This note illustrates using simple examples, how to evaluate the deformation gradient tensor and
derive its polar decomposition into a stretch and rotation tensors.
Diagrams are used to help illustrate geometrically the effect of applying the stretch and the
rotation tensors on a differential vector with the purpose of giving better insight into these
operations. For simplicity, only 2D shapes are used.
Starting by selecting some arbitrary differential vector in the undeformed shape. The shape is
then assumed to undergo a fixed form of deformation such that is constant over the whole body
(as opposed to being a field tensor where would be a function of the position). Then the tensor
is computed and shown using diagrams how the differential vector in the undeformed shape is
mapped to the vector in the deformed shape by successive application of the stretch tensor
followed by a parallel translation operation, and followed by the application of the rotation tensor
.
The point that is located at is labeled in the undeformed shape, and its image will be labeled in
the deformed shape. The coordinates in the undeformed shape will be upper case and in the
deformed shape will be lower case .
One observation found is that if the deformation is such that perpendicular lines in the
undeformed shape remain perpendicular to each others in the deformed shape, then this
implies that the rotation tensor will come out to be the identity tensor. The first 2
examples below illustrate this case. In the third example the rotation tensor is not
the identity tensor because lines do not remain perpendicular to each others after
deformation.
2 Examples
2.1 Square shape becomes longer with width fixed
The following diagram is the undeformed configuration.
Figure 1: undeformed configuration
In this shape, the vector extends from the point to the point . In this example, we assume a
deformation whereby the shape is pulled upwards by some distance, causing the shape to
become longer in the vertical direction and we assume the shape remain the same
width.
This is the simplest form of deformation. Let us assume for simplicity that the shape becomes 3
times as long as before.
Figure 2: shape becomes 3 times as long
We observe the following. The lines A,B,C have moved to new locations in the deformed
configuration. For instance, the line A started at and ended at in the undeformed shape
coordinates. While the same line now labeled lower case , starts from and ends at in the
deformed shape using the undeformed coordinates system.
The first step in finding is to determine the mapping between the coordinates in the undeformed
shape, and the coordinates in the deformed shape. In this example this mapping is constant over
any region of the shape. We see immediately that since the width of the shape did not change,
then and since the new shape is 3 times as long as before then And now we can calculate
Since then given that we obtain the numerical value for We note here that is
the same for any region of the deformed shape. This is because the deformation is
uniform.
Now we can find .
Since from the undeformed shape we see that Then
Hence Looking at the deformed shape we see that this agrees with the expected shape of the
deformed vector.
Now once is found, we can determine the stretch tensor and the rotation tensor . We will do this
algebraically first, then verify the result geometrically. Since by definition Once is known, we can
find using the relation
Now we take the square root of the matrix to find
1 and
now that is known, we can find
To verify this result algebraically, we write
Which agrees with earlier result.
To verify the result geometrically, we first apply the stretch tensor to , this results in a new
differential vector which we call , then we slide without changing its slope (i.e. parallel
translation) such that the vector starts at the point in the deformed configuration, where the
point is the image of the point in the undeformed shape, and then we apply the rotation tensor
to to obtain .
Hence
Now we apply the rotation of to , and since the rotation is a unit tensor, then this operation will
produce no effect.
Figure 3: rotation is a unit tensor
2.2 Square shape becomes both longer and wider
In this example we start with the same original shape as above, but we increase both
the length and the width of the shape and not just its length. Let the length be 3
times as long as the original length, and the width be 1.5 times as wide as the original
width.
Figure 4: Square shape becomes both longer and wider
As before, the first step in finding is to determine the mapping between the coordinates in the
undeformed shape, and the coordinates in the deformed shape. In this example, this mapping is
constant over any region of the shape. We see that and since the new shape is 3 times as
long as before then And now we can calculate Since then given that we obtain
numerical value for Now let us find . From the undeformed shape we see that Hence
hence, Looking at the deformed shape we see that this is indeed the case.
Now once is found, we can determine the stretch tensor and the rotation tensor .
We will do this algebraically first, then verify the result geometrically. Once is known, we can
find
Hence and now that is known, we can find
To verify the result geometrically, we first apply the stretch to , this results in a new differential
vector which we call , then we slide without changing its slope (i.e. parallel translation)
such that the vector starts at the point in the deformed configuration, where the
point is the image of the point and then we apply the rotation to to obtain . Hence
Now we apply the rotation of to , and since the rotation is a unit tensor, then no rotation will
occur.
Figure 5: after applying the rotation
2.3 square shape becomes wider and pulled at an angle.
In this example, the same undeformed shape shown in earlier examples will be deformed to cause
the rotation tensor to be something other than the identity tensor. We assume the following
deformation
Figure 6: deformation assumed
The above deformation is constructed such that
Now we can calculate Since then given that we obtain numerical value for Now we can find
.
From the undeformed shape we see that Hence
Therefore Looking at the deformed shape we see that this is indeed the case. Now once is found,
we can determine the stretch tensor and the rotation tensor .
We will do this algebraically first, then verify the result geometrically. Once is known, we can
find
Hence and now that is known, we can find
To verify the result geometrically, we first apply the stretch tensor to , this results in a new
differential vector which we call , then we slide without changing its slope (i.e. parallel
translation) such that the vector starts at the point in the deformed configuration, where the
point is the image of the point and then we apply the rotation tensor to to obtain
.
Hence
Now we apply the rotation to to to obtain
which agrees with the result obtained above.
The following diagram illustrates geometrically the action of and
Figure 7: final result
1To obtain the square root of a matrix, say matrix, follow these steps.
Determine the eigenvaluesof the matrix.
For each eigenvaluedetermine the correspending eigenvector
Construct the Matrixwhose columns are the eigenvectors. i.e. the first column will be the vectoretc...
Construct matrixwith diagonal elements that contains the. i.e.,, and so forth. (This is the Jordan
form for real distinct eigenvalues)
Now
In Matlab, the command expm() can be used to calculate sqrt of a matrix.