# Tensorial version of Hooke's law

It is well known that

$${\boldsymbol F} = k {\boldsymbol x}$$

for isotropic media. Also, according to Wikipedia

$$F_k = k_{jk} x_j$$ for some elastic tensor $k_{jk}$. I'm a bit confused as to how to relate the stiffness matrix $C_{ijkl}$ from

$$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$

to the 3x3 form $k_{jk}$. Ie if $x_j$ and $C_{ijkl}$ are known, how does one arrive at $F_k$?

• possibly helpful: gregegan.customer.netspace.net.au/SCIENCE/Rindler/… – user4552 Oct 13 '14 at 2:52
• Is there a particular example you are working with? The answer is geometry dependent. – John Alexiou Oct 15 '14 at 1:38
• Did you read Wikipedia on isotropic materials fully? Also look at en.wikipedia.org/wiki/Linear_elasticity – John Alexiou Oct 15 '14 at 1:45
• I have a anisotropic (hexagonally symmetric) elastic tensor $C_{ijkl}$. I am displacing the mesh by 1 nm in the z direction and am having trouble coming up with the force that this corresponds to. I don't want the stresses, I want the force vector. – John M Oct 15 '14 at 1:52
• Unfortunately, there is no easy answer. You can start with this page although you probably have read it already. If you are trying to relate elastic properties to spring constants, you will have a tough time. I just wrote my thesis on these very issues. You will find this paper helpful to show you how there is more than 1 way to get spring constants just from $E, G, \nu$, let alone the full $C$. And it is geometry dependent. – tpg2114 Oct 15 '14 at 1:59

If you're willing to make some approximations, here's a way to derive it.

Remember the relationship between the traction vector $$T$$ and the stress tensor $$\sigma$$:

$$T_{j} = n_{i}\sigma_{ij}$$

where $$n$$ is the unit normal vector of a differential area element being analyzed.

If we take the cross-sectional area $$A_0$$ of the object in question to not change under application of force (only true for very small deformations), and that the tractions are approximately uniform over the surface of application, we can multiply by this constant to get a physically relevant force on the left-hand side:

$$F_{j} = A_0n_{i}\sigma_{ij}$$

Inserting the linear stress-strain relationship involving the stiffness tensor $$C$$, we find:

$$F_{j} = A_0n_{i}C_{ijkl}\epsilon_{kl}$$

If we also approximate that the displacements are approximately linear, that is, that $$u_{i,j} \approx \frac{x_i}{(x_0)_j}$$ where $$x_i$$ is a displacement from rest and $$(x_0)_j$$ is a vector representing rest length in the $$j$$th coordinate, we can recall the definition of the infinitesimal strain tensor $$\epsilon_{ij} = \frac{1}{2}(u_{i,j} + u_{j,i})$$ to get a relationship between forces and displacements:

$$F_{j} = \frac{1}{2}A_0 n_i C_{ijkl}\left(\frac{x_k}{(x_0)_l} + \frac{x_l}{(x_0)_k}\right)$$

This looks hairy, but the linear relationship you're looking for between forces and displacements is in there! To make it look nicer, let's define a tensor of the form:

$$\boxed{k_{jk} = \frac{A_0 n_i C_{ijkl}}{\left(x_0\right)_{l}}}$$

Noting the symmetries of the stiffness matrix, we find that we can rewrite the relationship as:

$$F_i = k_{ij}x_j$$

where the previously defined second-order tensor $$k_{ij}$$ is precisely the tensor you're looking for!