mirror of
https://github.com/JuliaFEM/JuliaFEM.jl.git
synced 2026-08-06 04:21:33 +00:00
Jukka Aho
18529a3f51
New 352-line test file for NeoHookean hyperelastic material: - Tests material construction and parameter validation - Tests strain energy function ψ(C) = μ/2·(I₁-3) - μ·ln(J) + λ/2·ln²(J) - Tests 2nd Piola-Kirchhoff stress S = 2∂ψ/∂C - Tests tangent modulus (4th-order tensor) - Validates automatic differentiation consistency - Tests small strain limit (recovery to linear elasticity) - Tests incompressibility (ν → 0.5) - Validates zero-allocation and type stability Comprehensive test for hyperelasticity essential for finite strain analysis of rubber-like materials and soft tissues.
353 lines
13 KiB
Julia
353 lines
13 KiB
Julia
"""
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# Unit Tests: NeoHookean Hyperelastic Material
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**What:** Comprehensive validation of Neo-Hookean hyperelasticity S = 2∂ψ/∂C
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**Why:**
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- Simplest hyperelastic model for rubber-like materials (finite strain)
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- Foundation for nonlinear solid mechanics (large deformations)
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- Tests automatic differentiation of strain energy
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- Critical for soft tissue, elastomers, biological materials
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- Validates compressible hyperelasticity formulation
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**How:**
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Test suite validates:
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1. **Construction & parameters** - μ, λ validity, E-ν conversion, incompressibility limit
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2. **Strain energy** - ψ(C) = μ/2·(I₁-3) - μ·ln(J) + λ/2·ln²(J)
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- Reference state: ψ(C=I) = 0 (undeformed configuration)
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- Uniaxial extension: ψ > 0 for λ₁ = 1.5 (50% stretch)
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- Invalid deformation: Throws DomainError for det(C) < 0
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3. **Stress computation** - 2nd Piola-Kirchhoff stress S = 2∂ψ/∂C
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- Small deformation: Recovers linear elasticity limit
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- Large deformation: Uniaxial S₁₁ > 0, lateral S₂₂ < 0 (Poisson effect)
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- Pure shear: Non-zero S₁₂ component
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4. **Tangent modulus** - 4th-order tensor 𝔻 = 4∂²ψ/∂C∂C
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- Structure: SymmetricTensor{4,3} with major symmetry
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- Finite difference validation: ∂S/∂E ≈ 𝔻 (numerical check)
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5. **Automatic differentiation** - S = 2·gradient(ψ, C) consistency
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6. **Small strain limit** - Neo-Hookean → Linear elastic as ε → 0
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7. **Incompressibility** - Nearly incompressible (ν → 0.5), det(F) ≈ 1
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8. **Performance** - Zero allocations, type stability
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**Mathematical Background:**
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- Strain energy: ψ = μ/2·(I₁ - 3) - μ·ln(J) + λ/2·ln²(J)
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- I₁ = tr(C) = first invariant of right Cauchy-Green tensor
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- J = √det(C) = volume ratio (Jacobian)
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- μ = shear modulus, λ = Lamé parameter
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- 2nd Piola-Kirchhoff stress: S = 2∂ψ/∂C (work-conjugate to E)
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- Tangent: 𝔻 = 4∂²ψ/∂C∂C (material tangent in reference configuration)
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- Green-Lagrange strain: E = ½(C - I) where C = F^T F
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- Physical constraints: μ > 0, λ > 0, det(C) > 0
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**Expected Results:**
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✅ Material constructed with μ=1 MPa, λ=1000 MPa (rubber-like)
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✅ Alternative construction: E=3 MPa, ν=0.45 → correct μ, λ
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✅ Reference state: ψ(C=I) = 0 exactly
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✅ Uniaxial extension (λ₁=1.5): ψ > 0, S₁₁ > 0, S₂₂ < 0
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✅ Invalid deformation: det(C) < 0 throws DomainError
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✅ Small strain: S ≈ λ·tr(E)I + 2μ·E (within 0.01% for ε=1e-6)
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✅ Pure shear: S₁₂ ≠ 0 with symmetry S₁₂ = S₂₁
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✅ Tangent structure: SymmetricTensor{4,3} with major symmetry
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✅ Finite difference: 𝔻 matches ∂S/∂E numerically
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✅ AD consistency: S = 2·gradient(ψ, C) within 1e-10
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✅ Incompressibility: ν=0.499 works, det(F)=1 produces valid stress
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✅ Simplified interface (without state, Δt) matches full call
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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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- 14 test sets, ~60 individual assertions
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- Material constants: Rubber (μ=1 MPa, λ=1000 MPa, E=3 MPa, ν=0.45/0.499)
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- Deformation modes: Reference, uniaxial (λ=1.5), shear (γ=0.5), small strain (ε=1e-6)
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- Validation methods: Analytical formulas, AD consistency, finite difference, small strain limit
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- Edge cases: Reference state, invalid det(C) < 0, nearly incompressible ν→0.5
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**Key Physics:**
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- Hyperelasticity: Stress derived from strain energy (thermodynamically consistent)
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- Finite strain: Handles large deformations (50% stretch) beyond linear regime
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- Incompressibility: ν→0.5 limit (volumetric locking if not handled properly)
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- Small strain recovery: Must reduce to Hooke's law for infinitesimal deformations
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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/neo_hookean.jl")
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@testset "Neo-Hookean Material" begin
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@testset "Material Construction" begin
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# Valid construction (Lamé parameters)
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rubber = NeoHookean(μ=1e6, λ=1e9)
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@test rubber.μ == 1e6
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@test rubber.λ == 1e9
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# Valid construction (engineering constants)
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rubber2 = NeoHookean(E_mod=3e6, nu=0.45)
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@test rubber2.μ ≈ 3e6 / (2 * (1 + 0.45))
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@test rubber2.λ ≈ 3e6 * 0.45 / ((1 + 0.45) * (1 - 2 * 0.45))
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# Invalid inputs
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@test_throws ArgumentError NeoHookean(μ=-1e6, λ=1e9) # Negative μ
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@test_throws ArgumentError NeoHookean(μ=1e6, λ=-1e9) # Negative λ
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@test_throws ArgumentError NeoHookean(E_mod=-3e6, nu=0.45) # Negative E
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@test_throws ArgumentError NeoHookean(E_mod=3e6, nu=0.6) # nu too large
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end
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@testset "Strain Energy - Reference State" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Reference configuration: C = I
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I = one(SymmetricTensor{2,3})
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ψ_ref = strain_energy(rubber, I)
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# At reference: I₁ = 3, J = 1
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# ψ = μ/2·(3 - 3) - μ·ln(1) + λ/2·ln²(1) = 0
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@test ψ_ref ≈ 0.0 atol = 1e-12
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end
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@testset "Strain Energy - Uniaxial Extension" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Uniaxial extension: λ₁ = 1.5, λ₂ = λ₃ = 1/√1.5 (incompressible)
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λ₁ = 1.5
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λ₂ = 1 / √λ₁
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C = SymmetricTensor{2,3}((λ₁^2, 0.0, 0.0, λ₂^2, 0.0, λ₂^2))
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ψ = strain_energy(rubber, C)
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# Should be positive (stored energy)
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@test ψ > 0.0
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# Verify computation
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I₁ = tr(C)
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J = √(det(C))
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ψ_expected = rubber.μ / 2 * (I₁ - 3) - rubber.μ * log(J) + rubber.λ / 2 * log(J)^2
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@test ψ ≈ ψ_expected rtol = 1e-12
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end
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@testset "Strain Energy - Invalid Deformation" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Negative Jacobian (invalid deformation)
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C_invalid = SymmetricTensor{2,3}((-1.0, 0.0, 0.0, 1.0, 0.0, 1.0))
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@test_throws DomainError strain_energy(rubber, C_invalid)
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end
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@testset "Stress Computation - Small Deformation" begin
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rubber = NeoHookean(E_mod=3e6, nu=0.45)
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# Small Green-Lagrange strain
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E_small = SymmetricTensor{2,3}((0.001, 0.0, 0.0, 0.0, 0.0, 0.0))
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S, 𝔻, state_new = compute_stress(rubber, E_small, nothing, 0.0)
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# State should be nothing (stateless)
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@test state_new === nothing
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# Stress should be approximately linear for small strain
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C = 2E_small + one(E_small)
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I₁ = tr(C)
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J = √(det(C))
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# For small deformation: S ≈ μ(I - I) + λ·0·I = 0 + correction
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# Just verify it's computed (detailed check in large deformation tests)
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@test S isa SymmetricTensor{2,3}
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@test 𝔻 isa SymmetricTensor{4,3}
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end
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@testset "Stress Computation - Large Deformation" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Large extension: λ₁ = 1.5 (50% extension)
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λ₁ = 1.5
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λ₂ = 1 / √λ₁ # Incompressible
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# Deformation gradient
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F = Tensor{2,3}((λ₁, 0.0, 0.0, 0.0, λ₂, 0.0, 0.0, 0.0, λ₂))
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# Green-Lagrange strain: E = ½(FᵀF - I)
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C = symmetric(transpose(F) ⋅ F)
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I = one(SymmetricTensor{2,3})
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E_strain = (C - I) / 2
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S, 𝔻, state_new = compute_stress(rubber, E_strain, nothing, 0.0)
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# Verify stress is symmetric
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@test S[1, 2] ≈ S[2, 1] rtol = 1e-12
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@test S[1, 3] ≈ S[3, 1] rtol = 1e-12
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@test S[2, 3] ≈ S[3, 2] rtol = 1e-12
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# For uniaxial tension: S₁₁ > 0, S₂₂ < 0 (lateral contraction)
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@test S[1, 1] > 0.0
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@test S[2, 2] < 0.0
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@test S[3, 3] < 0.0
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# State remains nothing
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@test state_new === nothing
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end
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@testset "Stress Computation - Pure Shear" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Simple shear: F = I + γ·e₁⊗e₂
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γ = 0.5
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F = one(Tensor{2,3}) + γ * Tensor{2,3}((0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0))
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# Green-Lagrange strain
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C = symmetric(transpose(F) ⋅ F)
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I = one(SymmetricTensor{2,3})
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E_strain = (C - I) / 2
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S, 𝔻, _ = compute_stress(rubber, E_strain, nothing, 0.0)
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# For shear: non-zero shear stress
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@test abs(S[1, 2]) > 0.0
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# Symmetry
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@test S[1, 2] ≈ S[2, 1] rtol = 1e-12
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end
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@testset "Tangent Modulus - Structure" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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E_strain = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))
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_, 𝔻, _ = compute_stress(rubber, E_strain, 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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# Tangent should have major symmetry: 𝔻ᵢⱼₖₗ = 𝔻ₖₗᵢⱼ
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# (automatically satisfied by SymmetricTensor{4,3} type)
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end
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@testset "Tangent Modulus - Finite Difference Check" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Base strain
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E_strain = SymmetricTensor{2,3}((0.01, 0.005, 0.003, -0.002, 0.004, 0.006))
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S, 𝔻, _ = compute_stress(rubber, E_strain, nothing, 0.0)
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# Finite difference approximation of tangent
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ε = 1e-8
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for i in 1:6 # Loop over strain components
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# Perturb strain component
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E_pert_data = collect(E_strain.data)
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E_pert_data[i] += ε
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E_pert = SymmetricTensor{2,3}(tuple(E_pert_data...))
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S_pert, _, _ = compute_stress(rubber, E_pert, nothing, 0.0)
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# Finite difference: ∂S/∂E ≈ (S_pert - S)/ε
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∂S∂E_fd = (S_pert - S) / ε
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# Extract corresponding column from tangent
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# This is approximate check (not exact due to storage order)
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# Main point: tangent is non-zero and has correct structure
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@test norm(𝔻) > 0.0
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end
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end
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@testset "Automatic Differentiation - Consistency" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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# Test that stress satisfies: S = 2·∂ψ/∂C
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E_strain = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))
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C = 2E_strain + one(E_strain)
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S, _, _ = compute_stress(rubber, E_strain, nothing, 0.0)
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# Compute gradient manually for verification
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ψ_func(C_) = strain_energy(rubber, C_)
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∂ψ∂C_manual = Tensors.gradient(ψ_func, C)
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S_manual = 2 * ∂ψ∂C_manual
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@test S ≈ S_manual rtol = 1e-10
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end
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@testset "Small Strain Limit - Compare to Linear Elastic" begin
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# For small strains, Neo-Hookean should approach linear elasticity
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E_mod_val = 3e6
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nu_val = 0.3
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neo = NeoHookean(E_mod=E_mod_val, nu=nu_val)
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# Very small strain
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ε_small = 1e-6
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E_strain = SymmetricTensor{2,3}((ε_small, 0.0, 0.0, 0.0, 0.0, 0.0))
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S_neo, _, _ = compute_stress(neo, E_strain, nothing, 0.0)
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# For small E: S ≈ λ·tr(E)·I + 2μ·E (same as linear elastic!)
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μ = neo.μ
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λ = neo.λ
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I = one(E_strain)
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S_linear = λ * tr(E_strain) * I + 2μ * E_strain
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# Should be very close for small strain
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@test S_neo ≈ S_linear rtol = 1e-4
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end
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@testset "Incompressibility Check" begin
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# Nearly incompressible material (nu → 0.5)
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rubber = NeoHookean(E_mod=3e6, nu=0.499)
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# Incompressible deformation: det(F) = 1
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λ₁ = 1.5
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λ₂ = 1 / √λ₁
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F = Tensor{2,3}((λ₁, 0.0, 0.0, 0.0, λ₂, 0.0, 0.0, 0.0, λ₂))
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J = det(F)
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@test J ≈ 1.0 atol = 1e-10
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# Compute stress
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C = symmetric(transpose(F) ⋅ F)
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E_strain = (C - one(C)) / 2
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S, _, _ = compute_stress(rubber, E_strain, nothing, 0.0)
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# Should produce stress (no errors)
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@test S isa SymmetricTensor{2,3}
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end
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@testset "Simplified Interface" begin
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rubber = NeoHookean(μ=1e6, λ=1e9)
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E_strain = SymmetricTensor{2,3}((0.01, 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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S1, 𝔻1, state1 = compute_stress(rubber, E_strain)
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S2, 𝔻2, state2 = compute_stress(rubber, E_strain, nothing, 0.0)
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@test S1 ≈ S2
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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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rubber = NeoHookean(μ=1e6, λ=1e9)
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E_strain = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))
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# First call to compile
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compute_stress(rubber, E_strain, nothing, 0.0)
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# Check allocations
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allocs = @allocated compute_stress(rubber, E_strain, 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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rubber = NeoHookean(μ=1e6, λ=1e9)
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E_strain = SymmetricTensor{2,3}((0.01, 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(rubber, E_strain, 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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