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I calculated the stress $\sigma$ and strain $\varepsilon$ for a solid plate with dirichlet boundary conditions $u = g$, where $u$ is the displacement. With these I want to calculate $\nabla_n u = t$ on the boundary where $\nabla_n$ describes the normal derivative. Is there any easy way to do this?

I have the displacement $u$ but since it is a vector with real numbers I cannot calculate the normal derivative.

Edit: With some research I figured out that $\sigma = C\varepsilon = C \nabla u$ and with that $\nabla_n u = C^{-1}\sigma \cdot n$. My question is, how can I calculate a map between $u = g$ and $\nabla_n u$ which gives me $t$.

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  • $\begingroup$ Why do you need the normal derivative of the displacements? $\endgroup$ – nicoguaro May 29 at 14:10
  • $\begingroup$ Also, I don't think that your expression is correct. Due to the major symmetry in the $C$ tensor you are only getting the symmetric part of the gradient and projecting it. $\endgroup$ – nicoguaro May 29 at 16:19
  • $\begingroup$ @nicoguaro thank you for your answer. I am trying to calculate the so called Dirichlet-to-Neumann map. If my expression is incorrect, how could I define a correct way for a map which gives me the normal derivative of the displacement on the dirichlet boundary? $\endgroup$ – Kerem May 30 at 9:56
  • $\begingroup$ For elasticity the Neumann boundary conditions are given by $C\nabla_n u$, i.e., the normal projection of the stress tensor. $\endgroup$ – nicoguaro May 30 at 13:04
  • $\begingroup$ Thank you @nicoguaro. Where can I find the derivation of your expression? And is there a way to express $\nabla_n u$ with $\sigma$ or $\varepsilon$? $\endgroup$ – Kerem May 30 at 17:10
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Short answer

You can calculate the normal derivative of a vector, it is given by the projection of the gradient of the vector (a second order tensor) in the normal direction. In cartesian coordinates, this is i.e.,

$$ \nabla_n u = \nabla u \cdot n = \left[\begin{matrix}\frac{\partial}{\partial x} u_x & \frac{\partial}{\partial y} u_x & \frac{\partial}{\partial z} u_x\\ \frac{\partial}{\partial x} u_y & \frac{\partial}{\partial y} u_y & \frac{\partial}{\partial z} u_y\\ \frac{\partial}{\partial x} u_z & \frac{\partial}{\partial y} u_z & \frac{\partial}{\partial z} u_z\end{matrix}\right] \left[\begin{matrix}n_{x}\\n_{y}\\n_{z}\end{matrix}\right] = \left[\begin{matrix}n_{x} \frac{\partial}{\partial x} u_x + n_{y} \frac{\partial}{\partial y} u_x + n_{z} \frac{\partial}{\partial z} u_x\\n_{x} \frac{\partial}{\partial x} u_y + n_{y} \frac{\partial}{\partial y} u_y + n_{z} \frac{\partial}{\partial z} u_y\\n_{x} \frac{\partial}{\partial x} u_z + n_{y} \frac{\partial}{\partial y} u_z + n_{z} \frac{\partial}{\partial z} u_z\end{matrix}\right] $$

Long answer

In linear elasticity, the constitutive relation (Hooke's law) relates the strain (a second order tensor) to the stress (a second order tensor) by the stiffness tensor (a fourth order tensor), i.e.

$$\sigma = C \varepsilon\, ,$$

or, in index notation,

$$\sigma_{ij} = C_{ijkl} \varepsilon_{kl}\, ,$$

where summation over repeated indices is implied. The stiffness tensor has the following symmetries $C_{ijkl} = C_{ijlk} = C_{jikl} = C_{klij}$.

Furthermore, the strain tensor is given by the symmetric part of the gradient of the displacement vector:

$$\varepsilon = \frac{1}{2}\left[\nabla u + (\nabla u)^T\right]\, ,$$

or, in index notation,

$$\varepsilon = \frac{1}{2}(u_{i,j} + u_{j, i})\, ,$$

where the comma represents a spatial derivative.

Thus, due to symmetries in the stiffnes tensor we can say that

$$\sigma = C \varepsilon = C\nabla u\, .$$

But, that does not imply that we can invert the relation to directly obtain the gradient of $u$ from $\sigma$.

This is why:

$$\sigma = C\nabla u = C \varepsilon + \underbrace{C\omega}_{=0} \, ,$$

thus,

$$S\sigma = SC \varepsilon = \varepsilon \, ,$$

where $S$ is the compliance tensor, and $S_{ijkl}C_{klrs}=\frac{1}{2}(\delta_{ir}\delta_{js} + \delta_{is}\delta_{jr})$.

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  • $\begingroup$ Thank you for the detailed answer. So what is the problem now, since it is an equality, to take the inverse of $C$ and multiply the whole equation with the normal direction $n$? $\endgroup$ – Kerem May 31 at 14:39
  • $\begingroup$ If you multiply by the inverse of $C$ you would recover the symmetric part of the gradient. $\endgroup$ – nicoguaro May 31 at 15:14
  • $\begingroup$ Thank you, especially for your patience. But I have a big problem (as a mathematican) to understand "would recover the symmetric part of the gradient". I hope I understand correctly that $C \omega = 0 \, \Leftrightarrow \, \sigma = C \nabla u = C \varepsilon$. Is that correct? $\endgroup$ – Kerem May 31 at 18:54
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    $\begingroup$ @Kerem, well, I have never thought it that way. But, I would say that $C \omega = 0 \, \Leftrightarrow \, C \nabla u = C \varepsilon \equiv \sigma$. $\endgroup$ – nicoguaro May 31 at 18:56

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