AKCHES AND AKCHED H1HS 183 



When the equilibrium polygon has been drawn for the given system 

 of loads, the stress at any point of an arched rib can be calculated by 

 the method explained in Article 135. Thus, in Fig. 132 (A), let AGF 

 denote the arched rib, P lt P 2 , etc. the given loads, and ABCDEF the 

 corresponding equilibrium polygon. Then the stress on any section 

 mn is due to a force acting in the direction CD, of amount equal to 

 the corresponding ray OC 1 of the force diagram. 



Consequently, if the rib is composed of a solid web and flanges, 

 as shown in Fig. 132 (B) t the direct stress on the section is equal in 

 amount to the ray OC' of the force diagram, the bending stress on 



the upper flange is -* > the bending stress on the lower flange is , 

 d d 



and the shear normal to the rib is OC' sin a, where a is the angle 

 between CD and the tangent to the rib at the section. 



Similarly, for the trussed rib shown in Fig. 132 (C), by taking 

 moments about L and S the stresses in US and LK are found to be 



P z P z 1 



- and - - respectively, while the normal component of the stress 

 d d 



in LS is OC sin a. 



Arched ribs are usually constructed in one of three different ways : 

 (1) hinged at the abutments and at the crown; (2) hinged at the 

 abutments and continuous throughout; (3) fixed at the abutments 

 and continuous throughout. The method of constructing the equilib- 

 rium polygon differs for each of these three methods of support, and 

 will be treated separately in whut follows. 



150. Three-hinged arched rib. When a member is free to turn 

 at any point the bending moment at that point is zero, and con- 

 sequently the equilibrium polygon, or bending moment diagram, 

 passes through the point. For a three-hinged arched rib, therefore, 

 the equilibrium polygon must pass through the centers of the 

 three hinges and is therefore completely determined, as explained in 

 Article 134. 



151. Two-hinged arched rib. Consider an arched rib hinged at 

 the ends and continuous between these points. In this case the 

 equilibrium polygon must pass through the centers of both hinges, 

 but since there is no restriction on the vertical scale, this scale may 

 be anything whatever, depending on the choice of the pole in the 



