Introduction
This chapter on displacements and deformations is
the heart of continuum mechanics. And this page and the next, which cover the
deformation gradient, are the center of that heart. The deformation gradient
is used to separate rigid body translations and
rotations
from deformations, which are the source of
stresses.
The discussion below begins with a definition of the deformation gradient,
then proceeds in the following order: (i) rigid body translations,
(ii)
rigid body rotations,
and (iii) combined deformations and rotations.
At each step, a gradient of the displacement field is applied
to analyze the situation.
The examples are presented in 2-D to make it easier to
grasp the concepts.
As is the convention in continuum mechanics, the vector \({\bf X}\) is used
to define the undeformed reference configuration, and \({\bf x}\) defines
the deformed current configuration. In each case, imagine that the (almost)
square grid on the potato is displaced, rotated, and/or deformed.
Deformation Gradient
The deformation gradient \({\bf F}\) is the derivative of each component of the
deformed \({\bf x}\) vector with respect to each component of the
reference \({\bf X}\) vector. For \({\bf x} = {\bf x}({\bf X})\), then
\[
F_{ij} \quad = \quad x_{i,j} \quad = \quad \frac{\partial x_i}{\partial X_j} \quad = \quad
\left[ \matrix{
{\partial x_1 \over \partial X_1} & {\partial x_1 \over \partial X_2} & {\partial x_1 \over \partial X_3} \\
{\partial x_2 \over \partial X_1} & {\partial x_2 \over \partial X_2} & {\partial x_2 \over \partial X_3} \\
{\partial x_3 \over \partial X_1} & {\partial x_3 \over \partial X_2} & {\partial x_3 \over \partial X_3}
} \right]
\]
A slightly altered calculation is possible by noting that the displacement \({\bf u}\) of any point can be defined as
\[
{\bf u} = {\bf x} - {\bf X}
\]
and this leads to \({\bf x} = {\bf X} + {\bf u}\), and
\[
{\bf F} \quad = \quad { \partial \over \partial {\bf X} } ({\bf X} + {\bf u}) \quad = \quad
{ \partial {\bf X} \over \partial {\bf X} } + { \partial {\bf u} \over \partial {\bf X} } \quad = \quad
{\bf I} + { \partial {\bf u} \over \partial {\bf X} }
\]
In
tensor notation, this is written as
\[
F_{ij} = \delta_{ij} + u_{i,j}
\]
Rigid Body Displacements
An example of a rigid body displacement is
\[
\begin{eqnarray}
x & = & X & + & 5 \\
y & = & Y & + & 2
\end{eqnarray}
\]
In this case, \({\bf F} = {\bf I}\), is indicative of a lack of deformations.
As will be shown next, it is also indicative of a lack of rotations.
Clearly rigid body displacements do not appear in the deformation gradient.
This is good because rigid body displacements don't contribute to
stress,
strain, etc.
Rigid Body Rotations
An example of a
rigid body rotation is
\[
\begin{eqnarray}
x & = & X \cos \theta & - & Y \sin \theta \\
y & = & X \sin \theta & + & Y \cos \theta
\end{eqnarray}
\]
These equations rotate an object counter-clockwise about the \(z\) axis.
Notice how the minus sign is now in the \((1,2)\) slot instead of the
\((2,1)\) slot as was the case for
coordinate transformations.
See this page on
rotation matrices
for an explanation.
In this case, \({\bf F}\) is
\[
{\bf F} = \left[ \matrix{
\cos \theta & -\sin \theta \\
\sin \theta & \;\;\;\; \cos \theta
} \right]
\]
Rotations alter the value of \({\bf F}\) so that it is no longer equal to
\({\bf I}\) even though no deformations are present. This can lead to
misinterpretations in which
rigid body rotations
are mistaken for deformations and
strain.
This is
the main complicating factor in finite deformation
mechanics. Unfortunately, it will get even worse when rotations and deformations
are present at the same time.
Simple Deformations
Case 1 - Stretching
Start with stretching in the x and y directions. These equations describe
a 100% elongation in the x-direction and a 50% elongation in the y-direction.
\[
\begin{eqnarray}
x & = & 2.0 X \, & + & 0.0 \, Y \\
y & = & 0.0 X \, & + & 1.5 \, Y
\end{eqnarray}
\]
The deformation gradient is
\[
{\bf F} = \left[ \matrix{
2.0 & 0.0 \\
0.0 & 1.5
} \right]
\]
Note that all off-diagonal components are zero. \(F_{11}\) reflects
stretching in the x-direction and \(F_{22}\) reflects stretching
in the y-direction.
Case 2 - Shear (with Rotation)
These equations shear the square as shown.
\[
\begin{eqnarray}
x & = & 1.0 X \, & + & 0.0 \, Y \\
y & = & 0.5 X \, & + & 1.0 \, Y
\end{eqnarray}
\]
The deformation gradient is
\[
{\bf F} = \left[ \matrix{
1.0 & 0.0 \\
0.5 & 1.0
} \right]
\]
The non-zero off-diagonal value reflects shear. The figure also
shows that the square tends to rotate counter-clockwise. This is
reflected in the deformation gradient by the fact that it is not
symmetric.
Case 3 - Pure Shear
These equations shear the square with zero net rotation.
\[
\begin{eqnarray}
x & = & 1.0 X \, & + & 0.5 \, Y \\
y & = & 0.5 X \, & + & 1.0 \, Y
\end{eqnarray}
\]
The deformation gradient is
\[
{\bf F} = \left[ \matrix{
1.0 & 0.5 \\
0.5 & 1.0
} \right]
\]
The non-zero off-diagonal values mean that shearing is present. The fact that
\({\bf F}\) is symmetric reflects that there is no net rotation. The zero net
rotation arises from the fact that while the lower right area of the square
tends to rotate counter-clockwise, the upper left area tends to rotate clockwise
at the same time. Therefore, the net rotation for the square as a whole is zero.
General Deformations
Consider the example where an object is transformed from a square to
the position shown in the figure. The equations to do this are
\[
\begin{eqnarray}
x & = & 1.300 X \, & - & 0.375 \, Y \\
y & = & 0.750 X \, & + & 0.650 \, Y
\end{eqnarray}
\]
and the corresponding deformation gradient is
\[
{\bf F} = \left[ \matrix{
1.300 & -0.375 \\
0.750 & \;\;\;0.650
} \right]
\]
The object has clearly been stretched and rotated. But by how much?
The
rotation
doesn't contribute to
stress, but the deformation does.
So it is necessary
to partition the two mechanisms out of \({\bf F}\) in order to determine
the
stress and
strain state.
The next page on
polar decompositions
will show how to do this. But for now, just accept
that the following two-step process of deformation followed by
rigid body rotation
gets you there...
The first step is a 50% stretch in the x-direction and a 25% compression in
the y-direction. This gets us from the \({\bf X}\)
reference coordinates to the intermediate \({\bf x'}\) coordinates.
\[
\begin{eqnarray}
x' & = & 1.50 X \, & + & 0.00 \, Y \\
y' & = & 0.00 X \, & + & 0.75 \, Y
\end{eqnarray}
\]
The second step is to rotate the \({\bf x'}\) intermediate configuration to the final
\({\bf x}\) coordinates.
\[
\begin{eqnarray}
x & \quad = \quad & x' \cos(30^\circ) \, & - & y' \sin(30^\circ) & \quad = \quad & 1.300 X \, & - & 0.375 \, Y \\
y & \quad = \quad & x' \sin(30^\circ) \, & + & y' \cos(30^\circ) & \quad = \quad & 0.750 X \, & + & 0.650 \, Y
\end{eqnarray}
\]
In matrix form,
\[
\left\{ \matrix{ x \\ y } \right\} =
\left[ \matrix{
0.866 & -0.500 \\
0.500 & \;\;\;0.866
} \right]
\left\{ \matrix{ x' \\ y' } \right\}
\qquad \text{and} \qquad
\left\{ \matrix{ x' \\ y' } \right\} =
\left[ \matrix{
1.50 & 0.00 \\
0.00 & 0.75
} \right]
\left\{ \matrix{ X \\ Y } \right\}
\]
and the deformation gradient can be written as the product of two
matrices:
a
rotation matrix, and a symmetric matrix describing the deformation.
\[
{\bf F} \quad = \quad
\left[ \matrix{
1.300 & -0.375 \\
0.750 & \;\;\;0.650
} \right]
\quad = \quad
\left[ \matrix{
0.866 & -0.500 \\
0.500 & \;\;\;0.866
} \right]
\left[ \matrix{
1.50 & 0.00 \\
0.00 & 0.75
} \right]
\]
It is easy to see that the
rotation matrix corresponds to a 30° rotation.
The matrix product is commonly written as
\[
{\bf F} = {\bf R} \cdot {\bf U}
\]
where \({\bf R}\) is the rotation matrix, and \({\bf U}\) is the right
stretch tensor that is responsible for all the problems in life:
stress,
strain, fatigue, cracks, fracture, etc. Note that the process is read
from right to left, not left to right. \({\bf U}\) is applied first,
then \({\bf R}\).
This partitioning of the deformation gradient into the product of a
rotation matrix and stretch tensor is
known as a
Polar Decomposition.
The next page on
Polar Decompositions
will show how to do this for the general 3-D case.
Or Alternatively...
The deformed and rotated state could equally-well be arrived at by rotating it
first, and then deforming it second. In this case, the reference configuration,
\({\bf X}\), is first rotated by the same 30° angle to arrive at an
intermediate configuration, \({\bf x'}\).
\[
\begin{eqnarray}
x' & \quad = \quad & X \cos(30^\circ) & - & Y \sin(30^\circ) \\
y' & \quad = \quad & X \sin(30^\circ) & + & Y \cos(30^\circ)
\end{eqnarray}
\]
And then the intermediate configuration is deformed to arrive at the final,
deformed state
\[
\begin{eqnarray}
x & \quad = \quad & 1.313 x' \, & + & 0.325 \, y' \\
y & \quad = \quad & 0.325 x' \, & + & 0.938 \, y'
\end{eqnarray}
\]
The off-diagonal values reflect that the object is being
sheared in addition to being stretched/compressed.
The deformation gradient can be written as
\[
{\bf F} \quad = \quad
\left[ \matrix{
1.300 & -0.375 \\
0.750 & \;\;\;0.650
} \right]
\quad = \quad
\left[ \matrix{
1.313 & 0.325 \\
0.325 & 0.938
} \right]
\left[ \matrix{
0.866 & -0.500 \\
0.500 & \;\;\;0.866
} \right]
\]
This is commonly written as
\[
{\bf F} = {\bf V} \cdot {\bf R}
\]
where \({\bf R}\) is the
rotation matrix (same as before),
and \({\bf V}\) is the left stretch tensor.
This is also a
polar decomposition.
Relationship Between V and U
It is relatively easy to develop a relationship between \({\bf V}\) and
\({\bf U}\).
Since \({\bf F} = {\bf V} \cdot {\bf R}\) and
\({\bf F} = {\bf R} \cdot {\bf U}\), then
\[
{\bf V} \cdot {\bf R} = {\bf R} \cdot {\bf U}
\]
and post-multiplying through by \({\bf R}^T\) gives
\[
{\bf V} \cdot {\bf R} \cdot {\bf R}^T = {\bf R} \cdot {\bf U} \cdot {\bf R}^T
\]
But since \({\bf R} \cdot {\bf R}^T = {\bf I}\), this leaves
\[
{\bf V} = {\bf R} \cdot {\bf U} \cdot {\bf R}^T
\]
as the relationship between \({\bf V}\) and \({\bf U}\).
Alternatively, solving for \({\bf U}\) gives
\[
{\bf U} = {\bf R}^T \cdot {\bf V} \cdot {\bf R}
\]
Verifying Relationship Between V and U
Recall that
\[
\qquad
{\bf V} =
\left[ \matrix{
1.313 & 0.325 \\
0.325 & 0.938
} \right]
\qquad
{\bf R} =
\left[ \matrix{
0.866 & -0.500 \\
0.500 & \;\;\;0.866
} \right]
\qquad
{\bf U} =
\left[ \matrix{
1.500 & 0.0 \\
0.0 & 0.750
} \right]
\]
Evaluating \( {\bf R} \cdot {\bf U} \cdot {\bf R}^T \) gives
\[
\begin{eqnarray}
{\bf R} \cdot {\bf U} \cdot {\bf R}^T
\quad & = & \quad
\left[ \matrix{
0.866 & -0.500 \\
0.500 & \;\;\;0.866
} \right]
\left[ \matrix{
1.500 & 0.0 \\
0.0 & 0.750
} \right]
\left[ \matrix{
\;\;\; 0.866 & 0.500 \\
-0.500 & 0.866
} \right]
\\
\\
\\
\quad & = & \quad
\left[ \matrix{
1.313 & 0.325 \\
0.325 & 0.938
} \right]
\end{eqnarray}
\]
which confirms that
\({\bf R} \cdot {\bf U} \cdot {\bf R}^T \) is indeed equal to
\({\bf V}\).
Coordinate Mapping Example - Hourglassing
Find the deformation gradient for the deformation shown in the figure.
After much trial and error... the equations relating undeformed and deformed states are
\[
\begin{eqnarray}
x & = & X + \frac{1}{4} X Y + 5 \\
\\
y & = & Y + 4
\end{eqnarray}
\]
These equations can be checked by substituting the coordinates of any of the
four corners of the undeformed square to arrive at the deformed coordinates.
Once the mapping equations are available, the deformation gradient is easy
\[
{\bf F} =
\left[ \matrix{
1 + \frac{1}{4} Y & \frac{1}{4} X \\
\\
0 & 1
} \right]
\]
The bottom row values of \(F_{21} = 0\) and \(F_{22} = 1\) mean that nothing
is happening in the \(Y\) direction. All the action is in the \(X\) direction.
Of special note is the condition at the center of the square, at \((0,0)\). Inserting
\((0,0)\) into the deformation gradient gives
\[
{\bf F} =
\left[ \matrix{
1 & 0 \\
0 & 1
} \right]
\]
which is the
identity matrix.
This indicates that there is no deformation at all
at the center of the square even though it is clearly deforming at every other location.
This effect is known as
hourglassing in
FE analyses
and arises in 2-D quadrilateral
and 3-D brick elements undergoing
reduced integration. In such cases, element
deformations are only evaluated at the center of the element. As a result, hourglass
deformations are not detected and can grow uncontrollably.
V and U for Different Rigid Body Rotations
For an object that is stretched by 100% along the x-axis, but not rotated,
the deformation gradient and polar decompositions are
\[
\begin{eqnarray}
{\bf F} & = & {\bf R} \cdot {\bf U} \\
\\
\left[ \matrix{ 2 & 0 \\ 0 & 1 } \right]
& = &
\left[ \matrix{ \cos 0^\circ & \!\!\!\!\text{-}\sin 0^\circ \\ \sin 0^\circ & \!\!\cos 0^\circ } \right]
\left[ \matrix{ 2 & 0 \\ 0 & 1 } \right]
\end{eqnarray}
\]
And the \({\bf V} \cdot {\bf R}\) polar decomposition is
\[
\begin{eqnarray}
{\bf F} & = & {\bf V} \cdot {\bf R} \\
\\
\left[ \matrix{ 2 & 0 \\ 0 & 1 } \right]
& = &
\left[ \matrix{ 2 & 0 \\ 0 & 1 } \right]
\left[ \matrix{ \cos 0^\circ & \!\!\!\!\text{-}\sin 0^\circ \\ \sin 0^\circ & \!\!\cos 0^\circ } \right]
\end{eqnarray}
\]
Yes, when there is no rotation, \({\bf U}\), \({\bf V}\), and \({\bf F}\) are all the same. They must be.
If the object is stretched 100% along the direction that was originally parallel to the
x-axis, and also rotated 90°, then the deformation gradient and
polar decompositions are
\[
\begin{eqnarray}
{\bf F} & = & {\bf R} \cdot {\bf U} \\
\\
\left[ \matrix{ 0 & \text{-}1 \\ 2 & \;0 } \right]
& = &
\left[ \matrix{ \cos 90^\circ & \!\!\!\!\text{-}\sin 90^\circ \\ \sin 90^\circ & \!\!\cos 90^\circ } \right]
\left[ \matrix{ 2 & 0 \\ 0 & 1 } \right]
\end{eqnarray}
\]
And the \({\bf V} \cdot {\bf R}\) polar decomposition is
\[
\begin{eqnarray}
{\bf F} & = & {\bf V} \cdot {\bf R} \\
\\
\left[ \matrix{ 0 & \text{-}1 \\ 2 & \;0 } \right]
& = &
\left[ \matrix{ 1 & 0 \\ 0 & 2 } \right]
\left[ \matrix{ \cos 90^\circ & \!\!\!\!\text{-}\sin 90^\circ \\ \sin 90^\circ & \!\!\cos 90^\circ } \right]
\end{eqnarray}
\]
The \({\bf U}\) stretch tensor hasn't changed between the 0° and 90° rotation example, however the
\({\bf V}\) stretch tensor has (as well as \({\bf F}\)). \({\bf V}\)'s components always correspond to
deformations that are described after the rotation takes place.
Discussion
So what have we learned here? That any deformed state can be described
in (at least) two different ways. In the first, the object is deformed,
then rotated. In the second, it is rotated, then deformed. So which is
more correct? The answer is neither. Both are perfectly valid. In fact,
it is more likely that the object rotates and deforms
simultaneously. But if all you have is the final deformed state
from a calculation, probably FEA, then all you can do is apply a
polar decomposition.
Everyone always wants to know, "But which one, \({\bf R}\cdot{\bf U}\) or
\({\bf V}\cdot{\bf R}\)
should I use?"
Again, neither is more or less correct than the other.
Both describe the exact same deformed state.
The usual determining factor is whether one
prefers to think of the deformations as occurring in the reference
orientation, or in the final rotated state. Nevertheless, the usual
preference is to use \({\bf F} = {\bf R}\cdot{\bf U}\) because
it is normally easier to relate the
strains back to the undeformed
orientation. This leads to the
Green strain definition that is
popular in tire mechanics and will be discussed in a few more pages.
One last example: One could assume that an object is stretched in the x-direction,
and then rotated 90°. In this case, the \(\epsilon_{xx}\) strain
is non-zero. Alternatively, one could just as well assume that the object
was rotated 90° first, then stretched in the vertical direction, giving
a non-zero \(\epsilon_{yy}\) strain. If all you've got is the final
deformed shape to work with, then either path works.
And one last note: All this hints at a larger issue. It is the question,
"Of all the definitions of strain, which is the best?" The answer is,
"Neither is necessarily more or less correct than any other." Some
are just more convenient than others for a given application.
We will come back to this topic later.
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