9 MultiFreedom Constraints I IFEM Ch 9 Slide 1 Department of - PDF document

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM 9 MultiFreedom Constraints I IFEM Ch 9 – Slide 1

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 2 u y 2 linear, homogeneous 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 9 – Slide 2

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Sources of Multifreedom Constraints "Skew" displacement BCs Coupling nonmatched FEM meshes Global-local and multiscale analysis Incompressibility Model reduction IFEM Ch 9 – Slide 3

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM MFC Application Methods Master-Slave Elimination Chapter 9 Penalty Function Augmentation   Chapter 10 Lagrange Multiplier Adjunction  IFEM Ch 9 – Slide 4

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 9 – Slide 5

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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  K 11 K 12 0 0 0 0 0   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 Ku = f or IFEM Ch 9 – Slide 6

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Master-Slave Method for Example Structure Recall: u 2 = u 6 u 2 − u 6 = 0 or Taking u as master: 2    1 0 0 0 0 0  u 1   u 1 u 2 0 1 0 0 0 0     u 2       0 0 1 0 0 0 u 3       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 9 – Slide 7

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 9 – Slide 8

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 9 – Slide 9

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 9 – Slide 10

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 9 – Slide 11

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 with T T 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             f 5 − 0 . 2 K 56 0 K 56 0 K 45 K 55 0 u 5       0 0 0 0 f 7 − 0 . 2 K 67 K 67 K 77 u 7 ˆ ˆ Solve for u , then recover u = T u + g IFEM Ch 9 – Slide 12

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 IFEM Ch 9 – Slide 13

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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  u 2 5 / 6 1 / 6         � u 1 u 3 4 / 6 2 / 6     �     = u 4 3 / 6 3 / 6 2 masters 5 slaves     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 9 – Slide 14