Stress normal components

The shear stresses are proportional to the viscosity, in accordance with experience and intuition. However, the normal stresses also have viscosity-dependent components, not an intuitively obvious result. For flow problems in which the viscosity is vanishingly small, the normal stress component is negligible, but for fluid of high viscosity, eg, polymer melts, it can be significant and even dominant. 

Independent of the sign convention used, the stress components can be classified into two types those that act tangentially to the face of the element and those that act normal to the face. Tangential components such as rxy, ryx, Tyz tend to cause shearing and are called shear stress components (or simply shear stresses). In contrast, the stress components rxx, Tyy, jzz act normal to the face of the element and are therefore called normal stress components (or normal stresses). Although there are six shear stress components, it is easily shown that ri = t, for t = j for example, ryx = rxy. Thus there are three independent shear stress components and three independent normal stress components. 

The pressure acting on a surface in a static fluid is the normal force per unit area, ie the normal stress. The pressure of the surrounding fluid acts inwards on each face of a fluid element. Consequently, with the negative sign convention the normal stress components may be identified with the pressure. With the positive sign convention, the normal stress components may be identified with the negative of the pressure positive normal stresses correspond to tension with this convention. 

In the case of a flowing fluid the mechanical pressure is not necessarily the same as the thermodynamic pressure as is the case in a static fluid. The pressure in a flowing fluid is defined as the average of the normal stress components. In the case of inelastic fluids, the normal stress components are equal and therefore, with the negative sign convention, equal to the pressure. It is for this reason that the pressure can be used in place of the normal stress when writing force balances for inelastic liquids, as was done in Examples 1.7-1.9. 

It has been assumed that the flow is incompressible so that there are no fluctuations of the density. Equation 1.91 shows that the momentum flux consists of a part due to the mean flow and a part due to the velocity fluctuation. The extra momentum flux is proportional to the square of the fluctuation because the momentum is the product of the mass flow rate and the velocity, and the velocity fluctuation contributes to both. The extra momentum flux is equivalent to an extra apparent stress perpendicular to the face, ie a normal stress component. As (v x)2 is always positive it produces a compressive stress, which is positive in the negative sign convention for stress. 

Figure 6. Stress analyses for an acom-shaped island with e = 0.01 under the infinite-torque condition (b) normal stress components, On ando22, along the y-axis passing through the island center as a function of the distance from the substrate base, (c) the stress components of the substrate along the substrate surface, and (d) the substrate stress field along the vertical line passing through the midpoint between two islands.

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. 

A graphic example of the consequences of the existence of in stress in simple steady shear flows is demonstrated by the well-known Weissenberg rod-climbing effect (5). As shown in Fig. 3.3, it involves another simple shear flow, the Couette (6) torsional concentric cylinder flow,3 where x = 6, x2 = r, x3 = z. The flow creates a shear rate y12 y, which in Newtonian fluids generates only one stress component 112-Polyisobutelene molecules in solution used in Fig. 3.3(b) become oriented in the 1 direction, giving rise to the shear stress component in addition to the normal stress component in. 

We therefore observe that unlike in the Power Law model solution with a single shear stress component, xn, in the case of a CEF model, we obtain, in addition, two nonvanishing normal stress components. Adopting the sign convention for viscometric flow, where the direction of flow z is denoted as 1, the direction into which the velocity changes r, is denoted as 2, and the neutral direction 8 is denoted as direction 3, we get the expressions for the shear stress in terms of the shear rate, the primary, and secondary normal stress differences (see Eqs. 3.1-10 and 3.1-11) ... 

From the foregoing, the two other normal stress components can be evaluated... 

T22 = normal stress component in the direction of the velocity gradient or of the shear T33 = normal stress component in the third direction and perpendicular to the third... 

When studying the stability of the steady-state, time-dependent calculations are needed (see [7]). It can also be used as a simple method to compute the steady-state solution. A time-dependent approach using the Lesaint-Ravian technique for the normal stress components and the Baba-Tabata scheme for the shear stress component is developed by Saramito and Piau ([34]). This method allows one to obtain rapidly stationary solutions of the PTT models. Convergence with mesh refinement is obtained as well as oscillation-free solutions. 

Fig. 7 Normal stress component cjyy [Pa] at time T = 75 min after the beginning of the test in the TDCB FV simulation.

For an oscillating interface the difference in the normal stress component over the, now curved, interface is counterbalanced by the Laplace pressure. This leads to the next stress boundary condition normal to the interface ... 

Figure 9.2 Variation of normal-stress component a with direction 6 and a set of magnitudes of chemical potential p. Each value of p is associated with a value of a by an act of the imagination as in Figure 9.1.

Here is a normal-stress component and 0 is a variable specifying the orientation of direction i. 

Differences between the stress fields. The essential point about the cylindrical slot. Figure 11.5, and the planar slot. Figure 11.6, is that the sequence of normal-stress components along the slot is the same in each but differences are certainly present, as is readily seen on considering other stress components. In particular, consider a series of planar elements across each slot. Figure 11.7 in diagram (b), the normal-stress components on these elements are all equal, whereas in diagram (a) they are not. 

Turn now to a nonhomogeneous stress field where, in addition to change of shape, there is diffusive mass transfer from sites of higher compression to sites of lower compression it is possible to devise a stress field as in Figure 11.6 such that successive points across a planar surface have the same sequence of normal-stress components as successive points across a cylindrical surface in the homogeneous constrictive field. 

If the material responds to a gradational sequence of normal-stress states, it is not reasonable to suppose that the response will be different according as the normal-stress components act on elements that form a cylindrical surface or a planar surface. Thus the shortening rate across a plane subject to <7 ax will be (ffmax by diffusivc mass transfer, for some small... 

Consider first the spatially uniform situation, diagram (d). Any normal-stress component in the profile, diagram (a), is given by... 

The train of thought is not restricted to principal planes of the stress state but applies to any plane. At some point in the material, for any direction through the point there is a normal-stress component hydrostatic pressure P" and an associated equilibrium potential p". The associated equilibrium potential is a direction-dependent scalar with an infinite number of magnitudes at a point, just like the normal-stress component of a stress state. The two quantities are linked by the factor Ij, the volume of one unit of component i (1 kg or 1 kg-mol or other unit) ... 

Oj, (T(i normal stress component on a plane normal to i (no summation... 

In the steady, unidirectional flow problems considered in this section, the acceleration of a fluid element is identically equal to zero. Both the time derivative du/dt and the nonlinear inertial terms are zero so that Du/Dt = 0. This means that the equation of motion reduces locally to a simple balance between forces associated with the pressure gradient and viscous forces due to the velocity gradient. Because this simple force balance holds at every point in the fluid, it must also hold for the fluid system as a whole. To illustrate this, we use the Poiseuille flow solution. Let us consider the forces acting on a body of fluid in an arbitrary section of the tube, between z = 0, say, and a downstream point z = L, as illustrated in Fig. 3-4. At the walls of the tube, the only nonzero shear-stress component is xrz. The normal-stress components at the walls are all just equal to the pressure and produce no net contribution to the overall forces that act on the body of fluid that we consider here. The viscous shear stress at the walls is evaluated by use of (3 44),... 

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