Stress tensor normal 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. 

Using the equilibrium equations of the elasticity theory enables one to determine the stress tensor component (Tjj normal to the plane of translumination. The other stress components can be determined using additional measurements or additional information. We assume that there exists a temperature field T, the so-called fictitious temperature, which causes a stress field, equal to the residual stress pattern. In this paper we formulate the boundary-value problem for determining all components of the residual stresses from the results of the translumination of the specimen in a system of parallel planes. Theory of the fictitious temperature has been successfully used in the case of plane strain [2]. The aim of this paper is to show how this method can be applied in the general case. 

The diagonal components of the Reynolds stress tensor (e.g., u U )) are referred to as the Reynolds normal stresses, while the off-diagonal components are referred to as the... 

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.

The stress tensor describes the forces transmitted to an element of material through its contacts with adjacent elements (78). Traction is the force per unit area acting outwardly on the material adjacent to a material plane, and transmitted through its contact with material across the plane. If the components of traction are known for any set of three planes passing through a point, the traction across any plane through the point can be calculated. The stress at a material point is determined by an assembly erf nine components of traction, three for each plane. If the orientations of the three planes are chosen to be normal to the coordinate directions of a rectangular Cartesian coordinate system, the Cartesian components of the stress are obtained ... 

The pressure P0 represents the arbitrary additive contribution to the normal components of stress in an incompressible system, 8i is the Kronecker delta, C[ j 1(t t) is the inverse of the Cauchy-Green strain tensor for the configuration of material at t with respect to the configuration at the current time t [a description of the motion (221)], and M(t) is the junction age distribution or memory function of the fluid. 

The force F and the stress r are both vectors, which are typically represented in components that align with a coordinate system. Since the stress vector at any surface whose orientation is represented by the outward normal n may be determined from the stress tensor, it follows that... 

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... 

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 ... 

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. 

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) ... 

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... 

The SI units are N Grouping nine components together into a 3 x 3 matrix does not automaticcdly lead to a tensor for that other criteria should hold, such as invariance under change of coordinates. In 13.6.1) the three diagonal terms (r, T, T ), known as normal stresses, act normal to the six faces of a cube of the... 

First level tensors are vectors. As explained above, they are indicated by one index. Second level tensors are characterised by two indices. The stress tensor is such a quantity. It has nine components rn, t12, r13, r21. .. r33. The abbreviation normally written is r, where j and i each assume the values 1, 2, 3. In calculations with tensors, the following rule is used. If an index only appears... 

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... 

By definition it is a symmetric second-rank tensor. The stress tensor ffy, i,j= 1,3), is also a symmetric second-rank tensor defined as follows (Landau and Lifchitz ) the element Oy is the i component of the force acting on the unit area normal to the axis x. The symmetry of the stress tensor is imposed by the condition of mechanical equilibrium. 

The stress tensor may be represented as a 3 x 3 matrix, with components Oy, where i and j both go from 1 to 3. The diagonal elements represent normal stresses, whereas the off-diagonal ones represent shear stresses. Positive normal stresses are tensile, while negative ones are compressive (but an opposite sign convention is sometimes used, most notably in the soil mechanics literature). Finally, from the balance of angular momentum (or torque in the static case), it follows that the stress tensor and its matrix representation are symmetric (ay = aji), meaning that only six out of the nine components are in fact independent. 

Not to burden the presentation with unnecessary mathematical complexities, we will, in what follows, nevertheless, mostly restrict ourselves to one-dimensional examples. In this case, the stress tensor has only one component, which represents a normal stress (a pressure), and the indices may be omitted. 

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