# Compute components of elasticity tensor for isotropic material

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 $$C[E]$$?

• 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." Jun 14, 2021 at 17:39

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 $$(\lambda I \otimes I) : E = \lambda \operatorname{tr}(E)\, I$$.
• 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 Jun 14, 2021 at 9:40