CHAP. H.] SUPPLEMENT TO CHAP. XIV. 281 



CHAPTER II. 



HINGED ARCH IN GENERAL. 



3. Notation The outer forces in general. We suppose the 

 ends of the arch to be hinged at the abutments at the centre of gravity of 

 the end cross-sections. Then the end reactions must pass through these 

 points. These end reactions and the loads constitute, then, the outer forces. 

 For equilibrium, then, the horizontal components of these reactions must 

 be equaL Each of these components we call the horizontal thrust. 



We use the following notation [PL 23, Fig. 91] : 



R and R', the reactions at ends A and B. 



V and V, their vertical components. 



H, their horizontal component, or the thrust. 



a, the half span. 



A, the rise of arch. 



a, the half central angle, if arch is circular. 



The origin of co-ordinates we take at crown, x horizontal, y verticaL The 

 angle of radius of curvature at any point with y, or of tangent to curve at 

 any point with x, we call <. 



THE OUTER FORCES IN GENERAL. 



Suppose a single load P to act at any point E. Let its horizontal dis- 

 tance from crown be z, the corresponding central angle E O O be ft. 

 Then the conditions of equilibrium are : 



V + V = P, 



Va V'a Pa=0. 



From these last, we have 



V = P 



(16) 



v- P a - z 

 ~2T 



For a circular arc, since a r sin a, = r sin /3, 



Tr _. p sina + sinff sin a - sin/3 



V *^ . V f ^: . . . . I J. II 



2 sin a 2 SID a 



We distinguish three segments in the arch [Fig. 91], viz., A E, or from end 

 to the load ; E O, or from load to crown ; O B, or from crown to right 

 end. Quantities referring to the second we indicate by primes, to the 

 third by double primes. Thus for the tangential component of the result- 

 ant at any point within A E, we put Q ; for the normal component, N. 

 In E O, then, we have O' and N' ; in O B, O' and N". We have then 



