giaotrinh_pppthh_v10.pdf

i

PHNG PHP PHN T HU HN

L thuyt Bi tp Chng trnh MATLAB

H NI 2007

TRN CH THNH NG NH KHOA

TRN CH THNH NG NH KHOA

H NI 2007


P p


GS, TS Trn ch Thnh TS. Ng Nh Khoa



H NI 2007

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M U Gio trnh Phng php Phn t hu hn (PP PTHH) c bin son

da trn ni dung cc bi ging v kinh nghim ging dy mn hc cng tn trong nhng nm gn y cho sinh vin khoa C kh, trng i hc Bch khoa H Ni v hc vin cao hc ngnh C hc K thut, trng i hc K thut Cng nghip - i hc Thi Nguyn. Ni dung gio trnh c mc ch trang b cho sinh vin cc ngnh k thut: Cng ngh ch to my, C tin k thut, K thut hng khng, K thut tu thu, My thu kh, t, ng c, To hnh bin dng, Cng ngh cht do & composite, Cng ngh & kt cu hn v.v.:

- Nhng kin thc c bn nht ca PP PTHH ng dng, - p dng phng php gii quyt mt s bi ton k thut khc

nhau, - Nng cao k nng lp trnh Matlab trn c s thut ton PTHH. Gio trnh bin son gm 13 chng.

Sau phn gii thiu phng php PTHH, mt s loi phn t thc v phn t qui chiu hay gp (Chng 1), gio trnh cp n mt s php tnh ma trn, phng php kh Gauss (Chng 2) v thut ton xy dng ma trn cng v vct lc nt chung cho kt cu (Chng 3). Phng php Phn t hu hn trong bi ton mt chiu chu ko (nn) c gii thiu trong Chng 4 v ng dng vo tnh ton h thanh phng (Chng 5). Tip theo, gio trnh tp trung vo m t phn t hu hn tam gic bin dng hng s trong bi ton phng ca l thuyt n hi (Chng 6) v ng dng vo tnh ton kt cu i xng trc (Chng 7). Chng 8 gii thiu phn t t gic km theo khi nim tch phn s. Chng 9 m t phn t Hermite trong bi ton tnh dm v khung. Chng 10 trnh by phn t hu hn trong bi ton dn nhit mt v hai chiu. Chng 11 xy dng thut ton PTHH tnh tm-v chu un. Phn p dng phn t hu hn trong tnh ton vt liu v kt cu composite c gii thiu trong chng 12. Chng 13 m t phn t hu hn trong tnh ton ng lc hc mt s kt cu.

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Cui mi chng (t chng 4 n chng 13) u c chng trnh Matlab km theo v mt lng bi tp thch ng ngi c t kim tra kin thc ca mnh.

Gio trnh c bin son bi: - GS. TS Trn ch Thnh (ch bin): Chng 1, 3, 4, 5, 6, 8 v 9. - TS Ng Nh Khoa: Chng 2, 7, 10, 11, 12, 13 v cc chng trnh

Matlab. Gio trnh c trnh by mt cch h thng v nht qun t u n cui

nh Nguyn l cc tiu ho th nng ton phn. Cc quan h c xy dng trong "khng gian qui chiu", do rt thun li trong tnh ton v lp trnh.

C th dng gio trnh ny lm ti liu tham kho cho sinh vin, hc vin Cao hc v nghin cu sinh cc ngnh k thut lin quan. Rt mong nhn c nhng gp xy dng ca bn c. Tp th tc gi

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MC LC Chng 1

GII THIU PHNG PHP PHN T HU HN 1. Gii thiu chung ................................................................................... 1 2. Xp x bng phn t hu hn ............................................................... 1 3. nh ngha hnh hc cc phn t hu hn ............................................ 2 3.1. Nt hnh hc .................................................................................................. 2 3.2. Qui tc chia min thnh cc phn t .............................................................. 2 4. Cc dng phn t hu hn .................................................................... 2 5. Phn t quy chiu, phn t thc ........................................................... 4 6. Mt s dng phn t quy chiu ............................................................ 5 7. Lc, chuyn v, bin dng v ng sut ................................................. 6 8. Nguyn l cc tiu ho th nng ton phn ......................................... 7 9. S tnh ton bng phng php phn t hu hn ........................... 8

Chng 2

I S MA TRN V PHNG PHP KH GAUSSIAN 1. i s ma trn .................................................................................... 11 1.1. Vct ............................................................................................................ 11 1.2. Ma trn n v ............................................................................................. 12 1.3. Php cng v php tr ma trn. ................................................................... 12 1.4. Nhn ma trn vi hng s ............................................................................ 12 1.5. Nhn hai ma trn .......................................................................................... 13 1.6. Chuyn v ma trn ........................................................................................ 13 1.7. o hm v tch phn ma trn ..................................................................... 14 1.8. nh thc ca ma trn .................................................................................. 14 1.9. Nghch o ma trn ...................................................................................... 15 1.10. Ma trn ng cho ................................................................................. 16 1.11. Ma trn i xng ..................................................................................... 16 1.12. Ma trn tam gic ...................................................................................... 16 2. Php kh Gauss .................................................................................. 17 2.1. M t ............................................................................................................ 17 2.2. Gii thut kh Gauss tng qut.................................................................... 18

Chng 3

THUT TON XY DNG MA TRN CNG V VCT LC NT CHUNG

1. Cc v d ............................................................................................ 21 1.1. V d 1 ......................................................................................................... 21 1.2. V d 2 ......................................................................................................... 24 2. Thut ton ghp K v F ...................................................................... 27

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2.1. Nguyn tc chung ........................................................................................ 27 2.2. Thut ton ghp ni phn t: ....................................................................... 28

Chng 4

PHN T HU HN TRONG BI TON MT CHIU 1. M u ............................................................................................... 30 2. M hnh phn t hu hn ................................................................... 30 3. Cc h trc to v hm dng ......................................................... 31 4. Th nng ton phn ............................................................................ 34 5. Ma trn cng phn t .................................................................... 35 6. Qui i lc v nt ............................................................................... 36 7. iu kin bin, h phng trnh phn t hu hn.............................. 37 8. V d ................................................................................................... 39 9. Chng trnh tnh kt cu mt chiu 1D ......................................... 45 10. Bi tp ................................................................................................ 49

Chng 5

PHN T HU HN TRONG TNH TON H THANH PHNG 1. M u ............................................................................................... 51 2. H to a phng, h to chung.............................................. 51 3. Ma trn cng phn t .................................................................... 53 4. ng sut ............................................................................................. 54 5. V d ................................................................................................... 54 6. Chng trnh tnh h thanh phng...................................................... 56 7. Bi tp ................................................................................................ 66

Chng 6

PHN T HU HN TRONG BI TON HAI CHIU 1. M u ............................................................................................... 70 1.1. Trng hp ng sut phng ......................................................................... 71 1.2. Trng hp bin dng phng ....................................................................... 71 2. Ri rc ho kt cu bng phn t tam gic ........................................ 72 3. Biu din ng tham s ...................................................................... 75 4. Th nng ............................................................................................. 78 5. Ma trn cng ca phn t tam gic ............................................... 78 6. Qui i lc v nt ............................................................................... 79 7. V d ................................................................................................... 82 8. Chng trnh tnh tm chu trng thi ng sut phng ...................... 87 9. Bi tp ................................................................................................ 98

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Chng 7 PHN T HU HN

TRONG BI TON I XNG TRC CHU TI TRNG I XNG 1. M u ............................................................................................. 102 2. M t i xng trc .......................................................................... 102 3. Phn t tam gic ............................................................................... 104 4. Chng trnh tnh kt cu i xng trc .......................................... 113 5. Bi tp .............................................................................................. 122

Chng 8

PHN T T GIC 1. M u ............................................................................................. 126 2. Phn t t gic ................................................................................. 126 3. Hm dng ......................................................................................... 127 4. Ma trn cng ca phn t ............................................................ 129 5. Qui i lc v nt ............................................................................. 131 6. Tch phn s ..................................................................................... 132 7. Tnh ng sut.................................................................................... 136 8. V d ................................................................................................. 136 9. Chng trnh .................................................................................... 138 10. Bi tp .............................................................................................. 150

Chng 9 PHN T HU HN TRONG TNH TON KT CU DM V KHUNG 1. Gii thiu ......................................................................................... 152 2. Th nng ........................................................................................... 153 3. Hm dng Hermite ........................................................................... 153 4. Ma trn cng ca phn t dm .................................................... 155 5. Quy i lc nt ................................................................................ 157 6. Tnh mmen un v lc ct .............................................................. 159 7. Khung phng .................................................................................... 159 8. V d ................................................................................................. 162 9. Chng trnh tnh dm chu un ...................................................... 166 10. Bi tp .............................................................................................. 175

Chng 10

PHN T HU HN TRONG BI TON DN NHIT 1. Gii thiu ......................................................................................... 178 2. Bi ton dn nhit mt chiu ............................................................ 178 2.1. M t bi ton ............................................................................................ 178

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2.2. Phn t mt chiu ...................................................................................... 178 2.3. V d .......................................................................................................... 180 3. Bi ton dn nhit hai chiu ............................................................. 182 3.1. Phng trnh vi phn qu trnh dn nhit hai chiu ................................... 182 3.2. iu kin bin ............................................................................................ 183 3.3. Phn t tam gic ........................................................................................ 184 3.4. Xy dng phim hm ................................................................................. 186 3.5. V d .......................................................................................................... 190 4. Cc chng trnh tnh bi ton dn nhit ......................................... 192 4.1. V d 10.1 .................................................................................................. 192 4.2. V d 10.2 .................................................................................................. 197 5. Bi tp .............................................................................................. 203

Chng 11

PHN T HU HN TRONG TNH TON KT CU TM - V CHU UN

1. Gii thiu ......................................................................................... 206 2. L thuyt tm Kirchhof .................................................................... 206 3. Phn t tm Kirchhof chu un ........................................................ 209 4. Phn t tm Mindlin chu un .......................................................... 216 5. Phn t v ........................................................................................ 219 6. Chng trnh tnh tm chu un ....................................................... 222 7. Bi tp .............................................................................................. 231

Chng 12

PHN T HU HN TRONG TNH TON VT LIU, KT CU COMPOSITE

1. Gii thiu ......................................................................................... 234 2. Phn loi vt liu Composite ........................................................... 234 3. M t PTHH bi ton trong trng thi ng sut phng ................... 236 3.1. Ma trn D i vi trng thi ng sut phng ............................................. 236 3.2. V d .......................................................................................................... 238 4. Bi ton un tm Composite lp theo l thuyt Mindlin ................. 241 4.1. M hnh ha vt liu composite nhiu lp theo l thuyt Mindlin ............ 241 4.2. M hnh ha PTHH bi ton tm composite lp chu un ........................ 246 5. Chng trnh tnh tm Composite lp chu un............................... 250 6. Bi tp .............................................................................................. 267

Chng 13

PHN T HU HN TRONG BI TON NG LC HC KT CU

1. Gii thiu ......................................................................................... 268

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2. M t bi ton .................................................................................. 268 3. Vt rn c khi lng phn b ......................................................... 270 4. Ma trn khi lng ca phn t c khi lng phn b .................. 272 4.1. Phn t mt chiu ...................................................................................... 272 4.2. Phn t trong h thanh phng .................................................................... 272 4.3. Phn t tam gic ........................................................................................ 273 4.4. Phn t tam gic i xng trc .................................................................. 274 4.5. Phn t t gic ........................................................................................... 276 4.6. Phn t dm ............................................................................................... 276 4.7. Phn t khung ............................................................................................ 277 5. V d ................................................................................................. 277 6. Chng trnh tnh tn s dao ng t do ca dm v khung ................... 278 6.1. Chng trnh tnh tn s dao ng t do ca dm ..................................... 278 6.2. Chng trnh tnh tn s dao ng t do ca khung .................................. 283 7. Bi tp .............................................................................................. 288

TI LIU THAM KHO

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Chng 1 GII THIU PHNG PHP PHN T HU HN

1. GII THIU CHUNG S tin b ca khoa hc, k thut i hi ngi k s thc hin nhng

n ngy cng phc tp, t tin v i hi chnh xc, an ton cao. Phng php phn t hu hn (PTHH) l mt phng php rt tng qut

v hu hiu cho li gii s nhiu lp bi ton k thut khc nhau. T vic phn tch trng thi ng sut, bin dng trong cc kt cu c kh, cc chi tit trong t, my bay, tu thu, khung nh cao tng, dm cu, v.v, n nhng bi ton ca l thuyt trng nh: l thuyt truyn nhit, c hc cht lng, thu n hi, kh n hi, in-t trng v.v. Vi s tr gip ca ngnh Cng ngh thng tin v h thng CAD, nhiu kt cu phc tp cng c tnh ton v thit k chi tit mt cch d dng.

Trn th gii c nhiu phn mm PTHH ni ting nh: NASTRAN, ANSYS, TITUS, MODULEF, SAP 2000, CASTEM 2000, SAMCEF v.v.

c th khai thc hiu qu nhng phn mm PTHH hin c hoc t xy dng ly mt chng trnh tnh ton bng PTHH, ta cn phi nm c c s l thuyt, k thut m hnh ho cng nh cc bc tnh c bn ca phng php.

2. XP X BNG PHN T HU HN Gi s V l min xc nh ca mt i lng cn kho st no (chuyn

v, ng sut, bin dng, nhit , v.v.). Ta chia V ra nhiu min con ve c kch thc v bc t do hu hn. i lng xp x ca i lng trn s c tnh trong tp hp cc min ve.

Phng php xp x nh cc min con ve c gi l phng php xp x bng cc phn t hu hn, n c mt s c im sau:

- Xp x nt trn mi min con ve ch lin quan n nhng bin nt gn vo nt ca ve v bin ca n,


- Cc hm xp x trong mi min con ve c xy dng sao cho chng lin tc trn ve v phi tho mn cc iu kin lin tc gia cc min con khc nhau.

- Cc min con ve c gi l cc phn t.

3. NH NGHA HNH HC CC PHN T HU HN

3.1. Nt hnh hc Nt hnh hc l tp hp n im trn min V xc nh hnh hc cc

PTHH. Chia min V theo cc nt trn, ri thay min V bng mt tp hp cc phn t ve c dng n gin hn. Mi phn t ve cn chn sao cho n c xc nh gii tch duy nht theo cc to nt hnh hc ca phn t , c ngha l cc to nm trong ve hoc trn bin ca n.

3.2. Qui tc chia min thnh cc phn t Vic chia min V thnh cc phn t ve phi tho mn hai qui tc sau:

- Hai phn t khc nhau ch c th c nhng im chung nm trn bin ca chng. iu ny loi tr kh nng giao nhau gia hai phn t. Bin gii gia cc phn t c th l cc im, ng hay mt (Hnh 1.1).

- Tp hp tt c cc phn t ve phi to thnh mt min cng gn vi min V cho trc cng tt. Trnh khng c to l hng gia cc phn t.

4. CC DNG PHN T HU HN C nhiu dng phn t hu hn: phn t mt chiu, hai chiu v ba chiu.

Trong mi dng , i lng kho st c th bin thin bc nht (gi l phn t bc nht), bc hai hoc bc ba v.v. Di y, chng ta lm quen vi mt s dng phn t hu hn hay gp.

bin gii bin gii

v2 v1

bin gii

v2 v1 v

1 v2

Hnh 1.1. Cc dng bin chung gia cc phn t


Phn t mt chiu

Phn t hai chiu

Phn t ba chiu

Phn t t din

Phn t lng tr

Phn t bc nht Phn t bc hai Phn t bc ba





5. PHN T QUY CHIU, PHN T THC Vi mc ch n gin ho vic xc nh gii tch cc phn t c dng

phc tp, chng ta a vo khi nim phn t qui chiu, hay phn t chun ho, k hiu l vr. Phn t qui chiu thng l phn t n gin, c xc nh trong khng gian qui chiu m t , ta c th bin i n thnh tng phn t thc ve nh mt php bin i hnh hc re. V d trong trng hp phn t tam gic (Hnh 1.2).

Cc php bin i hnh hc phi sinh ra cc phn t thc v phi tho mn cc qui tc chia phn t trnh by trn. Mun vy, mi php bin i hnh hc phi c chn sao cho c cc tnh cht sau: a. Php bin i phi c tnh hai chiu (song nh) i vi mi im trong

phn t qui chiu hoc trn bin; mi im ca vr ng vi mt v ch mt im ca ve v ngc li.

b. Mi phn bin ca phn t qui chiu c xc nh bi cc nt hnh hc ca bin ng vi phn bin ca phn t thc c xc nh bi cc nt tng ng.

Ch : - Mt phn t qui chiu vr c bin i thnh tt c cc phn t thc ve

cng loi nh cc php bin i khc nhau. V vy, phn t qui chiu cn c gi l phn t b-m.

- C th coi php bin i hnh hc ni trn nh mt php i bin n gin. - (, ) c xem nh h to a phng gn vi mi phn t.

vr

v3

v2

v1

1,0 0,0

y

x

(1)

(2)

(3)

(4)

(5)

r3

r2

r1

0,1

Hnh 1.2. Phn t quy chiu v cc phn t thc tam gic


6. MT S DNG PHN T QUI CHIU

Phn t qui chiu mt chiu

Phn t qui chiu hai chiu

Phn t qui chiu ba chiu Phn t t din


vr

0,1,0 0,0,0

0,0,1

vr 0,1,0

0,0,1

vr

1,0,0

1,0,0

0,1,0

1,0,0

0,0,1


vr

1 0,0

1

vr

1 0,0

1

vr

1 0,0

1

1/2,1/2

1/2

1/2

1/3,2/3

2/3,1/3

2/31/3

1/32/3

0 1 -1 0 1 -1 -1/2 1 -1 1/20 Phn t bc nht Phn t bc hai Phn t bc ba


Phn t su mt

7. LC, CHUYN V, BIN DNG V NG SUT C th chia lc tc dng ra ba loi v ta biu din chng di dng vct ct:

- Lc th tch f : f = f[ fx, fy , fz]T

- Lc din tch T : T = T[ Tx, Ty , Tz]T - Lc tp trung Pi: Pi= Pi [ Px, Py , Pz]T Chuyn v ca mt im thuc vt c k hiu bi:

u = [u, v, w] T (1.1) Cc thnh phn ca tenx bin dng c k hiu bi ma trn ct:

= [x , y, z, yz, xz, xy] T (1.2) Trng hp bin dng b:

T

xv

yu

xw

zu

yw

zv

zw

yv

xu

+

+

+

= (1.3)

Cc thnh phn ca tenx ng sut c k hiu bi ma trn ct: = [x , y, z, yz, xz, xy] T (1.4)

Vi vt liu n hi tuyn tnh v ng hng, ta c quan h gia ng sut vi bin dng:

vr


0,1,1

vr

1,1,0

0,1,1

1,1,0

vr

0,1,1

1,1,0


= D (1.5) Trong :

( )( )

+=

500000005000000050000000100010001

211

,,

,ED

E l mun n hi, l h s Poisson ca vt liu.

8. NGUYN L CC TIU HO TH NNG TON PHN

Th nng ton phn ca mt vt th n hi l tng ca nng lng bin dng U v cng ca ngoi lc tc dng W:

= U + W (1.6) Vi vt th n hi tuyn tnh th nng lng bin dng trn mt n v

th tch c xc nh bi: T21

Do nng lng bin dng ton phn:

=V

T dvU 21

(1.7)

Cng ca ngoi lc c xc nh bi:

=

=

n

ii

Ti

S

T

V

T PuTdSuFdVuW1

(1.8)

Th nng ton phn ca vt th n hi s l:

=

=n

ii

Ti

S

T

V

T

V

T PuTdSudVfudV12

1 (1.9)

Trong : u l vct chuyn v v Pi l lc tp trung ti nt i c chuyn v l ui


p dng nguyn l cc tiu th nng: i vi mt h bo ton, trong tt c cc di chuyn kh d, di chuyn thc ng vi trng thi cn bng s lm cho th nng t cc tr. Khi th nng t gi tr cc tiu th vt (h) trng thi cn bng n nh.

9. S TNH TON BNG PHNG PHP PHN T HU HN Mt chng trnh tnh bng PTHH thng gm cc khi chnh sau:

Khi 1: c cc d liu u vo: Cc d liu ny bao gm cc thng tin m t nt v phn t (li phn t), cc thng s c hc ca vt liu (mun n hi, h s dn nhit...), cc thng tin v ti trng tc dng v thng tin v lin kt ca kt cu (iu kin bin);

Khi 2: Tnh ton ma trn cng phn t k v vct lc nt phn t f ca mi phn t;

Khi 3: Xy dng ma trn cng tng th K v vct lc nt F chung cho c h (ghp ni phn t);

Khi 4: p t cc iu kin lin kt trn bin kt cu, bng cch bin i ma trn cng K v vec t lc nt tng th F;

Khi 5: Gii phng trnh PTHH, xc nh nghim ca h l vct chuyn v chung Q;

Khi 6: Tnh ton cc i lng khc (ng sut, bin dng, gradin nhit , v.v.) ;

Khi 7: T chc lu tr kt qu v in kt qu, v cc biu , th ca cc i lng theo yu cu.

S tnh ton vi cc khi trn c biu din nh hnh sau (Hnh 1.3);


Tnh ton ma trn cng phn t k Tnh ton vct lc nt phn t f

Gii h phng trnh KQ = F (Xc nh vct chuyn v nt tng th Q)

c d liu u vo - Cc thng s c hc ca vt liu - Cc thng s hnh hc ca kt cu - Cc thng s iu khin li - Ti trng tc dng - Thng tin ghp ni cc phn t - iu kin bin

Xy dng ma trn cng K v vct lc chung F

p t iu kin bin (Bin i cc ma trn K v vec t F)

Tnh ton cc i lng khc (Tnh ton ng sut, bin dng, kim tra bn, v.v)

In kt qu - In cc kt qu mong mun - V cc biu , th

Hnh 1.3. S khi ca chng trnh PTHH


Chng 2 I S MA TRN V PHNG PHP KH GAUSSIAN

p dng phng php PTHH trong cc bi ton k thut thng lin quan n mt lot cc php ton trn ma trn. V vy, cc php ton c bn trn ma trn v phng php kh Gaussian (Gauss) gii h phng trnh tuyn tnh s l 2 ni dung chnh c cp trong chng ny.

1. I S MA TRN Cc cng c ton hc v ma trn c cp trong phn ny l cc cng

c c bn gii bi ton tm nghim ca h phng trnh tuyn tnh, c dng nh sau:

nnnnnn

nn

nn

bxaxaxa

bxaxaxabxaxaxa

=++

=++

=++

LLLLLLLLLLLL

LL

2211

22222121

11212111

(2.1)

trong , x1, x2, , xn l cc nghim cn tm. H phng trnh (2.1) c th c biu din dng thu gn:

Ax = b (2.2) trong , A l ma trn vung c kch thc (n n), v x v b l cc vct (n1), c bin din nh sau:

=

nnnn

n

n

aaa

aaaaaa

A

LLLLL

LL

21

22221

11211

=

nx

xx

x M2

1

=

nb

bb

b M2

1

1.1. Vct Mt ma trn c kch thc (1 n) c gi l vct hng, ma trn c kch

thc (n 1) c gi l vct ct. V d mt vct hng (1 4): { }61222 =r


v vct ct (3 1):

=

342

11c

1.2. Ma trn n v Ma trn n v l ma trn ng cho vi cc phn t trn ng cho

chnh bng 1, v d:

=

100010001

I

1.3. Php cng v php tr ma trn. Cho 2 ma trn A v B, cng c kch thc l (m n). Tng ca chng l 1

ma trn C = A + B v c nh ngha nh sau: cij = aij + bij (2.3)

V d:

=

+

34

7521

5815

23

php tr c nh ngha tng t.

1.4. Nhn ma trn vi hng s Nhn 1 ma trn A vi hng s c c nh ngha nh sau:

cA=[caij] (2.4) V d:

=

100500

20030015

23102


1.5. Nhn hai ma trn Tch ca ma trn A kch thc (m n) vi ma trn B kch thc (n p) l 1

ma trn C kch thc (m p), c nh ngha nh sau: A B = C (2.5)

(m n) (n p) (m p) trong , phn t th (ij) ca C l (cij) c tnh theo biu thc:

=

=

n

kkjikij bac

1

(2.6)

V d:

=

36387054

465254

413582

Ch : - iu kin tn ti php nhn 2 ma trn AB l s ct ca ma trn A

phi bng s hng ca ma trn B. - Trong phn ln cc trng hp, nu tn ti tch 2 ma trn AB v

BA, th tch 2 ma trn khng c tnh cht giao hon, c ngha l AB BA.

1.6. Chuyn v ma trn Chuyn v ca ma trn A = [aij] kch thc (m n) l 1 ma trn, k hiu l

AT c kch thc l (n m), c to t ma trn A bng cch chuyn hng ca ma trn A thnh ct ca ma trn AT. Khi , (AT)T = A. V d:

=

465254

A th:

=

455624TA

Chuyn v ca mt tch cc ma trn l tch cc chuyn v ca ma trn thnh phn theo th t o ngc, c ngha l:

(ABC)T=CTBT AT. (2.7)


1.7. o hm v tch phn ma trn Trong nhiu bi ton k thut, cc phn t ca ma trn khng phi l 1

hng s, chng l cc hm s 1 bin hay nhiu bin. V d:

+

+

+

=

yxxyx

xyxyxA

462

52 2

Trong cc trng hp , cc ma trn c th c o hm hay tch phn. Php o hm (hay php tch phn) ca 1 ma trn, n gin l ly o hm (hay ly tch phn) i vi mi phn t ca ma trn:

=

dxxda

xAdxd ij )()( (2.8)

[ ] = dxdyaAdxdy ij (2.9) Chng ta s s dng thng xuyn biu thc (2.8) xy dng h phng

trnh PTHH trong cc chng sau. Xt ma trn vung A, kch thc (n n) vi cc h s hng, vct ct x = {x1 x2 ... xn}T cha cc bin. Khi , o hm ca Ax theo 1 bin xp s l:

p

p

aAxdxd

=)( (2.10)

trong , ap l vct ct v chnh l ct th p ca ma trn A.

1.8. nh thc ca ma trn Cho ma trn vung A = [aij], kch thc (n n). nh thc ca ma trn A

c nh ngha nh sau:

( )( )

=

+

+

=

+=n

jijij

ji

nnn

Aa

AaAaAaA

1

111

12121111

)det(1

)det(1)det()det()det( L (2.11)

trong , Aij l ma trn kch thc (n-1 n-1) thu c bng cch loi i hng i ct j ca ma trn A.


V d:

=

=

nnnn

n

n

nnnn

n

n

aaa

aaaaaa

A

aaa

aaaaaa

A

LLLLL

LL

LLLLL

LL

32

33332

22322

11

21

22221

11211

Cng thc (2.11) l cng thc tng qut. Theo cng thc ny, nh thc ca ma trn vung c kch thc (n n) c xc nh theo phng php truy hi t nh thc cc ma trn c kch thc (n-1 n-1). Trong , ma trn ch c 1 phn t (1 1) c:

det(apq) = apq (2.12)

1.9. Nghch o ma trn Cho ma trn vung A, nu det(A) 0, th A c ma trn nghch o v k

hiu l A-1. Ma trn nghch o tha mn quan h sau: A-1A = AA-1 = I (2.13)

Nu det(A) = 0, A l ma trn suy bin v khng tn ti ma trn nghch o. Nu det(A) 0 ta gi A l ma trn khng suy bin. Khi , nghch o ca A c xc nh nh sau:

AadjAAdet

1=

(2.14)

Trong , adjA l ma trn b ca A, c cc phn t ( ) )det(1 jijiij Aa += v Aji l ma trn thu c t A bng cch loi i hng th j v ct th i. V d:

Nghch o ca ma trn A kch thc (2 2) l:

=

=

1121

12221

2221

12111

det1

aaaa

Aaaaa

A


1.10. Ma trn ng cho Mt ma trn vung c cc phn t bng khng ngoi tr cc phn t trn

ng cho chnh c gi l ma trn ng cho. V d:

=

500030002

D

1.11. Ma trn i xng Ma trn i xng l mt ma trn vung c cc phn t tho mn iu kin:

aij = aji (2.15a) hay:

A = AT (2.15b) Nh vy, ma trn i xng l ma trn c cc phn t i xng qua ng cho chnh. V d, ma trn A sau y l ma trn i xng:

=

9011043

1132A

1.12. Ma trn tam gic Ma trn c gi l ma trn tam gic trn hay ma trn tam gic di,

tng ng l cc ma trn c tt c cc phn t nm di hay nm trn ng cho chnh bng khng.

V d, cc ma trn c minh ho di y tng ng l ma trn tam gic trn A v ma trn tam gic di B:

=

900040

1132A

=

9011043002

B


2. PHP KH GAUSS Xt h phng trnh tuyn tnh c biu din dng ma trn nh sau:

Ax = b trong , A l ma trn vung kch thc (n n). Nu detA 0, th ta c th thc hin php bin i phng trnh trn bng cch nhn 2 v vi A-1 v nhn c nghim: x = A-1b. Tuy nhin, trong hu ht cc bi ton k thut, kch thc ca ma trn A l rt ln v cc phn t ca A thng l s thc vi min xc nh rt rng; do , vic tnh ton ma trn nghch o ca A l rt phc tp v d gp phi sai s do vic lm trn trong cc php tnh. V vy, phng php kh Gauss l mt cng c rt hu ch cho vic gii h phng trnh i s tuyn tnh.

2.1. M t Chng ta s bt u m t phng php kh Gauss thng qua mt v d minh ho sau y; sau tm hiu gii thut kh Gauss tng qut. Xt h phng trnh:

152 321 =++ xxx (1)

2352 321 =++ xxx (2)

415 321 =+ xxx (3)

Bc 1: bng cc php bin i tng ng kh x1 trong cc phng trnh (2) v (3), ta c h:

152 321 =++ xxx (1)

470 321 =+ xxx (21)

5200 321 =++ xxx (31)

Bc 2: kh x2 trong phng trnh (31), ta c h: 152 321 =++ xxx (1)

470 321 =+ xxx (21)

92700 321 =++ xxx (32)

y, ta nhn c h phng trnh m ma trn cc h s lp thnh ma trn tam gic trn. T phng trnh cui cng (32), ta tm c nghim x3, ln


lt th cc nghim tm c vo phng trnh trn n, (21) v (1). S nhn

c cc n s cn tm nh sau: 38;

35;

31

123 === xxx . Phng php tm

nghim khi ma trn cc h s l ma trn tam gic trn ny c gi l phng php th ngc. Cc thao tc trn c th c biu din di dng ngn gn nh sau:

9270047101521

5201047101521

415112352

1521

bng phng php th ngc, cui cng ta nhn c cc nghim:

38;

35;

31

123 === xxx

2.2. Gii thut kh Gauss tng qut Gii thut kh Gauss tng qut s c biu din thng qua cc bc thc

hin i vi mt h phng trnh tuyn tnh tng qut nh sau:

=

n

i

n

i

nnnjnnn

inijiii

nj

nj

nj

b

b

bbb

x

x

xxx

aaaaa

aaaaa

aaaaaaaaaaaaaaa

M

M

M

M

LLMMMLMMM

LLMLMLMMM

LLLLLL

3

2

1

3

2

1

321

321

33333231

22232221

11131211

(2.16)

thc hin phng php kh Gauss, ta xt cc th tc tc ng n cc ma trn h s A v ma trn cc s hng t do b nh sau:


nnnjnnn

inijiii

nj

nj

nj

aaaaa

aaaaa

aaaaaaaaaaaaaaa

LLMMMLMMM

LLMLMLMMM

LLLLLL

321

321

33333231

22232221

11131211

n

i

b

b

bbb

M

M3

2

1

(2.17)

Bc 1. S dng phng trnh th nht (hng 1) loi x1 ra khi cc phng trnh cn li. Bc ny s tc ng n cc phn t nm trong vng nh du v lm cho cc phn t t hng 2 n hng th n ca ct 1 bng khng nh php bin i (2.18) sau:

( )

( )

==

=

njibaabb

aaaaa

iii

ji

ijij

,...,2,;111

11

111

11

(2.18)

Bc 2. S dng phng trnh th hai (hng 2) loi x2 ra khi cc phng trnh cn li. Bc ny s tc ng n cc phn t nm trong vng nh du di y v lm cho cc phn t t hng 3 n hng th n ca ct 2 bng khng.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1113

12

1113

12

13

13

133

132

12

12

123

122

11131211

0

0

00

nnnjnn

inijii

nj

nj

nj

aaaa

aaaa

aaaaaaaaaaaaa

LLMMMLMMM

LLMLMLMMM

LLLLLL

( )( )

( )

( )

1

1

13

12

1

n

i

b

b

bbb

M

M (2.19)

Cc bc nh trn s c lp li n khi trong vng nh du ch cn 1 phn t. Mt cch tng qut, ti bc th k ta c:


( ) ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

+

+

+

+

++

1,

1,

11,

1,

1,

11,

1,1

1,1

11,1

23

23

233

12

12

123

122

11131211

000

000

000

000

knn

kjn

kkn

kni

kji

kki

knk

kjk

kkk

nj

nj

nj

aaa

aaa

aaa

aaaaaaaaaaaa

LLLMMMMMMMMM

LLLMMMMMMMMM

LLLMMMMMMMMM

LLLLLLLLLLLL

( )( )

( )

( )

( )

+

1

1

11

33

12

1

kn

ki

kk

b

b

b

bbb

M

M

M (2.20)

bc ny, cc phn t trong min nh du c tc ng nh php bin i

( ) ( ) ( )( )

( )

( ) ( ) ( )( )

( )

+==

+==

nkjiba

abb

nkjiaa

aaa

kkk

kk

kikk

ik

i

kkjk

kk

kikk

ijk

ij

,...,1,;

,...,1,;

11

11

11

11

(2.21)

Cui cng, sau n-1 bc nh trn, chng ta nhn c h (2.16) di dng:

=

)1(

)3(4

)2(3

)1(2

1

4

3

2

1

)1(

)3(4

)3(44

)2(3

)2(34

)2(33

)1(2

)1(24

)1(23

)1(22

114131211

0n

nnn

nn

n

n

n

n

b

bbbb

x

xxxx

a

aaaaaaaaaaaaaa

MLLLL

(2.22)

T h (2.22) ny, bng phng php th ngc t di ln ta nhn c cc nghim ca h phng trnh (2.16) nh sau ( y, tin theo di, chng ta b qua k hiu ch s trn trong cc h s ca ma trn A v b):

1211 ,,n,ni;a

xabx,;

abx

ii

n

ijjiji

inn

nn KK =

==

+= (2.23)


Chng 3 THUT TON XY DNG MA TRN CNG CHUNG

V VCT LC NT CHUNG

Vic ghp cc ma trn cng k v cc vct lc f ca cc phn t to

ra ma trn cng K v vct lc nt F chung cho c h, t thit lp h phng trnh PTHH l mt vn quan trng.

Ta s cng cc s hng ca ma trn cng ca mi phn t vo cc v tr tng ng ca ma trn cng chung v cng cc s hng ca vct lc vo vct lc chung.

Cch d nht ghp cc phn t l gn s cho mi dng v ct ca ma trn cng phn t ng vi bc t do ca phn t y, sau chng ta s lm vic qua cc s hng ca ma trn phn t; tc l cng cc s hng ny vo ma trn chung m mi dng, mi ct cng c gn ng nhng s trn. Di y ta s xt hai v d.

1. CC V D

1.1. V d 1 Mt kt cu c chia ra 8 phn t tam gic nh Hnh 3.1. Mi phn t c

3 nt; mi nt c 1 bc t do (v d nhit ).

1 2 3

5 4 6

7 8 9

12

3 4

56

7 8

e

1 2

3

Hnh 3.1


M t qu trnh ghp ni ma trn cng chung K vi gi s ch xt 3 phn t u tin: 1, 2 v 3 vi cc ma trn cng bit nh sau:

=

521263137

1k ;

=

432371218

2k ;

=

501064149

3k

Li gii 1. Xy dng bng ghp ni phn t (ng n cc nt ngc chiu kim ng h)

Bc t do Phn t

1 2 3

1 1 2 4 2 4 2 5 3 2 3 5

2. Xt tng phn t Vi phn t 1, cc dng v ct c nhn dng nh sau:

421

521263137421

1

=k

Ma trn ny c cng vo ma trn cng chung v ta s c:

MMMMMMMLLLLLL

54321

000000502100000020630103754321

=K

Ma trn cng ca phn t 2 c gn s bi:


524

432371218524

2

=k

Cc s hng ca ma trn k2 c cng thm vo ma trn chung, cho ta

MMMMMMMLLLLLL

54321

4203028501210000031207630103754321

++

++

=K

Vi phn t 3:

532

501064149532

3

=k

Cc s hng ca ma trn k3 c cng tip vo ma trn chung trn, cho ta

MMMMMMMLLLLL

L

54321

54200130213031

00064013349133

0103754321

+++

+

++

=K

Vic cng cc vct lc phn t vo vct lc chung c tin hnh hon ton tng t.


1.2. V d 2 Gi s c hai phn t 1 v 4 trong bi ton hai chiu; mi phn t c 3 nt,

mi nt c 2 bc t do (Hnh 3.2). Hy m t qu trnh ghp ni ma trn cng chung K v vct lc nt chung, theo cc ma trn cng phn t v vct lc nt phn t k1, k4, f1 v f4 cho trc nh sau:

=

242857221643168431694536309771992932647322

1k ;

=

571463

1f

=

287475572728734225768787302657421915386123

4k ;

=

542679

4f

Cc nt ca phn t 1 l: (1, 2, 5). Bc t do tng ng ca phn t l:

{ } { }TTkkjjii qqqqqqqqqqqq 10943212122121212 = Cc hng v ct ca ma trn cng phn t 1 c gn cc s ng vi bc t do tng ng ca n v cc s hng ca ma trn c cng vo cc v tr tng ng ca ma trn cng chung.

i

1 2

3

14

5 6

2 1

Hnh 3.2

2


1094321

242857221643168431694536309771992932647322

1094321

1

=k

MMMMMMMMMMMMMMLLLLLLLLLLLL

L

121110987654321

0000000000000000000000000024200008572002160000431600000000000000000000000000000000000000000000000000840000316940053000063097007100009929300026000047322

121110987654321

=K

Tin hnh tng t i vi ma trn cng ca phn t 4. Cc nt ca phn t 4 l: (5, 2, 6). Bc t do tng ng ca phn t l:

{ } { }TTkkjjii qqqqqqqqqqqq 1211431092122121212 =

121143

109

287475572728734225768787302657421915386123

121143109

4

=k


Cc s hng ca ma trn c cng vo cc v tr tng ng ca ma trn cng chung, ta nhn c kt qu nh sau:

MMMMMMMMMMMMMMLLLLLLLLLLLLL

121110987654321

287550000470072773000028005743300001277253339000012916

000000000000000000000000000000000000000000000000421412000056139478790000166097

00710000992930026000047322

121110987654321

=K

Vct lc nt ca cc phn t 1 v 4 cng c cng vo vct lc nt chung theo cch tng t:

1094321

571463

1

=f

MM121110987654321

005700001463

=F ;

121143

109

542679

4

=f

MM121110987654321

54

121600003

1063

=F


2. THUT TON GHP K V F

2.1. Nguyn tc chung Qua hai v d trn ta thy ma trn cng chung K chnh l tng ca cc

ma trn m rng [ke] ca cc phn t. Vct lc chung F cng chnh l tng ca cc vct lc m rng {fe} ca cc phn t:

[ ] { } ==e

e

e

e fFkK ; (3.1)

chun ho cc bc ghp ni, ta xy dng bng nh v index cho mi phn t. Bng index s cho bit v tr ca mi s hng ca {qn} trong {Qn}. Kch thc ca bng index l (noe edof ), vi edof l k hiu cho s bc t do ca phn t v noe l k hiu cho tng s phn t. Mi nt c mt bc t do Tr li v d 3.1, bng index chnh l bng ghp ni phn t trn.

Khi y: { }TQQQQQQ L54321= - Vi phn t 1 (e =1)

{ }( )421:),1(

421

=

=

indexQQQq T

- Vi phn t 2 (e =2) { }( )524:),2(

524

=

=

indexQQQq T

- Vi phn t 3 (e =3) { }( )532:),3(

532

=

=

indexQQQq T

Ch : k hiu (:) trong index(e,:) ch cc phn t thuc hng e ca index. Mi nt c hai bc t do Tr li v d 3.2 xt trn, bng index l:

Bc t do Phn t

1 2 3 4 5 6

1 1 2 3 4 9 10 ... ... ... ... 4 9 10 3 4 11 12


Khi y:

{ }TQQQQQQQQQQQQQ L121110987654321= - Vi phn t s 1

{ }( )1094321:),1(

1094321

=

=

indexQQQQQQq T

- Vi phn t s 4

{ }( )121143109:),4(

121143109

=

=

indexQQQQQQq T

Vi s gip ca bng index, mi s hng kij ca ma trn ke c cng vo IJK ca [K] sao cho:

I = index(e,i), vi i = 1.. sdof J = index(e,j), vi j = 1.. sdof

hoc:

jie

jeieIJ kKK += ),(index),(index (3.2)

Tng t, mi s hng fi ca {fe}c chuyn sang FI ca {F} sao cho: i

eieI fFK += ),(index (3.3)

2.2. Thut ton ghp ni phn t: Bc 1: Khi to ma trn vung [K] c kch thc (sdof sdof) v vct ct {F} c kch thc (sdof 1), vi cc s hng bng khng. Trong , sdof l k hiu cho tng s bc t do ca cc nt trong ton h. Bc 2: Vi mi phn t e (e = 1:noe), cng ng v tr mi s hng kij ca ca ma trn phn t ke vo s hng IJK ca ma trn [K]:

),(),,(;:1,; jeindexJieindexIedofjikKK jieIJIJ ===+= (3.4)

Bc 3: Vi mi phn t e (e = 1:noe), cng ng v tr mi s hng fi ca ca vct lc phn t f vo s hng FI ca vct lc chung F:

),(;:1; ieindexIedofifFF ieII ==+= (3.5)

S thut ton c m t nh Hnh 3.3 sau:


K=zero(sdof,sodf); F=zero(sdof,1);

e =1; i = 1; j = 1;

( ) ( ) ( )jikieindexieindexKieindexieindexK e ,),(),,(),(),,( +=

j edof

j = j + 1;

i = i+1;

i edof

( ) ( ) ( )ifieindexFieindexF e+= ),(),(

e = e +1;

e noe

...

...

T

T

T

F

F

F

Hnh 3.3. S thut ton ghp ni phn t


Chng 4 PHN T HU HN TRONG BI TON MT CHIU

1. M U gii bi ton mt chiu (1D) bng phng php phn t hu hn,

chng ta s s dng nguyn l cc tiu ho th nng ton phn v cc quan h: quan h ng sut-bin dng, quan h bin dng-chuyn v. Vi bi ton hai chiu (2D) v ba chiu (3D), cch tip cn cng tng t.

Trong bi ton mt chiu, cc i lng: chuyn v, bin dng, ng sut ch ph thuc vo bin x. Ta biu din chng nh sau:

( );xuu =

( );x = ( )x = (4.1) Quan h gia ng sut vi bin dng, bin dng vi chuyn v:

dxduE == ; (4.2)

Vi bi ton mt chiu, vi phn th tch dv c vit di dng: dv=Adx (4.2)

trong , A l din tch mt ct ngang.

2. M HNH PHN T HU HN C th chia thanh ra nhiu phn t, mi phn t c tit din ngang khng

i (hoc thay i). Chng hn, ta chia thanh ra 5 phn t vi cc nt c nh s t 1 n 6, cc ch s nt ny l ch s nt ton cc (Hnh 4.1a); mi phn t c hai nt 1 v 2, cc ch s nt ny l ch s nt cc b (Hnh 4.1b).

Trong bi ton mt chiu, mi nt ch c mt chuyn v theo phng x . V vy mi nt ch c mt bc t do, n nt c n bc t do.

Chuyn v tng th c k hiu l Qi ; i = 1, n; chuyn v a phng ca mi phn t c k hiu l qj; j = 1, 2

Vct ct [ ]Q Qi T= c gi l vct chuyn v chung (tng th). Lc nt c k hiu l Fi ; i = 1, n.

Vct ct [ ]F Fi T= c gi l vct lc nt chung (tng th).


ghp ni cc phn t vi nhau, ta s dng bng ghp ni cc phn t

nh sau: Bng 3.1. Bng ghp ni cc phn t

Phn t Nt

1(u) 2(cui) 1 2 3 4 5

1 2 3 4 5

2 3 4 5 6

3. CC H TRC TO V HM DNG Kho st mt phn t e nh Hnh 4.2. Theo s nh s nt cc b:

Nt th nht l 1 Nt th hai l 2

e 1 2

x2 x

x1 = -1 = 1

(b)(a)

Hnh 4.2. Phn t trong h to x v

1 2 3 4 5x

1 2 3 4 5 6

Q1 Q2 Q3 Q4 Q5 Q6

e1 2

q1 q2

Hnh 4.1a. Ch s ton cc Hnh 4.1b. Ch s cc b

Ch s a phng

Ch s chung


Theo k hiu, x = x1 l ta ca nt th nht; x = x2 l ta ca nt th hai. Ta nh ngha h ta qui chiu (hay chun ho) c k hiu l nh sau:

( ) 12 112

= xxxx

====1

1

2

1

xxxx

(4.3)

Vy: [ ] [ ]21 :1:1 xxx Ta s dng h ta a phng ny xc nh hm dng vi mc ch ni suy ra trng chuyn v trong cc phn t.

By gi trng chuyn v cn tm cho mt phn t s c ni suy bng mt php bin i tuyn tnh (Hnh 4.3).

thc hin c php ni suy ny, cn a vo mt hm dng tuyn

tnh:

( ) ( )2

1;2

121

+== NN (4.4) Cc hm dng c minh ho trn Hnh 4.4. th ca hm dng N1 trn Hnh 4.4a c suy ra t phng trnh (4.4): N1 = 1 ti = -1 v N1 = 0 ti = 1. Tng t ta c th ca N2.

1

N1

Hnh 4.4. (a), (b). Hm dng N1, N2; (c). Ni suy tuyn tnh

-1 0 1

1

N2

u

1 2

21

1

=N

21

2+

=N

q1 q2

u=N1q1+N2q2

2 1 1 2

(a) (b) (c) -1 0 1

e 1 2

u2 u1

Hnh 4.3. Ni suy tuyn tnh trng chuyn v ca mt phn t

q2 q1

e1 2


Mt khi cc hm dng c xc nh, trng chuyn v ca phn t s c biu din qua cc chuyn v nt q1 v q2 nh sau:

2211 qNqNu += (4.5) Hoc di dng ma trn:

u = Nq (4.6)

Trong : [ ]N N N= 1 2, [ ]Tqqq 11= (4.7)

Trong biu thc trn, q l vct chuyn v ca phn t. T (4.5), ta thy u = q1 ti nt 1; u = q2 ti nt 2 v u bin thin tuyn tnh trong phn t (Hnh 4.4c).

Ta bit: 12

11 0

11 qu

NN

xx =

=

===

22

12 1

01 qu

NN

xx =

=

===

By gi ta ni suy ta x nh cc hm dng 21 , NN

2211 xNxNx += (4.8)

So snh:

+=

+=

2211

2211

qNqNuxNxNx

ta thy chuyn v u v to x c ni suy trn cng phn t nh cng cc hm dng N1 v N2. Trong trng hp ny, ta c php biu din ng tham s. Ch : Cc hm dng cn tho mn:

1) o hm bc nht phi hu hn, 2) Chuyn v phi lin tc trn cc bin ca phn t.

Mt khc:

dxd

ddu

dxdu

== (4.9) m:

12

2xxdx

d

=

(4.10)


suy ra

212211 21

21 qqqNqNu ++=+= (4.11)

221 qq

ddu +

= (4.12)

( )2112

1 qqxx

+

= (4.13)

do :

[ ]111;12

==

xxBBq (4.14)

Trong ma trn B c gi l ma trn bin dng-chuyn v ca phn t. Theo nh lut Hooke, ta c biu thc tnh ng sut:

= EBq (4.15) Ch :

B, , l cc i lng hng s; Cc biu thc u = Nq; = Bq; = EBq m t chuyn v, bin dng v ng sut qua cc gi tr chuyn v nt ca phn t. Ta s th cc biu thc ny vo biu thc th nng ca thanh thit lp ma trn cng v ma trn lc nt ca phn t.

4. TH NNG TON PHN p dng cng thc (1.3) - Chng 1, ta tnh c th nng ton phn ca

thanh:

=

=n

ii

Ti

L

T

L

T

L

T PuTdxuAdxfuxdA12

1 (4.16)

Khi vt th c chia ra lm nhiu phn t hu hn, th

=

=n

iii

e e

T

e e

T

e e

T PQTdxuAdxfuxdA12

1 (4.17)


5. MA TRN CNG PHN T Gi:

xdAUe

Te = 21

l th nng bin dng ca phn t, ta c:

xdAqBEBqUe

eeTT

e = 21 qxdABEBqU

eee

TTe

= 21 (4.18) Ch rng: Ae, Ee v B l cc i lng hng s, v

dldxdxxdx e22

12== , vi: 12;11 xxle =

Khi y, ta c biu thc ca nng lng bin dng ca phn t:

qdBBElAqU TeeeT

e

=

1

1221

vi:

[ ]11112

=

xxB

ta c:

qlEAqUe

eeTe

=

1111

21

Gi:

=

1111

e

eee

lEAk (4.19)

l ma trn cng ca phn t . Khi , biu thc th nng (4.18) c biu din dng thu gn nh sau:

qkqU eTe 21

= (4.20)


6. QUI I LC V NT Khi vt th c ri rc ha bng cc phn t hu hn vi cc nt xc

nh, ta phi qui i cc loi lc tc dng v nt. Ln lt xt tng thnh phn biu din cng ca ngoi lc trong biu thc

th nng (4.17), ta c: - Cng do lc khi:

= e

e

ee

T

e

T

dxNfA

dxNfAqAdxfu

2

1

m:

=

+=

=

=

221

2

221

21

12

1

11

ee

e

ee

e

ldldxN

ldldxN

eTeeT

e

T fqlfAqAdxfu =

= 112 Vi:

=

11

2eee lfAf (4.21)

l lc th tch quy i v nt ca phn t - Cng do lc din tch:

( ) eTe

eTT

e

T TqdxNT

dxNTqdxTqNqNdxTu =

=+=

2

1

2211

Vi:

=

11

2ee lTT (4.22)

c gi l lc din tch qui i v nt ca phn t


Cui cng, biu thc c vit gn di dng

FQKQQ TT =21 (4.23)

Trong : Q l vct chuyn v nt chung,

K l ma trn cng chung, c xc nh t cc ma trn cng ke ca cc phn t:

Kke

e F l vct lc nt chung, c xc nh t cc vct lc nt: fe, Te, P ca cc phn t:

( ) FPTfe

ee ++ Vi phn t mt chiu v mi nt c mt bc t do, ta vn s dng bng ghp ni phn t trn thit lp ma trn cng K v vct lc F.

7. IU KIN BIN, H PHNG TRNH PHN T HU HN Sau khi ri rc ha vt th nh phng php phn t hu hn, ta xc nh

c biu thc th nng ton phn (4.23). By gi phi xy dng phng trnh cn bng t xc nh cc

chuyn v nt, sau tnh ng sut, bin dng v cc phn lc lin kt. Bng cch cc tiu biu thc th nng i vi Q, tc l cho cho th

nng bin dng "chu" iu kin bin, ta s thu c phng trnh cn bng. Di y ta trnh by cch nhp iu kin bin. Phng php ny c p

dng khng ch cho bi ton mt chiu m cn cho c bi ton hai, ba chiu. iu kin bin thng c dng:

Qi = ai Biu thc trn c ngha l chuyn v Qi phi bng ai . y, chng ta s p dng phng php kh nhp cc iu kin bin. Kho st trng hp n gin: Q1 = a1. Vi mt kt cu c n bc t do, ta c


{ }TnQQQQ L21= { }TnFFFF L21=

Ma trn cng tng th c dng:

=

nnnn

n

n

KKK

KKKKKK

K

LMLMM

LL

21

22221

11211

(4.24)

K l ma trn i xng Ta vit biu thc ca th nng di dng khai trin nh sau:

( )nnnnnnnnnn

nn

nn

FQFQFQ

QKQQKQQKQ

QKQQKQQKQQKQQKQQKQ

+++

++++

++++

+++

= K

KKKKKKKKKKKKKKK

KK

2211

2211

2222221212

1121211111

21

(4.25)

Thay Q1 = a1 vo phng trnh trn, ta c:

( )nnnnnnnnnn

nn

nn

FQFQFa

QKQQKQaKQ

QKQQKQaKQQKaQKaaKa

+++

++++

++++

+++

= K

KKKKKKKKKKKKKKK

KK

2211

2211

2222221212

1121211111

21

(4.26)

Ch rng chng ta kh chuyn v Q1 trong biu thc ca th nng trn. p dng iu kin cc tiu th nng:

niQi

,...,2;0 ==

(4.27)

ta thu c:

=+++

=+++

=+++

113322

13133333232

12122323222

aKFQKQKQK

aKFQKQKQKaKFQKQKQK

nnnnnnn

nn

nn

KLLLLLLLLLLLLLL

KK

(4.28)


Khi y, h phng trnh PTHH c biu din nh sau:

=

11

1313

1212

3

2

32

33332

22322

aKF

aKFaKF

Q

QQ

KKK

KKKKKK

nnnnnnn

n

n

MML

MLMMLL

(4.29)

Nhn xt: Ma trn cng (n-1)(n-1) trn c nhn t ma trn cng (nn) ban u (4.23) bng cch b i hng th nht v ct th nht (v Q1 = a1). H phng trnh (4.28) c vit di dng c ng:

KQ = F (4.30) Ma trn K trong (4.30) l ma trn khng k d cn ma trn K ban u (4.24) l ma trn k d (det K=0). p dng phng php kh Gauss (xem chng 2) gii h phng trnh (4.30), ta s tm c chuyn v Q; Nh bng thng tin ghp ni phn t gii thiu phn u, ta s xc nh c chuyn v nt q ca phn t t chuyn v chung Q tm c trn. p dng cng thc EBq= ta tm c ng sut;

xc nh phn lc lin kt R1, ta vit phng trnh cn bng cho nt 1:

111212111 RFQKQKQK nn +=+++ K (4.31) Trong Qi c xc nh, F1 l lc tc dng ti ni t lin kt cng bit.

8. V D

V d 4.1. Cho mt trc bc chu tc dng ca lc P = 10 N (hnh 4.5a). Bit tit din

cc on: A1=20 mm2; A2 = 10 mm2; chiu di cc on l1 = l2 = 100 mm; v mun n hi: E1 = E2 = 200 gPa. Hy xc nh chuyn v ti B v C; bin dng, ng sut trong cc on trc AB, BC.


Li gii Chia trc lm hai phn t: 1 v 2, Hnh 4.5b. 1. Bng ghp ni phn t c thit lp nh sau:

Phn t Nt i Nt j

1 1 2

2 2 3

2. Xc nh ma trn cng ca phn t 1: k1 v 2: k2

mmN

lEAk 41

111 104444

1111

=

=

mmN

lEAk 42

222 102222

1111

=

=

3. Ma trn cng chung K:

mmNK 410

2202244

044

+

=

4. Vct lc nt chung F: F = [0 0 10]T 5. H phng trnh phn t hu hn:

A

1

Hnh 4.5. (a) Trc bc chu ko ng tm; (b) S phn t

P=10 kN x

B C

1 2

2 3

(a)

(b)


=

+

100

2202244

04410

3

2

14

R

QQQ

6. p t iu kin bin: Do Q1 = 0 (lin kt ngm ti A), do ta loi dng 1 v ct 1 trong h phng trnh trn. Cui cng ta thu c h phng trnh:

=

100

2226

103

24

QQ

7. Xc nh chuyn v, bin dng, ng sut Gii h phng trnh trn ta c:

Q2 = 0,25 10-3 mm Q3 = 0,75 10-3 mm

p dng cng thc (4.31), ta tm c phn lc lin kt: R1 =104 (-4 Q2 ) = -10 N

Bin dng c tnh cho mi phn t 1 = (-q1 + q2 )/l = 0,25 x10-5 /100 = 2,5 x10-6

2 = (-q2 + q3 )/l = 5 x10-6

ng sut c tnh cho mi phn t 1 = E 1 = 0,5 N/mm2 2 = E 2 = 1 N/mm2

V d 4.2. Cho mt trc bc chu lin kt ngm 2 u v tc dng ca lc P = 200 kN

(hnh 4.6a). Bit tit din cc on: A1=2400 mm2; A2 = 600 mm2; chiu di cc on l1 = 300mm, l2 = 400 mm; v mun n hi: E1 = 70 gPa, E2 = 200 gPa. Hy xc nh chuyn v ti B; ng sut trong cc on trc AB, BC v phn lc ti A v C.


Li gii Chia trc lm hai phn t: 1 v 2, nh hnh 4.5b v d 4.1. 1. Bng ghp ni phn t c thit lp nh sau:

Phn t Nt i Nt j

1 1 2

2 2 3

2. Xc nh ma trn cng ca phn t 1: k1 v 2: k2

mmN

lEAk

=

=

1111

30010702400

1111 3

1

111

mmN

lEAk

=

=

1111

40010200600

1111 3

2

222

3. Ma trn cng chung K:

mmNK 310

3003003008605600560560

=

4. Vct lc nt chung F:

F = [R1 200103 R3]T

5. H phng trnh phn t hu hn:

2

A Hnh 4.6. Trc bc chu ko ng tm

x

1 B C

P=200


=

3

31

3

2

13 10200

30030003008605600560560

10R

R

QQQ

6. p t iu kin bin:

Do Q1 = 0 (lin kt ngm ti A) v Q3 = 0 (lin kt ngm ti C) , do ta loi dng 1, ct 1 v dng 3, ct 3 trong h phng trnh trn. Cui cng ta thu c phng trnh:

860 Q2 = 200

7. Xc nh chuyn v, bin dng, ng sut

Gii phng trnh trn ta c: Q2 = 0,23257 mm

p dng cng thc (4.31), ta tm c cc phn lc lin kt: R1 =103 (-560 Q2 ) = -130,233 KN R3 =103 (-300 Q2 ) = -69,767 KN

Bin dng c tnh cho mi phn t 1 = (-q1 + q2 )/l1 = 0,23257 /300 = 7,752 10-4 2 = (-q2 + q3 )/l2 = -0,23257 /400 = 5,81410-4

ng sut c tnh cho mi phn t 1 = E11 = 54,26 N/mm2

2 = E2 2 = 116,28 N/mm2

V d 4.3. Cho mt trc trn chu lin kt ngm ti A, khe h gia u C v thnh

cng l 1,2mm, chu tc dng ca lc P = 60 kN ti B (hnh 4.7). Bit tit din ca thanh l A=250 mm2; v mun n hi: E = 2103N/mm2 Hy xc nh chuyn v ti B; v phn lc ti A v C.


Li gii y, ta xem nh thc hin bc kim tra kt lun rng, trong qu trnh bin dng, u C ca trc tip xc vi thnh cng v tip tc bin dng. Tng t cc v d trn, ta chia trc lm hai phn t (1) v (2). Khi , ma trn cng chung K c xc nh nh sau:

mmNK 3

3

10110121

011

1501020250

=

Vct lc nt chung F: F = [R1 60103 R3]T H phng trnh phn t hu hn:

=

3

31

3

2

13

1060110121

011

1501020250

R

R

QQQ

p t iu kin bin: Do Q1 = 0 (lin kt ngm ti A) v Q3 = 1,2 (khe h ti C) , do ta loi dng 1, ct 1. Cui cng ta thu c h phng trnh:

3,3333104(2 Q2 1,2) = 60103 3,3333104(- Q2 + 1,2) = R3

Xc nh chuyn v, bin dng, ng sut Gii phng trnh trn ta c:

Q2 = 1,5 mm; R3 =3,3333104 (-Q2 + 1,2) = - 10 kN

R1 =3,3333104 (- Q2) = -50 kN

2A

Hnh 4.7. Tnh kt cu bng phng php PTHH

x

1 B C150mm150mm 1,2mm

P=60 KN


9. CHNG TRNH TNH KT CU MT CHIU - 1D Chng trnh ngun

%---------------------------------------------------------------------------- % Chuong trinh so 1, chuong 4. (P4_1) %---------------------------------------------------------------------------- % Tinh chuyen vi nut trong cac ket cau 1-D % % Mo ta cac bien % k = ma tran do cung phan tu % f = vecto luc nut phan tu % kk = ma tran do cung tong the % ff = vecto luc nut tong the % gcoord = toa do nut % nodes = ma tran chi so nut cua moi phan tu % index = vecto chuyen vi nut chung o moi phan tu %---------------------------------------------------------------------------- %------------------------------------ % Cac tham so dau vao %------------------------------------

clear edof=1; % edof = so bac tu do tai nut noe=input('Nhap so phan tu:'); % noe = so phan tu % Nhap du lieu: cac thong so hinh hoc cua ket cau va co tinh vat lieu for i=1:noe Doan_truc=i los(i)=input('Nhap chieu dai (don vi mm) cua doan '); E(i)=input('Nhap modul dan hoi keo nen (N/mm2) cua doan (phan tu)'); A(i)=input('Nhap tiet dien mat cat ngang (mm2) cua doan (phan tu)'); end % Nhap du lieu: cac thong tin ve chi so nut phan tu tuong ung voi chi so % nut tong the, phuc vu cho viec ghep noi phan tu for i=1:noe


Phan_tu = i index(i,1)=input('Chi so nut toan cuc cua nut 1:'); index(i,2)=input('Chi so nut toan cuc cua nut 2:'); end % Nhap du lieu: cac thong tin ve tai trong tac dung. % 1. Tai trong tap trung nof=input('Nhap so luc tap trung:'); % nof=Number Of Force for i=1:nof Luc_thu =i temp_f(i)=input('Gia tri luc (don vi N): '); force_pos(i)=input('Vi tri dat luc (nut so): '); end % Thong tin ve lien ket noc=0; % noc=Number Of Clamp while ((noc==0)|(noc>2)) noc=input('So luong lien ket (1 hoac 2):'); end for i=1:noc c(i)=input('Vi tri dat lien ket (nut dat lien ket): '); end % Tinh ma tran do cung phan tu for i=1:noe k(1,1,i)=E(i)*A(i)/los(i); k(1,2,i)=-k(1,1,i); k(2,1,i)=-k(1,1,i); k(2,2,i)=k(1,1,i); end for e=1:noe % In ma tran do cung cac phan tu k(e,:) end


% Xay dung ma tran do cung tong the non=noe+1; % non = Number Of Nodes sdof=non*edof; kk=zeros(sdof,sdof); for row_indx=1:non for e=1:noe for n1=1:2 if (index(e,n1)==row_indx) for col_indx=1:non for n2=1:2 if (index(e,n2)==col_indx) kk(row_indx,col_indx)=kk(row_indx,col_indx)+ k(n1,n2,e); end end end end end end end kk % In ma tran do cung tong the % Tinh ma tran luc nut phan tu f=zeros(noe,2); for e=1:noe for i=1:nof if (index(e,1)==force_pos(i)) f(e,1)=temp_f(i); end if (index(e,2)==force_pos(i)) f(e,2)=temp_f(i); end


end end for i=e:noe % In vecto luc nut phan tu f(e,:) end % Xay dung vecto luc nut chung ff=zeros(sdof,1); for node=1:non for e=1:noe for n=1:2 if (index(e,n)==node) ff(node)=f(e,n); end end end end ff % In vecto luc nut chung % Ap dat dieu kien bien for node=1:noc kk(c(node),:)=0; kk(:,c(node))=0; ff(c(node))=0; kk(c(node),c(node))=1; end kk ff Q=kk\ff;


10. BI TP

4.1. Cho kt cu 1D c ri rc ho bi 2 phn t mt chiu nh Hnh 4.10.1 di y.

a. Hy chng t rng ma trn cng tng th K l ma trn k d. b. Ch ra mt vct chuyn v Q0 0 m tho mn KQ0 = F = 0. Bng cch

m t qua hnh v, hy phn tch ngha ca cc chuyn v ny. V ch ra nng lng bin dng n hi trong cu trc trng hp ny ?

c. Chng minh dng tng qut rng, vi bt k vct chuyn v Q 0 l nghim ca h phng trnh KQ = 0, vi K l ma trn k d.

4.2. Xt kt cu thanh bng thp, mun n hi E=200109N/m2. C

lin kt v chu lc nh Hnh 4.10.2. Xc nh cc chuyn v nt (cc chm en trn hnh), ng sut trong cc phn t v cc thnh phn phn lc ti ngm. Hy gii bi ton bng tay v nghin cu k Chng trnh cho, sa i li mt s im nu cn thit v b sung phn chng trnh tnh ng sut trong cc phn t; thc hnh tnh ton bng chng trnh v so snh kt qu.

4.3. Xt kt cu thanh bng thp, mun n hi E=200109N/m2. C

lin kt v chu lc nh Hnh 4.10.3. Xc nh cc chuyn v nt, ng sut trong cc phn t v cc thnh phn phn lc ti ngm.

150mm150mm 300mm

x P=300 kN250mm2 400mm

2

Hnh 4.10.2

1 2 3x

Hnh 4.10.1


4.4. Xt kt cu lin kt v chu lc nh Hnh 4.10.4. Thanh nm ngang

c xem nh l tuyt i cng, cc thanh treo c lm bng thp v nhm, c mun n hi nh ch ra trn hnh v. Tnh ng sut trong mi thanh treo.

Hnh 4.10.4

thp 22 cm

E=200109 N/m2

Nhm 24 cm

50cm 40 cm 30 cm 20 cm

60 KN

Thanh tuyt i cng, trng lng khng ng k

E=70109 N/m2

150mm150mm 200mm 200mm

250mm2 400mm2

P=300 kN P=600 kN x

3.5mm

Hnh 4.10.3


Chng 5 PHN T HU HN TRONG TNH TON H THANH PHNG

1. M U Trong chng ny chng ta s p dng phng php phn t hu hn

tnh ton h thanh phng (h gm n thanh lin kt vi nhau bi cc khp quay). H thanh phng in hnh c trnh by trn Hnh 5.1.

Trong h thanh, ti trng hoc phn lc lin kt t cc khp ni; b qua

ma st trong cc khp ni. R rng, mi phn t ca h thanh hoc chu ko, hoc chu nn.

Ta c th gp h thanh tnh nh hoc siu tnh.

2. H TO A PHNG, H TO CHUNG H thanh khc vi cc kt cu mt chiu xt trong Chng 4 ch:

trong h thanh, cc phn t (cc thanh) c cc phng khc nhau. c th tnh n s khc nhau v phng ca cc phn t trong h, ta cn phi a ra khi nim h to a phng v h to chung.

Mt phn t thanh c m t trong h to a phng v h to chung nh trong Hnh 5.2.

2 3 4 5

6 7 8

1

Q2

Q1 Q5 Q7 Q9

Q4

Q3

Q6 Q8 Q10

Q15

Q16

Q13

Q14Q11

Q12

Hnh 5.1. H thanh phng


Trong s nh s nt a phng, hai nt ca phn t c nh s 1 v s 2. H to a phng hng theo trc x, chy t nt 1 n nt 2. Tt c cc i lng trong h to a phng c k hiu bi du (). H to chung (x,y) l c nh v khng ph thuc vo phng ca cc phn t. Trong h to chung, mi nt cng c hai bc t do. Chng hn, nt j s c hai chuyn v l Q2j-1 v Q2j. Gi q1 v q2 l cc chuyn v ca nt 1 v 2 tng ng trong h to a phng. Ta k hiu vct chuyn v trong h to a phng bi:

q = [q1 , q2]T (5.1) Trong h to chung, vct chuyn v c 4 thnh phn:

q = [q1, q2 , q3 , q4 ]T (5.2) Ta i tm quan h gia q v q. D thy

q1 = q1 cos + q2 sin (5.3a) q2 = q3 cos + q4 sin (5.3b)

K hiu l = cos (5.4a) m = sin (5.4b)

x

y

x

1

(a)

q1q2

q4

q3

q1cosq2sin

q3cos

q4sin

q1

q2

(b)

Hnh 5.2. Phn t thanh trong h to a phng (a) v trong h to chung (b)


Ta c th vit q = L q (5.5)

Trong L l ma trn chuyn v, c vit di dng:

=

mlml

L00

00 (5.6)

3. MA TRN CNG PHN T Cc phn t trong h thanh u l cc phn t mt chiu. V vy, ta p dng nhng kt qu ca chng 4 vo h thanh. Trong h to a phng, ta xc nh c ma trn cng ca phn t

=

1111

'e

ee

lAEk (5.7)

thit lp ma trn cng ca phn t trong h to chung, ta ch ti biu thc nng lng bin dng ca phn t

'''21 qkqU Te = (5.8)

Thay q = Lq vo biu thc trn, ta c

[ ]qLkLqU TTe '21

= (5.9)

Cui cng, nng lng bin dng trong h to chung c vit di dng:

qkqU Te 21

= (5.10)

Trong k l ma trn cng ca phn t trong h to chung v k = LT k' L (5.11)

Thay biu thc ca L t (5.6) v ca k' t (5.7) vo (5.11), ta c

=

22

22

22

22

mlmmlmlmllmlmlmmlmlmllml

lAEke

ee (5.12)


T cc ma trn cng ca cc phn t v nh bng ghp ni phn t, ta s thu c ma trn cng chung ca c h thanh.

4. NG SUT Nh lu trn, mi phn t trong h thanh hoc chu ko, hoc chu nn. Do , ng sut trong thanh c xc nh bi:

= Ee Hoc

[ ] [ ]LqlE

qq

lE

lqqE

e

e

e

e

ee 11'

'11''

2

112=

=

=

Th biu thc ca L t (5.6) vo biu thc trn ta c:

[ ]qmlmllE

e

e= (5.13)

Nh vy, sau khi tm c chuyn v, ta s xc nh c ng sut trong mi phn t ca h thanh.

5. V D Kho st h gm hai thanh chu lc P nh hnh di. Cc thanh c cng din tch mt ct ngang v cng vt liu. Xc nh chuyn v ti im t lc.

x 300

y

L, A, E 1 2 300

P

3

2

1

(b)(a) Hnh 5.3. (a) Kt cu bng chu lc, (b) s phn t


Li gii 1. M hnh. Ta m hnh ho h thanh bi 2 phn t hu hn; mi nt phn t c 2 bc t do. 2. Xc nh ma trn cng ca cc phn t

p dng cng thc (5.12), ta tnh c cc ma trn cng ca cc phn t. Vi phn t 1: LLml ===== 1;0sin;1cos

=

0000010100000101

1

LEAk

Vi phn t 2: LLml3

2;21sin;

23cos 2 ===

==

=

83

83

83

83

83

833

83

833

83

83

83

83

83

833

83

833

2

LEAk

T y, ta thit lp c ma trn cng chung K v h phng trnh:


=

+

6

5

2

1

4

3 0

00

00

83

83

83

8300

83

833

83

83300

83

83

83

8300

83

833

83

833101

000000000101

RR

P

RR

QQ

LEA

p dng iu kin bin: Q1 = Q2 = Q5 = Q6 =0, ta thu c h phng trnh PTHH:

=

+

PQQ

LEA 0

83

83

83

8331

4

3

Gii h phng trnh trn, ta c:

+

=

EALP

EALP

QQ

33

8

3

4

3

Thay cc gi tr chuyn v trn vo (5.14), ta tm c phn lc lin kt: { } { }PRRRRR 13036511 ==

6. CHNG TRNH TNH H THANH PHNG

V d Kho st h thanh chu lc nh sau (Hnh 5.4). Cc thanh c cng din tch

mt ct ngang A = 2,5cm2 v cng vt liu, vi E = 2105N/cm2. Xc nh chuyn v ti im t lc. Xc nh ma trn cng ca cc phn t v ma trn cng chung; chuyn v ti im t lc v ng sut trong cc thanh v cc phn lc lin kt.


Chng trnh ngun %---------------------------------------------------------------------------- % Chuong trinh so 1, chuong 5 - (P5_1) %---------------------------------------------------------------------------- % Tinh toan chuyen vi nut, ung suat trong cac thanh cua he thanh phang % tinh phan luc lien ket tai cac lien ket cua he thanh phang chiu luc % su dung phan tu thanh % (Hinh. 5.4 mo ta mo hinh PTHH tinh he thanh phang) % % Mo ta cac bien % k = ma tran do cung phan tu % f = vecto luc nut phan tu % kk = ma tran do cung tong the % ff = vecto luc nut tong the % disp = vecto chuyen vi nut tong the % eldisp = (element_disp) vecto chuyen vi nut phan tu % stress = ma tran ung suat % strain = ma tran bien dang % gcoord = toa do nut % nodes = ma tran chi so nut cua moi phan tu % index = vecto chuyen vi nut chung o moi phan tu %----------------------------------------------------------------------------

x

y 1

2

P=100KN

2

1

Hnh 5.4. Tnh kt cu bng phng php PTHH

3

4 34

Q2Q3

Q4

Q6Q5Q7

Q8

100 cm

75 cm

Q1


%------------------------------------ % Cac tham so dau vao %------------------------------------ clear % Type of geometric construction type_geometric =1; switch (type_geometric) case 1 length=1000; % mm emodul=100e3; % MPa (N/mm^2) area=(2.5^2)e2; % mm^2 force=100e3; % N noe=4; % noe = Number Of Elements(segments) non=4; % non = Number Of Nodes lcoord(1,1,1)=0; lcoord(1,2,1)=0; lcoord(2,1,1)=length; lcoord(2,2,1)=0; lcoord(1,1,2)=lcoord(2,1,1); lcoord(1,2,2)=lcoord(2,2,1); lcoord(2,1,2)=lcoord(2,1,1); lcoord(2,2,2)=length*3/4; lcoord(1,1,3)=0; lcoord(1,2,3)=0; lcoord(2,1,3)=length; lcoord(2,2,3)=length*3/4; lcoord(1,1,4)=0; lcoord(1,2,4)=length*3/4; lcoord(2,1,4)=length; lcoord(2,2,4)=length*3/4; % Chi so nut phan tu theo chi so nut chung index(1,1)=1; index(1,2)=2; index(2,1)=2; index(2,2)=3;


index(3,1)=1; index(3,2)=3; index(4,1)=4; index(4,2)=3; end % Tinh chieu dai cac thanh l(e) va ma tran chuyen doi he co so: % trans_mat(e). for i=1:noe L(i)=sqrt((lcoord(2,1,i)-lcoord(1,1,i))^2+(lcoord(2,2,i)-lcoord(1,2,i))^2); l(i)=(lcoord(2,1,i)-lcoord(1,1,i))/L(i); m(i)=(lcoord(2,2,i)-lcoord(1,2,i))/L(i); % Ma tran chuyen doi he toa do trans_mat(1,1,i)=l(i); trans_mat(1,2,i)=m(i); trans_mat(1,3,i)=0; trans_mat(1,4,i)=0; trans_mat(2,1,i)=0; trans_mat(2,2,i)=0; trans_mat(2,3,i)=l(i); trans_mat(2,4,i)=m(i); % Ma tran chuyen doi he toa do ung suat stress_trans(i,1)=-l(i); stress_trans(i,2)=-m(i); stress_trans(i,3)=l(i); stress_trans(i,4)=m(i); % Modul dan hoi cua cac thanh E(i)=emodul; A(i)=area; % Tiet dien ngang cua cac thanh end % Tinh ma tran do cung phan tu trong he toa do dia phuong for i=1:noe k_local(1,1,i)=(E(i)*A(i)/L(i)); k_local(1,2,i)=-k_local(1,1,i); k_local(2,1,i)=-k_local(1,1,i); k_local(2,2,i)=k_local(1,1,i);


end % Tinh ma tran do cung phan tu trong he toa do chung trans_trans_mat=permute(trans_mat,[2,1,3]); for i=1:noe k(:,:,i)=trans_trans_mat(:,:,i)*k_local(:,:,i)*trans_mat(:,:,i); end k % In ma tran do cung phan tu % Xay dung ma tran do cung tong the edof=2; %edof: so bac tu do cua 1 node sdof=non*edof; kk=zeros(sdof,sdof); for row_indx=1:non for e=1:noe for n1=1:2 if (index(e,n1) = = row_indx) for col_indx=1:non for n2=1:2 if (index(e,n2)==col_indx) for i=1:2 for j=1:2 kk((row_indx-1)*edof+i,(col_indx-1)*edof+j)=... kk((row_indx-1)*edof+i,(col_indx-1)*edof+j)+... k((n1-1)*edof+i,(n2-1)*edof+j,e); end end end end end end end end end kkk=kk; kk % In ma tran do cung tong the % Tinh ma tran luc nut phan tu


f=zeros(noe,2*edof); f(2,1)=20000 f(2,4)=-25000; f % In ve to luc nut phan tu % Xay dung ve to luc nut chung ff=zeros(sdof,1); for row_indx=1:non for e=1:noe for n=1:2 % 2:so node/phan tu if (index(e,n)==row_indx) for i=1:2 ff((row_indx-1)*edof+i)=ff((row_indx-1)*edof+i)... +f(e,(n-1)*edof+i); end end end end end % In vec to luc nut chung ff % Ap dat dieu kien bien for i=1:sdof disp(i)=1; end disp(1)=0; disp(2)=0; disp(4)=0; disp(7)=0; disp(8)=0; for i=1:sdof if (disp(i)==0) kk(i,:)=0; kk(:,i)=0; ff(i)=0; kk(i,i)=1;


end end kk ff % Giai he PT PTHH xac dinh chuyen vi nut disp=kk\ff; % In vec to chuyen vi nut chung disp % Xac dinh chuyen vi nut trong cac thanh for e=1:noe for i=1:2 % 2 nut for j=1:edof % edof=2: 2 bac tu do/nut eldisp(e,(i-1)*edof+j)=disp((index(e,i)-1)*edof+j); end end end eldisp % Tinh Ung suat trong cac thanh stress=zeros(noe,1); for e=1:noe stress(e)=(E(e)/L(e))*stress_trans(e,:)*eldisp(e,:)'; end stress % Tinh Phan luc lien ket tai cac goi R=zeros(sdof,1); R=kkk*disp; R


Kt qu s

k(:,:,1) =

1.0e+011 *

1.2916 0 -1.2916 0

0 0 0 0

-1.2916 0 1.2916 0

0 0 0 0

k(:,:,2) =

1.0e+011 *

0 0 0 0

0 1.7221 0 -1.7221

0 0 0 0

0 -1.7221 0 1.7221

k(:,:,3) =

1.0e+010 *

6.6128 4.9596 -6.6128 -4.9596

4.9596 3.7197 -4.9596 -3.7197

-6.6128 -4.9596 6.6128 4.9596

-4.9596 -3.7197 4.9596 3.7197

k(:,:,4) = 1.0e+011 *

1.2916 0 -1.2916 0 0 0 0 0 -1.2916 0 1.2916 0 0 0 0 0


kk = 1.0e+011 *

1.9529 0.4960 -1.2916 0 -0.6613 -0.4960 0 0 0.4960 0.3720 0 0 -0.4960 -0.3720 0 0 -1.2916 0 1.2916 0 0 0 0 0 0 0 0 1.7221 0 -1.7221 0 0 -0.6613 -0.4960 0 0 1.9529 0.4960 -1.2916 0 -0.4960 -0.3720 0 -1.7221 0.4960 2.0941 0 0 0 0 0 0 -1.2916 0 1.2916 0 0 0 0 0 0 0 0 0

disp = 1.0e-006 * Q1 0 Q2 0 Q3 0.1549 Q4 0 Q5 0.0323 Q6 -0.1270 Q7 0 Q8 0


eldisp=

Phn t Nt 1 Nt 2

Phng x 1.0e-006 *

Phng y 1.0e-006 *

Phng x 1.0e-006 *

Phng y 1.0e-006 *

1 0 0 0.1549 0

2 0.1549 0 0.0323 -0.1270

3 0 0 0.0323 -0.1270

4 0 0 0.0323 -0.1270

stress=

Phn t ng sut (mPa)

1 31.0001

2 -33.9063

3 -8.0729

4 6.4583

R =

Ri Phn lc lin kt (N) 1.0e+004 *

1 -1.5833

2 0.3125

3 2.0000

4 2.1875

5 -0.0000

6 -2.5000

7 -0.4167

8 0


7. BI TP

5.1. Mt kt cu thanh ging nh trn Hnh 5.7.1. Vt liu cc thanh bng thp, c mun n hi E=200gPa. Xc nh ma trn cng tng th ca h.

5.2. Mt kt cu gin gm 3 thanh c nh s (nt v thanh) nh trn Hnh 5.7.2. Vt liu ca cc thanh I v II l nhm, vt liu ca thanh III l thp. Tit din ca thanh I l 15cm2 v tit din ca thanh II v III l 8cm2. Xc nh chuyn v ca nt 2 v ng sut trong cc thanh. Gii bi ton bng tay v bng cch s dng phn mm tnh ton kt cu tng ng. Khi gii bng tay yu cu biu din ma trn cng tng th di dng ton b v di dng rt gn. Cho Enhm = 70gPa, Ethp = 210gPa.

Hnh 5.7.1

1000 mm2

1250 mm2

P

500 mm

750 mm

Q2i

Q2i-1 1

2

3

i

0,7m

1 2

8 kN

3 4

5 kN

y

x

I

III II

0,5m

1m

Hnh 5.7.2. Dn chu lc


5.3. Mt kt cu gin gm 5 thanh c nh s (nt v thanh) nh trn Hnh 5.7.3. Vt liu ca cc thanh u l thp v c mun n hi Ethp = 210gPa. Tit din ca thanh I, II v III l 15cm2 v tit din ca thanh IV v V l 8cm2. Xc nh chuyn v ca cc nt v ng sut trong cc thanh. Gii bi ton bng tay v bng cch s dng chng trnh tnh ton kt cu tng ng. Khi gii bng tay nn ch n tnh i xng ca kt cu. Cho a = 0,5 m; = 600; P = 2kN; Q = 4kN.

5.4. Mt cy cu ng st c ghp t cc thanh thp, tit din ca cc thanh thp bng nhau v bng 3250 mm2. Mt on tu dng trn cu, cu phi chu ti trng ca on tu (Hnh 5.7.4). Tnh chuyn v theo phng ngang gi di ng R di tc dng ca cc ti trng. Xc nh chuyn v ti cc nt v ng sut trong mi thanh cu.

5.5. Mt kt cu cu c tnh ton thit k theo m hnh dn thanh

nh trn Hnh 5.7.5. Kt cu ny c cu thnh t 6 nhp. Ti trng biu din trn hnh v m t trng thi lm vic nguy him nht ca kt cu. Vt liu s dng trong kt cu l thp vi mun n hi Ethp = 210gPa. Xc nh tit din

280 kN 210 kN 280 kN 360 kN 3.118m

3.6m 3.6m 3.6m

600 600

Hnh 5.7.4. M hnh mt nhp cu chu lc

R

y

a a

x 1

23

4 5

IIV V

III

II

Q

P

Hnh 5.7.3. Dn chu lc

Q

P


cho cc thanh sao cho ti u theo iu kin bn. Cho a=5,5m; b=4,5m; c=1m; P1=25kN; P2=15kN; P3=40kN v P4=20kN. Ch : Kt cu ny s c tnh ton, thit k li theo m hnh khung (xem bi tp Chng 8).

5.6. S kt cu ca mt chic cn cu c th hin trn Hnh 5.7.6, ti trng thit k l 10 tn. Chn loi thp ph hp v s dng h s an ton bng 4, xc nh tit din cho tt c cc thanh. Cho a=3m; b=9m; P=10000kg.

5.7. Mt kt cu gin cng xn phi chu ti nh trn Hnh 5.7.7; cc

thanh u bng thp v c tit din 8cm2, ng sut cho php ca vt liu l 600mPa. Kim tra xem thit k c tha mn iu kin bn hay khng? Thit k li (thit k tinh) vi iu kin s dng cng loi vt liu v gi nguyn ng bao ca kt cu. Thit k li y c th hiu l thay i cch sp xp cc

P1 P2 P3 P4

ac

b

Hnh 5.7.5. M hnh dm cu chu lc

a

b

b

a

P

Hnh 5.7.6. M hnh cn cu


thanh, loi b mt s thanh, hoc thay i tit din ca cc thanh. Mt trong cc mc ch ca thit k tinh v d l tm cch lm gim khi lng tng th ca kt cu. Cho a=0,5m ; b=0,9m; c=0,4m; d=0,6m; =600; P = 30kN; Q = 40kN.

5.8. Cho kt cu gin nh Hnh 5.7.8. Vt liu v tit din ca cc

thanh ging nh bi 5.7. Hy phn tch bi ton ging nh lm vi bi ton 5.7. Cho a=0,4m; b=6,5m; c=0,4m; = 300; P = 40kN; Q = 60kN.

a

a

c d b

P

Q

Hnh 5.7.7. M hnh dn cng xn chu lc

a

a

b c b P

Q

Hnh 5.7.8. M hnh dn cng xn chu lc


Chng 6 PHN T HU HN TRONG BI TON HAI CHIU

1. M U Trong chng ny, chng ta p dng phng php PTHH tnh kt cu

phng (2D) ca bi ton n hi. Cc bc c tin hnh ging nh bi ton mt chiu xt trong chng 4. Vct chuyn v u c xc nh bi:

u = [u v]T (6.1) Trong : u, v l cc chuyn v theo phng x v y tng ng (Hnh 6.1).

ng sut v bin dng c k hiu bi:

= [x, y, xy]T (6.2) = [x, y, xy] T (6.3)

Lc th tch, lc din tch v vi phn th tch c xc nh nh sau: f = [fx fy]T T = [Tx Ty]T (6.4) dv = tdA .

trong : t l dy theo phng z.

x

u

v

fx

fy

i

(x,y)

A L

y

T

u=0 v=0

Hnh 6.1. Bi ton hai chiu


Quan h gia bin dng v chuyn v: T

xv

yu

yv

xu

+

= (6.5)

Xt quan h ng sut vi bin dng cho hai trng hp:

1.1. Trng hp ng sut phng

=

xy

y

x

xy

y

x E

2100

0101

1 2 (6.6)

Hoc: = D (6.7) Trong

=

2100

0101

1 2

ED (6.8)

1.2. Trng hp bin dng phng

( )( )

+=

xy

y

x

xy

y

x E

22100

0101

211 (6.9)

Hoc: = D (6.10)

Trong

( )( )

+=

22100

0101

211

ED (6.11)

E l mun n hi; l h s Poisson ca vt liu.


2. RI RC KT CU HO BNG PHN T TAM GIC Min hai chiu c ri rc ho bng cc phn t tam gic nh hnh 6.2. Mi phn t tam gic c 3 nt, mi nt c 2 chuyn v (theo phng x v y).

Ta k hiu vct chuyn v nt chung bi:

{ }TnQQQQ L21= (6.12) tin tnh ton, cc thng tin v vic chia min thnh cc phn t tam

gic s c th hin qua cc bng to nt v bng nh v cc phn t. Bng nh v cc phn t c thit lp nh sau:

Bc t.do Phn t 1 2 3 4 5 6

1 1 2 3 4 11 12 2 3 4 13 14 11 12 3 3 4 5 6 13 14 ... 11 13 14 9 10 21 22

Qui c: ng i t nt u n nt cui trong mi phn t theo chiu ngc chiu kim ng h. Bng nh v phn t m t tnh tng ng gia chuyn v a phng v chuyn v chung ca phn t. Cc thnh phn chuyn v trong

x

1

y

T

Hnh 6.2. Ri rc kt cu bng phn t tam gic

Q1

Q2

1 3

42

5

6

7

8

9

1011

2 3

4

5

6

7 8

9

1011

Q3

Q4Q5

Q6

Q2j-1 j

Q2j


h to a phng ca nt i c k hiu l q2i-1 v q2i theo phng x v y tng ng. Ta k hiu vct chuyn v ca phn t bi:

{ }Tqqqq 621 L= (6.13) R rng, t bng nh v cc phn t trn, sau khi tm c vct chuyn

v chung Q, ta s tm c vct chuyn v nt ca tng phn t ri t i xc nh cc i lng khc nh ng sut, bin dng trong mi phn t.

Chuyn v ti mt im bt k trong phn t c biu din qua cc thnh phn chuyn v ca nt phn t. i vi phn t tam gic c bin dng l hng s, cc hm dng bin thin tuyn tnh trong phn t. Ta c th biu din cc hm dng N1, N2, N3 nh trn Hnh 6.3.

=1

2

N1=1

1

3

=1

N1

=1

2

1

3

=1

N2

N2 =1

=1

2

1

3

=1

N3 =1N3

Hnh 6.3. Biu din hnh hc cc hm dng


Nhn xt: - Hm dng N1=1 nt 1, gim tuyn tnh n 0 ti nt 2 v nt 3. Tng t

i vi N2 v N3. - Bt k mt t hp tuyn tnh no ca cc hm dng trn cng u biu din

mt mt phng. - Tng N1+ N2+ N3 biu din mt mt phng c chiu cao l mt n v

cc nt 1, 2 v 3; mt phng ny song song vi mt phng (1, 2, 3). V vy, vi N1, N2 v N3 bt k, ta c:

N1+ N2+ N3 = 1 Trong ba hm dng, c hai hm l c lp. Cc hm dng c biu din

qua v nh sau: N1= 1- - ; N2 = ; N3 =. (6.14)

Trong v c gi l cc to chun ho hay to t nhin. Tng t nh trong bi ton mt chiu (to x c bin i qua to

, cc hm dng l hm s ca ), trong bi ton hai chiu, cc to x, y cng c biu din qua cc to v .

V mt vt l, cc hm dng c biu din bi cc to din tch. Khi ni mt im nm trong mt tam gic vi ba nh, tam gic s c chia ra ba tam gic c din tch A1, A2, A3 nh Hnh 6.4.

AAN

AAN

AAN 332211 ;; ===

Trong A l din tch ca phn t.

(x,y)A1

A3

A2

1

2

3

Hnh 6.4. To din tch


3. BIU DIN NG THAM S Ta biu din chuyn v trong phn t qua cc hm dng v cc chuyn v

nt ca n nh sau:

++=

++=

634221

533211

qNqNqNvqNqNqNu

(6.15)

hay u = Nq (6.16)

Trong

=

321

321

000000

NNNNNN

N (6.17)

Thay (6.14) vo (6.15), ta c biu thc xc nh chuyn v qua chuyn v nt xt trong h to quy chiu nh sau:

( ) ( )( ) ( )

++=

++=

22624

11513

qqqqqvqqqqqu

(6.18)

i vi phn t tam gic, nh php m t ng tham s, ta c th biu din to (x,y) qua to nt phn t vi cng cc hm dng trn:

++=

++=

332211

332211

yNyNyNyxNxNxNx

(6.19)

Hay

( ) ( )( ) ( )

++=

++=

11312

11312

yyyyyyxxxxxx

(6.20)

Ta k hiu: xij = xi - xj yij = yi - yj

T (6.20), suy ra:

++=

++=

13121

13121

yyyyxxxx

(6.21)

y l mi lin h gia (x, y) vi (, ). xc nh cc thnh phn bin dng, ta cn tnh cc o hm ring u v v theo x v y.


Ta c: u = u(x(, ), y(, )). v = v(x(, ), y(, )).

p dng qui tc o hm hm hp:

+

=

+

=

yyux

xuu

yyux

xuu

(6.22)

Hoc di dng ma trn:

=

yuxu

yx

yx

u

u

(6.23)

Trong ma trn vung (2x2) c gi l Jacobian ca php bin i, k hiu l J:

=

3131

2121

yxyx

J ( ((

Trin khai ly o hm ca x v y theo v , ta c:

=

u

u

J

yuxu

1 (6.24)

1J l ma trn nghch o ca J

=

2131

21311

det1

xxyy

JJ (6.25)

Trong : det J = x21 y31 x31 y21 Ta cng bit rng, det J chnh bng hai ln din tch tam gic.

|det J|= 2A (6.26)


Ch : Nu cc nt 1, 2, 3 c xp t theo chiu ngc chiu kim ng h, th det J lun c du dng. T (6.24), (6.25), ta c th vit:

+

=

uxux

uyuy

Jyuxu

2131

2131

det1 (6.27)

Thay vai tr ca u bi v, ta cng c biu thc tng t:

+

=

vxvx

vyvy

Jyvxv

2131

2131

det1 (6.28)

Khi y, cc thnh phn bin dng c xc nh bi:

+++++

++

++

=

+

=

612521431313223132

621413232

512331123

det1

qyqxqyqxqyqxqxqxqxqyqyqy

J

xv

yu

yvxu

(6.29)

Hoc di dng ma trn: = B q (6.30)

Trong :

=

122131132332

211332

123123

000000

det1

yxyxyxxxx

yyy

JB (6.31)

y l ma trn lin h gia bin dng v chuyn v nt, trong cc s hng u l cc hng s v c xc nh qua cc to nt ca cc phn t.


4. TH NNG Th nng ca h c xc nh bi:

=i

iT

iL

T

A

T

A

T PudlTtdAftudAtD 21 (6.32)

Trong : T: lc din tch; f: lc th tch; t: chiu dy phn t Pi: lc tp trung, Pi = [Px, Py

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