While average strain generally looks at the strain of a volume, we will not consider how a point on an elastic body moves and how points near it do also.
<FIGURE> "2D Strain at a Point" (Description)
Let's begin in 2D. Say there is a point () on an elastic body that is located at coordinate . (This notation will be more convenient when we want to work with tensors.) If we deform the body, then is displaced to which has the coordinates . <FIGURE> We call the displacement vector.
Now let's say that there is a point infinitesimally close to , called , with coordinates . When is displaced to by the deformation, is similarly displaced to with coordinates . Thinking critically, the displacement experienced on a body depends on the position on the body. Therefore . This allows us to use the chain rule to express infinitesimal displacements.
Now define the following terms:
This allows us to write our infinitesimal displacements using Einstein Notation:
What is the physical significance of this? This is easier to see looking in special directions. Considering the points , where and where . <FIGURE> Then after the deformation:
How do we interpret this? In the case of , we have . Thus, based on our initial equations expressing the infinitesimal displacements, we can infer that , and . This tells us that is an expression of uniaxial extension in the direction and is a rotation of around the point . Similarly, in the case of , we can again combine equations which yields , and . Thus is a uniaxial extension in the direction, and is a rotation of around point . These are our displacement tensors.
Let's return to , displacing to . What is the relationship between , and ?
In a similar fashion we can also prove that .
Looking at our picture <FIGURE>, we can see that our deformations also have translations and rotations. We're not interested in these because they do not tell us about material response such as dilatation (change in volume) or distortion (change in shape). Translations and rotations are a part of the field of mechanics called dynamics. Here we are interested in small scale elastic deformations. Our , but we know that our stress tensor is symmetric as . We can therefore rewrite our displacement tensor as a combination of a symmetric and antisymmetric tensor.
<FIGURE> "Visual representation of the displacement tensor ." (Here the displacement tensor has been broken up into the strain tensor , and the rotation tensor .)
Here is the strain tensor and is the rotation tensor. Schematically this looks like <FIGURE>. In the scope of this text, we are only interested in , but it is generally still worth remembering that displacement includes both shear and rotation components:
The strain tensor maps the irrotational displacement at a point to an imaginary plane, with normal in any direction that cuts through the point. Because the strain tensor is a tensor it must transform in the same manner as the stress tensor did in earlier sections of this text. As a reminder:
Note that when we first started to look at this subject we defined shear strain as , which is asymmetric. In terms of our strain tensor, this would be . (It must be rotated back so each side had an angle of .) You may frequently see a matrix written as:
It is sometimes useful to write it this way. However, it is not a tensor, because it does not transform the same as the above equations do, due to its asymmetry. Textbooks generally like this "average engineering strain" , but we will not be using this here unless absolutely necessary.
The results we found for our 2D strain tensor can easily be generalized to 3D by writing them in Einstein Notation and using "3" in the place of "2" in the implicit sums.
Which in 3D express 3 equation. , and expands to:
The displacement tensor is:
The strain tensor is:
The rotation tensor is:
Which gives us the displacement:
The new displaced coordinates:
Now that we have a symmetric strain tensor with properties analogous to stress, we can examine the other properties these other properties using similar methods as when we analyzed stress. For small strains where , we define the mean stress as:
The total strain tensor can be broken into dilatation and deviatoric components.
In a similar fashion we also have deviatoric stresses and hydrostatic stresses which are analogous to the deviatoric and hydrostatic stresses which are analogous to the deviatoric and dilatation strains. The hydrostatic, or mean, stress is:
Therefore, the deviatoric stress can be reduced because:
The principal components of break down to:
And we know that these are just the maximum shear stresses:
Keep in mind that we took this from:
<FIGURE> "Title" (Strain Gauges only measures uniaxial strain.)
Strain Gauges only measures uniaxial strain, no stress and no shear.