Derivation of elastic energy per unit volume So I basically asked this question a little while back and didn't get much help, but I really need help, so I'm coming back and asking again. 
Looking at the section on Continuum Systems on the wikipedia page for elastic energy, the first equation in the section is 
$$ U = \frac{1}{2}C_{ijkl}\epsilon_{ij}\epsilon_{kl} $$ 
And remember, this is in Einstein notation, so U is really a sum over $i,j,k,l$. 
I CANNOT figure out how to derive this equation. Literally everywhere I go, this relation is just HANDED out with either no justification at all or a derivation so notationally byzantine I cannot read it. 
If someone knowledgeable enough/smarter than me could PLEASE give me a quick but clear derivation of U from the elastic strain tensor I would REALLY, REALLY appreciate it. 
Thank you. 
 A: I needed this too so I worked out a different derivation that works for shear as well.
Consider a block of material with side legths $2\delta x$, $2\delta y$, and $2\delta z$.
We deform it from a strain state of $\varepsilon_{ij}=0$ to $\varepsilon_{ij}=\varepsilon_{ij}^0$ linearly.
That is,
$$
\varepsilon = t\,\varepsilon_{ij}^0
$$
A point $\vec x^0$ travels along
$$
  x_i = t\,\varepsilon_{ij}^0 x_j^0 + tu_i
$$
Where $u_i$ is a rigid body displacement (this will not contribute).
The traction on a surface is
$$
T_i = n_j\sigma_{ij} = tn_jC_{ijkl}\varepsilon_{kl}^0.
$$
The work done on the top surface is
$$
W _{z}
  = \int_{-\delta x}^{\delta x}\int_{-\delta y}^{\delta y}
   \int_0^1
   T_i\frac{dx_i}{dt}
   \,dt \, dx \,dx
$$
If you sub it all in and work out the integrals then you'll find that
$$
W_z = 2C_{i3kl}\varepsilon_{kl}^0 (\varepsilon_{i3}^0\,\delta x\,\delta y\,\delta z-2u_i \,\delta x\,\delta y).
$$
Do the same for the other 5 surfaces, add them together and you get the total work done is
$$
W = 4C_{ijkl}\varepsilon_{kl}^0 \varepsilon_{ij}^0\,\delta x\,\delta y\,\delta z
$$
Which you then divide by the volume of $8\,\delta x\, \delta y\,\delta z$ to obtain the average energy put into your block.
A: Here is a derivation in one dimension for squashing by $\Delta x$ a block of material of dimensions $x,y,z$.  The block of material in this direction is like a spring (Force=SpringConstant x Distance).  The potential energy stored in a spring with spring constant k is:
$$
[Joules]=[{{Newtons}\over {meters}}][meters]^2
$$
$$
V={1\over 2}k(\Delta x)^2
$$
$$
{V\over x}={1\over 2}kx({{\Delta x} \over x})^2
$$
$$
{V\over xyz}={1\over 2}({{kx}\over {yz}})({{\Delta x} \over x})^2
$$
$$
U={1\over 2}C_{1111}\epsilon_{11}\epsilon_{11}
$$
$$
[{{Joules}\over {meters^3}}]=[{{Force_x}\over {meters^2}}][Frac][Frac]
$$
The equation
$$
U={1\over 2}C_{ijkl}\epsilon_{ij}\epsilon_{kl}
$$
is covarient (invariant under rotation).  Since it was demonstrated true in one frame, it is therefore true in all frames.
