LAWS OF BONE ARCHITECTURE 199 



Then z- = unit stress at any distance z from the neutral 

 c 



surface. Designating an elementary area by {Da) and its dis- 

 tance from the neutral axis by z then the resisting moment of 

 the elementary area (Da) of the cross section, is equal to Da 



times the unit-stress iz--) on that area times the distance z from 



c 



the neutral axis to this area or {Do) z--z. Designating the 



c 



resisting moment of a single elementary area {Da) by the symbol 



(Di¥J, then 



(Dikffi) = {Da) Z-- z 



sz'- 

 = (Da) - 



The resisting moment for all elementary areas may be similarly 

 expressed. It therefore follows that the total resisting mo- 

 ment for the entire area of the cross section will be the sum of all 

 the resisting moments of all the elementary areas. This may 

 be expressed by the following formula, in which S represents 

 the adding together, or summation of all the resisting moments 

 of the elementary areas, 



sz^ 

 Sum of all (DMi) = sum of all (Da) — 







y. SZ" 



Q9. Since s and c are constants for any given case the above 

 formula may be WTitten, 



Resisting moment of the section =2) {Da)-z^ 



The expression 2 {Da).z^ is generally called the moment of 

 inertia of the section considered, and is usually represented 



by the symbol I. Hence, the sum of the internal moments - I, 



must equal the sum of the external bending moments about 



