Given:
$$ \sigma_{ij}= \left[ {\begin{array}{cc} -20 & 60 \\ 60 & 90 \\ \end{array} } \right],\quad i,j=x,y $$
I want to find the principle stress tensor $\sigma_{ij}^{pr}$. Using the Mohr's Cirlce, I get:
$$\sigma_{max}=116.39,\sigma_{min}=-46.39$$
the points where the circle intersects with the $x(\sigma_{xx},\sigma_{yy})$ axis.
From there, how do these points make up the principle stress tensor?
Is
$$ \sigma_{ij}^{pr}= \left[ {\begin{array}{cc} 116.39 & 0 \\ 0 & -46.39 \\ \end{array} } \right],\quad i,j=x,y $$ correct?