146 HOMOGENEOUS BEAMS 



EXAMPLE. Prove the formula for the modulus of a section 

 as shown in Fig. 5 to be correct as given in the table on 

 page 133. 



SOLUTION. According to the table, the moment of inertia 



of the section is - X d 3 . It is a question whether 



However, on examining the values 

 given for these two quantities 

 in the table, it is seen that they 

 are the same, with the exception 

 of the first part of the numerator. 

 For d, this part of the numer- 

 ator is b + 2bi, while for c it is 

 bi + 2b. In Fig. 5, b is greater 



'* "* than bi; therefore, bi + 2b is 



FlG - 5 greater than b + 2bi, and it fol- 



lows that c is greater thanci. Accordingly, the formula is 



__ ^ a 



} ~ c~ 36 (b + bi) ~"~ b + bi 3 



which is the same as the one given in the table. 



PROPERTIES OF ROLLED-STEEL SHAPES 



The table of values for standard rolled sections already 

 given shows various cross-sections of steel beams. Each 

 one of these beams is made in various sizes and weights. 

 On account of the confusion that ensued from using beams 

 of different sizes from different manufacturers, the latter 

 agreed to adopt standard sizes, which are known as American 

 standard. Therefore, knowing the dimensions of the section 

 of any beam that is rolled according to the American stand- 

 ard, the properties of the section may be found by using the 

 table already given. This table, however, while quite accu- 

 rate, is inconvenient to use, because the formulas it con- 

 tains are long and complicated. The following tables of 

 properties of standard sections are therefore given to diminish 

 the labor required in beam design. 



