Strain energy density in index notation The strain energy density is defined as
$$dU = \int_0^{\epsilon_{ij}} \sigma_{ij} d \epsilon_{ij}$$
(see Reddy "Energy Principles and Variational Methods in Applied Mechanics", 2nd Ed, 4.11). Assuming a linear stress-strain relationship, I get 
$$U = \frac12 \sigma_{ij} \epsilon_{ij}$$
 which is consistent with the literature. Now, if I do the summation and assume symmetry of the stress and strain tensors, I get 
$$ U = \frac12 (\sigma_{11} \epsilon_{11} + \sigma_{22} \epsilon_{22} + \sigma_{33} \epsilon_{33} + 2 \sigma_{23} \epsilon_{23} + 2 \sigma_{13} \epsilon_{13} + 2 \sigma_{12} \epsilon_{12})$$
However, in Soedel "Vibration of Shells and Plates", 2nd Ed, 2.6.1, all the "shear terms" (i.e. terms with indices 23, 13, 12) are not multiplied by two. Did I make a mistake, is there some assumption in the Soedel book I might have missed or is this possibly a typo in the book?
Edit: There is still some confusion on my part. For elastic materials the stresses can be derived from the strain energy
$$ \sigma_{ij} =  \frac{\partial U}{\partial \epsilon_{ij} }$$
For orthotropic materials the constitutive relation is
$$ \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{13} \\ \sigma_{12} \end{bmatrix} = 
\begin{bmatrix}
  C_{11} & C_{12} & C_{13} & 0 & 0 & 0 \\
C_{12} & C_{22} & C_{23} & 0 & 0 & 0 \\
C_{13} & C_{23} & C_{33} & 0 & 0 & 0 \\
0 & 0 & 0 & C_{44} & 0 & 0 \\
0 & 0 & 0 & 0 & C_{55} & 0 \\
0 & 0 & 0 & 0 & 0 & C_{66} \end{bmatrix}
\begin{bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{13} \\ \epsilon_{12} \end{bmatrix} $$
So when I try to get $\sigma_{11}$, that works fine
$$ \begin{align}
\sigma_{11} = \frac{\partial U}{\partial \epsilon_{11} } &= \frac12 (\frac{\partial \sigma_{11}}{\partial \epsilon_{11}} \epsilon_{11} + \sigma_{11} + \frac{\partial \sigma_{22}}{\partial \epsilon_{11}} \epsilon_{22} + \frac{\partial \sigma_{33}}{\partial \epsilon_{11}} \epsilon_{33}) \\
 & = \frac12 ( C_{11} \epsilon_{11} + \sigma_{11} + C_{12} \epsilon_{22} + C_{13} \epsilon_{33}) \\
& = \frac12 (\sigma_{11} + \sigma_{11} )
\end{align}
$$
However, for $\sigma_{12}$ I get
$$ \begin{align}
\sigma_{12} = \frac{\partial U}{\partial \epsilon_{12} } &= \frac12 \cdot 2 (\frac{\partial \sigma_{12}}{\partial \epsilon_{12}} \epsilon_{12} + \sigma_{12} ) \\
 & = ( C_{66} \epsilon_{12} + \sigma_{12} ) \\
& = 2 \sigma_{12} 
\end{align}
$$
which is obviously wrong. This works out fine if you leave out the 2s of the strain energy, as Soedel did. Any clarification would be greatly appreciated.
 A: Your original math was correct. The strain energy density should have those factors of two in your original answer, when defined in terms of the tensorial definitions of the shear strains. 
$$
U=\frac{1}{2}\sigma_{ij}\epsilon_{ij}=\frac{1}{2} (
\sigma_{11}\epsilon_{11} + \sigma_{22}\epsilon_{22} +  \sigma_{33}\epsilon_{33} + 2\sigma_{12}\epsilon_{12} + 2\sigma_{13} \epsilon_{13}+
 2\sigma_{23}\epsilon_{23}
)
$$ 
The key is to realize that in switching from tensorial notation:
$$
U=\frac{1}{2}\sigma_{ij}\epsilon_{ij}=\frac{1}{2} \left[ \begin{array}{ccc}
\sigma_{11} & \sigma_{12} & \sigma_{13} \\
\sigma_{21} & \sigma_{22} & \sigma_{23}\\
\sigma_{31} & \sigma_{32} & \sigma_{33} \end{array} \right]\
 \left[ \begin{array}{ccc}
\epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\
\epsilon_{21} & \epsilon_{22} & \epsilon_{23}\\
\epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{array} \right]\
$$
to engineering (i.e. Voigt) notation, one must account for a change in definition of the shear strains. Explicitly, in engineering notation, we write:
$$
\left[ \begin{array} {ccc}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6} \end{array} \right]\ = C_{ij}
\left[ \begin{array} {ccc}
\epsilon_{1} \\
\epsilon_{2} \\
\epsilon_{3} \\
\epsilon_{4} \\
\epsilon_{5} \\
\epsilon_{6} \end{array} \right]\, where \left[ \begin{array} {ccc}
\epsilon_{1} \\
\epsilon_{2} \\
\epsilon_{3} \\
\epsilon_{4} \\
\epsilon_{5} \\
\epsilon_{6} \end{array} \right]\ = \left[ \begin{array} {ccc}
\epsilon_{11} \\
\epsilon_{22} \\
\epsilon_{33} \\
2\epsilon_{23} \\
2\epsilon_{13} \\
2\epsilon_{12} \end{array} \right]\ = \left[ \begin{array} {ccc}
\epsilon_{11} \\
\epsilon_{22} \\
\epsilon_{33} \\
\gamma_{23} \\
\gamma_{13} \\
\gamma_{12} \end{array} \right]\
$$
In engineering notation, we can also write: $U=\frac{1}{2}\sigma_i \epsilon_i$, and this is equivalent to the tensorial definition; you get the same factors of two when you switch from $\epsilon_i$ to $\epsilon_{ij}$ notation. The various defintions of shear strains can get confusing!
A: Notation varies by book. Not only with the stress/strain tensor cross terms, but also with the mass moment of inertia cross terms. Some books have the multiple of 2 and some don't. Some have the terms negative and some positive. 
It all works out in the end, but what differs is the definition of strain. Strain is ratio of displacement over domain, and in some cases it may be defined across the section of an element, and in others only from the neutral axis (middle of element).
So in your question you are actually mixing up two different notations from two different books. I bet if you look carefully at the sketches where they define strain as a function of displacements there are going to be different.
A: You have to remember to express $U$ as
$$ U = \frac12 (\sigma_{11} \epsilon_{11} + \sigma_{22} \epsilon_{22} + \sigma_{33} \epsilon_{33} + \sigma_{23} \epsilon_{23} + \sigma_{32} \epsilon_{32} + \sigma_{13} \epsilon_{13} + \sigma_{31} \epsilon_{31}+  \sigma_{12} \epsilon_{12}+\sigma_{12} \epsilon_{12}) $$
to get the correct expressions for $\sigma_{ij}$.  Even though $\sigma_{ij} = \sigma_{ji}$ in value, that does not mean they are the same object.
