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```-da .F
P'
(1.4.17)
For incompressible materials p/p' = l.*
Combining eq. 1.4.17 with 1.4.14 gives
da . F) . (da - F) u = = (
(1.4.18) da . da
da . ( F . FT) .da da . B -da
The unit normal to the surface around Q is just
(1.4.19)
So we can write eq. 1.4.18
Thus, physically the Finger tensor describes the area change
around a point on a plane whose normal is a. B can give the
deformation at any point in terms of area change by operating on
the normal to the area defined in the present or deformed state.
Because area is a scalar, we need to operate on the vector twice.
We can also express deformation in terms of length change.
This comes from the Green or Cauchy-Green tensor
axk axk
ax; ax;
c = F ~ . F or cij = F ~ ~ F ~ ~ = -- (1.4.21)
Note that we have merely switched the order of the tensor product
from that given in the Finger tensor, but as we will see, in general,
this switch gives us different results. By a similar derivation, (see
Exercise 1.10.4) as given for p, the relative area change, we can
*Note that plp' can be calculated from the determinant of F. Since plp' is a scalar
multiplier, it can be readily carried along if desired (Malvern, 1969).
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use C to calculate the extension ratio Q at any point in the material
by the equation
(1.4.22)
Length, area, and volume change can also be expressed in terms
of the invariants of B or C (see eqs. 1.4.45-1.4.47). Note that the
Cauchy tensor operates on unit vectors that are defined in the past
state. In the next section we will see that the Cauchy tensor is not
as useful as the Finger tensor for describing the stress response at
large strain for an elastic solid. But first we illustrate each tensor
in Example 1.4.2. This example is particularly important because
we will use the results directly in the next section with our neo-
Hookean constitutive equation.
Example 1.4.2 Evaluation of B and C
For the deformations illustrated in Figure 1.4.2 and Example 1.4.1,
evaluate the components of the Finger and the Cauchy deformation
tensors.
Solutions
This is straightforward using the definitions of B (eq. 1.4.13) and
C (eq. 1.4.21) and the results we obtained for Fij in Example 1.4.1.
(a) Uniaxial Extension
Since F = FT, then B = C = F.F = F2 and it does not matter
which deformation measure is used.
(b) Simple Shear
l y 0 1 0 0 1+y2 y 0
B i j = [ O 1 1 O ] = [ yI(1.4.24)
0 0 1 0 0 1
Here we see that B and C do have different components for a
shear deformation. Note that both tensors are symmetric, as they
must be.
ELASTICSOLID / 31
(c) Solid Body Rotation
cos8 -sin8 0 cos8 sin8 0 1 0 0
Bij = sin8 cos8 O ] [ -sin8 cos8 O ] = [0 1 0 1 (1.4.26)
[ o 0 1 0 0 1 0 0 1
Cij also gives the same result, just the identity tensor I.
The last result in Example 1.4.2 says that there is no area or
length change in the sample for the solid body rotation (Le., there
is no deformation). This is what we expect; deformation tensors
should not respond to rotation. They are useful candidates for
constitutive equations, for predicting stress from deformation.
1.4.2 Strain Tensor
So far we have defined deformation in terms of extension, the ra-
tio of deformed to undeformed length, a! = L/L’. Thus when
deformation does not occur, the extensions are unity and B = I.
Frequently deformation is described in terms of strain, the ratio of
change in length to undeformed length
L - L’
L’ c=- = f f - 1 (1.4.27)
When there is no deformation, the strains are zero. A fi-
nite strain tensor can be defined by subtracting the identity tensor
from B
E = B - I (1.4.28)
Thus from Example 1.4.2 the strain tensor in uniaxial extension
becomes
0 ff+l
(2:-1 0
Y O
and in simple shear Eij = [
0 0 0 0 0 1
Since it is often simpler to write the Finger deformation ten-
sor, and since it only differs from the strain tensor by unity, we
usually write constitutive equations in terms of B.
1.4.3 Inverse Deformation Tensors *
As we noted earlier, the stress tensor depends only on the current
state, while the deformation gradient term depends on two states.
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the present x and the past x'. It is a relative tensor. We have defined
the deformation gradient as the change of the present configuration
with respect to the past, dx = F + dx'. However, we can reverse
the process and describe the past state in terms of the present. This
tensor is called the inverse of the deformation gradient; it is like
reversing the tensor machine.
dx' = F-' - dx (1.4.29)
where FIT' = a x ; / a x j
define inverses of the deformation tensors B and C
Using the inverse of the deformation gradient, we can also
and
Physically B-' is like B; it operates on unit vectors in the
deformed or present state n, but instead of area change it gives the
inverse of length change at any point in a material
Similarly C-' operates on unit vectors in the undeformed or past
state of the material to give the inverse of the area change (as defined
in eq. 1.4.14).
(1.4.33)
1 - = 3 . c-' .h'
P2
These results are proven in Exercise 1.10.5.
Thus we have four deformation tensor operators that can de-
scribe local length or area change, eliminating any rotation involved
in the deformation. It is also possible to derive these four tensors
directly from F by breaking it down into a pure deformation and a
pure rotation (Astarita and Marrucci, 1974; Malvern, 1969). Other
deformation tensors can also be defined, but clearly all are derived
from the same information so they are not independent. We can
convert from one to another, although this operation may be diffi-
cult. Which deformation tensor we use in a particular constitutive
equation depends on convenience and on predictions that compare
favorably with real materials.
Example 1.4.3 Evaluation of B-l and C-'
Evaluate the inverse deformation tensors for the deformations given
in Example 1.4.1.
ELASTICSOLID / 33
Solutions
(a) Uniaxial fitension. The inverse of a diagonal matrix is simply
the inverse of the components. Thus,
Verify this by inverting the displacement functions (xi = al 'x l ,
etc.), and determining the components ( B ; ' , etc.) directly from
the definition of B;' in eq. 1.4.30.
(b) Simple Shear: The inverted displacement functions are
x; = X I - yx2
x; = x2
x; = x3
(1.4.35)
(1.4.36)
Cij' = F ; ' ( q y = F , p , ; r ' = (1.4.37)
(c) Solid Body Rotation. N o change: B;' = C-' = Z i j .
I J
1.4.4 Principal Strains
For any state of deformation at a point, we can find three planes on
which there are only normal deformations (tensile or compressive).
As with the stress tensor, the directions of these three planes are
called principal directions and the deformations are called princi-
pal deformations oLi , or principal extensions. Determining of the
principal extensions is an eigenvalue problem comparable to de-
termining the principal stresses in the preceding section. All the
same equations hold. Thus from eq. 1.3.5 principal extensions are
the three roots or eigenvalues of
ct3 - I F C I ~ + I I F U - I I I F = 0 (1.4.38)
where the invariants of the deformation gradient tensor F are the
same as those defined for T, eqs. 1.3.6-1.3.8. To help illustrate
the principal extensions and invariants, consider the following
example.
34 / RHEOLOGY
Example 1.4.4 Principal Extensions and Invariants
of B and C
Determine the principal extensions and the invariants for each of
the tensors B and C given in Example 1.4.2.
Solutions
(a)
obtain
Uniaxial Extension. Taking eq. 1.4.23 and setting a1 = a, we
Here the components of B and C form a diagonal matrix; the princi-
pal directions are already lined up with the laboratory coordinates.
There is no rotation in this deformation. We can find the invariants
by using eqs. 1.3.6-1.3.8
2
Zg = trB = a 2 +```