Nonlinear finite elements/Kinematics - spectral decomposition

Infinitesimal line segments in the material and spatial configurations are related by

${\displaystyle d\mathbf {x} ={\boldsymbol {F}}\cdot d{\boldsymbol {X}}={\boldsymbol {R}}\cdot ({\boldsymbol {U}}\cdot d{\boldsymbol {X}})={\boldsymbol {V}}\cdot ({\boldsymbol {R}}\cdot d{\boldsymbol {X}})~.}$

So the sequence of operations may be either considered as a stretch of in the material configuration followed by a rotation or a rotation followed by a stretch.

Also note that

${\displaystyle {\boldsymbol {V}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}~.}$

Let the spectral decomposition of ${\displaystyle {\boldsymbol {U}}}$ be

${\displaystyle {\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}}$

and the spectral decomposition of ${\displaystyle {\boldsymbol {V}}}$ be

${\displaystyle {\boldsymbol {V}}=\sum _{i=1}^{3}{\hat {\lambda }}_{i}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}~.}$

Then

${\displaystyle {\boldsymbol {V}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {R}}\cdot ({\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i})\cdot {\boldsymbol {R}}^{T}=\sum _{i=1}^{3}\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})\otimes ({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})}$

Therefore the uniqueness of the spectral decomposition implies that

${\displaystyle \lambda _{i}={\hat {\lambda }}_{i}\quad {\text{and}}\quad \mathbf {n} _{i}={\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i}}$

The left stretch (${\displaystyle {\boldsymbol {V}}}$) is also called the spatial stretch tensor while the right stretch (${\displaystyle {\boldsymbol {U}}}$) is called the material stretch tensor.

The deformation gradient is given by

${\displaystyle {\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}}$

In terms of the spectral decomposition of ${\displaystyle {\boldsymbol {U}}}$ we have

${\displaystyle {\boldsymbol {F}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {R}}\cdot ({\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i})=\sum _{i=1}^{3}\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})\otimes {\boldsymbol {N}}_{i}=\sum _{i=1}^{3}\lambda _{i}~\mathbf {n} _{i}\otimes {\boldsymbol {N}}_{i}}$

Therefore the spectral decomposition of ${\displaystyle {\boldsymbol {F}}}$ can be written as

${\displaystyle {\boldsymbol {F}}=\sum _{i=1}^{3}\lambda _{i}~\mathbf {n} _{i}\otimes {\boldsymbol {N}}_{i}}$

Let us now see what effect the deformation gradient has when it is applied to the eigenvector ${\displaystyle {\boldsymbol {N}}_{i}}$.

We have

${\displaystyle {\boldsymbol {F}}\cdot {\boldsymbol {N}}_{i}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {N}}_{i}={\boldsymbol {R}}\cdot \left(\sum _{j=1}^{3}\lambda _{j}~{\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j}\right)\cdot {\boldsymbol {N}}_{i}}$

From the definition of the dyadic product

${\displaystyle (\mathbf {u} \otimes \mathbf {v} )\cdot \mathbf {w} =(\mathbf {w} \cdot \mathbf {v} )~\mathbf {u} }$

Since the eigenvectors are orthonormal, we have

${\displaystyle ({\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j})\cdot {\boldsymbol {N}}_{i}={\begin{cases}0&{\mbox{if}}~i\neq j\\{\boldsymbol {N}}_{i}&{\mbox{if}}~i=j\end{cases}}}$

Therefore,

${\displaystyle \left(\sum _{j=1}^{3}\lambda _{j}~{\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j}\right)\cdot {\boldsymbol {N}}_{i}=\lambda _{i}~{\boldsymbol {N}}_{i}{\text{no sum on}}~i}$

That leads to

${\displaystyle {\boldsymbol {F}}\cdot {\boldsymbol {N}}_{i}=\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})=\lambda _{i}~\mathbf {n} _{i}}$

So the effect of ${\displaystyle {\boldsymbol {F}}}$ on ${\displaystyle {\boldsymbol {N}}_{i}}$ is to stretch the vector by ${\displaystyle \lambda _{i}}$ and to rotate it to the new orientation ${\displaystyle \mathbf {n} _{i}}$.

We can also show that

${\displaystyle {\boldsymbol {F}}^{-T}\cdot {\boldsymbol {N}}_{i}={\cfrac {1}{\lambda _{i}}}~\mathbf {n} _{i}~;~~{\boldsymbol {F}}^{T}\cdot \mathbf {n} _{i}=\lambda _{i}~{\boldsymbol {N}}_{i}~;~~{\boldsymbol {F}}^{-1}\cdot \mathbf {n} _{i}={\cfrac {1}{\lambda _{i}}}~{\boldsymbol {N}}_{i}}$

Recall that the Lagrangian Green strain and its Eulerian counterpart are defined as

${\displaystyle {\boldsymbol {E}}={\frac {1}{2}}~({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}={\frac {1}{2}}~({\boldsymbol {\mathit {1}}}-\left({\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\right)^{-1})}$

Now,

${\displaystyle {\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}={\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}}={\boldsymbol {U}}^{2}~;~~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {V}}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {V}}={\boldsymbol {V}}^{2}}$

Therefore we can write

${\displaystyle {\boldsymbol {E}}={\frac {1}{2}}~({\boldsymbol {U}}^{2}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}={\frac {1}{2}}~({\boldsymbol {\mathit {1}}}-{\boldsymbol {V}}^{-2})}$

Hence the spectral decompositions of these strain tensors are

${\displaystyle {\boldsymbol {E}}=\sum _{i=1}^{3}{\frac {1}{2}}(\lambda _{i}^{2}-1)~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} =\sum _{i=1}^{3}{\frac {1}{2}}\left(1-{\cfrac {1}{\lambda _{i}^{2}}}\right)~\mathbf {n} _{i}\otimes \mathbf {n} _{i}}$