Two- and Three Dimensional Solid Elements; Plane Stress, Plane Strain, and Axisymmetric Conditions
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1 Topic 7 Two- and Three Dimensional Solid Elemens; Plane Sress, Plane Srain, and Axisymmeric Condiions Conens: soparameric inerpolaions of coordinaes and displacemens Consisency beween coordinae and displacemen inerpolaions Meaning of hese inerpolaions in large displacemen analysis, moion of a maerial paricle Evaluaion of required derivaives The Jacobian ransformaions Deails of srain-displacemen marices for oal and updaed Lagrangian formulaions Example of 4-node wo-dimensional elemen, deails of marices used Texbook: Example: Secions 6.3.2,
2 Topic Seven 7-3 J=TE ELfHEAlrs CA 5EE'KAl ~E GORZE]) AS CATE - COTlA~ M ELE.METs ( ~oc..'j» - S~U(TLfRAL ELEMETS. THS LEC.TlARE We COSlbE'R T\-\E 2-b C.O' la,u M ~ 0 PA'RA HETT<\C. ELEMETS. T 11E- ElE METS.A'RE. VE~~ ~Ee~AL ELE METS FoR Q ~o- M ET~C Ab H ATJ;R.AL OLEAR. COl:>TOS WE Al<;o 'POT O\AT How bferal 5-]) ElEHETS ARE CALClALATEb (As b TH : SA~E '?~O(E- )) U~E S.. T4~~ E ELf HE:T~ ARE- US~'> VE'1C' WDE.l'( Markerboard 7-1
3 7-4 Two- and Three-Dimensional Solid Elemens 7-1 TWO- AD THREE-DMESOAL SOLD ELEMETS Two-dimensional elemens comprise - plane sress and plane srain elemens - axisymmeric elemens The derivaions used for he wodimensional elemens can be easily exended o he derivaion of hreedimensional elemens. Hence we concenrae our discussion now firs on he wo-dimensional elemens. 7-2 TWO-DMESOAL AXSYMMETRC, PLAE STRA AD PLAE STRESS ELEMETS X1
4 Topic Seven 7-5 Because he elemens are isoparameric, 0 L hk x~ L hk x~ X1 = k=1, X2 = k=1 and L h k x~ X1 =, X2 = L h k k=1 k=1 x~ 7-3 where he hk's are he isoparameric inerpolaion funcions. Example: A four-node elemen s 4 Xi = L hk x~ k=1 4 Xi = 2: hk x~ k=1 7-4 ime 0 where 1 h1 = 4: (1 + r)(1 + s) 1 h 2 = - (1 - r)(1 + s) 4 1 h 3 = - (1 - r)(1 - s) 4 1 h4 = 4 (1 + r)(1 - s)
5 7-6 1\vo- and Three-Dimensional Solid Elemens 7-5 x x; = hkl k-1 r;0.5 5=0.5 r x~ A major advanage of he isoparameric 7-6 finie elemen discreizaion is ha we may direcly wrie U1 ~ h k u~ U2 = k=1 k=1 and ~ h k u~ U1 ~ hk u~ U2 = ~ hk u~ k=1 k=1
6 Topic Seven 7-7 This is easily shown: for example, ~ h k Xi = J k Xi k=1 o ~ h 0 k Xi = J k Xi k=1 7-7 Subracing he second equaion from he firs equaion gives 0 ~ h ( k 0 k) Xi - Xi = J k Xi - Xi,. k=1 ',. The elemen marices require he following derivaives: 7-8
7 7-8 '\vo- and Three-Dimensional Solid Elemens 7 9 These derivaives are evaluaed using a Jacobian ransformaion (he chain rule): ahk _ ahk aox1 + ahk aox2 ar - aox1 ar aox2 ar ahk ahk aox1 ahk aox2 as - aox1 as + aox;. as ~REaURED n marix form, ~ DERVATVES r A, ahk ilxl aox2 ahk as as as aox The required derivaives are compued using a marix inversion: ahk a O x1 ahk a Ox 2 = 0J-1 ahk - ar ahk - as The enries in oj are compued using he inerpolaion funcions. For example, h a X1 = L a k 0x~ ar k=1 ar
8 Topic Seven 7-9 The derivaives aken wih respec o he configuraion a ime can also be evaluaed using a Jacobian ransformaion ah k a 1 x1 a l X2 ah - k = - - ar ar ar a l X1 ah k a 1 x1 a 1 x2 ah - k - - as as as a 1 x2 J f ah k X~ ah k ah k k=1 as a 1 x1 ah k a l X2 = J-1 - ar ah - k as We can now compue he required elemen marices for he oal Lagrangian formulaion: 7-12 Elemen Marix Marices Required oc, dbl ds, dbl A os, ob L
9 7-10 Two- and Three-Dimensional Solid Elemens We define oc so ha analogous o OSij. = oc~,s oe,s For example, we may choose (axisymmeric analysis), 1 v 1 - v v v 0 _ E(1 - v) 1 - v 1 - v oq - (1 + v)(1-2v) 1-2v (1 - v) v v 1 - v 1 - v o We noe ha, in wo-dimensional analysis, Oe11 = OU1,1 +lou1,1 OU1,1 + OU2,1 OU2,1, Oe22 = OU2,2 + OU 1,2 ou 1,2 + OU2,2 OU2,2, + f ) U2,2 OU2,1 2 Oe12 = (OU1,2 + OU2,1) +,(JU 1,1 OU1,2 ~e:~: ;~2: +(~~~1)2O:~" X1 X1 X1 ""-TAL DSPLACEMET EFFECT
10 Topic Seven 7-11 and (U1)2 01")33 = 2 Ox1 Derivaion of 0833, o~: X2./' ime 0 /ime +d axis of r-\ C revolu~ L-:=1 ~ Transparenc) 7-16 '" X1 We see ha
11 7-12 '\vo- and Three-Dimensional Solid Elemens 7-17 Hence H.1 _ 1 [(H.1dS)2 ] oe :;-: ds 7-18 We consruc rib l so ha ~ Oe22 oe11] 2 Oe12 = o~ = (06l0 + 06L1) a De.. ~ JBL J Oe33 is only included for axisymmeric analysis conains iniial displacemen effec
12 'bpic Seven 7-13 Enries in 6Sl0: node k ohk,1 0 0 ohk,2 ohk,2 ohk,1 hkfx included only for axisymmeric analysis u~ u~ node k l 7-19 This is similar in form o he B marix used in linear analysis. Enries in 6Bl1: - u~ node k.. u~. - 6U1,1 ohk,1 6U2,1 ohk,1 6U1,20hk,2 6U2,20hk, U1,1 Ohk,2 6U2,10hk,2 h + 6U2,2 ohk,1 + OU1,2 0 k,1 U1 hk X1 X1 0 Th... d' ff included only e nlla SP acemen e ec u is conained in he erms 6Ui,~, OX:. analyss ~ - for axisym~eric 7-20
13 7-14 1\vo- and Three-Dimensional Solid Elemens 7-21 We consruc dsl and ds so ha Enries in ds: s:: ~ T ST s s ~ s s:: u!! O_L 0_ O_L!! = i}uolli} r- - ds11 ds ds21 ds ds11 ds ds21 ds ds33 - included only for axisymmeric analysis Enries in dsl: 7-22 f node k u~, u~ ohk,1 0 r- - ohk, Ohk ohk,2 hk/ox1 0 L- - included only for axisymmeric analysis u~ --- u~ node k 1
14 Topic Seven 7-15 os A is consruced so ha 7-23 Enries in JS: JS11 J822 J8 12 J833 ~included only for axisymmeric analysis
15 7-16 '\vo- and Three-Dimensional Solid Elemens Exam~: Calculaion of JBl JBl X Plane srain condiions l-.l"'"--ime 0 Exam~: Calculaion of JBl JBl 7-26 X Plane srain condiions.2 r-:======:i.j-j-...~ime (0.1,0.1 ) 4 1 maerial fibers have only ranslaed rigidly
16 Topic Seven 7-17 Example: Calculaion of db L db L Plane srain condiions 0.2 (0.1,0.1) 0.2 x, Exam~: Calculaion of db Ll db L X Plane srain condiions ime 0 maerial fibers have sreched and roaed (0.1,0.1) x,
17 7-18 Two- and Three-Dimensional Solid Elemens Exam~: Calculaion of JBL, JBL X Plane srain 2 condffions s ~~rime ( ) 1 A ime 0, 7-30 We can now perform a Jacobian ransformaion beween he [, s) coordinae sysem and he ( X1,OX2) coordinae sysem: By inspecion, Hence oj - = [0.1 a x1 = 1 a X2 = and a a a a :;CC-a = 1 -a ':;CC-a = 1 -a ar., ar aox1 = aox2 as 'as = 0.1 0~1], 1 41 = 0.01 X1 r X2 s
18 'bpic Seven 7-19 ow we use he inerpolaion funcions o compue JU1,1, JU1,2: 7-31 node ahk ahk u~ ahk k ahk k aox1 aox2 r U1 - U1 k X1 aox (1 + 5) 2.5(1 + r) (1 + 5) 0.25(1 + r) 2-2.5(1 + 5) 2.5(1 - r) (1 + 5) 0.25(1 - r\ 3-2.5(1-5) -2.5(1 - r) (1-5) -2.5(1 + r) Sum: 0.0 " 4,/ OU1,1 0.5 " 4 j OU1,2 For his simple problem, we can compue he displacemen derivaives by inspecion: 7-32 From he given dimensions, Hence Jx = [ ] JU1,1 = JX11-1 x OU1,2 = 0 12 x OU2,1 = 0 21 JU2,2 = JX22-1 =0 = 0.5 =0 = 0.5
19 7-20 Two- and Three-Dimensional Solid Elemens 7-33 We can now consruc he columns in rib L ha correspond o node 3: ['0-2.5(1-5) o -2.5(1 - r)! -2.5~1 - r)..j -2.5(1-5) 0 : (1 - r) : -1.25(1 - r) [ -1.25(1-5)1-1.25(1-5) ooj 7-34 Similarly, we consruc he columns in rib L ha correspond o node 3: -2.5(1-5) 0-2.5(1 - r), 0 o -2.5(1-5) o -2.5(1 - r)
20 Thpic Seven 7-21 Consider nex he elemen marices required for he updaed Lagrangian formulaion: 7-35 Elemen Marix Marices Required C, ~Bl 'T, ~Bl A 'T, Bl We define C so ha Sl1 S22 S12 S33 = e el1 e22 2 e12 e33 analogous o S = Crs ars 7-36 For example, we may choose (axisymmeric analysis), 1 v 1 - v o v v 1 0 C - E(1 - v) 1 - v 1 - V _ - (1 + v)(1-2v) 1-2v (1 - v) v v v 1 - v
21 7-22 1\\'0- and Three-Dimensional Solid Elemens 7-37 We noe ha he incremenal srain componens are, in wo-dimensional analysis, au1 e11 = -a = U1,1 X1 e22 = U2,2 2 e12 = U1,2 + U2,1 e33 = U1/X and 'T11 = ~ ((U1,1)2 + (U2,1)2) 1 (U1)2 T)33 = 2 X1
22 Topic Seven 7-23 We consruc ~BL so ha e11 e22 2 e12 e 33 = e = B A L U only included for axisymmeric analysis Enries in ~BL: node k h k,1 0 0 h... k,2... hk,2 hk,1 hk/x u~ u~ only included for axisymmeric analysis node k 1 This is similar in form o he B marix used in linear analysis.
23 7-24 Two- and Three-Dimensional Solid Elemens 7-41 We consruc ~SL and T so ha ~ A TsT,.,. s A,.,. ~ u~ _L...!.- _L ~ = y.u'tly. Enries in 'T: r-- - rr 'T11 'T21 'T 'T11 'T 'T21 'T 'T33 included only for axisymmeric analysis 7-42 Enries. n s L: node k ' u~!, u~ hk,1 hk,2 0 0 hk, hk,2 hk/x1 0 l-. - u~ u~ node k l ~included only for axisymmeric analysis
24 Topic Seven 7-25 " 'T is consruced so ha 7-43 Enries in f: included only for axisymmeric analysis Three-dimensional elemens 7-44 X2 ~ node k (ox~, 0x x~)
25 7-26 Two- and Tbree-Dimeusional Solid Elemens 7-45 Here we now use o ~ h k X1 = J k X1 k=1 0X3 = ~ h k O~, k=1 o ~ h k,x2= J kx2 k=1 where he hk's are he isoparameric inerpolaion funcions of he hreedimensional elemen Also 1<1= ~ hk 1<~, 1<2 = ~ hk 1<~ k=1 k=1 1cs = ~ h k x~ k=1 and hen all he conceps and derivaions already discussed are direcly applicable o he derivaion of he hree-dimensional elemen marices.
26 MT OpenCourseWare hp://ocw.mi.edu Resource: Finie Elemen Procedures for Solids and Srucures Klaus-Jürgen Bahe The following may no correspond o a paricular course on MT OpenCourseWare, bu has been provided by he auhor as an individual learning resource. For informaion abou ciing hese maerials or our Terms of Use, visi: hp://ocw.mi.edu/erms.
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