known from the works of Emerson, Hutton, and Whewell. It is impossible to conceive any arrangement of the parts of an arch by which its stability can be more effectually secured, so far as the tendency of its voussoirs to slide upon one another is concerned: there is, however, probably, no practical case in which this tendency really affects the equilibrium. So great is the limiting angle of resistance in respect to all the kinds of stone used in the construction of arches, that it would perhaps be difficult to construct an arch, the resultant pressure upon any of the joints of which above the springing should lie without this angle, or which should yield by the slipping of any of its voussoirs. Traced to the abutment of the arch, the line of resistance ascertains the point where the direction of the resultant pressure intersects it, and the line of pressure determines the inclination to the vertical of that resultant; these elements determine all the conditions of the equilibrium of the abutments, and therefore of the whole structure; they associate themselves directly with the conditions of the loading of the arch, and enable us so to distribute it as to throw the points of rupture into any given position on the intrados, and give to the line of resistance any direction which shall best conduce to the stability of the structure; from known dimensions, and a known loading of the arch, they determine the dimensions of piers which will support it; or conversely from known dimensions of the piers they ascertain the dimensions and loading of the arch, which may safely be made to span the space between them. The inclination of the resultant pressure at the springing to the vertical may be determined independently of the line of pressure, as will hereafter be shown. 336. TO DETERMINE THE LINE OF RESISTANCE IN AN ARCH WHOSE INTRADOS IS A CIRCLE, AND WHOSE LOAD IS COLLECTED OVER TWO POINTS OF ITS EXTRADOS SYMMETRICALLY PLACED IN RESPECT TO THE CROWN OF THE ARCH. Let ADBF represent any portion of such an arch, P a D PE pressure applied at its extreme voussoir, and X and Y the horizontal and vertical components of any pressure borne upon the portion DT of its extrados, or of the resultant of any number of such pressures; let, moreover, the co-ordinates, from the centre C, of the point of application of this pressure, or of this resultant pressure, be x and y. Let the horizontal force P be applied in AD at a vertical distance p from C; also let CT represent any plane which, passing through C, intersects the arch in a direction parallel to the joints of its voussoirs. resultant of the These pressures Let this plane be intersected by the pressures applied to the mass ASTD in R. are the weight of the mass ASTD, the load X and Y, and the pressure P. Now if pressures equal and parallel to these, but in opposite directions, were applied at R, they would of themselves support the mass, and the whole of the subjacent mass TSB might be removed without affecting the equilibrium. (Art. 8.) Imagine this to be done; call M the weight of the mass ASTD, and h the horizontal distance of its centre of gravity from C, and let CR be represented by p, and the angle ECS by 0, then the perpendicular distances from C of the pressures M + Y and P-X, imagined to be applied to R, are p sin. and p cos. 9; therefore, by the condition of the equality of moments, P (M + Y)p sin, 0+ (P − X)p cos. ◊ = Mh+ Yx − Xy + Pp; Mh+Yx-Xy+Pp ́(M+ Y) sin. 6+(P−X)cos. which is the equation to the line of resistance. (453), M and h are given functions of ; as also are X and Y, if the pressure of the load extend continuously over the surface of the extrados from D to T. B R 6 A E It remains from this equation to determine the pressure P, being that supplied by the opposite semiarch. As the simplest case, let all the voussoirs of the arch be of the same depth, and let the inclination. ECP of the first joint of the semiarch to the vertical be represented by, and the radii of the extrados and intrados by R and r. Then, by the known principles of statics, Mh=ffr2sin. Oddr = — } (R3 — r”) (cos. 6 — cos. ©) ; also, M=1(R2— p2)(0 — ©) ; •*• p {¦(R2 — r2)(0 — ☺) sin. + Y sin.-X cos.+Pcos.} =(R3 3)(cos. -cos.)+Yx-Xy+Pp. — ✪ which is the general equation to the line of resistance. THE ANGLE OF RUPTURE. 337. At the points of rupture the line of resistance meets the intrados, so that there p=r: if then be the corresponding value of 0, r {} (R2 — r2) (¥—O)sin. V+ Ysin. Y-X cos. Y + Pcos. V} - = }(R3 — μ3) (cos. —cos. Y) + Yx-Xy+Pp Also at the points of rupture the line of resistance touches simplify the results, that the pressure of the load is wholly in a vertical direction, so that X=0, and that it is collected dY over a single point of the extrados, so that 0, and dif r{}(R2-r2)(¥-0) cos.+ (R2-2) sin.+Y cos. -Psin. Y}=(R3—73) sin. Y; hence, assuming R=r(1+a), +a2 (2a+3) tan. Y= {6X−3a (a+2)0} +3a (a+2)¥ (456). Eliminating (-O) between equations (455) and (456), we have Eliminating P between equations (455) and (456), and re ducing, Y* (p cos. Y+r sin. v −1}=(\a2 +a) (1 —2, cos. ¥) (¥—0)+”, (ja2+fa3) sin. ¥ — {(a+a2+}a3) cos. O—(† a2+a) cos. ¥} sin. v (458). * This equation might have been obtained by differentiating equation (454) in respect to P and 0, and assuming =0 when r and Y are substi dP ρ tuted for p and 0; for if that equation be du dP + dP do represented by u=0, u being a = function of P, p and 0, du sult =0 is therefore obtained, whether we assume do which last supposition is that made in equation (456), whence equation dP (458) has resulted. The hypotheses=0,p=r, determine the minimum of the pressures P, which being applied to a given point of the key-stone will prevent the semi-arch from turning on any of the successive joints of its voussoirs. B Let AP=λr; therefore ?= (1+λ) cos. . Substituting this value of 2, Y sin. +(1+) cos. O cos. Y-1 by which equation the angle of rupture ¥ is determined. has a If the arch be a continuous segment the joint AD is vertically above the centre, and CD coinciding with CE, O=0; if it be a broken segment, as in the Gothic arch, given value determined by the character of the arch. pure or equilateral Gothic arch, ✪=30°. Assuming ✪=0, and reducing, Y In the {;-(tan.λ cot. ¥) } = (4a2+a) { (tan.—2 cot¥) Y-vers. ¥ } +λ(ja2+jæ3) · · · (460). It may easily be shown that as increases in this equation Y increases, and conversely; so that as the load is increased the points of rupture descend. When Y=0, or there is no load upon the extrados, (tan. 3+2a -Acot. VV-vers.+2+a = = 0...... (461). When x=0, or the load is placed on the crown of the |