the piers or abutments were sufficiently strong to resist the horizontal thrust. And, although this property cannot safely be imputed to any cement (strong as many cements are known to be), yet, in a structure, whose component parts are united with a very powerful cement, the matter above an arch will not yield, as when the whole is formed of simple wedges, or as when it would give way in vertical columns, but by the separation of the entire mass into three, or at most, into four pieces: that is, either into the two piers, and the whole mass between them, or into the two piers, and the including mass splitting into two at its crown. It may be advisable, therefore, to investigate the conditions of equilibrium for both these classes of dislocations. 125. PROP. XII. Suppose that the arch Fff'F' tend to fall vertically in one mass, by thrusting out the piers at the joints of fracture, Ff, F'f'; it is required to investigate the equations by which the equilibrium may be determined. Let 2A denote the whole weight of the arch lying between Ff, and Ff", G the centre of gravity of one half of that arch, the centre of gravity of the whole lying on CV; let P be the weight of one of the piers, reckoned as high as Ff, and G' the place of its centre of gravity. I F H Now, FV, FV, being respectively perpendicular to Ff, F'f', the weight 2A may be understood to act from V, in the directions VF, VF, and pressing upon the two joints Ff, F'f'. The horizontal thrust which it exerts on F, will be = A tan. FVI = A cot. FCI =A. ; and at the same time the vertical effort will = A. CI ADBE EB'A Now, the first of these forces tends to thrust out the solid AF horizontally, an effort which is resisted by friction; and since it is known that, cæteris paribus, the friction varies as the pressure, that is, here, as the weight, we shall have for the resisting force, f. A +f. P. Equating this with the above exCI pression, A. we obtain for the first equation of equilibrium FI' Moreover, the horizontal thrust that tends to overturn the pier AF about the angle A, must be regarded as acting at the arm of lever FE, and, therefore, as CI exerting altogether the energy, A .FE. This is counteracted by the ver tical stress A, operating at the horizontal distance AE, and by the weight P, acting at the distance AD; DG' being the vertical line passing through the centre of gravity, G', of the pier. Hence we have and, after a little reduction, there results for the second equation of equilibrium : 126. PROP. XIII. Suppose that each of the two halves kF, kF', of the arch, tend to turn about the vertex k, removing the points F, and F': it is required to investigate the conditions of equilibrium in that case. Referring the weight, A, of the semi-arch from its centre of gravity to the FH direction of the vertical joint kK, its energy is represented by A. and the FH FI FI' FH = Ꭺ . kI' The resulting horizontal thrust at A is, evidently, A. vertical stress is = PA; and therefore the friction is represented by ƒ. P+ f. A. Equating this with the above value of the horizontal thrust, that the pier AF may not move horizontally, we have Then, considering the arch and piers as a polygon capable of moving about the angles A, F, k, F', A', we must, in order to equilibrium, balance the joint action of P and the semi-arch A at the point F, with the horizontal thrust beforementioned, acting at the arm of lever EF. Thus we shall have P. AD + A. FH AEA. . EF: from which, after due reduction, there results kl P. AD = A (TH FH AE (II.) Corol. Hence it will be easy to examine the stability of any arch whose parts are cemented as in the hypotheses of these two propositions. Assume different points such as F, in the arch, for which let the numerical values of the equations (I.) and (II.) be computed. To ensure stability, the first members of those equations, which represent the resistance to motion, must exceed the second members; the weakest points will be those in which the excess of the first above the second member is the least. If the dimensions of the arch were given, and the thickness of the pier required, the same equations would serve for its determination *. 127. PROP. XIV. To determine the magnitude of the piers, or abutments, that they may sustain the arch in equilibrio, independently of other arches. In order to give a solution to this problem we must assume some particulars as having been determined by adequate experiments and admeasurement: for we do not consider the piers as prisms standing upon their bases and resisting the pressure of the arches, though upon such an hypothesis it would be easy to lay down rules for the determination of the centres of gravity both of the arch and the piers; but, since the stones, &c. of the wall above the voussoirs are * The principles adopted in the two last propositions are due to De la Hire, and Coulomb, respectively. For a more comprehensive view of this interesting subject, the student may consult Hutton's Tracts, vol. i., the Appendix to Bossut's Mechanics, and Berard's Treatise on the Statics of Vaults and Domes. bonded in with those of the pier, the pier will by these means become augmented, and the weight of the arch diminished. We, therefore, regard the piers as extending to the joints of fracture (art. 117), and that portion of the arch which is comprised between those joints as a ponderating body resting in a state of equilibrium upon those joints as upon two inclined planes. Let, then, FS, F'S', be the joints of fracture, G the centre of gravity of the pier QTSFBP, g that of the half arch DEFS, and let go the perpendicular from g upon FS be produced till it meets the horizontal line Q'Q in I: draw GH perpendicular to QQ' and Qh perpendicular to gI: and let the mass of the semi-arch DESF be represented by A, that of the pier by P, and the force of gravity by g: the weight of the former will then be gA, and of the latter gP. The magnitude of the pier is generally computed on the supposition that the pressure of the arch has a tendency to make the pier turn upon Q as a centre or fulcrum; and this hypothesis is often consistent with fact: but when the height CD is small compared with the span, the weight of the arch has a strong tendency to make the pier slide along the horizontal line PI; we shall, therefore, state the conditions of equilibrium on this supposition also. First, supposing the pier solely capable of turning upon Q as a centre of rotation: then will the case be the same as if the body DESF whose weight is 9A, by pressing upon the face ES, tended to move the mass FSTQP upon the fulcrum Q. But the weight gA is to its pressure upon FS, as sine of angle included between ED and FS, to sine of angle ERQ, that is sin. I rad. :: gA: gA = pressure of half arch upon the joint of sin. I fracture. Now g being the centre of gravity of the half arch, the pressure it occasions is exerted in the direction gI: and G being the centre of gravity of the pier, the force resulting from its weight acts in the vertical direction GH; therefore in the case of equilibrium, we must, by the nature of the lever, have, pressure on SF × Qh = weight of pier × QH, that is, gA Qh=gP. QH, sin. I This equation comprises the conditions of the equilibrium of rotation about the point Q; and we may find by its means any one of the five quantities it contains, when the other four are given. When the arch springs vertically from rectangular piers, whose height and breadth are H and B respectively, the preceding theorem reduces to In the second case, in which we suppose the pier may slide along in the horizontal direction, let ƒ be a force which is exerted horizontally in opposition to the motion of translation: then ƒP acting in the direction Ik must countergA balance acting along hI. Here hI being to Ik as radius to cos. I, we shall sin. I tion including the conditions of the equilibrium of translation we have As to the position of the joints of rupture, and of the centres of gravity of the semi-arch and pier, they may in most cases be determined with tolerable accuracy, thus having drawn on pasteboard the arch and proposed pier, upon a pretty large scale, and described the voussoirs of the arch, of the intended thickness, draw from the middle of the key voussoir a tangent to the intrados, and produce it till it again meets the middle of a voussoir, as at F, from which point draw FS perpendicular to the intrados; it will be nearly the position of a joint of fracture. Next, cut the pasteboard through at the several outlines, and find by some of the methods described in art. 92, the centres of gravity of the two parts DESF, STQPF. With regard to the ratio of A to P, it may always be found pretty nearly, either by weighing or measuring the pieces of pasteboard which represent them; and the distances QH, Qh, and angle I, will be ascertained by the construction. If, when these values of A, P, &c. are introduced into the equations, the first members are less than the second, the piers will be large enough to ensure the equilibrium: if otherwise, some of these particulars must be changed until that takes place. : This mode of considering the subject suggests, that to diminish the thrust of the arch, or increase the stability of the pier, the commencement of the flanks ought to be loaded; and that the thickness of the voussoirs near the key ought to be lessened considerably in short, to make the arch, instead of having a uniform thickness throughout its whole extent, to be very thick at its origin, and at the key to be no thicker than is necessary to resist the pressure of the flanks for by such a procedure a part of the force which tends to move the pier is thrown upon that which resists being overturned, and the latter will gain a great advantage in point of stability. 128. PROP. XV. PRESSURE OF EARTH AGAINST WALLS. Lemma. A weight W, being placed on a plane, inclined to the vertical in angle i, to find a horizontal force, H, sufficient to sustain it, so that it shall not run down the plane, taking friction into the account. Each of the forces, W, H, being H resolved into two, the one parallel, the other perpendicular, to the plane; there will result, parallel to the plane, a force = W cos. i H sin. i. In order to an equilibrium, the first of these forces ought to be precisely equal to the friction down the plane. Corol. Hence, if instead of a horizontal force, the weight W were sustained by a wall, or by any obstacle whatever, weight against the obstacle would be W. the horizontal effort exerted by the 129. PROP. XVI. To determine the horizontal stress of the terrace whose vertical section is BCEF, against the wall whose section is ABCD, and the momentum of the pressure to overturn the wall about the angle A. Considering, first, the stress of a triangle CBE, whose sloping side BE makes the angle i with the vertical; let be, b'e', be each parallel to BE, limiting the The fluent of sxxM, when x = a, gives a2sM, for the whole horizontal thrust of the triangle CBE. Referring the momentum of the thrust of the elementary portion bb'ee, to the length of lever bB = a — x, we have for that momentum Ms (a — x)xx. The fluent of this when x = a, is = {a3sM. 130. It remains to determine the angle i. Now, it is evident that 1 - f tan. i = M, vanishes, and consequently, both the horizontal thrust and its momentum vanish, whether tan. i = 0, or = Between these two values, therefore, there is one which gives both the greatest thrust and the greatest momentum. This value is found by making Substituting this value of tan. i for it in the above expression for M, we have for the horizontal thrust }a2s { −ƒ + √(1 +ƒ2)}2 = ja2s tan.' I. while the momentum of the stress is found to be ¿a3s { −ƒ + √(1 +ƒ2)2} which was to be found. 131. The angle which has for its tangent = fa3s tan.' I, is the angle of the slope, which the earth would, of itself, naturally take, if it were not sustained by any wall. For a body has a tendency to descend along a plane (inclination to vertical =i) with a force=g cos. i, and it presses the plane with a force = g sin. i. |