Stress tensor equilibrium

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. 

In integrated photoelasticity it is impossible to achieve a complete reconstruction of stresses in samples by only illuminating a system of parallel planes and using equilibrium equations of the elasticity theory. Theory of the fictitious temperature field allows one to formulate a boundary-value problem which permits to determine all components of the stress tensor field in some cases. If the stress gradient in the axial direction is smooth enough, then perturbation method can be used for the solution of the inverse problem. As an example, distribution of stresses in a bow tie type fiber preforms is shown in Fig. 2 [2]. 

The situation is more complex for rigid media (solids and glasses) and more complex fluids that is, for most materials. These materials have finite yield strengths, support shears and may be anisotropic. As samples, they usually do not relax to hydrostatic equilibrium during an experiment, even when surrounded by a hydrostatic pressure medium. For these materials, P should be replaced by a stress tensor, <3-j, and the appropriate thermodynamic equations are more complex. 

We shall consider an equilibrium problem with a constitutive law corresponding to a creep, in particular, the strain and integrated stress tensor components (IT ), ay(lT ) will depend on = (lT, w ), where (lT, w ) are connected with (IT, w) by (3.1). In this case, the equilibrium equations will be nonlocal with respect to t. 

Different equilibrium, hydrodynamic, and dynamic properties are subsequently obtained. Thus, the time-correlation function of the stress tensor (corresponding to any crossed-coordinates component of the stress tensor) is obtained as a sum over all the exponential decays of the Rouse modes. Similarly, M[rj] is shown to be proportional to the sum of all the Rouse relaxation times. In the ZK formulation [83], the connectivity matrix A is built to describe a uniform star chain. An (f-l)-fold degeneration is found in this case for the f-inde-pendent odd modes. Viscosity results from the ZK method have been described already in the present text. 

The first ingredient in any theory for the rheology of a complex fluid is the expression for the stress in terms of the microscopic structure variables. We derive an expression for the stress-tensor here from the principle of virtual work. In the case of flexible polymers the total stress arises to a good approximation from the entropy of the chain paths. At equilibrium the polymer paths are random walks - of maximal entropy. A deformation induces preferred orientation of the steps of the walks, which are therefore no longer random - the entropy has decreased and the free energy density/increased. So... 

Consider first the trivial case of a static fluid. Here there can only be normal forces on a fluid element and they must be in equilibrium. If this were not the case, then the fluid would move and deform. Certainly any valid relationship between stress and strain rate must accommodate the behavior of a static fluid. Hence, for a static fluid the strain-rate tensor must be exactly zero e(/- = 0 and the stress tensor must reduce to... 

For ionic crystals, electrochemical potentials have to be used in place of Eqn. (10.10). Mechanical equilibrium (in the absence of body forces) is attained if for the stress tensor... 

Therein, D-1 is the positive definite, isotropic, fourth order viscous compliance, where r/ s and (s are the macroscopic viscosity parameters, D, , is the inelastic solid deformation rate and = F, 1 rfEQ F 71 is the corresponding non-equilibrium stress tensor. Furthermore, the superscript ( ) indicates the belonging to the intermediate configuration. 

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... 

For the momentum conservation of a single-phase fluid, the momentum per unit volume / is equal to the mass flux pU. The momentum flux is thus expressed by the stress tensor i/r = (pi — t). Here p is the static pressure or equilibrium pressure / is a unit tensor and r is the shear stress tensor. Since <1> = —pf where / is the field force per unit mass, Eq. (5.12) gives rise to the momentum equation as... 

Surface tractions or contact forces produce a stress field in the fluid element characterized by a stress tensor T. Its negative is interpreted as the diffusive flux of momentum, and x x (—T) is the diffusive flux of angular momentum or torque distribution. If stresses and torques are presumed to be in local equilibrium, the tensor T is easily shown to be symmetric. 

Expression (4.17) for non-equilibrium moments allows us to determine the stress tensor for a dilute suspension of macromolecular coils in the linear viscoelastic liquid... 

To calculate the dynamic modulus, we turn to the expression for the stress tensor (6.46) and refer to the definition of equilibrium moments in Section 4.1.2, while memory functions are specified by their transforms as... 

However, in this case, the stresses in entangled systems can also be related to the tensor of mean orientation of the segments with a relation similar to equation (7.45), but with other coefficient of proportionality, because we deal with non-equilibrium situation in this case. In this way one can correspond the two expressions for the stress tensor to each other and relate the introduced variables xft and u"k to the tensor of mean orientation of the segments... 

In an equilibrium state, the stress tensor is determined by the form and the volume of a deformed body. To determine the stress tensor of the deformed body at arbitrary (not small) deformation, we follow the method demonstrated by Landau and Lifshitz (1987b) for the calculation of the stress tensor for small deformation. 

Any stationary state is seen to be brought about by the force balance W = —Wq. The quantum force is a function of the pressure potential, or stress tensor, f dp/p that produces inner forces in the continuum. At equilibrium, in stationary states, the potential energy remains constant, i.e. 

After a sufficiently long run in the network mode, the equilibrium value of the stress tensor ty was determined by application of the virial formula [5, 6]... 

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. 

The state of mechanical equilibrium is characterized by vanishing acceleration dy/dt = 0. Usually, mechanical equilibrium is established faster than thermodynamic equilibrium, for example, in the initial state when diffusion is considered. In the case of diffusion in a closed system, the acceleration is very small, and the corresponding pressure gradient is negligible the viscous part of the stress tensor also vanishes t = 0. The momentum balance, Eq. (3.97), is limited to the momentum conservation equation... 

Force equilibrium considerations show that, in the absence of an applied torque, the strain and stress tensors are symmetric o-y = o, and Sy = s, . Consequently, there are really only six independent stresses for three directions that can be applied to strain a body. The mathematical representation of stress is thus ... 

The Rayleigh line drawn in Fig. 20 assumes an equilibrium steady flow from the detonation front to the CJ point. However, this cannot be true because, although the flow at the front could be steady, it certainly is not in equilibrium as the stress tensor is not equal the hydrostatic pressure. However, if P in Eq. (4b) is replaced with P x, the component perpendicular to the shockfront, Eq. (4b) will be valid as long as the flow is steady, even... 

The stress tensor in this expression has been defined such that at equilibrium [Pg.173]

Note first that if the fluid is at a state of equilibrium with no flow, then the time derivative d is equal to zero, and the velocity gradient Vv is also zero. This implies from the above equations that = G8. Hence cr, i = <7 2 = ct t, = G at equilibrium, and aj = 0, for i j. Thus, although the diagonal stress components are not zero at equilibrium, they are all equal to each other, and the nondiagonal components are all equal to zero. Hence, the stress tensor is isotropic, but nonzero at equilibrium. (If one redefines the stress tensor as H = a — G8, then S " = 0 at equilibrium. The upper-convected Maxwell equation can then be rewritten in terms of Z .)... 

Not all the components of the stress tensor are independent, as can easily be shown by taking the moments around the different coordinate axes in Figure 4.3. A balance of forces at equilibrium gives... 

The substitution of the components of the stress tensor, given in the preceding section, into the equilibrium equations [Eq. (4.14) with = 0] leads to... 

In sec. 1.6b we have used a similar matrix notation for the stress tensor t imd in sec. I.app.yf we did so for the polarizability tensor a. We assume the system to be at mechanical equilibrium, i.e. the fluids are at rest. This means that the isotropic pressures in the bulk phases must the same everywhere (p = p ). Then... 

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