Question 27.13: Figure P.27.13 shows a portion of the undersurface of a wing......
Figure P.27.13 shows a portion of the undersurface of a wing comprising two outer longerons, two central stringers, and skin panels. The wing includes a cut-out for the undercarriage bay and may be assumed to be attached to a rigid bulkhead at its left-hand end. Aerodynamic calculations show that for a particular loading case, the end loads in the outer longerons are equal to P_{0}, as is the end load on the central stringer. If the wing skin is effective only in shear and has a shear modulus G while the direct stresses are carried by the longerons and stringer, each having a Young’s modulus E, determine the distribution of load in the stringer 9 10. What are the values of the loads in portions 32 and 67 of the longerons?
Referring to Fig.S.27.13(a), the panel is symmetrical about a horizontal axis so that the shear flows q_{1} may be chosen as shown.
Consider the element \delta z δz of the longeron 34 shown in Fig.S.27.13(b).
For equilibrium of the element in the z direction,
from which
{\frac{\partial P_{\mathrm{A}}}{\partial x}}=q_{1} (i)
Similarly, from Fig.S.27.13(c) for the stringer 9 10,
\frac{\partial P_{\mathrm{B}}}{\partial z}=-2q_{1} (ii)
Now, from the overall equilibrium of a length z of the end panel,
2P_{\mathrm{{A}}}+P_{\mathrm{{B}}}=3P_{\mathrm{{o}}} (iii)
The compatibility condition is shown in Fig.S.27.13(d).
Then
Since γ is a function of z, only the partial derivative may be replaced by the full derivative. Also, γ=q/Gt, so that simplifying and rearranging the above equation we have
{\frac{\mathrm{d}q_{1}}{\mathrm{d}z}}={\frac{G t}{d E}}\left({\frac{P_{\mathrm{A}}}{A}}-{\frac{P_{\mathrm{B}}}{B}}\right) (iv)
From Eq.(iii),
P_{\mathrm{{A}}}={\frac{1}{2}}(3P_{\mathrm{{o}}}-P_{\mathrm{{B}}})Substituting in Eq. (iv) for q_{1} from Eq. (ii) and for P_{\mathrm{{A}}} and rearranging gives
\frac{\partial^{2}P_{\mathrm{B}}}{\partial z^{2}}-\frac{G t(2A+B)}{d E A B}P_{\mathrm{B}}=\frac{-3G t P_{\mathrm{o}}}{d E A}the solution of which is
P_{\mathrm{B}}=C\cosh\mu z+D\sinh\mu z+{\frac{3B}{2A+B}}P_{\mathrm{o}}where \displaystyle{\mu}^{2}=\frac{G t(2A+B)}{d E A B}
When z=0,P_{\mathrm{B}}=P_{\mathrm{o}}, which gives
C={\frac{2(A-B)}{2A+B}}P_{0}When z=L,P_{\mathrm{B}}=0, which gives
D=\frac{1}{\sinh\mu L}\Bigg[\frac{-2P_{\mathrm{o}}(A-B)}{2A+B}{\cosh \mu L-\frac{3B}{2A+B}P_{\mathrm{o}}\Bigg]}Therefore,
P_{\mathrm{B}}={\frac{P_{\mathrm{o}}}{24+B}}\bigg\{2(A-B)\cosh\mu z-\big[2(A-B)\cosh\mu L+3B\big]{\frac{\sinh\mu z}{\sinh\mu L}}+3\mathrm{B}\bigg\}From simple equilibrium, the load in each of the longerons 32 and 67 is 3P_{\circ}{/2}.
