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chore(test): remove legacy perfect plasticity regression
Delete outdated plasticity tests pending `test_plasticity_integration.jl`. - Drop `test/materials/test_perfect_plasticity.jl`.
This commit is contained in:
Jukka Aho
@@ -1,352 +0,0 @@
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"""
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# Unit Tests: Perfect Plasticity (J2 Plasticity with Hardening)
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**What:** Comprehensive validation of rate-independent J2 plasticity with return mapping
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**Why:**
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- Foundation of metal plasticity (steel, aluminum, etc.)
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- Tests critical algorithms: radial return mapping, consistency enforcement
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- Validates state evolution: plastic strain ε_p, backstress α, hardening κ
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- Critical for nonlinear structural analysis (beyond elastic limit)
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- Demonstrates stateful material model (history-dependent)
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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** - PlasticityState(ε_p, α, κ) with zero default
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3. **Elastic loading** - f < 0: stress σ = λ·tr(ε)I + 2μ·ε, no plastic strain
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4. **Plastic loading** - f = 0: yield criterion √(3/2·s:s) = σ_y enforced
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5. **Radial return mapping** - Projects trial stress back to yield surface
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- Consistency: von Mises stress = σ_y (within tolerance)
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- Plastic strain: ε_p deviatoric (tr(ε_p) = 0)
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- Stress reduction: ||σ|| < ||σ_elastic|| after return
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6. **Hardening behavior** - Kinematic hardening via backstress α
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- H > 0: Stress increases with plastic strain
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- H = 0: Perfect plasticity (constant yield)
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- Backstress: norm(α) > 0 for hardening
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7. **Incremental loading** - Monotonic tension: stress and κ increase
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8. **Bauschinger effect** - Cyclic loading: early yield in reverse direction
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9. **Pure shear** - Validates τ_yield = σ_y/√3 for shear stress
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10. **Consistency** - Yield criterion f ≤ 0 satisfied for all strain levels
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11. **Performance** - Zero allocations (elastic), minimal (plastic), type stability
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**Mathematical Background:**
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- Yield criterion: f = √(3/2·s:s) - σ_y ≤ 0 (von Mises)
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- s = dev(σ - α) = deviatoric relative stress
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- Flow rule: ε̇_p = Δγ·∂f/∂σ = Δγ·(3/2·s/||s||) (associative plasticity)
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- Hardening: α̇ = H·ε̇_p (kinematic hardening, models Bauschinger effect)
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- Plastic work: κ̇ = √(2/3·ε̇_p:ε̇_p) (accumulated plastic strain)
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- Radial return: σ_n+1 = σ_trial - 2μ·Δγ·n where n = s/||s||
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- Consistency: Kuhn-Tucker conditions (f ≤ 0, Δγ ≥ 0, Δγ·f = 0)
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- Physical constraints: E > 0, -1 < ν < 0.5, σ_y > 0, H ≥ 0
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**Expected Results:**
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✅ Material constructed: E=200 GPa, ν=0.3, σ_y=250 MPa, H=1 GPa
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✅ Perfect plasticity: H=0 valid (no hardening)
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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: ε_p=0, α=0, κ=0 (virgin material)
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✅ Elastic (ε=1e-5): σ = elastic, ε_p=0, α=0, κ=0
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✅ Plastic (ε=0.003 > ε_y=0.00125): ε_p > 0, κ > 0, von_mises = σ_y ± 1e-6
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✅ Radial return (ε=0.01): f = 0 enforced, ||σ|| < ||σ_elastic||, ε_p > 1e-4
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✅ Hardening: H=1 GPa → ||α|| > 0, ||σ_hard|| > ||σ_perfect||
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✅ Incremental: Monotonic stress and κ increase (10 steps to ε=0.005)
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✅ Bauschinger: Tension then compression → κ increases, α evolves
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✅ Pure shear: τ_yield ≈ σ_y/√3 = 144 MPa for σ_y=250 MPa
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✅ Consistency: f ≤ 1e-6 for all strain levels (0.1% to 2%)
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✅ Simplified interface (without state, Δt) matches full call
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✅ Zero allocations (elastic path), ≤256 bytes (plastic path for state)
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✅ Type-stable: returns Tuple{SymmetricTensor{2,3}, SymmetricTensor{4,3}, PlasticityState}
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**Test Coverage:**
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- 13 test sets, ~80 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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- Loading: Elastic (ε=1e-5), yield (ε=0.003), large (ε=0.01), shear (γ=0.005)
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- Validation methods: Analytical yield criterion, stress comparison, state evolution
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- Algorithms: Radial return, consistency enforcement, incremental loading, cyclic loading
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- Edge cases: Perfect plasticity (H=0), high hardening (H=10 GPa), pure shear
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**Key Physics:**
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- J2 plasticity: Pressure-independent yield (metals)
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- Radial return: Closest-point projection to yield surface
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- Kinematic hardening: Backstress α shifts yield surface (Bauschinger effect)
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- Consistency: Active constraint f=0 during plastic loading
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- History dependence: Current stress depends on loading path via state
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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 implementation
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include("../src/materials/perfect_plasticity.jl")
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@testset "Perfect Plasticity Material" begin
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@testset "Material Construction" begin
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# Valid construction
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steel = PerfectPlasticity(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 = PerfectPlasticity(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 PerfectPlasticity(E=-200e9, ν=0.3, σ_y=250e6, H=1e9) # Negative E
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@test_throws ArgumentError PerfectPlasticity(E=200e9, ν=0.6, σ_y=250e6, H=1e9) # ν too large
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@test_throws ArgumentError PerfectPlasticity(E=200e9, ν=0.3, σ_y=-250e6, H=1e9) # Negative σ_y
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@test_throws ArgumentError PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=-1e9) # Negative H
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end
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@testset "State Construction" begin
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# Default state (zero)
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state0 = PlasticityState()
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@test state0.ε_p == zero(SymmetricTensor{2,3})
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@test state0.α == zero(SymmetricTensor{2,3})
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@test state0.κ == 0.0
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# Custom state
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ε_p = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))
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α = SymmetricTensor{2,3}((1e8, 0.0, 0.0, 0.0, 0.0, 0.0))
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state = PlasticityState(ε_p, α, 0.01)
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@test state.ε_p == ε_p
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@test state.α == α
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@test state.κ == 0.01
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# Invalid state (negative κ)
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@test_throws ArgumentError PlasticityState(ε_p, α, -0.01)
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end
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@testset "Elastic Loading (Small Strain)" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Small strain (well below yield)
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ε_small = SymmetricTensor{2,3}((1e-5, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔻, state_new = compute_stress(steel, ε_small, nothing, 0.0)
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# Should remain elastic
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@test state_new.ε_p == zero(SymmetricTensor{2,3}) # No plastic strain
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@test state_new.α == zero(SymmetricTensor{2,3}) # No backstress
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@test state_new.κ == 0.0 # No plastic work
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# Stress should be elastic
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μ = steel.μ
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λ = steel.λ
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I = one(ε_small)
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σ_elastic = λ * tr(ε_small) * I + 2μ * ε_small
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@test σ ≈ σ_elastic rtol = 1e-12
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# Tangent should be elastic
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@test 𝔻 isa SymmetricTensor{4,3}
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end
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@testset "Plastic Loading (Yield)" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Strain beyond yield (uniaxial tension)
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# Yield strain: ε_y = σ_y / E ≈ 0.00125
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ε_plastic = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔻, state_new = compute_stress(steel, ε_plastic, nothing, 0.0)
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# Should have plastic strain
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@test norm(state_new.ε_p) > 0.0
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@test state_new.κ > 0.0
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# Check yield criterion (should be satisfied)
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s = dev(σ - state_new.α)
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von_mises = √(3 / 2) * √(s ⊡ s)
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@test von_mises ≈ steel.σ_y rtol = 1e-6 # On yield surface
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# Plastic strain should be deviatoric
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@test abs(tr(state_new.ε_p)) < 1e-12
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end
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@testset "Radial Return Mapping" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Large strain (far beyond yield)
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ε_large = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔻, state_new = compute_stress(steel, ε_large, nothing, 0.0)
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# Check yield criterion (must be satisfied)
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s = dev(σ - state_new.α)
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von_mises = √(3 / 2) * √(s ⊡ s)
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@test von_mises ≈ steel.σ_y rtol = 1e-6
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# Stress should be less than elastic prediction
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μ = steel.μ
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λ = steel.λ
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I = one(ε_large)
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σ_elastic = λ * tr(ε_large) * I + 2μ * ε_large
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@test norm(σ) < norm(σ_elastic)
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# Plastic strain should be significant
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@test norm(state_new.ε_p) > 1e-4
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end
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@testset "Hardening Behavior" begin
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# Compare hardening vs perfect plasticity
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steel_hard = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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steel_perf = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=0.0)
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ε_test = SymmetricTensor{2,3}((0.005, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ_hard, _, state_hard = compute_stress(steel_hard, ε_test, nothing, 0.0)
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σ_perf, _, state_perf = compute_stress(steel_perf, ε_test, nothing, 0.0)
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# Hardening material should have backstress
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@test norm(state_hard.α) > 0.0
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@test norm(state_perf.α) == 0.0
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# Hardening material should have higher stress
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@test norm(σ_hard) > norm(σ_perf)
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end
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@testset "Incremental Loading" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Load in increments
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n_steps = 10
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ε_max = 0.005
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state = PlasticityState()
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stresses = []
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plastic_strains = []
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for i in 1:n_steps
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ε = SymmetricTensor{2,3}((i * ε_max / n_steps, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, _, state = compute_stress(steel, ε, 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 monotonically (hardening)
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@test all(diff(stresses) .≥ 0)
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# Plastic strain should increase monotonically
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@test all(diff(plastic_strains) .≥ 0)
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# Final plastic strain should be positive
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@test plastic_strains[end] > 0.0
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end
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@testset "Bauschinger Effect (Cyclic Loading)" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=10e9) # High H for visibility
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# Step 1: Tension to plastic regime
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ε_tension = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ_t, _, state_t = compute_stress(steel, ε_tension, nothing, 0.0)
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# Step 2: Reverse to compression
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ε_compression = SymmetricTensor{2,3}((-0.002, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ_c, _, state_c = compute_stress(steel, ε_compression, state_t, 0.0)
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# Should yield in compression earlier (Bauschinger effect from backstress)
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@test state_c.κ > state_t.κ # Additional plastic strain
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@test norm(state_c.α) > 0.0 # Backstress present
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end
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@testset "Pure Shear" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Pure shear strain
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γ = 0.005
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ε_shear = SymmetricTensor{2,3}((0.0, γ / 2, 0.0, 0.0, 0.0, 0.0))
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σ, _, state = compute_stress(steel, ε_shear, nothing, 0.0)
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# Check shear stress
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@test abs(σ[1, 2]) > 0.0
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# Check yield in shear
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# For pure shear: τ_yield = σ_y / √3
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s = dev(σ - state.α)
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von_mises = √(3 / 2) * √(s ⊡ s)
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if von_mises > steel.σ_y - 1e-3 # Plastic
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@test von_mises ≈ steel.σ_y rtol = 1e-6
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end
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end
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@testset "Consistency Check" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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# Multiple strain levels
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strain_levels = [0.001, 0.002, 0.005, 0.01, 0.02]
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for ε_mag in strain_levels
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ε = SymmetricTensor{2,3}((ε_mag, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, _, state = compute_stress(steel, ε, nothing, 0.0)
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# Check yield criterion
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s = dev(σ - state.α)
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von_mises = √(3 / 2) * √(s ⊡ s)
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# Must satisfy: f = von_mises - σ_y ≤ 0
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f = von_mises - steel.σ_y
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@test f ≤ 1e-6 # On or inside yield surface
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end
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end
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@testset "Simplified Interface" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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ε = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))
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# Test with and without explicit state/Δt
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σ1, 𝔻1, state1 = compute_stress(steel, ε)
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σ2, 𝔻2, state2 = compute_stress(steel, ε, nothing, 0.0)
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@test σ1 ≈ σ2
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@test 𝔻1 ≈ 𝔻2
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@test state1.κ ≈ state2.κ
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end
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@testset "Zero Allocation" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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state = PlasticityState()
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# Test elastic path (no state change)
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ε_elastic = SymmetricTensor{2,3}((1e-5, 0.0, 0.0, 0.0, 0.0, 0.0))
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# First call to compile
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compute_stress(steel, ε_elastic, state, 0.0)
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# Check allocations on elastic path
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allocs_elastic = @allocated compute_stress(steel, ε_elastic, state, 0.0)
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@test allocs_elastic == 0 # Elastic path should have zero allocations
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# Test plastic path (state changes)
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ε_plastic = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))
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# First call to compile
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compute_stress(steel, ε_plastic, state, 0.0)
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# Check allocations on plastic path
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allocs_plastic = @allocated compute_stress(steel, ε_plastic, state, 0.0)
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# Note: Plastic path allocates ~128 bytes for PlasticityState struct
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# This is acceptable for stateful materials
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@test allocs_plastic ≤ 256 # Allow some allocation for state
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end
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@testset "Type Stability" begin
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steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
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ε = SymmetricTensor{2,3}((0.002, 0.0, 0.0, 0.0, 0.0, 0.0))
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state = PlasticityState()
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# Infer return types
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result = @inferred compute_stress(steel, ε, 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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PlasticityState}
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end
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end
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