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

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a new constant 
of proportionality. Thus his constant was not uniquely a material 
property but also depended on the particular geometry of the sam- 
ple. To find the true material constant-the elastic modulus-of 
his wires, Hooke needed to develop the concepts of stress, force 
per unit area, and strain. Stress and strain are key concepts for 
rheology and are the main subjects of this chapter. 
If crosslinked rubber had been available in 1678, Hooke 
might well have also tried rubber bands in his experiments. If 
so he would have drawn different conclusions. Figure 1.1.2 shows 
results for a rubber sample tested in tension and in compression. 
We see that for small deformations near zero the stress is linear 
with deformation, but at large deformation the stress is larger than 
is predicted by Hooke's law. A relation that fits the data reasonably 
well is 
1 TII = G(aZ - -) 
( 1.1.2) 
where T1 I is the tensile force divided by the area a, which it acts 
f Figure 1.1.2. 
stress versus extension ra- a 
tio for a rubber sample. (b) 
Schematic diagram of the de- 
formation, Data from Treloar 
(1975) on sulfur-vulcanized 
natural rubber. Solid line is 
( 1.1.3) 
The extension ratio (Y is defined as the length of the deformed 
sample divided by the length of the undeformed one: 
(a) Tensile and compressive Ti1 = - 
( Y = - 
eq. 1.1.2 with G = 3.9 x lo5 
undefonned 15 - 
.- 8 
I t - L ' +I 
compression T I 1 -bTll 
4 L b - 
(a> (b) 
Figure 1.13. 
(a) Shear and normal stresses 
versus shear strain for a sili- 
cone rubber sample subject to 
simple shear shown schemati- 
cally in (b). The open points 
indicate the normal stress 
difference TII - T22 neces- 
sary to keep the block at con- 
stant thickness x2, while the 
solid points are for the shear 
stress. Notice that the nor- 
mal stress stays positive when 
the shear changes sign. Data 
are for torsion of a cylinder 
(DeGroot, 1990; see also 
Example 1.7.1). 
Figure 1.1.3 shows the results of a different kind of experi- 
ment on a similar rubber sample. Here the sample is sheared be- 
tween two parallel plates maintained at the same separation x2. We 
see that the shear stress is linear with the strain over quite a wide 
range; however, additional stress components, normal stresses TI I 
and T22, act on the block at large strain. In the introduction to this 
part of the text, we saw that elastic liquids can also generate nor- 
mal stresses (Figure 1.3). In rubber, the normal stress difference 
depends on the shear strain squared 
T I I - T22 = G y 2 (1.1.5) 
where the shear strain is defined as displacement of the top surface 
of the block over its thickness 
S y = - 
( 1.1.6) 
y = s / x z 
while the shear stress is linear in shear strain with the same coeffi- 
T21 = GY (1.1.7) 
These apparently quite different results for different defor- 
mations of the same sample can be shown to come from Hooke’s 
law when it is written properly in three dimensions. We will do this 
in the next several sections of this chapter, calling on a few ideas 
from vector algebra, mainly the vector summation and the dot or 
scalar products. For a good review of vector algebra Bird et al. 
(1987a, Appendix A), Malvern (1969) or Spiegel (1968) is help- 
ful. In the following sections we develop the idea of a tensor and 
some basic notions of continuum mechanics. It is a very simple 
y= 0 
T22 - 0.4 - 0.2 0.0 0.2 0.4 
y= -0.4 
development, yet adequate for the rest of this text and for start- 
ing to read other rheological literature. More detailed studies of 
continuum mechanics can be found in the references above and in 
books by Astarita and Marmcci (1974), Billington and Tate (198 I), 
Chadwick (1976), and Lodge (1964, 1974). 
1.2 The Stress Tensor * 
To help us see how both shear and normal stresses can act in a 
material, consider the body shown in Figure 1.2.la. Let us cut 
through a point P in the body with a plane. We identify the direction 
of a plane by the vector acting normal to it, in this case the unit vector 
fi. If there are forces acting on the body, a force component f,, will 
act on the cutting plane at point P. In general f,, and fi will have 
different directions. If we divide the force by a small area d a of the 
cut surface around point P, then we have the stress or traction vector 
tn per unit area acting on the surface at point P. Figure 1.2.1 b shows 
a cut that leacjs to a normal stress, while Figure 1.2. l c shows another 
that gives a shear stress tm. Note that Figure 1.2.1 shows two stress 
vectors of the same magnitude acting in opposite directions. This is 
required by Newton’s law of motion to keep the body at rest. Both 
vectors are manifestations of the same stress component. In the 
discussion that follows we usually show the positive vector only. 
As we have seen in Figures 1.1.2 and 1.1.3, materials may 
respond differently in shear and tension, so it is useful to break 
the stress vector tn into components that act normal (tensile) to the 
plane fi and those that act tangent or shear to the plane. If we pick 
a Cartesian coordinate system with one direction fi, the other two 
directions m and 6 will lie in the plane. Thus, t , is the vector sum 
of three stress components. 
t, = fiT,,,, + mTnm + 6Tn0 (1.2.1) 
We designate the magnitude of these stress components with a capi- 
tal T and use two subscripts to identify each one. The first subscript 
refers to the plane on which the components are acting; the sec- 
ond indicates the direction of the component on that plane. If we 
take another cut, say with a normal vector m, through the same 
point in the body, then the stress vector acting on m will be tm with 
components T m m , Tmo, and T m n . 
So what we have now is a logical notation for describing 
the normal and shear stresses acting on any surface. But will it 
be necessary to pass an infinite number of planes through P to 
*Many students with engineering or physics backgrounds are already familiar with 
the stress tensor. They may skip ahead to the next section. The key concepts in this 
section are understanding ( 1 ) that tensors can operate on vectors (eq. 1.2.10), ( 2 ) 
standard index notation (eq. 1.2.21), (3) symmetry of the stress tensor (eq. 1.2.37), 
( 4 ) the concept ofpressure (eq. 1.2.44), and (5) normal stress differences (eq. 1.2.45). 
Figure 1.2.1. 
(a) A force acting on a body. 
(b) A cut through point P 
nearly perpendicular to the 
direction of force . The nor- 
mal to the plane of this cut 
is ii. The stress on this plane 
is t,, = flu, where a equals 
the area of the cut. (c) An- 
other cut nearly parallel to 
the force direction. The equal 
and opposite forces acting at 
point P are represented by 
the single component t,,,. 
Figure 1.2.2. 
(a) Three mutually perpen- 
dicular planes intersecting at 
the point P with their associ- 
ated stress vectors. (b) Stress 
components acting on each of 
these planes. (c) A plane ii 
is cut across the three planes 
to form a tetrahedron. As in 
Figure 1.2.1, t,, is the stress 
vector acting on this plane 
with area a,,. For any plane 
6, tIz can be determined from 
the components on the three 
perpendicular planes. 
characterize the state of stress at this point? No, because in fact, 
the stresses acting on all the different planes are related. The stress 
on any plane through P can be determined from a quantity called the 
stress tensor. The stress tensor is a special mathematical operator 
that can be used to describe the state of stress at any point in the 
To help visualize the stress tensor, let us set up three mutually 
perpendicular planes in the body near point P, as shown in Figure 
1.2.2a. Let the normals to each plane be f, f , and 2 respectively. 
On each plane there will be a stress vector. These planes will form 
the Cartesian coordinate system f, 9 , i. As shown in Figure