# Relation between Stress/Strain and normal derivative of displacement

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 $$t$$.

• Why do you need the normal derivative of the displacements? – nicoguaro May 29 '19 at 14:10
• 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. – nicoguaro May 29 '19 at 16:19
• @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? – user29088 May 30 '19 at 9:56
• For elasticity the Neumann boundary conditions are given by $C\nabla_n u$, i.e., the normal projection of the stress tensor. – nicoguaro May 30 '19 at 13:04
• 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 $\varepsilon$? – user29088 May 30 '19 at 17:10

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

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

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 stiffness 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 $$S_{ijkl}C_{klrs}=\frac{1}{2}(\delta_{ir}\delta_{js} + \delta_{is}\delta_{jr})$$.

• 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 $n$? – user29088 May 31 '19 at 14:39
• If you multiply by the inverse of $C$ you would recover the symmetric part of the gradient. – nicoguaro May 31 '19 at 15:14
• 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? – user29088 May 31 '19 at 18:54
• @Kerem, well, I have never thought it that way. But, I would say that $C \omega = 0 \, \Leftrightarrow \, C \nabla u = C \varepsilon \equiv \sigma$. – nicoguaro May 31 '19 at 18:56