222 Kansas Academy of Science, 



This formula may be deduced very simply in another way from 

 the well-known theorem that, with both stresses and strains propor- 

 tional to the corresponding distances from the neutral axis, the 

 neutral axis must pass through the center of gravity of a homo- 

 geneous section. To bring our non- homogeneous section under 

 this theorem, we must replace the steel by concrete, and also add 

 enough concrete at the same distance from the top and bottom of 

 the beam to take the excess of stress in steel, over that taken by 

 the concrete in its place. (See fig. 2.) The unit stresses in the 



steel are fs^ = e ^fe^ and fs = e^^ ^fc. The total stresses taken 



by the steel are fs^As^ = e ^fg^Ag^ and fsAs = e^^ — -^fcAg. 



^ y 



The unit stresses in the concrete at this same position are 

 ^-Z^f,} and L-iife 



So the stresses taken by the concrete replacing the area of the 

 steel will be 



^^Z^fe^A,^ and ^-^-^f,A3 



Hence, denoting by Ag^ and Ac the additional area of concrete 

 in compression and tension, we have 



X— di' J J x-dii J ^ 



e 'a -^s fc -^s 



A.^= ^ ^ = ( e-1 )A. 



X— dii j^ 



'c 



X 



and Ac= (e — l)As. Taking moments about the bottom of the sec- 

 tion, we have, for the position of the center of gravity of this area,. 





+ (e-lJA,i(h-da^) + (e-i)A,di 



or. 



bh+(e-l)A,i+(e-l)As 

 A,di+A,i(hi-dii)j(e_l) 



bh-^ 



A+(A, + V)(e-1) 



