# How to introduce the CONSTITUTIVE equation into structure mechanics [closed]

Suppose I have stress and strain relationship and it can be simplified to the equation $$\sigma=Function(\epsilon)$$. This is the CONSTITUTIVE equation of the material I want to do research.

Assume that I need to do the finite element analysis and it has dynamic(nonlinear) deformation, I have the model and I have the CONSTITUTIVE equation which I mentioned the 1st paragraph.
How can we put those things into a structural equation to do the simulation and what structural equation I need to use?

Can we put this CONSTITUTIVE equation into the following equation $$\nabla\cdot \sigma =Force$$? Then we can get $$\epsilon$$, we get displacement and $$\sigma$$ explicitly.

Also, I think the above is the linear equation, what about nonlinear equation?

Thank you very much.

## closed as unclear what you're asking by OpticalResonator, Wasabi♦Dec 7 '18 at 0:56

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• I'm not sure what you are doing which is "new" here. For a general material you already have $\sigma_{ij} = C_{ijkl}\epsilon_{kl}$ in tensor notation where because of the symmetry of $\sigma$ and $\epsilon$, the 81 terms of $C$ contain 21 independent parameters in general, all of which can be functions of $\epsilon$ if you like. (In general, they will be functions of other things as well). You can solve the nonlinear equations by any method that works - numerically, this will require an iterative method. – alephzero Dec 6 '18 at 9:44

## 1 Answer

The balance of linear momentum is $$\nabla \cdot \boldsymbol{\sigma} + \rho \mathbf{b} = \rho \mathbf{a}$$ where $$\boldsymbol{\sigma}$$ is the Cauchy stress, $$\rho$$ is the mass density, $$\mathbf{b}$$ is the body force, and $$\mathbf{a}$$ is the acceleration.

Given a nonlinear stress-strain relation $$\boldsymbol{\sigma} = g(\boldsymbol{F})$$ where $$\boldsymbol{F}$$ is the deformation gradient, we can write the linear momentum equation as $$\nabla \cdot [g(\boldsymbol{F})] + \rho \mathbf{b} = \rho \mathbf{a}$$ Note that various strain definitions can be derived from $$\boldsymbol{F}$$ (see, for example, https://en.wikipedia.org/wiki/Finite_strain_theory#Seth%E2%80%93Hill_family_of_generalized_strain_tensors).

The deformation gradient tensor ($$\boldsymbol{F}$$) is related to the displacement vector ($$\mathbf{u}$$) as explained in great detail in https://en.wikipedia.org/wiki/Finite_strain_theory#Material_coordinates_(Lagrangian_description). For some problems (but not all) we can write $$\nabla \cdot [g(\nabla \mathbf{u} + \boldsymbol{I})] + \rho \mathbf{b} = \rho \ddot{\mathbf{u}}$$ where $$\boldsymbol{I}$$ is the second-order identity tensor. This is clearly a highly nonlinear problem that can be solved for the displacement $$\mathbf{u}$$ given the appropriate boundary conditions.

So, the answer to your question is that the balance of linear momentum holds whether the problem is linear or not.