Recently I have obtained a csv value form and experiment I have computed. However, I was trying to understand how each individual component being calculated. The image attached shows sample point deformation with time as each row represent a time step (I am not sure if I can upload the csv file here as there is no upload option). I have managed to calculate the deformations (U,V,W) which is for example U = Xi - Xo. However, I am trying to calculate the lagrangian strains using the equations provided by the software it is not giving me similar results. I would appreciate if someone can show me how the strains are being computed at each time steps using the provided data.

Deformation Gradient Data

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    $\begingroup$ The strains would depend on the distribution of points and the interpolation used. The latter depends on the method. Were the data obtained using a finite element or finite difference method? $\endgroup$
    – nicoguaro
    Oct 25, 2022 at 12:09
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    $\begingroup$ If each line corresponds to a time step, then you track a single point from which you cannot compute strains. Or do you have further tables for more points? Or does the program process simulated/measured surrounding points not listed in the table? $\endgroup$ Oct 27, 2022 at 9:08
  • $\begingroup$ @DominikKern Yes, this is just a point in a larger dataset. The csv file that the data came from is too big (0.5 GB) so I am not sure if I can share that. The reason why I want to know how these being computed is that I want to compute the strain in the z direction (Ezz). Would this be possible for the dataset given that they all have the location (X,Y,Z) at each time step. $\endgroup$
    – Abdullah
    Nov 2, 2022 at 3:10

1 Answer 1


If you have $x$,$y$,$z$-coordinates and displacements $u$,$v$,$w$ for a set of points, then you could approximate the displacement gradient by finite differences (or possibly other interpolations). So for derivatives with respect to $x$, you would sort the points with respect to the $x$-coordinate, such that they are neighbors in this direction and then compute ($n$ corresponds to spatial position on a row in $x$-direction) \begin{align} \frac{\text{d}u}{\text{d}x} &\approx \frac{u_{n}-u_{n-1}}{x_{n}-x_{n-1}} ,\\ \frac{\text{d}v}{\text{d}x} &\approx \frac{v_{n}-v_{n-1}}{x_{n}-x_{n-1}} ,\\ \frac{\text{d}w}{\text{d}x} &\approx \frac{w_{n}-w_{n-1}}{x_{n}-x_{n-1}}. \end{align} Then sort the rows with respect to $y$, repeat the steps for the derivatives with respect to $y$.

If you want to calculate $\varepsilon_{zz}$ then do the same for $z$, required that there are same neighbored points with offset $z_{n}-z_{n-1}\ne 0$, i.e. the points are not complanar, which may be, since the shown formulas are for two dimensions. In 3D you would have \begin{align} \varepsilon_{xx} &= \frac{\text{d}u}{\text{d}x} +\frac{1}{2}\left( \left(\frac{\text{d}u}{\text{d}x}\right)^2 + \left(\frac{\text{d}v}{\text{d}x}\right)^2 + \left(\frac{\text{d}w}{\text{d}x}\right)^2 \right)\\ \dots& \\ \varepsilon_{zz}&= \frac{\text{d}w}{\text{d}z} +\frac{1}{2}\left( \left(\frac{\text{d}u}{\text{d}z}\right)^2 + \left(\frac{\text{d}v}{\text{d}z}\right)^2 + \left(\frac{\text{d}w}{\text{d}z}\right)^2 \right) \end{align} Note that with geometric linearization you neglect the quadratic terms in the strains, but if they are relevant for your problem, account for them.


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