Nonlinear finite elements/Balance of angular momentum

The balance of angular momentum in an inertial frame can be expressed as:

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

We assume that there are no surface couples on ${\displaystyle \partial {\Omega }}$ or body couples in ${\displaystyle \Omega }$. Recall the general balance equation

${\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }f(\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\partial {\Omega }}f(\mathbf {x} ,t)[u_{n}(\mathbf {x} ,t)-\mathbf {v} (\mathbf {x} ,t)\cdot \mathbf {n} (\mathbf {x} ,t)]~{\text{dA}}+\int _{\partial {\Omega }}g(\mathbf {x} ,t)~{\text{dA}}+\int _{\Omega }h(\mathbf {x} ,t)~{\text{dV}}~.}$

In this case, the physical quantity to be conserved the angular momentum density, i.e., ${\displaystyle f=\mathbf {x} \times (\rho ~\mathbf {v} )}$. The angular momentum source at the surface is then ${\displaystyle g=\mathbf {x} \times \mathbf {t} }$ and the angular momentum source inside the body is ${\displaystyle h=\mathbf {x} \times (\rho ~\mathbf {b} )}$. The angular momentum and moments are calculated with respect to a fixed origin. Hence we have

${\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {v} )~{\text{dV}}\right]=\int _{\partial {\Omega }}[\mathbf {x} \times (\rho ~\mathbf {v} )][u_{n}-\mathbf {v} \cdot \mathbf {n} ]~{\text{dA}}+\int _{\partial {\Omega }}\mathbf {x} \times \mathbf {t} ~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}$

Assuming that ${\displaystyle \Omega }$ is a control volume, we have

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

Using the definition of a tensor product we can write

${\displaystyle [\mathbf {x} \times (\rho ~\mathbf {v} )][\mathbf {v} \cdot \mathbf {n} ]=[[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]\cdot \mathbf {n} ~.}$

Also, ${\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }$. Therefore we have

${\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\partial {\Omega }}[[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]\cdot \mathbf {n} ~{\text{dA}}+\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}$

Using the divergence theorem, we get

${\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]~{\text{dV}}+\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}$

To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

${\displaystyle \left[\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}\right]_{i}=\int _{\partial {\Omega }}e_{ijk}~x_{j}~\sigma _{kl}~n_{l}~{\text{dA}}=\int _{\partial {\Omega }}A_{il}~n_{l}~{\text{dA}}=\int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}}$

where ${\displaystyle [~]_{i}}$ represents the ${\displaystyle i}$-th component of the vector. Using the divergence theorem

${\displaystyle \int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}=\int _{\Omega }{\boldsymbol {\nabla }}\bullet {\boldsymbol {A}}~{\text{dV}}=\int _{\Omega }{\frac {\partial A_{il}}{\partial x_{l}}}~{\text{dV}}=\int _{\Omega }{\frac {\partial }{\partial x_{l}}}(e_{ijk}~x_{j}~\sigma _{kl})~{\text{dV}}~.}$

Differentiating,

${\displaystyle \int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}=\int _{\Omega }\left[e_{ijk}~\delta _{jl}~\sigma _{kl}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{kl}}{\partial x_{l}}}\right]~{\text{dV}}=\int _{\Omega }\left[e_{ijk}~\sigma _{kj}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{kl}}{\partial x_{l}}}\right]~{\text{dV}}=\int _{\Omega }\left[e_{ijk}~\sigma _{kj}+e_{ijk}~x_{j}~[{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}]_{k}\right]~{\text{dV}}~.}$

Expressed in direct tensor notation,

${\displaystyle \int _{\partial {\Omega }}{\boldsymbol {A}}\cdot \mathbf {n} ~{\text{dA}}=\int _{\Omega }\left[[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}]_{i}+[\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})]_{i}\right]~{\text{dV}}}$

where ${\displaystyle {\mathcal {E}}}$ is the third-order permutation tensor. Therefore,

${\displaystyle \left[\int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}\right]_{i}==\int _{\Omega }\left[[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}]_{i}+[\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})]_{i}\right]~{\text{dV}}}$

or,

${\displaystyle \int _{\partial {\Omega }}\mathbf {x} \times ({\boldsymbol {\sigma }}\cdot \mathbf {n} )~{\text{dA}}==\int _{\Omega }\left[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}+\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\right]~{\text{dV}}~.}$

The balance of angular momentum can then be written as

${\displaystyle \int _{\Omega }\mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]~{\text{dV}}+\int _{\Omega }\left[{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}+\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\right]~{\text{dV}}+\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}~.}$

Since ${\displaystyle \Omega }$ is an arbitrary volume, we have

${\displaystyle \mathbf {x} \times \left[{\cfrac {\partial }{\partial t}}(\rho ~\mathbf {v} )\right]=-{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}+\mathbf {x} \times ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})+\mathbf {x} \times (\rho ~\mathbf {b} )}$

or,

${\displaystyle {\mathbf {x} }\times {\left[{\frac {\partial }{\partial t}}(\rho ~\mathbf {v} )-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-{\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}~.}$

Using the identity,

${\displaystyle {\boldsymbol {\nabla }}\bullet (\mathbf {u} \otimes \mathbf {v} )=({\boldsymbol {\nabla }}\bullet \mathbf {v} )\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )\cdot \mathbf {v} }$

we get

${\displaystyle {\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]=({\boldsymbol {\nabla }}\bullet \mathbf {v} )[\mathbf {x} \times (\rho ~\mathbf {v} )]+({\boldsymbol {\nabla }}[\mathbf {x} \times (\rho ~\mathbf {v} )])\cdot \mathbf {v} ~.}$

The second term on the right can be further simplified using index notation as follows.

${\displaystyle {\begin{aligned}\left[({\boldsymbol {\nabla }}[\mathbf {x} \times (\rho ~\mathbf {v} )])\cdot \mathbf {v} \right]_{i}=\left[({\boldsymbol {\nabla }}[\rho ~(\mathbf {x} \times \mathbf {v} )])\cdot \mathbf {v} \right]_{i}&={\frac {\partial }{\partial x_{l}}}(\rho ~e_{ijk}~x_{j}~v_{k})~v_{l}\\&=e_{ijk}\left[{\frac {\partial \rho }{\partial x_{l}}}~x_{j}~v_{k}~v_{l}+\rho ~{\frac {\partial x_{j}}{\partial x_{l}}}~v_{k}~v_{l}+\rho ~x_{j}~{\frac {\partial v_{k}}{\partial x_{l}}}~v_{l}\right]\\&=(e_{ijk}~x_{j}~v_{k})~\left({\frac {\partial \rho }{\partial x_{l}}}~v_{l}\right)+\rho ~(e_{ijk}~\delta _{jl}~v_{k}~v_{l})+e_{ijk}~x_{j}~\left(\rho ~{\frac {\partial v_{k}}{\partial x_{l}}}~v_{l}\right)\\&=[(\mathbf {x} \times \mathbf {v} )({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )+\rho ~\mathbf {v} \times \mathbf {v} +\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )]_{i}\\&=[(\mathbf {x} \times \mathbf {v} )({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )+\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )]_{i}~.\end{aligned}}}$

Therefore we can write

${\displaystyle {\boldsymbol {\nabla }}\bullet [[\mathbf {x} \times (\rho ~\mathbf {v} )]\otimes \mathbf {v} ]=(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )+({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )]~.}$

The balance of angular momentum then takes the form

${\displaystyle {\mathbf {x} }\times {\left[{\frac {\partial }{\partial t}}(\rho ~\mathbf {v} )-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )-\mathbf {x} \times (\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}}$

or,

${\displaystyle {\mathbf {x} }\times {\left[{\frac {\partial }{\partial t}}(\rho ~\mathbf {v} )+\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} -{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}}$

or,

${\displaystyle {\mathbf {x} }\times {\left[\rho {\frac {\partial \mathbf {v} }{\partial t}}+{\frac {\partial \rho }{\partial t}}~\mathbf {v} +\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} -{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}}$

The material time derivative of ${\displaystyle \mathbf {v} }$ is defined as

${\displaystyle {\dot {\mathbf {v} }}={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} ~.}$

Therefore,

${\displaystyle {\mathbf {x} }\times {\left[\rho ~{\dot {\mathbf {v} }}-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right]}=-\mathbf {x} \times {\cfrac {\partial \rho }{\partial t}}~\mathbf {v} +-(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )-({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )+{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}~.}$

Also, from the conservation of linear momentum

${\displaystyle \rho ~{\dot {\mathbf {v} }}-{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} =0~.}$

Hence,

${\displaystyle {\begin{aligned}0&=\mathbf {x} \times {\cfrac {\partial \rho }{\partial t}}~\mathbf {v} +(\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times ~\mathbf {v} )+({\boldsymbol {\nabla }}\rho \cdot \mathbf {v} )(\mathbf {x} \times \mathbf {v} )-{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}\\&=\left({\frac {\partial \rho }{\partial t}}+\rho {\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} \right)(\mathbf {x} \times \mathbf {v} )-{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}~.\end{aligned}}}$

The material time derivative of ${\displaystyle \rho }$ is defined as

${\displaystyle {\dot {\rho }}={\frac {\partial \rho }{\partial t}}+{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} ~.}$

Hence,

${\displaystyle ({\dot {\rho }}+\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} )(\mathbf {x} \times \mathbf {v} )-{\mathcal {E}}:{\boldsymbol {\sigma }}^{T}=0~.}$

From the balance of mass

${\displaystyle {\dot {\rho }}+\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} =0~.}$

Therefore,

${\displaystyle {\mathcal {E}}:{\boldsymbol {\sigma }}^{T}=0~.}$

In index notation,

${\displaystyle e_{ijk}~\sigma _{kj}=0~.}$

Expanding out, we get

${\displaystyle \sigma _{12}-\sigma _{21}=0~;~~\sigma _{23}-\sigma _{32}=0~;~~\sigma _{31}-\sigma _{13}=0~.}$

Hence,

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