# Can I determine a transformation matrix for a stress tensor with respect to the axes $O x_1 x_2 x_3$ to define new axes of maximum shear stresses?

Is it possible to determine a transformation matrix for a given stress tensor with respect to the axes $$O x_1 x_2 x_3$$ to define new axes $$O x'_1 x'_2 x'_3$$ of maximum shear stresses?

To make my question more clear, take for example a stress tensor with respect to $$O x_1 x_2 x_3$$ axes: $$\sigma_{ij} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & -6 & -12 \\ 0 & -12 & 1 \end{pmatrix}$$ Now I want to determine a transformation matrix to define new axes $$O x'_1 x'_2 x'_3$$ of maximum shear stresses (with respect to $$O x_1 x_2 x_3$$ axes). Is it possible to determine that?

I was able to determine the transformation matrix to determine principal axes $$O x^*_1 x^*_2 x^*_3$$ (with respect to $$O x_1 x_2 x_3$$ axes) by finding principal stresses $$\sigma_1 = 10$$; $$\sigma_2 = 5$$; $$\sigma_3 = -15$$ and principal directions of the stress tensor. My result was: $$\begin{pmatrix} 0 & 1 & 0 \\ -\frac{3}{5} & 0 & \frac{4}{5} \\ \frac{4}{5} & 0 & \frac{3}{5} \end{pmatrix}$$

I tried to find maximum shear stresses values (which i am not sure, whether they are correct) (=$$12.5$$; $$2.5$$ and $$10$$) but I am not able to determine the transformation matrix nor the stress tensor with respect to axes $$O x_1 x_2 x_3$$, how would I do that?

Thank you.

• Max shear stress is along the three slip planes, which in turn are always 45° from the principal axes. Commented Jan 1, 2020 at 20:17
• But aren't there three maximal shear stresses (one for each plane)? Commented Jan 1, 2020 at 20:34

You calculated the the transformation matrix to the principal axes by means of eigenvalues (the principal stresses) and eigenvectors (which if normalised give the corresponding transformation matrix) correctly.

The corresponding maximum shear stress can be found by rotating the coordinate system by 45° in the plane spanned by the maximum and minimum principle stress, in your case $$\sigma_1$$ and $$\sigma_3$$.

• Thank you. I have two additional questions. My task was to find a coordinate system in which the tangential stresses (plural) will be maximal. Is it correct to assume that tangential stresses and shear stresses are in this case synonyms? And also if the task says stresses not stress will rotating the system by 45 degrees be sufficient? Commented Jan 1, 2020 at 21:02
• @LaurenSin Tangential and shear stresses are synonyms. I assume that in that case they might want you to list the maximum shear for every single shear stress, meaning you have to rotate principal stress system about all possible axis by an angle of 45°. Apply $R_x$, $R_y$ and $R_z$ (en.wikipedia.org/wiki/Rotation_matrix#Basic_rotations) on the principal stress matrix for $\theta = 45°$. Keep in mind that the stress tensor transforms according to $\sigma' = R \cdot \sigma \cdot R^T$ (continuummechanics.org/stressxforms.html).
– 2b-t
Commented Jan 1, 2020 at 21:15
• @LaurenSin In case you need an example to counter-check: continuummechanics.org/principalstressesandstrains.html (Note that the $\tau_{max}$ in the example is rounded to the closest integer, it should be $\tau_{max} = \frac{409}{2} = 204.5$) Additionally WolframAlpha comes in handy: e.g. wolframalpha.com/input/…
– 2b-t
Commented Jan 1, 2020 at 21:20
• @LaurenSin E.g. the example transformation given on the continuum mechanics page: wolframalpha.com/input/… If you change $R$ from $\{0,1,0\}$ (overall maximum shear stress in the plane spanned by $\sigma_1$ and $\sigma_3$ with normal $\sigma_2$) to $\{1,0,0\}$ (normal = $\sigma_1$) and $\{0,0,1\}$ (normal = $\sigma_3$) you find the other two maximum shear stresses which are smaller in amplitude.
– 2b-t
Commented Jan 1, 2020 at 21:41
• @LaurenSin I obtained the same values as you did for your example.
– 2b-t
Commented Jan 1, 2020 at 21:46