HOMOGENEOUS BEAMS 167 



Sometimes, the sections are not symmetrical and the neutral 

 axis itself must be found. As an example, consider the 

 beam shown in Fig. 6 (a). It is composed of an I beam and 

 a channel riveted together. Such a beam is often used to 

 carry the track of a traveling crane. 



The first problem is to locate the neutral axis cd, Fig. 6 (6). 

 Assume any axis, as ab, about which to take ordinary 

 moments of the areas. The moment of each area about ab 

 is equal in each case to the product of the area and the dis- 

 tance from its center of gravity to ab. The areas and loca- 

 tion of the centers of gravity of the beam and column 

 sections can be obtained from the tables. Adding these 

 moments, the work is as follows: 



Moment of I beam = 20.59X9 = 185.31 



Moment of channel = 11.76 X (18 + .78) =220.85 

 Total 406.16 



The total area of the section is 20.59 + 11.76 = 32.35 sq. in. 

 Therefore, the distance from the line ab to the neutral 



axis cd is ^^=12.56 in. 

 32.o5 



It now remains, by the method given, to find the moments 

 of inertia of the channel and the I beam about the axis cd 

 and to add them together. This can be done with the help 

 of the tables, as follows: The moment of inertia of the 

 channel about cd is 9.39 + 11. 76 X (5.44 + . 78) 2 = 464.36. 

 Likewise, the moment of inertia of the I beam about the 

 axis cd is 921.2 + 20.59X (12.56-9) 2 = 1,182.15; therefore, 

 the total moment of inertia is 464.36+1,182.15 = 1,646.51 in 4 . 



FORMULAS FOR DESIGN 



BENDING-MOMENT FORMULAS 



In any homogeneous beam the maximum stress at any 

 section of the beam developed by the loads may be found 

 by the following formula: 



M-i/, 



c 



in which M is the bending moment at that point, in inch- 



