BAR & TRUSS FINITE ELEMENT. Direct Stiffness Method

Transcription

1 BAR & TRUSS FINITE ELEMENT Drect Stness Method FINITE ELEMENT ANALYSIS AND APPLICATIONS INTRODUCTION TO FINITE ELEMENT METHOD What s the nte element method (FEM)? A technqe or obtanng approxmate soltons o derental eqatons. Partton o the doman nto a set o smple shapes (element) Approxmate the solton sng pecewse polynomals wthn the element F xx xy bx 0 x y xy yy Strctre by 0 x y Element Pecewse-Lnear Approxmaton x

2 INTRODUCTION TO FEM cont. How to dscretze the doman? Usng smple shapes (element) All elements are connected sng nodes Nodes Elements Solton at Element s descrbed sng the vales at Nodes,, 6, and 5 (Interpolaton). Elements and share the solton at Nodes and 6. INTRODUCTION TO FEM cont. Methods Drect method: Easy to nderstand, lmted to D problems Varatonal method Weghted resdal method Obectves Determne dsplacements, orces, and spportng reactons Wll consder only statc problem

3 -D SYSTEM OF SPRINGS F 5 F 6 5 F Bodes move only n horzontal drecton External orces, F, F, and F, are appled No need to dscretze the system (t s already dscretzed!) Rgd body (ncldng walls) NODE Sprng ELEMENT 5 SPRING ELEMENT Element e Consst o Nodes and, Sprng constant k (e) Force appled to the nodes: e e, () e e, () e Dsplacement and Elongaton: e e e e Force n the sprng: P k k Relaton b/w sprng orce and nodal orces: Eqlbrm: ( e ) e e e e 0 or e P e 6

4 Sprng Element e SPRING ELEMENT cont. Relaton between nodal orces and dsplacements e e e e k k Matrx notaton: k: stness matrx q: vector o DOFs : vector o element orces e e k e e k k k k [ k ] q k q 7 Stness matrx SPRING ELEMENT cont. It s sqare as t relates to the same nmber o orces as the dsplacements. It s symmetrc. It s snglar,.e., determnant s eqal to zero and t cannot be nverted. It s postve sem-dente Observaton For gven nodal dsplacements, nodal orces can be calclated by [ k ] q ( e ) ( e ) ( e ) For gven nodal orces, nodal dsplacements cannot be determned nqely e k k e k k e e 8

5 Element eqaton and assembly SYSTEM OF SPRINGS cont. F F 5 F 6 5 () k k () k k () k k () k k () k k () k k () k k () () k 0 k 0 k k () 0 k 0 k SYSTEM OF SPRINGS cont. () k k () () () () k k 0 k k k k k k () k k () 0 k k 0 0 ( ) 0 k 0 k () () kk k k 0 0 () () () () k k 0 k k k k k k () k k () () k k kk 0 0 () 0 k 0 k (5) k5 k5 (5) k5 k 5 () () kk k k 0 0 () () () k k 0 k k k k () () (5) k k k k k5 k5 0 () (5) 0 k k5 k k

6 (6) k6 k6 (6) k6 k SYSTEM OF SPRINGS cont. () () kk k k 0 0 () () () k k k k k k 0 () () (5) k k k k5 k k5 0 () (5) (6) 0 k k5 k k5 k6 k 6 (6) k6 k6 5 5 F 5 F 6 5 F Relaton b/w element orces and external orce Force eqlbrm F e e e e e e SYSTEM OF SPRINGS cont. 0 F,,... ND At each node, the smmaton o element orces s eqal to the appled, external orce F () () F 5 F 6 5 (5) F R () () () () () F () () (5) F () (5) (6) F (6) R 5 5 6

7 SYSTEM OF SPRINGS cont. Assembled System o Matrx Eqaton: k k k k 0 0 R k k k k k k 0 F k k k k5 k k5 0 F 0 k k5 k k5 k6 k 6 F k6 k6 5 R5 [ ]{ Q } { F } s s s [ s ] s sqare, symmetrc, snglar and postve sem-dente. When dsplacement s known, orce s nknown 5 0 R and R 5 are nknown reacton orces SYSTEM OF SPRINGS cont. Imposng Bondary Condtons Ignore the eqatons or whch the RHS orces are nknown and strke ot the correspondng rows n [ s ]. Elmnate the colmns n [ s ] that mltply nto zero vales o dsplacements o the bondary nodes. k k k k 0 0 R k k k k k k 0 F k k k k5 k k5 0 F 0 k k k k k k F k6 k6 5 R5 7

8 Global Matrx Eqaton SYSTEM OF SPRINGS cont. k k k k k F k k k k5 k 5 F k k5 k k5 k 6 F [ ]{ Q} { F} Global Stness Matrx [] sqare, symmetrc and postve dente and hence non-snglar Solton { Q} [ ] { F} Once nodal dsplacements are obtaned, sprng orces can be calclated rom e e e e P k k 5 UNIAXIAL BAR For general naxal bar, we need to dvde the bar nto a set o elements and nodes Elements are connected by sharng a node Forces are appled at the nodes (dstrbted load mst be converted to the eqvalent nodal orces) Assemble all elements n the same way wth the system o sprngs Solve the matrx eqaton or nodal dsplacements p(x) F Calclate stress and stran sng nodal dsplacements Statcally ndetermnate p(x) Statcally determnate F x 6 8

9 Two-orce member Only constant cross-secton Element orce s proportonal to relatve dspl Frst node: second code: Force-dsplacement relaton D BAR ELEMENT () e Node x L =EA/L Node A () e AE ( ) L AE ( ) L Smlar to the sprng element 7 Matrx notaton D BAR ELEMENT cont. AE L Ether orce or dsplacement (not both) mst be gven at each node. Example: = 0 and = 00 N. What happens when and are gven? Nodal eqlbrm Eqlbrm o orces actng on Node I () e Node =EA/L ( ) ( ) ( ) { e } [ k e ]{ q e } Node () e F 0 F In general F e F e Element e (e) Node (e+) Element e+ 8 9

10 Assembly D BAR ELEMENT cont. Smlar process as sprng elements Replace all nternal nodal orces wth External Appled Nodal Force Obtan system o eqatons [ ]{ Q } { F } s s s Property o [ s ] [ s ]: Strctral stness matrx {Q s } Vector o nodal DOFs {F s }: Vector o appled orces Sqare, symmetrc, postve sem-dente, snglar, non-negatve dagonal terms Applyng bondary condtons Remove rgd-body moton be xng DOFs Strkng-the-nodes and strkng-the-colmns (Reer to sprnt elements) [ ]{ Q} { F} []: Global stness matrx {Q} Vector o nknown nodal DOFs {F}: Vector o known appled orces 9 D BAR ELEMENT cont. Applyng bondary condtons cont. [] s sqare, symmetrc, postve dente, non-snglar, nvertble, and postve dagonal terms Can obtan nqe {Q} Element orces Ater solvng nodal dsplacements, the element orce can be calclated AE P L P A Element stress Reacton Forces P AE P L Note P = P Use [ s ]{Q s } = {F s }: the rows that have been deleted (strke-the-rows) Or, se e F e 0 0

11 EXAMPLE elements and nodes At node : F () () () Eqaton or each element: F Element Element x Element () () () () () () N F Element Element N N Element N F F EXAMPLE cont. How can we combne derent element eqatons? (Assembly) Frst, prepare global matrx eqaton: Dsplacement vector Stness matrx Appled orce vector Wrte the eqaton o element n the correspondng locaton 0 0 () ()

12 Wrte the eqaton o element : EXAMPLE cont. () 0 0 () Combne two eqatons o elements and 0 0 () () () 0 () Wrte the eqaton o element EXAMPLE cont () () 0 0 Combne wth other two elements () F 0 0 () () () F ( ) () F 0 0 () F 0 0 Strctral Stness Matrx

13 EXAMPLE cont. Sbsttte bondary condtons and solve or the nknown dsplacements. Let = 50 N/cm, = 0 N/cm, = 70 N/cm and = 0 N. F F 50 ( ) 0 70 F F nowns: F, F,, and Unknowns: F, F,, and ( ) 0 70 F F EXAMPLE cont. Remove zero-dsplacement colmns: and F 0 0 F 0 70 Remove nknown orce rows: F and F Now, the matrx shold not be snglar. Solve or and. Usng and, Solve or F and F.. cm 0. cm F 0 0 N F N 6

14 EXAMPLE cont. Recover element data () () Element orce () () () () N F = 0 N. cm 0. cm -8 N 7 EXAMPLE Statcally ndetermnate bars E = 00 GPa R L F = 0,000 N A = 0 m, A = 0 m Element stness matrces: () [ k ] A B C 0.5 m () [ k ] Assembly 0 F , F F 0. m R R 8

15 Applyng BC EXAMPLE cont ,000.0 m 7 Element orces or Element stresses AE P L Reacton orces () 7 P 0, N () 7 P 50 5,556N () RL P N () RR P 5,556N,, 9 PLANE TRUSS ELEMENT What s the derence between D and D nte elements? D element can move x- and y-drecton ( DOFs per node). However, the stness can be appled only axal drecton. Local Coordnate System D FE ormlaton can be sed a body-xed local coordnate system s constrcted along the length o the element Y The global coordnate system (X and Y axes) s chosen to represent the entre strctre X The local coordnate system (x and y axes) s selected to algn the x-axs along the length o the element x EA x L cm 8 cm 50 N 0 5

16 PLANE TRUSS ELEMENT cont. Element Eqaton (Local Coordnate System) Axal drecton s the local x-axs. D element eqaton x 0 0 y v EA x L 0 0 y v { } [ k]{ q} [ k] s sqare, symmetrc, postve sem-dente, and non-negatve dagonal components. How to connect to the neghborng elements? Cannot connect to other elements becase LCS s derent Use coordnate transormaton y x Local coordnates v y x v x Global coordnates x COORDINATE TRANSFORMATION Transorm to the global coord. and assemble cos sn v sn cos v cos sn v sn cos v v v Transormaton matrx cos sn 0 0 v sn cos 0 0 v 0 0 cos sn v 0 0 sn cos v local global { q} [ T]{ q} Transormaton matrx 6

17 COORDINATE TRANSFORMATION cont. The same transormaton or orce vector x cos sn 0 0 x y sn cos 0 0 y 0 0 cos sn x x y 0 0 sn cos y local global Property o transormaton matrx T T { } [ T]{ } { } [ T] T { } [ ] [ ] T { } [ T]{ } ELEMENT STIFFNESS IN GLOBAL COORD. Element x 0 0 y EA v L 0 0 x y v element stness matrx { } [ k]{ q} Transorm to the global coordnates [ T]{ } [ k][ T]{ q} y x v N { } [ T] [ k][ T] { q} global global v N x x [ k] [ T] [ k][ T] { } [ k]{ q} 7

18 ELEMENT STIFFNESS IN GLOBAL COORD. cont. Element stness matrx n global coordnates [ k] T [ k] T [ k] EA L T cos cos sn cos cos sn cos sn sn cos sn sn cos cos sn cos cos sn cos sn sn cos sn sn Depends on Yong s modls (E), cross-sectonal area (A), length (L), and angle o rotaton () Axal rgdty = EA Sqare, symmetrc, postve sem-dente, snglar, and non-negatve dagonal terms 5 EXAMPLE N 50 N Two-bar trss Dameter = 0.5 cm E = 00 6 N/cm Element 8 cm Element In local coordnate () () () { } [ k ]{ q } N cm Element N x 0 0 y EA v L 0 0 x y v y x v N v N x =.7 o E = 0 x 0 6 N/cm A = pr = 0.09 cm L =. cm x 6 8

19 Element cont. EXAMPLE cont. Element eqaton n the global coordnates () x () y v 050 () x () y v y () () () { } [ k ]{ q } x Element () x () 0 0 y v 85 () x () y 0 0 v = 90 o E = 0 x 0 6 N/cm A = pr = 0.09 cm L = 8 cm N v N v x x 7 EXAMPLE cont. Assembly Ater transormng to the global coordnates Element F x F y v F x F y v F x F y v Bondary Condtons Nodes and are xed. Node has known appled orces: F x = 50 N, F y = 0 N Element 8 9

20 EXAMPLE cont. Bondary condtons (strkng-the-colmns) F x F y v F x F y Strkng-the-rows v Solve the global matrx eqaton v 8.80 cm.80 cm 9 Spport reactons EXAMPLE cont. F x F y 8.80 N F x F y The reacton orce s parallel to the element length (two-orce member) Element orce and stress (Element ) Need to transorm to the element local coordnates v v v

21 EXAMPLE cont. Element orce and stress (Element ) cont. Element orce can only be calclated sng local element eqaton x EA N y x L y There s no orce components n the local y-drecton In x-drecton, two orces are eqal and opposte The orce n the second node s eqal to the element orce Normal stress = 60. / 0.09 = 8 N/cm. 60. N 60. N OTHER WAY OF ELEMENT FORCE CALCULATION Element orce or plane trss AE AE P L L Wrte n terms o global dsplacements AE P l mv l mv L l mv v AE L l cos m sn

22 EXAMPLE Drectly assemblng global matrx eqaton (applyng BC n the element level) Element property & drecton cosne table Elem AE/L -> l = cos m = sn > > > Snce and v wll be deleted ater assembly, t s not necessary to keep them v v l lm l lm () () EA lm m lm m v L l lm l lm lm m lm m v k F 5 () () EA l lm L lm m v k v SPACE TRUSS ELEMENT A smlar extenson sng coordnate transormaton DOF per node, v, and w x, y, and z Y v Y X w v N x Element stness matrx s 6x6 x w N Z X Z FE eqaton n the local coord. x AE x L { } [ k]{ q}

23 SPACE TRUSS ELEMENT cont. Relaton between local and global dsplacements Each node has DOFs (, v, w ) v l m n 0 0 0w { q} [ T] { q} l m n () ( 6) (6) v w Drecton cosnes x x y y z z l cos x, m cos y, n cosz L L L L x x y y z z 5 SPACE TRUSS ELEMENT cont. Relaton between local and global orce vectors Stness matrx x l 0 y m 0 z n 0 x x 0 l x y 0 m z 0 n { } [ T] T { } T T { } [ k]{ q } [ T] { } [ T] [ k][ T]{ q } { } [ k]{ q} l lm ln l lm ln m mn lm m mn v EA n ln mn n w [ k] L l lm ln sym m mn v n w [ k] [ T] T [ k][ T] 6

24 THERMAL STRESSES Temperatre change cases thermal stran L L L No stress, no stran No stress, thermal stran (a) at T = T re Thermal stress, no stran (b) at T = T re + T Constrants case thermal stresses Thermo-elastc stress-stran relatonshp = E T = + T E Thermal expanson coecent 7 THERMAL STRESSES cont. Force-dsplacement relaton L L P = AE T AE AET L L Fnte element eqaton Thermal orce vector ( ) ( ) ( ) ( ) { e } [ k e ]{ q e } { e T } ( ) 0 e v { T } AET For plane trss, transorm to the global coord. 0 v { } [ k]{ q} { } [ k]{ q} { } { } T T l m v { T} AET l mv [ ]{ Q } { F } { F } s s s Ts 8