Problems with Special Symmetries

The geometric characteristics of some interesting viscoelastic problems have special symmetries. For example, spherical shells have wide applications in pressure vessels, heat exchangers, and nuclear reactors. Loading of these structures can occur not only in accidental conditions but also in normal situations, e.g., during overpressurization. These spherical shells are radially symmetrical. 

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 

From the result of Problem 4.7 in Chapter 4, the following differential equation for the displacement u (assuming = ) is obtained  

The nonzero strain tensor components are easily obtained as 

The radial and transverse stresses can be determined from the stress-strain relationships. Owing to the orthogonality of the spherical coordinates, the formal structure of the generalized Hooke s law, given by Eq. (P4.11), is preserved, so that the nonzero components of the stress tensor are expressed in terms of the strain tensors as 

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Physicists are familiar with many special functions that arise over and over again in solutions to various problems. The analysis of problems with spherical symmetry in P often appeal to the spherical harmonic functions, often called simply spherical harmonics. Spherical harmonics are the restrictions of homogeneous harmonic polynomials of three variables to the sphere S. In this section we will give a typical physics-style introduction to spherical harmonics. Here we state, but do not prove, their relationship to homogeneous harmonic polynomials a formal statement and proof are given Proposition A. 2 of Appendix A. 

The compositions of the solutions to the 9 and 4> equations, by virtue of their relevance to all problems with spherical symmetry, are so important as to merit a special name, the spherical harmonics. These solutions bear the same relation to spherical problems that the Bessel functions do to those with cylindrical symmetry. The spherical harmonics, Yim 9, (/>), may be written as... 

We need to point out that, if the wavelengths of laser radiation are less than the size of typical structures on the optical element, the Fresnel model gives a satisfactory approximation for the diffraction of the wave on a flat optical element If we have to work with super-high resolution e-beam generators when the size of a typical structure on the element is less than the wavelengths, in principle, we need to use the Maxwell equations. Now, the calculation of direct problems of diffraction, using the Maxwell equations, are used only in cases when the element has special symmetry (for example circular symmetry). As a rule, the purpose of this calculation in this case is to define the boundary of the Fresnel model approximation. In common cases, the calculation of the diffraction using the Maxwell equation is an extremely complicated problem, even if we use a super computer. 

Symmetry is the easiest to apply. It is based on the correct selection of the coordinate system for a given problem. For example, a temperature field with circular symmetry can be described using just the coordinates (r, z), instead of (x, y, z). In addition, symmetry can help to get rid of special variables that are not required by the conservation equations and interfacial conditions. For example, the velocity field in a tube, according to the Navier-Stokes and continuity equations, can have the functional form uz(r). 

The structures with two R carbons and two S carbons (J, N, and O) have special symmetry. J is a meso structure it has chirality centers and is superimposable on its mirror image—see Problem 5-20(h). N and O are enantiomers, and are diastereomers of all of the other structures. Give yourself a gold star if you got this correct ... 

The coherent states demonstrate most clearly the limiting transition from quantum to classical mechanics. However, this approach has a shortcoming The coherent states (54) can be constructed only for very special models. On the other hand, the nodeless states (and the states with p,q n) are easily constructed with the help of the 1/n expansion for an arbitrary potential, V(r), as well as in a number of problems without spherical symmetry (for example, for the problem of two Coulomb centers [8]). Such states minimize the uncertainty relations at n —> oo, and in that respect they are similar to (54). However, there is also a difference the states a t)) are nonstationary, while the states discussed above are stationary. This is due to the fact that we are now considering the motion of a wave... 

There is a problem with the rotational term. A rigid body (the equilibrium atomic positions Ooi are used), such as the benzene molecule, rotates, but due to symmetry, it may have some special axes characterizing the moments of inertia. The moment of inertia represents a tensor of rank 3 with the following components ... 

In the case of ethylene, because of 2-fold symmetry, odd terms drop out of the series, V3, V5,... = 0. In the case of ethane, because of 3-fold symmeti-y, even temis drop out, V2, V4,... = 0. Terms higher than three, even though permitted by symmetry, are usually quite small and force fields can often be limited to three torsional terms. Like cubic and quaitic terms modifying the basic quadratic approximation for stretching and bending, terms in the Fourier expansion of Ftors (to) beyond n = 3 have limited use in special cases, for example, in problems involving octahedrally bound complexes. In most cases we are left with the simple expression... 

Fig. D.5 The mesh network to solve the momentum equation for the axial velocity distribution in a rectangular channel. As illustrated, the control volumes are square. However, the spreadsheet is programmed to permit different values for dx and dy. Because of the symmetry in this problem, only one quadrant of the system is modeled. The upper and left-hand boundary are the solid walls, where a zero-velocity boundary condition is imposed. The lower and right-hand boundaries are symmetry boundaries, where special momentum balance equations are developed to represent the symmetry. As illustrated, there is an 12 x 12 node network corresponding to a 10 x 10 interior system of control volumes (illustrated as shaded boxes). The velocity at the nodes represents the average value of the velocity in the surrounding control volume. There are half-size control volumes along the boundaries, with the corresponding velocities represented by the boundary values. There is a quarter-size control volume in the lower-left-hand corner.

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