Normal component of the stress tensor

To derive a molecular expression for the stress tensor component which is the basic quantity if one wishes to compute pseudo-experimental data F (/i) /R, we start from the thermodynamic expression for the exact differential given in Eq. (1.59), where the strain tensor r is given in Eq. (1.41). From these two expressions, it follows that 

the symmetry of the force expression provides another useful dieck on the simulations. The accuracy to be expected is illustrated by entries in Table 5.1. 

To illustrate the relation between microscopic structure and experimentally accessible information, we focus on the computation of pseudo-experimental solvation-force curv es F h) /R [see Eqs. (5.57), (5.59), (5.63), and (E.46)] as they would be determined in SEA experiments. However, here these curves are computed from computer simulation data for and where P, is 

Plots of / ( z) aiid F (/ .) /R versus Sz and h, respectively, are shown in Fig. 5.3. The oscillatory decay of both quantities is a direct consequence of the oscillatory dependence of on Sz, which has also been investigated by integral equations of varying degree of sophistication [157-161]. As can be seen in Fig. 5.3, zeros of f (s ) correspond to successive extrema of F (h) /R 

Alternatively, one may employ colloidal probe atomic force microscopy (AFM) to measure force distance curves such as the ones plotted in Fig. 5.1 [162]. The important difference between SFA and colloidal probe AFM experiments is that in the latter the entire force distance curve is accessible rather than only that portion satisfying Eq. (5.66) [163, 164]. In Ref. 164 a comparison is presented between theoretical and experimental data for confined poly-electrolyt.e systems. 

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The above governing equations are supplemented by initial conditions and boundary conditions on the cluster-macro void interface T js. Denote N the unit normal exterior to Cls. continuity of mass, concentrations, streaming potentials, total flux of the species and the normal component of the stress tensor give (where tpf = (RTtp /F))... 

For isotropic materials, a general relationship correlating the three normal components of the stress tensor and strain is expressed as... 

To solve the equations for the pressure and velocity of the fluid, one must specify boundary conditions. Usually one assumes that the fluid sticks to a solid wall, so that v(r, t) = 0 when r is on the solid surface. (This no-slip boundary condition may not be completely accurate at microscopic length scales.) The other boundary condition that is often important is that flows at infinity are unperturbed by the boundaries. Finally, at surfaces or interfaces, there is continuity of the normal and tangential forces. The force on a surface is related to the normal component of the stress tensor, Eq. (1.143) the ith component of the force per unit area, fs = dFs/dS, obeys... 

Significant wall force field effects are found on the velocity, density, shear stress distributions in the near-wall region that extends approximately three molecular diameters from each surface for van der Waals interactions. Within this wall force penetration depth, a density buildup with a single peak point is observed. Exactly matching the earlier static case, normal components of the stress tensor are anisotropic and... 

We want to work this out in three simple cases. First we consider a homogeneous dilatation of a cubic volume V = LxLyLz-We also assume that the shear components of the stress tensor vanish, i.e. Oafi = 0 for a In such a system the normal components of the stress tensor should all be the same, i.e. ct = ctxx = [Pg.5]

Fig. F.l. Normal stress and normal strain, (a) The normal force per unit area in the X direction is the x component of the stress tensor, (b) The normal stress causes an elongation in the x direction and a contraction in the y, and z directions.

For an element in equilibrium with no body forces, the equations of equilibrium were obtained by Lame and Clapeyron (1831). Consider the stresses in a cubic element in equilibrium as shown in Fig. 2.3. Denote 7y as a component of the stress tensor T acting on a plane whose normal is in the direction of e and the resulting force is in the direction of ej. In the Cartesian coordinates in Fig. 2.3, the total force on the pair of element surfaces whose normal vectors are in the direction of ex can be given by... 

Hence, only six independent components of the stress tensor are needed to fully define the state of the stress at point P, where n u are the normal stress components, and it (i / j) are the shear-stress components. 

Then, equation (9.5) defines the non-zero components of the stress tensor, which makes it possible to formulate expressions for the shear viscosity and the differences between the normal stresses ... 

Here are the components of the stress tensor as defined in rheology Tn—T22 is the first normal stress difference and T21 the shear stress, equal to Nt and rxsh, respectively. Hence, from dynamic mechanical measurements it is possible to determine the zero shear first normal stress coefficient Fq0 and zero shear viscosity y0. 

In a shearing flow (Fig. 1-16) of an incompressible isotropic liquid, the stress tensor contains at least two nonzero components, cti2 = oix, as well as an isotropic pressure term. If the fluid is Newtonian, these are the only nonzero components of the stress tensor. However, a non-Newtonian liquid in general has other nonzero components of a, namely the normal stresses, C, 022, and (T33 (Weissenberg 1947). Since the stress tensor is only determined to within an additive isotropic tensor, only the normal stress dijferences N = — 022... 

This scaling law, Eq. (9-48), implies that all components of the stress tensor are linear in the shear rate. Consider for example, a constant-shear-rate experiment. At steady state, not only is the shear stress predicted to be proportional to the shear rate, but so also is the first normal stress difference N This prediction has been nicely confirmed in recent experiments by Takahashi et al. (1994), who studied mixtures of silicon oil and hydrocarbon-formaldehyde resin. Both these fluids are Newtonian, and have the same viscosity, around 10 Pa s. Figure 9-18 shows that both the shear stress o and the first normal stress difference N = shear rate, so that the shear viscosity rj = aly and the so-called normal viscosity rjn = N /y are constants. The first normal stress difference in this mixture must be attributed entirely to the presence of interfaces, since the individual liquids in the mixture have no measurable normal stresses. A portion of the shear stress also comes from the interfacial stress. Figure 9-19 shows that the shear and normal viscosities are both maximized at a component ratio of roughly 50 50. At this component ratio, the interfacial term accounts for roughly half the total shear stress. 

Normal stress differences can be observed in Couette flow, cone-plate and plate-plate geometries, and capillary flow. The only nonzero components of the stress tensor in coaxial cylinders are a. e(r), cSrr(f), CTee(r), and... 

If the cross-sectional area of the macroscopic network is doubled, then twice as large a force is required to obtain the same deformation. This leads naturally to a definition of stress as the ratio of force and cross-sectional area. Both the force and the cross-sectional are have direction and magnitude (the direction of the cross-sectional area being described by the unit vector normal to its surface), making the stress a tensor. The (/-component of the stress tensor is the force applied in the i direction per unit cross-sectional area of a network perpendicular to the j axis. For... 

To evaluate the normal component of the stress balance, we must first evaluate V n. This is slightly more subtle than it may seem. Because we are calculating everything in general vector/tensor form, we note that the surface gradient operator in (8-61) can be expressed in the form... 

Equation 31a is the steady state form of the kinematic condition which also is used to describe wave motion as discussed on page 595 of Levich s book, Physicochemical Hydrodynamics. One also must equate the normal and tangential components of the forces in each phase at the free surface. Since we consider a gas-liquid interface, we neglect gas phase resistance due to its viscosity and include only the pressure it imposes on the interface on the gas side. Therefore, at y - -h(x) the tangential component of the stress tensor for the liquid phase is equal to the tangential force created by the change in surface tension with temperature in the x direction. Thus the tangential force balance at the interface becomes... 

In a dynamic experiment, a small-amplitude oscillatory shear is imposed to a molten polymer confined in the rheometer. The shear stress response of the polymeric system can be expressed as in Equation 22.14. In this equation, G and G" are dynamic moduli related to the elastic storage energy and dissipated energy of the system, respectively. For a viscoelastic fluid, two independent normal stress differences, namely, first and second normal stress differences can be defined. These quantities are calculated in terms of the differences of the components of the stress tensor, as indicated in Equation 22.15a and 22.15b, and can be obtained, for instance, from the radial pressure distribution in a cone-and-plate rheometer [5]. Some other experiments used in the determination of the normal stress differences can be found elsewhere [9, 22] ... 

In our case of two infinite surfaces which are separated by a medium that transmits stress (e.g., the charged fluid discussed earlier), the force per unit area in the z direction between two rigid surfaces of area A whose normals are in the z and —z directions, is therefore given by /7zz(z = D) = ITc(D) where the subscript i denotes that this is the longitudinal component of the stress tensor. The transverse component of the stress tensor, /7,(z) = TIxx(z) = Hyy(z) (for a system that is isotropic in the plane perpendicular to the z direction), is related to the force in the Jc, y direction for displacements that increase the cross-sectional area of the system. The relevant normal vectors are also in... 

For the gas velocity u, the wall boundary conditions can be of free slip, no-slip, or turbulent law-of-the-wall type. The free-slip condition demands that the normal velocity components of the fluid and the wall coincide, while the tangential components of the stress tensor satisfy o = slip boundary conditions require the fluid velocity to coincide with the wall velocity. No-slip boundary conditions impose large velocity gradients, which, because of lack of sufficient computational resources, cannot be resolved. Therefore, wall functions are used. The following discussion reflects the exhibition given by Amsden et al. [4]. 

One component of stress is normal to the surface and represents the normal stress. The other two components are tangential to the surface and represent the shear stresses. In Figure 7.1 the stress tensor Otj for a body of a cubic shape is shown. The first subscript i indicates direction of the surface normal, along which the stress acts. The second subscriptj indicates direction of the stress component. Thus, Oxy means the component of stress apphed along the normal to y Oz plane and acting along y axis, o x, Oyy, o x are the normal components of the stress, the rest components are the shear ones. 

Fig. 8.2 Vector of force F acting on a body and its tangential (f, fy, shear) and normal (/z, pressure) components (a) and the nine components of the stress tensor (b). Note that the surface element vector is directed outward from the bulk...

In Eqs. (1-5), the vector x and the scalar t denote spatial coordinates (x ) and time, respectively. The vector u signifies the displacement vector with contravariant components u, the components s denote the contavariant components of the stress tensor, and the vectors , Uo, and vo are prescribed functions. The vectors u(x, f) and t(x, t) represent time dependent prescribed boundary conditions on the parts T and of the boundary F, respectively, and p denotes mass density. Finally, n, signify the components of the outward unit normal to F(. It should be noted that this set of equations is supplemented by the equilibrium of angular momentum (a generalized symmetry condition on the stress tensor), the material law, and the kinematic relationships between strains/rigid rotations and the spatial derivatives of displacements. 

We already saw in section 2.2.2 that a load that causes a normal strain in its direction also causes transversal normal strains. For example, a stress in xi direction, cth, causes the following strains, according to equations (2.13) and (2.14) n = a /E, 22 = 33 = —vawjE. One component of the stress tensor a thus acts on several components of the strain tensor e. Similarly, a prescribed strain in one direction may change the stresses in other directions. If we restrict ourselves to small deformations, the relation between stress and strain is linear. Mathematically, an arbitrary linear relation between two tensors of second order can be described using a double contraction ... 

These analyses allowed us to present a discussion on how the equUibrimn equations of state become modified when the system is subject to sufficiently strong flows, because not only the non-diagonal components of the stress tensor become modified, but also the normal components. In addition, through the discussion we have shown that the chemical potential becomes corrected by a term proportional to the focal kinetic energy of diffusion, and therefore, is a flow dependent quantity. This correction was analyzed in the context of nucleation, were it was found that both the nucleation barrier and consequently the nucleation rate can be written in terms of the kinetic energy of the flow. 

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