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Spectral decompositions

Many numerical algorithms use spectral decompositions to compute material behavior.

Spectral decompositions of stretch tensors

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

 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

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

Let the spectral decomposition of \boldsymbol{U} be

 \boldsymbol{U} = \sum_{i=1}^3 \lambda_i~\boldsymbol{N}_i\otimes\boldsymbol{N}_i

and the spectral decomposition of \boldsymbol{V} be

 \boldsymbol{V} = \sum_{i=1}^3 \hat{\lambda}_i~\mathbf{n}_i\otimes\mathbf{n}_i ~.


 \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

 \lambda_i = \hat{\lambda}_i \quad \text{and} \quad \mathbf{n}_i = \boldsymbol{R}\cdot\boldsymbol{N}_i

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


Spectral decompositions of deformation gradient

The deformation gradient is given by

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

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

 \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 \boldsymbol{F} can be written as

 \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 \boldsymbol{N}_i.

We have

 \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

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

Since the eigenvectors are orthonormal, we have

 (\boldsymbol{N}_j\otimes\boldsymbol{N}_j)\cdot\boldsymbol{N}_i = \begin{cases} 0 & \mbox{if}~ i \ne j \ \boldsymbol{N}_i & \mbox{if}~ i = j \end{cases}


 \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

 \boldsymbol{F}\cdot\boldsymbol{N}_i = \lambda_i~(\boldsymbol{R}\cdot\boldsymbol{N}_i) = \lambda_i~\mathbf{n}_i

So the effect of \boldsymbol{F} on \boldsymbol{N}_i is to stretch the vector by λi and to rotate it to the new orientation \mathbf{n}_i.

We can also show that

 \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

Spectral decompositions of strains

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

 \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})


 \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

 \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

 \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

Generalized strain measures

We can generalize these strain measures by defining strains as

 \boldsymbol{E}^{(n)} = \cfrac{1}{n}~(\boldsymbol{U}^n - \boldsymbol{\mathit{1}}) ~;~~ \boldsymbol{e}^{(n)} = \cfrac{1}{n}~(\boldsymbol{\mathit{1}} - \boldsymbol{V}^{-n})

The spectral decomposition is

 \boldsymbol{E}^{(n)} = \sum_{i=1}^3 \cfrac{1}{n}(\lambda_i^2 - 1)~\boldsymbol{N}_i\otimes\boldsymbol{N}_i ~;~~ \mathbf{e}^{(n)} = \sum_{i=1}^3 \cfrac{1}{n}\left(1 - \cfrac{1}{\lambda_i^n}\right)~ \mathbf{n}_i\otimes\mathbf{n}_i

Clearly, the usual Green strains are obtained when n = 2.

Logarithmic strain measure

A strain measure that is commonly used is the logarithmic strain measure. This strain measure is obtained when we have n \rightarrow 0. Thus

 \boldsymbol{E}^{(0)} = \ln(\boldsymbol{U}) ~;~~ \boldsymbol{e}^{(0)} = \ln(\boldsymbol{V})

The spectral decomposition is

 \boldsymbol{E}^{(0)} = \sum_{i=1}^3 \ln\lambda_i~\boldsymbol{N}_i\otimes\boldsymbol{N}_i ~;~~ \mathbf{e}^{(0)} = \sum_{i=1}^3 \ln\lambda_i~\mathbf{n}_i\otimes\mathbf{n}_i


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