A40 E

where the product qA40 is negative.

• VTD 2 - Central force and simply supported edge (A3 = 0): With A3 = 0, the substitution of A1,a1 and A2,a2 in the bending moments (3.24a) provides Ai = —a2(1-v)/(3+v)A2 since Mr {a} = 0. From (3.22), this result becomes C1+C2 = 0; the third term of the rigidity in (3.28) vanishes. Choosing D0 = —q/64A40, we derive the net shearing force from (3.24b) as Vr = —((1 —v)qa2C2/8r. This is obtained by central force F in reaction to the edge if C2 = —4/( 1 — v) and F = na2q, then A2 = —4D0a2/(1 —v). The mirror and its deforming configuration are summarized as follows:

^ A tulip-shaped mirror provides a Sphe 3 deformation mode z = A40 r4, if a central force F applied to its surface is in reaction at its edge. The characterizing features are

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