Stress tensor shear component

We recall here the physical interpretation of the stress tensor in Cartesian coordinates [161, pp. 131-132]. Let tj be the stress vector (surface force) representing the force per unit area exerted by the material outside the coordinate surface upon the material inside (where the unit outward normal to this surface is in the direction e ). The component Uj then represents the component of this stress vector at a point on the coordinate surface. For example, if the x coordinate surface has unit outward normal i/ = (1,0,0) then the stress vector at a point on this coordinate surface is simply ti = BiUjUj = eita = (tii, 21, 3i)- A similar interpretation arises for the couple stress tensor. The components tn, 22 and 33 are called the normal stresses or direct stresses and the components ti2, t2i> i3, 3i, 23, 32 are called the shear stresses. 

A strength value associated with a Hugoniot elastic limit can be compared to quasi-static strengths or dynamic strengths observed values at various loading strain rates by the relation of the longitudinal stress component under the shock compression uniaxial strain tensor to the one-dimensional stress tensor. As shown in Sec. 2.3, the longitudinal components of a stress measured in the uniaxial strain condition of shock compression can be expressed in terms of a combination of an isotropic (hydrostatic) component of pressure and its deviatoric or shear stress component. 

The shear stress Is uniform throughout the main liquid slab for Couette flow ( ). Therefore, two Independent methods for the calculation of the shear stress are available It can be calculated either from the y component of the force exerted by the particles of the liquid slab upon each reservoir or from the volume average of the shear stress developed Inside the liquid slab from the Irving-Kirkwood formula (JA). For reasons explained In Reference (5) the simpler version of this formula can be used In both our systems although this version does not apply In general to structured systems. The Irvlng-Klrkwood expression for the xy component of the stress tensor used In our simulation Is... 

With the gel equation, we can conveniently compute the consequences of the self-similar spectrum and later compare to experimental observations. The material behaves somehow in between a liquid and a solid. It does not qualify as solid since it cannot sustain a constant stress in the absence of motion. However, it is not acceptable as a liquid either, since it cannot reach a constant stress in shear flow at constant rate. We will examine the properties of the gel equation by modeling two selected shear flow examples. In shear flow, the Finger strain tensor reduces to a simple matrix with a shear component... 

A similar KR approach can be employed for the intrinsic viscosity. In this case it is necessary to assume the presence of a small amount of shear rate, which cancels out in the calculations. Now, the linear equations are used to evaluate an averaged crossed component of the stress tensor on unit i, .The final result for a long linear non-draining chain is... 

For shear flow we utilize the component of the stress tensor and in this coordi-... 

Fig. F.2. Shear stress and shear strain, (a) The shear force per unit area is a component of the stress tensor, (b) The shear stress causes a shear strain.

Fig. 26. Images of the liquid stress tensor derived from the data shown in Fig. 23. Data are shown for (a) (b) and (c) ez with slices taken in the xy, yz, and xz planes for each of the shear components.

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 ... 

Further on, we shall consider the case of shear stress when one of the components of the velocity gradient tensor has been specified and is constant, namely V12 0. This situation occurs in experimental studies of polymer solutions (Ferry 1980). In order to achieve such a flow, it is necessary that the stresses applied to the system should be not only the shear stress a 12, as in the case of a linear viscous liquid, but also normal stresses, so that the stress tensor is... 

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. 

Another method to calculate viscoelastic quantities uses measurements of flow birefringence (see Chap. 10). In these measurements, two quantities are determined as functions of the shear rate y the birefringence An and the extinction angle y. The following relationships exist with the stress tensor components ... 

Newtonian shear flow of polymer melts is a stable process. This means that small disturbances in the flow conditions, caused by external effects, are readily suppressed. As the rate of shear increases, however, the elastic response of the melt becomes more pronounced relative to the viscous response. In other words, components of the stress tensor in directions different from the direction of the shear stress become more important. As a result, small disturbances are not so readily compensated and may even be magnified. 

To conclude this subsection, we expose an interesting paradox arising from the time dependence of the particle configuration. As discussed in Section III, Frankel and Acrivos (1967) developed a time-independent lubrication model for treating concentrated suspensions. Their result, given by Eq. (3.7), predicts singular behavior of the shear viscosity in the maximum concentration limit where the spheres touch. Within the spatially periodic framework, the instantaneous macroscopic stress tensor may be calculated for the lubrication limit, e - 0. The symmetric portion of its deviatoric component takes the form (Zuzovsky et al, 1983)... 

The remaining six quantities are called shear stresses. They have two subscripts associated with the coordinates, and are referred to as the components of the molecular momentum flow tensor, or the components of the molecular stress tensor, as they are associated with molecular motion. Usually, the viscous stress tensor, t, and the molecular stress tensor, it, are simply referred to as stress tensors. For a Newtonian fluid, we may express the stresses in terms of velocity gradients and viscosities in Cartesian coordinates as follows ... 

The mass forces may be the gravitational force, the force due to the rotational motion of a system, and the Lorentz force that is proportional to the vector product of the molecular velocity of component i and the magnetic field strength. The normal stress tensor a produces a surface force. No shear stresses occur (t = 0) in a fluid, which is in mechanical equilibrium. 

Equation 10.1 is a second-rank tensor with transpose symmetry. The normal components of stress are the diagonal elements and the shear components of stress are the nondiagonal elements. Although Eq. 10.1 has the appearance of a [3 x 3] matrix, it is a physical quantity that, for one set of axes, is specified by nine components, whereas a transformation matrix is an array of coefficients relating two sets of axes. The tensor coefficients determine how the three components of the force vector, /, transmitted across a small surface element, vary as different values are given to the components of a unit vector / perpendicular to the face (representing the face orientation) ... 

The final step to obtain shear viscosity is to calculate the xy component of the stress tensor which is expressed, on the one hand, in terms of t) and 7 as... 

CTjj Stress tensor component Octahedral shear stress... 

Rheological properties of food materials over a wide range of phase behavior can be expressed in terms of viscous (viscometric), elastic and viscoelastic functions which relate some components of flie stress tensor to specific components of the strain or shear rate response. In terms of fluid and solid phases, viscometric... 

In Sections 1.3.1 and 1.3.2, we discussed the shear stress and the extensional stress in shearing flows and extensional flows, respectively. These are components of the three-dimensional state-of-stress tensor T. The ith row of T is the force per unit area that material exterior to a unit cube exerts on a surface perpendicular to the tth coordinate axis (see Fig. 1-18). In general, if F is the force per unit area acting on a surface perpendicular to an arbitrary outward-directed unit vector, n, then... 

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. 

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