Navier equations

For many purposes it is convenient to express the equilibrium equations in terms of the displacements. In the absence of body forces and inertial terms, Eq. (4.14) can be written as 

The relationship between stress and strain in terms of the tensile modulus and the Poisson ratio is given by 

Expressing the tensor strain in terms of the components of the vector displacement, Eq. (4.103) becomes 

This expression is known as the Navier equation. It should be noted that since 

Finally, it is worth noting that taking the divergence and Laplacian in Eq. (4.108) one finds 

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The transport equation (441) has been studied in another context j30.35.36 it is out of place to give this analysis here and we shall merely quote the conclusion in the case of small gradients, Eq. (440) is entirely equivalent to the well-known Stokes-Navier equation of hydrodynamics.f More precisely, the average velocity w, which from Eqs. (424), (426), and (439) is given by ... 

Relativistic Effects. Consider the relatively simple case of a screw dislocation moving along x at the constant velocity v (see Fig. 11.3). The elastic displacements, Mi, U2, and U3, around such a dislocation may be determined by solving the Navier equations of isotropic linear elasticity [3]. 5 For this screw dislocation, the only nonzero displacements are along 2, and for the moving dislocation the Navier equations therefore reduce to... 

If the displacements on the surface of the body are given (second boundary problem), the stress-displacement relationships are obtained first, and their substitution into the equilibrium equations permits us to eliminate the stress variables and thus to obtain the three equilibrium equations in terms of the displacements (see Navier equations in... 

To start with, let us determine the stress and the deformation of a hollow sphere (outer radius J 2, inner radius R ) under a sudden increase in internal pressure if the material is elastic in compression but a standard solid (spring in series with a Kelvin-Voigt element) in shear (Fig. 16.1). As a consequence of the radial symmetry of the problem, spherical coordinates with the origin in the center of the sphere will be used. The displacement, obviously radial, is a function of r alone as a consequence of the fact that the components of the strain and stress tensors are also dependent only on r. As a consequence, the Navier equations, Eq. (4.108), predict that rot u = 0. Hence, grad div u = 0. This implies that... 

Let us consider now the deformation and stresses of a cylindrical pipe under two different boundary conditions (Fig. 16.2). In both eases the length of the pipe is considered constant according to the requirements for a plane strain problem. The external and internal radii are R2 and R, respectively. If the applied forces and the displacements are also uniform, the deformation is purely radial, and in cylindrical coordinates = u r). According to the Navier equations, rot u = 0. Hence, Vdiv u = 0, which implies... 

The starting point is once more the Navier equations. At equilibrium, the gravitational force corresponding to the inertial term is included in the linear momentum equation [Eq. (4.35), where bi = pg,], so that... 

Isotropic Elasticity and Nervier Equations Use the constitutive equation for an isotropic linear elastic solid given in eqn (2.54) in conjunction with the equilibrium equation of eqn (2.84), derive the Navier equations in both direct and indicial notation. Fourier transform these equations and verify eqn (2.88). 

In the present setting p is the mass density while the subscript i identifies a particular Cartesian component of the displacement field. In this equation recall that Cijki is the elastic modulus tensor which in the case of an isotropic linear elastic solid is given by Ciju = SijSki + ii(5ikSji + SuSjk). Following our earlier footsteps from chap. 2 this leads in turn to the Navier equations (see eqn (2.55))... 

We begin with the special case in which it is assumed that the displacements resulting from the presence of the point defect are spherically symmetric. We know that the fields must satisfy the Navier equations derived in section 2.4.2, namely... 

For many purposes, we will find that antiplane shear problems in which there is only one nonzero component of the displacement field are the most mathematically transparent. In the context of dislocations, this leads us to first undertake an analysis of the straight screw dislocation in which the slip direction is parallel to the dislocation line itself. In particular, we consider a dislocation along the X3-direction (i.e. = (001)) characterized by a displacement field Usixi, X2). The Burgers vector is of the form b = (0, 0, b). Our present aim is to deduce the equilibrium fields associated with such a dislocation which we seek by recourse to the Navier equations. For the situation of interest here, the Navier equations given in eqn (2.55) simplify to the Laplace equation (V ms = 0) in the unknown three-component of displacement. Our statement of equilibrium is supplemented by the boundary condition that for xi > 0, the jump in the displacement field be equal to the Burgers vector (i.e. Usixi, O" ") — M3(xi, 0 ) = b). Our notation usixi, 0+) means that the field M3 is to be evaluated just above the slip plane (i.e. X2 = e). 

In fluid mechanics, it is common to separate the stress tensor as a sum of pressure and viscous stress, that is, ay = —pSy + xy, where 8y is the Kronecker delta function. Thus, for the components jc, y, and z, the Navier equations are written as ... 

The first approach is based upon direct solution involving the displacements. In the most basic sense, a strategy can be found to solve the 15 coupled differential equations directly. However, other approaches are more expedient. The most classical approach is to develop the Navier equations by putting the strain-displacement equations (Elq. 9.26) into the constitutive equations (Eq. 9.27) to obtain the stresses, a, in terms of the displacements, Uj. The result is then inserted into the equilibrium equations (Eq, 9.25), yielding three, coupled, second order partial differential equations on the three displacements, Uj. These three equations can then be solved for the displacements. Upon solution the stresses and strains can be found by substitution of the displacements in to the appropriate expressions. 

IPMC material is modeled via a multi-physics coupled problem, consisting of the PNP system of equations coupled to the Navier equation. Finite element methods of solving are typically utilized in simulating the solution. 

For the Navier equations 9 and 10, the following Dirichlet boundary conditions can be applied ... 

If no external forces are considered, zero Neumann boundary conditions can be applied on 9Q for the Navier equations. 

Rheological Models for Structured Fluids 2. Navier equation ... 

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