Question 29.4: An initially untwisted rectangular wing of semi-span s and c......
An initially untwisted rectangular wing of semi-span s and chord c has its flexural axis normal to the plane of symmetry and is of constant cross-section with torsional rigidity GJ. The aerodynamic center is ec ahead of the flexural axis, the lift-coefficient slope is a, and the pitching moment coefficient at zero lift is C_{m,0}. At speed V in air of density ρ, the wing-root incidence from zero lift is a_0. Using simple strip theory—that is, ignoring the downwash effects—show that the incidence at a section distant y from the plane of symmetry is given by
\alpha _{0}+\theta =\left(\frac{C_{m,0}}{ea}+\alpha _{0} \right) \frac{\cos \lambda (s-y)}{\cos \lambda s} -\frac{C_{m,0}}{ea}where
\lambda ^{2}=\frac{ea\frac{1}{2}\rho V^{2}c^{2} }{GJ}Hence, assuming C_{m,0} to be negative, find the condition giving the speed at which the lift would be reduced to zero.
Answer: V_{d}={\sqrt{\frac{\pi^{2}G J}{2\rho e c^{2}s^{2}α}}}
The solution is obtained directly from Eq. (29.9) in which \partial c_{1}/\partial\alpha=a,\alpha=\alpha_{0}\;\mathrm{and}\;c_{\mathrm{m}},0=C_{M,\,0}. Thus,
\theta =\left[\frac{c_{m,0}}{e(\partial c_{1}/\partial \alpha )}+\alpha \right] \left[\frac{\cos \lambda (s-z)}{\cos \lambda s}-1 \right] (29.9)
\theta=\left({\frac{C_{M,0}}{e a}}+\alpha_{0}\right)\left[{\frac{\cos\lambda(s-y)}{\cos\lambda s}}-1\right]
which gives
\theta=\left(\frac{C_{M,0}}{e a}+\alpha_{0}\right)\frac{\cos\lambda(s-y)}{\cos\lambda s}-\frac{C_{M,0}}{e a}-\alpha_{0}Thus,
\theta+\alpha_{0}=\left(\frac{C_{\mathrm{M,0}}}{e a}+\alpha_{0}\right)\frac{\cos\lambda(s-y)}{\cos\lambda s}-\frac{C_{\mathrm{M,0}}}{e a}where
\lambda^{2}{=}\frac{e a\frac{1}{2}\rho V^{2}c^{2}}{G J}Also, from Eq. (29.11), the divergence speed V_{d} is given by
V_{d}={\sqrt{\frac{\pi^{2}G J}{2\rho ec^{2}s^{2}(\partial c1/\partial α )}}} (29.11)
V_{\mathrm{d}}={\sqrt{\frac{\pi^{2}G J}{2\rho e c^{2}s^{2}a}}}