102 MECHANICS 



direction; the sum of their moments, in foot-pounds, with 



respect to the assumed center, may be computed as follows: 



8.000X 5= 40000 



6,000X19 = 1 14000 



2,000 X27= 54000 



Total, 2~0 8 



This is the total negative moment about the point b. The 

 positive moment about this point is of course RiX'60. 

 Since the resultant moment is zero, the positive moment 

 must equal the negative moment, or 208,000=30 Rj or 

 RI= 208,000 H- 30 = 6,933$ Ib. The sum of all the loads is 

 2,000 + 6,000 + 8,000 = 16,000 Ib. This is also the sum of 

 the reactions. Therefore, R 2 = 16,000- 6,933 J =9,066$ Ib. 

 In this problem, the weight of the beam itself has been 

 neglected. 



EXAMPLE 1. What is the reaction at R 2 in Fig. 7? 

 SOLUTION. In computing the moment due to a uniform, 

 or evenly distributed, load, as at a, the lever arm is always 

 considered as the distance from the center of moments to 



l ^3SO(Tr6~j>ef]runnina ft. * 



1 Cftltcr of Gravity 

 'i of Uniform Load. 



lOft. 



30ft. 



FIG. 7 



the center of gravity of the load. The amount of the uni- 

 form load a is 3, 000 X 10 = 30,000 Ib., and the distance of its 

 center of gravity from RI is 13 ft. Therefore, the moments 

 of the loads on this beam about RI, in foot-pounds, are as 

 follows: 



30,000X13 = 390000 



4.000X 4= 16000 



9,000X20 = 180000 



Total, 586000 





