# Nonlinear finite elements/Kinematics - volume change and area change
## Volume change

Consider an infinitesimal volume element in the reference configuration.
If a right-handed orthonormal basis in the reference configuration is $(\mathbf {E} _{1},\mathbf {E} _{2},\mathbf {E} _{3})$ then the vectors representing the edges of the element are
    $d\mathbf {X} _{1}=dX_{1}~\mathbf {E} _{1}~;~~d\mathbf {X} _{2}=dX_{2}~\mathbf {E} _{2}~;~~d\mathbf {X} _{3}=dX_{3}~\mathbf {E} _{3}$
The volume of the element is given by
    $dV=d\mathbf {X} _{1}\cdot (d\mathbf {X} _{2}\times d\mathbf {X} _{3})=dX_{1}~dX_{2}~dX_{3}~\mathbf {E} _{1}\cdot (\mathbf {E} _{2}\times \mathbf {E} _{3})=dX_{1}~dX_{2}~dX_{3}$
Upon deformation, these edges go to $(d\mathbf {x} _{1},d\mathbf {x} _{2},d\mathbf {x} _{3})$ where
    $d\mathbf {x} _{1}={\boldsymbol {F}}\cdot d\mathbf {X} _{1}={\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}\cdot d\mathbf {X} _{1}~;~d\mathbf {x} _{2}={\boldsymbol {F}}\cdot d\mathbf {X} _{2}={\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}\cdot d\mathbf {X} _{2}~;~d\mathbf {x} _{3}={\boldsymbol {F}}\cdot d\mathbf {X} _{3}={\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}\cdot d\mathbf {X} _{3}$
or,
    $d\mathbf {x} _{1}=dX_{1}~{\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}\cdot \mathbf {E} _{1}=dX_{1}~{\cfrac {\partial \mathbf {x} }{\partial X_{1}}}~;~~d\mathbf {x} _{2}=dX_{2}~{\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}\cdot \mathbf {E} _{2}=dX_{2}~{\cfrac {\partial \mathbf {x} }{\partial X_{2}}}~;~~d\mathbf {x} _{3}=dX_{3}~{\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}\cdot \mathbf {E} _{3}=dX_{3}~{\cfrac {\partial \mathbf {x} }{\partial X_{3}}}$
Therefore, the deformed volume is given by
    $dv=d\mathbf {x} _{1}\cdot (d\mathbf {x} _{2}\times d\mathbf {x} _{3})=dX_{1}~dX_{2}~dX_{3}~{\cfrac {\partial \mathbf {x} }{\partial X_{1}}}\cdot \left({\cfrac {\partial \mathbf {x} }{\partial X_{2}}}\times {\cfrac {\partial \mathbf {x} }{\partial X_{3}}}\right)$
Now,
    $J=\det({\boldsymbol {F}})=\det \left({\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}\right)={\cfrac {\partial \mathbf {x} }{\partial X_{1}}}\cdot \left({\cfrac {\partial \mathbf {x} }{\partial X_{2}}}\times {\cfrac {\partial \mathbf {x} }{\partial X_{3}}}\right)$
Hence,
Recall that from conservation of mass we have
    $\rho _{0}~dV=\rho ~dv$
Therefore, an alternative form of the conservation of mass is

## Distortional component of the deformation gradient

For many materials it is convenient to decompose the deformation gradient in a volumetric part and a distortional part. This is particularly useful when there is no volume change in the material when it deforms - for example in muscles, rubber tires, metal plasticity, etc.
Let us assume that the deformation gradient can be decomposed into a volumetric part and a distortional part, i.e,
    ${\boldsymbol {F}}={\boldsymbol {F}}_{v}\cdot {\boldsymbol {F}}_{d}$
Then
    $J=\det({\boldsymbol {F}})=\det({\boldsymbol {F}}_{v})~\det({\boldsymbol {F}}_{d})$
Since there is no volume change due to the pure distortion, we have
    $\det({\boldsymbol {F}}_{d})=1$
If ${\boldsymbol {F}}_{d}=J^{-1/3}~{\boldsymbol {F}}$ we have
    $\det({\boldsymbol {F}}_{d})=(J^{-1/3})^{3}~\det({\boldsymbol {F}})=J^{-1}~J=1$
Therefore the distortional component of the deformation gradient is given by
We can use this result to find the distortional components of various strain and deformation tensors. For example, the right Cauchy-Green deformation tensor is given by
    ${\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}$
If we define its distortional component as
    ${\boldsymbol {C}}_{d}={\boldsymbol {F}}_{d}^{T}\cdot {\boldsymbol {F}}_{d}$
we have

## Area change - Nanson's formula

Nanson's formula is an important relation that can be used to go from areas in the current configuration to areas in the reference configuration and vice versa.
This formula states that
where $da$ is an area of a region in the current configuration, $dA$ is the same area in the reference configuration, and $\mathbf {n}$ is the outward normal to the area element in the current configuration while $\mathbf {N}$ is the outward normal in the reference configuration.
Proof:
   To see how this formula is derived, we start with the oriented area elements in the reference and current configurations:
       $d\mathbf {A} =dA~\mathbf {N} ~;~~d\mathbf {a} =da~\mathbf {n}$
   The reference and current volumes of an element are
       $dV=d\mathbf {L} \cdot d\mathbf {A} ~;~~dv=d\mathbf {l} \cdot d\mathbf {a}$
   where $d\mathbf {l} ={\boldsymbol {F}}\cdot d\mathbf {L}$.
   Therefore,
       $d\mathbf {l} \cdot d\mathbf {a} =dv=J~dV=J~d\mathbf {L} \cdot d\mathbf {A}$
   or,
       ${\boldsymbol {F}}\cdot d\mathbf {L} \cdot d\mathbf {a} =dv=J~dV=J~d\mathbf {L} \cdot d\mathbf {A}$
   or,
       $d\mathbf {L} \cdot ({\boldsymbol {F}}^{T}\cdot d\mathbf {a} )=d\mathbf {L} \cdot (J~d\mathbf {A} )~.$
   So we get
       $d\mathbf {a} =J~{\boldsymbol {F}}^{-T}\cdot d\mathbf {A}$
   or,
       $da~\mathbf {n} =J~dA~{\boldsymbol {F}}^{-T}\cdot \mathbf {N}$