Nonlinear finite elements/Overview of balance laws

Governing equations in the deformed configuration
The governing equations for a continuum (apart from the kinematic relations and the constitutive laws) are


 * 1) Balance of mass
 * 2) Balance of linear momentum
 * 3) Balance of angular momentum
 * 4) Balance of energy
 * 5) Entropy inequality

Two important results from calculus
There are two important results from calculus that are useful when we use or derive these governing equations. These are


 * 1) The Gauss divergence theorem.
 * 2) The Reynolds transport theorem

These are useful enough to bear repeating at this point in the context of second order tensors.

The Gauss divergence theorem
The divergence theorem  relates volume integrals to surface integrals.

Let $$\Omega$$ be a body and let $$\partial \Omega $$ be its boundary with outward unit normal $$\mathbf{n}$$. Let $$\boldsymbol{A}(\mathbf{x})$$ be a second order tensor valued function of $$\mathbf{x}$$. Then, if $$\boldsymbol{A}$$ is differentiable at least once (i.e., $$\boldsymbol{A} \in C^0$$) then

\int_{\Omega} \boldsymbol{\nabla} \cdot \boldsymbol{A}(\mathbf{x})~\text{dV} = \int_{\partial \Omega} \boldsymbol{A}^T(\mathbf{x})\cdot\mathbf{n}~\text{dA} $$ It is often of interest to consider the situation where $$\boldsymbol{A}$$ is not continuously differentiable, i.e., when there are jumps in $$\boldsymbol{A}$$ within the body. Let $$\partial \Omega _{\text{int}}$$ represent the set of surfaces internal to the body where there are jumps in $$\boldsymbol{A}$$. In that case, the divergence theorem is written as

\int_{\Omega} \boldsymbol{\nabla} \cdot \boldsymbol{A}~\text{dV} = \int_{\partial \Omega} \boldsymbol{A}^T\cdot\mathbf{n}~\text{dA} + \int_{\partial \Omega_{\text{int}}} \left[\boldsymbol{A}^T_{+} - \boldsymbol{A}^T_{-}\right]\cdot\mathbf{n}_{+}~\text{dA} $$ where the subscripts $$+$$ and $$-$$ represents the values on the two sides of the jump discontinuity with normal $$\mathbf{n}_{+}$$.

Reynold's transport theorem
The transport theorem shows you how to ''' calculate the material time derivative of an integral. ''' It is a generalization of the Leibniz formula.

Let $$\Omega$$ be a body in its current configuration and let $$\partial \Omega $$ be its surface. Also, let $$\mathbf{v} = \dot{\mathbf{x}}$$.

If $$\phi(\mathbf{x},t)$$ is a scalar valued function of $$\mathbf{x}$$ and $$t$$ then

\cfrac{D}{Dt}\left[\int_{\Omega} \phi(\mathbf{x},t)~\text{dV}\right] = \int_{\Omega} \frac{\partial \phi}{\partial t}~\text{dV} + \int_{\partial \Omega} (\mathbf{v}\cdot\mathbf{n})~\phi~\text{dA} = \int_{\Omega} \left[\frac{\partial \phi}{\partial t} + \boldsymbol{\nabla} \cdot \left( \phi~\mathbf{v}\right) \right]~\text{dV} $$

If $$\mathbf{f}(\mathbf{x}, t)$$ is a vector valued function of $$\mathbf{x}$$ and $$t$$. Then

\cfrac{D}{Dt}\left[\int_{\Omega} \mathbf{f}(\mathbf{x}, t)~\text{dV}\right] = \int_{\Omega} \frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega} (\mathbf{v}\cdot\mathbf{n})~\mathbf{f}~\text{dA} = \int_{\Omega} \left[\frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\right]~\text{dV} $$

To refresh your memory, recall that the material time derivative is given by

\begin{align} \cfrac{D\phi}{Dt} & = \frac{\partial \phi}{\partial t} + \boldsymbol{\nabla} \phi\cdot\mathbf{v} \\ \cfrac{D\mathbf{f}}{Dt} & = \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} \\ \cfrac{D\boldsymbol{F}}{Dt} & = \frac{\partial \boldsymbol{F}}{\partial t} + (\boldsymbol{\nabla} \boldsymbol{F})^T\cdot\mathbf{v} \end{align} $$

Conservation of mass
For the situation where a body does not gain or lose mass, the balance of mass is written as

\cfrac{D\rho}{Dt} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v} = 0 $$ Sometimes, this equation is also written in  conservative form as

\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \cdot (\rho~\mathbf{v}) = 0 $$ If the material is  incompressible then the density does not change with time and we get

\boldsymbol{\nabla} \cdot \mathbf{v} = 0 $$ For Lagrangian descriptions we can show that

\rho~J = \rho_0 $$ where $$\rho_0$$ is the initial density.

Conservation of linear momentum
The balance of linear momentum is essentially Newton's second applied to continua. Newton's second law can be written as

\cfrac{D\mathbf{p}}{Dt} = \mathbf{f} $$ where the linear momentum $$\mathbf{p}$$ is given by

\mathbf{p}(t) = \int_{\Omega} \rho(\mathbf{x},t)~\mathbf{v}(\mathbf{x},t)~\text{dV} $$ and the total force $$\mathbf{f}$$ is given by

\mathbf{f}(t) = \int_{\Omega} \rho(\mathbf{x},t)~\mathbf{b}(\mathbf{x},t)~\text{dV} + \int_{\partial \Omega} \mathbf{t}(\mathbf{x},t)~\text{dA} $$ where $$\rho$$ is the density, $$\mathbf{v}$$ is the spatial velocity, $$\mathbf{b}$$ is the body force and $$\mathbf{t}$$ is the surface traction. Therefore, the balance of linear momentum can be written as

\cfrac{D}{Dt}\left[\int_{\Omega} \rho(\mathbf{x},t)~\mathbf{v}(\mathbf{x},t)~\text{dV}\right] = \int_{\Omega} \rho(\mathbf{x},t)~\mathbf{b}(\mathbf{x},t)~\text{dV} + \int_{\partial \Omega} \mathbf{t}(\mathbf{x},t)~\text{dA} $$ Now using the transport theorem, we have

\cfrac{D}{Dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\Omega} \left[\rho~\cfrac{D\mathbf{v}}{Dt} + \left(\cfrac{D\rho}{Dt} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v}\right)\right]~\text{dV} $$ From the conservation of mass, the second term on the right hand side is zero and we are left with

\cfrac{D}{Dt}\left[\int_{\Omega} \rho(\mathbf{x},t)~\mathbf{v}(\mathbf{x},t)~\text{dV}\right] = \int_{\Omega} \rho~\cfrac{D\mathbf{v}}{Dt}~\text{dV} $$ Therefore, the balance of linear momentum can be written as

\int_{\Omega} \rho~\cfrac{D\mathbf{v}}{Dt}~\text{dV} = \int_{\Omega} \rho~\mathbf{b}~\text{dV} + \int_{\partial \Omega} \mathbf{t}~\text{dA} $$ Using Cauchy's theorem ($$\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$$) and the divergence theorem we can show that the balance of linear momentum can be written as

\boldsymbol{\nabla} \cdot \boldsymbol{\sigma} + \rho~\mathbf{b} = \rho~\cfrac{Dv}{Dt} = \rho~\mathbf{a} $$ In index notation,

\sigma_{ij,j} + \rho~b_i = \rho~\cfrac{Dv_i}{Dt} $$

Conservation of angular momentum
We also have to make sure that the moments are balanced. This requirement takes the form of the conservation of angular momentum and can be written as

\cfrac{D}{Dt}\left[\int_{\Omega} \mathbf{x} \times (\rho~\mathbf{v})~\text{dV}\right] = \int_{\Omega} \mathbf{x} \times (\rho~\mathbf{b})~\text{dV} + \int_{\partial \Omega} \mathbf{x} \times \mathbf{t}~\text{dA} $$ We can show that this equation reduces down to the requirement that the Cauchy stress is symmetric, i.e.,

\boldsymbol{\sigma} = \boldsymbol{\sigma}^T $$

Conservation of energy
The balance of energy can be expressed as

\rho~\cfrac{De}{Dt} = \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s $$ where $$e$$ is the internal energy per unit mass, $$\mathbf{q}$$ is the heat flux vector and $$s$$ is the heat source per unit volume.

We may also write this equation as

\rho~\cfrac{De}{Dt} = \boldsymbol{\sigma}:\boldsymbol{d} - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s $$ where $$\mathbf{d}$$, the rate of deformation tensor, is the symmetric part of the velocity gradient. We can do this because the contraction of the skew symmetric part of the velocity gradient with the symmetric Cauchy stress gives us zero.

For purely mechanical problems, $$\mathbf{q} = 0$$ and $$s = 0$$. So we can write

\rho~\cfrac{De}{Dt} = \boldsymbol{\sigma}:\boldsymbol{d} $$ This shows that $$\boldsymbol{\sigma}$$ and $$\boldsymbol{d}$$ are  conjugate in power.

Entropy inequality
The entropy inequality is useful in determining which forms of the constitutive equations are admissible. This inequality is also called the dissipation inequality. In its Clausius-Duhem form, the inequality may written as

\cfrac{d}{dt}\left(\int_{\Omega} \rho~\eta~\text{dV}\right) \ge -\int_{\partial \Omega} \rho~\eta~\mathbf{v}\cdot\mathbf{n}~\text{dA} - \int_{\partial \Omega} \cfrac{\mathbf{q}\cdot\mathbf{n}}{T}~\text{dA} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~. $$ where $$\eta$$ is the specific entropy (entropy per unit mass) and $$T$$ is the temperature. In differential form the Clausius-Duhem inequality can be written as

\rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. $$

In terms of the specific internal energy, the entropy inequality can be expressed as

\rho~(\dot{e} - T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$ If we define the  Helmholtz free energy (the energy that is available to do mechanical work) as

\psi = e - \eta~T $$ we can also write,

\rho~(\dot{\psi} + T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{d} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$

Governing equations in the reference configuration
The Lagrangian form of the governing equations can be obtained using the relations between the various measures in the deformed and reference configurations.

Balance of mass
The Lagrangian form of the balance of mass is

\rho(\mathbf{X},t)~J(\mathbf{X},t) = \rho_0(\mathbf{X}) $$

Balance of linear momentum
The Lagrangian form of the balance of linear momentum is

\boldsymbol{\nabla}_{\circ}\cdot\boldsymbol{N} + \rho_0~\mathbf{b} = \rho_0~\frac{\partial \mathbf{v}(\mathbf{X},t)}{\partial t} $$ where $$\boldsymbol{N}$$ is the nominal stress and $$\boldsymbol{\nabla}_{\circ}\cdot(\bullet)$$ indicates that the derivatives are with respect to $$\mathbf{X}$$. In terms of the first Piola-Kirchhoff stress $$\boldsymbol{P}$$ we can write

\boldsymbol{\nabla}_{\circ}\cdot\boldsymbol{P}^T + \rho_0~\mathbf{b} = \rho_0~\frac{\partial \mathbf{v}(\mathbf{X},t)}{\partial t} $$ In index notation,

P_{ji,j} + \rho_0~b_i = \rho_0 \dot{v}_i $$

Balance of angular momentum
The balance of angular momentum in Lagrangian form is

\boldsymbol{N}^T\cdot\boldsymbol{F}^T = \boldsymbol{F}\cdot\boldsymbol{N} $$ In terms of the first Piola-Kirchhoff stress

\boldsymbol{P}\cdot\boldsymbol{F}^T = \boldsymbol{F}\cdot\boldsymbol{P}^T $$ In terms of the second Piola-Kirchhoff stress

\boldsymbol{S} = \boldsymbol{S}^T $$

Balance of energy
In the material frame, the balance of energy takes the form

\rho_0~\frac{\partial e(\mathbf{X},t)}{\partial t} = \dot{\boldsymbol{F}}^T:\boldsymbol{N} - \boldsymbol{\nabla}_{\circ}\cdot\mathbf{Q}(\mathbf{X},t) + \rho_0~s $$ where $$\mathbf{Q}$$ is the heat flux per unit reference area.

In terms of the first Piola-Kirchhoff stress tensor we have

\rho_0~\frac{\partial e(\mathbf{X},t)}{\partial t} = \dot{\boldsymbol{F}}^T:\boldsymbol{P}^T - \boldsymbol{\nabla}_{\circ}\cdot\mathbf{Q}(\mathbf{X},t) + \rho_0~s $$

Entropy inequality
The entropy inequality in Lagrangian form is

\rho_0~(\dot{\psi} + T~\dot{\eta}) - \boldsymbol{N}:\dot{\boldsymbol{F}} \le - \cfrac{\mathbf{\mathbf{Q}}\cdot\boldsymbol{\nabla}_0 T}{T} ~. $$ In terms of the first P-K stress we have

\rho_0~(\dot{\psi} + T~\dot{\eta}) - \boldsymbol{P}^T:\dot{\boldsymbol{F}} \le - \cfrac{\mathbf{\mathbf{Q}}\cdot\boldsymbol{\nabla}_0 T}{T} ~. $$

Conjugate work measures
Whatever measures we choose to use to represent stress and strain (or a rate of strain), their product should give us a measure of the work done (or the power spent). This measure should not depend on the chosen measures.

Therefore, the correct combination of stress and strain should be  work conjugate or  power conjugate. Three commonly used power conjugate stress and rate of strain measures are


 * 1) The Cauchy stress ($$\boldsymbol{\sigma}$$) and the rate of deformation ($$\boldsymbol{d}$$).
 * 2) The nominal stress ($$\boldsymbol{N}$$) and the rate of the deformation gradient ($$\boldsymbol{F}$$).
 * 3) The second P-K stress ($$\boldsymbol{S}$$) and the rate of the Green strain ($$\boldsymbol{E}$$).

We can show that, in the absence of heat fluxes and sources,

\dot{e} = \cfrac{1}{\rho}~\boldsymbol{\sigma}:\mathbf{d} = \cfrac{1}{\rho_0}~\mathbf{N}:\dot{\boldsymbol{F}} = \cfrac{1}{\rho_0}~\mathbf{P}^T:\dot{\boldsymbol{F}} = \cfrac{1}{\rho}~\boldsymbol{S}:\dot{\boldsymbol{E}} $$ Many more work/power conjugate measures can be found in the literature.