Nonlinear finite elements/Kinematics - polar decomposition

Polar decompositionEdit

The w:Polar decomposition theorem states that any second order tensor whose determinant is positive can be decomposed uniquely into a symmetric part and an orthogonal part.

In continuum mechanics, the deformation gradient   is such a tensor because  . Therefore we can write

 

where   is an orthogonal tensor ( ) and   are symmetric tensors (  and  ) called the right stretch tensor and the left stretch tensor, respectively. This decomposition is called the polar decomposition of  .

Recall that the right Cauchy-Green deformation tensor is defined as

 

Clearly this is a symmetric tensor. From the polar decomposition of   we have

 

If you know   then you can calculate   and hence   using  .

How do you find the square root of a tensor?Edit

If you want to find   given   you will need to take the square root of  . How does one do that?

We use what is called the spectral decomposition or eigenprojection of  . The spectral decomposition involves expressing   in terms of its eigenvalues and eigenvectors. The tensor product of the eigenvectors acts as a basis while the eigenvalues give the magnitude of the projection.

Thus,

 

where   are the principal values (eigenvalues) of   and   are the principal directions (eigenvectors) of  .

Therefore,

 

Since the basis does not change, we then have

 

Therefore the   can be interpreted as principal stretches and the vectors   are the directions of the principal stretches.

Exercise:Edit

If

 

show that

 

Example of polar decompositionEdit

Let us assume that the motion is given by

 

The adjacent figure shows how a unit square subjected to this motion evolves over time.

 
An example of a motion.

Deformation gradientEdit

The deformation gradient is given by

 

Therefore

 

At   at the position   we have

 

You can calculate the deformation gradient at other points in a similar manner.

Right Cauchy-Green deformation tensorEdit

We have

 

Therefore,

 

To compute   we have to find the eigenvalues and eigenvectors of  . The eigenvalue problem is

 

where

 

To find the eigenvalues we solve the characteristic equation

 

Plugging in the numbers, we get

 

or

 

This equation has two solutions

 

Taking the square roots we get the values of the principal stretches

 

To compute the eigenvectors we plug into the eigenvalues into the eigenvalue problem to get

 

Because this system of equations is not linearly independent, we need another equation to solve this system of equations for   and  . This problem is eliminated by using the following equation (which implies that   is a unit vector)

 

Solving, we get

 

We can do the same thing for the other eigenvector   to get

 

Therefore,

 

and

 

Therefore,

 

We usually don't see any problem to calculate   at this point and go straight to the right stretch tensor.

Right stretchEdit

The right stretch tensor   is given by

 

or

 

We can invert this matrix to get

 

RotationEdit

We can now find the rotation matrix by using th relation

 

In matrix form,

 

You can check whether this matrix is orthogonal by seeing whether  .

You thus get the polar decomposition of  . In an actual calculation you have to be careful about floating point errors. Otherwise you might not get a matrix that is orthogonal.