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test(materials): add linear elastic material test
New 319-line test file for LinearElastic material model: - Tests material construction and parameter validation - Tests Lamé parameters (λ, μ) computation - Tests stress computation: uniaxial, pure shear, hydrostatic - Tests tangent modulus (4th-order elasticity tensor) - Validates symmetry, isotropy, and physical properties - Tests simplified interface (without state, Δt) - Validates zero-allocation and type stability Comprehensive test for Hooke's law implementation essential for linear static and dynamic analysis.
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Jukka Aho
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"""
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# Unit Tests: LinearElastic Material Model
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**What:** Comprehensive validation of isotropic linear elastic material σ = C:ε
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**Why:**
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- Foundation of structural mechanics (Hooke's law in 3D)
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- Most common material model in engineering FEA
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- Validates correct implementation of elasticity tensor C
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- Critical for linear static/dynamic analysis
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**How:**
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Test suite validates:
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1. **Construction & parameters** - E, ν validity, Lamé parameters λ and μ
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2. **Stress computation** - Hooke's law σ = λ·tr(ε)I + 2μ·ε for various load cases:
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- Uniaxial extension: σ₁₁ = (λ + 2μ)·ε₁₁, lateral: σ₂₂ = σ₃₃ = λ·ε₁₁
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- Pure shear: σ₁₂ = 2μ·ε₁₂ (shear modulus definition)
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- Hydrostatic: σ = K·ε_vol·I where K = E/(3(1-2ν)) is bulk modulus
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- General strain: validates full 3D constitutive law
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3. **Tangent modulus** - 4th-order tensor 𝔻 = λ·I⊗I + 2μ·𝕀ˢʸᵐ
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- Structure: SymmetricTensor{4,3}
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- Consistency: strain-independent (linear elasticity)
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- Double contraction: σ = 𝔻 ⊡ ε
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4. **Physical properties** - Symmetry, isotropy, positive-definiteness
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5. **Performance** - Zero allocations, type stability
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**Mathematical Background:**
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- Lamé parameters: λ = Eν/((1+ν)(1-2ν)), μ = E/(2(1+ν)) = G
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- Bulk modulus: K = E/(3(1-2ν)) = λ + 2μ/3
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- Elasticity tensor: C_{ijkl} = λ·δ_{ij}δ_{kl} + μ·(δ_{ik}δ_{jl} + δ_{il}δ_{jk})
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- Physical constraints: E > 0, -1 < ν < 0.5 (0 ≤ ν < 0.5 for stable materials)
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**Expected Results:**
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✅ Material constructed with valid E, ν
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✅ Lamé parameters computed correctly: λ ≈ 115.4 GPa, μ ≈ 76.9 GPa for steel
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✅ Uniaxial stress: (λ+2μ)·ε₁₁ ≈ 269 GPa × 0.001 = 269 MPa
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✅ Shear stress: 2μ·ε₁₂ ≈ 77 GPa × 0.002 = 154 MPa
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✅ Hydrostatic: σ = K·ε_vol·I with correct bulk modulus
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✅ General strain: σ = λ·tr(ε)I + 2μ·ε matches analytical
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✅ Tangent 𝔻 has correct structure, constant for all strains
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✅ Stress symmetry: σ_{ij} = σ_{ji}
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✅ Isotropy: same strain magnitude → same stress magnitude in any direction
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✅ Simplified interface (without state, Δt) works
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✅ Zero allocations after compilation
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✅ Type-stable: returns Tuple{SymmetricTensor{2,3}, SymmetricTensor{4,3}, Nothing}
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**Test Coverage:**
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- 12 test sets, ~70 individual assertions
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- Material constants: Steel (E=200 GPa, ν=0.3), Aluminum (E=70 GPa, ν=0.33)
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- Numerical validation: Analytical formulas + physical constraints
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- Edge cases: Zero strain, pure modes, combined loading
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"""
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using Test
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using Tensors
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# Load implementation
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include("../src/materials/linear_elastic.jl")
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@testset "Linear Elastic Material" begin
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@testset "Material Construction" begin
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# Valid construction
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steel = LinearElastic(E=200e9, ν=0.3)
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@test steel.E == 200e9
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@test steel.ν == 0.3
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# Keyword constructor
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aluminum = LinearElastic(E=70e9, ν=0.33)
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@test aluminum.E == 70e9
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@test aluminum.ν == 0.33
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# Invalid inputs
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@test_throws ArgumentError LinearElastic(E=-100e9, ν=0.3) # Negative E
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@test_throws ArgumentError LinearElastic(E=200e9, ν=0.6) # ν too large
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@test_throws ArgumentError LinearElastic(E=200e9, ν=-1.1) # ν too small
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end
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@testset "Lamé Parameters" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# First Lamé parameter: λ = E·ν/((1+ν)(1-2ν))
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λ_expected = 200e9 * 0.3 / ((1 + 0.3) * (1 - 2 * 0.3))
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@test λ(steel) ≈ λ_expected rtol = 1e-12
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@test λ(steel) ≈ 115.38461538461539e9 rtol = 1e-10
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# Shear modulus: μ = E/(2(1+ν))
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μ_expected = 200e9 / (2 * (1 + 0.3))
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@test μ(steel) ≈ μ_expected rtol = 1e-12
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@test μ(steel) ≈ 76.92307692307693e9 rtol = 1e-10
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# Test inline optimization (should compile to constants)
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@test @inferred λ(steel) isa Float64
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@test @inferred μ(steel) isa Float64
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end
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@testset "Stress Computation - Uniaxial Extension" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# Uniaxial extension in x-direction: ε = [ε₁₁, 0, 0; 0, 0, 0; 0, 0, 0]
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ε₁₁ = 0.001
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ε = SymmetricTensor{2,3}((ε₁₁, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔻, state_new = compute_stress(steel, ε, nothing, 0.0)
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# Expected stress: σ₁₁ = (λ + 2μ)·ε₁₁, σ₂₂ = σ₃₃ = λ·ε₁₁
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λ_val = λ(steel)
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μ_val = μ(steel)
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σ₁₁_expected = (λ_val + 2μ_val) * ε₁₁
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σ₂₂_expected = λ_val * ε₁₁
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@test σ[1, 1] ≈ σ₁₁_expected rtol = 1e-12
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@test σ[2, 2] ≈ σ₂₂_expected rtol = 1e-12
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@test σ[3, 3] ≈ σ₂₂_expected rtol = 1e-12
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@test σ[1, 2] ≈ 0.0 atol = 1e-15
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@test σ[1, 3] ≈ 0.0 atol = 1e-15
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@test σ[2, 3] ≈ 0.0 atol = 1e-15
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# State should be nothing (stateless material)
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@test state_new === nothing
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# Numerical check: σ₁₁ = (λ + 2μ)·ε₁₁ ≈ 269.2 MPa
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@test σ[1, 1] ≈ 269.2e6 rtol = 1e-2
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@test σ[2, 2] ≈ 115.4e6 rtol = 1e-2 # λ·ε₁₁ (positive for extension)
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end
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@testset "Stress Computation - Pure Shear" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# Pure shear: ε₁₂ = γ/2 (engineering shear strain γ = 0.002)
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γ = 0.002
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ε₁₂ = γ / 2 # Tensor shear strain
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ε = SymmetricTensor{2,3}((0.0, ε₁₂, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔻, state_new = compute_stress(steel, ε, nothing, 0.0)
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# Expected stress: σ₁₂ = 2μ·ε₁₂
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μ_val = μ(steel)
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σ₁₂_expected = 2μ_val * ε₁₂
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@test σ[1, 2] ≈ σ₁₂_expected rtol = 1e-12
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@test σ[1, 1] ≈ 0.0 atol = 1e-15
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@test σ[2, 2] ≈ 0.0 atol = 1e-15
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@test σ[3, 3] ≈ 0.0 atol = 1e-15
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# Numerical check: σ₁₂ = 2μ·(γ/2) = μ·γ ≈ 77 GPa × 0.002 = 154 MPa
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@test σ[1, 2] ≈ 154e6 rtol = 1e-2
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@test state_new === nothing
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end
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@testset "Stress Computation - Hydrostatic Pressure" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# Hydrostatic strain: ε = ε_vol/3 · I
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ε_vol = 0.003 # Volumetric strain
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ε_iso = ε_vol / 3
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ε = SymmetricTensor{2,3}((ε_iso, 0.0, 0.0, ε_iso, 0.0, ε_iso))
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σ, 𝔻, state_new = compute_stress(steel, ε, nothing, 0.0)
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# Expected stress: σ = (λ + 2μ/3)·ε_vol·I = K·ε_vol·I
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# Bulk modulus: K = λ + 2μ/3 = E/(3(1-2ν))
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λ_val = λ(steel)
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μ_val = μ(steel)
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K = λ_val + 2μ_val / 3
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σ_expected = K * ε_vol
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@test σ[1, 1] ≈ σ_expected rtol = 1e-12
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@test σ[2, 2] ≈ σ_expected rtol = 1e-12
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@test σ[3, 3] ≈ σ_expected rtol = 1e-12
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@test σ[1, 2] ≈ 0.0 atol = 1e-15
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@test σ[1, 3] ≈ 0.0 atol = 1e-15
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@test σ[2, 3] ≈ 0.0 atol = 1e-15
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# Bulk modulus check
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K_expected = steel.E / (3 * (1 - 2 * steel.ν))
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@test K ≈ K_expected rtol = 1e-12
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@test state_new === nothing
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end
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@testset "Stress Computation - General Strain" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# General strain tensor (all components non-zero)
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ε = SymmetricTensor{2,3}((0.001, 0.0005, 0.0003, -0.0002, 0.0004, 0.0006))
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σ, 𝔻, state_new = compute_stress(steel, ε, nothing, 0.0)
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# Verify Hooke's law: σ = λ·tr(ε)·I + 2μ·ε
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λ_val = λ(steel)
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μ_val = μ(steel)
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I = one(ε)
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σ_expected = λ_val * tr(ε) * I + 2μ_val * ε
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@test σ ≈ σ_expected rtol = 1e-12
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# Check each component explicitly
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@test σ[1, 1] ≈ σ_expected[1, 1] rtol = 1e-12
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@test σ[2, 2] ≈ σ_expected[2, 2] rtol = 1e-12
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@test σ[3, 3] ≈ σ_expected[3, 3] rtol = 1e-12
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@test σ[1, 2] ≈ σ_expected[1, 2] rtol = 1e-12
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@test σ[1, 3] ≈ σ_expected[1, 3] rtol = 1e-12
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@test σ[2, 3] ≈ σ_expected[2, 3] rtol = 1e-12
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@test state_new === nothing
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end
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@testset "Tangent Modulus - Structure" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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ε = SymmetricTensor{2,3}((0.001, 0.0, 0.0, 0.0, 0.0, 0.0))
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σ, 𝔻, _ = compute_stress(steel, ε, nothing, 0.0)
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# Verify tangent is 4th order symmetric tensor
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@test 𝔻 isa SymmetricTensor{4,3}
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# Verify 𝔻 = λ·I⊗I + 2μ·𝕀ˢʸᵐ
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λ_val = λ(steel)
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μ_val = μ(steel)
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I = one(ε)
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𝕀ˢʸᵐ = one(SymmetricTensor{4,3,Float64})
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𝔻_expected = λ_val * (I ⊗ I) + 2μ_val * 𝕀ˢʸᵐ
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@test 𝔻 ≈ 𝔻_expected rtol = 1e-12
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end
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@testset "Tangent Modulus - Consistency" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# Tangent should be constant (independent of strain)
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ε1 = SymmetricTensor{2,3}((0.001, 0.0, 0.0, 0.0, 0.0, 0.0))
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ε2 = SymmetricTensor{2,3}((0.005, 0.002, 0.001, -0.003, 0.0, 0.0))
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_, 𝔻1, _ = compute_stress(steel, ε1, nothing, 0.0)
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_, 𝔻2, _ = compute_stress(steel, ε2, nothing, 0.0)
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@test 𝔻1 ≈ 𝔻2 rtol = 1e-12
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end
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@testset "Tangent Modulus - Double Contraction" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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ε = SymmetricTensor{2,3}((0.001, 0.0005, 0.0003, -0.0002, 0.0004, 0.0006))
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σ, 𝔻, _ = compute_stress(steel, ε, nothing, 0.0)
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# Verify σ = 𝔻 ⊡ ε (double contraction)
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σ_from_tangent = 𝔻 ⊡ ε
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@test σ ≈ σ_from_tangent rtol = 1e-12
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end
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@testset "Symmetry Properties" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# Stress tensor should be symmetric
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ε = SymmetricTensor{2,3}((0.001, 0.0005, 0.0003, -0.0002, 0.0004, 0.0006))
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σ, _, _ = compute_stress(steel, ε, nothing, 0.0)
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@test σ[1, 2] ≈ σ[2, 1] rtol = 1e-15
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@test σ[1, 3] ≈ σ[3, 1] rtol = 1e-15
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@test σ[2, 3] ≈ σ[3, 2] rtol = 1e-15
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end
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@testset "Isotropy Verification" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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# Same strain magnitude in different directions → same stress magnitude
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ε_x = SymmetricTensor{2,3}((0.001, 0.0, 0.0, 0.0, 0.0, 0.0))
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ε_y = SymmetricTensor{2,3}((0.0, 0.0, 0.0, 0.001, 0.0, 0.0))
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ε_z = SymmetricTensor{2,3}((0.0, 0.0, 0.0, 0.0, 0.0, 0.001))
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σ_x, _, _ = compute_stress(steel, ε_x, nothing, 0.0)
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σ_y, _, _ = compute_stress(steel, ε_y, nothing, 0.0)
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σ_z, _, _ = compute_stress(steel, ε_z, nothing, 0.0)
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# σ₁₁(ε_x) should equal σ₂₂(ε_y) and σ₃₃(ε_z)
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@test σ_x[1, 1] ≈ σ_y[2, 2] rtol = 1e-15
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@test σ_x[1, 1] ≈ σ_z[3, 3] rtol = 1e-15
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end
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@testset "Simplified Interface" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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ε = SymmetricTensor{2,3}((0.001, 0.0, 0.0, 0.0, 0.0, 0.0))
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# Test simplified call (without state and 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 === nothing
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@test state2 === nothing
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end
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@testset "Zero Allocation" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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ε = SymmetricTensor{2,3}((0.001, 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, nothing, 0.0)
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# Check allocations
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allocs = @allocated compute_stress(steel, ε, nothing, 0.0)
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@test allocs == 0
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end
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@testset "Type Stability" begin
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steel = LinearElastic(E=200e9, ν=0.3)
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ε = SymmetricTensor{2,3}((0.001, 0.0, 0.0, 0.0, 0.0, 0.0))
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# Infer return types
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result = @inferred compute_stress(steel, ε, nothing, 0.0)
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@test result isa Tuple{SymmetricTensor{2,3,Float64},SymmetricTensor{4,3,Float64},Nothing}
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end
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end
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