## Lpql Clpq1 Cpay 1108a toyvtoVto i

1/2 / t \-p dy where q is an integer and ¡5 a dimensionless constant. Starting from a given value of the relative thickness ti/t0 at the edge, a numerical integration allows determining the constant ¡5 and then the thickness distribution t/t0. This resolution has been carried out by computer for the case p = 3, q = 2, of an equal curvature rod; the solution is almost a truncated cone (cf. Fig. 1.45).

The degenerated cases of null thickness at the edge, ti = 0, correspond to ¡5 = 1. After integration we obtain the general form t = to

which, for rods and beams with a=constant (p — q = 0) gives a parabolic thickness, and for rods and beams with parabolic flexure (p - q = 1) gives a cone and a wedge respectively.12

7. Special cases of beam cantilevers with constant thickness: Beams of equal constraint or with parabolic flexure can be obtained, when the thickness t=constant, if now the width a(y) in the x-direction varies with the position y of the cross section. We assume that a(y)|max C i. The inertia moment Ix = at3/12 is varying with a and, for the loading cases considered in Table 1.9, the shearing forces are ri

After substitutions of the bending moment My and of Qy into the equilibrium equation (1.101), we obtain, for the equal constraint cases,

12 For a null thickness end, ti = 0, Clebsch [34] derived the solution of equal constraint when a bar is flexed by its own weight by considering homothetic cross-sections. In the cases of both rod and beam he found parabolas. For a beam of square cross-sections, his result can be easily demonstrated: the inertia moment and shearing force become Ix = t4/12 and Qy = ¡lg jf t2 dy which, after substitution into the equilibrium equation (1.105), entails p = q = 2. Hence p - q = 0, so the Clebsch parabola is the exact answer to Galileo's problem and to his drawing of Fig. 1.42 which shows homothetic cross-sections.

In his annotations of Clebsch's book, Saint-Venant ([138] p. 359, cf. Thodhunter and Pearson [161] vol.II, part II, pp. 263-164) comments that the "paradox of a null thickness" at the suspension point leads to an infinite shearing force, but this can be solved by giving the vertex a finite thickness. He did not consider Clebsch's integration constant that was set to zero and which would have led to cantilevers with finite thickness, the above truncated class.

6f ta fe

and since R = Et/2a, the substitutions of a give expressions of the same form for parabolic flexure beams.

Considering the first case of a concentrate force F and denoting a0 and at the widths at y = 0 and t, the solution is represented by

Hence for constant thickness beams of equal constraint or of parabolic flexure, the geometry is the same: a trapeze which degenerates into a triangle when at = 0 (Fig. 1.46).

From (1.111), whether a truncated or a sharp end, the constant constraint and parabolic flexure are satisfied if the angle p formed by the lateral faces is

In practice, these solutions have been developed for improving suspension systems by adding to the cantilever a symmetric part with respect to the x, z plane, so a force applied at y = 0 is in reaction with two opposite forces at the ends ±t. The trapeze plate is sliced into parallel strips which are symmetrically rearranged on the central strip with freedom to slide. This arrangement is known as a multiple-leaf spring system (cf. Fig. 1.46).

• Compressed beams and buckling critical load: Euler [52, 53] investigated the important case of a compressed beam (or rod) of circular and constant section. Considering one end clamped at the origin whilst the other is free (Fig. 1.44-Right), the force F must be substituted by —F in Eq. (98), the boundary conditions are 9 = 0 for s = 0, and d9/ds = 0 for s(90) = t. This entails p = 2arctan I I = 2arctan t2a

6F i 112RF

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