Archdeacon Pratt on the Stability of Arches. 

 Fig. 2. 



263 



ZBAM = ai , ZCB6 = « 2 , ZCDd=« 3 , ZDEe=«, 



ZG 1 AB= / e i , ZG 2 BC 



:/? 2 , ZG 3 DC=/3 3 . ZG 4 Ee=/3 4 , 



AB = «j, 



BC: 



BG< 



<2> 



CD = « 3 , DE = « 4J AM = m, ME=^ 



DG 3 =6 3 , 



EG 4 =^ 



ftp rip ^ 2 , ri 2 ,/t 3 , ii 



A Q , H M> A 4 , H 4 are the lengths of the projections of AGj, 



AB, BG 2 , BC, DG 3 , DC, EG 4 , ED upon the horizontal line AM. 



I. Suppose a load, L (fig. 1), to be placed on the bridge over the 

 arch, and not in the centre. B is the point of the extrados under 

 the middle of the load. Draw BAX, BCD, DEY tangents to the 

 intrados. Imagine the arch at the points A, B, C, D, E to be 

 divided into four parts, and to be forcibly held in the position 

 shown in the diagram, in which the arch is slightly open at 

 joints through those points. If the equilibrium of the arch is 

 stable, the parts will fall back into their places when left to 

 themselves ; if unstable, the arch will fall. 



Let Wj, W 2 , W 3J W 4 be the weights of these four portions of 

 the arch, including the total load each carries (not omitting L). 

 Gj, G 2 , G 3 , G 4 (in fig. 2) the centres where these weights may be 

 supposed to be collected. Let V be the height of the centre of 

 gravity of these four weights above the horizontal AM. Then 



(W 1 + W 2 + W 3 + W 4 )V=W 1 6 1 sin(« 1 + ft) 



+ W 2 («! sin a x + b 2 sin (a 2 + /? 2 ) ) + W 3 (?2 + a 4 sin a 4 + 6 3 sin (/3 3 — a 3 ) \ 



+ W 4 (ra-f 6 4 sina 4 ). 



Also the angles are connected by the following relations 

 (N.B. a 3 =« 2 , but I shall at first reason on the more general 

 supposition that they are not necessarily so), 



a x cos aj + a 2 cos a 2 + a 3 cos a 3 + a 4 cos « 4 = m, 

 a x sin a, l -f a 2 sin a 2 + a 3 sin a 3 — a 4 sin a 4 = n. 



Differentiating these with respect to the angles, which alter when 

 the arch is slightly moved out of its position, 



«! sin a Y 8a x + a 2 sin a 2 Sa 2 + a 3 sin a 3 Bot 3 + a 4 sin a 4 Sa 4 = 0, 

 «! cosaj 8a x -f # 2 cos a 2 Sa 2 + a 3 cos a 3 Bcc 3 —a 4 cos a 4 Sa 4 = 0. 



