184 RESISTANCE OF MATERIALS 



Now, if the equilibrium polygon is to pass through N, the pole of 

 the force diagram must lie somewhere on a line HK drawn through 

 H parallel to MN, as explained in the preceding article. The next 

 step, therefore, is to determine the position of the pole on this line 

 HK, so that the equilibrium polygon through M and N shall also 

 pass through L. This is done by drawing a vertical LS through L 

 and treating the points M and L exactly as M and N were treated. 

 Thus OABCD is the force diagram for this portion of the original 

 figure, and MA'B'C' S is the corresponding equilibrium polygon, the 

 reactions for this partial figure being H'A and DH'. If, then, the 

 equilibrium polygon is to pass through L, its closing side must be 

 the line ML, and consequently the pole of the force diagram must 

 lie on a line H'K* drawn through H' parallel to ML. The pole is 

 therefore completely determined as the intersection 0' of the lines 

 HK and H'K'. If, then, a new force diagram is drawn with 0' as 

 pole, the corresponding equilibrium polygon starting from the point 

 M will pass through both the points L and N. 



Since there is but one position of the pole 0', only one equi- 

 librium polygon can be drawn through three given points. In other 

 words, an equilibrium polygon is completely determined by three 

 conditions. 



113. Application of equilibrium polygon to calculation of stresses. 

 Consider any structure, such as an arch or arched rib, supporting 

 a system of vertical loads, and suppose that the force diagram and 

 equilibrium polygon are drawn as shown in Fig. 123. Then each 

 ray of the force diagram is the resultant of all the forces which 

 precede it and acts along the segment of the equilibrium polygon 

 parallel to this ray. For instance, OC is the resultant of all the 

 forces on the left of P z and acts along C'D f . Consequently, the 

 stresses acting on any section of the structure, say mn, are the same 

 as would result from a single force OC acting along C'D'. 



Let 6 denote the angle between the segment C'D' of the equi- 

 librium polygon and the tangent to the arch at the point S. Then 

 the stresses acting on the section mn at S are due to a tangential 

 thrust of amount OC cos 6 ; a shear at right angles to this, of amount 

 OC sin 6 ; and a moment of amount OC d, where d is the perpen- 

 dicular distance of C'D' from S. 



