bin i cc phng trnh c s (5.1)(5.4) v dng thch hp vi tng bi ton bin nu trn. Cc phng trnh c s (5.1)(5.4) cha 15 Nn s c bn, bao gm cc thnh phn ng sut, cc thnh phn bin dng v cc thnh phn chuyn v. Bng cch s dng nh lut Hooke biudinccthnhphnngsutvdngquanhngsut-bindngloitrccthnhphnbin dng, ta thu c h 3 phng trnh ch vi 3 Nn s l cc thnh phn chuyn v. l dngchuyn v ca cc phng trnh c s. Dng chuyn v ca cc phng trnh c s chung c s dng rt rng ri v c bit thch hp cho cc bi ton bin chuyn v, khi m cc iu kin bin ch chachuyn v. Mt L Thuyt n Hi 75 cchkhc,cngdngccquanhngsut-bindngnhngloitrccthnhphnchuynvri dngnhlutHookebiudinccthnhphnbindngtheoccthnhphnngsut,ccphng trnh c s s tr thnh cc phng trnh ch cha cc Nn s c bn l cc thnh phn ng sut.Ta c dng ngsutcaccphngtrnhcs.Tronghaidngcaphngtrnhcstrny,dngngsutt thngdnghnvchtrathchhpvibitonbinlc,trong,cciukinbinchchacc thnh phn ng sut. Di y ta tin hnh bin i cc ng sut tng qut v hai dng ni trn, ch hn ch trong trng hp vt th n hi n hi ng hng.5.4.1 Dng chuyn v ca cc phng trnh c s chung a cc ng sut v dng chuyn v, trc tin, s dng quan h ng sut- bin dng biu din cc thnh phn bin dng trong quan h Hooke (cho vt th n hi ng hng) di dng (4.43) . ; ;; 2; 2; 2||

\|+=|||

\|+=|||

\|+=+|||

\|++=+|||

\|++=+|||

\|++=zuxwGywzvGxvyuGzwGzwyvxuyvGzwyvxuxuGzwyvxuzx yz xyzyx (5.14) Tip n, thay cc kt qu ny vo cc phng trnh cn bng (5.1), biu din tt c cc thnh phn ng suttheoccthnhphnchuynv,tathuch3phngtrnhcnbngvittheoccthnhphn chuyn v. H phng trnh kt qu c th vit gn di dng k hiu ch s nh sau: ( ) 0, ,= + + +i ki k kk iF u G Gu . (5.15) Phng trnh(5.15) c bit n di tn gi phng trnh Lame hoc phng trnh Navier. Hay, nu s dng ton t Del v Laplacian i vi vector, c th vit phng trnh (5.15) di dng ( ) ( ) 0 .2= + + + F u u G G , (5.16) trong,( ) w v u , , u u = v( )z y xF F F , , F F = .Phngtrnh(5.16)cthvitdidngvhngnh sau: ( )( )( ) . 0; 0; 0222= +|||

\|+++ + = +|||

\|+++ + = +|||

\|+++ + zyxFzwyvxuzG w GFzwyvxuyG v GFzwyvxuxG u G (5.17) Ton t Laplacian c xc nh bi: 2222222z y x ++= . (xem 1.8) PhngtrnhNavierldngphhpvibitonbinchuynv.Thccht,ylh3phng trnh cn bng vi 3 Nn l 3 thnh phn chuyn v. Cc phng trnh c s cn li u c s dngtrong qu trnh bin i hnh thnh c kt qu trn. Tuy dng thc ca phng trnh Navier l kh n gin nhng vic tm nghim ca bi ton bin chuyn v trn s s ca phng trnh ny vn khng h n gin. gii bi ton loi ny, ngi ta phi pht trin thm cc k thut ton hc b sung. Mt trong cc phng php thng dng nht l s dng hm th chuyn v. L Thuyt n Hi 76 5.4.2 Dng ng sut ca cc phng trnh c s chung c c cc ng sut di dng ch cha Nn l cc thnh phn ng sut, cng xut pht t quan h Hooke cho vt th n hi ng hng thay th cc thnh phn bin dng trong phng trnh tng thch bi cc thnh phn ng sut ri kt hp vi cc phng trnh cn bng s thu c cc phng trnh c th vit gn theo cc k hiu ch s nh sau: i j j i k k ij ij kk kk ijF F F, , , , ,1 11 =++ .(5.18) Phng trnh (5.18) c tn l phng trnh tng thch Beltrami-Michell. Trong trng hp khng c lc khi, cc quan h trn c th biu din di dng v hng sau y: ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) . 0 1; 0 1; 0 1; 0 1; 0 1; 0 1222222222222222= + + + += + + + += + + + += + ++ += + ++ += + ++ +z y x zxz y x yzz y x xyz y x zz y x yz y x xx zz yy xzyx (5.19) Lu rng, trong 6 phng trnh trn ch c 3 quan h c lp nhau (xem 2.). V th cho nn, c c s phng trnh xc nh 6 thnh phn ng sut, cn kt hp (5.19) vi cc phng trnh 3 cn bng (5.1). Phng trnh Beltrami-Michell, vi cc Nn s l cc thnh phn ng sut, ph hp vi bi ton bin lc,trongcciukinbinchaccthnhphnngsuttrnbin.Saukhigiibitonbinlc bng phng trnh Beltami-Michell, xc nh c trc tip cc thnh phn ng sut. Vi cc thnh phn ng sut tm c, s dng nh lut Hooke (4.40), d dng tm c cc thnh phn bin dng, trong khi cc thnh phn chuyn v s l kt qu ca vic tch phn phng trnh (5.2).TuyrngphngtrnhBeltrami-Michellnginhnnhiusovidngbanucaccphng trnh c s chung ca l thuyt n hi, nhng vic tm li gii ca bi ton bin lc vn cn ht sc kh khn. Mt cng c ton hc b tr cho cng vic ny l hm ng sut. Khi nim ny to nn mt phng thcbiuthccthnhphnngsutmtheoccphngtrnhcnbngtngcthamn. Trong bi ton 2D, vi khi nim ny, cc thnh phn ng sut c xc nh qua duy nht mt hm s ca hai bin khng gian v bi ton ban u dn n ch mt phng trnh o hm ring (phng trnh iuhakp)cahmngsut.Nhqutrnhnginhanymtacthgiibitonbngcc phng php gii tch. Vn ny s c tho lun tip sau ny khi gii quyt bi ton phng. Vi bi ton bin lc, phng trnh dng ng sut thng c s dng v ta ni bi ton l thuyt n hi c thit lp di dng ng sut, cn vi bi ton binchuyn v, phng trnh dngchuyn v thng c s dng, v khi ta ni l bi ton l thuyt n hi c thit lp theochuyn v. 5.5 Hai nguyn l thng dng

Haitrongscccngchudnggiiquytccbitonnhitrongkthutlnguynl cng tc dng v nguyn l Saint-Venant. L Thuyt n Hi 77 5.5.1 Nguyn l cng tc dng Nguynlcngtcdnglkthutkhphbin,pdngchoccbitoncdnbicc phng trnh tuyn tnh. Vi gi thit bin dng b, cc ng sutchuyn v v ng sut ng sut u l tuyn tnh. Hn na, cc iu kin bin(5.6) v (5.7) thng cng l tuyn tnh. V nh vy, vi tt c cc phng trnh dn u l tuyn tnh, c th p dng nguyn l cng tc dng, c chng minh v pht biu bi Chou PC v Pagano NJ,1967, nh sau [3]:Nguyn l cng tc dng: Vi mt bi ton bin cho trc, nu nh trng thi () () (){ }1 1 1, ,i ij iju e l ligiicaccphngtrnhnhics,ngvilckhi () 1iF vlcmt () 1iT vtrngthi ( ) ( ) ( ){ }2 2 2, ,i ij iju e l li gii ca cc phng trnh n hi c s, ng vi lc khi ( ) 2iFv lc mt ( ) 2iTth trng thi () ( ) ( ) ( ) ( ) ( ){ }2 2 2 2 2 1, ,i i ij ij ij iju u e e + + + l li gii ca cc phng trnh n hi c s, ng vi lc khi ( ) ( ) 2 1i iF F +v lc mt () ( ) 2 1i iT T +Hnh (H5.4) minh ha cho vic p dng nguyn l trn vo bi ton 2D khi khng c lc khi. 5.5.2 Nguyn l Saint-Venant Mtnguynlkhc,cxydngtrncsthcnghim,nhngcnghudngvphbin khng km nguyn l cng tc dng l nguyn l Saint-Venant. Nguyn l c th c pht biu nh sau: Nguyn l Saint-Venant: Hai h lc khc nhau nhng tng ng tnh hc vi nhau, gy ra cc trng thai ng sut, bin dng v chuyn v gn ging nhau ti vng xa v tr tc dng lc. Trong pht biu ca nguyn l Saint-Venant cha cc cm t v t mang tnh nh tnh nh: gn ging, xa m khng cp n cc nh lng phn bit no i vi hai trng thi trn.Nghin cu nh lng v vn ny l ti c rt nhiu nh khoa hc thc hin [4]. 5.6 Phng php thun, phng php ngc v phng php na ngc gii cc bi ton l thuyt n hi phi vn dng nhiu chin thut (phng php) khc nhau. y ta ch nu ra 3 phng php gii tng qut l phng php thun, phng php ngc vphng php na ngc. 5.6.1 Phng php thun Phng php thun tm li gii bi ton l thuyt n hi thng qua tch phn trc tip h phng trnh c s (5.1)(5.4) hoc cc dngchuyn v v dng ng sut tng ng vi h ny, vi vic tha mn mt cch chnh xc cc iu kin bin. Phng php ny gp phi nhng kh khn ton hc nghimtrng v chi v th ch hn ch cho cc bi ton vi cc dng hnh hc n gin. L Thuyt n Hi 78 V d 5.1: Phng php tch phn trc tip. Ko dm lngtr di tc dng ca ti trng bn thn minh ha cho phng php tch phn trc tip, ta xt mt v d n gin, kho st trng hp mt thanh lng tr b ko di tc dng ca trng lng bn thn (H5.5).

Trong trng hp kho st, lc khi lg F F Fz y x = = = ; 0 , trong -l t trng cn g l gia tc trng trng. Trn c c gi thit rng mi mt tit din u chu lc ca phn dm pha di, ko phn b u, c th xc nh cc thnh phn ng sut nh sau:( ) zz z zx yz xy y x = = = = = = ; 0(5.20) Khi , phng trnh cn bng s lg Fzzz= = (5.21) Vi iu kin bin khng chu lc ti tit din di cng0 =z , bng vic tch phn trc tip phng trnh trn, ta c kt qu( ) gz zz = .(5.22) xc nh cc thnh phn bin dng, cn s dng nh lut Hooke, vic ny dn n kt qu sau: 0 ; ; = = = = = =zx yz xy y x ze e eEgze eEgze (5.23) Tip n, cc thnh phnchuyn v c xc nh nhtch phn quan hchuyn v- bin dng (5.2), ng vi iu kin bin khng chuyn vv iu kin khng xoay ti im A (x = y = 0;z = l ), c kt qu cui cng nh sau: ( ) [ ]2 2 2 22; ; l y x v zEgwEgyzvEgxzu + + = = = .(5.24) 5.6.2 Phng php ngc Theochinthutngc,ccthnhphnchuynvhocngsutcthcchntrc.Vnd cnlilxcnhxemccthnhphnchuynvhocngsuttrnlnghimcabitonno.Thc cht l tm dng hnh vt th, cc iu kin bin cng nh lc khi m cc thnh phn chuyn v hoc ng sut trn nghim ng. V nh vy nghim ca mt bi ton c ch ra trc cn bn thn bi ton c xc nh sau. Tr ngi chnh ca vic p dng k thut ny kh xc lp nghim ca cc bi ton c th m thc t cng ngh t ra. V d 5.2: Phng php ngc: Un thun ty dmXt trng hp ca bi ton n hi khng lc khi, vi cc thnh phn ng sut cho bi: L Thuyt n Hi 79 0 ; = = = = = =zx yz xy z y xAy , (5.25) trong , A l hng s. Vi trng ng sut tuyn tnh nh trn, cc phng trnh tng thch v phng trnh cn bng t tha mn v trng ng sut ny chnh l nghim ca mt bi ton n hi no . Vn t ra l bi ton no nhn trng ng sut trn lm nghim? Cch thc chung tr li cu hi trn l kho nghim cc dng hnh khc nhau cng cc iu kin bin thc t khc nhau, tm hiu xem trng hp no th xut hin trng ng sut cho trc trn y. V th, ta kho st khng gian 2D dng hnh ch nht nh trn H5.6 tha mn trng ng sut (5.25), h lc trn cc mt bin bn tri v trn mt bin bn phi cn phi c vector chnh bng 0 v ng sut php phi bin i tuyn tnh nh trn H5.6. D thy rng cc iu kin bin ny l iu kin ca bi ton ung thun ty dm, khi m hp lc ti hai bin u dm khng c vector chnh m ch c hai momen chnh . y chnh l iu kin un thun ty i vi vt th dng dm lng tr. 5.6.3 Phng php na ngcMt trong cc phng php, cho php thu c li gii ca bi ton L thuyt n hi trong mt s trng hp ring, l phng php na ngc St. Venan. Ni dung ca phng php ny l:mt phn ca nghim bi ton c gn trc, trong khi phn cn litm c bng cch tch phn phng trnh vi phn, l phng trnh c n gin ho nhiu nh thay vo phn nghim gn trc ni trn. nh l v tnh duy nht nghim chnh l c s cho phng php ny. Nh p dng nguyn l Saint-Venant, phng php na ngc c kh nng thay nhiu bi ton phc tp bng cc bi ton vi iu kin bin n tng ng tnh hoc, gin hn. V d 5.3: Phng php na ngc: Xon dm lng tr Bi ton xon dm lng tr s c trnh by chi tit trong phn di y minh ha cho phng php na ngc. Trong v d ny, ta ch trch dn cc kt qu chnh m khng i vo chi tit. Bt u vi trngchuyn v c cho trc nh sau: ( ) y x w w xz v yz u , ; ; = = = , (5.26) trong , l hng s. Mt phng tit din dm song song vi mt phng ta x-y cn trc z song song vi trc dm. S dng quan h ng sut- bin dng v nh lut Hooke, c cc thnh phn ng sut tng ng vi cc thnh phn chuyn v (5.26): .;; 0|||

\|=||

\|== = =xywGyxwGyzzxz y x (5.27) L Thuyt n Hi 80 Thay kt qu trn vo phng trnh cn bng, thu c phng trnh thuc dng Navier cho trng hp kho st: . 02222=+ywxw (5.28) y chnh l phng trnh gip ta xc nh phn cn li ca nghim bi ton ban u. iu kinchuyn v ti hai mt bin cng phi c xc nh thun theo qui lut (5.26). Cn nhn ra rng, iu kin bin trn, d c b p t min cng cng khng lm sai lch kt qu i bao nhiu, c bit l nhng vng xa hai bin, tc vng gia dm. Nh vy l, nh n nh ngay t u cc chuyn v di dng (5.26),bi ton c n gin ha rt nhiu. Vic gii phng trnh Laplace (5.28) tha mn iu kin bin t do mt bn ca dm l khng my kh khn. 5.7 Cc phng php gii phng trnh bi ton n hi C rt nhiu cc phng phphng php ton hc khc nhau phc v cho vic gii cc h phng trnh ca bi ton n hi. Mt s trong cho php thu c cc li gii chnh xc, s cn li cho cc li gii gn ng. Nhm th 3, ngy cng pht trin mnh m, l cc phng php s. Di y l s lc v mt s trong cc phng php ny. 5.7.1 Cc phng php gii tchCc phng php thuc loi gii tch bao gm: 1. Phng php chui ly tha: Gii bi ton 2D, thng qua vic biu din hm ng sut( ) y x, di dng chui ly tha( ) =n mn mmny x C y x, Cc h s mnC c xc nh theo cc iu kin bin.2. Phng php Fourier: Nhiu bi ton n hi ln, trong , cc phng trnh dn l phng trnh o hm ring, c gii nh k thut tch bin s, cng tc dng v chui Fourier hoc tch phn Fourier. Trong nhiu ni dung ca ti liu ny, phng php Fourier c s dng. 3. Phng php bin i tch phn: Lcngctonhchudngtrongvicgiiccphngtrnhohmring,datrnkthut bin hnh tch phn. Nh php bin i tch phn tuyn tnh, mt s dng vi phn c n gin ha hoc b loi tr. iu ny cho php tm c li gii n gii cho cc bin s bin i. Sau khi thc hin tip php bin i ngc, tm c cc bin s ban u.Cc php bin i thng dng bao gm: php bin i Laplace, Php bin i Fourier v Php bin i Hankel.4. Phng php bin s phc: Nhiu bi ton n hi nh bi ton phng, bi ton xon v mt s bi ton bi ton nhit n hi c th thit lp c di dng s dng cc hm bin phc. y l mt cng c mnh v hu dng ngay c trong nhng bi ton c coi l ht sc kh khn, nu theo cc k thut khc. Cc nh khoa hc Nga l nhng ngi c nhiu ng gp trong lnh vc ny [5]. 5.7.2 Cc phng php xp x Nhiu phng php xp x c pht trin, phc v cho vic gii cc bi ton n hi, c l v tnh phc tp ca vic gii cc bi ton ny khi dng cc phng php chnh xc. a s cc phng php xp x l cc phngphp bin phn , ni ring, v phng php nng lng, ni chung. Phng php ny hnh thnh nh vic tm ra mi lin h gia vic gii cc phng trnh dn ca bi ton n hi vi vic cc tr ha mt mt phim hm tch phn. Mt trong nhng i din ca phng php hm xp x l phng php Ritz. Trong phng php Ritz, mt nhm cc hm xp x c dng gii gn ng bi ton n hi, thng qua vic tm gi tr dng ca mt tch phn nng lng c th no . Cc hm xp x L Thuyt n Hi 81 c chn sao cho iu kin bin c tha mn cn tch phn nng lng ch t xp x cc tr. S hm xp x chn cng nhiu, chnh xc ca li gii, ni chung, cng c ci thin. Do kh khn trong vic chn cc hm xp x tha mn cc iu kin bin trong cc bi ton c dng hnh hc phc tp nn hin nay, phng php bin phn hm xp xt c s dng. Tuy vy, Cc phng php bin phn ni chung ng vai tr v cng quan trng trong vic p dng vo phng php phn t hu hn, l mt trong nhng k thut tnh ton mnh nht hin nay, c bit l trong vic gii cc bi ton n h 5.7.3 Cc phng php s

Lch s pht trin ca phng php s khng lu i bng cc phng php khc nhng, ngc li, cng vi s pht trin mnh m ca cng ngh thng tin, vai tr ca cc phng php nytrong thc t hin nay khng phng php no so snh c. Cc phng php s c bit hu hiu trong cc bi ton n hi vi dng hnh hc phc tp. Chnh cc phng php s l c s cho vic lp trnh ca cc chng trnh tnh ton thng dng hin nay. Ta ch im qua mt cch ngn gn mt s phng php quan trng nht hin nay. 1. Phng php sai phn hu hn (FDM): Phng php sai phn hu hn da trn vic thay th cc o hm trong cc phng trnh dn bng cc bc nhy tng i ca cc hm ti cc im li ri rc ha khng gian kho st. Kt qu ca vic thay th ny dn n h phng trnh i s theo cc gi tr ti cc im li ca hm cn tm. Phng php ny c thit lp t hn mt th k trc y. y cng l mt cng c c ng dng gii cc bi ton n hi [6]. Phng php sai phn hu hn thng gp phi kh khn trong vic gii cc bi ton n hi vi dng hnh hc phc tp, tuy k thut bin i h ta khc phc c mt phn kh khn ny.2. Phng php phn t hu hn (FEM): L mt phng php s thng dng nht hin nay, phng php phn t hu hn da trn tng thay th khng gian kho st lin tc bng mt tp hp hu hn cc khng gian con, gi l cc phn t c kch thc nh nhng hu hn. Gi tr cc hm cn tm ti cc im nt ca cc phn t v cc hm dng, xc nh trong phm vi ca tng phn t, c dng thay th xp x hm cn tm. Bng k thut thay th ny, phim hm nng lng tr thnh hm nhiu bin, l gi tr ca hm Nn ti cc im nt. p dng iu kin cc tr ca hm nhiu bin, tng ng vi gi tr dng ca phim hm, tm c cc gi tr trn bin ca hm Nn. Kt qu , vi bi ton n hi tuyn tnh, dn n mt h phng trnh i s tuyn tnh m vic gii khng cn l vn ng quan tm. Bn thn cc hm Nn tm c qua qu trnh thay ngc da vo cc gi tr ti nt ca hm Nn c c nh gii h phng trnh i s tuyn tnh v cc hm dng ni trn. 3.Phng php phn t bin (BEM)Phng php phn t bin da trn vic thay th cc phng trnh dn ca l thuyt n hi bng mt phng trnh tch phn vi Nn s xc nh trn mt bin ca min kho st. Phng trnh tch phn bin c th gii trn c s ca phng php phn t hu hn, theo , mt bin cri rc ha ra thnh mt tp hp cc phn t hu hn ri p dng khi nim xp x ni suy biu din cc hm cn tm. Qu trnh thc hin li dn n mt h phng trnh i s tuyn tnh cho php gii ra cc gi tri trn bin ca hm cm tm. Bng khi nim tch phn bin, c th gii bi ton n hi ban u vi s Nn t hn nhiu so vi vic gii bi ton ban u theo phng php phn t hu hn. Nh trnh c vic ri rc ha bn trong vt th, phng php phn t bin t ra c nhiu u th so vi phng php phn t hu hn trong cc bi ton lin quan n cc khng gian ln, ti cc min v hn hoc trong cc trng hp ch cn cc thng tin trn bin [7]. L Thuyt n Hi 82 Chng VI BI TON PHNG Nh c nhc n, cho ti nay vn cha tm c li gii trc tip ca bi ton n hi cho trng hp tng qut, v th cho nn, li gii trong cc trng hp ring c mt gi tr ht sc to ln. Cc li gii trongcctrnghpnyccnhhnchbt,bngmtcchno,tnhtngqutcabiton c t ra. Vic bin i gn ng bi ton ba chiu (3D) v bi ton hai chiu (2D), dn n vic hnh thnh cc bi ton phng, l mt v d. y l mt loi bi ton m li gii ca n c ng dng thc t rng ri. 6.1 Thit lp bi ton phng BitonphngcaLthuytnhicthphnthnhhainhm:ccbitonvbindng phngv cc bi ton v ng sut phng. Trc khi tin hnh gii bi ton, ta hy xc nh cc phng trnh c bn cng vi cc cng thc thit yu ca hai bi ton ni trn, cho vt th trc hng.6.1.1 Trng thi bin dng phng. Khinimbindngphngdngchmttrngthicavtth,mtheo,mttrongcc chuyn v bng 0 cn hai chuyn v cn li khng ph thuc vo to tng ng vi chuyn v bng 0 ni trn. Trc tng ng vi thnh phn cv bng 0, gi s, l z. Gi thit thm rng, mt phng x-y l mt phng n hi i xng. nh ngha ca trng thi bin dng phng c th c biu din nh sau: ( ) ( ) y x v v y x u u w , ; , ; 0 = = = . (6.1) Trn c s ca quan hbin dng - chuyn v(5.2) v nh lut Hooke tng qut (4.31), t (6.1) c th suy ra: ( ) ( ) ( ) . 0 ; , ; , ; , = = = = = =zx yz z xy xy y y x xy x y x y x .(6.2) Trng thi bin dng phng xy ra trong vt th hnh lng tr di, chu ti trng tc dng vung gc vi trcca lng tr v khng i dc theo trc ny (H6.1). C th nhn thy rng cc tit in ngang R ca L Thuyt n Hi 83 hnh lng tr trn chuyn v ging ht nhau v nh vy, bi ton 3D c th a v 2D, xc lp trong min R (mt phng x-y). Da trn cc iu kin (6.1) v (6.2), c th thu c cc phng trnh c s v cc cng thc ch yu ca bi ton n hi nh sau: 1. Phng trnh cn bng: Phng trnh cn bng (5.1) trong trng hp kho st c th vit:; 0 ; 0 =+=+y x y xy xy xyx (6.3) (Phng trnh th 3 dn n s ng nht 0 = 0 gia hai v; Lc khi c b qua: 0 = = =z y xF F F ) 2. Quan h bin dng-cv (5.2) s c dng: . 0 ;; 0 ; ;= =|||

\|+====zx yz xyz y xyuxvyvxu (6.4) 3. Phng trnh tng thch:T(5.3) v (6.2), c: .22222y x x yxy yx =+ (6.5) 4. nh lut Hooke: a. Cho vt th trc hng: c c cng thc ca nh lut Hooke cho vt th trc hng trong trng hp kho st, c th xut pht t cng thc (4.31), c kt qu:( )( ). 0 ;1; 1 1; 1 1= = = =|||

\| =|||

\| =zx yz z xyxyxyxyxz zyxyxxyz zyyyyyxyZ zxyxyyxz zxxxxGE EE E (6.6) b. Cho vt th ng hng: Trng hp vt th ng hng, cng tin hnh nh trn, nhng t cng thc (4.40): . 0 ;1;11;1122= = = =||

\|=||

\|=zx yz z xy xyy x xy x xGEE (6.7) Ngoi ra, t cng thc (4.41) ta c quan h o ca (6.7): L Thuyt n Hi 84 ( ) ( ) . 0 ;2 12; 0 ;2 112 12;212 ;2 1 2 112= + = +== ||

\|+=||

\|= ||

\|+=zx y x y x zyz y x yxy xy y x xGGG G (6.8) 5. Qui lut bin i cc thnh phn ng sut v cc thnh phn bin dng khi xoay h to :Trn c s cc cng thc (2.21) v (3.22) (6.2) c cc cng thc bin i thnh phn ng sut v thnh phn bin dng s nh sau (H6.2): (vi cos, sin, sin, cos2 2 1 1= = = = m l m l ) ( ) ; 2 cos 2 sin21; 2 sin cos sin; 2 sin sin cos' '2 2'2 2' xy x y y xxy y x yxy y x x+ = + =+ + =(6.9) ( ) . 2 cos 2 sin21; 2 sin21cos sin; 2 sin21sin cos' '2 2'2 2' xy x y y xxy y x yxy y x x+ = + =+ + = (6.10) 6. Cng thc th nng n v cho vt liu ng hng:( ) .12212 2 22((

++ +=y x xy y xEW (6.11) 7. Cc bin dng chnh : ( ) , 0;21232 22 , 1)`=+ += xy y xy x (6.12) (trc Oz trng vi mt trong cc trc chnh ca bin dng). 6.1.2 Trng thi ng sut phng. Trnghpthhaicabitonnhicrtgnv2Dlbitonvtrngthingsutphng. Trng thi ng sut c gi l phng khi ng sut tc dng trn mt, vung gc vi mt trong cc trc to , bng 0, ng thi, cc thnh phn ng sut cn li khng ph thuc vo to ng vi trc ny. Cng gi thit rng, mt phng x-y l mt phng n hi i xng.Trn thc t, trng thi ng sut phng xy ra trong cc tm mng (chiu dy 2h), chu tc dng ca cc lc t vo vnh tm, theo phng song song vi cc mt khng chu lc.Ta chn mt phng khng chu lc (mt y) lm mt ta x-y cn trc z tt nhin l vung gc vi mt ta ny. T nh ngha ca bi ton, ta c:( ) ( ) ( ). 0 ; , ; , ; , = = = = = =zx yz z xy xy y y x xy x y x y x (6.13)Ngoi ra cng c th kt lun rng cc thnh phn chuyn vkhc 0 cng khng ph thuc vo to z. tha mn iu kin ccthnh phn ng sut v thnh phn chuyn vkhng ph thuc vo ta z cn phi trit tiu cc thnh phn lc khi v lc mt theo phng z. Cng c th cho php cc lc mt v lc khi ny khc 0 khi chng phn b i xng qua mt phng chia i chiu dy vt th kho st. Trng L Thuyt n Hi 85 hp ny, vi gi thit chiu dy, 2h, b, c th s dng gi tr trung bnh (bng 0) lm c s xc nh gn ng. 1. Phng trnh cn bng: S dng (6.14) v tnh cht khng ph thuc ca cc thnh phn ng sut cn li vo to z, t phng trnh cn bng (5.1) ta c kt qu: ; 0 ; 0 =+=+y x y xy xy xyx (6.3) y cng chnh l phng trnh cn bng (6.3) ca bi ton bin dng phng (vi X = Y = Z = 0). 2. nh lut Hooke: a. Trn c s ca nh lut Hooke cho vt liu trc hng (4.31) v cc quan h (4.24), ta c: ( )( ). 0 ;1;;1;1xy= = =|||

\|+ = = =zx yz xyxyyyyzxxxzzx yx yyyy xy xxxGE EEE (6.14) b. Trn c s ca nh lut Hooke cho vt liu ng hng (4.40) v cc quan h (4.27), ta c:( )( )( ) ( ). 0 ;1;1;1;1= = =+ = + = = =yz zx xy xyy x y x zx y yy x xGEEE (6.15) 3. Quan h bin dng - chuyn v :Cng vi c s trn y, c kt qu ca quan hbin dng - chuyn vtrong trng hp kho st: . 021; 021;21; ; ;= ||

\|+==|||

\|+=|||

\|+====xwzuywzvxvyuzwyvxuzxyzxyz y x (6.4) Quanh(6.14)v(6.4)trnykhnghontongingviccphngtrnh(6.6),(6.7)v(6.4)ca trnghptrngthibindngphng.Khcbitlscmt(khc0)cathnhphnbindng z L Thuyt n Hi 86 trong bi ton trng thi ng sut phng v cho bi ton l 2D, cn phi x l gn ng nh trnh by trong phn di y. 4. Phng trnh tng thch: S c mt khng mong mun ca bin dng z trong bi ton v trng thi ng sut phng lm cho cc b mt khng chu lc b vnh cht t. Tuy nhin, vi cc tm mmg, vnh ni trn l khng ng k v c thbqua.Vthchonnkhilygnngz =0,trncsphngtrnhtngthch(5.3),tac phngtrnhtngthchtrongtrngthingsutphngtrngviktqunhtrongtrnghptrng thi bin dng phng, l phng trnh: .22222y x x yxy yx =+ (6.5) 5. Qui lut bin i cc thnh phn ng sut v cc thnh phn bin dng khi xoay h to :Da trn cc cng thc (2.21) v (3.22) ta thu c cc cng thc bin i cc thnh phn ng sut, bin dng cho trng th ng sut phng.Cc cng thc ny cng c dng ging ht cc cng thc tng ng sca trng hp bin dng phng. 6. Th nng bin dng n v th tch (cho vt liu ng hng): ( ) [ ]. 2 2212 2 2 2y x xy xy y xEW + + + =(6.16) 7. Cc ng sut chnh trong trng thi ng sut phng c th xc nh theo cc cng thc ( ). 0; 421232 22 , 1=+ += xy y xy x (6.17) (trc Oz trng vi mt trong cc trc ng sut chnh) 6.1.3 Cc iu kin bin iu kin bin ca c hai bi ton ng sut phng v bin dng phng l ging ht nhau, l iu kin cho trc lc v/hoc chuyn vtrn bin ca hnh phng (H6.2) a. Trn phn mt bin cv, Su: cho trc cc cv: ( ) ( ) y x v v y x u ub b, ; , = =. (6.19) b. Trn mt bin chu lc, ST (H6.3):. sin cos; sin cos y xy y xybyxy x xy xbxm l Tm l T+ = + =+ = + =(6.20) L Thuyt n Hi 87 6.1.3. Gii bi ton phng. Hm ng sut. Phng trnh iu ha kp Bygi,tatinhnhbiniccphngtrnhcschungcatrngthibindngphngv trng thi ng sut phng, l cc phng trnhcn bng (6.3) v phng trnhtng thch (6.5) c c mt phng trnh duy nht cho c hai trng hp. Cc phng trnh c s (6.3) v (6.5) c th bin i v dng mt phng trnh cp 4, nu nhta a vo mt hm mi, c tn l hm ng sut (hay hm Airy) c nh ngha nh sau: Ga s tn ti hm) , ( y x ca hai bin khng gian, sao cho y x x yx y x ===22222; ;.(6.21) Khi,ccphngtrnhcnbng(6.3)stthomn.SdngnhlutHookevsaukhithaybiu thc (6.18) vo phng trnh tng thch (6.5), ta thu c phng trnh i vi hm F(x,y) 0 24412 243442=+ +y y x x . (6.22) Do s khc nhau ca biu thc quan h Hooke trong bi ton bin dng phng v trong bi ton trng thi ng sut phng nn cch s1, 2 v3 trong phng trnh (6.19) nhn ccgi tr khc nhau cho hai loi bi ton ny, nh sau : cho trng thi bin dng phng:|||

\|+ ===yyz zyxxyxy yyz zyxxz zxE E G E E 21;1;13 2 1; (6.23) cho trng thi ng sut phng: xxyxy y xE G E E = = =21;1;13 2 1 .(6.24) i vi vt th ng hng, vic xc nh hm ng sut, phng trnh tng thch cho c trng thi bin dng phng ln trng thi ng sut phng, cng c dng iu ho kp nh sau 0 2442 2444=+ +y y x x (6.25) 6.1.4 Bin i cc iu kin binCciukinbin,mhm(x,y)phithomn, c th tm c nh bin i quan h (6.20) trn y. Nu trn vnh bao (ng bao) ca tm kho st c tc dng ca ti trng, m hnh chiu ca n ln cc trc to l nxT v nyTth trn c s (6.20) v (6.21) ta thu c . sin cos; sin cos22 2222x y xTy xyTnynx+ = =(6.26) trong ( )( ) n yn x, sin sin; , cos cos==. L Thuyt n Hi 88 Ta qui c di chuyn trn ng bao theo cch sao cho phn din tch c bao k cn lun nm v bn tri ca hng di chuyn, nh minh ha trn hnh v (H6.4), v khi nysxnxsy= ===sin; cos(6.27) trong , svn - tng ng,l cc hng tip tuyn vphp tuyn ca ng bao. S dng cng thc(6.27), c th vit li cng thc (6.26) di dng.;.22 2222||

\| = =|||

\|= +=x s sxx syy xTy s sxy x syyTnynx (6.28) Chn mt im A bt k trn ng bao lm gc to , v gi s ti im ny, gi tr cc o hm v gi tr ca chnh hml:A A Ay x| ; | ; | . (6.29) Khi , gi tr ca cc o hm ring y x vti im Bno trn ng bao c th tnh c, trn c s tch phn cng thc (6.28), nh sau +==AB ABnx Any Ads Ty yds Tx x| ; | (6.30) Cng thc (6.30) cho thy rng, gia s ca cc o hm ring x v y khi di chuyn t im A n im B, bng hnh chiu ln cc trc y v x, tng ng, ca cng lc (lc ri) tc dng trn cungAB . T suy ra, nu nh vect chnh ca cc lc tc dng ln ng bao bng 0, th tch phn ng trong (6.30)khidichuyntrnvng(kn)theongbao,sbng0.Nicchkhc,trongtrnghpny, x v y l cc hm n tr trn ng bao. iu pht biu trn y lun ng cho vt th n lin. Khi bit cc o hm ring ca hm ng sut theo cc to x v y, c th tnh cc o hm ring ca hm ny theo tip tuyn v php tuyn ca ng bao nh cc cng thc: nyy nxx nsyy sxx s+=+= ;(6.31) T c th d dng xc nh gi tr ca bn thn hm F(x,y) trn ng bao ( ) ( )||

\|+||

\| |||

\|+ ||

\|+ =|||

\|++ =yByA ABnxxBxA ABny A B A A B A AABAdy ds T dx ds T y yyx xxdssyy sxxF | | | | (6.32), trong : xA , yA , xB, yB -to cc im A v B. Lu : cc tch phn trong (6.32) chnh l momen ca cc lc tc dng ln cung AB, ly i vi im B. Tchphnngtrong(6.32)khidichuyntrnvng(kn)theongbao,sbng0nunh momen chnh cc lc tc dng trn ton ng baobng 0. Trong trng hp ny, hm (x,y) l n tr L Thuyt n Hi 89 trnngbao.Cccngthc(6.30)v(6.32)chotakhnngxcnhgitrtrnbincabnthn hm ng sut v o hm ca n . Cn nh rng gi tr ca cc hng s A A Ay x| ; | ; | khng nh hng n trng thi ng sut trong trng hp kho st vt th n lin, v do , c th gn cho chng gi tr bt k, bng 0 chng hn. Khi vt th kho st l a lin, cc hng s ny khng th ly bt k trn mi ng bao: Trn mt ng bao no c th ly bng 0 v trn cc ng bao cn li chng phi c xc nh theo iu kin n tr ca ca hm. Tphngtrnh(6.25)vcciukinbin(6.30),(6.32)suyrarng,ivivtthng hng, khi cho trc iu kin bin, hm ng sut khng ph thucg vo cc hng s n hi,v nh vy: Nuccvtthccnghnhdngvchutcdngcatitrngngoi nhnhauthhmng sut l ging ht nhau, cho d chng c lm t cc vt liu khc nhau (nh l M. Lewis) iu pht biu trn y ch ng cho min n lin. i vi min a lin, hm ng sut ch khng phthucvocchngsnhikhimvectchnhvmomenchnhcacclctcdngtrnmi ng bao u bng 0. Cc iu kin bin (6.30) v (6.32) nu ra trn y xc nh tnh cht chung ca hm ng sut trn bin. Chng thng c s dng gii cc bi ton trong trng hp cc ng bin khng phi l cc on thng song song vi cc trc to .Cc phng trnh vi phn (6.22) v (6.25) thun tin cho vic xc nh cc ng sut trn cctm c ng bao hnh ch nht. i vi cc tm ch nht ny, phng trnh iu kin bin (6.26) s rt n gin. 6.2 Thit lp bi ton phng trong ta cc H to cc c trnh by trong chngI nh l mt trng hp ring ca h ta cong (1.9, v d 1.3) cng vi cc quan h ton t vi phn vector c bn. Gii bi ton n hi phng trong h taccchnhlxcnhccthnhphnchuynv,ngsut,bindng { } r r r r re e e u u , , ; , , ; ,trong min R (H6.4).Quanhbindng-chuynvtrongtacccthxcnhctrncsphttrincc cng thc trong mc 3.7: b qua bin z trong cng thc (3.43), thu c kt qu: .121;1;||

\|+=||

\|+ ==ruru ureuureruerrrrr (6.33) Kt qu trn cng c th thu c nh p dng qui lut bin i cc thnh phn bin dng khi thay i ta . Da trn khi nim vt th n hi ng hng, c th kt lun rng, dngca nh lut Hooke cho vt th n hi ng hng khng thay i khi chuyn sang h ta cong trc giao bt k. Do , dng ca nh lut Hooke trong trng hp ca bi ton ng sut phng v bin dn phng cng khng thay i trong h ta cc m ch n gin i cc ch s x y sang r - , c th nh sau: L Thuyt n Hi 90 Bin dng phngng sut phng ( )( )( ) ( ). 0 ; 2;; 2; 2= = =+ = + =+ + =+ + =rz z r rr r zrr r rGee eGe e eGe e e (6.34) ( )( )( ) ( ). 0 ;1;1;1;1= =+=+ = + = = =rz z r rr r rr rr re eEee eEeEeEe (6.35) Mt cch tng t, t (2.42) ta c th vit dng tng ng ca phng trnh cn bng ( ). 02 1; 01= + ++= +++ Fr r rFr r rr rrr r r (6.36) Biu din quan h trn y theo cc chuyn v , ta c dng tng ng ca phng trnh Navier: Bin dng phng ( )( ) . 01 1; 0122= + ||

\|+ ++ + = + ||

\|+ ++ + Fur rururG u GFur rururG u Gr rrr rr(6.37) nsut phng ( )( ), 01 11 2; 011 222= + ||

\|+ ++ = + ||

\|+ ++ Fur rururEu GFur rururEu Gr rrr rr (6.38) trong , kt qu ca v d 1.3 mc 1.9 c vn dng v Laplacian hai chiu c cho bi: 222 2221 1 ++= r r r r (6.39). Cngtrncsktqunhnct(1.9,vd1.3),cngvitnhcht: + = +r y x,cc phng trnh tng thch c th bin i v dng tng ng: Bin dng phng ( ) ;1112||

\|+ + = + Fr rFrFr rr (6.40) ng sut phng( ) ( ) ||

\|+ ++ = + Fr rFrFr rr112(6.41) S dng inh ngha ca hm Airy v qui tc bin i ccthnh phn ng sut cho bi ton 2D, ta c quan h sau gia ccthnh phn ng sut trong ta c cc v hm Airy: L Thuyt n Hi 91 .1;;1 12222||

\|==+= r rrr r rrr(6.42) C th thyrng viccthnh phnng sut xc nh theo (6.42),cc phng trnh cn bng (6.36) tr thnh ng nht thc vi mi hm lin tc n o hm cp 2 ca n, trong iu kin vng mt lc khi. Khi , vi (6.42), phng trnhtng thch c th bin i v dng iu ha kp nh sau . 01 1222 224=|||

\|++= r r r r(6.43) Nh vy, bi ton n hi 2D trong ta cc c thit lp ch vi mt phng trnh dn dng iu ha kp, theo mt hm ng sut (r, ), ca hai bin s xc nh trong min hai chiu, R (H6.4). tm li gii cho bi ton, cn cn c iu kin bin thch hp. Cc iu kin ny s c xc lp khi xt cc ng dng c th. 6.2 Gii cc bi ton phng

Bi ton phng, bao gm bin dng phng v ng sut phng, c thit lp trong 6.1 v, nh thy,vicgiiccbitonnysrttinlinungdnghmngsutAiry.Nhngdnghm Airy, hphng trnh ca bi ton n hi 2D, trong trng hp khng c lc khi, a c v mt phng trnh iu ha kp, xc nh trong minkho st. Nh vy, bi ton n hi 2D trong iu kin khngclckhidnnbitontmnghimcaphngtrnhiuhakptrongminkhost. Nghim ny ng nhin l phi tha mn iu kin bin tng ng ca bi ton n hi c th cn gii quyt. Ta hn ch trnh by cch li gii cho trng hp vt liu ng hng. C nhiu gii php c th chovictmnghimnitrn.Haitrongsccgiiphpnylvicngdngchuilythavchui Fourier s c trnh by di y. 6.2.1 ng dng chui ly tha tm li gii trong to -Cc Xtbitoncthitlptheota-Cc,khikhngclckhi,vivtliunghng. Bi ton loi ny thch hp hn c cho trng hp bin ch nht v s n gin trong vic m t v p ng iu kin bin. gii bi ton, ta cn n nh trc, cn c trn mt lp lun no , dng nghim ca phng trnh iu ha kp0 2442 2444=+ +y y x x (6.25) vtm xembitonnothamndngnghimtngngvihmngsuttrn.ylcchgiida trn quan im v li gii ngc. Hm ng sut Airy c cho di dng chui ly tha, c Neou gii thiu u tin, vo nm 1957, ( ) === 0 9,m nn mmny x A y x ,(6.44) trong , cc h sAnm l cc hng s cn xc nh. Trn c s ca biu thc nh ngha hm ng sut (6.21) v cng thc (6.25) c th thy rng: 3 s hng cp thp nht ca chui trn, tha mn iu kin1 + n mkhng tham gia to nn s khc bit v ng sut nn c b qua; Cc s hng cp 2 to nn trng ng sut khng i; Cc s hng cp 3 to nn trng ng sut bin i tuyt tnh v ; L Thuyt n Hi 92 Cc s hng c ly tha3 + n ms t ng tha mn phng trnhiu ha kp vi Anm bt k; Cc s hng c ly tha3 > + n mmun tha mn phng trnhni trn th gia cc h s Anm phi tn ti mt quan h xc nh no . V d: 4042 222440y A y x A x A + +s ch tha mn phng trnhiu ha kp vi iu kin0 3 304 22 40= + + A A A v khi ch cn 2 trong 3 hng s l c lp, cn xc nh. Xt trng hp tng qut. Sau khi thay (6.44) vo (6.25), thu c ( )( )( )( )( ) ( )( )( )( ) . 0 3 2 11 2 1 23 2 10 042 22 20 04= + + ==== ==m nn mmnm nn mmnm nn mmny x A n n n ny x A n n m m my x A m m m m (6.45) t tha s chung theo ly tha ca x v y, c th vit li phng trnh trn di dng ( )( ) ( ) ( ) ( ) ( )( ) ( ) [ ] = + + + + + +== + +2 22 22 2 2 , 20 , 1 1 2 1 1 2 1 1 2m nn mn m mn n my x A n n n n A n n m m A m m m m .(6.46) V quan h trn phi tha mn vi cc gi tr ty ca x v y nn biu thc trong du ngoc vung phi trit tiu, tc ( )( ) ( ) ( ) ( ) ( )( ) ( ) . 0 1 1 2 1 1 2 1 1 22 , 2 2 , 2= + + + + + ++ + n m nm n mA n n n n A n n m m A m m m m (6.47) Vi mi cp gi tr ca m v m, (6.47) chnh l quan h tngqutphithamnchoathctothnhl iu ha kp. y cng chnh l iu kin m ta tho lunbntrnvi(m=2,n=2).Cthnhxtthy ngayrng,vphngphpkhostdnnphnb ngsutdcdngathcnnkhngthhyvng rngiukinbinscthamniukinbin didngtngqut.Tuynhin,hnchnycth vt qua c nh vic thay i iu kin bin ca bi ton kho st trn c s ca nguyn l Saint Venant v do , kt qu thu c ch chnh xc cho vng xa bin. Phng php tho lun trn y thng p dng cho bi ton coa min ch nht, nht l cc bi ton mt chiu nh trong tnh ton cc dm chng hn. Ta xt mt v d loi ny. V d 6.1: Dm chu lc ct ti du mt Xt mt dm chu lc ct ngang ti du mt (mt bin) nh trn hnh H6.5, vi iu kin bin xc nh nh sau: Tih y = :; 0 = =y xy (6.48) Ti 0 = x : ; 0 =x F dyhhxy= (6.49) Nhn xt trc gic: ng sut php theo phng dc trc s l ch yu (trn mt tit din bt k); Momen un v ng sut php theo phng dc trc s ph thuc vo bin y. Nhn xt trc gic trn l c s cho suy lun rng ng sut php dc trc x, x , l hm ca hai bin x, y: ( ) y xx x, = . T chn dng hm ng sut th : 313xy A = . (6.50) Cc thnh phn ng sut tng ng: L Thuyt n Hi 93 )` === =. 3; 0; 621213313y Axy Axy Axyyx(6.51) Vi qui lut trn, ng sut tip s l khc 0 ti cc binh y = . Cn b sung vo hm ng sut mt s hng trit tiu gi tr ng sut khng mong mun ny. Tin hnh chn li hm ng sut: ( )213 11 11213 113133 3 , h A A A y A xy A xy A y xxy = = + = . ( )2 2133 y h Axy = . S dng iu kin bin xc nh h s cha bit 13A : 3134hFA F dyhhxy= = . Kt qu cui cng: Hm ng sut:( ) ( ) y h yhFxyy x2 3334, = .(6.52) Cc thnh phn ng sut thu c:( ).43; 0;2332 22||||||||

\|===hy h FhFxyxyyx(6.53) C th nhn xt thy rng c cc kt qu trn y, khng cn s dng n gi thit tit din phng quen thuc trong l thuyt un dm. Trn c s cc thnh phn ng sut trn y, v nh lut Hooke, xc nh c cc bin dng tng ng: ( )( )( ) ( )( ).431 12121;23 1;23 132 233Ehy h FyuxvEhFxyE yvEhFxyE xuxy xyx y yy x x+ = + =|||

\|+= = === == (6.54) Tch phn cc thnh phn bin dng cho php thu c cc thnh phn chuyn v : ( )( )||||||

\|+ =+ =.43;433232x gEhFxyvy fEhy Fxu(6.55) trong ,( ) y fv( ) x gl cc hm xc nh theo iu kin bin dng ct. Sau khi thay cc tcv vo quan h bin dng ct cv, c ( )( )32 23232431434321Ehy h FdxdgEhFydydfEhFx + =|||

\|+ + Biu thc trn c th sp xp li nh sau L Thuyt n Hi 94 (6.56) T , c (6.57) Cc hng s A, B, C trong li gii biu th cc cv vt rn:A v B cv tnh tin vt rn , C cv xoay vt rn. Cc hng s ny c th xc nh t cc iu kin bin ci bin theo nguyn l Saint Venant (iu kin bin yu). Vi bi ton kho st, tn ti3 s la chn iu kin bin sau y: 0 ) 1 = = v utiL x =v 323343;2; 0 0EhFLCEhFLB Ayu= = = = 0 ) 2 = = v u tiL x =v( ) ( )((

+ + =((

+ + = = =223222331 2 143; 1 3 12; 0 0LhEhFLCLhEhFLB Ayu 3) iu kin bin tt nht:T iu kin bin th 3, c: ti L x = 0 ; 0 ; 0 === hhhhhhyudy vdx udy( ) ( )((

++ =((

++ = =2233223259 8143;511 1212; 0hLEhFLCLhEhFLB A (6.58) V d 6.2 Dm chu tc dng ca ti trng ngang ri uHnh (6.6) biu th dm chu ti trng ngang ri u dc theo mt trn ca n. Dy l bi ton phng vi min kho st l hnh ch nht cao 2c,rng 2l. iu kin bin chnh xc c p t cho mt trn v mt y, trong khi ti hai u mt s dng iu kin bin yu vi momen un bng 0 cn lc ct ch i hi tha mn iu kin cn bng tnh ton dm.iu kin bin ca bi ton c biu th bi: L Thuyt n Hi 95 Ti mt bin trn v bin di: ( )( )( ) = == p c xc xc xyyxy,; 0 ,; 0 ,(6.59)Ti hai mt (iu kin bin yu) ( )( )( )===. ,; 0 ,; 0 ,wl dy y lydy y ldy y lccxyccxccxm (6.60) Nhn xt trc gic v phn b ng sut trn cc tit din ngang dm cng tng t nh trong v d 6.1.Hm ng sut th chn chn dng cha cc s hng cp 2 , cp 3 v cp 5: ( )5 23 3 2233032212205, yAy x A y A y x A x A y x + + + = . (6.61) Ch rng s hng cp 5 c chn tha mn phng trnhiu ha kp. Hm ng sut trn tng ng vi cc thnh phn ng sut: . 6 2; 2 2 2;326 6223 21323 21 203 223 03xy A x Ay A y A Ay y x A y Axyyx =+ + =||

\| + = (6.62) p dng 3 iu kin (6.59) dn n 3 phng trnhxc nh 3 hng s21 20, A A v 23A . Sau khi gii h phng trnhny, c kt qu: 323 21 208;83;4 cpAcpApA = = =. (6.63) Cc kt qu trn cn tha mn c cc iu kin th 4 v th 6. iu kin bin th 5 v momen bng 0 dn n kt qu |||

\| =||

\| =528 52222 223 03clcpc l A A . (6.64) Vi 4 hng s ca hm Airy th c xc nh nh trn, ccthnh phn ng sut tng ng s l .4343;4 432;524352432333 33 23 22xycpxcpycpycp py y xcpyclcpxyyx+ = + =||

\| |||

\| =(6.65) tin so snh kt qu thu c vi li gii trong sc bn vt liu, ta a vo k hiu momen qun tnh tit din ngang 3 / 23c I =v vit li (6.65) di dng: L Thuyt n Hi 96 ( )( ).2;323 2;22 23 232 2y c xIpc y cyIpy y x lIpxyyx =+|||

\| = = (6.66) Kt qu trong sc bn vt liu nh sau: ( )( ),2; 0;22 22 2y c xIpItVQy x lIpIMyxyyx = == = =(6.67) trong , men ( ) ; 2 /2 2x l M =lc ctpx V = ; momen tnh din tch tit din ngang ( )22 2y cQ= ; chiu dy t ly bng n v.So snh hai kt qu thy rng: ng sut tip ging ht nhau; Cc ng sut php khc nhau; L thuyt s cp (sc bn vt liu) cho kt qu ng sut xthay i tuyn tnh trong khi l thuyt n hi cho thy quan h phi tuyn (theo ta y); Sai lch ln nht xut hin ti mp ngoi cng catitdinngang(trnnhvdiy),ttrs5 / p ,khngphthucvokchthcca dm. Vi dm thng dng trong thc t (c chiu di ln hn nhiu so vi chiu rng,c l >>), s sai khc ny tr thnhnh (trong v d trn, khi4 / = c l , sai khc ch khong 1%). S sai khc v gi tr ng sut php ygia hai l thuyt l p ti mt trn ca dm v sai lch ny l nh khic l >> . Cc nhn xt trn v nh tnh vn cn ng cho cc trng hp ti trng khc. C ngha l, vi dm c chiu di ln hn chiu cao nhiu, cc kt qu t l thuyt s cp sc bn vt liu ni chung l rt gn vi kt qu t l thuyt n hi. Tipn,xcnhcccvbngcchtngtnhtrongvd6.1:Trctin,sdngnhlut Hooke biu din v quan hbin dng - chuyn v biu din cv theo ng sut. Tip n, thay ng sut t (6. 66) v cui cng l dng php tch phn xc nh cc cv. Qu trnh trn dn n: ( )( ) ( ).5 6 2 3 2 12 2;323 52323 22 2 4 22 23 2 2 4323 2 3 32x gy c yvyx l vy c y c yEIpvy fcy cyvxy c yx yxx lEIpu+(((

|||

\| + +|||

\|+ =+(((

|||

\|+ +|||

\| +|||

\| = (6.68) Trong ,( ) y fv( ) x gl cc hm s ty ca tch phnThay kt qu trn vo quan h bin dng ct-cv, thu c ( ) ( ) ( ) ( )2 2 2 2 222322'2'5223 2y cGlpx g vxyEIpy f c y vxcy xxx lEIp = + + +(((

+|||

\| + (6.69) C th vit li phng trnhtrn di dng tch bin sL Thuyt n Hi 97 ( )( ) .584 24;0 02 2 2 40 0v x q x c v lEIpxEIpx gu y q y f+ ((

||

\|+ =+ =(6.70) Chn iu kin bin( ) 0 0 , , 0 = l v y u , c th xc nh c cc hngs q0 , u0v v0nh sau: ((

||

\|+ + = = =22 40 0 02 545121245; 0lc vEIplv q u (6.71) Vi kt qu trn, c cng thc cui cng xc nh cv: ( )((

||

\|+ + +|||((

||

\|+ + +

\|((

+ + + =(((

|||

\|+ +|||

\| +|||

\| =22 42 22 42 2 4 22 23 2 2 4323 2 3 322 5451212452 542 123 6 2 322 12 2;323 52323 2lc vEIplx cv l xy c y yx l vy c y c yEIpvcy cyvxy c yx yxx lEIpu (6.72) vng cc i c xc nh theo cng thc sau ( )((

||

\|+ + = =22 4max2 5451212450 , 0lc vEIplv v (6.73) Kt qu t sc bn vt liu lEIplv2454max= (6.74) Sai lch gia hai l thuyt xc nh c t (6.72) v (6.73) l 22 42 5424lc vEIpl||

\|+ . y l thnh phn xut hindoscmtcalcct.Vidmdi:c l >> ,sailchtrnlrtb.Mtlnnatachngt c rng, vi m di, kt qu t s bn vt liu khc trng khp vi kt qa ca l thuyt n hi. Cng thc cv (6.72) cho php rt ra nhn xt: Trncctitdin(ngvitaxkhngi)chuynvkhngcnltuyntnhvnhvytit din ngang ca dm khng cn l phng; Quan h ( )220 ,dxx v dEI M=ca l thuyt Euler-Bernoulli c s dng trong sc bn vt liu khng khng nghim ng li gii ca bi ton n hi ang xt. 6.2.2 ng dng phng php Fourier tm li gii trong h ta -Cc Mt s gii phng trnhiu ha kp tng qut hn c th tm c nh s dng phng php Fourier. ng dng ca phng php ny xut hin t gia th k 20, v c m t chi tit trong cc cng trnh ca Pickett (1944),Timoshenko v Goodier (1970) [6]. Theo ppp ny, hm ng sut ci tm di dng tch bin kt hp vi ng dng chui Fourier hoc tch phn Fourier. Hm Airy c tm trong ta -Cc di dng tch bin sau nh sau ( ) ( ) ( ) y Y x X y x = , .(6.75) Trn nguyn tc, cc hm X v Y c th chn di dng bt k, tuy nhin, tin li v thng dng hn c l dng hm s my xe Y e X = = , . Sau khi thay cc dng ny vo phng trnhiu ha kp (6.41), thu c phng trnh( ) 0 24 2 2 4= + +y xe e . (6.76) L Thuyt n Hi 98 Buc biu thc trong ngoc bng 0 ta thu c phng trnhc trng ( ) 022 2= + .(6.77) Nghim ca phng trnhc trng trn c dng kp: i =(6.78) Nghim tng qut ca bi ton t ra c tm theo pp chng nghim, bao gm nghim zero v nghim tng qut ca phng trnhiu ha kp. Vi nghim zero 0 = (nghim chp 4): 3322 1 0 0x C x C x C C + + + == (6.79) trong khi nghim zero ng vi0 = cho nghim bi ton di dng 2928 73625 4 0xy C y x C xy C y C y C y C + + + + + == .(6.80) D thy trn y chnh l nghim di dng a thc ca phng trnhiu ha kp. Trong trng hp tng qut, nghim ca phng trnhny l[ ] [ ]y y y y x i y y y y x iye D ye C e B e A e Dye Cye Be Ae e + + + + + = ' ' ' '(6.81) Cc hng s ty A, B, C, D v A, B, C, D c xc nh t cc iu kin bin ca bi ton. Nghim kt qu ca bi ton tm c bng pp chng nghim, t cc nghim (6.79) , (6.80) v (6.81). S dng dng lng gic v dng hyperbolic ca hm s m, c th vit dng nghim thc ca bi ton nh sau ( ) ( ) [ ]( ) ( ) [ ]( ) ( ) [ ]( ) ( ) [ ]0 0cosh ' ' sinh ' ' sincosh sinh sincosh ' ' sinh ' ' coscosh sinh sin= =+ ++ + + ++ + + ++ + + ++ + + = x x H F x x G E yx x H F x x G E yy y D B y y C A xy y D B y y C A x (6.82) Dng nghim trn kt hp vi chui Fourier cho php gii nhiu bi ton vi cc ti trng bin phc tp. V d 6.3 Tm ch nht chu ti bt k trn bin (pp Fourier vi vic p dng chui Fourier )

Xt tm ch nht chu ti trng ngoi tc dng tc dng lnccbin.Tuypptngquttrnychophpgii quyt bi ton theo iukin bin tng qut vi cc lc tcdnglnc4mtbin,nhng,trongvdkthpp dng chui Fourier ny, ta xt trng hp ch hai bin trnvdichulcixngquatrcynhtrnhnh H6.7.Cccnhtmlcngc(tckhngchnhlch nhau qu nhiu). p dng iu kin bin chnh xc: ( )( )( )( ) ( ) ) 86 . 6 ( . ,) 85 . 6 ( ; 0 ,) 84 . 6 ( ' 0 ,) 83 . 6 ( ; 0 ,x p b xb xy ay ayxyxyx = = = = Trn c s gi thit ti trng i xng, ng sut cng i xng (i vi trc y), v hm Airy c th chon di dng ( ) [ ][ ] . sinh cosh cossinh cosh cos ,2011x C x x G x F yy y C y B x y xmm m m m mnn n n n n+ + + + === (6.87) L Thuyt n Hi 99 Cc ng sut c tnh t hm Airy trn y s l: ( ) [ ][ ][ ]( ) [ ]( ) [ ]( ) [ ] + + + + + =+ + + + =+ + + =======1212120121212sinh cosh sinh sinhsinh cosh sinh sinh; cosh 2 sinh cosh cos2 sinh cosh cos; sinh cosh coscosh 2 cosh cosnm m m m m m m mnn n n n n n n n xym m m m m m mmmn n n n n nnn xm m m m m mmmn n n n n nnn xx x x C x F yy y y C y B xx x x G x F yC y y C y B xx x G x F yy y C y B x (6.88) T cc iu kin bin (6.83), (6.84), (6.85) b mm/ = v a nn/ =T iu kin bin (6.84) ( ) a a G Fm m m m coth 1+ = ,(6.89) Cn iu kin (6.85) dn n ( ) b b C Bn n n n coth 1+ = .(6.90) p dng diu kin bin (6.83)( ) [ ][ ] + = + +==1212sinh cosh coscosh 2 cosh cosmm m m m m m mnn n n n n n na a G a F yy y C y B a . Kt qu ny c th vit li di dng ( ) [ ] + + ===+1 11 2cosh 2 sinh cosh 1 cosm nn n n n n nnn m my y y C y B y A (6.91) trong ( )m m m mmmmG a a aaA cosh sinhsinh2+ =. (6.92) p dng qui tc khai trin Fourier, Thc hin tch phn 2 v ca (6.91), xc nh c cc h s Am theo cng thc sau: ( ) ( ) [ ] + + ==+1 0.1 2. cos cosh 2 sinh cosh 12nbm n n n n n nnn md C BbA (6.93) Thc hin tch phn v so snh vi (6.91) cho kt qu ( )( )( ).sinh 1cosh sinhsinh 412 22++== nn mnmnnnm m mmmbCa a a baG (6.94) iu kin bin cui cng (6.86) cho php thu c quan h[ ]( ) [ ] ( ) x p x x x G x F bC b b C b B xm m m m m mmm mn n n n nnn n= + + ++ + == cosh 2 sinh cosh cos2 sinh cosh cos12012 Quan h trn c th c vit li di dng sau ( ) ( ) ( ) [ ] + + + = =+= 11 20*cosh 2 sinh cosh 1 cosmm m m m m mmmnn nx x x G x F x p x A (6.95) L Thuyt n Hi 100 trong ,( ). 2; cosh sinhsinh0*02*C AC b b bbAn n n nnnn=+ = (6.96) Mt ln na, thc hin khai trin Fourier i vi v bn phi ca (6.95) v so snh h s ca cc s hng cng c cc tha s l cc hm cosine ging nhau 2 v ca phng trnhkt qu, thu c: ( )( )( )( )( )( )( ) .21... , 3 , 2 , 1coscosh sinhsinh 2 sinh 1cosh sinhsinh 40002122 23 d paCnd pb b b ab aGb b b abCaann n n nnmn mmn mmmn n nnn ==+ ++==+(6.97) Cc phng trnhcng knh (6.94) v (6.97) c th vit li di dng gn gh hn: = + = +==11;; 0mn n mn nmn mn mT G S CC R G (6.98) vi mn mnS R , v nThon ton xc nh. y l h v s cp phng trnhvi v s cp Nn s nCv nG . Trong li gii gn ng, c th ct h trn c h hu hn phng trnhvi hu hn Nn s. L Thuyt n Hi 101 Chng VII XON THANH LNG TR Mt trong nhng ng dng thc t quan trng, ca phng php na ngc Saint Venantl vic gii bi ton thanh lng tr chu tc dng ca hai momen xon ti hai u, khilc khi c b qua (xon thun ty). Mt bn ca thanh hon ton t do. Vt liu thanh l ng hng. H ta Oxyz l h ta thun, trc giao, vi cc trc Oy v Oz l tu trn tit din ngang. 7.1 Li gii tng qut theo phng php na ngc Bi ton t ra c gii theo phng php na ngc: mt phn nghim c n nh trc (trn mt c s no ), phn cn li c tm theo cch gii thun: thchin vic tch phn phng trnh c s chung, c n gin ha. 7.1.1 Dng nghim ban u Trn c s quan st cc kt qu th nghim xon thanh lng tr thng, tit din ngang bt k ta c th n nh chuyn v ca im trn tit din ngang lm dng nghim ban u nh sau: xy wzx v= = ; (7.1) ( ) z y u u , =(7.2) Vi cc chuyn v xc nh theo (10.1), tit din ngang ca thanh xoay mt gc , trong khi vn bo ton hnh dng (ca hnh chiu trn mt phng ta y-z) (H7.1) L Thuyt n Hi 102 Trng hp thanh trn, trn c s gi thit Coulom, ta xc nh c ngay, 0 ) , ( = z y u tc khi b xon, tit din ngang thanh trn khng b vnh. Khi thanh l lng tr, chuyn v theo phng trc Ox ca cc im trn tit din ngang, xc nh theo cng thc (7.2), l hm ca hai ta (y,z). Nh vy, gi thit rng, khi b xon thun tu, tit din ngang thanh lng tr khng cn phng na. 7.1.2 Hm ng sut v phng trnh c s Xutphttdngnghimbanu(7.1)v(7.2),tasgiitipbitontheongsut.Sdng phng trnh ca nh lut Hooke tng qut v quan h Cauchy, c th chng t rng, vi chuyn v ca cc im c xc nh s b bi cc cng thc(7.1) v (7.2) th trong cc thnh phn ng sut,ch c ng sut zx xy vl khc 0, tc: 0 ; 0 ; 0 = = = = yz z y x zx xy (7.3) Bi ton dn n vic tm hai thnh phn ng sut cha bit trong (7.3). u tin, phng trnh cn bng phi c tha mn. Sau khi thay (7.3) vo phng trnh cn bng (2.16), vi X = Y = Z = 0, ta thu c . 0 ; 0; 0===+x xz yzxxyzxxy(7.4) T cc ng thc (7.4), c th suy ra rng, cc ng sut zx xy vkhng ph thuc vo to x. tho mn (7.4), ta ch cn tm c hmca hai bin y, z c nh ngha nh sau: ( )( ),,;,yz yzz yzxxy ==(7.5) trong ,(y,z) l hm bt k (hai ln kh vi) ca cc to yv z. Hm (y,z) c gi l hm ng sut. Hm ng sut (y,z) trn y do Prandt a ra u tin 7.1.3 iu kin tng thch -Phng trnh PoissonCc iu kin cn bng (7.4) s t tho mn vi mi hm (y,z). tm ra phng trnh xc nh hm ny, ta s dng cc iukin tng thch (3.32).Trncscccngthc(7.4),(7.5),nhlutHooketngqut(4.40)vphngtrnhtng thch(3.35*), ta thu c , 0; 02 22 2= == =z zy y (7.6) trong ,.22222z y += T cc ng thc trn suy ra rng, hm (y,z) cn tho mn phng trnh Poisson( ) c z y = ,2 ,(7.7) trong ,c = const. xc nh hng sccn s dng kt hp:cng thc (7.5); L Thuyt n Hi 103 nh lut Hooke tng qut (4.40);cc quan h ng sut-bin dng (3.14); cng vi dng nghim ban u (7.1). Kt qu, thu c.;||

\|+ = =|||

\|+ ==zuy Gyyuz Gzzxxy (7.8) Ly o hm phng trnh u ca (7.8) theo z, phng trnh th 2 theo y, ri trkt qu th nhtcho kt quth 2, ta thu c. 222222 Gz y = =+(7.9) T ,. 2 G c = Cng ly o hm cc phng trnh (7.8) nhng theo th t ngc li: ly o hm phng trnh u theo y, phng trnh th 2 theo z ri cng cc kt qu, ta c: 02= u .(7.10) hay, hm chuyn vu phi l iu ha, (7.10). 7.1.4 iu kin bin. Qu o ng sut tip xc nh hon ton hm (y,z) cn xc lp cc gi tr trn bin ca n (tc trn ng bao tit dinngang)ngvimtbncathanhkhngchuti( 0 = = =nznynxT T T ).Kthpccphngtrnh (2.16)vcngthc(7.3),cthnhnthyrng,ccphngtrnhth2vth3cah(2.16)ttho mn, cn phng trnh th nht s c dng(H7.2) ( ) ( ) 0 , cos , cos = + z n y n szx xy (7.11) ngha ca (7.11) kh r rng: hnh chiu ca ng sut tip ton phn ln php tuyn vi ng bao tit din ngang nht thit phi bng 0. Trn c s ca (7.5), cc iu kin bin (7.11) c th vit li L Thuyt n Hi 104 ( ) ( ) . 0 , cos , cos =z nyy nz (7.12 ) Mt khc, c: (xem H7.2)( ) ( )( ) ( ) , , cos , cos; , cos , cossynzy s z nsznyz s y n == === =(7.13) trong ,sv n- tng ng, l tip tuyn v php tuyn vi ng bao ti im A. Da trn (7.13), c th vit li biu thc (7.12)di dng. 0 ==+s syy szz (7.14) T (7.14) suy ra rng, gi tr ca hm (y,z) dc theo ng bao l khng i, hay( ) . , const B z y = = (7.15) Khi thay i gi tr ca hng s B, phng trnh (7.15) xc nh mt h ng cong v ti mi im trn ngcongknthuchngcongny,ngsuttiptonphnlunsongsongvitiptuyncang bao ti im kho stCc ng cong ny c tn l qu o ng sut tip. Nunhtitdinnganglmtminnlin,thhngsbiuthgitrcahmngsuttrn bin ca tit din ngang c th nhn mt gi tr bt k, c th bng 0. Vic chn mt cch bt kgi tr cahngsnykhngnhhnggnktqucaligii.Thcvy,gis0(y,z)lmthmng sut nhn gi tr bng 0 trn ng bao, cn hm( ) ( ) B z y z y + = , ,0 (7.16)cng s l hm ng sut ca bi ton vi gi tr B bt k, khng i trn ng bao. Thy ngay rng, cc ng sut tip zx xy ,tnh c t hm (y,z), theo cng thc (7.5)s khng ph thuc g vo gi tr B.Nu tit din ngang ca thanh xc nh mt min a lin, th ch trn mt trong cc ng bao c th chn hng s ni trn l bt k. Cc gi tr ca hng s trn cc bin khc c th tm c theo iu kin n tr ca chuyn v u (v tt nhin ph thuc vo ga tr ca hng s c chn bt k ni trn). Gitrcahmngsuttrnngbaongoicngthngcchnbtk,trongkhigitr hm ny trn cc ng bao khc, bn trong, c xc nh trn c s ca iu kin n tr ca chuyn v u. 7.1.5 nh l lu s ng sut tip iu kin n tr ca chuyn v u c vit di dng =|||

\|+, 0 dzzudyyu (7.17) trong , tch phn c ly theo cc ng bao kn trong t