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How can I obtain the below formulas of infinitesimal strain in cylindrical coordinates using matrix calculation given the first formula? I find it hard to study them because I still don't know how to derive them.

$$ \epsilon_{ij}=\frac{1}{2}\left(u\otimes\nabla+\nabla\otimes u\right)\\ \,\\ \begin{align} u\otimes\nabla &=\begin{bmatrix}u_r\\u_{\vartheta}\\u_z\end{bmatrix}\begin{bmatrix}\dfrac{\partial}{\partial r}&\dfrac{1}{r}\dfrac{\partial}{\partial\vartheta}&\dfrac{\partial}{\partial z}\end{bmatrix}\\\\ &=\begin{bmatrix}\dfrac{\partial U_r}{\partial r}&\dfrac{1}{r}\dfrac{\partial U_r}{\partial\vartheta}&\dfrac{\partial U_r}{\partial z}\\\\\dfrac{\partial U_{\vartheta}}{\partial r}&\dfrac{1}{r}\dfrac{\partial U_{\vartheta}}{\partial\vartheta}&\dfrac{\partial U_{\vartheta}}{\partial z}\\\\\dfrac{\partial U_z}{\partial r}&\dfrac{1}{r}\dfrac{\partial U_z}{\partial\vartheta}&\dfrac{\partial U_z}{\partial z}\end{bmatrix}\end{align} $$

Above, I show my try in deriving the first part of the tensor, but I didn't know how to derive the second part.

\begin{align} \varepsilon_{ij} &= \frac{1}{2} (U_{i,j} + U_{j,i})\\ \varepsilon_{rr} & = \cfrac{\partial u_r}{\partial r} \\ \varepsilon_{\theta\theta} & = \cfrac{1}{r}\left(\cfrac{\partial u_\theta}{\partial \theta} + u_r\right) \\ \varepsilon_{zz} & = \cfrac{\partial u_z}{\partial z} \\ \varepsilon_{r\theta} & = \cfrac{1}{2}\left(\cfrac{1}{r}\cfrac{\partial u_r}{\partial \theta} + \cfrac{\partial u_\theta}{\partial r}- \cfrac{u_\theta}{r}\right) \\ \varepsilon_{\theta z} & = \cfrac{1}{2}\left(\cfrac{\partial u_\theta}{\partial z} + \cfrac{1}{r}\cfrac{\partial u_z}{\partial \theta}\right) \\ \varepsilon_{zr} & = \cfrac{1}{2}\left(\cfrac{\partial u_r}{\partial z} + \cfrac{\partial u_z}{\partial r}\right) \end{align}

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  • $\begingroup$ Hello! Please read How do I ask homework questions on Physics Stack Exchange? and edit your question accordingly. Thanks! $\endgroup$ – Jonas Mar 25 at 20:16
  • $\begingroup$ It is not a homework... I just have a well known formula used in all textbooks of continuum mechanics but I'm not finding its derivation..... $\endgroup$ – user134613 Mar 25 at 20:17
  • $\begingroup$ The homework-and-exercises-tag does not only apply to actual homework assignments, but also to homework-like questions. Please show what you have tried so far or if there is a specific step or concept that you are having troubles with. $\endgroup$ – Jonas Mar 25 at 20:22
  • $\begingroup$ I have edited my question........ $\endgroup$ – user134613 Mar 25 at 20:36
  • $\begingroup$ The first line in your formula $\varepsilon_{ij}=(1/2)(U_{i,j}+U_{j,i})= (1/2)(\partial_j U_I+ \partial_j U_)j$ applies only to Cartesian coordinates, unless by the comma you mean the covariant derivative. $\endgroup$ – mike stone Mar 25 at 21:08
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Te strain is not defined by a simple adaptation of the cartesian formula. The essential tool is the Lie derivative.

In a coordinate system with metric $g_{\mu\nu}$ the strain due to an infinitesimal displacement $\eta^\mu$ is $$ e_{\mu\nu} =\frac 12 ( \eta^\alpha \partial_\alpha g_{\mu\nu}+ g_{\mu \alpha}\partial_\nu \eta^\alpha + g_{\alpha \nu}\partial_\nu \eta^{\alpha})\\ \equiv \frac 12 [{\mathcal L}_\eta g]_{\mu\nu} $$ when $g_{\mu\nu}=\delta_{\mu\nu}$ this reduces to the Cartsian expression.

You can also write $$ [{\mathcal L}_\eta g]_{\mu\nu}= \nabla_\mu \eta_\nu+ \nabla_\nu \eta_\mu $$ where the $\nabla_\mu$ are the covariant derivatives in your chosen coordinates.

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  • $\begingroup$ I haven't mentioned anything about Cartesian coordinates. I just like to know the derivation of it in cylindrical coordinates $\endgroup$ – user134613 Mar 25 at 20:41
  • $\begingroup$ In other words, the mentioned formulas in my question, I'm sorry but you have confused me more $\endgroup$ – user134613 Mar 25 at 20:43
  • $\begingroup$ The first line in your formula $\varepsilon_{ij}=(1/2)(U_{i,j}+U_{j,i})= (1/2)(\partial_j U_I+ \partial_j U_j)$ applies only to Cartesian coordinates, unless by the comma you mean the covariant derivative. Try Googling strain and Lie derivative $\endgroup$ – mike stone Mar 25 at 21:09
  • $\begingroup$ Here is a pedagogical account of the derivation for spherical polars: Linear strain tensor and differential geometry E. de Prunelé, doi.org/10.1119/1.2750376 $\endgroup$ – mike stone Mar 25 at 21:31
  • $\begingroup$ I thought my question was simple. Now it is easier for me to just study them as they are or just wait for someone to continue my derivation $\endgroup$ – user134613 Mar 25 at 21:43

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