Producing be till it meets Ct in v, we see that abl and ube are similar triangles, and al: uv :: horizontal distance of horizontal distance of v, from the point b. Hence, we may take the point v instead of the point 7, by which to establish the position of the line bc, and thereby secure greater relative accuracy of measure meut. So may we also take vv', or ll' instead of bm, to determine the line cd. By this means we multiply the small spaces al, bm, &c., and diminish the amount of error in measurement, and if the angular points, or nodes be at uniform horizontal distances, the process is very simple. LXXII. We have assumed, in describing the arch a, b, c, d, &c., a uniform distribution of load, horizontally. But the general process is obviously the same for an unequal distribution, after locating the first segment ab; which we may do by first ascertaining the amount of bearing at a, due to the load of the arch. This will be to the whole load, as the distance of the centre of gravity of load from k, horizontally, to the whole chord ak. For instance, if the centre of gravity be half way between Cand k, one quarter of the load bears at a. The weight bearing at a, whatever it be, may be represented by A; and supposing it to exert the same horizontal thrust at a as half the load (W), would do when uniformly distributed, we take u' in ft, so that W: A:: uC: u' C.* Then au' gives the direction of ab', and we proceed in the same manner *We may assume any amount of horizontal thrust, and the greater the assumed thrust, with a given load, the less will be the depth of the arch, and vice versa. It is proposed here to construct an equilibrated arch a, b, c, d, &c., of about the same rise at the crown, as the normal curve, a, b c, d, &c., has. as before using the weights given for the several nodes of the arch, to determine the points c', d', &c. These being connected by straight lines, we have an equilibrated arch adapted to the given distribution of load. LXXIII. But of course, this arch will not stand under any other disposition of the load. To obviate this difficulty, and to construct an arch which will stand under a variable load, without the chord and counter-bracing of the arch truss, the device has been adopted, of constructing the arch of such vertical width that all the equilibrated arches or curves, required by all possible distributions of load; may be embraced within the width of the arched rib. Then, if there be sufficient material to oppose and withstand the forces liable to act in the lines of said several equilibrated curves, complete vertical stability must result. The proper width, or depth of the arched rib, will depend upon the length and versed sine of the arch, as well as the amount and distribution of load; and the material will act most efficiently, when mostly disposed in the outer and inner edges, or members of the rib, and connected, either by a full, or an open web, to distribute the action between the outer and inner members, according as the resultant line of action approaches the one or the other of those members. The normal form of the arch should be such as to be in equilibrio under a uniform load,* and hence it will be parabolic, as to the movable load, and the weight of road-way, and catenarian, as to the weight * The method above explained, for describing an equilibrated arch, is applicable to all cases where the load, both constant and variable acting on the several parts of the arch, is known, whether it be the normal curve, adapted to a full load, or a distorted curve, suited to an irregular distribution of load. of arches (as far as they are uniform in section), and should approach the one or the other form, according to the weight of arches, as compared with the other weight to be supported thereby. The distance between outer and inner members, or the width of web, reckoned from centre to centre of those members, should be such that no condition of unequal and partial load, could throw greater action at any point of either member, than the extreme uniform load would throw upon both. Let us suppose that the curve a, b, c, d, etc., be centrally between the two members and that dd', and hh' be the greatest vertical departures, inward and outward, of any equilibrated curve, from the normal curve a, b, c, etc. Let us further suppose that the thrust of the arch at the points d and h, be as great under the load acting in the curve of greatest departure from the normal as the extreme uniform load produces at those points. Then, if the outer and inner members of the rib, be placed at a distance of six times the greatest departure of the distorted from the primary, or central curve, one member will be twice as far from the line of action (at the point of greatest departure), as the other, and the latter will sustain two-thirds of the action, equal to one-half the action of the full load, and the same as in the latter case. If the width of web be less than six times the greatest aberration of the distorted curve, the action, under the suppositions above, will be greater upon one member than that due to a full uniform load; a condition altogether to be avoided. A few trials at constructing curves adapted to assumed possible distributions of load, may determine satisfactorily what condition gives the curve of greatest distortion and the greatest departure from the normal; and the amount of action under that condition, can be readily calculated with sufficient nearness, whence the proper width of web may be deduced. LXXIV. The points of the equilibrated curve may be located by calculation, and perhaps with as much ease, and greater accuracy than by construction. Suppose Fig. 22 to have a vertical depth, Cf, equal to one of ten equal sections of the chord ak. Having found the length of fu, in the manner already explained [LXXI], it is known that for a uniform load at each angle, the vertical reaches of the several segments, begining at the centre, are as the odd numbers, 1, 3, 5, 7 and 9; and, if we conceive Cf to be divided into 25 equal parts (25 being the sum of these numbers), each of these parts will be equal to 0.04 Cf, or .04v; and this factor, multiplied by the numbers 1, 3, 5, &c., give the vertical reaches of the respective straight segments, which vertical reaches being substracted successively from v, and successive remainders, show the several verticals to be as follows: At the centre, f, vertical = Cfv. At e, vertical v-.04v.96v. At d, verti= cal = (.96—.12)v = .84v; at c, vertical = (.84—.2)v .640, and at b, vertical = (.64—.28)v, .36v. This establishes the normal curve for uniform load. = Now, supposing the weight of structure to be equal to 1w at each of the angles of the arch, and also, that a movable load of a like weight, w, be acting at each of the five points f, g, h, i, j; the permanent weight of structure gives a bearing of 4.5w at a, and the movable weights at f, g, h, &c., give respectively .5w, .4w, .3w, .2w, and .1w, together, equal to 1.5w; making the whole bearing at a, equal to 6w, which is 4th less than if the same weight were distributed uniformly. = Then taking Cu' Cu, and drawing the line au', (not shown in the diagram), we have the inclination of ab', the first segment of the required curve, which gives the same horizontal thrust at a, as the normal curve would exert under the same load uniformly distributed. We find fu (=fs), by the proportion. = Cf: fs:: Cu2 (= 5v2): sr2 (= 4.5v2 :: 25v2 : 20.25v2 : : v :.81; and, reducing 1.81v (=Cu), by one-seventh, we obtain Cu' 1.5514v. This length is to aC :: A (=6w), : horizontal thrust of ab'; that is (making v=1), 1.5514 : 5:: 6w: 30w 1.5514 = 19.33w, horizontal thrust ab. = = Now, if this thrust be represented by aC, v=1, then w will be represented by a space equal to 1 19.33' = .05173, which is equal to the vertical departure (D), of b'c' from ab'u'. Knowing the value of this departure which, of course, is directly as v, and inversely as a C), we can locate the points c', d', e' and f', by their vertical distances from au', as follows: The vertical at b', is evidently equal to ×1.5514, = .31026; consequently, the vertical at c' 2x.31026.05173.56879. Vertical at d'= 3x.31026-3x.05173.77559. Vertical = = e' = 4×.31026—6×.05173 = .93068, and the vertical at f', equals 5x.31026-10x.05173 1.034, showing that the new curve crosses the normal, between e and ƒ, and f' is above f, but not shown. Then, if each of the segments b'c', c'd', &c., be produced to meet the indefinite vertical drawn through a, they will evidently cut that line at intervals of D, 2D, 3D and 4D together, equal to 10 D, .5173. Then, |