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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 - -) (Y ( 1.1.2) where T1 I is the tensile force divided by the area a, which it acts upon. 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 Pa. ( 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 = - (1.1.4) L L' ( 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) 6 / RHEOLOGY 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 = - x2 ( 1.1.6) Strain y = s / x z while the shear stress is linear in shear strain with the same coeffi- cient 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=0.4 y= 0 T22 - 0.4 - 0.2 0.0 0.2 0.4 y= -0.4 ELASTICSOLID I 7 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). 8 / RHEOLOGY 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. X 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 body. 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