Symmetry properties of the tangent modulus (strain energy function)

Question

For a strain energy function W depending on the Lagrangian strain E,the Piola-Kirchhoff stress of the second kind can be obtained by

S_ij = ∂W/∂E_ij

and the tangent modulus by

C_ijkl = ∂S_ij/∂E_kl

The symmetry conditions apply

C_ijkl = C_jilk , C_ijkl = C_ijlk , C_ijkl = C_klij

If I check the second symmetry condition for the tangent modulus of the St.Venant-Kirchhoff material with

C_ijkl = λ δ_ijδ_kl + 2μ δ_ikδ_jl

with i,j,k,l = 1,2,1,2 and i,j,l,k = 1,2,2,1 I obtain

C_ijkl = C₁₂₁₂ = λ δ₁₂δ₁₂ + 2μ δ₁₁δ₂₂ = 2μ

C_ijlk = C₁₂₂₁ = λ δ₁₂δ₂₁ + 2μ δ₁₂δ₂₁ = 0

I see this representation of the tangent module in various textbooks. However, I only get the same results if I symmetrize the expression in the second term with

0.5(δ_ikδ_jl + δ_ilδ_jk)

My question is:Why are the two calculated expressions not equal if I choose the non-symmetrized representation? I thought the symmetry of the strain tensor alone should make this verified symmetry condition valid.

Answer 1

The representation $C_{ijkl}=\lambda\delta_{ij}\delta{kl}+2\mu\delta{ik}\delta{jl}$ is probably a writing simplification meant only to act on the symmetrized strain tensor (i.e. $\epsilon_{kl}=\frac{1}{2}(u_{k,l}+u_{l,k})$).

This works out because the second term of $C$ can be written as the sum of a symmetric part (your $\frac{1}{2}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})$) plus a skew part. However, the product of this skew part with the symmetrized strain tensor will always be zero, so it does not matter.

If you plan to act $C_{ijkl}$ directly on $u_{k,l}$ you need to use the version with the symmetrized second term.

This is just a guess though, I'd need to see the specific source to really know what's actually going on.