S= By a concurrent cyclic interchange of the subscripts 1, 2, 3; of έ, n, ; of P, Q, R; and of S, T, U and addition we complete the e-solution. To complete, for example, the value of P, we take one cyclic step forward from the value of R, and one such step retrograde from the value of Q. To find a solution from the -triad, we start from where again the triad may be completed by cyclic interchange. Each of these solutions is a complete general solution of the stress-equations. First deduction. 2. If we express the elements of any stress whatever according to the e-solution, and the elements of the reverse of that stress according to the y-solution, and add the two solutions, the sum of the expressions on the right in each of the six cases will be null. Stating this result otherwise. If we add the 6-solutions and solutions in full for each of the six elements of stress, there will be certain related values of 1, 2, 3, 41, 42, 43 which will make the value of each element of stress zero. Both by analogy with the results for Cartesians, and from general considerations, it will appear that the relations between 01, 02, 03, 1, 2, 3 in this case are the same as those between e, f, g, a/2, b/2, c/2. We therefore obtain the six identical relations between the elements of strain, by writing the sum of the e-solutions, and the-solutions in full, for each element of stress, and then replacing by e, 02 by ƒ, 03 by g, 1 by a/2, 1⁄2 by b/2, 3 by c/2, and replacing each element of stress by zero. The leading terms of the first and fourth of the set have the form The complete expressions are lengthy, and require the expenditure of a little labour to obtain. That the expressions on the right, expressed in terms of u, v, w, vanish identically may be verified to any desired extent, but the verification is really only of the accuracy of the work. It is certain that there is one set, and only one set, of identical relations between the differential coefficients of the elements of strain not exceeding the second degree. An illustration, ab initio, in polar cylindrical coordinates is simple. The case of polar spherical coordinates is not much. simpler than the general case. Second deduction. 3. To illustrate these six equations of identity, I proceed to find the six Elastic Stress Relations for this coordinate system. In the first place, we must replace the elements of strain in these equations by their equivalents in terms of stress. To explain the further processes, I shall suppose that, for the moment, the inertia terms are included in the stressequations in the form pü, pë, pw. Then we must proceed to obtain from those equations the values of pë, etc., pä, etc., after which these inertia terms may be omitted. Then double the expression for pë and subtract the sum of the second and third of the six equations of identity. The details of the simplification are laborious and of little interest. There is again a triad of analogous relations from which we may verify that ▼2(P+Q+R) = 0. 486 Two Solutions of Stress Equations under Tractions only. The final result for the first relation of the first triad is To form the first relation of the second triad, we add to the expression for pä the fourth of the six equations of identity, and find LIII. The Crystalline Structures of Silver Iodide. Na previous communication + giving the results of X-ray analyses of the crystal structures of the silver halides, the writer reported that silver iodide gave the diffraction pattern (powdered crystal) of the diamond cubic or "zinc-sulphide type" of structure. As stated then, crystals showing evidence of the hexagonal structure had not * Communication No. 184 from the Research Laboratory of the Eastman Kodak Company. + R. B. Wilsey, Phil. Mag. xlii. p. 262 (1921). |