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In linear elasticity we have, for an isotropic material, $$C[E]=2 \mu E + \lambda \operatorname{tr}(E)I$$ where $\mu,\lambda$ are called Lamè moduli and $E=\frac{\nabla u + \nabla{u}^T}{2}$

I've seen that we can write $$C=\lambda I \otimes I + 2 \mu \mathbf{I}$$ where $\mathbf{I}$ is the fourth order identity tensor (since $C$ is a fourth order tensor). My question is: how can one derive the latter expression for $C$, starting from the one for $C[E]$?

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  • $\begingroup$ I'm drafting a note that explains how this relationship (known as generalized Hooke's Law) is assembled from basic normal and shear stress–strain relations; perhaps you'll find it useful. I quote: "Remarkably, we can derive [this law] using only two assumptions: (1) All stable materials stretch when pulled and contract when pushed and (2) The lateral dimensions may also change." $\endgroup$
    – Chemomechanics
    Commented Jun 14, 2021 at 17:39

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By definition, the fourth-order identity tensor $\mathbf{I}$ is the tensor such that $\mathbf{I} : E = E$ for every second-order symmetric tensor $E$. Next, the meaning of the tensor product $\otimes$ is that $(A \otimes B) : E = A \operatorname{tr}(BE)$. It follows that $(2\mu\mathbf{I}) : E = 2\mu E$ and $(\lambda I \otimes I) : E = \lambda \operatorname{tr}(E)\, I$.

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  • $\begingroup$ Yes, that's clear. But how can I see that $C$ has precisely that form, starting from $C[E]$? Is there a way to find out the components $C_{ijkl}$ ? @nanoman $\endgroup$
    – bobinthebox
    Commented Jun 14, 2021 at 9:40

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