Art. 110. COLUMNS ECCENTRICALLY LOADED. 



179 



upon the maximum allowed unit stress, the eccentricity, and the 

 form and size of the cross section. s c may be applied to the total 

 stress to find the required area just as s w is applied in tension or 

 in blocks in compression when the load is concentric. 



Equation (47) can best be solved by trial because r depends 

 upon the section which is to be found. Both A and r are un- 

 known. 



In Fig. 134 it is evident that one side of the block will be 

 compressed more than the other, and if there is any tension, one 

 side may even be extended. This is similar to the deformation 

 of a beam (68) and the axis must bend. The point a moves to- 

 ward the left so that the moment is really greater than Pe since 

 the final lever arm is greater than e by the -amount of the de- 

 flection of the axis between a and the section under consideration. 

 In a block, hoAvever, this deflection is so small compared with e 

 that it is negligible. 



110. Columns Eccentrically Loaded. 



eccentrically loaded, the deflection is not 

 negligible as it is in a Mock. As in Fig. 

 134, the stress at any section is composed 

 of two parts, direct compression and 

 bending. The lever arm of the couple 

 is, however, not e, but e-\-y mSLK y. Fig. 

 137. The moment of the couple is 

 M=P ( e -f- |/ m a x y). y may be found 

 from the differential equation of the 

 elastic line just as for a beam, (81), but 

 in this case the moment is in terms of y 

 and not of x. 



Putting e-f-2/max^e! for convenience, 



If a column is 



-M= P 

 2P, 



Fig. 137. 



dy 

 ti 



Integrating : 



=0 when y=0 hence 0=0 





