Art. 100. TWO CONCENTRATED MOVING LOADS. 



105 



out disregarding the fact that R should be as near the middle as 

 possible. These two factors make it impossible 1o have the maxi- 

 mum moment at the middle; the center line of the brain must be 

 half way between P 1 and R, as will now be proven. 



dMc 





dx ' L ^~ ^ x ~ &)=0 wh en M C = M. 



x = yjj^b, that is, 1\ is \Ab to the left of the center and 

 therefore R must be %6 to right of it. The load under which thfi 

 maximum moment occurs must be as far OH OIK side, of the center 

 of ike beam as the resultant of the loads in on the other side of 

 the center. 



If P = 36000 Ibs., P 2 28000 Ibs., L 40 ft. and a = 8 

 ft., 36000 6 28000 (a 6) 28000X8-280006 and 6 31/2 

 ft., P t must be placed 1% ft. to the left of the center line so tlia*. 

 R will be 13/4 ft. to the right of it or 18*4 ft. from B. Now 



1== ^ (36000+28000) 29200 Ibs. and 



^l/max = jB a; = 29200 X18*4 = 533000 ft. Ibs. 



In a similar manner that value of x may be found which 

 will give a maximum shear, but it is evident that the maximum 

 shear occurs at the end when the heavier load is there and tho 

 other load is on the beam. In the above example, when P is at A, 



max == 36000+ f| 28000 36000+22400 58400 Ibs. 



101. Any Number of Con= 

 centrated Moving Loads Fixed 

 Distances Apart. Fig. 118 shows 

 six loads which move together. 

 The distances between the loads 

 being known, the location of 

 the resultant is found at a dis- 

 tance 6 from P 4 . Exactly the 

 same reasoning will apply as for 

 the case of two loads, and the 

 same rule for the position giv- 

 ing maximum moment will ap- 

 ply; the load must be so placed 

 that the center line of the beam 

 will bisect 6, when the maximum 



moment occurs under P 4 . The proof is as follows, R' being the 

 resultant of the loads to the left of C. 



