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In elasticity, there is are static relations of the form $R(\sigma,\epsilon)=0$. In fluid dynamics, there is a dynamics relation with the conservation of momentum leading to Navier-Stokes equation.

My question is : how to describe the dynamics of the deformation in elasticity, since all the relations are static ?

I know why way to do it is to introduce a viscosity, and to make the material viscoelastic. But it seems to me a bit ad hoc. Are there more rigorous ways to introduce time ?

For example, defining an energy and writing not far from equilibrium $\frac{\partial u}{\partial t}=-\frac{\delta F}{\delta u}$, with $u$ the displacement ? Are there models for that ?

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    $\begingroup$ I suggest you search the web for "lattice dynamics" or phonons. Or read a classic text such as "Dynamical theory of crystal lattices" by M Born and K Huang, or any of a wide variety of more modern texts on solid state physics. $\endgroup$
    – user197851
    Commented Nov 28, 2018 at 14:35

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In fluid dynamics, we substitute the relationship between the stress tensor and the rate of deformation tensor (expressed in terms of velocity gradients) for a Newtonian fluid into the conservation of momentum equation to obtain the time-dependent Navier-Stokes equations. For solid dynamics, we substitute the relationship between the stress tensor and the strain tensor (expressed in terms of displacement gradients) for a Hookean solid into the conservation of momentum equation to obtain the time-dependent equations for a linearly elastic solid.

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  • $\begingroup$ Got it. But does it make sense ? I mean, let's take an example : if we stretch a material (a cube of size L) in the direction $x$ with a stress $\sigma$, so at $t=0$, on one hand we have $\epsilon_{xx}(0,L)=0$ since we actually want to describe the relaxation from $\epsilon_{xx}(0,L)=0$ to $\epsilon=\sigma/E$, but on the other hand, we should write $\epsilon_{xx}(0,L)=\sigma/E$, since the formalism that we use prescribe that equality ?? So we're mixing statics and dynamics... ? $\endgroup$
    – J.A
    Commented Nov 28, 2018 at 15:24
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    $\begingroup$ In this case, there is a discontinuity at time zero at the end where the stress is suddenly applied. If you work this out with the time-dependent momentum equations, this results in an extension wave that travels along the rod. So, whereas the viscous equations are dissipative and parabolic, in the case of the transient solid dynamics equations, the behavior is elastic, and the equations are hyperbolic (wave equations). $\endgroup$ Commented Nov 28, 2018 at 15:34
  • $\begingroup$ What do u mean by "working out" ? That I should remove the discontinuity by making a fast increasing stress from $0$ to $\sigma$ ? $\endgroup$
    – J.A
    Commented Nov 28, 2018 at 15:47
  • $\begingroup$ By "working out," what I mean is deriving the partial differential equation for the axial displacement or axial strain as a function time and position. It will typically be $$\frac{\partial ^2 u}{\partial t^2}=\frac{E}{\rho}\frac{\partial ^2u}{\partial x^2}$$ where E is Young's modulus and $\rho$ is the density. The same equation applies to the axial strain. $\endgroup$ Commented Nov 28, 2018 at 15:54
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The dynamic equations for the components of the stress tensor $\sigma_{ij}(x_1,x_2,x_3,t)$ are not static, they are of the form:

$$\sum_{j} \frac{\partial \sigma_{ij}}{\partial x_j} + b_i = \rho \frac{\partial v_i}{\partial t}$$

where $v_i(x_1,x_2,x_3,t)$ is the field of velocities. So, the stress components are functions of time and the strain relations can be computed algebraically from:

$$R(\sigma_{ij}(\boldsymbol{x},t),\varepsilon_{ij}(\boldsymbol{x},t)) = 0$$

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