Stiffness matrix of an orthotropic transversely isotropic material I am studying the generalized Hooke's law for an orthotropic transversely isotropic material (the same behaviour along the directions $x_2=y$ and $x_3=z$).
The general elastic law by Hooke states that
$$\sigma_{ij}=\sum_{k,l}E_{ijkl}\epsilon_{kl}$$
For an orthotropic material the law reduces to
$${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{12}\\\sigma _{13}\\\sigma _{23}\end{bmatrix}}\,=\,{\begin{bmatrix}E_{1111}&E_{1122}&E_{1133}&0&0&0\\E_{2211}&E_{2222}&E_{2233}&0&0&0\\E_{3311}&E_{3322}&E_{3333}&0&0&0\\0&0&0&E_{1212}&0&0\\0&0&0&0&E_{1313}&0\\0&0&0&0&0&E_{2323}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\\varepsilon _{12}\\\varepsilon _{13}\\\varepsilon _{23}\end{bmatrix}}}$$
The transverse isotropy means the same behaviour along directions 2 and 3, so the index 2 and 3 can be exchanged in the stiffness matrix and nothing changes.
$$E_{1122}=E_{1133},\,E_{2222}=E_{3333},\, E_{1212}=E_{1313},$$
Additionally, I have read that $$E_{2323}=\frac{E_{2222}-E_{2233}}{2}$$
I can't understand how to obtain the last formula, can you help me?
 A: What is $E_{2323}$? It's the amount of shear stress $\sigma_{23}$ we have to apply (in the 2–3 plane) to obtain a certain amount of shear strain $\varepsilon_{23}$ in the same plane:

$~~~~~~~~$
But Nature doesn't know or care how we chose to orient our axes; we just as well could have obtained the same stress state by applying a tensile stress $\sigma_{2^\prime2^\prime}$ and a symmetric compressive stress $\sigma_{3^\prime3^\prime}=-\sigma_{2^\prime2^\prime}$ in the 2'–3' plane:

(Exercise: Show using Mohr's circle that for the stress states to indeed be equivalent, $\sigma_{2^\prime2^\prime}$ and $\sigma_{3^\prime3^\prime}$ must each have a magnitude equal to $\sigma_{23}$: $\sigma_{2^\prime2^\prime}=|\sigma_{3^\prime3^\prime}|=\sigma_{23}$.)
Symmetry considerations (corresponding to the zeros in the stiffness tensor) reassure us that these replacement normal stresses don't produce any shear strains that would further distort our infinitesimal element. All they do is individually each produce a normal strain
$$\varepsilon_{2^\prime2^\prime}=\frac{1}{E}\sigma_{2^\prime2^\prime};~~~\varepsilon_{3^\prime3^\prime}=\frac{1}{E}\sigma_{3^\prime3^\prime}$$
and individually each produce a lateral strain
$$\varepsilon_{3^\prime3^\prime}=-\frac{\nu}{E}\sigma_{2^\prime2^\prime};~~~\varepsilon_{2^\prime2^\prime}=-\frac{\nu}{E}\sigma_{3^\prime3^\prime},$$
where $\nu$ is the relevant Poisson ratio.
Now consider the changes in length of some internal features:

Along line bc, based on the equations linking normal stress and normal strain, we expect an elongational strain from the tensile stress $\sigma_{2^\prime2^\prime}$  and another one (a Poisson-mediated one) from the compressive stress $\sigma_{3^\prime3^\prime}$ of the same magnitude:
$$\varepsilon_{2^\prime2^\prime}=\frac{1}{E}\sigma_{2^\prime2^\prime}-\frac{\nu}{E}\sigma_{3^\prime3^\prime}=\frac{1+\nu}{E}\sigma_{2^\prime2^\prime}.$$
For line ad, we expect an contractile strain from $\sigma_{3^\prime3^\prime}$ and another one (a Poisson-mediated one) from $\sigma_{2^\prime2^\prime}$:
$$\varepsilon_{3^\prime3^\prime}=\frac{1}{E}\sigma_{3^\prime3^\prime}-\frac{\nu}{E}\sigma_{2^\prime2^\prime}=-\frac{1+\nu}{E}\sigma_{2^\prime2^\prime}=-\varepsilon_{2^\prime2^\prime}.$$
Now consider the angle $\theta$, which was initially 45° or $\frac{\pi}{4}$ radians. It can be expressed as $$\tan^{-1}\frac{|ad|}{|bc|}=\tan^{-1}\frac{1-\varepsilon_{2^\prime2^\prime}}{1+\varepsilon_{2^\prime2^\prime}}.$$ For small $\varepsilon_{2^\prime2^\prime}$, this is approximately $\frac{\pi}{4}-\varepsilon_{2^\prime2^\prime}$. (Exercise: prove this using a Taylor series expansion.)
The shear strain $\varepsilon_{23}$ is defined as reduction in a corner angle in the 2–3 plane in radians. But we've just shown that this reduction is $2\theta=2\varepsilon_{2^\prime2^\prime}$.
Now let's put it all together. From the matrix you wrote,
$$\sigma_{23}=E_{2323}\varepsilon_{23};$$
$$\require{cancel}\sigma_{2^\prime2^\prime}=\cancelto{0}{E_{2^\prime2^\prime11}\varepsilon_{11}}+E_{2^\prime2^\prime2^\prime2^\prime}\varepsilon_{2^\prime2^\prime}+E_{2^\prime2^\prime3^\prime3^\prime}\varepsilon_{3^\prime3^\prime}.$$
(Note that the $E_{2^\prime2^\prime11}\varepsilon_{11}$ term is zero because there's no strain in the 1 direction, as the lateral or Poisson strains from $\sigma_{2^\prime2^\prime}$ and $\sigma_{3^\prime3^\prime}$ cancel out.)
We've already shown that $\sigma_{23}=\sigma_{2^\prime2^\prime}$, that $\varepsilon_{23}=2\varepsilon_{2^\prime2^\prime}$, and that $\varepsilon_{3^\prime3^\prime}=-\varepsilon_{2^\prime2^\prime}.$ Therefore,
$$E_{2323}=\frac{E_{2222}-E_{2233}}{2},$$
QED.
