Advanced elasticity/Polar decomposition

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Polar decomposition[edit]

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 decomposition[edit]

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 gradient[edit]

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 tensor[edit]

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 stretch[edit]

The right stretch tensor is given by

or

We can invert this matrix to get

Rotation[edit]

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.