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```diflerence and N2 the second
normul stress diflerence. Some authors use the difference TII -
T33. However, there are only two independent quantities because
TI - T33 is just the sum of the other two. The reader should also
be aware that other notations for stress are common: P or II for T
and u or T’ for T. Also, several authors use the opposite sign for
T and T (See Bird et al., 1987a, p. 7, who consider compression,
eq. 1.2.40, to be positive).
It is perhaps consoling to the student struggling with the stress
tensor to learn that although Hooke wrote his force extension law
before 1700, it took many small and painful steps until Cauchy in
the 1820s was able to write the full three-dimensional state of stress
at a point in a material.
1.3 Principal Stresses and Invariants *
Later in this and subsequent chapters we will want to make consti-
tutive equations independent of the coordinate system. In particular
we will need to make scalar rheological parameters like the modu-
lus or viscosity a function of a tensor.
How can a scalar depend on a tensor? Let us start by con-
sidering a simpler but similar problem: How does a scalar depend
on a vector? In particular, consider how scalar kinetic energy de-
pends on the vector velocity. Recall the equation for kinetic energy
E K = 1/2mu2, where u2 = v v. Kinetic energy is a function
of the dot or scalar product of the velocity vector, the magnitude
of the vector squared. Thus, v v is independent of the coordinate
system; it is the invariant of the vector v.
There is only one commonly used invariant of a vector: its
magnitude. However there are three possible invariant scalar func-
tions of a tensor. For the stress tensor we can give these three
invariants physical meaning through the principal stresses.
It is always possible to take a special cut through a body such
that only a normal stress acts on the plane through the point P. This
is called a principal plane, and the stress acting on it is a principal
stress u. As demonstrated below, there are three of these planes
through any point and three principal stresses.
We can visualize the principal stresses in terms of a stress
ellipsoid. The surface of th is ellipsoid is found by the locus of
the end of the traction vector t, from P when fi takes all possible
directions. The three axes of the ellipsoid are the three principal
used in Section 1.6.
20 / RHEOLOGY
Figure 1.3.1.
(a) A section of the stress
ellipsoid at P through two
principal axes a,& and a&2.
(b) The stress ellipsoid for a
hydrostatic state of stress.
stresses and their directions the principal directions. A section of
such an ellipsoid through two of the axes is shown in Figure 1.3.1.
Note that in the simplest case all the principal stresses are
equal: u1 = u2 = 0 3 = u. This equivalence represents the hydro-
static pressure p = -0. As we saw at the end of the Section 1.2.3,
a hydrostatic state is the only kind of stress that can exist in a fluid
at rest.
If we line up our coordinate system with the three principal
stresses, all the shear components in the stress tensor will vanish.
This is nice because it reduces the stress tensor to just three diagonal
components:
components of the principal stress tensor = T;
0 0
0 0 u3
(1.3.1)
However, in practice it is often difficult to figure out the rota-
tions of the coordinate system at every point in the material, so as
to line it up with the principal directions. Furthermore, it is usually
more convenient to leave the coordinates in the laboratory frame.
Thus, we normally do not measure the principal stresses (except for
purely extensional deformations) but rather calculate them from the
measured stress tensor.* We show this next.
Because a principal plane is defined as one on which there is
only a normal stress, the traction vector and the unit normal to that
plane must be in the same direction:
t, =ah (1.3.2)
*An exception ispow birefringence where differences in the principal stresses and
their angle of rotation are measured directly; see Section 9.4.
ELASTICSOLID / 21
Thus, a is the magnitude of the principal stress and h its
direction. As we saw earlier (eq. 1.2.10), the stress tensor is the
machine that gives us the traction vector on any plane through the
dot operation. Thus,
t ,= f i .T=af i (1.3.3)
This equation can be rearranged to give
fi . (T - 01) = 0 or ni(Tij - aZij) = 0 (1.3.4)
Since fi is not zero, to solve this equation we need to find values
of a such that the determinant of T - a1 vanishes. This is usually
called an eigenvalue problem.
TI1 - 0 TI2 TI 3
[ T31 T32 T33 - det(T-aI) = det T2l T22 -a T23 ] = 0
Expanding t h i s determinant yields the characteristic equation of the
matrix
a3 - I T a 2 + IITCJ - IIIT = 0 (1.3.5)
where the coefficients are
IT is called the first invariant of the tensor T, IIT the second invari-
ant, and IIIT the third invariant. They are called invariants because
no matter what coordinate systems we choose to express T, they
will retain the same value. We will see that this property is par-
ticularly helpful in writing constitutive equations. Note that other
combinations of I;:j can be used to define invariants (cf. Bird et al.,
1987a, p. 568).
Equation 1.3.5 is a cubic and will have three roots, the eigen-
values 01, a2, and 0 3 . If the tensor is symmetric all these roots will
be real. The roots are then the principal values of Tij and ni, the
principal directions. With them T can be transformed to a new
tensor such that it will have only three diagonal components, the
principal stress tensor, eq. 1.3.1.
To help illustrate the use of eq. 1.3.5 to determine the principal
stresses, consider Example 1.3.1.
22 I RHEOLOGY
Example 1.3.1 Principal Stresses and Invariants
Determine the invariants and the magnitudes and directions of the
principal stresses for the stress tensor given in Example 1.2.2.
Check the values for the invariants using the principal stress mag-
nitudes.
Solution
For eq. 1.2.32 we obtain
Using eqs. 1.3.6-1.3.8 we can calculate the invariants
IT = trT = 7
I ZZT = -(Z; - trT2) = 14
2
111~ = detT = 8
( 1.3.10)
From eq. 1.3.5 we can find the principal stress magnitudes:
a3 - 7a2 + 14a - 8 = 0, which factors into
(a - l ) (a - 2)(a - 4) = 0
Thus
a1=1 a 2 = 2 a 3 = 4 (1.3.11)
Clearly most cases will not factor so easily, but the cubic can be
solved by simple numerical methods. We can check the values for
the invariants using these ai :
IT = + f a3 = 7
IIT = a102 + a1a3 -b 0 2 0 3 = 14 (1.3.12)
IIIT = a l ~ 2 ~ 3 = 8
To obtain the principal directions, we seek r,(i), which are
solutions to
For each principal magnitude eq. 1.3.13 results in three equations
for the three components of each principal direction.
These three sets of equations are solved for directions of unit length
as follows:
Thus the principal directions are
where n(’) is rotated + 45” from the i 2 axis.*
1.4 Finite Deformation Tensors
Now that we have a way to determine the state of stress at any point
in a material by using the stress tensor, we need a similar meas-
ure of deformation to complete our three-dimensional constitutive
equation for elastic solids.
Consider the small lump of material shown in Figure 1.4.1.
We have drawn a cube, but any lump will do. P is a point embed-
ded in the body and Q is a neighboring point separated by a small
distance dx’. Note that dx’ is a vector. The area vector da‘ repre-
sents a small patch of area around Q. We use the ’ to denote the rest
or reference state of the material; or, if the material is continually
deforming, the ’ denotes the state of the material at some past time,
t’. From here on we concentrate on deformations from a rest state.
In the following chapters we treat continual deformation with time.
Now let the body```