Numerical Solution

The aerodynamic coefficients for the wings undergoing simple harmonic oscilla­tions, the integro-differential equation 4.51 can be solved to obtain the amplitude of the reduced circulation as we did for the steady case. For this purpose, expanding the reduced circulation into Fourier like series will give us the algebraic system of equations. Before expanding into the series, let us first transform the spanwise coordinates with y* = l* cos / ve g* = l* cos в. The series form of the reduced circulation in series can be expressed as follows

Here, a denotes rotation, h vertical displacement and b flap motion. With this notation Eq. 4.51 becomes the following set of equations

Si (kol*; /1 )S3 (k0l*, /1).. .S2N-1 (kol*; /1 ) S1(kJ*, /2ШКІ*, /2). . – S2N-1(kol*, /2)

S1(kol*, /N)S1(kol*, /N).. .S2N-1(kol*, /n)

The entries of the matrix Sn and the right hand side of Eq. 4.55 are complex. Therefore, coefficients Knj are obtained as N complex numbers. These coefficients help us to find the reduced circulation values at each station. From the reduced circulation values we obtain the amplitude of the circulation. Integrating the cir­culation along the span gives us the amplitude of the total lift. The total lift value being complex gives us the phase difference between the simple harmonic motion of the wing.

For a rectangular planform with a constant chord 2b the reduced frequency along the span remains the same. Therefore, for a given frequency and the mode shape the reduced circulation becomes proportional with the amplitude of the motion. Hence, the right hand side of 4.54 is simplified as follows.

While computing the coefficients Knj from 4.54 the right hand side of the equation may become real. If we denote the new coefficients with K’nh, we can write

K’nh = Jew h to have 4.54 as follows

kHl12)(k)b

Similarly, knowing the coefficients K’nh, we can calculate the amplitude of circulation at spanwise stations from the reduced circulation values as follows.

Example 3: A rectangular wing with an aspect ratio 6 undergoes vertical oscil­lation with k = 2/3 and amplitude h. Find the spanwise distribution of lift.

Solution: Using the Reissner’s tables and the 2-D lift value: L(2)/2pU2h = -0.425 + 1.19г we find

/= 0.0; 0.4; 0.8; 1.0

L/2pU2h =-0.441 + . 195i; -0.455 + 1.18г, -0.461 + 1.071г, -0.042 + 0.23г