MECHANICS 



269 



SOLUTION. Let I x , r x , I y , r y be, respectively, the moments 

 of inertia and radii of gyration of a single C about the axes 

 XX and YY; a its area and CG, its 

 center of gravity, then, from the figure, 

 b = dc, and I x = ar x 2 ; also, I y 

 -\-ad-. Hence, by the condition of 

 the problem, ar^ = ar y 2 + ad 2 , or r^ 



= rj?+&. Whence, d= Vr^-r/. The 

 values of rx, ry, and c for any L may 

 be taken from a steel manufacturer's 

 handbook. For instance, for a 15" 

 C of 33 Ib. r x = 5.62, ^ = .912, and c=> 

 .794; hence, d= Vi!62*-. 9122= 5.545, 



A practical rule giving good approximate results for a channel 

 column or strut is to subtract r y from r x ; the result is b. 

 Applying this rule for the 15" E of 33 Ib. column or strut, b = 5.62 

 -.912 = 4.708. 



SECTION 



MODULUS AND MOMENT 

 RESISTANCE 



OF 



The expression , in which / is the moment of inertia and c 

 c 



the distance of the outermost fiber of the section from the 

 neutral axis, is called the section modulus. For a given material, 

 this quantity is a measure of the capacity of the section to 

 resist bending. Multiplied by the unit stress to which the 

 outermost fibers are subjected under given loads, the product 

 gives the amount of bending moment the section is resisting, 

 and is therefore called moment of resistance. If / is the unit 

 stress that certain loads develop in the outermost fibers of the 

 section, the moment of resistance is 



EXAMPLE 1. What is the section modulus of a 20-in. I beam 

 at 75 Ib. whose moment of inertia is 1,268.9? 



