8 MultiFreedom Constraints I IFEM Ch 8 Slide 1 Introduction to - PDF document

Introduction to FEM 8 MultiFreedom Constraints I IFEM Ch 8 – Slide 1

Introduction to FEM Multifreedom Constraints Single freedom constraint examples u x 4 = 0 linear, homogeneous u y 9 = 0 . 6 linear, non-homogeneous Multifreedom constraint examples u x 2 = 1 linear, homogeneous 2 u y 2 u x 2 − 2 u x 4 + u x 6 = 0 . 25 linear, non-homogeneous ( x 5 + u x 5 − x 3 − u x 3 ) 2 + ( y 5 + u y 5 − y 3 − u y 3 ) 2 = 0 nonlinear, homogeneous IFEM Ch 8 – Slide 2

Introduction to FEM Sources of Multifreedom Constraints "Skew" displacement BCs Coupling nonmatched FEM meshes Global-local and multiscale analysis Incompressibility Model reduction IFEM Ch 8 – Slide 3

Introduction to FEM MFC Application Methods Master-Slave Elimination Chapter 9 Penalty Function Augmentation   Chapter 10 Lagrange Multiplier Adjunction  IFEM Ch 8 – Slide 4

Introduction to FEM Example 1D Structure to Illustrate MFCs u 1 , f 1 u 2 , f 2 u 3 , f 3 u 4 , f 4 u 5 , f 5 u 6 , f 6 u 7 , f 7 (1) (2) (3) (4) (5) (6) x 1 2 3 4 5 6 7 Multifreedom constraint: u 2 − u 6 = 0 u 2 = u 6 or Linear homogeneous MFC IFEM Ch 8 – Slide 5

Introduction to FEM Example 1D Structure (Cont'd) u 1 , f 1 u 2 , f 2 u 3 , f 3 u 4 , f 4 u 5 , f 5 u 6 , f 6 u 7 , f 7 (1) (2) (3) (4) (5) (6) x 1 2 3 4 5 6 7 Unconstrained master stiffness equations  0 0 0 0 0      K 11 K 12 u 1 f 1 K 12 K 22 K 23 0 0 0 0 u 2 f 2             0 K 23 K 33 K 34 0 0 0 u 3 f 3             = 0 0 K 34 K 44 K 45 0 0 u 4 f 4             0 0 0 K 45 K 55 K 56 0 u 5 f 5             0 0 0 0 K 56 K 66 K 67 u 6 f 6       0 0 0 0 0 K 67 K 77 u 7 f 7 or Ku = f IFEM Ch 8 – Slide 6

Introduction to FEM Master-Slave Method for Example Structure Recall: u 2 − u 6 = 0 u 2 = u 6 or Taking u as master: 2  u 1   1 0 0 0 0 0    u 1 u 2 0 1 0 0 0 0     u 2       u 3 0 0 1 0 0 0       u 3       = u 4 0 0 0 1 0 0       u 4       u 5 0 0 0 0 1 0       u 5       u 6 0 1 0 0 0 0     u 7 u 7 0 0 0 0 0 1 u = T ˆ u . or IFEM Ch 8 – Slide 7

Introduction to FEM Forming the Modified Stiffness Equations Unconstrained master Ku = f stiffness equations: u = T ˆ u Master-slave transformation: K = T T KT ˆ Congruential transformation: f = T T f ˆ ˆ u = ˆ K ˆ f Modified stiffness equations: IFEM Ch 8 – Slide 8

Introduction to FEM Modified Stiffness Equations for Example Structure ˆ ˆ ˆ K u = f In full  K 11 K 12 0 0 0 0   u 1   f 1  K 22 + K 66 f 2 + f 6 K 12 K 23 0 K 56 K 67 u 2             0 K 23 K 33 K 34 0 0 u 3 f 3       =       0 0 K 34 K 44 K 45 0 u 4 f 4             0 K 56 0 K 45 K 55 0 u 5 f 5       0 K 67 0 0 0 K 77 u 7 f 7 ˆ ˆ Solve for u , then recover u = T u IFEM Ch 8 – Slide 9

Introduction to FEM Multiple MFCs Suppose u 2 − u 6 = 0 , u 1 + 4 u 4 = 0 , 2 u 3 + u 4 + u 5 = 0 Pick 3, 4 and 6 as slaves: u 4 = − 1 u 3 = − 1 2 ( u 4 + u 5 ) = 1 8 u 1 − 1 u 6 = u 2 , 4 u 1 , 2 u 5 Put in matrix form: 1 0 0 0     u 1 0 1 0 0 u 2     u 1     1 − 1   0 0 u 3     8 2 u 2     = − 1 u 4     0 0 0   u 5  4      u 5     0 0 1 0   u 7   u 6     0 1 0 0   u 7 0 0 0 1 ˆ This is u = T u - then proceed as before IFEM Ch 8 – Slide 10

Introduction to FEM Non-homogeneous MFCs u 2 − u 6 = 0 . 2 Pick again u as slave, put into matrix form: 6  u 1   1 0 0 0 0 0   0    u 1 u 2 0 1 0 0 0 0 0       u 2         u 3 0 0 1 0 0 0 0         u 3 +         = u 4 0 0 0 1 0 0 0         u 4         u 5 0 0 0 0 1 0 0         u 5 −0.2         u 6 0 1 0 0 0 0       u 7 u 7 0 0 0 0 0 1 0 IFEM Ch 8 – Slide 11

Introduction to FEM Nonhomogeneous MFCs (cont'd) ˆ u = T u + g g = "gap" vector T Premultiply both sides by T K , replace K u = f and pass data to RHS. This gives ˆ ˆ ˆ K u = f K = T K T and f = T (f − K g) ˆ ˆ T T with a modified force vector For the example structure       K 11 K 12 0 0 0 0 u 1 f 1 K 22 + K 66 f 2 + f 6 − 0 . 2 K 66 K 12 K 23 0 K 56 K 67 u 2             0 K 23 K 33 K 34 0 0 u 3 f 3       =       0 0 K 34 K 44 K 45 0 u 4 f 4             0 0 0 f 5 − 0 . 2 K 56 K 56 K 45 K 55 u 5       f 7 − 0 . 2 K 67 0 K 67 0 0 0 K 77 u 7 ˆ ˆ Solve for u , then recover u = T u + g IFEM Ch 8 – Slide 12

Introduction to FEM Model Reduction Example u 1 , f 1 u 2 , f 2 u 3 , f 3 u 4 , f 4 u 5 , f 5 u 6 , f 6 u 7 , f 7 (1) (2) (3) (4) (5) (6) x 1 2 3 4 5 6 7 2 master DOFs to be retained u 1 , f 1 u 7 , f 7 x 1 7 5 slave DOFs to be eliminated Master Master u 1 , f 1 u 7 , f 7 Reduced model x 1 7 IFEM Ch 8 – Slide 13

Introduction to FEM Model Reduction Example (cont'd) Lots of slaves, few masters. Only masters are left. Example of previous slide:  u 1   1 0  5 / 6 1 / 6 u 2         � u 1 u 3 4 / 6 2 / 6     �     = u 4 3 / 6 3 / 6 5 slaves 2 masters     u 7     u 5 2 / 6 4 / 6         u 6 1 / 6 5 / 6     u 7 0 1 Applying the congruential transformation we get the reduced stiffness equations � ˆ � ˆ � � u 1 ˆ � � K 11 K 17 f 1 = ˆ ˆ ˆ K 17 K 77 u 7 f 7 where ˆ 1 K 11 = 36 ( 36 K 11 + 60 K 12 + 25 K 22 + 40 K 23 + 16 K 33 + 24 K 34 + 9 K 44 + 12 K 45 + 4 K 55 + 4 K 56 + K 66 ) ˆ 1 K 17 = 36 ( 6 K 12 + 5 K 22 + 14 K 23 + 8 K 33 + 18 K 34 + 9 K 44 + 18 K 45 + 8 K 55 + 14 K 56 + 5 K 66 + 6 K 67 ) ˆ 1 K 77 = 36 ( K 22 + 4 K 23 + 4 K 33 + 12 K 34 + 9 K 44 + 24 K 45 + 16 K 55 + 40 K 56 + 25 K 66 + 60 K 67 + 36 K 77 ) ˆ ˆ f 1 = 1 f 7 = 1 6 ( 6 f 1 + 5 f 2 + 4 f 3 + 3 f 4 + 2 f 5 + f 6 ), 6 ( f 2 + 2 f 3 + 3 f 4 + 4 f 5 + 5 f 6 + 6 f 7 ). IFEM Ch 8 – Slide 14

Introduction to FEM Model Reduction Example: Mathematica Script (* Model Reduction Example *) ClearAll[K11,K12,K22,K23,K33,K34,K44,K45,K55,K56,K66, f1,f2,f3,f4,f5,f6]; K={{K11,K12,0,0,0,0,0},{K12,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,K56,0},{0,0,0,0,K56,K66,K67}, {0,0,0,0,0,K67,K77}}; Print["K=",K//MatrixForm]; f={f1,f2,f3,f4,f5,f6,f7}; Print["f=",f]; T={{6,0},{5,1},{4,2},{3,3},{2,4},{1,5},{0,6}}/6; Print["Transformation matrix T=",T//MatrixForm]; Khat=Simplify[Transpose[T].K.T]; fhat=Simplify[Transpose[T].f]; Print["Modified Stiffness:"]; Print["Khat(1,1)=",Khat[[1,1]],"\nKhat(1,2)=",Khat[[1,2]], "\nKhat(2,2)=",Khat[[2,2]] ]; Print["Modified Force:"]; Print["fhat(1)=",fhat[[1]]," fhat(2)=",fhat[[2]] ]; Modified Stiffness: (Some print output removed so slide fits) Khat � 1,1 � � 1 36 � 36 K11 � 60 K12 � 25 K22 � 40 K23 � 16 K33 � 24 K34 � 9 K44 � 12 K45 � 4 K55 � 4 K56 � K66 � ������� Khat � 1,2 � � 1 36 � 6 K12 � 5 K22 � 14 K23 � 8 K33 � 18 K34 � 9 K44 � 18 K45 � 8 K55 � 14 K56 � 5 K66 � 6 K67 � ������� Khat � 2,2 � � 1 36 � K22 � 4 K23 � 4 K33 � 12 K34 � 9 K44 � 24 K45 � 16 K55 � 40 K56 � 25 K66 � 60 K67 � 36 K77 � ������� Modified Force: fhat � 1 � � 1 fhat � 2 � � 1 6 � 6 f1 � 5 f2 � 4 f3 � 3 f4 � 2 f5 � f6 � 6 � f2 � 2 f3 � 3 f4 � 4 f5 � 5 f6 � 6 f7 � ���� ���� IFEM Ch 8 – Slide 15

Introduction to FEM Assessment of Master-Slave Method ADVANTAGES exact if precautions taken easy to understand retains positive definiteness important applications to model reduction DISADVANTAGES requires user decisions messy implementation for general MFCs hinders sparsity of master stiffness equations sensitive to constraint dependence restricted to linear constraints IFEM Ch 8 – Slide 16