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The strain energy density is: $$\frac{1}{2}\sigma_{ij}\epsilon_{ij}$$ Where $\sigma$ is the Cauchy stress tensor ($\sigma_{ij}=T_j(\mathbf{e}_i))$ and $\epsilon^e$ is the infinitesimal strain tensor ($\epsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right))$ and summation notation was used.

Is there a way to derive this relationship? Intuitively, the stress tensor represents the force per unit area and the strain tensor represents the displacement per unit length, so their product represents the "work" done by the internal stresses per unit volume, but I can't find a way to state that in a more rigorous derivation.

Any advice would be appreciated

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Infinitesimal stress–strain work is $\sigma V\,d\varepsilon=C\varepsilon V\,d\varepsilon$, where $C$ is the stiffness tensor. Integrate from initial strain 0 to final strain $\varepsilon$, replace an $\varepsilon$ with an equivalent $\sigma/C$, and normalize by volume to get the strain energy density: $\frac{1}{2}\sigma \varepsilon$.

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  • $\begingroup$ Where does the formula for infinitesimal work done come from (in one dimension I understand it, but how does it work in general)? If $\epsilon$ is a tensor rather than a scalar quantity, how does integrating w.r.t to it make sense (and why can you apply the fundamental theorem of calculus?) I understand why the relationships hold when the stress and strain are in a single dimension, but I don't get how this generalizes when they're tensors. $\endgroup$
    – QED
    Nov 27, 2021 at 0:24
  • $\begingroup$ Hint: tensors are vector spaces (i.e. if $\epsilon$ is a tensor, then so is $s\epsilon$, where $s$ is a real number.) $\endgroup$
    – TLDR
    Nov 27, 2021 at 0:31
  • $\begingroup$ Rather than retyping lengthy content, I’d like to point you to Nye’s Physical Properties of Crystals, which covers the tensor math associated with linear elasticity (and various other phenomena). $\endgroup$ Nov 27, 2021 at 0:36

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