Thesis_Dang Trung Hau

8/13/2019 Thesis_Dang Trung Hau

1/70

1

I HC QUC GIA THNH PH H CH MINHTRNG I HC KHOA HC T NHIN

----- -----

NG TRUNG HU

PHN TCH NG X V REISSNER MINDLIN

C DM TIMOSHENKOGIA C NG

BNG PHN T MITC4

Khoa: Ton Tin hc

Chuyn ngnh: C hc

LUN VN TT NGHIP I HC

TP. H CH MINH,thng 6 nm 2013


2/70

I HC QUC GIA THNH PH H CH MINHTRNG I HC KHOA HC T NHIN

KHOA TON-TIN HC

----- -----

LUN VN TT NGHIP

PHN TCH NG X V REISSNER MINDLIN

C DM TIMOSHENKO GIA C NG

BNG PHN T MITC4

GVHD: TS. NGUYN TH I TRUNG

SVTH: NG TRUNG HU

MSSV: 0911044

TP. H CH MINH, thng 6 nm 2013


3/70

i


4/70

ii

LI CM N

Trong sut th i gian hc t p ti khoa Ton-Tin hc tr ng i hc Khoa hc T nhin HQGTP.HCM, em nhn c nhiu kin thc cng nh s dy d qu bu t cc Thy c. V i lngknh tr ng su sc, em xin gi l i cm n chn thnh nht n ban gim hiu, cc phng ban cngton th cc Thy c, cn b v ang cng tc ti tr ng.

c bit hn c, em xin gi l i cm n n Thy TS. Nguyn Th i Trung. Mc d cng vicnghin cu cng nh ging dy ca Thy lun bn r n nhng Thy khng qun ngi, Thykhng nhng tn ty gip em trong sut giai on chuyn ngnh cng nh h ng dn em thchin lun vn ny m cn truyn t cho em nhng trit l sng cng nh k nng lm vic emlm hnh trang b c vo cuc sng sau ny.

Em xin gi l i cm n n Thy TS. Tr nh Anh Ngc, ngoi nhng kin thc chuyn mn,Thy cn truyn t nhng kinh nghim sng gip em t tin hntrong hc tp cng nh trongcuc sng hin ti v tng lai.

Em cng xin gi l i cm n n Thy ThS. Bi Xun Thng, Thy gn b v i em ngay t nhng ngy u tin em hc t p ti tr ng, Thy cng dnh nhiu thi gian gip em trongqu trnh thc hin lun vn ny.

Ngoi ra, em cng xin gi l i cm n n qu thy c trong b mn C hc v khoa Ton-Tinhc, n cc anh ch v cc bn trong nhm nghin cu Fosat gip em trong qu trnh hct p v thc hin lun vn.

Bn cnh , em cng xin gi l i cm n n gia nh, lun st cnh v to iu kin emhc t p v thc hin nhng c m ca bn thn.

D c gng r t nhiu nhng chc chn s khng trnh khi nhng sai st trong lun vn ny.V vy, em mong nhn c nhng kin ng gp ca qu Thy c lun vn ca em honthin hn.

Em xin chn thnh cm n.

TP. H Ch Minh thng 6 nm 2013

ng Trung Hu


5/70

iii

MC LC L I CM N.................................................................................................................

MC LC..................................................................................................................

DANH SCH HNH V ..................................................................................................... DANH SCH BNG BIU ................................................................................................ CHNG 1 TNG QUAN........................................................................................... CHNG 2 L THUYT V DY REISSNER -MINDLIN GIA C NG DMTIMOSHENKO ................................................................................................................

2.1 L thuyt tm dy Reissner - Mindlin ............................................................................ 2.1.1 Khi nim chung .....................................................................................................

2.1.2 Gi thit tm dy Reissner-Mindlin ......................................................................... 2.1.3 Quy c du ............................................................................................................. 2.1.4 Tr ng chuyn v, ng sut, bin dng ca tm ........................................................... 2.1.5 Nng l ng bin dng n hi ca tm ...................................................................... 1 2.1.6 ng nng ca tm ..................................................................................................

2.2 L thuyt v dy Reissner-Mindlin ................................................................................ 2.2.1 Chuyn i h ta .................................................................................................

2.2.2 Nng l ng bin dng v ng nng ca v............................................................... 1 2.3 L thuyt dm Timoshenko ............................................................................................

2.3.1 Tr ng chuyn v, ng sut, bin dng ca dm ........................................................ 1 2.3.2 Nng l ng bin dng n hi ca dm ..................................................................... 2 2.3.3 ng nng ca dm ..................................................................................................

2.4 L thuyt v dy Reissner-Mindlin gia c ng b i dm Timoshenko ................................ 2

CHNG 3 PHNG PHP PHN T H U HN CHO BI TON V GIAC NG DM 29

3.1 Phng php phn t hu hn cho bi ton v ................................................................. 2 3.1.1 Cng thc phn t hu hn ca v trn min t gic bn nt bt k .......................... 29

3.1.2 Cng thc phn t hu hn ca v khi s dng phn t MICT4 ................................ 33 3.2 Phng php phn t hu hn cho bi ton dm ............................................................... 3


6/70

iv

3.2.1 Cng thc phn t hu hn ca dm trn phn t mt chiu bt k ........................... 37 3.2.2 Cng thc phn t hu hn ca dm trn phn t ng tham s mt chiu ................ 40

3.3 Phng php phn t hu hn cho bi ton v gia c ng dm ......................................... 42

3.4 iu kin bin................................................................................................................

CHNG 4 V D S .................................................................................................. 4.1 Bi ton tnh hc .........................................................................................................

4.2 Bi ton ng hc ........................................................................................................ CHNG 5 KT LUN V H NG PHT TRIN ....................................................

5.1 Cc k t qu t c ....................................................................................................... 5.2 Hn ch v tn ti ............................................................................................................

5.3 H ng pht trin trong tng lai ...................................................................................... TI LIU THAM KHO ..................................................................................................


7/70

v

DANH SCH HNH V

Hnh 1. K t cu v trong xy dng .............................................................................................

Hnh 2. R i r c min bi ton thnh cc min con....................................................................... Hnh 3. Phn t v thoi............................................................................................................ Hnh 4. Mt phng trung ha ca tm. ....................................................................................... Hnh 5. Quy c du ca tm. ................................................................................................... Hnh 6. a) Quy c du trong mt phng Oxz; b) Quy c du trong mt phng Oyz. .................. 7 Hnh 7. Bin dng ca tm. ....................................................................................................... Hnh 8. Chuyn h ta cho phn t t gic [6]....................................................................... 1 Hnh 9. Dm trong h ta a phngOrsz ........................................................................... 1 Hnh 10. Quy c du cho dm. ................................................................................................ Hnh 11. Bin dng ca dm. .................................................................................................... Hnh 12: i bin dm. ............................................................................................................ Hnh 13. Phn t v gia c ng .................................................................................................. Hnh 14. Phn t t gic 4 nt. ................................................................................................. Hnh 15. Phn t gic trong min tham chiu ............................................................................. 3 Hnh 16. Phn t MITC4.......................................................................................................... Hnh 17. Phn t dm 2 nt. ..................................................................................................... Hnh 18: Phn t dm trn min tham chiu ............................................................................... 4 Hnh 19. Mt s loi iu kin bin. ......................................................................................... Hnh 20. V hnh tr v i lc tp trung v iu kin bin mng. ................................................. 4 Hnh 21: vng ca v tr khng gia c ng dm chu lc t p trung. ...................................... 50 Hnh 22. Tc hi t nghim ca phn t MITC4 cho bi ton v tr chu lc t p trung. ........ 51 Hnh 23. V cu gia c ng b i hai dm ng tm. ..................................................................... Hnh 24: vng ca v cu chu lc tp trung c gia c ng b i hai dm ng tm. ............ 52 Hnh 25: Tc hi t nghim ca phn t MITC4 cho bi ton v cu chu lc t p trung giac ng hai dm ng tm. 53 Hnh 26. V tr gia cng 5 dm lch tm. ................................................................................ Hnh 27. Hnh dng tm mode dao ng t nhin ca v tr gia cng nm dm lch tm chuiu kin bin ngm. ............................................................................................................... Hnh 28: Gi tr tn s ca tm mode dao ng u tin............................................................. 5 Hnh 29. Hnh dng mi mode dao ng t nhin u tin ca v cu gia c ng hai dm ngtm vi iu kin bin ta n. .................................................................................................. Hnh 30. Gi tr mi mode dao ng t nhin u tin. ............................................................ 5


8/70

vi

DANH SCH BNG BIU

Bng 1: K t qu phn tch vng ca v tr khng gia c ng dm. ........................................ 50 Bng 2: K t qu phn tch vng ti tm ca v cu gia c ng dm ng tm. ....................... 52 Bng 3: K t qu phn tch tn s dao ng ca v tr gia c ng dm lch tm. ......................... 56 Bng 4: K t qu phn tch tn s dao ng ca v cu gia c ng dm ng tm. ....................... 58


9/70

1

CHNG 1 TNG QUAN

Ngy nay, v i s tin b ca nhn loi i hi chng ta phi gii quyt nhng bi ton ngy cng phc tp hn trong khoa hc, k thut, mt trong s l nhng bi ton lin quan n cc k t cuv. V l loi k t cu c s dng r ng ri trong nhiu ngnh k thut nh c kh ch to my,xy dng dn dng, ch to tu thuyn, my bay, v.v. V i tham vng pht trin v vn ln canhn loi i hi nhng cu trc ny phi bn hn, nh hn, chi ph r hn, v.v. Chnh v vy, ccnh nghin cu cng khng ngng tm ti, ci tin nhng cu trc ny p ng c yu cu, ph h p v i tng mc ch khc nhau. Mt s loi v th ng dng c th k n nh v composite,

v gia c ng gn, v c c tnh thay i, v p in, v.v. V compositec c bng cch dn ccl p v khc nhau li v i nhau hoc tr n thm cc s i bn trong v. u im ca loi vt liu nyl c tnh c -l r t tt nh khi l ng nh, kh nng chu ti, chu nhit cao, tuy nhin, vic ch to loi vt liu ny g p r t nhiu kh khn, do dn n chi ph sn xut cao. V gia c ng gn,v bn cht ging nh v composite, tuy nhin, chng c cu trc n gin hn rt nhiu. V giacng gn c c bng cch gn thm cc dm lm cng ln b mt ca v. Loi vt liu ny cnhc im l tnh thm m khng cao, khng nh nh v composite, tuy nhin, do kh nng chuti cao, chi ph sn xut thp nn y l mt trong nhng loi vt liu c s dng r ng ri nhthin nay. Do c tnh c -l khc bit nn v c c tnh thay i v v p in ch c s dngtrong mt s ngnh k thut ring bit nh ch to tu v tr, tn la, bng truyn vn ti, v.v.

Hnh 1. K t cu v trong xy d ng


10/70

2

Vic phn tchng x cho k t cu v gia cng, c bit nhng k t cu c hnh hc phc t p bng phng php gii tch hoc bn gii tchl iu v cng kh khn. Trong th i gian gn y,cc nh nghin cu p dng nhng phng php s phn tch nhng k t cu ny, ph bin

l phng php phn t hu hn. Khi phn tch k t cu v gia c ng, ta coi v v dm l hai k tcu ring bit, cc nt phn t ca dm trng v i nt phn t ca v, do ta s phn tch c l phai k t cu ny v sau cng s s dng iu kin tng thch[ HYPERLINK \l "LXP06"1 ] ghp ni chng li v i nhau. C hai loi v gia cng th ng c s dng l v mng Kirchhofoffk t h p dm Euler v v dy Reissner-Mindlin k t h p dm Timoshenko. L thuyt v dyReissner-Mindlin c xy dng a trn l thuyt tm dy Reissner-Mindlin.

Qu trnh phn tchng x v dy Reissner-Mindlin da trn l thuyt tm dy Reissner-

Mindlin c thc hin thng qua phn t ng tham s. Tuy nhin, phn t ng tham s g p phi hin t ng kha ct (shear locking) khi b dy tm tin v khng, ng th i, k t qu bi tonthay i kh nhiu khi s l ng mt li thay i. Nguyn nhn do nghim bi ton c x p x thng qua phn t tuyn tnh bc th p. khc phc hin tng ny, c nhiu phng phc xut nhng n gin hn c l phng php ni suy cc thnh phn chu ko (MITC)2] [HYPERLINK \l "DCh00"3 ].

Hnh 2. R i rc min bi ton thnh cc min con.


11/70

3

Trong ti ny, phng php Phn t hu hn c p dng phn tch bi ton v dyReissner- Mindlin c gia c ng b i dm Timoshenko. phn tch nhng bi ton v, chngta c th s dng 3 l thuyt ring bit: l thuyt v thoi, l thuyt v cong v l thuyt v suy

bin, trong l thuyt v thoi c s dng ph bin b i r t nhiu nhng u im nhd dngtrong vic xy dng cng thc, hiu qu trong qu trnh tnh ton v trong l p trnh, linh hot trongvic p dng cho c bi ton tm v v, chi ph tnh ton th p.

L thuyt v thoi tha nhn r ng, phn t v thoi l s k t h p gia phn t tm chu un v phn t mng tng ng.

Lun vn gm nm chng. Chng 1 gi i thiu tng quan v cu trc v, v gia c ng v phng php phn t hu hn. Chng 2trnh by l thuyt v Reissner-Mindlin c pht trint l thuyt tm Reissner-Mindlin, l thuyt dm Timoshenko v l thuyt v Reissner-Mindlingia c ng b i dm Timoshenko. Chng 3 trnh by phng php phn t hu hn cho v giac ng. V s dng phn t MITC4. Dm s dng phn t mt chiu hai nt. Chng 4 trnh bk t phn tch tnh hc v ng hc ca mt s bi ton v v v gia cng. Chng 5 trnh bnhng k t qu t v nhng hn ch cn tn ti trong lun vn, ng thi a ra h ng phttrin trong tng lai.


12/70

4

CHNG 2 L THUYT V DY REISSNER -MINDLIN GIA C NG DM TIMOSHENKO

2.1L thuyt tm dy Reissner - Mindlin

Nh trnh by trn, phn t v thoi c to thnh t s k t h p gia phn t tm chu unv i phn t mng.

Hnh 3. Phn t v thoi.

Nh vy, l thuyt v thoi Reissner-Mindlin c pht trin da trn l thuyt tm Reissner-Mindlin.

2.1.1Khi nim chung Tm l nhng cu trc hnh lng tr ng, c kch th c mt phng nh hn rt nhiu so

v i cc phng cn li, kch thc c gi l chiu dy ca tm. Mt phng qu tch cch

u hai mt ca tm c gi l mt phng trung ha ca tm.


13/70

5

Hnh 4. Mt phng trung ha ca tm.

Cn c vo t l gia kch thc dy v kch th c cnh xung quanh ca tm, ng i ta phn

ra tm dy v tm mng. Nu 1 5mint l th c gi l tm dy vi l thuyt tm Reissner-Mindlin. Khi1 80 1 5t l min/ th c gi l tm mng vi l thuyt tm c din Kirchhof[4],

trong ,t l dy ca tm, minl l kch thc cnh nh nht ca tm.

2.1.2 Gi thit tm dy Reissner-Mindlin

L thuyt tm Reissner-Mindlin tha nhn r ng [5]

(1) Cc on thng vung gc v i mt phng trung ha ca tm tr c khi bin dng vn cnthng nhng khng nht thit cn vung gc v i mt phng trung ha sau bin dng.

(2) Cc on thng vung gc v i mt phng trung ho ca tm c di khng i tr c vsau bin dng.

(3) B qua s tng tc gia cc l p song song v i mt phng trung ha.

Nh vy, ng x ca tm c phn tch thng qua mt phng trung ha ca tm.

2.1.3 Quy c du Xt mt phn t tm trong h ta Oxy v i th tch V. Mt phng trung ha ca tm nm trnmt phng Oxy v chu ti vung gc v i b mt, h ng xung d i. tr c Oz h ng ln trn.D itc dng ca lc ti trn b mt, phn t tm b bin dng. Chiu dng ca gc xoay v chuynv c quy c theo hnh v sau y.


14/70

6

Hnh 5.Quy c du ca tm.

Cc thnh phn 0 0,u v l chuyn v trong mt phng phng trung ho ca tm, 0w l vng,

, x y l gc xoay ln lt quanh trcOy, Ox.

Nh vy, theo l thuyt tm Reissner-Mindlin, tr ng chuyn v ca mi phn t tm trong h ta Oxyz c m t y b i nm thnh phn ca vc-t sau

0 0 0 T

e p x yu v wu . (2.1.1)

Tuy nhin, d dng hn trong vic thnh l p cng thc ta thm thnh phn z vo tr ng

chuyn v ca tm, z c xem nh l gc xoay quanh trc Oz . Nhvy, tr ng chuyn v ca

mi phn t tm c m t b i su thnh phn vc-t l

0 0 0 .

T e p x y z u v wu (2.1.2)

Theo nhng gi thit c cp trong mc (2.1.2), ta c mi lin h gia cc thnh phchuyn v v bin dng nh sau.

Theo Hnh 6 a), sau khi bin dng, gc xoay x quanh trcOy l tng ca gc xoay do bi

dng un 0w x

v gc xoay do bin dng ct xz .


15/70

7

a) b)

Hnh 6. a)Quy c du trong mt phng Oxz; b) Quy c du trong mt phng Oyz.

B i v thnh phn vng 0w tng dn theo chiu tng ca tr c ta Ox, do

0 00 0

0

lim

x

w x x w x w x x

, (2.1.3)

v vy, ta c c gc xoay do bin dng ct quanh trc Oy l

0 xz x

w x

. (2.1.4)

Theo Hnh 6 b), sau khi bin dng, gc xoay y quanh trcOx l tng ca gc xoay do bi

dng un 0wy

v gc xoay do bin dng ct yz .

B i v thnh phn vng 0w tng dn theo chiu tng ca tr c ta Oy, do

0 00 0

0

lim

y

w y y w yw

y y

, (2.1.5)

v vy, ta c c gc xoay do bin dng ct quanh trcOx l

0 yz y

wy

. (2.1.6)


16/70

8

2.1.4 Trng chuyn v, ng sut, bin dng catm

Theo gi thit (2) trong mc (2.1.2), chiu dy ca tm khng thay i sau khi bin dng nn ta

c 0 . z

w

z

Hnh 7. Bin dng ca tm.

D i tc dng ca lc ti c phng vung gc v i mt ca phn t tm, chiu hu ng xungd i, tm b bin dng nh Hnh 7. T ta c chuyn v theo ba phng ca mi phn t tm l

0

0

0

, , ,

, , ,

, .

x

y

u u x y z x yv v x y z x y

w w x y (2.1.7)

Theo quan h chuyn v- bin dng, trng bin dng ca tm c biu din nhsau

0

0

00 0

0

0

0

000 0

0 00 0

.

x

x y

y

z

xyy x x

xz

yz y

u x x

vy y

wu v x y x w

z

y x

y

(2.1.8)


17/70

9

Khi bin dng mng, bin dng un, bin dng ct ca tm ln lt l

00

0

00

0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0

em m p x

y

z

uuv x x w

vy yu v

y x y x

L u , (2.1.9)

0

0

0

0 0 0 0 0

0 0 0 0 0

0 0 0 0

x

y eb b p

x

yy x

z

uv x x w

z z y y

y x y x

L u , (2.1.10)

0

00

0

0

0 0 1 0 0

0 0 0 1 0

x xz e

s s p x yz

yy

z

uvww x x

wyy

L u . (2.1.11)

Theo nh lut Hooke cho bi ton ng sut phng ta c

1

.ij ij kk ij E E (2.1.12)

T cng thc (2.1.8) v cng thc (2.1.12)ta c mi lin h gia bin dng vng sutca tm l


18/70

10

1 0 0 0 0

1 0 0 0 0

0 0 0 0 0 010 0 0 2 1 0 0

0 0 0 0 2 1 00 0 0 0 0 2 1

,

xx xx

yy yy

zz zz

xy xy

xz xz

yz yz

E (2.1.13)

ta tnh c cc thnh phn ng sut ca tm nh sau

2

0

0

0 0

1 0

1 0

1 10 0

2

,m

u x v

y

E

yu v

x

(2.1.14)

2

1 0

1 01

10 0

2

,

x

by

y x

x

y

y

Ez

x

(2.1.15)

0

0

1 0

2 1 0 1. x

x

s

y

z

yz

w x wy

E

(2.1.16)

y, cc thnh phn ng sut , ,m b s ln l t ng v i cc thnh phn bin dng mng,

bin dng un v bin dng ct ca tm.2.1.5 Nng lng bin dng n hi ca tm

T cng thc(2.1.9) - (2.1.16), ta c nng lng bin dng trn mi phn t tm l

1 d2 ,T p p

i i V

e pU V (2.1.17)


19/70

11

1 d2

T T T e p m m b b s s

V

U V . (2.1.18)

Ly tch phn trn b dy ca tm, ta c

1 d2 ,T T T e e e

p p m m m b b b s s s pT

A

U Au L D L L D L L D L u (2.1.19)

vi A l din tch b mt ca phn t, , ,m b sD D D xc nh bi

2

1 0

1 01

10 0 2

,m Et D (2.1.20)

3

2

1 01 0

12 110 0

2

,( )b Et D (2.1.21)

1 0

2 1 0 1

,

( )s Ekt

D (2.1.22)

trong t l chiu dy ca tm, E l m-un n hiv l h s poisson vk l h s iu chnh bin dng ct v thng c gi tr bng 5/6.

Cng ca ngoi lc tc dng ln mi phn t tm l

d ,e e p p A

W Apu (2.1.23)

v i p l vc-t lc tc dng ln tm.

2.1.6 ng nng ca tm

ng nng ca mi phn t tm c tnh theo cng thc


20/70

12

21 d2

,e pV

T m V v (2.1.24)

trong ,m l khi l ng ca mi phn t tm v v l o hm theo th i gian ca chuyn v

.T u v w v (2.1.25)

Thay cng thc (2.1.25) vo cng thc (2.1.24) ta tnh c ng nng ca mi phn t tmnh sau

1 d2

,V

e p

uT u v w v V

w

(2.1.26)

2 2 21 d2 ,e

pV

T u v w V (2.1.27)

222 2 20 0 0 0 02 2 d , x ye

p x yV

T u z zu v z zv w V (2.1.28)

Ly tch phn theo chiu dy ca tm, ta thu c

3

2 2 2 2 20 0 0 d12

,e p x y A

t T u v w t A (2.1.29)

vit d i dng ma tr n ta c

1 d2 ,T e e e

p p p p A

T Au m u (2.1.30)

trong vc-t e pu v ma tr n khi l ng pm c xc nh nh sau

0 0 0 ,

T x y z

e p u v wu (2.1.31)


21/70

13

3

3

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 12 0 0

0 0 0 0 12 0

0 0 0 0 0 0

,/

/

p

t t

t

t

t

m (2.1.32)

v 0 0 0 , , , , , x y z u v w l cc o hm theo th i gian ca cc chuyn v, l mt khi ca

vt liu.

2.2L thuyt v dy Reissner-Mindlin

2.2.1 Chuyn i h ta V l mt cu trc ba chiu, xc nh trong mt h ta bt k . Khng mt tnh tng qut, ta gi s r ng, phn t v ang xt c xc nh bt k trong h ta ton cc OXYZ . Tuy nhin, nhtrnh by trong phn (2.1), v c xy dng thng qua phn t tm, xc nh hon ton trn mt phng Oxy trong h ta a phng Oxyz . Do , phn tchng x ca v, ta cn xy dng c ma tr n chuyn i gia h ta ton cc OXYZ v h ta a phng Oxyz .

Xt phn t t gic phng bt k e trong h ta ton cc OXYZ nh Hnh 8

Hnh 8. Chuyn h ta cho phn t t gic [6].


22/70

14

S th t cc nh ca phn t t gic c nh s ng c chiu kim ng h. Cc im , , ,i j k l ln lt l trung im ca cc cnh 4-1, 1-2, 2-3, 3-4.

Gi s r ng, tr c Ox ca h ta a phng song song v i vc-t i qua hai im ,i k . Tac vc-t i qua hai im ,i k l

.k i

x k i

k i

x x y y z z

V (2.2.1)

Tng t ta c vc-t i qua hai im , j l l

.l j

r l j

l j

x x y y

z z

V (2.2.2)

Gi z V l vc-t cng phng v i tr c Oz , z V c c bng cch ly tch c h ng ca hai

vc-t x V v r V

. z x r V V V (2.2.3)

Vc-t y V cng phng v i tr c Oy, c c bng cch ly tch c h ng ca hai vc-t z V

v x V ,

.y z x V V V (2.2.4)

Cc vc-t n v , , x y z ca cc tr c ta Ox, Oy, Oz c c bng cch chun ha cc

vc-t , , x y z V V V

1 1 1 ; ; , x x y y z z x y z l l l

V V V (2.2.5)

trong , , x y z l l l l di cc vc-t , , x y z V V V .


23/70

15

Theo lut chuyn h ta , ta c ma tr n chuyn t h ta OXYZ sang h ta Oxyz l

3 3 , x y z (2.2.6)

Nh vy, mi lin h gia h ta a phngOxyz v h ta ton cc OXYZ l

.T xyz XYZ (2.2.7)

Gi T e

s s s s sx sy sz u v wu l tr ng chuyn v ca mi phn t v trong h

ta ton cc OXYZ , ta c mi lin h gia tr ng chuyn v ca v trong h ta a phngv h ta ton cc nh sau

0

0

0 0

0

,

s

sT

s

T sx x

syy

sz z

uuvvww

(2.2.8)

t0

0

,

T

T

T cng thc (2.2.8) tr thnh

,e e p su = Tu (2.2.9)

V i c tnh theo cng thc (2.2.6).

2.2.2 Nng lng bin dng v ng nng ca v T cng thc (2.1.19), (2.1.23), (2.1.30) v (2.2.9), ta c nng l ng bin dng, cng ngai lc vng nngtrn mi phn t trong h ta ton cc OXYZ ln l t l

1 d2 ,T T T e e eT

s s m m m b b b s s sT

s A

U Au T L D L L D L L D L Tu (2.2.10)


24/70

16

d ,e es s A

W ApTu (2.2.11)

1 d2 .e e eT s s sT

p

A

T AuT Tu m (2.2.12)

2.3L thuyt dm Timoshenko

Dm l vt th lng tr hoc hnh tr c chiu di l ln hn rt nhiu so vi kch th c ca hai phng cn li. ng thng dc theo chiu di ca dm c gi l tr c dm.

Theo l thuyt dm Timoshenko (Timoshenko beam theory -TBT) cc on thng vung gcv i tr c dm tr c v sau khi bin dng s vn l thng nhng khng cn vung gc v i tr c dm

khi bin dng nhng di ca chng khng i [5].2.3.1Trng chuyn v, ng sut, bin dng ca dm

Xt mt phn t dm bt k nm trong h ta ton cc OXYZ nh Hnh 9, tr ng chuyn v cami phn t dm c cho b i

uT

b b b b bx by bz u v w (2.3.1)

Trong h ta Oxyz Tr ng chuyn v ca mi phn t dm c m t b i cc thnh phnca vc-t nh sau

0 0 0 u .T p

b x y z u v w (2.3.2)

T cng thc (2.2.9), ta c mi lin h gia tr ng chuyn v ca dm trong hai h ta nhsau

. p

bbu Tu (2.3.3)

Gi s r ng, tr c dm to v i tr c Ox gc . Khi ta gn vo mi phn t dm mt h taa phngOrsz nh Hnh 9, trng chuyn v ca mi phn t dm trong h ta a phOrsz l


25/70

17

0 .T b r s z r s z u u uu (2.3.4)

Hnh 9. Dm trong h ta a phng Orsz

Gi im P bt k nhhnh v. Trong h ta Ors , im P c ta 0,o P r s v i 0 0,r s l

cc i lng bit, ta c 0 PD s v 0OD r . Trong h ta Oxy, im P c ta 0 0, P x y

v i 0 0, x y l cc i lng cha bit. ta s i tm mi lin h gia hai h ta Oxy v Ors

p dng t s l ng gic ln l t trong cc tam gic vung PDF v FDC ta c

0 .tan DF s , (2.3.5)

0

0

cos ,tan cos ,

sin .

FC DF ss

(2.3.6)

p dng t s l ng gic trong cc tam gic vungODB ta c

0

cos ,

cos .

OB OD

r (2.3.7)

T cng thc (2.3.6), (2.3.7)ta c

0 0

,,

cos sin .

OA OB ABOB FC r s

(2.3.8)


26/70

18

Nh vy ta c

0 0 0 cos sin . x OA r s (2.3.9)

p dng t s l ng gic cho tam gic vungOMH ta c0

.

cos cossOM OH (2.3.10)

p dng t s l ng gic cho tam gic vung PNH ta c

0 0 tan tan tan cos sin NH NP OA r s . (2.3.11)

T (2.3.10), (2.3.11) ta c

0 0 02

0 0 0

20 0

0 0

tan cos sin ,cos

cos sin sin ,cos

cos cos sin ,cos

sin cos .

sON OH HN r s

s r s

s r

r s

(2.3.12)

Nh vy ta c

0 0 0 sin cos .y r s (2.3.13)

T cng thc (2.3.9), (2.3.13) ta c lut chuyn i gia hai h ta l

0 0

0 0

cos sin

,sin cos

x ry s

(2.3.14)

0 0

0 0

cos sin

.sin cos

r x s y

(2.3.15)


27/70

19

0 0 0

0 0 0

0 0 0

0

0

0 0 1

cos sinsin cos .

r x x s y yw w w

Q (2.3.16)

T cng thc (2.3.16), ta c mi lin h gia tr ng chuyn v ca mi phn t dm trong h ta a phngOrsz v h ta Oxyz l

0

0

0

cos 0 0 0 0

cos 0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 cos 0

0 0 0 0 0 1

sinsin

.cos sinsin

r

s

z

x r

ys

z z

uuvuwu (2.3.17)

Thay cng thc (2.3.2) v cng thc (2.3.4) vo cng thc (2.3.17)ta c

00

0

0

. p pb b bQ

u u T uQ

(2.3.18)

D i tc dng ca lc ti tc dng trn b mt dm, dm b bin dng. Chiu dng ca chuynv v gc xoay ca dm c th hin trong Hnh 10.

Hnh 10.Quy c du cho dm.


28/70

20

p dng l thuyt dm Timoshenko, xt trong mt phng Osz , sau khi chu lc ti, dm b vng xung theo phngOz vi vng z u v dch chuyn theo phngOs mt khong su .

ng th i, dm v xoay thm gc s quanh tr cOr . Ngoi ra, dy ca dm khng thay i sau

khi bin dng.Do ta c chuyn v theo phngOs v Oz ca mi phn t dm ln l t l

,s sv u r z r (2.3.19)

. z sw u r s r (2.3.20)

Hnh 11. Bin dng ca dm.

Trong mt phng Orz , theo phngOr sau khi chu lc ti, dm b bin dng mt khong

r r u z so vi ban u. K t h p v i (2.3.19), (2.3.20) ta c chuyn v theo ba phng ca dm

mi phn t dm l

, ,

.

r r

s s

z s

u u r z rv u r z r

w u r s r

(2.3.21)

Theo quan h chuyn v- bin dng, trng bin dng ca dm c biu din nh sau


29/70

21

0

0

0

.

r r

r

s

z bs s

rs

rz s z

rsz

ur u

z v r rsw z

uu v z r rs r

uu w sr r z r

v w z s

(2.3.22)

Theo cng thc nh lut Hooke, tr ngng sut ca dm c biu din nh sau

0

0

0

.

r r

r r

s s

z z bs s

rs rs

rz rz s z

sz sz r

u E z

r

G

r E E E

u z G

r rG

uG sr r

G

(2.3.23)

2.3.2Nng lng bin dng n hi ca dm

Nng l ng bin dng n hi ca mi phn t dm trong h ta Orsz c cho b i cng thcsau

0 1 d2 ,T b b

b

V

U V (2.3.24)

T cng thc (2.3.22) v cng thc (2.3.23) ta c

2 20 2 d ,r rsV

rz bU G E V G (2.3.25)


30/70

22

2 2 20 0 01 d

2

,s srb rV

u v wU E z G z G s V

r r r r r r (2.3.26)

2 20

2 2

22

12 d

2

12 d

2

12 d

2

,

r r r r

b V

s s s s

V

s s z z r r

V

u uU E z z V

r r r r

u uG z z V

r r r r

u uG s s V

r r r r

(2.3.27)

trong V l th tch ca mi phn t dm. Ly tch phn theo din tch thit din ca dm,

cng thc (2.3.27) tr thnh2 2

2

2 22

22

0

2

12

2

12

2

12

2

r r r r L A A

s s s s

L A A

s s z z r r

b

A A

u uU E A zdA z dA dr

r r r r

u uG A zdA z dA dr

r r r r

u uG A sdA s dA

r r r r

. L

dr

(2.3.28)

vi L l chiu di ca mi phn t dm.V 0 A

sdA nn

2 22

2 2

22 2

0

2

12

2

12

2

12

.

.

b A

r r r r

L A

s s s s

L

s

L A

A A

z r

u uU E A zdA z dA dr

r r r r

u uG A zdA z dA dr

r r r r

uG A s dA drr r

(2.3.29)

tnh 2d d ; A A

z A z A ta p dng cng thc i bin cho dm nh hnh v sau


31/70


32/70

24

2 2

2

0

2

22

1 1d d

2 2

1 1d d

2 2

1 1d d

2 2

,

r r rs

r r

s s ss

r r

s z r z

r

b

r

uU EA e r EI r

r r r

uGA e r GI r

r r r

uGA r GI r

r r

(2.3.33)

2 20

2 2

2

1 1d d

2 2

1 1d d

2 2

1 d2

.

r r rs

r r

s s ss z

r r

rr

b

z

uU EA e r EI r

r r r

uGA e r G I I r

r r r

uGA rr

(2.3.34)

Cch khc, vit d i dng ma tr n, ta c

0 1

d2

,T b

bb b

r

U r D (2.3.35)

trong cc thnh phn ,b b D xc nh nh sau

0

0 0 0 0

0 0 0 0 0

0 0 0 0

0 0 1 0 0

0 0 0 0 0

,

r r

rr

s

z s s

r

z

b

sr

z

s

b

eu

er r r r

uur ruu

er rr r

urr

rr

e bL u (2.3.36)


33/70

25

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 00 0 0 0

.s

b

s z

EA EI

GA

GAG I I

D (2.3.37)

Thay cng thc (2.3.36) vo cng thc (2.3.35)ta thu c phng trnh nng l ng bindng n hi ca mi phn t dm trong h ta Orsz l

0 00 1

d2

,T T b b b

b br

bU ru L D L u (2.3.38)

T cc cng thc (2.2.9), (2.3.18), (2.3.38)ta tnh c nng l ng bin dng ca mi phn t dm trong h ta ton cc OXYZ l

1 d2 .T T T T b b b

b b bl

U l u T Q L D L QTu (2.3.39)

2.3.3 ng nng ca dm

Ta c cng thc tnh ng nng ca mi phn t dm l

21d

2,b

V

T m V v (2.3.40)

trong ,m l khi l ng ca mi phn t dm vv l o hm theo th i gian ca cc chuyn v

.u v w v (2.3.41)

T cng thc (2.3.21), (2.3.40), (2.3.41)ta c ng nng ca mi phn t dm trong h ta Orsz l

0 1 d2

,bV

uT u v w v V

w (2.3.42)


34/70

26

0 2 2 21 d2 .b V T u v w V (2.3.43)

Thay cng thc (2.3.21) vo cng thc (2.3.43)ta c

2 2 22 2 20 2 2 2 d ,r r r r s s s s z s z sbV

u z zu u z zu u s suT V (2.3.44)

22 2 2 2 2 20 2 2 d .r s z r s r r s s s z sV

bT u u u z z u u s su V (2.3.45)

p dng (2.3.30), (2.3.31) vo (2.3.45)ta c

2 2 2 2 2 2 20

2 d ,r s z r s r r s s sl

s z bT A u u u e A I I eA u u l (2.3.46)

2 2 2 2 2 2 2 2 20 2 2 d

,s z r r r r s s s s z s rl

sbT A u eu e u eu e A

I A

I I u l (2.3.47)

vit d i dng ma tr n ta c

0 0 01 d2

,T

b b b b

l

T l u m u (2.3.48)

trong vc-t 0bu v ma tr n khi l ng bm c xc nh nh sau

0 ,T

b r s z r s z u u uu (2.3.49)

2

2

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0

,sb

s z

ee

I A e e A

I I e e

A

m (2.3.50)


35/70

27

v cc thnh phn r s z r s z u u u , , , , , l cc o hm theo th i gian ca cc chuyn v, l mt

khi ca vt liu.

Thay cng thc (2.2.9), (2.3.18) vo cng thc (2.3.48)ta tnh c ng nng ca mi phnt dm trong h ta ton cc OXYZ l

d ,T T T b b b bl

T l u T Q m Q Tu (2.3.51)

trong .T

b b b b bx by bz u v wu

2.4L thuyt v dy Reissner-Mindlin gia cng bi dm Timoshenko

Trong lun vn ny, v gia c ng dm l v ng hng, c gn cht thm cc dm, nh ,v s cng hn d i tc dng ca lc ti. Gi thit ca m hnh v gia c ng dm nh sau [1]

1) V v dm c tch r i khi phn tchng x.

2) iu kin tng thch chuyn v: v v dm c hn cht vo nhau, do , chuyn v vgc xoay ca v v dm ti cc v tr ghp ni l bng nhau.

3) Theo nguyn l chng chp nng lng, nng l ng ca v c cc dm gia c ng bng tngnng l ng ca v v cc dm.

Do , nng l ng bin dng v ng nngca v c gia c ng dm l

,b

e es b

N U U U (2.4.1)

.b

e es b

N T T T (2.4.2)

v i b N l tng s dm gia c ng trn mi phn t v.

Cng ngoi lc tc dng ln mi phn t v gia c ng chnh l cng tc dng ln mi phn t v v xc nh b i cng thc (2.2.11)nh sau


36/70

28

1 d2

,e

V

esW V pTu (2.4.3)

v i P l vc-t lc tc dng ln mt v gia c ng.

Th nng ton phn trn mi phn t v gia c ng chu un l

+ .U W (2.4.4)

Theo nguyn l cc tiu ha phim hm nng l ng, ta c 0 . V vy, cng thc (2.4.4)

tr thnh

0

0

,

.e eU W

U W (2.4.5)


37/70

29

CHNG 3 PHNG PHP PHN T H U HN CHO BITON V GIA C NG DM

Trong phng php phn t hu hn, min gi i hn c r i r c thnh e N min con. Nh vy,

e

e N

v i j , i j . Nghim bi ton s c tnh x p x tr n mi min con.

Theo l thuyt c cp trong muc (2.4), phn tchng x cho cu trc v gia c ng,chng ta s i phn tch ln l t cho cu trc v v cu trc dm. Vic x p x tr ng chuyn v cho mi phn t v gia c ng trong h ta tng th OXYZ c thc hin trn phn t tmtng ng trong h ta a phng Oxyz .

Hnh 13. Phn t v gia c ng

V s c r i r c thnh t p h p cc phn t t gic 4 nt, cn dm c r i r c thnh cc phn t 2 nt. Cc nt ca dm trng kh p v i cc nt phn t ca v.

3.1 Phng php phn t hu hn cho bi ton v

3.1.1 Cngthc phn t hu hn ca v trn min t gic bn nt bt k

Tr ng chuyn v 0 0 0 T

e p x y z u v wu ca mi phn t tm Reissner-Mindlin c x p x trn phn t t gic bn ntnh sau


38/70

30

Hnh 14. Phn t t gic 4 nt.

4

0 1 1 2 2 3 3 4 4

1

4

0 1 1 2 2 3 3 4 4

1

4

0 1 1 2 2 3 3 4 4

1

4

1 1 2 2 3 3 4 4

1

4

1 1 2 2 3 3 4 4

1

1 1 2 2 3 3 4

,

,

,

,

,

i i i

i i i

i i i

x x x x x i xi i

y y y y y i yi i

z z z z z

u N u N u N u N u N u

v N v N v N v N v N v

w N w N w N w N w N w

N N N N N

N N N N N

N N N N 4

4

1

.i zi i

N

(3.1.1)

Vit d i dng ma tr n, ta c

4

1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 00 0 0 0 0

.

i i

i i n

i ee i p i pi

xi i i i

yi i

i zi

u N v N w N

N

N N

u N d (3.1.2)

Tng t nh trn ta c


39/70

31

4

1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 00 0 0 0 0

.

i i

i i n

i ee i p i pi

xi i i i

yi i

i zi

u N v N w N

N

N N

u N d (3.1.3)

Trong ,cc thnh phn , , , , ,i i i x i yi zi u v w l chuyn v v gc xoay ti cc nt ca phn

t tm trong h ta da phng, ,i N x y l hm dng ti cc nt.

Thay cng thc (3.1.2) vo cng thc (2.1.19), (2.1.23), (2.1.30)ta c phng trnh nngl ng, cng ngoi lc v ng nngca tm trn mi phn t tm l

1 d2 ,e e e

p p m m b b

T T T T m

Ab s s s pU Ad B D B B D B B D B d (3.1.4)

d ,e e p p A

W ApNd (3.1.5)

1 d2 ,T e e eT

p p p p A

T Ad N m Nd (3.1.6)

trong , ,m b sB B B c c bng cch thaycng thc(3.1.2) vo cc cng thc(2.1.9), (2.1.10)

, (2.1.11)v xc nh nh sau

4 4

1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0

,

i

i m m i

i i

i i

x

y

N

N

N y x

N

N

B L (3.1.7)


40/70

32

4 4

1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0

,

i

i b b i

i i

i i

x N

N N

N y x

N

yB L (3.1.8)

4 4

1 1

0 0 0 0

0 0 0 0.

i i

s s i i i i

i

N N

x N N

N y

B L (3.1.9)

Gi

T e

si si si sxi syi syi si u v wd l chuyn v ti mi nt ca phn t v trong h

ta ton cc, ma tr n chuyn h ta T c c bng cch thay cng thc (3.1.2) vo cngthc (2.2.8)v xc nh nh sau

0

0

0

24 1 24 1

0

0

,

i si

i si T

i si i T

xi sxi i

yi syi

zi szi

u uv vw w

(3.1.10)

6 6

6 6

6 6

6 6 6 6 6 66 6

6 6 6 6 6 6

6 6 6 6 6 6

6 6 6 6 6 6

24 24

00 0 0

0

00 0 0

0

00 0 00

00 0 0

0

.

T

T

T

T

T

T

T

T

T

(3.1.11)

Nh vy, ta tnh c nng l ng bin dng, cng ngoi lc, ng nng ca mi phn t v l


41/70

33

1 d2 ,T

m m b b b s s s

T T T T e e es s m s

A

U Ad B D B B dT TD B B D B (3.1.12)

d ,e e

A

s sW ApNTd (3.1.13)

1 d2 .T

T T e e es s

As pT Ad T N m NTd (3.1.14)

3.1.2Cng thc phn t hu hn ca v khi s dng phn t MICT4 thun li hn trong vic tnh ton cc cng thc (3.1.12), (3.1.13), (3.1.14), mi min t giccon e s c tham chiu n min hnh vung n v 1 1 , ; trong h ta a phng

O nh Hnh 15. Trn mi min tham chiu ny, ta nt v tr ng chuyn v c x p x thng qua cng mt b hm dng, do n cn c gi l phn t ng tham s.

Hnh 15. Phn t gic trong min tham chiu

Ta c mi lin h hnh hc gia hai phn t v ni suy tr ng chuyn v ca mi phn t nhsau

4 4

1 1

, ,i i i i i i

x N x y N y

(3.1.15)

4

1

.e e p i pi i

N u d (3.1.16)

trong ,i i x y l ta ti cc nh, i N l cc hm s ph thuc , v xc nh b i


42/70

34

1

2

3

4

1 1

1 114 1 1

1 1

( )( )( )( )

.( )( )( )( )

N N N N

(3.1.17)

o hm cc hm dng theo v l:

1 1

2 2

3 3

4 4

1 11 1

4 4

1 11 1

4 4

1 11 1

4 4

1 11 14 4

( ) , ( ),

( ) , ( ),

( ) , ( ),

( ) , ( ).

N N

N N

N N

N N

(3.1.18)

o hm ca x v y theo v l

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

1 1 1 11 1 1 1

4 4 4 41 1 1 1

1 1 1 14 4 4 41 1 1 1

1 1 1 14 4 4 4

1 1 1 11 1 1 1

4 4 4 4

( ) ( ) ( ) ( ) ,

( ) ( ) ( ) ( ) ,

( ) ( ) ( ) ( ) ,

( ) ( ) ( ) ( ) .

x x x x x

x x x x x

yy y y y

yy y y y

(3.1.19)

Ta c

3

,

x x x y I

y y

x x

(3.1.20)

do


43/70

35

1

1

.s

x x x y

y y x x

J (3.1.21)

T cc cng thc (3.1.18), (3.1.21), ta tnh c o hm cc hm dng theo x v y l

. . ,

. . .

i i i

i i i

N N N x x x

N N N y y y

(3.1.22)

Thay cng thc (3.1.17), (3.1.22) vo cng thc(3.1.7), (3.1.8), (3.1.9),ta tnh c cc thnh

phn , ,m b sB B B .Ta thy r ng, khi chiu dy ca tm tin dn v khng, thnh phn bD ca tm s tin v vc-

t khng nhanh hn thnh phn sD , dn n nghch l l bin dng ct ca tm ln hn bin dng

un ca tm khi chiu dy tm nh. Hin tng ny c gi l hin t ng kha ct (shearlooking)[7].

khc phc ny, tr ng bin dng ct ca tm c x p x li thng qua phn t MITC4.

Phn t MITC4 c biu din trong h ta tham chiu thng qua hnh vung n v nh Hnh16.

Hnh 16. Phn t MITC4


44/70

36

Trn phn t MITC4, tr ng bin dng mng v bin dng un ca tm vn c x p x tng t nh trn. Tr ng bin dng ct ca tm c x p x li nh sau

1 2

3 4

,

,

s C s A xz xz xz

s B s D

yz yz yz

N N

N N

(3.1.23)

trong cc thnh phn , N c xc nh nh sau

1 2

3 4

1 11 1

2 21 1

1 12 2

, ,

, ,

s s

s s

N N

N N (3.1.24)

3 4 3 4 1 22 1

3 2 1 43 2 4 1

2 2

2 2

, ,

, ,

A C x x x x xz xz

y y y y D Byz yz

w w w w

a aw w w w

b b

(3.1.25)

V i a, b ln l t l di cc cnh ni cc nh 1-2 v 1-4.Do ta c bin dng ct ca tm c x p x li l

1 2

3 4

0 0

0 0,

C C xz xz A A xz xz

B Byz yz

D

s s xz s s

s s syz

Dyz yz

N N N

N N

(3.1.26)

1 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 21 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 02 2

1 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 2

1 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 2

ss

s

a a

a a

b b

b b

dN


45/70

37

1 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 21 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 02 2

1 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 21 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 02 2

.ss

a a

a a

b b

b b

B N

(3.1.27)

Bng cch thay th cc thnh phn , ,m b sB B B va tnh c vo cng thc (3.1.12), ta tnh

c nng l ng bin dng trn mi phn t v nh sau.

1 d d2 det .

T m m b b

T T T T e e e ss s m sb s s sU d B D B B D B B d JTD BT

(3.1.28)

ng nng v cng ngoi lc trn mi phn t v c c bng cch thay th cng thc (3.1.17)vo cng thc (3.1.13) v cng thc (3.1.14).

det d d

,e e ss sW JpNTd (3.1.29)

1 dde d2 t .T

T T p

e e e ss s sT d T N m NT Jd (3.1.30)

3.2 Phng php phn t hu hn cho bi ton dm

3.2.1 Cng thc phn t hu hn ca dm trn phn t mt chiu bt k Tr ng chuyn v 0 T b r s z r s z u u uu ca mi phn t dm c x p x trn phn

t mt chiu haint nh sau


46/70

38

Hnh 17. Phn t dm 2 nt.2 2

1 1 2 2 1 1 2 2

1 1

2 2

1 1 2 2 1 1 2 2

1 1

2 2

1 2 1 1 21 22

1 1

,

,

.

,

,

,

r r r i ri r r i ri i i

s s s i si s s i si i i

r

s

z z z zi i z z i z zi i i

u M u M u M u M M M

u M u M u M u M M M

u u u M M M M u M M

(3.2.1)

Vit d i dng ma tr n ta c

20 0

1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

u M d .

i ri

i si

ni zi

b i bi i i i ri

i si

i zi

M u M u

M u M

M M

(3.2.2)

Tng t ta c

20 0

1 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 00 0 0 0 0

0 0 0 0 0

.

i ri

i si n

i zi b i bi

i ri i i

i si

i zi

M u M u

M u

M M

M

u M d (3.2.3)

Trong cc thnh phn , , , , ,ri si zi ri si zi u u u l chuyn v v gc xoay ti cc nt, ,i M r s

l hm dng ti cc nt.


47/70

39

Thay cng thc (3.2.2) v cng thc (3.2.3) vocng thc(2.3.38)v cng thc(2.3.48) tatnh c nng lng bin dng v ng nng ca dm nh sau

0 00 1

d2

,T T b b b

b br

bU rd B D B d (3.2.4)

0 0 01 d2 ,T T

b b b bl

T l d M m Md (3.2.5)

trong , ma trn bB c c bng cc thay th cng thc(3.2.2) vo cng thc (2.3.36)

2 2

1 1

0 0 0 0

0 0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0

.

i i

i

b i i

i i

i i

b

i

i

e

e

M M r r

M r

M M M r r

M M

r M

r

LB (3.2.6)

Gi

T

bi bi bi bi bxi byi bzi u v wd l chuyn v ti cc nt ca mi phn t dm trong

h ta ton cc. T cc cng thc (2.3.18), (2.2.8) ta c mi lin h v chuyn v ca dm giah ta a phngOrsz v h ta ton cc OXYZ l

0 00 0 , ,b b b bd T Td d T Td (3.2.7)

trong ma trn 0T , T xc nh nh sau


48/70

40

6 66 6

0

6 66 6

6 66 6

6 66 6

00

0

00

0

00

0

00

0

,

.

T

T

T

T

QQ

TQ

Q

T

(3.2.8)

Nh vy, ta tnh c nng l ng bin dng v ng nng ca mi phn t dm trong h ta ton cc theo cng thc sau

0 01 d2 ,T T bT

bb

b bb

l

U l T T T dd TB D B (3.2.9)

0 01 d2 .T T T

b b b bl

T l Td T M m MT Td (3.2.10)

3.2.2 Cng thc phn t hu hn ca dm trn phn t ng tham s mt chiu tnh nng l ng ca dm trn phn t mt chiu hai nt, ta s tnh chng trn min tham chiunh hnh v.

Hnh 18: Phn t dm trn min tham chiuMi lin h hnh hc gia hai phn t l

2

1

,i i i

r M r (3.2.11)


49/70

41

v i i r l gi tr ti mi nt v i M l hm dng trn phn t tham chiu v c tnh theo cng thc

1 21 1

1 12 2

, . M M (3.2.12)

Ly o hm ca hm dng i M theo ta thu c

1 21 12 2

, . M M (3.2.13)

Ly o hm ca r theo ta thu c

2 112

.r r r (3.2.14)

T (3.2.14) ta tnh c o hm ca theo r l

2 1

2

.r r r

(3.2.15)

Bng cch ly o hm ca hm hp, ta tnh c c o hm ca M theo r nh sau

1 1

2 1

2 2

2 1

1

1

. ,

. .

M M r r r r

M M r r r r

(3.2.16)

Thay cng thc (3.2.12), (3.2.16) vo cc cng thc (3.2.3), (3.2.6), ta tnh c cc thnh phn,bB M . Thay ,bB M va tnh c vo cng thc (3.2.9), (3.2.10), ta tnh c nng l ng bin

dng v ng nng trn mi phn t dm nh sau

0 0 d d2

et1

,T T bT bb

bb

b bU d B D BT T T Td J (3.2.17)

0 01 det d2 ,

T T T bb b b bT Td T M m MT Td J (3.2.18)

trong 2 1 2 det / .b r rJ


50/70

42

3.3 Phng php phn t hu hn cho bi ton v gia cng dm

Theo l thuyt 2.3, tr ng chuyn v ca dm c biu din trn mt trung ha ca v. Ti v trghp ni, tr ng chuyn v ca dm bng tr ng chuyn v ca v. Ti cc v tr khc,tr ng

chuyn v ca dm bng 0.Do ,ei j b ij sd d (3.3.1)

v i ij l ton t kronecker,i,j l v tr cc nt. Nh vy ta c

, ,b be e

s sd d d d (3.3.2)

v i l ma tr n m cc thnh gm 0 hoc 1.

Nng l ng bin dng v ng nngca mi phn t v gia c ng dm c tnh b i

,b

e es b

N U U U (3.3.3)

.b

e es b

N T T T (3.3.4)

T cc cng thc (3.1.12), (3.1.14), (3.2.9), (3.2.10), (3.3.2), (3.3.3), (3.3.4)ta tnh c nngl ng bin dng v ng nng ca mi phn t v gia c ng dm l

0 0

1d

2

1d

2

,b

T T T T e e eT s m m m b b b s s s s

AT T e eT b b b

s s N

T T

l

U A

l T

d T B D B B D B B D B Td

d B D B dT T T (3.3.5)

0 0

1d

21

d2 .

b

T e e eT T

s s s AT e eT

sT T T

sb N l

T A

l T T M m MT T

d T N m NTd

d d (3.3.6)

v i l l phn t chiu di ca mi phn t dm trong h ta a phng Orsz .

thun tin hn khi tnh ton, ta t


51/70

43

0 0

0 0

d

d

d

d

d

,

,

,

b

b

T T T e T m m m b b b s s s

AT T b b b

N l

e T T s A

T

N l

e

T T

T T T b

A

A

l

A

A

l

T T

K T B D B B D B B D B T

B D B

M T N

T

m NT

T

T T M m MT T

pNT

F

(3.3.7)

khi e e eK , M , F l ma tr n cng v ma tr n khi l ng v ma tr n lc ti ca mi phnt v gia c ng, nng l ng bin dng, ng nng, cng ngoi lc ca mi phn t v gia c ng

c biu din theo e e eK , M , F nh sau

1212

,

,

.

T e es s

T e es s

e e

e e

e

s

e

e

U

T

W

d K d

d M d

F d

(3.3.8)

Sau khi ly bin phn, ta thu c cng thc

,

,

.

e e e es s

e e e es s

e e es

U

T

W

K d d

M d d

F d

(3.3.9)

V bin phn l ty , do sau khi thay cng thc (3.3.9) vo cng thc (2.4.5) tathu ch phng trnh cn bng cho mi phn t v gia c ng l

.e e esK d F (3.3.10)

p dng phng trnhLagrange cho bi ton phn tch dao ng t do ca v gia c ng, ta c phng trnh cn bng


52/70


53/70

45

B i v ed l ty , do phng trnh(3.3.15) tr thnh

0 .e e e es sM d K d (3.3.16)

p dng phng php x p x Rayleigh-Ritzcho bi ton dao ng iu ha, nghim ca h phng trnh(3.3.16) c x p x d i dng

cos ,e es t d d (3.3.17)

trong , ed l bin dao ng ca v gia c ng, ed khng ph thuc vo th i gian t . ltn s gc ca dao ng, c n v l rad/s, l pha ban u ca dao ng.

Thay cng thc (3.3.17) vo cng thc(3.3.16)ta c

2

2

0

0

cos cos ,

cos .

e e e e

e e e

t t

t

M d K d

K M d (3.3.18)

Khi , bin dao ng ed v tn s gc l nghim ca h phng trnh

2 0 .e e eK M d (3.3.19) h phng trnh(3.3.19) c v s nghim th

2 2 0 det | | .e e e eK M K M (3.3.20)

H phng trnh(3.3.20) cho ta s l ng nghim tng ng v i s phng trnh ca h. ng v i mi gi tr va tm c, thay vo h phng trnh(3.3.19), ta s tm c bin daong tng ng cho v gia c ng.

T cc cng thc (3.3.7), ta thu c ma tr n cng, ma tr n khi l ng v ma tr n lc ticho mi phn t v gia c ng l


54/70


55/70

47

3.4 iu kin bin

Tr c khi gii cc phng trnh (3.3.25), ta phi i kh cc iu kin bin m bi ton a ra. Mts loi iu kin bin th ng g p trong nhng bi ton phn tchng x ca v l:

Hnh 19. Mt s loi iu kin bin.

(1) iu kin bin ngmBin chu iu kin ngm tha mn iu kin tt c chuyn v v gc xoay u bng

0, tc l 0

. x y z u v w (2) iu kin bin gi ta (diaphragm)

Trn Hnh 19, bin chu iu kin gi ta tha mn iu kin chuyn v theo phng x v phng z bng 0, ngoi ra, gc xoay quang tr c Oy cng bng 0, tc l 0 . x u w

(3) iu kin bin ta n Trn Hnh 19, bin chu iu kin ta n tha mn iu kin chuyn v theo ba phng x,

y, z bng 0, tc l 0 .u v w (4) iu kin drilling [6]

Khi thnh l p cng thc cho bi ton v, ta t ng thm vo gc xoay z , do trong

cc ma tr n cng phn t eK v ma tr n khi l ng phn t eM , gi tr ng ng tiv tr gc xoay z bng khng, iu ny dn n hin t ng suy bin khi gii phng trnh


56/70

48

(3.3.25). khc phc hin t ng ny, ti cc v tr tng ng v i gc xoay z trong ma

tr n cng v ma tr n khi l ng phn t, ta x p x li mt gi tr khc xc nh nh sau

3

3

10

10

, max ,

, max ,

e e

e e

in in diag

in in diag

K K

M M (3.3.26)

v i 6 12 18 24in v ediag K , ediag M l cc vc-t cha ng cho chnhca ma tr n cng phn t v ma tr n khi l ng phn t.


57/70

49

CHNG 4 V D S

Trong chng ny, mt s v s v v v v gia cng c phn tch bng phng php phn t hu hn dng phn t MITC4. K t qu phn tch ca cc bi ton s c so snh v nh giv i cc k t qu c ca cc tc gi khc.

4.1 Bi ton tnh hc Bi ton 1. v tr khng gia c ng.

Xt cu trc v hnh tr vi iu kin bin c mng hai u, lc p=1 tp trung im A. 1/4 cutrc v tr nh Hnh 20. vng tham kho ti A l 51 8248 10. [8].

Hnh 20. V hnh tr v i l c tp trung v iu kin bin mng.

K t qu vng ca v d i tc dng ca lc c th hin trong Hnh 21.


58/70

50

Hnh 21: vng ca v tr khng gia c ng dm chu l c tp trung.

Chia li vng (x10-5)MITC42x2 0.05764x4 0.73266x6 1.14768x8 1.3820

10x10 1.522512x12 1.610314x14 1.667816x16 1.707218x18 1.735220x20 1.755622x22 1.769024x24 1.7806

Tham kho[8] 1.8248

Bng 1. K t qu phn tch vng ca v tr khng gia c ng dm.

0

100

200

300

0

100

200

3000

50

100

150

200

250

300

X

do vong cua vo tru

Y

Z


59/70

51

Hnh 22. Tc hi t nghim ca phn t MITC4 cho bi ton v tr chu l c tp trung.

Hnh 22cho ta tc hi t ca vng ti im t lc, khi s l ng phn t tng ln th vng ti im A cng gn nghim chnh xc hn.

Bi ton 2: v cu gia c ng dm

Xt v cu c gia c ng b i hai dm ng tm vi kch th c hnh hc nh Hnh 23.

Hnh 23. V cu gia c ng b i hai dm ng tm.

0 100 200 300 400 500 6000

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

So phan tu

Do vong *10-5

Toc do hoi tu cua do von g

MITC4Ket qua tham khao


60/70

52

V cu chu lc t p trung ti tm p=45KN, vi iu kin ta n bn bin. vng ti tmca v c th hin trong

Hnh 24: vng ca v cu chu l c tp trung c gia c ng b i hai dm ng tm.

S phn t vng ti tm v (mm) Sai s

(%)MITC4Trc gia cng MITC4

Sau gia cng Kt qu

tham kho 16 64.98 34.84 42.26 -21.2936 73.83 40.00 42.95 -7.3764 76.29 42.15 43.19 -2.47100 78.75 43.31 43.3 0.01144 79.66 43.94 43.37 1.3

Bng 2: K t qu phn tch vng ti tm ca v cu gia c ng dm ng tm.

-1-0.5

00.5

1

-1

-0.5

0

0.5

12.25

2.3

2.35

2.4

2.45

2.5

X

Do vong tai tam vo gia cuong

Y

Z


61/70

53

Hnh 25: Tc hi t nghim ca phn t MITC4 cho bi ton v cu chu l c tp trunggia c ng hai dm ng tm.

Sau khi c gia cng, cng ca v tngln r t nhiu, ti mc l i 144 phn t, vngca v gim t 76.67 mm xung 43.95 mm. Sai s gia phn t MITC4 v tc gi kh nh 2% .

4.2 Bi ton ng hc Bi ton 4: Phn tch dao ng ca v tr gia c ng dm

Xt mt v tr c gia c ng 5 dm b ngm 1 cnh vi kch th c hnh hc nh Hnh 26.

20 40 60 80 100 120 140

35

40

45

50

55

60

65

70

75

80

So phan tu

Do vong taitam (mm)

Toc do hoi tu do vong tai tam

Ket qua tham khaoMITC4-Sau gia cuongMITC4-Truoc gia cuong


62/70


63/70

55

Mode 3 Mode 4

Mode 5 Mode 6

Mode 7 Mode 8

Hnh 27. Hnh dng tm mode dao ng t nhin ca v tr gia cng nm dm lch tmchu iu kin bin ngm.


64/70

56

mode

Tn s dao ng t nhin(Hz) Sai s

(%)MITC4(24*24)

Thamkho 1 105.0751596 115 8.63

2 155.2606013 160 2.963 320.3040897 313 -2.334 467.7738806 479 2.345 516.6032379 530 2.536 773.1143817 714 -8.287 809.1875498 803 -0.778 989.4732511 947 -4.49

Bng 3. K t qu phn tch tn s dao ng ca v tr gia c ng dm lch tm.

Hnh 28: Gi tr tn s ca tm mode dao ng u tin.

K t qu phn tch tn s dao ng t do ca v tr gia c ng so v i k t qu thc nghim lkh gn. Tuy nhin, ti mode 1 v mode 6, tn s ca bi ton c phn tch bng phn t MITC4 cn xa so v i k t qu thc nghim ( ~ 8%).

1 2 3 4 5 6 7 8100

200

300

400

500

600

700

800

900

1000

Mode dao dong

Tan so dao dong (Hz)

Tan so dao dong cua vo tru gia cuong dam

Tham khaoMITC4


65/70

57

Bi ton 5: Phn tch dao ng ca v cu gia cng dm.

Xt v cu vi kch th c hnh hc v vt liu trong bi ton 3, ta c k t qu gi tr tn s gc(rad/s) ca 10 mode u tin nh sau.

Mode 1 Mode 2

Mode 3 Mode 4

Mode 5 Mode 6


66/70

58

Mode 7 Mode 8

Mode 9 Mode 10

Hnh 29. Hnh dng mi mode dao ng t nhin u tin ca v cu gia c ng hai dmng tm vi iu kin bin ta n.

Mode Tn sgc (rad/sec) Sai s (%) MITC4 Tham kho[10] 1 45.63842 41.68 8.672 75.5371 74.38 1.533 75.7006 74.49 1.64 99.5975 100.5 -0.915 115.65 105.44 8.836 115.671 105.55 8.75

7 128.3576 126.87 1.168 139.0469 134.69 3.139 151.179 146.47 3.1110 151.4326 146.51 3.25

Bng 4. K t qu phn tch tn s gc ca v cu gia c ng dm ng tm.


67/70


68/70

60

CHNG 5 KT LUN V HNG PHT TRIN

5.1 Cc kt qu t c

Lun vn thnh lp c cc cng thc phn t hu hn cho k t cu v thoi gia c ng dm.ng th i, p dng phng php MITC4 trong vic phn tch tnh hc v ng hc cho bi tonv, v gia c ng dm ng tm v v gia c ng dm lch tm. Trong v c m hnh theo lthuyt v thoi v i gi thit Reissner-Mindlin. Dm c m hnh theo l thuyt dm Timoshenko.Sai s trong cc bi ton kh th p 5% . Vic l p trnh m phng bi ton c thc hin

bng phn mm Matlab.

5.2 Hn ch v tn ti V iu kin th i gian v phm vi kin thc c hn nn lun vn m i ch dng li vic s dngl thuyt v thoi phn tch bin dng ca cu trc v v v gia c ng chu lc t p trung v iiu kin bin n gin, phn tchdao ng t do ca v v v gia c ng,cha cp n nhngcu trc v v v gia c ng chu dao ng c ng bc, dao ng tt dn. Nhng cu trc c trnh by trong lun vnc min hnh hc cn kh n gin, cha phn nh c nhng cu trc thctrong thc t. Cc nghin cu v s hi t v tc hi t cn kh s si.

5.3 Hng pht trin trong tng lai Trong tng lai, phn t MITC4 c th c p dng phn tch nhng cu trc c min hnhhc phc t p hn nh cu trc v yn nga, v hnh nn, v hnh cu, v.v. Phn t MITC4 cngc th c s dng phn tch dao ng cng bc, dao ng tt dn ca v v v gia c ng,v.v. Ngoi ra, c th s dng l thuyt v suy bin, v 3D, v cong phn tch nhng cu trcv v v gia c ng.ng th i, c th ti u ha cho cu trc v cng nhv gia c ng.y l bi ton mang tnh th i s v kh m i m.


69/70

61

TI LIU THAM KHO

[1] K.M. Liew, S. Kitipornchai L.X. Peng, "Buckling and free vibration analyses of stiffen plates using the FSDT mesh-free method," Journal of Sound and Vibration , 2006.

[2] F. Brezzi K.J Bathe, "One the convergence of a four-node plate bending element basedmindlin plate theory and a mixed interpolation," In proceedings of the conference onmathematics of finite elements and applications v, whiteman j(ed.). Academic press: NewYork , pp. 491-503, 1985.

[3] K.J Bathe, A. Iosilevich D. Chaplle, "An Evanluation of the MITC Shell Element,"Computers and Structures , 2000.

[4] Chu Quc Thng, "Bi ging l thuyt tm v".

[5] J.N. Readdy and K.K. Lee C.M. Wang, "Shear deformable beams and plates relationshwith classical solutions," Elsevier , 2000.

[6] Kaushalkumar Kansara, "Development of Membrane, Plate and Flat Shell Elements in," pp. 69-71, 2004.

[7] Timon Rabczuk, H. Nguyen Xuan, Stephane P. A Bordas N. Nguyen Thanh, "A smoot

finite element method for shell analysis," 2008.[8] W. Fluge, "Stress and shells,"Springer. Berlin , 1960.

[9] R. Ati B. A. J. Mustafa, "Frediction of natural frequency of vibration of stiffenedcylindrical shells and orthogonally stiffenend curved panels,"Sound and Vibration , 1986.

[10] Madhujit Mukhopadhyay Asokendu Samanta, "Free vibration analysis of stiffened shethe finite element technique," European Journal of Mechanics A/Solids , 2003.

[11] B. Gangadhara, S. K. Satsangi Prusty, "Analysis of stiffened shell for ships and oceanstructures by finite element method,"Ocean Engineering , pp. 636-637, 1999.

[12] Nguyn Xun Hng Nguyn Thi Trung, "Phng php phn t hu hn," 2012.

[13] Tr nh Anh Ngc, "Bi ging phng php phn t hu hn," 2012.


70/70

Documents

Thesis_Dang Trung Hau