182 



RESISTANCE OF MATERIALS 



Now the closing side A' G' of the equilibrium polygon determines 

 the line of action of the resultants P' and P" at A' and G 1 respec- 

 tively. For a simple beam, however, the reactions are vertical. 

 Therefore, in order to find these reactions, each of the forces P 1 and 

 P" must be resolved into two components, one of which shall be 

 vertical. To accomplish this, suppose that a line OH is drawn from 

 the pole in the force diagram parallel to the closing side G'A' of 

 the 'equilibrium polygon. Then HO (or P') may be replaced by 

 its components HA and AO, parallel to R l and A'B' respectively; 



FIG. 120 



and similarly, OH may be replaced by its components FH and OF, 

 parallel to R Z and F 1 G' respectively. HA and FH are therefore the 

 required reactions. 



111. Equilibrium polygon through two given points. Let it be 

 required to pass an equilibrium polygon through two given points, 

 say M and ^ (Fig. 121). 



To solve this problem a trial force diagram is first drawn with 

 any arbitrary point as pole, and the corresponding equilibrium 

 polygon MA'B'C'D'E' constructed, starting from one of the given 

 points, say M. The reactions are then determined by drawing a 

 line OH parallel to the closing side ME 1 of the equilibrium polygon, 

 as explained in the preceding article. 



The reactions, however, are independent of the choice of the pole 

 in the force diagram, and, consequently, they must be of amount AH 

 and HE, no matter where is placed. Moreover, if the equilibrium 



