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

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fields to characterize complex materials. 
(b) From eq. 1.2.10 the stress vector t,, acting on the plane whose 
normal is fi is just t,, = fi T, where ii = cos 821 + sin O i 2 . 
Using matrix multiplication gives 
f/a 0 0 f / a cos e 
t,, =fi-T=(cose,sine,O) 0 0 0 
[ o o J = [ 8 1 
= (i) coseil (1.2.26) 
The normal stress on the fi plane is just the projection of tn on fi 
The shear stress comes from vector subtraction 
t,, - Tnni = Tn,i (1.2.28) 
where i is the vector in the % I f 2 plane tangent to the plane 6. 
Substituting on the left-hand side gives 
(~cose ,o ,o ) - ( f . cos3e , -cos28sin8,0 f ) = (f cos 8 sin2 8 , -- f cos2 8 sine, 0 
a a a 
Since i = sin 812, - cos e i 2 , then 
T,,, = (i) cos 8 sin 8 (1.2.30) 
This shear stress can be important in failure. For example, if a 
certain crystal plane in a material has a lower slip or yield stress, 
this stress may be exceeded although the tensile strength between 
the planes may not. 
Example 1.2.2 Stress on a Surface 
Measurements of force per unit area were made on three mutually 
perpendicular test surfaces at point P , (Figure 1.2.2a), with the 
following results: 
Direction of Vector ti 
Normal to Test Surface Measured Force/Area { P a = k N / m 2 ) 
322 - 23 
- %* + 3%3 
(a) What is the state of stress at P? 
(b) Find the magnitude of the stress vector acting on a surface 
whose normal is 
(c) What is the normal stress acting on this interface? 
(a) The state of stress at a point is determined by the stress tensor 
(eq. 1.2.24) 
T = f i t , + %2t2 + i 3 t 3 
where the measured tractions & are 
(b) From q. 1.2.10 we know that the stress vector tn acting on 
the plane whose normal is fi is just 
& =ri .T (1.2.10) 
Using matrix multiplication (eq. 1.2.14) we find 
Remembering the omitted unit vectors, we write 
The magnitude of this stress vector is 
(c) The normal stress Tnn is just the projection of &, onto the unit 
Ten = i ( l , 1, 0)’- L:,] = 2Pa (1.2.35) Jz 
1.2.2 Symmetry 
Notice that the stress tensor in each of the examples above is sym- 
metric; that is, the rows and columns of the matrix for the compo- 
nents of T can be interchanged without changing T. The compo- 
nents of the traction vectors ti were picked that way intentionally. 
The symmetry of the stress tensor can be shown by considering the 
Figure 1.2.5. 
To balance angular momen- 
tum about the 21 axis, the 
two shear stress components 
T32 and T23 acting on the 
tetrahedron must be equal. 
shear stresses acting on the small tetrahedron sketched in Figure 
1.2.5. The component T23 gives rise to a moment about the XI axis. 
To conserve angular momentum, this moment must be balanced 
by the one caused by T32. Thus, T32 = T23, and by similar argu- 
ments the other pairs of shear components T12 = T21 and TI3 = T31 
are equal. Thus, the stress tensor is symmetric with only six in- 
dependent components, which when written in matrix form (again 
leaving out the dyads), become 
In Gibbs notation, we show that a tensor is symmetric by 
T = TT (1.2.37) 
where TT is called the rrunspose of T. In the tensor TT the scalar 
components of the rows and columns of T have been interchanged. 
This interchange may be clearer when TT is written in index nota- 
TT = xi ( f i % j T ~ j ) ~ = xi cj 12jkjTji (1.2.38) 
The transpose has wider utility in tensor analysis. For exam- 
ple, we can use it to reverse the order of operations in the vector 
product of eq. 1.2.10 
tn = TT .ii (1.2.39) 
This result can be verified by using matrix multiplication. Try it 
yourself. Follow eq. 1.2.14, switch rows with columns in T, and 
make 6 a column vector on the other side. Of course, in the end 
this manipulation does not matter for T because it is symmetric, 
but we will find the operation useful later. 
i 2 i 
Figure 1.2.6. 
A hydrostatic state of stress 
on a tetrahedron of fluid. 
The possibility of a nonsymmetrical stress tensor is discussed 
by Dahler and Scriven (1961, 1963). Asymmetry has not been 
observed experimentally for amorphous liquids. Body torques do 
exist on suspension particles, but these can be treated by calculating 
the stress distribution over the particle surface for each orientation 
(see Chapter 10). 
1.2.3 Pressure 
One particularly simple stress tensor is that of uniform pressure. A 
fluid is a material that cannot support a shear stress without flowing. 
When a fluid is at rest, it can support only a uniform normal stress, 
TI I = T22 = T33, as indicated in Figure 1.2.6. This normal stress 
is called the hydrostatic pressure p . Thus, for a fluid at rest, the 
stress tensor is 
0 1 0 0 
0 0 1 
Tij = [ -p - p :] = - p [ 0 1 0 1 (1.2.40) 
0 -P 
where the minus sign is used because compression is usually con- 
sidered to be negative. 
The matrix or tensor with all ones on the diagonal given in 
eq. 1.2.40 has a special name. It is called the identity or unit tensor. 
When multiplied by another tensor, it always generates the same 
tensor back again: 
T . L = T (1.2.41) 
The Gibbs notation for the identity tensor is I. Its components are 
Thus, the stress tensor for a fluid at rest is 
T = -PI (1 -2.43) 
When dealing with fluids in motion, it is convenient to retain 
p . Thus, we write the total stress tensor as the sum of two parts 
T = - p I + r (1.2.44) 
where r is known as the extra or viscous stress tensor.* Often T is 
referred to as the toral stress tensor and r as just the stress tensor. 
In rheology we generally assume that a material is incom- 
pressible, The deviations from simple Hookean or Newtonian be- 
havior due to nonlinear dependence on deformation or deformation 
history are usually much greater than the influence of compress- 
ibility. We discuss the influence of pressure briefly in Chapters 2 
and 6. For incompressible materials the overall pressure cannot 
influence material behavior. In other words, increasing the baro- 
metric pressure in the room should not change the reading from a 
rheometer. For incompressible materials the isotropic pressure is 
determined solely by the boundary conditions and the equations of 
motion (see Sections 1.7 and 1.8). 
Thus, it makes sense for incompressible materials to subtract 
p . The remaining stress tensor I contains all the effects of defor- 
mation on a material. Constitutive equations are usually written in 
terms of z. However, experimentally we can measure only forces 
which, when divided by the area, give components of the total stress 
T. Since T includes p and r , we would like to remove the pressure 
term from r . This presents no problem for the shear stress compo- 
nents (because T12 = 712, etc.), but the normal stress components 
will differ by p; T;; = - p + ~ i ; . As we said, determination of 
p requires boundary conditions for the particular problem. Thus, 
normal stress difSerences are used to eliminate p since 
TII - T22 = t i 1 - 722 
T22 - T33 =r22 - 733 
As an example of how we use the normal stress difference, 
consider the simple uniaxial extension shown in Figure 1.1.2b. The 
figure shows that tension acting on the f 2 l faces of the rubber cube 
will extend it. However, as shown in Figure 1.2.7, a compression, 
-T22 = -T33, on the f22 and f23 faces could generate the same 
deformation. The rubber cube does not know the difference. The 
deformation is caused by Ti, - T22, the net difference between 
tension in the 21 direction and the compression in the 22 direction. 
Figure 1.2.7. 
Uniaxial extension generated 
by a uniform compression. 
- - J 7-22 
*r is an exception to the general rule for using boldface Latin capital letters for 
tensors. Howeves it is in such common use in rheology that we retain it here, 
For the special case of simple shear we use 
We call NI thejrs t normul stress