EQUILIBRIUM POLYGON AS A FRAMED STRUCTURE 



only known point on the reaction R 2 , and draw the polygon as prev- 

 iously described. A line drawn through point o in the force polygon 

 parallel to the closing line of the equilibrium polygon will meet R lt 

 drawn parallel to reaction R lt in the point y, which is also a point on R 2 . 

 The reactions R^ and R are therefore completely determined in direc- 

 tion and amount. 



The method just given is the one commonly used for finding the re- 

 actions in a truss with one end on rollers (see Chapter VII). 



Equilibrium Polygon as a Framed Structure. In (a) Fig. 

 17, the rigid triangle supports the load PL Construct a force polygon 



T7T 



(a) 



FIG. 17. 



by drawing rays a I and c i in (b) parallel to sides a I and c I, respec- 

 tively, in (a), and through pole i draw I b parallel to side I b in (a). 

 The reactions R and R 2 will be given by the force polygon (b), and 

 the rays I a, i c and i b represent the stresses in the members i a, I c 

 and i b, respectively, in the triangular structure. The stresses in I a 

 and i c are compression and the stress in i b is tension, forces acting 

 toward the joint indicating compression and forces acting away from 

 the joint indicating tension. Triangle (a) is therefore an equilibrium 

 polygon and polygon (b) is a force polygon for the force /V 



From the preceding discussion it will be seen that the internal 

 stresses at any point or in any section hold in equilibrium the external 

 forces meeting at a point or on either side of the section. 



