Question

Using the definition of the line element, for the reference configuration and for the deformation. Say how is defined λ_{m} associated with m's direction. Describe the different cases for this deformation (stretching and shortening) and from this write the definition of the unitary elongation ε_{m}.

Answer 1

The deformation gradient F is the derivative of each component of the deformed x vector with respect to each component of the reference X vector. For x=x(X), then

A slightly altered calculation is possible by noting that the displacement u of any point can be defined as

u=x−X

and this leads to x=X+u, and

In tensor notation,also written as

Fij=δij+ui,j

Rigid Body Displacements:

An example of a rigid body displacement is

x=X+5

y=Y+2
In this case, F=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

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

.

Rotations alter the value of F so that it is no longer equal to 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.

The deformation gradient is

all off-diagonal components are zero. F11 reflects stretching in the x-direction and F22 reflects stretching in the y-direction.

Case 2 - Shear (with Rotation):-

These equations shear the square as shown.

The deformation gradient is

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.

The deformation gradient is

The non-zero off-diagonal values mean that shearing is present. The fact that 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

and the corresponding deformation gradient is

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 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 X reference coordinates to the intermediate ′x′ coordinates.

The second step is to rotate the x′ intermediate configuration to the final x coordinates.

and the deformation gradient can be written as the product of two matrices: a rotation matrix, and a symmetric matrix describing the deformation.

the rotation matrix corresponds to a 30° rotation. The matrix product is commonly written as

where Ris the rotation matrix, and 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. U is applied first, then 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.