Nonlinear finite elements/Overview of balance laws

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.

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

${\displaystyle {\cfrac {D\rho }{Dt}}+\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} =0}$

Sometimes, this equation is also written in conservative form as

${\displaystyle {\frac {\partial \rho }{\partial t}}+{\boldsymbol {\nabla }}\cdot (\rho ~\mathbf {v} )=0}$

If the material is incompressible then the density does not change with time and we get

${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} =0}$

For Lagrangian descriptions we can show that

${\displaystyle \rho ~J=\rho _{0}}$

where ${\displaystyle \rho _{0}}$ is the initial density.

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

${\displaystyle {\cfrac {D\mathbf {p} }{Dt}}=\mathbf {f} }$

where the linear momentum ${\displaystyle \mathbf {p} }$ is given by

${\displaystyle \mathbf {p} (t)=\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {v} (\mathbf {x} ,t)~{\text{dV}}}$

and the total force ${\displaystyle \mathbf {f} }$ is given by

${\displaystyle \mathbf {f} (t)=\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {b} (\mathbf {x} ,t)~{\text{dV}}+\int _{\partial \Omega }\mathbf {t} (\mathbf {x} ,t)~{\text{dA}}}$

where ${\displaystyle \rho }$ is the density, ${\displaystyle \mathbf {v} }$ is the spatial velocity, ${\displaystyle \mathbf {b} }$ is the body force and ${\displaystyle \mathbf {t} }$ is the surface traction. Therefore, the balance of linear momentum can be written as

${\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {v} (\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {b} (\mathbf {x} ,t)~{\text{dV}}+\int _{\partial \Omega }\mathbf {t} (\mathbf {x} ,t)~{\text{dA}}}$

Now using the transport theorem, we have

${\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\rho ~\mathbf {v} ~{\text{dV}}\right]=\int _{\Omega }\left[\rho ~{\cfrac {D\mathbf {v} }{Dt}}+\left({\cfrac {D\rho }{Dt}}+\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} \right)\right]~{\text{dV}}}$

From the conservation of mass, the second term on the right hand side is zero and we are left with

${\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {v} (\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\Omega }\rho ~{\cfrac {D\mathbf {v} }{Dt}}~{\text{dV}}}$

Therefore, the balance of linear momentum can be written as

${\displaystyle \int _{\Omega }\rho ~{\cfrac {D\mathbf {v} }{Dt}}~{\text{dV}}=\int _{\Omega }\rho ~\mathbf {b} ~{\text{dV}}+\int _{\partial \Omega }\mathbf {t} ~{\text{dA}}}$

Using Cauchy's theorem (${\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }$) and the divergence theorem we can show that the balance of linear momentum can be written as

${\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}+\rho ~\mathbf {b} =\rho ~{\cfrac {Dv}{Dt}}=\rho ~\mathbf {a} }$

In index notation,

${\displaystyle \sigma _{ij,j}+\rho ~b_{i}=\rho ~{\cfrac {Dv_{i}}{Dt}}}$

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

${\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {v} )~{\text{dV}}\right]=\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}+\int _{\partial \Omega }\mathbf {x} \times \mathbf {t} ~{\text{dA}}}$

We can show that this equation reduces down to the requirement that the Cauchy stress is symmetric, i.e.,

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

The balance of energy can be expressed as

${\displaystyle \rho ~{\cfrac {De}{Dt}}={\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}$

where ${\displaystyle e}$ is the internal energy per unit mass, ${\displaystyle \mathbf {q} }$ is the heat flux vector and ${\displaystyle s}$ is the heat source per unit volume.

We may also write this equation as

${\displaystyle \rho ~{\cfrac {De}{Dt}}={\boldsymbol {\sigma }}:{\boldsymbol {d}}-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}$

where ${\displaystyle \mathbf {d} }$, 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, ${\displaystyle \mathbf {q} =0}$ and ${\displaystyle s=0}$. So we can write

${\displaystyle \rho ~{\cfrac {De}{Dt}}={\boldsymbol {\sigma }}:{\boldsymbol {d}}}$

This shows that ${\displaystyle {\boldsymbol {\sigma }}}$ and ${\displaystyle {\boldsymbol {d}}}$ are conjugate in power.

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

${\displaystyle {\cfrac {d}{dt}}\left(\int _{\Omega }\rho ~\eta ~{\text{dV}}\right)\geq -\int _{\partial \Omega }\rho ~\eta ~\mathbf {v} \cdot \mathbf {n} ~{\text{dA}}-\int _{\partial \Omega }{\cfrac {\mathbf {q} \cdot \mathbf {n} }{T}}~{\text{dA}}+\int _{\Omega }{\cfrac {\rho ~s}{T}}~{\text{dV}}~.}$

where ${\displaystyle \eta }$ is the specific entropy (entropy per unit mass) and ${\displaystyle T}$ is the temperature.

In differential form the Clausius-Duhem inequality can be written as

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

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

${\displaystyle \rho ~({\dot {e}}-T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}$

If we define the Helmholtz free energy (the energy that is available to do mechanical work) as

${\displaystyle \psi =e-\eta ~T}$

we can also write,

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

The Lagrangian form of the balance of mass is

${\displaystyle \rho (\mathbf {X} ,t)~J(\mathbf {X} ,t)=\rho _{0}(\mathbf {X} )}$

The Lagrangian form of the balance of linear momentum is

${\displaystyle {\boldsymbol {\nabla }}_{\circ }\cdot {\boldsymbol {N}}+\rho _{0}~\mathbf {b} =\rho _{0}~{\frac {\partial \mathbf {v} (\mathbf {X} ,t)}{\partial t}}}$

where ${\displaystyle {\boldsymbol {N}}}$ is the nominal stress and ${\displaystyle {\boldsymbol {\nabla }}_{\circ }\cdot (\bullet )}$ indicates that the derivatives are with respect to ${\displaystyle \mathbf {X} }$. In terms of the first Piola-Kirchhoff stress ${\displaystyle {\boldsymbol {P}}}$ we can write

${\displaystyle {\boldsymbol {\nabla }}_{\circ }\cdot {\boldsymbol {P}}^{T}+\rho _{0}~\mathbf {b} =\rho _{0}~{\frac {\partial \mathbf {v} (\mathbf {X} ,t)}{\partial t}}}$

In index notation,

${\displaystyle P_{ji,j}+\rho _{0}~b_{i}=\rho _{0}{\dot {v}}_{i}}$

The balance of angular momentum in Lagrangian form is

${\displaystyle {\boldsymbol {N}}^{T}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {F}}\cdot {\boldsymbol {N}}}$

In terms of the first Piola-Kirchhoff stress

${\displaystyle {\boldsymbol {P}}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {F}}\cdot {\boldsymbol {P}}^{T}}$

In terms of the second Piola-Kirchhoff stress

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

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

${\displaystyle \rho _{0}~{\frac {\partial e(\mathbf {X} ,t)}{\partial t}}={\dot {\boldsymbol {F}}}^{T}:{\boldsymbol {N}}-{\boldsymbol {\nabla }}_{\circ }\cdot \mathbf {Q} (\mathbf {X} ,t)+\rho _{0}~s}$

where ${\displaystyle \mathbf {Q} }$ is the heat flux per unit reference area.

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

${\displaystyle \rho _{0}~{\frac {\partial e(\mathbf {X} ,t)}{\partial t}}={\dot {\boldsymbol {F}}}^{T}:{\boldsymbol {P}}^{T}-{\boldsymbol {\nabla }}_{\circ }\cdot \mathbf {Q} (\mathbf {X} ,t)+\rho _{0}~s}$

The entropy inequality in Lagrangian form is

${\displaystyle \rho _{0}~({\dot {\psi }}+T~{\dot {\eta }})-{\boldsymbol {N}}:{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {\mathbf {Q} } \cdot {\boldsymbol {\nabla }}_{0}T}{T}}~.}$

In terms of the first P-K stress we have

${\displaystyle \rho _{0}~({\dot {\psi }}+T~{\dot {\eta }})-{\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {\mathbf {Q} } \cdot {\boldsymbol {\nabla }}_{0}T}{T}}~.}$