60 



REACTIONS FOR A SINGLE FORCE. 



Art. 46. 



is in equilibrium with P and 7? 2 its line of action passes through 

 D. The force polygon gives the magnitudes of Ri and R 2 . 



Fig. 40. 



In like manner, reactions may be found for a load on the 

 right hand half. By finding the resultant of the two reactions] 

 at A (for example) we get the reaction for both loads. If there 

 are a number of loads, the case may be reduced to the above by 

 first finding the resultant of all the loads on each half. 



Algebraically, we have from 5 hor. comps.= 

 Hor. comp. R! = hor. comp. R 2 H 



From S vert, comps. = 0, 



p = vert. comp. J^+vert. comp. R 2 = V 1 -{-V 2 . 



Prom S moms, about C = (considering only half the arch), 

 Hr = F 2 X $L or Br+P(L li) = Vi X%L. 



Three of these equations determine the unknows H, "F,, and 

 V 2 . R! is the resultant of H and T^ and R 2 of H and V 2 . Equa- 

 tions may, of course, be written with centers at A or B. 



In this case, just as in that of Fig. 31, 



Substituting V 2 in the equation above, 



Hr = \Ph find, 11= \ pA. 



47. Reactions for Any Number of Loads. The reaction 

 for a number of loads may be gotten by finding the reaction for 

 each load separately and adding these reactions, or by finding 



