BEAMS AND STRUCTURES 81 



To determine the location of the neutral axis, proceed in the conventional man- 

 ner, as follows : 



1. Divide the corrugated sheet chord sections, upper and lower, into convenient 

 short lengths L as indicated. L must be short enough so that the moment of inertia 

 of the section of length L, about its own neutral axis, will be small compared with its 

 moment of inertia about the neutral axis of the whole section of the structure. 



2. Determine the areas A of the unit sections of length L, and locate the centroids 

 or centers of gravity of these sections. 



3. Choose any convenient horizontal reference line. 



4. Determine the distance R from the centroids to the arbitrarily chosen hori- 

 zontal reference hne. 



5. Tabulate in adjacent columns the areas A with their corresponding R, and 

 calculate and tabulate the products AR. 



6. Add all the AR values. 



7. Divide the summation of AR values by A, and the result wdll be D, which, as 

 indicated in the figure, locates the neutral axis. 



To calculate 7^, the moment of inertia, proceed as follows: 



1. Determine and tabulate the y values, i.e., the distances from the centroid of 

 each short length element to the neutral axis. It is necessary to do this only for the 

 elements lying to one side of the axis of symmetrj-. 



2. Tabulate in the adjacent column the square of each y value. 



3. Multiply each elemental area A by the square of its centroid distance y. 



4. Add the Ay'^ values. 



5. Multiply this summation by two if the elemental areas on only one side of the 

 axis of symmetry have been tabulated. 



6. The result 21,Ay'^ will be the moment of inertia I^ of the section about the 

 XX axis. 



This method is applicable only when the section is symmetrical and the bending 

 moment is normal to the neutral axis. 



Unsymmetrical Pure Monocoque Sections 



An example of an unsymmetrical box beam is shown in the accompanying 

 Fig. 11. The fiber stress at any point on the beam cross section can be expressed bj' 

 the equation 



. {MyH - MJy)y + {MM + M,L)x ... 



^'' - ■ 7j;^ip . ^^^ 



XX and YY are any convenient set of rectangular axes passing through the cen- 

 troid of the section, which is located by using the same method as described above for 

 the symmetrical section. 



Ix and ly are calculated by the same method as used for the symmetrical section, 

 Ix being the moment of inertia about the XX axis and ly the moment of inertia about 

 the YY axis. 



M^ is the component of the bending moment perpendicular to the XX axis. 



My is the component of moment perpendicular to FF axis. 



