100 



LOCATION OF THE NEUTRAL AXIS. 



Art. 71. 



distributed over the area; it is the point of application of the 

 resultant of all the elements of an area treated like forces and 

 may, therefore, be found algebraically or graphically. 



It follows that the center of gravity is always on an axis of 

 symmetry. Thus in Fig. 67, the center of gravity is on VV; it, 

 is only necessary to find v or v s . 



In equation (13) v is the distance of any element of area 

 from an axis through the center of gravity. But since it is the 

 location of this axis that is to be found, moments can not b; 

 taken about it. Moments may be taken about any parallel axis 



d\ fi'i +1-2 

 as QQ, Fig. 67. Now / v / dA = /L(t; 2 +d 1 ), in which v' is 



measured from QQ. 

 may be zero, whence 



The center of gravity of a rectangular 

 cross section lies, of course, at the middle of 

 its height, but applying equation (14) 



A 



v'dA 



may be any assumed distance, it 



(14) 



Cv'dA Cv 



J J 



i 

 + 



5 



I-*- 



1 



Fig. 71. 



bh - bh - bh - 



Since the location of the center of gravity of a rectangle 

 is known, the center of gravity of any section ivhich may bo 

 divided into rectangles, is easily calculated. Any irregular sec- 

 tion may be thus treated, the accuracy depending upon the num- 

 ber of divisions made. It will sometimes be simpler to consider 

 a section as a combination of rectangles and triangles. 



Fig. 72 shows an area which may be divided 

 into two rectangles whose areas are 

 6X 1 /2= = 3.00 sq. in. 

 8X1 =8.00 sq. in. 



A =11.00 sq. in. 

 Taking moments about QQ 

 Ad = ILOOd = 3X8.25+8X4.0 = 56.75 

 From which d = 5 . 16 inches. 



The center of gravity may be found with fewer figures if 

 we take moments about an axis through the center of gravity of 



