160 CONSOLIDATED STEEL CORPORATION 



The relations existing between bending moment, moment of resistance, 

 section modulus and stress per square inch are expressed thus: 



o m 

 m=R ^ = T 



, c f m 



m=/S /=g 



When the bending moment is in foot-pounds, the following relations are 

 useful : 



C = 8M M= 



o 



If W is a uniformly distributed load in pounds, and the span, L, is taken in 

 feet, then: 



C=WL W = 



L 



The last two formulas are convenient. To find the safe uniformly distrib- 

 uted load in pounds for any section, it is only necessary to divide its coefficient 

 of strength by the span in feet- If the uniformly distributed load in pounds is 

 given, multiply it by the span in feet and the result is the coefficient of strength 

 required by the section. 



On the next page formulas are given for finding bending moments, safe 

 loads and deflections for beams loaded and supported in usual ways. Bending 

 moments will be in foot-pounds or inch-pounds according as the lengths are 

 taken in feet or inches. To obtain deflection in inches the lengths must be taken 

 in inches. 



For illustration, take a center load of 30,000 pounds on a span of 20 feet: 



M = 3(M)0 4 * 20 = 150,000 foot-pounds 

 C = 8M = 8 X 150,000 = 1,200,000 



The nearest beam is a 20-inch Bethlehem I-Beam, weighing 59 pounds per 

 foot, which has a coefficient of 1,250,300. 



If the bending moment has been taken in inch-pounds, then 



S =j = l,800,000-f-16,000 = 112.5 



The beam selected by the first method has a section modulus of 117.2, which 

 is the nearest to that required. Both methods of calculation give identical 

 results. 



