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I know $E_{xx}$, $E_{yx}$, and $E_{xy}$ but I can't figure out how to find the value for $E_{yy}$. I think it has something to do with the phrase 'homogenous $2D$ strain' as a big hint since there's no trigonometry or geometry method I know of to find two missing sides of a triangle given a single side and no angles. I'm only concerned with the first portion of the problem; I've attached a photo along with a diagram I've made of the plot before and after the earthquake. Original Problem

I would very much appreciate a nudge in the right direction, thank you!

Diagrams

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  • $\begingroup$ For 2D problems it's common to neglect the out of plane components of strain. Now, depending on the approximation made (plane stress/strain) you can know the value for $E_{zz}$. $\endgroup$
    – nicoguaro
    Commented Feb 16, 2020 at 17:53
  • $\begingroup$ I just realized that you might have a typo in your question since it does not match with the title. $\endgroup$
    – nicoguaro
    Commented Feb 16, 2020 at 17:58
  • $\begingroup$ Yes @nicoguaro you are correct, I need Eyy. Sorry! $\endgroup$
    – dogman
    Commented Feb 16, 2020 at 18:02

3 Answers 3

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As you mentioned, the word "homogenous" does gives information about what you want. In this context, homogeneous means that the tensor is the same for all the points.

So, if you compute the strain at the right or left of your shape you would get the same value. In you case it would suffice you use the change in length on the left side to compute $E_{yy}$.

Update

Geometrically, we can say that the components are the following:

  • $E_{xx}$ is the change in the horizontal line divided by the original length of this side. Thus, it would be an elongation.

  • $E_{yy}$ is the change in the vertical line divided by the original length of the side. And it would be a contraction.

  • $E_{xy}$ is half the angular change between two originally orthogonal lines. In your case, the change in the angle.

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  • $\begingroup$ In that case, would the strain be 1/182.482 since the change in length for Exx and Eyx were both 1 over their original values as well? $\endgroup$
    – dogman
    Commented Feb 16, 2020 at 18:14
  • $\begingroup$ I think that it would be something like $1 - 149/150$. $\endgroup$
    – nicoguaro
    Commented Feb 16, 2020 at 18:18
  • $\begingroup$ How did you derive that? Wouldn't the change in Exx have just as much of an effect on Eyy but Eyx and Exy are the only ones that can relate to Eyy in your scenario $\endgroup$
    – dogman
    Commented Feb 16, 2020 at 21:06
  • $\begingroup$ Sorry, it would be negative because it represents a contraction: $149/150 - 1$. The components are evaluated independently. I would suggest you to check the section "Geometric derivation" on en.wikipedia.org/wiki/Infinitesimal_strain_theory?wprov=sfla1 $\endgroup$
    – nicoguaro
    Commented Feb 16, 2020 at 21:11
  • $\begingroup$ Makes sense; so the just for clarity Exx = 1/100, Eyy = (149/150-1), Eyx = 1/149 and Exy = -1/149? Or is that one positive? $\endgroup$
    – dogman
    Commented Feb 16, 2020 at 21:58
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The $\epsilon_{yy}$ component of the strain is simply the change in length of the originally vertical left boundary divided by its original length.

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For small strains, the strain in an arbitrary direction is given by $$\epsilon=E_{xx}\cos^2{\theta}+E_{xy}\sin{2\theta}+E_{yy}\sin^2{\theta}$$ where $\theta$ is the original angle that a material line makes with the x axis. From the given data, for $\theta=0$, $\epsilon=E_{xx}=(101-100)/100=0.01$ and for $\theta=90\ degrees$, $\epsilon=E_{yy}=(149-150)/150=-0.00667$. In addition, we can show for the angle going to 89 degrees, $E_{xy}=(\cos{89})/2=0.00875$.

The principal directions are given by the equation $$\tan{2\theta}=\frac{2E_{xy}}{(E_{xx}-E_{yy})}$$ and the principal strains in these directions are given by $$\epsilon=\frac{(E_{xx}+E_{yy})}{2}\pm \sqrt{\left(\frac{E_{xx}-E_{yy}}{2}\right)^2+E_{xy}^2}$$

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