On numerical modeling of coupled magnetoelastic problem Anouar - PowerPoint PPT Presentation

On numerical modeling of coupled magnetoelastic problem Anouar Belahcen 1 , Katarzyna Fonteyn 1 , Antti Hannukainen 2 and Reijo Kouhia 3 firstname.lastname@tkk.fi Helsinki University of Technology 1 Department of Electrical Engineering 2 Institute of Mathematics 3 Department of Structural Engineering and Building Technology NSCM-21, Trondheim, October 16-17, 2008 – p.1/10

CONTENTS � Magnetoelasticity � Balance equations � Constitutive model • Helmholtz free energy • Experimental verification � Numerical solution � Finite element procedures • Potential approach • Least-square formulations � Example NSCM-21, Trondheim, October 16-17, 2008 – p.2/10

MAGNETOELASTICITY � Balance equations − div σ = f + f em curl H = J , div B = 0 , div J = 0 NSCM-21, Trondheim, October 16-17, 2008 – p.3/10

MAGNETOELASTICITY � Balance equations − div σ = f + f em curl H = J , div B = 0 , div J = 0 � Constitutive equations B = µ 0 ( H + M ) σ = ρ∂ψ M = − ρ∂ψ ∂ ε , ∂ B NSCM-21, Trondheim, October 16-17, 2008 – p.3/10

MAGNETOELASTICITY � Balance equations − div σ = f + f em curl H = J , div B = 0 , div J = 0 � Constitutive equations B = µ 0 ( H + M ) σ = ρ∂ψ M = − ρ∂ψ ∂ ε , ∂ B τ = σ + µ − 1 0 [ BB − 1 2 ( B · B ) I ] + ( M · B ) I − BM − div τ = f NSCM-21, Trondheim, October 16-17, 2008 – p.3/10

CONSTITUTIVE MODEL - Helmholtz free energy � For isotropic magnetoelastic solid: ψ ( ε , B ) � Integrity basis I 2 = 1 2 tr ε 2 , I 3 = 1 3 tr ε 3 I 1 = tr ε , I 6 = B · ε 2 · B I 4 = B · B , I 5 = B · ε · B , � Suitable choice for ρψ = � 4 � 1 � 2 λI 2 1 1 + 2 GI 2 + 1 i + 1 g i ( I 1 ) I i +1 + 1 2 γ 5 I 5 + 1 2 γ 6 I 6 4 2 i =0 NSCM-21, Trondheim, October 16-17, 2008 – p.4/10

CONSTITUTIVE MODEL - Helmholtz free energy � For isotropic magnetoelastic solid: ψ ( ε , B ) � Integrity basis I 2 = 1 2 tr ε 2 , I 3 = 1 3 tr ε 3 I 1 = tr ε , I 6 = B · ε 2 · B I 4 = B · B , I 5 = B · ε · B , � Suitable choice for ρψ = �� 4 � 2 λI 2 1 1 + 2 GI 2 + 1 i +1 g i ( I 1 ) I i +1 1 + 1 2 γ 5 I 5 + 1 2 γ 6 I 6 i =0 4 2 NSCM-21, Trondheim, October 16-17, 2008 – p.4/10

CONSTITUTIVE MODEL - Helmholtz free energy � For isotropic magnetoelastic solid: ψ ( ε , B ) � Integrity basis I 2 = 1 2 tr ε 2 , I 3 = 1 3 tr ε 3 I 1 = tr ε , I 6 = B · ε 2 · B I 4 = B · B , I 5 = B · ε · B , � Suitable choice for ρψ = � 4 � 1 � 2 λI 2 1 1 + 2 GI 2 + 1 i + 1 g i ( I 1 ) I i +1 + 1 2 γ 5 I 5 + 1 2 γ 6 I 6 4 2 i =0 NSCM-21, Trondheim, October 16-17, 2008 – p.4/10

CONSTITUTIVE MODEL - Helmholtz free energy � For isotropic magnetoelastic solid: ψ ( ε , B ) � Integrity basis I 2 = 1 2 tr ε 2 , I 3 = 1 3 tr ε 3 I 1 = tr ε , I 6 = B · ε 2 · B I 4 = B · B , I 5 = B · ε · B , � Suitable choice for ρψ = � 4 � 1 � 2 λI 2 1 1 + 2 GI 2 + 1 i + 1 g i ( I 1 ) I i +1 + 1 2 γ 5 I 5 + 1 2 γ 6 I 6 4 2 i =0 NSCM-21, Trondheim, October 16-17, 2008 – p.4/10

CONSTITUTIVE MODEL - functions g i ( I 1 ) � Deformation isochoric under pure magnetic excitation = ⇒ g i ( I 1 ) � Result γ 6 = 2 γ 5 g 0 = ( γ (0) + 1 4 µ − 1 0 − 1 4 γ 5 ) exp( 4 3 I 1 ) − 1 4 µ − 1 0 + 1 4 γ 5 4 g i = γ ( i ) 4 exp( 4 3 I 1 ) when i = 1 , . . . , 4 where γ ( i ) 4 = g i (0) are unknown parameters NSCM-21, Trondheim, October 16-17, 2008 – p.5/10

CONSTITUTIVE MODEL - experimental verification Epstein frame (Belahcen 2004) 1–4: load cells 5: screw system for loading NSCM-21, Trondheim, October 16-17, 2008 – p.6/10

CONSTITUTIVE MODEL - experimental verification Magnetization curves ...experimental results, − proposed model 2 1.8 1.6 1.4 1.2 B1 (Tesla) 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14 16 18 H1 (kA/m) NSCM-21, Trondheim, October 16-17, 2008 – p.6/10

CONSTITUTIVE MODEL - experimental verification Magnetostriction different compressive pre-stress 2 - 7 MPa −6 x 10 6 6 5 5 7 4 4 eps 1 : −−− model *** experiments 8 3 3 9 2 2 10 1 1 0 −1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 B1 (Tesla) NSCM-21, Trondheim, October 16-17, 2008 – p.6/10

NUMERICAL SOLUTION - potential approach � Vector potential A , such that B = curl A . � FE-problem: find { A h , p h } ∈ R h 0 × G h 0 0 ⊂ H 1 R h 0 ⊂ H 0 (curl , Ω) and G h 0 (Ω) (curl ˆ A h , H h ) + ( ˆ A h , grad p h ) = ( ˆ A h , J ) − [ ˆ A h , ¯ ∀ ˆ A h ∈ R h H ]; 0 p h ∈ G h (grad ˆ p h , A h ) = 0; ∀ ˆ 0 Magnetic field strength from the constitutive model f 4 = ρ ∂ψ H = ( µ − 1 + 2 f 4 ) B + γ 5 B · ε + γ 6 B · ε 2 , 0 ∂I 4 NSCM-21, Trondheim, October 16-17, 2008 – p.7/10

NUMERICAL SOLUTION - potential approach � Vector potential A , such that B = curl A . � FE-problem: find { A h , p h } ∈ R h 0 × G h 0 0 ⊂ H 1 R h 0 ⊂ H 0 (curl , Ω) and G h 0 (Ω) (curl ˆ A h , H h ) + ( ˆ A h , grad p h ) = ( ˆ A h , J ) − [ ˆ A h , ¯ ∀ ˆ A h ∈ R h H ]; 0 p h ∈ G h (grad ˆ p h , A h ) = 0; ∀ ˆ 0 Magnetic field strength from the constitutive model f 4 = ρ ∂ψ H = ( µ − 1 + 2 f 4 ) B + γ 5 B · ε + γ 6 B · ε 2 , 0 ∂I 4 � Interelement continuity for the tangential component required for fields in H (curl , Ω) � Nedelec edge elements NSCM-21, Trondheim, October 16-17, 2008 – p.7/10

NUMERICAL SOLUTION - least-square approach � B-formulation L B = 1 2 (div B , div B ) − ( p , curl H − J ) FEM: B h ∈ D h p ∈ R h 0 ⊂ H 0 (div , Ω) , 0 ⊂ H 0 (curl , Ω) Continuity for the normal component of B h required Raviart-Thomas elements for B h NSCM-21, Trondheim, October 16-17, 2008 – p.8/10

NUMERICAL SOLUTION - least-square approach � B-formulation L B = 1 2 (div B , div B ) − ( p , curl H − J ) FEM: B h ∈ D h p ∈ R h 0 ⊂ H 0 (div , Ω) , 0 ⊂ H 0 (curl , Ω) Continuity for the normal component of B h required Raviart-Thomas elements for B h � H-formulation L H = 1 2 (curl H − J , curl H − J ) − ( p, div B ) FEM: H h ∈ R h ⊂ H (curl , Ω) , p ∈ G h ⊂ H 1 (Ω) Continuity for the tangential component of H h required Nedelec elements for H h NSCM-21, Trondheim, October 16-17, 2008 – p.8/10

NUMERICAL SOLUTION - least-square approach � Minimizing the error in the constitutive model B = µ H : � � 1 2 | B − µ H | 2 + q div B + p · (curl H − J ) � L BH = dV Ω � Piecewise continuous interpolation for q, p � Penalty approach ⇒ element level elimination of q, p NSCM-21, Trondheim, October 16-17, 2008 – p.9/10

EXAMPLE Plane -strain problem, 36 × 18 bilinear element mesh J 1 J 2 z = 100 A , z = − 100 A , L = 1 m E = 183 . 62 GPa , ν = 0 . 34 , µ 0 = 4 π · 10 − 7 N / A 2 γ (0) 0 , γ (1) 0 , γ (2) = − 0 . 99974858 µ − 1 = 0 . 00076054 µ − 1 = − 0 . 00089881 µ − 1 4 4 4 0 γ (3) 0 , γ (4) = 0 . 00008964 µ − 1 = 0 . 00012688 µ − 1 0 , γ 5 = − 0 . 028 µ − 1 4 4 0 y ✛ ✲ 2 L � ✻ � � � � � � � � � � � ❞ J 1 ❅ J 2 � � L ❅ � � � r ❞ � z z A z = 0 on ∂ Ω � � � � � � � � � � � � ❄ � � ❞ r z x NSCM-21, Trondheim, October 16-17, 2008 – p.10/10

EXAMPLE Magnetic flux density NSCM -21, Trondheim, October 16-17, 2008 – p.10/10

EXAMPLE Deformations NSCM -21, Trondheim, October 16-17, 2008 – p.10/10