mirror of
https://github.com/JuliaFEM/JuliaFEM.jl.git
synced 2026-09-11 06:08:07 +00:00
Jukka Aho
e9ead51f24
- Tests LinearElastic material construction with parameter validation - Validates Lamé parameter computation (λ and μ from E and ν) - Stress computation tests: uniaxial extension, pure shear, hydrostatic pressure, general strain - Verifies Hooke's law: σ = λ·tr(ε)·I + 2μ·ε - Tangent modulus validation: 𝔻 = λ·I⊗I + 2μ·𝕀ˢʸᵐ (4th-order tensor) - Double contraction consistency: σ = 𝔻 ⊡ ε - Symmetry and isotropy property verification - Tests both full and simplified compute_stress() interfaces - Zero allocation and type stability checks - 279 lines validating fundamental elasticity operations with Tensors.jl
280 lines
9.6 KiB
Julia
280 lines
9.6 KiB
Julia
"""
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Unit tests for LinearElastic material model.
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Tests cover:
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1. Material construction and validation
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2. Lamé parameter computation
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3. Stress computation (uniaxial, shear, hydrostatic, general)
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4. Tangent modulus verification
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5. Symmetry and isotropy
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6. Zero allocation verification
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7. Type stability
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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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