CHAP. XVI.] THE INVERTED ARCH. 333 



If this also passes within the middle third of the arch, as 

 represented by the dotted line, the arch is, under all circum- 

 stances, stable, and can fully resist the tension of the rear 

 chains. 



We can now, finally, so dispose the joints as to prevent 

 sliding. 



15. Stiffened Suspension System. We have already re- 

 ferred to the methods of stiffening the cable or chain so as to 

 prevent the changes of shape due to partial loading. Of these 

 methods, it only remains to notice particularly the last, viz., by 

 means of an auxiliary truss. The office of this truss is to dis- 

 tribute a partial load over the whole length. We have now to 

 investigate the forces which act upou the truss. 



In PI. 27, Fig. 107, let the chain be acted upon by the truss 

 represented by A B, which is called into action only by a par- 

 tial load, and not at all by a total uniform load. We can neg- 

 lect then the weight of the truss itself, as this is borne by the 

 cable. At the apices 3, 4, 5, 6, 7 let us suppose partial loads 

 indicated by the small arrows pointing down. Then, at every 

 point of connection with the chain, we have the reactions 1', 2', 

 3', etc., acting upwards. Now the truss must prevent deforma- 

 tion, and hence these forces are dependent upon the form of 

 the cable itself. Indeed, if we take any point, as O, as a pole, 

 and draw lines parallel to the respective sides of the cable, 

 these lines will cut off upon a vertical P' these forces. The 

 absolute value of these forces will, it is true, vary according to 

 the position of the pole assumed, but their relative proportions 

 remain always the same. The resultant P' of all these forces 

 passes then through the intersection of the two outer sides of 

 the catenary. 



Since the truss distributes its load P upon the cable, the reac- 

 tion B at the right support is here zero. The reaction, however, 

 at A cannot be zero unless P and P' coincide, as is the case for 

 total uniform load. These, then, are all the forces which are 

 kept in equilibrium by the truss. If P is given, P' and the re- 

 action at A can be easily found, and if we then divide P', ac- 

 cording to the form of the chain, into the portions 1', 2', 3', 

 etc., we have the forces at each apex. 



Thus we lay off to scale the given forces 3, 4, 5, 6, 7 = P, and 

 with a pole distance any convenient multiple of the height of 

 truss draw lines to these points of division, and then construct 



