#### Installation — business terrible - 1 part

September 8th, 2015

Consider the differential harmonic operator in F. q. (13.1.13). Then (13.1.21) reduces to the equation f ( A 2 d2d

£ 42‘w ш = Гл for (13л-23)

whose general solution is

є? — Та I – C cos f + ф for x Є A (13.1.24)

where C and ф are arbitrary constants. To determine these constants and the possible size of the localiz. alion zone, the jump conditions between the regions A and В must be determined. These conditions are obtained easily if the solution is analyzed in the sense of the theory of distributions. Then, since є? — T — 0 at the interior points of В and T — Y л at the interior points of A, the solution for T has C~1 continuity, therefore, the solution for for є? must have Cn~l continuity, where n is the differential order of the operator (a jump in the n-lh derivative exists). In our case n — 2, and so є2 must have C continuity, i. e., it must be continuous, with a continuous first derivative.

Taking the x origin to lie at the left interface between parts A and B, as shown in Fig. 13.1.3a, and requiring that at this point both and its first derivative must vanish, we get C = –T/t and ф — 0, from which the possible solution takes the form

є1 = Тл Гі – cos^) 2Тл sin2-у (13.1.25)

Writing now the continuity conditions at the right interface between zl and B, we find that the siz, e h of the localization zone can take only discrete values h — m, with m = 1,2, ■ • Thus, a periodic solution with an integer number of wavelengths is possible. However, we immediately see that the inelastic displacement and energy requirements are minimum for the smallest possible size, i. e., for a single wavelength. Figs. 13.1.3b—d show the resulting distributions for T, and, where td — Jx єf dx is the displacement associated with the fracturing strain.

Note that in solving this problem we assume that there is a region В in which the material unloads, and that the softening region A is continuous. Obviously, there also exist solutions in which (a) the strain is uniform along the bar and є? = Та everywhere, or (b) there exist various nonoverlapping distributions identical in shape to that in Fig. 13.1.3c. However, it is easy to see that the single wavelength solution in this figure is energetically preferred.

Figure 13.1.4 Examples of weight functions: (a) Rectangular; (h) second-degree parabola, a(x) = 1 — (x/0.75f)2; (c) fourth-degree parabola, Eq. (13.1.4) with /i0 = 15/16; (d) cosine, a = 1 + cos(7rs/f). (Adapted from Planas, Elices and Guinea 1993.) |