My strength of materials book provides the following equation to determine the x-component of the normal stress acting on all possible planes of a two dimensional element.

$$ \sigma_{x^{\prime}} = \frac{1}{2}(\sigma_x + \sigma_y ) + \frac{1}{2}(\sigma_x - \sigma_y )\cos(2\theta) + \tau_{xy}\sin(2\theta)$$

My question is -- in what orientation are the non-primed variables assumed? My instinct is that the initial orientation that the non-primed quantities are given in is irrelevant, only the numerical value of the components matter and the angle from which you measure the new components (primed quantities).

For example, if you are given a 2-D element that is initially inclined at 60 degrees and you have the values of $\sigma_x$, $\sigma_y$, and $\tau_{xy}$ in that orientation and you want to find $\sigma_{x^{\prime}}$ at the 0 degree orientation you would compute the equation above using a value of $\theta = -60$ because you have to measure relative to where you "start".

Is that correct or do I have this completely confused?


1 Answer 1


The transformation relation for 2D stresses in matrix form is $$ \boldsymbol{\sigma}' = \mathbf{Q}^T\, \boldsymbol{\sigma}\, \mathbf{Q} $$ where $$ \boldsymbol{\sigma} = \begin{bmatrix} \sigma_{xx} & \tau_{xy} \\ \tau_{xy} & \sigma_{yy} \end{bmatrix} ~,~~ \mathbf{Q} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} $$ In writing the stresses in matrix form, we assume that the stress components have been expressed in terms of a coordinate system with base vectors ($\mathbf{e}_x, \mathbf{e}_y$) and that the angle $\theta$ is measured counter-clockwise with respect to the $\mathbf{e}_x$ vector.

Therefore, the angle $\theta$ is always zero when you are looking along the $\mathbf{e}_x$ vector, by definition.

If you work out the algebra, you will find that (assuming I've made no mistakes) $$ \begin{bmatrix} \sigma_{xx}' \\ \sigma_{yy}' \\ \tau_{xy}' \end{bmatrix} = \begin{bmatrix} \cos^2\theta & \sin^2\theta & \sin2\theta \\ \sin^2\theta & \cos^2\theta & -\sin2\theta \\ -\tfrac{1}{2}\sin 2\theta & \tfrac{1}{2}\sin 2\theta & \cos 2\theta \\ \end{bmatrix} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \tau_{xy} \end{bmatrix} $$

The inverse relation is $$ \mathbf{Q} \,\boldsymbol{\sigma}' \, \mathbf{Q}^T= \boldsymbol{\sigma} $$ In expanded form, $$ \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \tau_{xy} \end{bmatrix} = \begin{bmatrix} \cos^2\theta & -\sin^2\theta & -\sin2\theta \\ \sin^2\theta & \cos^2\theta & \sin2\theta \\ \tfrac{1}{2}\sin 2\theta & -\tfrac{1}{2}\sin 2\theta & \cos 2\theta \\ \end{bmatrix} \begin{bmatrix} \sigma_{xx}' \\ \sigma_{yy}' \\ \tau_{xy}' \end{bmatrix} $$ If the initial stress components are with respect to a set of initial axes that are not aligned with ($\mathbf{e}_x, \mathbf{e}_y$) and you want to compute the components along $\mathbf{e}_x$ etc., you will have to use the inverse relationship.

  • $\begingroup$ Thanks for the response. The bit about how we define it really cleared it up for me. Another way to do it I realized is you could solve the 2-D transformation equations for the stresses at the 0 degree orientation (3 equations and 3 unknowns). I suppose you could try to do it visually using Mohrs Circle as well. Regarding your orthogonal transformation matrix -- I'm curious, is it the same as the 3-D version (9 elements, 3 by 3) just reduced by the third row and third column (but exact same elements)? $\endgroup$
    – Nukesub
    Commented May 12, 2018 at 13:18
  • $\begingroup$ I have spent a bit of time on this and I am still unclear. In almost any resource I find, it never addresses how to solve these types of problems for elements ALREADY rotated. I am confused because I can't get the correct answer using Mohr's circle (solving graphically). I AM able to get the right answer using the transformation equations or the matrix form you show above. $\endgroup$
    – Nukesub
    Commented May 12, 2018 at 17:01
  • $\begingroup$ The 3D version of the rotation matrix will have 0 off-diagonal elements in the third row/column and 1 in the third diagonal element (for rotation around the $z$-axis). The stress matrix will have zeros in the corresponding elements for your problem. For general 3D rotation things get more complex. $\endgroup$ Commented May 12, 2018 at 21:06
  • $\begingroup$ For a deeper understanding of Mohr's circles see mech.utah.edu/~brannon/public/Mohrs_Circle.pdf (I'd start with the appendix). For maximum benefit, this write-up needs to be read carefully and the examples worked out. Once you have mastered the ideas you should be able to apply them to your case. $\endgroup$ Commented May 12, 2018 at 21:09
  • $\begingroup$ Thanks for the resource. I'll do what you say and see if I can make sense of applying Mohr's circle to already rotated element. $\endgroup$
    – Nukesub
    Commented May 13, 2018 at 15:14

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