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Macosko, Christopher W - Rheology - Principles, Measurements and Applications-John Wiley & Sons (1994)

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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). 
30 / RHEOLOQY 
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. 
*The reader may skip to Section 1.5 on aprst reading. 
32 / RHEOLOGY 
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 +