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

On numerical modeling of coupled magnetoelastic problem

Anouar Belahcen1, Katarzyna Fonteyn1, Antti Hannukainen2 and Reijo Kouhia3

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

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CONTENTS

Magnetoelasticity Balance equations Constitutive model

• Helmholtz free energy
• Experimental verification

Numerical solution Finite element procedures

• Potential approach
• Least-square formulations

Example

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MAGNETOELASTICITY

Balance equations

−div σ = f + fem curl H = J, div B = 0, div J = 0

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MAGNETOELASTICITY

Balance equations

−div σ = f + fem curl H = J, div B = 0, div J = 0

Constitutive equations

B = µ0(H + M) σ = ρ∂ψ ∂ε , M = −ρ∂ψ ∂B

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MAGNETOELASTICITY

Balance equations

−div σ = f + fem curl H = J, div B = 0, div J = 0

Constitutive equations

B = µ0(H + M) σ = ρ∂ψ ∂ε , M = −ρ∂ψ ∂B τ = σ + µ−1

0 [BB − 1 2(B · B)I] + (M · B)I − BM

−div τ = f

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CONSTITUTIVE MODEL - Helmholtz free energy

For isotropic magnetoelastic solid: ψ(ε, B) Integrity basis

I1 = tr ε, I2 = 1

2tr ε2,

I3 = 1

3trε3

I4 = B · B, I5 = B · ε · B, I6 = B · ε2 · B

Suitable choice for ρψ = 1 2λI2 1 + 2GI2 + 1 2

4

• i=0

1 i + 1gi(I1)Ii+1

4

• + 1

2γ5I5 + 1 2γ6I6

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CONSTITUTIVE MODEL - Helmholtz free energy

For isotropic magnetoelastic solid: ψ(ε, B) Integrity basis

I1 = tr ε, I2 = 1

2tr ε2,

I3 = 1

3trε3

I4 = B · B, I5 = B · ε · B, I6 = B · ε2 · B

Suitable choice for ρψ = 1 2λI2 1 + 2GI2 + 1 2

4

i=0 1 i+1gi(I1)Ii+1 4

• + 1

2γ5I5 + 1 2γ6I6

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CONSTITUTIVE MODEL - Helmholtz free energy

For isotropic magnetoelastic solid: ψ(ε, B) Integrity basis

I1 = tr ε, I2 = 1

2tr ε2,

I3 = 1

3trε3

I4 = B · B, I5 = B · ε · B, I6 = B · ε2 · B

Suitable choice for ρψ = 1 2λI2 1 + 2GI2 + 1 2

4

• i=0

1 i + 1gi(I1)Ii+1

4

• + 1

2γ5I5 + 1 2γ6I6

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CONSTITUTIVE MODEL - Helmholtz free energy

For isotropic magnetoelastic solid: ψ(ε, B) Integrity basis

I1 = tr ε, I2 = 1

2tr ε2,

I3 = 1

3trε3

I4 = B · B, I5 = B · ε · B, I6 = B · ε2 · B

Suitable choice for ρψ = 1 2λI2 1 + 2GI2 + 1 2

4

• i=0

1 i + 1gi(I1)Ii+1

4

• + 1

2γ5I5 + 1 2γ6I6

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CONSTITUTIVE MODEL - functions gi(I1)

Deformation isochoric under pure magnetic

excitation = ⇒ gi(I1)

Result

γ6 = 2γ5 g0 = (γ(0)

4

+ 1

4µ−1 0 − 1 4γ5) exp(4 3I1) − 1 4µ−1 0 + 1 4γ5

gi = γ(i)

4 exp(4 3I1)

when i = 1, . . . , 4 where γ(i)

4 = gi(0) are unknown parameters

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CONSTITUTIVE MODEL - experimental verification Epstein frame (Belahcen 2004) 1–4: load cells 5: screw system for loading

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CONSTITUTIVE MODEL - experimental verification Magnetization curves

2 4 6 8 10 12 14 16 18 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 H1 (kA/m) B1 (Tesla) ...experimental results, − proposed model

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CONSTITUTIVE MODEL - experimental verification Magnetostriction different compressive pre-stress 2 - 7 MPa

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −1 1 2 3 4 5 6 x 10

−6

10 B1 (Tesla) eps1: −−− model *** experiments 9 8 7 6 5 4 3 2 1

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NUMERICAL SOLUTION - potential approach

Vector potential A, such that B = curl A. FE-problem: find {Ah, ph} ∈ Rh

0 × Gh

Rh

0 ⊂ H0(curl, Ω) and Gh 0 ⊂ H1 0(Ω)

(curl ˆ Ah, Hh) + (ˆ Ah, grad ph) = (ˆ Ah, J) − [ˆ Ah, ¯ H]; ∀ ˆ Ah ∈ Rh (grad ˆ ph, Ah) = 0; ∀ ˆ ph ∈ Gh Magnetic field strength from the constitutive model H = (µ−1 + 2f4)B + γ5B · ε + γ6B · ε2, f4 = ρ ∂ψ ∂I4

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NUMERICAL SOLUTION - potential approach

Vector potential A, such that B = curl A. FE-problem: find {Ah, ph} ∈ Rh

0 × Gh

Rh

0 ⊂ H0(curl, Ω) and Gh 0 ⊂ H1 0(Ω)

(curl ˆ Ah, Hh) + (ˆ Ah, grad ph) = (ˆ Ah, J) − [ˆ Ah, ¯ H]; ∀ ˆ Ah ∈ Rh (grad ˆ ph, Ah) = 0; ∀ ˆ ph ∈ Gh Magnetic field strength from the constitutive model H = (µ−1 + 2f4)B + γ5B · ε + γ6B · ε2, f4 = ρ ∂ψ ∂I4

Interelement continuity for the tangential component required

for fields in H(curl, Ω)

Nedelec edge elements

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NUMERICAL SOLUTION - least-square approach

B-formulation

LB = 1

2(div B, div B) − (p, curl H − J)

FEM: Bh ∈ Dh

0 ⊂ H0(div, Ω),

p ∈ Rh

0 ⊂ H0(curl, Ω)

Continuity for the normal component of Bh required Raviart-Thomas elements for Bh

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NUMERICAL SOLUTION - least-square approach

B-formulation

LB = 1

2(div B, div B) − (p, curl H − J)

FEM: Bh ∈ Dh

0 ⊂ H0(div, Ω),

p ∈ Rh

0 ⊂ H0(curl, Ω)

Continuity for the normal component of Bh required Raviart-Thomas elements for Bh

H-formulation

LH = 1

2(curl H − J, curl H − J) − (p, div B)

FEM: Hh ∈ Rh ⊂ H(curl, Ω), p ∈ Gh ⊂ H1(Ω) Continuity for the tangential component of Hh required Nedelec elements for Hh

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NUMERICAL SOLUTION - least-square approach

Minimizing the error in the constitutive model B = µH:

LBH =

• Ω

1

2|B − µH|2 + qdiv B + p · (curl H − J)

• dV

Piecewise continuous interpolation for q, p Penalty approach ⇒ element level elimination of q, p

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EXAMPLE

Plane-strain problem, 36×18 bilinear element mesh J1

z = 100 A,

J2

z = −100 A, L = 1 m

E = 183.62 GPa, ν = 0.34, µ0 = 4π · 10−7N/A2 γ(0)

4

= −0.99974858µ−1

0 , γ(1) 4

= 0.00076054µ−1

0 , γ(2) 4

= −0.00089881µ−1 γ(3)

4

= 0.00008964µ−1

0 , γ(4) 4

= 0.00012688µ−1

0 , γ5 = −0.028µ−1

r ❞

z x y

✛ ✲

2L

❄ ✻

L

r ❞J1 z ❞

• ❅

❅J2 z

• Az = 0 on ∂Ω

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EXAMPLE Magnetic flux density

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EXAMPLE Deformations

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