20 PRACTICAL STRUCTURAL DESIGN 



Checking by taking moments about the right end. 

 (4 x 250) + (8 X 300) + (12 x 200) 



K\ = r-= - = oou.U/ 



15 



R 2 = 750 = 386.67 = 363.33 Ibs. 



The same results may be obtained by common proportion, but 

 upon examination the method by moments is seen to be exactly 



the same thing, but with 

 less work. If the reactions 

 were found by the common 

 school arithmetical process 

 of proportion the work 



would be longer and mis- 

 Fig. 11 Several Concentrated Loads on takeg more t to occur 



Beam Having Two Supports _., , . , , 



The method of moments 



is the shortest and neatest way of working. 



In Fig. 12 three concentrated loads are shown on a beam of 

 which the weight is uniformly distributed. First find the reactions 

 due to the concentrated loads. Then add to each reaction half 

 the weight of the uniformly distributed load. 



Example. Assume a beam having a weight of 50 Ibs. per 

 lineal foot on a 15-ft. span, carrying the concentrated loads given 

 in the last example. What are the reactions? 



Answer. Weight of beam = 15 X 50 = 750 Ibs. 

 The reaction at each end = 



375 Ibs. 



Li 



R!= 386.67 + 375 = 761.671bs. 

 R 2 = 363.33 +375= 738.33 " 



Total load 1500.00 " 



When the bending moment Fig. 12 Uniform Load and Several 

 is wanted at any section of a Concentrated Loads on Beam 



Having Two Supports 

 beam we assume the beam at 



the section to be supported at the section and moments are taken 

 about it as if it were a cantilever beam. The reaction is an up- 

 ward force creating an upward moment and the loads are down- 

 ward forces creating a downward moment. The difference between 

 the moments in opposite directions is the bending moment at the 

 chosen section. When a beam is freely supported on two or more 



