# How to derive Infinitesimal Strain Tensor in Cylindrical Coordinates [closed]

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}

• Hello! Please read How do I ask homework questions on Physics Stack Exchange? and edit your question accordingly. Thanks! – Jonas Mar 25 at 20:16
• 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..... – user134613 Mar 25 at 20:17
• 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. – Jonas Mar 25 at 20:22
• I have edited my question........ – user134613 Mar 25 at 20:36
• 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. – mike stone Mar 25 at 21:08

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.
• 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 – mike stone Mar 25 at 21:09