MECHANICS 



267 



Formulas for the values of / about an axis passing through 

 the center of gravity of the section are given for various forms 

 of sections in the accompanying table. For any other section, 

 it can be computed by means of the following principles: 



Principle I. The moment of inertia of a section about any axis 

 is equal to the algebraic sum of the moments of inertia about the 

 same axis, of the separate parts of which the figure may be con- 

 ceived to consist. 



Principle II. The moment of inertia of any figure about an 

 axis not passing through the center of gravity, is equal to the moment 

 of inertia about a parallel axis through the center of gravity, plus 

 the product of the entire area of the section by the square of the 

 distance between the two axes. 



EXAMPLE 1. Find the moment of inertia about the neutral 

 axis XX of the Bethlehem I column section having dimen- 

 sions as shown in Fig. 1. 



FIG. 2 



SOLUTION. Conceive the section to consist of the square 

 A BCD minus twice the rectangle abed. Then, by applying 

 principle I and the formulas of the table for moments of inertia, 

 12 2X5.75X10.5* 



I= T 2 IT" 



NOTE. This result can be obtained directly by the I beam 

 formula, given in the same table. 



EXAMPLE 2. Find the moment of inertia of the section 

 shown in Fig. 2 about the neutral axis parallel to the cover- 

 plate. 



SOLUTION. The neutral axis passes through the center of 

 gravity, which has been found to be 9.63 in. from the back of the 



