SIMPLE STRUCTURES 



193 



making the intercept mn on the equilibrium polygon. Then, since 

 the triangles fmn and OA C are similar, we have 





or R'r' = H f x mn, 



which is the expression required by the theorem. 



For a system of parallel forces the pole distance H is constant 

 and hence the equilibrium polygon is similar to the moment 

 diagram for the forces on 

 either side of any given point. 

 Therefore the moment of all 

 the forces on one side of a 

 given point, taken with re- 

 spect to this point, is equal 

 to the constant pole distance 

 If multiplied by the intercept 

 made by the equilibrium poly- 

 gon on a vertical through the 

 point in question. 



To apply this method to the 

 roof truss shown in Fig. 131, 

 for example, draw the force 

 polygon and the correspond- 

 ing equilibrium polygon, as 

 shown in the figure. Now take 

 any section of the truss, such 

 as xy in the figure, and take 



moments of the stresses in the -i 6 FlG 131 



members cut about one of 

 the joints, say B. Then the condition of equilibrium 



V moments about B = 

 may be written 



Moment of stress in AF+ moments of P x , P 2 , E^ 

 But by the above theorem 



2) moments of P 1? P a , R# about B = bb f X Oh, 



bb' x Oh 

 Hence Stress in AF = . 



0. 



