Mr {a} = 0, Vr = - —, to = -2nr a2\4/(3+v) a2 r2 J

16 A40 E

where F = na q and the products qA40 and FA40 are negative.

• VTD 3 - Uniform load and free edge: The input of A1, A2, and A3 in the load (3.19c) provides A3 = q/64A40 and after substituting the Ai in (3.24b), the force Vr is

If the second term on the right is considered as qa2/2r or F/2nr with F = na2q, the expression of Vr represents the net shearing forces generated by a uniform load q reacting with a central force —F which corresponds to a free edge condition Vr{a} = 0; so, retaining this case, we set A2 = —qa2/16(1 — v)A40. Substituting A2 and A3 in (3.24a), we obtain Mr {a} = 0 if A1 = (3+v)qa8/(3+v) / 64(1 — v)A40. Normalizing with D0 = q/64A40 as for the previous configurations, the Ci coefficients, derived from (3.22), are finally

^ A tulip-shaped mirror provides a Sphe 3 deformation mode z = A40 r4, if a uniform load q applied to its surface is in reaction at its center. The characterizing features are where the product qA40 is positive.

These three configurations show a central thickness T40(0) ^ ^ which locally respects a null variation of mirror curvature at its center. For practical applications, the mirror central thickness will be given a finite value by considering an enough purity of the diffraction images, i.e. the central cut could provide a slight amount of curvature deformation but laying within the quarter-wave Rayleigh criteria or the Marechal criteria [3(a)]. For comparisons, the normalized thickness distributions are displayed by Fig. 3.4 and associated external forces by Fig. 3.5.

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