Recall from eq. 1 in Section 2: Strain, that

Since ,

or, in index notation:

, since

and , by definition.

The tensor, , is typically referred to as the “deformation gradient.” To see how is found in 2D and 3D in an actual FEA application, see Finite Element Coordinate Mapping [McGinty].

In matrix form:

(1)

:

(2)

is sometimes called the “stretch” tensor

Now, consider how the length of any element or fiber within the continuum may change under deformation. To find such length “magnitudes” we can take dot products as follows:

( = length )

( = length )

note: you can’t do this transpose manipulation as easily if multiplying two tensors, but it works for two vectors or a vector and a tensor (use indices to easily prove)

note: “Right C – G” (Cauchy – Green) deformation tensor. The reason for this name will become clear once we begin discussion our on “polar decomposition” theory.

(3)

where the Lagrangian Strain Tensor (or Green-Lagrange Strain Tensor) is:

or

(4)

strain vector on a plane whose normal vector is (the actual location in space is specified within )

strain (scalar) in direction

note:

From eq. 3,

Suppose (infinitesimal deformation):

*i.e.*

what about ?

e.x.

Suppose and are orthogonal in the undeformed configuration.

Since is the strain vector on a plane whose unit normal is ,

is the component of in direction .

*i.e.* if is *in* the plane of interest (orthogonal to ), then is the shear strain

So,

From eq. 3 and recalling that , we get:

, where and are unit vectors

note: is “zero” on the LHS of the above equation because and are originally orthogonal

where is the “stretch ratio”

note: As seen in the figure below,

Infinitesimal engineering shear strain =

Does our shear strain reduce to this value for infinitesimal deformation?

For infinitesimal deformation, ; ;

for linear infinitesimal deformation

We can also see from the following equation (eq. 5) that, in general, for linear infinitesimal deformation (higher order terms are neglected).

For large (“finite”) strain:

(5)

To prove, first consider:

Now, from eq. 3,

Recall the definition of from the figure at the beginning of this chapter, and recall that is the transpose of

Eulerian strain:

Here, “Eulerian strain” is simply referring to a measure of strain that is defined in spatial coordinates. Under rigid body rotation, the Eulerian strain values will change, whereas the Lagrangian strain tensor is invariant to rigid body rotation. In other words, the coordinate system in which is calculated (the “material” coordinate system) rotates with rigid body rotation. The coordinate system in which is calculated (the “spatial” coordinate system) remains constant.

Similar to the way that we derived , let’s consider the difference in lengths of any particular element, or fiber, within our strain potato, before and after deformation.

where the Eulerian Strain Tensor (sometimes called the Almansi Strain Tensor) is:

or

(6)

where Left C-G Tensor =

, which is easy to prove using indices

This is analogous to the previously derived

Proof comes from:

We’ll see the physical meaning of and when we discuss “polar decomposition.”

How is related to ?

;

(To convince yourself that these subscripts are correct, simply write out the matrix multiplication long-hand, summing only the dummy index “k”)

Since ,

(7)

If only infinitesimal deformation, and so long as no significant rigid body rotations are present, then

(this formula may be familiar from undergrad, for example)