Question 29.5: The rectangular wing shown in Fig. P.29.5 has a constant tor......
The rectangular wing shown in Fig. P.29.5 has a constant torsional rigidity GJ and an aileron of constant chord. The aerodynamic center of the wing is at a constant distance ec ahead of the flexural axis, while the additional lift due to operation of the aileron acts along a line a distance hc aft of the flexural axis; the local, two-dimensional lift-curve slopes are a_{1} for the wing and a_{2} for aileron deflection. Using strip theory and considering only the lift due to the change of incidence arising from aileron movement, show that the aileron reversal speed is given by
\tan \lambda k s \int_0^{k s} y \sin \lambda y d y-\tan \lambda s \int_0^s y \sin \lambda y d y-\int_{k s}^sy \cos \lambda y d y=\frac{(e+h)}{2 h \cos \lambda ks}\left[(k s)^2-s^2\right]where
\lambda^2=\frac{1}{2} \rho V^2 a_1 e c^2 / G J
Since the additional lift due to operation of the aileron is at a distance hc aft of the flexural axis, the moment equilibrium equation (29.25) for an elemental strip becomes
\frac{ d T}{ d z} \delta z+\Delta L e c+\Delta M_0=0 (29.25)
{\frac{\mathrm{d}T}{\mathrm{d}y}}\delta y-\Delta L e c-\Delta L_{\xi}h c=0 (i)
in which, from Eq. (29.23),
\Delta L=\frac{1}{2} \rho V^2 c \delta z\left[\frac{\partial c_1}{\partial \alpha}\left(\theta-\frac{p z}{V}\right)+\frac{\partial c_1}{\partial \xi}f_a(z) \xi\right] (29.23)
\Delta L=\frac{1}{2}\rho V^{2}c\delta y\Bigl[a_{1}\Bigl(a_{1}-\frac{p y}{V}\Bigr)+a_{2}f_{a}(y)\xi\Bigr]where f_{a}\left(y\right)=0\;\mathrm{for}\;0\leq y\leq k s\;\mathrm{and}f_{a}\left(y\right)=1\;\mathrm{for}\;k s\leq y\leq s. Also,
\Delta L_{\xi}=\frac{1}{2}\rho V^{2}c\delta y a_{2}f_{a}(y)\xiThen, substituting for T(= GJ dθ/dy), ΔL and \Delta L_{\xi} in Eq. (i) and writing \lambda^{2}=\rho V^{2}e c^{2}a_{1}/2G J,
\frac{\mathrm{d}^{2}\theta}{\mathrm{d}y^{2}}+\lambda^{2}\theta=\lambda^{2}\frac{p y}{V}+\lambda^{2}\frac{h a_{2}}{e a_{1}}f_{\mathrm{a}}(y)\xi (ii)
The solution of Eq. (ii) is obtained by comparison with Eq. (29.29). Thus,
\theta_1(0-k s)=\frac{p}{V}\left(y-\frac{\sin\lambda y}{\lambda \cos \lambda s}\right)-\frac{h a_2 \xi}{e a_1}(\tan \lambda s \cos \lambda k s-\sin \lambda k s) \sin \lambda y (iii)
and
\theta_2(k s-s)=\frac{p}{V}\left(y-\frac{\sin\lambda y}{\lambda \cos \lambda s}\right)+\frac{h a_2 \xi}{e a_1}(1-\cos \lambda y \cos\lambda k s-\tan \lambda s \cos \lambda k s \sin\lambda y) (iv)
Then, from Eq. (29.32),
\int_0^s \frac{\partial c_1}{\partial \alpha}\left(\theta-\frac{p z}{V}\right) z d z=-\xi\int_0^s \frac{\partial c_1}{\partial \xi} f_a(z) z d z (29.32)
\int_0^{k s} a_1\left(\theta_1-\frac{p y}{V}\right)y d y+\int_{k s}^s a_1\left(\theta_2-\frac{p y}{V}\right) y d y=-a_2 \xi \int_{k s}^s y d y (v)
Substituting for \theta_1 \text { and } \theta_2 in Eq. (v) from Eqs (iii) and (iv) gives
-\tan \lambda s \int_0^s y \sin \lambda y d y+\tan\lambda k s \int_0^{k s} y \sin \lambda y d y-\int_{k s}^s y \cos \lambda y dy +\frac{(e+h)}{h \cos\lambda k s} \int_{k s}^s y d y =\frac{p e a_1}{h a_2\xi \lambda V \cos \lambda s \cos \lambda k s}\int_0^s y \sin \lambda y d yHence, the aileron effectiveness is given by
-\tan \lambda s \int_0^s y \sin \lambda y d y+\tan\lambda k s \int_0^{k s} y \sin \lambda y d y
\frac{(p s / V)}{\xi}=\frac{-\int_{k s}^s y \cos\lambda y d y+\frac{(e+h)}{2 h \cos \lambda ks}\left[s^2-(k s)^2\right]}{\frac{e a}{h a_2 \lambda s\cos \lambda s \cos \lambda k s} \int_0^s y \sin\lambda y d y} (vi)
The aileron effectiveness is zero, i.e., aileron reversal takes place, when the numerator on the righthand side of Eq. (vi) is zero, i.e., when
\tan \lambda k s \int_0^{k s} y \sin \lambda y d y-\tan \lambda s \int_0^s y \sin \lambda y d y-\int_{k s}^sy \cos \lambda y d y=\frac{(e+h)}{2 h \cos \lambda ks}\left[(k s)^2-s^2\right]