STRENGTH OF BEAMS 51 



Since the normal, or bending, stresses are the only horizontal forces 

 acting on the portion of the beam considered, in order to satisfy 

 the condition of equilibrium V horizontal forces = we must have 



Resultant tensile stress = resultant compressive stress. 



Therefore, since the tensile and compressive stresses act in opposite 



directions (that is, are of opposite sign), the algebraic sum of all the 



normal stresses acting on the section must 



be zero. Thus, if A^4 denotes an element of 



area of the section, and p the intensity of the 



normal stress acting on it, the total stress on 



this area is p&A, and consequently 



= 0. FIG. 48 



Now, if the normal stress at a variable distance y from the neutral 

 axis is denoted by jo, and that at some fixed distance y' is denoted 



by p 1 , then, from the straight-line law, . = . or 



P V 



Inserting this value of p in the above condition of equilibrium, 

 it becomes / 



Therefore, since p' and y' are definite quantities different from zero, 

 we have V/A^ = 0. But, from article 13, the distance of the centroid 

 from the neutral axis is given by 



and if T^A^ = 0, then also y Q = 0. Therefore the neutral axis 

 passes through the centroid of the cross section ; that is, the neu- 

 tral axis coincides with the horizontal centroidal axis. 



34. Fundamental formula for beams. For equilibrium the result- 

 ant moment of the normal stresses acting on any cross section must 

 be equal to the resultant moment of the external forces on one 

 side of the section, taken with respect to the neutral axis of the 



