MultiFreedom Constraints I

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1 Introdction to FEM 8 MltiFreedom Constraints I IFEM Ch 8 Slide

2 Introdction to FEM x4 = 0 Mltifreedom Constraints Single freedom constraint examples linear, homogeneos y9 = 0. linear, non-homogeneos Mltifreedom constraint examples x2 = 2 y2 linear, homogeneos x2 2 x4 + x = 0.25 linear, non-homogeneos (x 5 + x5 x 3 x3 ) 2 +(y 5 + y5 y 3 y3 ) 2 = 0 nonlinear, homogeneos IFEM Ch 8 Slide 2

3 Introdction to FEM Sorces of Mltifreedom Constraints "Skew" displacement BCs Copling nonmatched FEM meshes Global-local and mltiscale analysis Incompressibility Model redction IFEM Ch 8 Slide 3

4 Introdction to FEM MFC Application Methods Master-Slave Elimination Chapter 8 Penalty Fnction Agmentation Lagrange Mltiplier Adjnction Chapter 9 IFEM Ch 8 Slide 4

5 Introdction to FEM Procedre Smmary in Static Analysis Valid for the Three Methods Unmodified master stiffness eqations K = f before applying MFCs Apply MFCs master slave penalty fnction Lagrange mltiplier ^ Modified stiffness eqations K ^ = ^ f Eqation solver gives ^ Recover if necessary IFEM Ch 8 Slide 5

6 Introdction to FEM Example D Strctre to Illstrate MFCs, f 2, f 2 3, f 3 4, f 4 5, f 5, f 7, f7 () (2) (3) (4) (5) () x Mltifreedom constraint: = or 2 2 = 0 Linear homogeneos MFC IFEM Ch 8 Slide

7 Introdction to FEM Example D Strctre (Cont'd), f 2, f 2 3, f 3 4, f 4 5, f 5, f 7, f7 () (2) (3) (4) (5) () Unconstrained master stiffness eqations K K K 2 K 22 K K 23 K 33 K K 34 K 44 K = K 45 K 55 K K 5 K K K 7 K 77 7 or K = f x f f 2 f 3 f 4 f 5 f f 7 IFEM Ch 8 Slide 7

8 Introdction to FEM Master-Slave Method for Example Strctre Recall: = or 2 2 = 0 Taking 2 as master and as slave = or = T ^ IFEM Ch 8 Slide 8

9 Forming the Modified Stiffness Eqations Introdction to FEM Unconstrained master stiffness eqations: Master-slave transformation: K = f = T ^ Replace & premltiply T both sides by T : K T = f T ^ ^ T T K T = T f ^ T On renaming K = T K T and ^ T f = f we get the modified stiffness eqations: ^ ^ ^ K = f IFEM Ch 8 Slide 9

10 Introdction to FEM Modified Stiffness Eqations for Example Strctre ˆ ˆ In fll K K K 2 K 22 + K K 23 0 K 5 K K 23 K 33 K K 34 K 44 K K 5 0 K 45 K K K 77 7 ˆ K = f = f f 2 + f f 3 f 4 f 5 f 7 ˆ Solve for, then recover = T ˆ IFEM Ch 8 Slide 0

11 Sppose Pt in matrix form: = 5 7 Mltiple MFCs 2 = 0, = 0, = 0 Pick 3, 4 and as slaves while, 2, 5 and 7 are masters Introdction to FEM = 2 4 = 4 3 = 2 ( ) = This is = T ˆ - then proceed as before IFEM Ch 8 Slide

12 Non-homogeneos MFCs Introdction to FEM 2 = 0.2 Pick again as slave, pt into matrix form: = IFEM Ch 8 Slide 2

13 Nonhomogeneos MFCs (cont'd) Introdction to FEM = T ˆ + g g: "gap" vector T Premltiply both sides by T K, replace K = f and pass data to RHS. This gives K ˆ ˆ = fˆ with ˆ T ˆ T K = T K T and f = T (f K g) a modified force vector IFEM Ch 8 Slide 3

14 Nonhomogeneos MFCs (cont'd) Introdction to FEM For the example strctre K K K 2 K 22 + K K 23 0 K 5 K K 23 K 33 K K 34 K 44 K K 5 0 K 45 K K K 77 7 = f f 2 + f 0.2K f 3 f 4 f 5 0.2K 5 f 7 0.2K 7 Solve for, ˆ then recover = T ˆ + g IFEM Ch 8 Slide 4

15 Model Redction Example Introdction to FEM, f 2, f 2 3, f 3 4, f 4 5, f 5, f 7, f7 () (2) (3) (4) (5) () x,f 2 master DOFs to be retained 7,f slave DOFs to be eliminated Master,f 7,f 7 Redced model 7 Master x x IFEM Ch 8 Slide 5

16 Model Redction Example (cont'd) Introdction to FEM Lots of slaves, few masters. Only masters are left. Example of previos slide: / / 4/ 2/ 3/ 3/ 2/ 4/ / 5/ 0 Applying the congrential transformation we get the redced stiffness eqations where 5 slaves = 2 masters ˆK ˆK 7 ˆK 7 ˆK 77 7 ˆK = 3 (3K +0K 2 +25K K 23 +K K 34 +9K 44 +2K 45 +4K 55 +4K 5 +K ) ˆK 7 = 3 (K 2+5K 22 +4K 23 +8K 33 +8K 34 +9K 44 +8K 45 +8K 55 +4K 5 +5K +K 7 ) ˆK 77 = 3 (K 22+4K 23 +4K 33 +2K 34 +9K K 45 +K K 5 +25K +0K 7 +3K 77 ) ˆ f = ( f +5 f 2 +4 f 3 +3 f 4 +2 f 5 + f ); = 7 ˆ f ˆ f 7 ˆ f 7 = ( f 2+2 f 3 +3 f 4 +4 f 5 +5 f + f 7 ): IFEM Ch 8 Slide

17 Introdction to FEM Model Redction Example: Mathematica Script (* Model Redction by Kinematic Constraints - Example *) ClearAll[K,K2,K22,K23,K33,K34,K44,K45,K55,K5,K,K7,K77, f,f2,f3,f4,f5,f,f7]; K={{K,K2,0,0,0,0,0},{K2,K22,K23,0,0,0,0}, {0,K23,K33,K34,0,0,0},{0,0,K34,K44,K45,0,0}, {0,0,0,K45,K55,K5,0},{0,0,0,0,K5,K,K7}}; {0,0,0,0,0,K7,K77}; Print["K=",K//MatrixForm]; f={f,f2,f3,f4,f5,f,f7}; Print["f=",f]; T={{,0},{5,},{4,2},{3,3},{2,4},{,5},{0,}}/; Print[" T (transposed to save space)=",transpose[t]//matrixform]; Khat=Simplify[Transpose[T].K.T]; fhat=simplify[transpose[t].f]; Print["Modified stiffness entries:"]; Print["Khat(,)=",Khat[[,]],"\nKhat(,2)=",Khat[[,2]], "\nkhat(2,2)=",khat[[2,2]] ]; Print["Modified force entries:"]; Print["fhat()=",fhat[[]]," fhat(2)=",fhat[[2]] ]; K K K2 K22 K K23 K33 K K= 0 0 K34 K44 K K45 K55 K K5 K K K7 K77 f={f,f2,f3,f4,f5,f,f7}; T (transposed to save space)= Modified stiffness entries: Kkat(,)= (3 K + 0 K K K23 + K K K K K55 + 4K5 + K) 3 Khat(,2)= ( K2 + 5 K K K K K K K K5 + 5 K + K7) 3 Khat(2,2)= (K K K K K K45 + K K K + 0 K7 + 3 K77) 3 Modified force entries: fhat()= ( f + 5 f2 + 4 f3 + 3 f4 + 2 f5 + f) fhat(2)= (f2 + 2 f3 + 3 f4 + 4 f5 + 5 f + f7) IFEM Ch 8 Slide 7

18 Introdction to FEM Assessment of Master-Slave Method ADVANTAGES exact if precations taken easy to nderstand retains positive definiteness important applications to model redction DISADVANTAGES reqires ser decisions messy implementation for general MFCs hinders sparsity of master stiffness eqations sensitive to constraint dependence restricted to linear constraints IFEM Ch 8 Slide 8

19 Introdction to FEM Psedo Code for Compter Exercises (sample) procedre MasterStiffnessOfSixElementBar(kbar); real K=nllmatrix(7,7)); K(,)=K(7,7)=kbar; for i=2,i<=,i++, K(i,i)=2*kbar; endfor; for i=,i<=,i++, K(i,i+)=K(i+,i)= kbar; endfor; retrn(k); endprocedre; procedre FixLeftEndOfSixElementBar(Khat,fhat); Kmod=Khat; fmod=fhat; fmod()=0; Kmod(,)=; Kmod(,2)=Kmod(,2)=0; retrn((kmod,fmod)); endprocedre; // Main script follows K=MasterStiffnessOfSixElementBar(00); print K; // constrct master K f=vector(,2,3,4,5,,7); print f; // constrct force vector T=matrix((,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),(0,0,0,0,0,)); print T; // constrct transformation matrix T g=vector(0,0,0,0,0, /5,0); print g; // constrct gap vector Khat=T'.K.T; fhat=t'.(f K.g); // modified stiffness and force (Kmod,fmod)=FixLeftEndOfSixElementBar(Khat,fhat); // fix left end print Kmod; print fmod; mod=inverse(kmod).fmod; print mod; // modified soltion vector =T.mod+g; print ; // complete soltion f=k.; print f; // recovered forces inclding reactions Simplifies implementation into high-level langage of yor choice IFEM Ch 8 Slide 9

Multi-Point Constraints

Multi-Point Constraints Multi-Point Constraints Multi-Point Constraints Single point constraint examples Multi-Point constraint examples linear, homogeneous linear, non-homogeneous linear, homogeneous