Analytical Solutions

A number of closed-form solutions of Eq. 3 are possible. We will consider here only the case of the squared and the triangular plate. We leave to the reader the integration of other simple geometrical forms.

A solution for the squared plate is given by Eq. 4. Fig. 4 shows the map of the admittance in the imaginary plane for different gust speed ratios for a squared figure. With increasing A the system tends tends asymptotically to a closed loop. The range of the reduced frequency is k = 1 — 10.

The case of the circular plate is somewhat more complicated, since the ad­mittance contains trigonometric functions in implicit form. Therefore this case was solved using a numerical method, as described in Filippone, 2003.

For a generic triangular plate with a sweep angle a and height h, the geo­metrical functions are:

ф) = 2^1-2^, <^(~)=8M. (18)

Therefore, the admittance becomes

1 Г h/2

H = – r(z)H[r(z)k] е~акф) dz.

h – h/2

z

By operating a change of variables z = – j—- we hnd

2 f 2(1 – Z1)HX [2к(I – Ы)] e~iXk4^dz.

Therefore, H does not depend on the sweep angle. Similar conclusion is found for the pitching moment. We plotted the admittance for the triangular plate at A = 1 in Fig. 5 and we compared with the admittance of squared plate Hi (k) for frequencies up to k = 10. The oscillations of the admittance have smaller amplitude and show no critical damping.