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chore(test): delete finite-strain plasticity tests
Remove coverage for deleted `finite_strain_plasticity.jl` material prototype. - Drop `test/materials/test_finite_strain_plasticity.jl`.
This commit is contained in:
Jukka Aho
@@ -1,328 +0,0 @@
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"""
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# Unit Tests: Finite Strain Plasticity (Multiplicative Decomposition)
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**What:** Comprehensive validation of finite strain J2 plasticity with F = F_e F_p decomposition
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**Why:**
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- Geometrically exact plasticity for large deformations (>10% strain)
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- Tests multiplicative decomposition F = F_e F_p (not additive ε = ε_e + ε_p)
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- Validates plastic incompressibility det(F_p) = 1 (fundamental constraint)
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- Critical for metal forming, impact, crashworthiness (extreme deformations)
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- Demonstrates objective stress update (rotation-independent)
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**How:**
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Test suite validates:
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1. **Construction & parameters** - E, ν, σ_y, H validity, computed μ and λ
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2. **State management** - FiniteStrainPlasticityState(F_p, α_bar, κ) with F_p=I default
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3. **Small strain limit** - Should recover small-strain plasticity for F ≈ I + ∇u
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4. **Identity deformation** - F = I gives σ = 0, F_p = I, κ = 0
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5. **Pure rotation** - Rigid body rotation (no stretch) should give σ ≈ 0 (objectivity)
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6. **Uniaxial extension** - Elastic (λ=1.01) and plastic (λ=1.10) regimes
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7. **Simple shear** - Validates shear response, det(F) = 1
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8. **Incremental loading** - Monotonic loading: stress and κ increase
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9. **Plastic incompressibility** - det(F_p) ≈ 1 for all stretches λ ∈ [1.02, 1.20]
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10. **Hardening behavior** - H > 0: higher stress, backstress α_bar ≠ 0
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11. **State persistence** - Unloading: plastic strain κ does not decrease
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12. **Performance** - Type stability
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**Mathematical Background:**
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- Multiplicative decomposition: F = F_e F_p (Lee decomposition)
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- F: Total deformation gradient
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- F_e: Elastic part (recoverable on unloading)
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- F_p: Plastic part (permanent deformation)
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- Plastic incompressibility: det(F_p) = 1 (volume preservation in plastic flow)
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- Mandel stress: M = C_e S_e (intermediate configuration)
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- Yield criterion: f = √(3/2·dev(M):dev(M)) - σ_y ≤ 0 (von Mises)
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- Flow rule: Ḟ_p F_p⁻¹ = Δγ·n (exponential map integration)
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- Hardening: α̇_bar = H·ε̇_p (backstress evolution in intermediate config)
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- Objectivity: σ(Q·F) = Q·σ(F)·Q^T for rotation Q (frame-invariance)
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- Physical constraints: det(F) > 0, det(F_e) > 0, det(F_p) = 1
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**Expected Results:**
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✅ Material constructed: E=200 GPa, ν=0.3, σ_y=250 MPa, H=0-10 GPa
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✅ Perfect plasticity: H=0 valid
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✅ Invalid inputs rejected: E<0, ν>0.5, σ_y<0, H<0, κ<0
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✅ Default state: F_p=I (det=1), α_bar=0, κ=0
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✅ Small strain (ε=1e-5): F_p≈I, κ=0, ||σ|| < 1 MPa
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✅ Identity (F=I): σ=0 exactly
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✅ Pure rotation (45° around z): ||σ|| < 1 MPa (objectivity), F_p≈I
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✅ Uniaxial elastic (λ=1.01): F_p≈I, κ=0, σ_xx > 0
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✅ Uniaxial plastic (λ=1.10): ||F_p-I|| > 1e-6, κ > 0, |det(F_p)-1| < 0.001
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✅ Simple shear (γ=0.1): σ_xy ≠ 0, det(F)=1
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✅ Incremental (5 steps to λ=1.05): Monotonic stress and κ
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✅ Incompressibility: |det(F_p)-1| < 0.01 for λ ∈ [1.02,1.20]
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✅ Hardening: H=10 GPa → σ > σ_perfect, ||α_bar|| > 0
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✅ State persistence: Load λ=1.08 then unload λ=1.02 → κ doesn't decrease
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✅ Simplified interface (without state, Δt) matches full call
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✅ Type-stable: returns Tuple{SymmetricTensor{2,3}, SymmetricTensor{4,3}, FiniteStrainPlasticityState}
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**Test Coverage:**
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- 14 test sets, ~70 individual assertions
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- Material constants: Steel (E=200 GPa, ν=0.3, σ_y=250 MPa, H=0-10 GPa)
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- Deformations: Identity, small (ε=1e-5), rotation (45°), uniaxial (λ=1.01-1.20), shear (γ=0.1)
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- Validation methods: Plastic incompressibility, objectivity, state persistence, hardening comparison
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- Algorithms: Multiplicative decomposition, exponential map, return mapping in intermediate config
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- Edge cases: Perfect plasticity (H=0), pure rotation, incremental loading, unloading
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**Key Physics:**
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- Multiplicative decomposition: Geometrically exact (not linearized)
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- Plastic incompressibility: Fundamental for metals (no volume change in plastic flow)
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- Objectivity: Stress independent of observer reference frame (essential for large rotations)
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- Lee decomposition: Separates elastic (lattice stretch) from plastic (slip) deformations
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- Intermediate configuration: Where plasticity lives (stress-free but plastically deformed)
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- Exponential map: Preserves det(F_p) = 1 during integration (unlike additive schemes)
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"""
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using Test
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using Tensors
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using LinearAlgebra
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# Load implementations
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include("../src/materials/abstract_material.jl")
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include("../src/materials/finite_strain_plasticity.jl")
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@testset "Finite Strain Plasticity Material" begin
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@testset "Material Construction" begin
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# Valid construction
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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@test steel.E == 200e9
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@test steel.ν == 0.3
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@test steel.σ_y == 250e6
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@test steel.H == 1e9
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@test steel.μ ≈ 200e9 / (2 * (1 + 0.3))
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@test steel.λ ≈ 200e9 * 0.3 / ((1 + 0.3) * (1 - 2 * 0.3))
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# Perfect plasticity (H=0)
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perfect = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=0.0)
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@test perfect.H == 0.0
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# Invalid inputs
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@test_throws ArgumentError FiniteStrainPlasticity(E=-200e9, ν=0.3, σ_y=250e6, H=1e9)
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@test_throws ArgumentError FiniteStrainPlasticity(E=200e9, ν=0.6, σ_y=250e6, H=1e9)
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@test_throws ArgumentError FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=-250e6, H=1e9)
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@test_throws ArgumentError FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=-1e9)
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end
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@testset "State Construction" begin
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# Default state (identity F_p)
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state0 = FiniteStrainPlasticityState()
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@test state0.F_p == one(Tensor{2,3})
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@test state0.α_bar == zero(SymmetricTensor{2,3})
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@test state0.κ == 0.0
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@test det(state0.F_p) ≈ 1.0
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# Custom state
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F_p = one(Tensor{2,3}) + 0.01 * Tensor{2,3}((0.0, 0.01, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0))
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F_p = F_p / det(F_p)^(1 / 3) # Enforce det = 1
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α_bar = SymmetricTensor{2,3}((1e8, 0.0, 0.0, 0.0, 0.0, 0.0))
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state = FiniteStrainPlasticityState(F_p, α_bar, 0.01)
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@test state.F_p ≈ F_p
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@test state.α_bar == α_bar
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@test state.κ == 0.01
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# Invalid state (negative κ)
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@test_throws ArgumentError FiniteStrainPlasticityState(F_p, α_bar, -0.01)
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end
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@testset "Small Strain Limit" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Small deformation: F ≈ I + ∇u
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ε_small = 1e-5
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F_small = one(Tensor{2,3}) + ε_small * Tensor{2,3}((1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔸, state = compute_stress(steel, F_small, nothing, 0.0)
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# Should remain elastic
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@test state.F_p ≈ one(Tensor{2,3})
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@test state.α_bar == zero(SymmetricTensor{2,3})
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@test state.κ == 0.0
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# Stress should be small
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@test norm(σ) < 1e6 # Less than 1 MPa
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end
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@testset "Identity Deformation" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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F_identity = one(Tensor{2,3})
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σ, 𝔸, state = compute_stress(steel, F_identity, nothing, 0.0)
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# Zero stress for no deformation
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@test norm(σ) < 1e-10
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@test state.F_p == one(Tensor{2,3})
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@test state.κ == 0.0
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end
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@testset "Pure Rotation (Elastic)" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# 45-degree rotation around z-axis (no stretching)
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θ = π / 4
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c = cos(θ)
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s = sin(θ)
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R = Tensor{2,3}((c, s, 0.0, -s, c, 0.0, 0.0, 0.0, 1.0))
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σ, 𝔸, state = compute_stress(steel, R, nothing, 0.0)
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# Pure rotation should give zero stress (if formulation is objective)
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# Note: May not be exactly zero due to numerical precision
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@test norm(σ) < 1e6 # Should be small
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@test state.F_p ≈ one(Tensor{2,3}) rtol = 1e-6
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end
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@testset "Uniaxial Extension (Elastic)" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# 1% extension in x-direction
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λ = 1.01
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F_ext = Tensor{2,3}((λ, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ, 𝔸, state = compute_stress(steel, F_ext, nothing, 0.0)
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# Should remain elastic (small extension)
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@test state.F_p ≈ one(Tensor{2,3}) rtol = 1e-6
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@test state.κ == 0.0
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# Check that σ_xx > 0 (tension)
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@test σ[1, 1] > 0.0
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end
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@testset "Uniaxial Extension (Plastic)" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Large extension (10%)
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λ = 1.10
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F_ext = Tensor{2,3}((λ, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ, 𝔸, state = compute_stress(steel, F_ext, nothing, 0.0)
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# Should have plastic deformation
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@test norm(state.F_p - one(Tensor{2,3})) > 1e-6
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@test state.κ > 0.0
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# Plastic incompressibility: det(F_p) ≈ 1
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@test abs(det(state.F_p) - 1.0) < 1e-3
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end
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@testset "Simple Shear" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Shear deformation: γ = 0.1
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γ = 0.1
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F_shear = Tensor{2,3}((1.0, γ, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ, 𝔸, state = compute_stress(steel, F_shear, nothing, 0.0)
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# Check shear stress exists
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@test abs(σ[1, 2]) > 0.0
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# det(F) should be 1 for simple shear
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@test abs(det(F_shear) - 1.0) < 1e-10
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end
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@testset "Incremental Loading" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Load in increments
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n_steps = 5
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λ_max = 1.05
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state = FiniteStrainPlasticityState()
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stresses = Float64[]
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plastic_strains = Float64[]
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for i in 1:n_steps
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λ = 1.0 + (λ_max - 1.0) * i / n_steps
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F = Tensor{2,3}((λ, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ, 𝔸, state = compute_stress(steel, F, state, 0.0)
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push!(stresses, σ[1, 1])
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push!(plastic_strains, state.κ)
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end
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# Stress should increase (with hardening)
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@test all(diff(stresses) .≥ -1e-6) # Allow small numerical errors
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# Plastic strain should increase monotonically
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@test all(diff(plastic_strains) .≥ 0.0)
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end
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@testset "Plastic Incompressibility" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Various deformation levels
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stretches = [1.02, 1.05, 1.10, 1.15, 1.20]
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for λ in stretches
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F = Tensor{2,3}((λ, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ, 𝔸, state = compute_stress(steel, F, nothing, 0.0)
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# Check plastic incompressibility
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det_Fp = det(state.F_p)
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@test abs(det_Fp - 1.0) < 0.01 # Within 1% (relaxed due to exponential map approximation)
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end
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end
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@testset "Hardening Behavior" begin
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steel_hard = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=10e9)
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steel_perf = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=0.0)
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F_test = Tensor{2,3}((1.08, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ_hard, _, state_hard = compute_stress(steel_hard, F_test, nothing, 0.0)
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σ_perf, _, state_perf = compute_stress(steel_perf, F_test, nothing, 0.0)
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# Hardening material should have higher stress
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@test σ_hard[1, 1] > σ_perf[1, 1]
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# Hardening material should have backstress
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@test norm(state_hard.α_bar) > 0.0
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@test norm(state_perf.α_bar) == 0.0
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end
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@testset "State Persistence" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# First load
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F1 = Tensor{2,3}((1.08, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ1, _, state1 = compute_stress(steel, F1, nothing, 0.0)
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# Unload to smaller deformation
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F2 = Tensor{2,3}((1.02, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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σ2, _, state2 = compute_stress(steel, F2, state1, 0.0)
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# Plastic strain should not decrease
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@test state2.κ ≥ state1.κ
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# F_p should not go back to identity
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@test norm(state2.F_p - one(Tensor{2,3})) > 1e-6
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end
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@testset "Simplified Interface" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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F = Tensor{2,3}((1.05, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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# Test with and without explicit state/Δt
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σ1, 𝔸1, state1 = compute_stress(steel, F)
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σ2, 𝔸2, state2 = compute_stress(steel, F, nothing, 0.0)
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@test σ1 ≈ σ2
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@test state1.κ ≈ state2.κ
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end
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@testset "Type Stability" begin
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steel = FiniteStrainPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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F = Tensor{2,3}((1.05, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0))
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state = FiniteStrainPlasticityState()
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# Infer return types
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result = @inferred compute_stress(steel, F, state, 0.0)
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@test result isa Tuple{SymmetricTensor{2,3,Float64},
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SymmetricTensor{4,3,Float64},
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FiniteStrainPlasticityState}
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end
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end
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