User:Eas4200c.f08.carbon.orear/HW5/MATLAB

=HW5 MATLAB Code=

Case 1 Results
Neglect the bending effects of the skin and the partition walls (spar webs).

location of the centroid of the stringers; plot the stringers with black solid circles, the centroid as a crosshair, with the skin and spar webs in dotted lines.


 * Stringer Centroid = (0 , 0.2083   , 0.0081) m



all components of the moment of inertia tensor: I_22, I_33, I_23


 * I_22 = 6.3913e-007
 * I_33 = 8.3333e-006
 * I_23 = -7.6055e-008

slope beta = - alpha of the neutral axis (see classnotes); plot the neutral axis.


 * TanAlpha = -0.0216 m/m

highest normal bending stress sigma_11 and location


 * stress_11Max = 85,134,000 Pa


 * LocationMax = (0, 0.1538, 0.0470) m (See Red Cross Hair)

let the ratio between M_y and M_z to be constant based on the above data, set the maximum bending stress sigma_11 to the ultimate stress for steel 300 M (see Table 1.1, p.14), find the ultimate bending moments M_y and M_z.


 * stress_11Ult =  300,000,000 Pa


 * M_y_ult = -7,176.2 N m


 * M_z_ult = 2,870.5 N m

Case 1 Results
location of the centroid; plot the NACA airfoil in solid lines, with the stringers in black solid circles, this centroid and the centroid in Part I (as crosshairs with different colors and line styles) in the same figure. You may need to magnify the figure in case the two centroids are very close to each other.


 * Entire Centroid  = (0  , 0.2344   , 0.0071) m (See Black Cross Hair)
 * Stringer Centroid = (0 , 0.2083   , 0.0081) m (See Blue X)



all components of the moment of inertia tensor: I_22, I_33, I_23


 * I_22 = 1.0893 x 10-4
 * I_33 = 1.7658 x 10-4
 * I_23 = -3.4076 x 10-7

slope beta = - alpha of the neutral axis (see classnotes); plot the neutral axis.


 * TanAlpha = -0.2450 m/m

highest bending stress sigma_11 and location.


 * stress_11Max = 836,720 Pa


 * LocationMax = (0,  0.5000, 0) m (See Red Cross Hair)

'''compare the results of the above full-blown NACA airfoil to the results obtained in Part I, and draw some conclusions. '''

Next, since the thin skin would buckle under high normal bending stress acting in the plane of the skin, and thus the skin would not participate in resisting bending at high bending moments; to estimate the ultimate bending moments as done in Part I, do the following:

find the highest normal bending stress sigma_11 in any of the 4 stringers, and identify that stringer(s).


 * stress_11B = 763,870 Pa
 * stress_11F = 170,880 Pa
 * stress_11H = 630,850 Pa
 * stress_11E = 108,530 Pa

Stringer B has the highest bending stress.

let the ratio between M_y and M_z to be constant based on the above data, set the maximum bending stress sigma_11 in the above stringer(s) to the ultimate stress for steel 300 M (see Table 1.1, p.14), find the ultimate bending moments M_y and M_z for the full-blown NACA 2415 airfoil.


 * stress_11_B_Ult =  300,000,000 Pa


 * M_y_ultB = -490,920 N m

 compare the results to those obtained in Part I. 
 * M_z_ultB = 196,370 N m