Transforming inertia tensor from corner of a beam to the centre So I have a beam with mass M and sides a (x-direction) ,b (y-direction) and c (z-direction). I have figured out that the inertia tensor in the corner is 
$$I = M\begin{bmatrix} \frac{b^2 + c^2}{3} & -\frac{ab}{4} & -\frac{ac}{4}\\ -\frac{ab}{4} & \frac{a^2+c^2}{3} & -\frac{bc}{4} \\ -\frac{ac}{4} & -\frac{bc}{4} & \frac{a^2+b^2}{3}\end{bmatrix}$$
and in the middle a nice diagonal matrix 
$$I = \frac{M}{12}\begin{bmatrix} b^2+c^2 & 0 & 0\\ 0 & a^2 + c^2 & 0\\ 0 & 0 & a^2 + b^2 \end{bmatrix}$$
and I know the vector you should use to transform from the corner to the middle is
$$ R = \begin{bmatrix} \frac{a}{2} \\ \frac{b}{2} \\ \frac{c}{2} \end{bmatrix}$$
but I don't really get how I would for example transform the $I_{zz}$ component of the corner to the middle with the transform formula $I_a = I_b + MR^2$

The 3D form of the parallel axis theorem is defined as follows. 

$$ \mathbf{I}_A = \mathbf{I}_C + m \left( -[\boldsymbol{c} \times][\boldsymbol{c} \times] \right) $$
where $[\boldsymbol{c}\times] = \pmatrix{0 & -z & y\\z & 0 & -x\\-y & x & 0}$ is the 3×3 cross product operator matrix. Combined the above expression is
$$ \mathbf{I}_A = \mathbf{I}_C + m \begin{vmatrix} y^2+z^2 & -x y & -x z \\ -x y & x^2 + z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{vmatrix} $$
Do you see the pattern in the above? The diagonal elements are the sum of squares of the vector components excluding the row of the diagonal (1st row has 2nd and 3rd components, etc.) and the non diagonals contains the negative product of the vector components of the row and column (the 2nd row, 3rd column contains $-y z$).
Using the above you can solve for $\mathbf{I}_C$ at the center given the mass moment of inertia tensor at the corner $\mathbf{I}_A$ and the position vector of the center $\boldsymbol{c} = \pmatrix{ a/2 \\ b/2 \\ c/2 }$.
