ARCHES. By B. A. SMITH, M.C.E. AT the last meeting of the Association the writer contributed a Paper on Circular Arches; a slight modification of the method followed in that paper enables us to obtain the formal solution for the general case, in which the intrados is any given curve. The method is applied in the present paper to obtain the solution for an Elliptic Arch under a uniform partial load with an additional concentrated load at the end of the partial load. The assumptions usually made in engineering textbooks as to the position of the "line of pressure are discarded, and instead we merely express the conditions that the ends of the arch are fixed in position and direction, and that at the point below the end of the partial load the arch remains unbroken, i.e., the displacements of two points beside one another (one in each segment) at the end of the partial load are the same, and the tangents to the two segments at this point remain in the same straight line. The pressure at each point is assumed to be normal to the arch ring and the notation is the same as in the former paper. Considering the equilibrium of a small element PQ of the arch ring we have where T, L, M are the tension shear and bending moment at P w is the normal pressure at P due to the load, K is the curvature at P, and is the inclination of the tangent at P to the horizon. We have since the bending moment is proportional to the change of curvature where 2 is the thickness of the arch ring at P (not necessarily uniform) and ƒ is the elongation of the element at P. Young's modulus for the material. normal displacement, outwards, at P tangential displacement towards the right at P. Sev From (4) and (5) calling = + we have δψ From (1) & (2) eliminating I we have A Particular Integral of this is (Forsyth, Diffl. Equations, pp. 86-7) To = f(4) = cos 4 (-) cos 4.dif cos24. (9) So that the complete solution is (A and B being arbitrary constants) TA cos + B sin+f() (10) From (1) & (10) L= A sin - B cos-f' (4) (11) Asin ds-B cos 4 ds - [f' (4) ds + C or, M=- Ay−Bx+ C −ƒ ƒ' (4)ds (12) dy (14) v R cos +S' sin + cos cold (15) ↓ ƒ F(V). d↓ (15) u is then known from (5) & (10) K u, v, are thus known throughout; the 6 arbitrary constants are determined by the end conditions at each abutment. which furnish six equations for the arbitrary constants. This is the formal solution of the problem. To apply this to the elliptic arch of uniform thickness, observe that we have where ρ is the height from the major axis to the roadway (or to an imaginary line representing the partial load treated as of the same density as the filling above the arch) p is the density of the material, |