490 PROFESSOR MOSELEY, ON THE THEORY OF 



-i=M¥['^9"---'a^i'i-F)i- «■ 



^=4L--9-i>-if.e-?)-^i--7)- «")• 



It has been supposed that the load X is collected over a single point 

 in the arch, or rather in a single line stretching across it in a direction 

 parallel to the axis of the cylinder of which it forms part. Let it 

 now be imagined to be distributed in any way over the extrados, but 

 symmetrically on the two opposite semi-arches. Find the center of gravity 

 of the load on either semi-arch, and let x be its distance from the vertical 

 which passes through the center of the circle of which the semi-arch 

 forms part. Imagine the whole load X to be collected in this point, 

 and on this hypothesis determine ^ and P ; the values thus determined will 

 evidently be their true values. To find the line of resistance, substitute 

 in equation (46) the value of P, and for X and x substitute their values 

 in terms of 6; that is, for X substitute the load incumbent upon that 

 portion of the arch which subtends the angle 9, and for x the distance 

 from the vertical through the center, of the center of gravity of that 

 portion of the load. The resulting equation will be the true equation 

 to the line of resistance. Thus the point where the resultant pressure 

 of the arch intersects the supporting surface of the abutment will become 

 known ; and its direction being found, as in equation (57), all the cir- 

 cumstances which determine the equilibrium of the uhutment will be 

 known, and the conditions of its equilibrium may be determined by 

 the equation given in section 5 of this paper. The analytical discussion 

 of these conditions, and of that case in which the arch being overloaded 

 at the haunches, its rupture takes place by the elevation of the crown, 

 is yet wanting to complete the theory of the arch. 



A very simple expression for P offers itself in the case in which 

 a = 0, or in which the thickness of the arch is considered evanescent in 

 comparison with its radius. In this case 



P=Xcot>i'„ = x{g+\/p-sin^e)"-4^sec^|)...(68). 



