When performing rheological tests on biological materials, it is important to take into consideration that these kinds of materials usually cannot be deformed evenly in the entire shear gap due to their mostly inhomogeneous and anisotropic structures and superstructures, related to the size of the macroscopic structures in the range of around 0.1 mm to 1 mm. This may lead to conditions at which the reproducibility of test results will be poor, and therefore, these kinds of materials often cannot be analyzed scientifically in terms of absolute values (see also Chapter 3.3.4.2c: inhomogeneous, “plastic” behavior).
5.1.2.1.20h) Haptic perception (haptics) of the consistency of a material, or of a system
Having direct contact to a material, e. g. by holding it in hands or in the mouth, consciously, but also unconsciously, we expect a certain sense of touch, a certain sensory impression or even excitation.
Examples: soft materials in the interior of a car showing the soft touch effect, handshake, keystroke; chewing gum, bread, curd, ice cream. More about haptics, see the Note in Chapter 8.3.2 (structural character, amplitude test); and about the tactile perception of a surface, see application example (d) in Chapter 10.8.4.3 (tribology).
Summary: Behavior of viscoelastic solids
Desired behavior of viscoelastic materials can be obtained only if the viscous and elastic portions are in a well-balanced ratio.
Example: Bamboo
Required behavior is given for bamboo bars only if both, flexibility (viscous behavior) and rigidity (elastic behavior) are present, showing the right mixture of both components. Only in this case, the material may show the optimal (viscoelastic) behavior. By the way: Also in East Asian nature philosophy, bamboo is an illustrative example to characterize balanced behavior or properties.
5.3Normal stresses
This chapter is intended for giving merely in a nutshell some basic information on normal stress tests and on the corresponding terminology in this special field of rheology.
If a viscoelastic material is deformed, there are not only one-dimensional forces or stresses acting in the main direction of shear deformation. Due to its elastic proportion, a material will permanently show a certain ability of molecular or structural re-formation or recovery, already during the deformation process. Besides the shear stress τ = F / A (see Two-Plates model, Figures 2.1 and 4.1), so-called normal stresses are occurring, which have an effect in perpendicular direction to the shear stress. The result is a state of three-dimensional deformation. This can be illustrated using a (3 × 3) tensor (see also Chapter 14.2: A. L. Cauchy, 1827 [5.19], and J. C. Maxwell, 1855 [5.1]). It contains nine values, which can be displayed in a three-dimensional Cartesian coordinate system. In order to explain this, we can imagine the behavior of an infinitely (infinitesimal) small cube-shaped volume element of our test material being within the shear gap (Figure 5.8).
Taking the shear direction or deformation direction as x-direction; in the Two-Plates model the x-axis is pointing to the right (see Chapter 4.2, Figure 4.1). The y-direction is the direction of the shear gradient; in the Two-Plates model the y-axis is pointing upwards. The z-direction is called the indifferent or neutral direction; in the Two-Plates model the z-axis is pointing out of the writing plane, towards the reader.
Figure 5.8: An infinitesimal small volume element of the sample, and the effective directions in the Cartesian coordinate system:
x as shear direction, y as direction of the shear gradient, and z as neutral direction (with the z-axis pointing perpendicular out of the page)
Figure 5.9: An infinitesimal small volume element of the sample, shear stress τ and the three
normal stresses τxx, τyy and τzz
(the latter works vertically out of the picture
surface)
The stress tensor indicates the following stress values:
1 Into x-direction:τxx, τyx, τzx
2 Into y-direction:τxy, τyy, τzy
3 Into z-direction:τxz, τyz, τzz
The first index of the stress tensor values indicates the position of the area of the cube-shaped volume element on which the stress is acting. The term normal direction is used in mathematics and physics to determine the position of an area. For example here, the area with the x-direction as the normal direction means, it is the area on which the x-coordinate is standing in a right-angled position. The normal force or normal stress, respectively, acts in the direction of the surface-normal standing vertically on the regarded area, and it is presented in the form of a vector (arrow). The second index of the stress sensor values indicates the direction of the force or stress acting on the area which is described by the first index. Thus, for the x-direction counts here, it is pointing to the right side.
The tensor contains the three normal stresses τxx, τyy and τzz (see Figure 5.9). For most of the viscoelastic liquids, the τxx component clearly shows the highest value of the three normal stresses. The other six tensor stresses are shear stresses. If a liquid is flowing, i. e. if viscous behavior clearly dominates, these six stress values can be reduced to the τyx component only, and the other five stress values can be ignored. This is the reason why in this textbook, beside of the chapter at hand, there is mentioned this one τ-value only, and therefore here, it can be written without any index.
The shear stress τyx = F / A = τ is acting in x-direction on an area which is parallel to the plates of the Two-Plates model, and the normal direction of the corresponding area is the y-direction. We can imagine this area as an individual flowing layer. An illustrative example is a stack of beer mats (see Experiment 2.1 and Figure 2.2 of Chapter 2.2). Here, each individual beer mat represents an area with the y-direction as the normal direction, and each beer mat is moving into x-direction when applying a shear force F.
Usually, the following curve functions are presented in a diagram, typically on a double-
logarithmic scale.
1 The 1 st normal stress difference N1( γ ̇ ), as a function of the shear rate
Equation 5.3
N1 [Pa] = τxx – τyy
1 The 2 nd normal stress difference, as N2( γ ̇ )
Equation 5.4
N2 [Pa] = τyy – τzz
1 The 1 st normal stress coefficient, as ψ1( γ ̇ ); psi, pronounced: “psee” or “sy”
Equation 5.5
ψ1 [Pa ⋅ s2] = N1 / γ ̇ 2
showing the 1st zero-normal stress coefficient as a plateau value in the low-shear range
Equation 5.6
ψ1,0 [Pa ⋅ s2] = lim γ ̇ → 0 ψ1( γ ̇ ) = const
1 The 2 nd normal stress coefficient, as ψ2( γ ̇ )
Equation 5.7
ψ2