Hey Bob and Rene,

First of all, thanks Bob for your reply and the generalization! It helped me quite a bit

I have worked with Cedric (big thanks to him!) on a test example to see whether the 2 should or shouldn’t be there.

Let’s start with a 3d system in 2d simple shear:

\sigma = \begin{pmatrix}0 & \tilde\sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}

Since we can choose any u and v, let’s start with

u = e_x = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \text{ and }
v = e_y = \begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}

Given these vectors we expect that we need to rotate u and v 45 degrees, because of the state of stress. The tilde 2 indicates that the 2 might be there or not.

\tan(2\alpha) = \frac{\tilde 2 u^T \sigma v}{u^T \sigma u - v^T \sigma v} \\
= \frac{\tilde 2 \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}^T \begin{pmatrix}0 & \tilde \sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}}{\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}^T \begin{pmatrix}0 & \tilde\sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} - \begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}^T \begin{pmatrix}0 & \tilde\sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}} \\
= \frac{\tilde 2 \sigma }{0 - 0} \\
= \inf \\
\alpha = 0.5*\arctan(\inf) = \pi/4 = 45 \deg

Which is what we expect for this case, but doesn’t test whether the factor 2 is correct. Now let’s take two slightly different vectors:

u = \begin{pmatrix}\frac{1}{2} \\ 1 \\ 0 \end{pmatrix} \text{ and }
v = \begin{pmatrix}-1 \\ \frac{1}{2} \\ 0 \end{pmatrix}

The angle between these vectors and the axes are (between v and e_x)

\theta = \arctan(\frac{1}{\frac{1}{2}}) = \arctan(2) \approx 63.4349 \deg \\
\alpha = 45-\theta = -18.4349 \deg

So we expect alpha to be 18.4349 degrees.

\tan(2\alpha) = \frac{\tilde 2 u^T \sigma v}{u^T \sigma u - v^T \sigma v} \\
= \frac{\tilde 2 \begin{pmatrix}\frac{1}{2} \\ 1 \\ 0 \end{pmatrix}^T \begin{pmatrix}0 & \tilde\sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \begin{pmatrix}-1 \\ \frac{1}{2} \\ 0 \end{pmatrix} }{\begin{pmatrix}\frac{1}{2} \\ 1 \\ 0 \end{pmatrix}^T \begin{pmatrix}0 & \tilde\sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \begin{pmatrix}\frac{1}{2} \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix}-1 \\ \frac{1}{2} \\ 0 \end{pmatrix} ^T \begin{pmatrix}0 & \tilde\sigma & 0 \\
\tilde\sigma & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \begin{pmatrix}-1 \\ \frac{1}{2} \\ 0 \end{pmatrix} } \\
= \frac{\tilde 2 \frac{3}{4}\sigma}{\sigma - \sigma} \\
= \frac{\tilde 2 \frac{3}{4}\sigma}{2 \sigma} \\
= \tilde 2 \frac{3}{8} \\
\alpha_1 = 0.5\arctan(-\frac{3}{8}) = -10,2750 \\
\alpha_2 = 0.5\arctan(-2\frac{3}{8}) = -18.4349

This shows that we need the 2 to be there.