ARCHES AND ARCHED RIBS 



167 



corresponding equilibrium polygon starting from the point M will 

 pass through both the points L and N. 



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

 polygon can be drawn through three given points. In other words, an 

 equilibrium polygon is completely determined by three conditions. 



135. 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. 124. Then each 

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

 cede 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 



K 



FIG. 124 



D 



left of P 9 , and acts along C'D'. 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 jp'D'. 



Let a denote the angle between the segment C'D' of the equilibrium 

 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 ; a shear at right angles to this, of amount OC sin ; and a 

 moment of amount OC-d, where d is the perpendicular distance of 

 C'D 1 from S. 



From Fig. 124, it is evident that the horizontal component of any 

 ray of the force diagram is equal to the pole distance Off. There- 

 fore if OC is resolved into its vertical and horizontal components, the 

 moment of the vertical component about S is zero, since it passes 



