A40 E

where the product qE40 is positive.

The uniform bending moment at the contour, Mr{a}, exactly cancels the curvature component of the flexure (see also Sect. 7.3). Examples of such configurations are displayed by Fig. 3.3.

3.3.2 Configurations in the VTD Class

Interesting solutions in the VTD class can be derived by assuming a null rigidity at the contour, D{a} — 0. This condition directly entails that Mr{a} — 0 which is easy to satisfy in practice and corresponds to a simply supported edge or to a free edge. From (3.21), the condition D{a} — 0 is realized if

After substitution of C3, these configurations are all of the form [13]

With setting Mr{a} = 0 and depending on the Ai selection, we obtain three solutions (Lemaitre [14]).

• VTD 1 - Uniform load and simply supported edge (A2 = 0): With A2 = 0, the substitution of A1,a1 and A3,a3 in the bending moments (3.24a) provides Ai = -a8/(3+v)A3 since Mr {a} = 0. From (3.22), this result becomes C3 = -Ci which is compatible with (3.27) only if C2 = 0. Choosing D0 = —q/64A40, we derive the net shearing force from (3.24b) as Vr = —qrC1/2. This is a uniform load in reaction to edge if C1 = 1, then A3 = -D0 . This case is summarized as follows:

^ 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 edge. The characterizing features are

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