483 PROFESSOR MOSELEY, ON THE THEORY OF 



.■.p = ^\/a{a + 2)[^+a%ia + l)\-a(i,a^ + a + l)C0Se-^] (54). 



4', P, p are thus completely determined, and all the circumstances of 

 the equilibrium of the circular arch, thus loaded, are known. 



If there be no loading, and the two semi-arches be parts of the same 

 continuous cylindrical mass, X = 0, and 6 = 0. 



Therefore, by equation (52), ^ = 0. 



In this case, therefore, the point of rupture is in the crown of the 

 arch (Fig. 18), at the intrados, also by equation (53), 



therefore, by equation (54), p = r. 



Substituting these values of P and p in equation (46), we obtain 

 for the equation to the line of resistance in the unloaded circular arch, 



a + 2 - {\a + a + 1) COS g 



= r 



(ia+l ) sin + (1 - \a) cos Q 



.(55). 



Let y\r be the angular distance from the crown, at which the line 

 of resistance meets the extrados, (Fig. 17), as * is that at which it 

 meets the intrados. Therefore, by the preceding equation, 



g + 2 - {\a + g + 1) cos x// 

 ■^ ■^" ^ (lg + l)v//Sinx/, + (l-ia')C0S>|r' 



... (1 + a)(ig + 1) x// sin x|. + (2 + 2a -^a') COS r// = a + 2 (56). . 



x|/ determined from this equation will measure the greatest semi-arch, 

 which being unloaded, can be made to stand. 



To determine the inclination ^ of the resultant P, to the vertical, 

 corresponding to the angle d, we have 



P, sin * = horizontal force on segment = P = y* { g - ^ a' } , 



P, cos <t> = vertical =mass of segment = ^r'!g= + 2a} 0; 



