268 MECHANICS 



bottom angles. The distances of the centers of gravity of the 

 subdivisions of this section from the axis XX, Fig. 2, are: 



For the cover-plate 15.25-9.63 =5.62 



For the web-plates 9.63-7.50 =2.13 



For the 3i"X3i"X*"l_'s, 15.00- 1.06-9.63. =4.31 



For the4"X4"Xi" L's, 9.63-1.18 =8.45 



The moments of inertia of the respective parts about their 

 own neutral axes parallel to XX are: 



OA\x ( 1^3 



For the cover-plate = .25 



o vx l vx -I rr 3 



For the web-plates = 281.25 



From a steel manufacturer's handbook, the value of / for a 

 3i"X3J"X$" L is found to be 3.64; and for a 4" X 4" Xi" L 

 it is 5.56. Applying principle II, the moment of inertia of the 

 entire section is, / = .25+24XiX5.622+281.25+2X15X JX 

 2.132 + 2 X 3.64 + 2 X 3.25 X 4.312 + 2 X 5.56 +2X3.75X8.452 

 - 1,403.22. 



RADIUS OF GYRATION 



Let I denote the moment of inertia of any section and a its 

 area; then, the relation between / and a is expressed in the 

 formula, / = or 2 , in which r is a constant depending on the shape 

 of the section and is called the radius of gyration of the section 

 referred to the same axis as /. Then, 



EXAMPLE 1. What is the radius of gyration of the section 

 shown in Fig. 1 about the axis XX? 



SOLUTION. The moment of inertia of this section has been 

 found to be 618.6 and its area is 2X12X1 + 10.5X1 = 23.25 

 sq. in. Substituting in the formula, 

 H8.6 



- = 5.16 



/ 23.25 



EXAMPLE 2. Determine the distance b in the strut made 

 up of two latticed channels, as shown in Fig. 3, so that the radii 

 of gyration about the axes XX and YY will be equal. 



