By using, transformation of the stress tensor property, the stresses can be calculated for transformed orientations. The properties are unique in case of isotropic in all directions. But what will be the case for which the orthotropic material is transformed? How the transformes stress matrix can be calculated for it?

  • $\begingroup$ If you actually want an answer, see engineering.stackexchange.com/questions/14291/… for an example of a better way to ask your question. $\endgroup$ – Brian Drummond Mar 17 '17 at 11:48
  • $\begingroup$ If the stress tensor is known, its transformation is unrelated to the underlying material properties. $\endgroup$ – agentp May 16 '17 at 19:28

I assume you want to know what the stress tensor given in one cartesian coordinate system looks like in another cartesian coordinate system. If not, please be more specific in your question.

In general, tensor coordinates transform with an orthogonal matrix. So if you have an order 1 tensor (a vector) $\underline{a}$ with coordinates $a_i$ in one cartesian coordinate system, and a transformation matrix $\alpha_{ij}$ to a second cartesian coordinate system, then the coordinates of $\underline{a}$ in the second cartesian coordinate system are $\tilde a_i = \alpha_{ji} a_j$ (assuming Einstein summation convention, which is usually applied in tensor algebra).

For an order 2 tensor (like the stress tensor), every index transforms individually: $\tilde a_{ij} = \alpha_{mi} \alpha_{nj} a_{mn}$.


  1. Tensor coordinates transform with the transposed matrix, while base vectors transform with the original matrix: $\underline{\tilde e}_i = \alpha_{ij} \underline{e}_j$.

  2. Since the transform matrix is orthogonal, its inverse is identical to its transposed. All this can be deduced. See the Wikipedia article on tensors for more details.

  3. The 3D orthogonal matrix that rotates the 1,2 plane (or $x$,$y$ plane) is $\begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$ with rotation angle $\theta$.


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