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


2 Answers 2


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: $\frac{1}{2}\sigma \varepsilon$.

  • $\begingroup$ 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
    Commented 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
    Commented 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). $\endgroup$ Commented Nov 27, 2021 at 0:36

For quasistatic reversible elasticity, conservation of energy can be written as, $$\int \limits_V \dot u dV = \int\limits_V b_i v_i dV + \int \limits_S t_i v_i dS$$

Where $u$ is the strain energy density, $b_i$ is the body force density, $t_i$ is the surface traction, and $v_i$ is the velocity. Conservation of linear and angular momentum, which for quasi-static processes are simply equilibrium conditions are, $$\sigma_{ij,j} + b_i=0$$ and $$\sigma_{ij}=\sigma_{ji}$$ and the traction-stress relationships on any surface are, $$t_i=\sigma_{ij}n_j$$

Using the traction-stress relationship in the conservation of energy equation yields,

$$ \int \limits_V \dot u dV = \int\limits_V b_i v_i dV + \int \limits_S \sigma_{ij}n_j v_i dS$$

Apply the divergence theorem to the last term and add the two resulting volume integrands,

$$ \int \limits_V \dot u dV = \int \limits_V (b_i +\sigma_{ij,j})v_i + \sigma_{ij}v_{i,j} dV$$

The term in parentheses is zero due to mechanical equilibrium and due to the symmetry of the stress the last term is,


Since the integrals must be valid for any arbitrary volume we then have,

$$\dot u=\sigma_{ij}\dot\epsilon_{ij}$$

Taking the strain energy density as a function of the strain, $u = u(\epsilon_{ij})$, we then have,

$$\dot u=\frac{\partial u}{\partial \epsilon_{ij}} \dot\epsilon_{ij}=\sigma_{ij}\dot\epsilon_{ij}$$

which must hold for arbitrary strain increments leaving us with,

$$\frac{\partial u}{\partial \epsilon_{ij}}=\sigma_{ij}$$


$$u=\int \frac{\partial u}{\partial \epsilon_{ij}} d\epsilon_{ij} = \int \sigma_{ij} d\epsilon_{ij} $$

Edit: Sorry, that should have said "almost finally". That result is valid for any linear or nonlinear elastic material. For a linear elastic material, $\sigma_{ij}=c_{ijkl}\epsilon_{kl}$, and the integral on the right evaluates to $\frac{1}{2}c_{ijkl}\epsilon_{ij}\epsilon_{kl}=\frac{1}{2}\sigma_{ij}\epsilon_{ij}$.


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