$$ \sigma_{ij}= \left[ {\begin{array}{cc} -20 & 60 \\ 60 & 90 \\ \end{array} } \right],\quad i,j=x,y $$

I want to find the principle stress tensor $\sigma_{ij}^{pr}$. Using the Mohr's Cirlce, I get:


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?


$$ \sigma_{ij}^{pr}= \left[ {\begin{array}{cc} 116.39 & 0 \\ 0 & -46.39 \\ \end{array} } \right],\quad i,j=x,y $$ correct?


The values you determined for $\sigma_{\mathit{max}}$ and $\sigma_\mathit{min}$ are correct. However, your answer is only partially correct.

The value of $\sigma^\mathrm{pr}$ has to be expressed in the principal coordinate system. You can calculate the angle of the principal system to your original coordinate system using: $$ tan(2\varphi) = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2 * 60}{-20-90} $$ $$ \varphi_1 \approx 23.74° $$

So yes the values are correct, but your coordinate system $x^\mathrm{pr}, y^\mathrm{pr}$ is rotated by $\varphi_1$ degrees, compared to your original coordinate system $x, y$.

  • $\begingroup$ So, the final answer is the matrix $\sigma_{ij}^{pr}$ I found plus the angle of rotation $\varphi_1$ ? $\endgroup$ – Andrew Jun 16 '18 at 12:07
  • $\begingroup$ Yes, and call the axes something different. For example $x^\mathrm{pr}, y^\mathrm{pr}$. $\endgroup$ – bgro Jun 16 '18 at 12:39

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