ing w, with length = 1, the two require material = M, making the whole amount of thrust material = 5M. Thus we see that the two plans require each the same precise amount of material for sustaining both tension and thrust, upon the supposition that the material is capable of sustaining the same stress to the square inch of crosssection, in the one plan ast in the other. This is true k e FIG. 4. as to tension material; but with regard to thrust material, the power of withstanding compression, varies with the ratio of length to diameter of pieces, as well as with the form of cross-section; and it will hereafter be seen, that in this respect, plan 3 has the advantage in having the compression sustained mostly by shorter pieces, unless ki be supported vertically and laterally by a stiff connection with the beam at f, which would increase the amount of material. XII. Plan Fig. 5 has three members (In, mp and mq) exposed to tension, and the remaining three exposed, to а FIG. 5. m Р compression. With the same length and depth of truss, and the same load w, at m, and with obliques equally inclined n (at 45°), it is manifest that the vertical and horizon tal reaches, each for each, is equal to 1, and the length, equal to 2; while the weight sustained by each equals w. Hence, the action (of tension or compression), equals w√2, and the material equals }✔2×✔2.м =1м; making for the four pieces, 2м for tension, and 2м for compression. The tie or chord In, suffers tension equal to the horizontal constituent of the thrust of ql, manifestly equal to the weight sustained by ql, or equal tow. There fore, the length being equal to 4, the material required in its construction, equals 2м. The remaining member pq (2) sustains compression equal to the combined horizontal constituents of the tension of mq, and the compression of ql, each of said constituents equal to w, making compression of pq, equal to w, and length being 2, material = 2M We have therefore, for this plan of truss, 4M, for thrust material, and 4м for tension material, which is less than in case of Figs. 3 and 4. Consequently, this plan is decidedly more economical than either of the others, unless the compression material acts with better advantage in the latter than the former; that is, unless the thrust members in 3 and 4, have a greater power of resistance to the square inch of cross-section, than those in Fig. 5. XIII. As to this, both theory and experiment prove, as will be shown in a subsequent part of this work, that the long thrust members in bridge trusses, are liable to be broken by deflection, rather than by a crushing of the material; that in pieces with similar cross-sections, with the same ratio of length to diameter, power of resistance to the square inch is the same. That, since the cross-section is as the square of the diameter, and the diameters (in similar pieces), as the lengths, the absolute powers of resistance (being as the cross sections), are as the squares of the lengths. the Hence, if the compressive forces acting upon two pieces of different lengths, be to one another as the squares of the lengths of pieces respectively, and the diameters be as the lengths, the forces are as the crosssections, and proportional to the power of resistance in each case, and the material in the two pieces, acts with equal advantage, as far as regards cross-section, so that the products of stress into length of pieces, are the true exponents of amount of material required in the two pieces respectively. It follows, that, if on dividing the forces acting upon the pieces in question respectively, by the squares of the lengths, the quotient be the same in both cases, the two pieces have the same power of resistance to the square inch, and in general, the greater the value of such quotient, the greater the power per inch, and the greater the economy, though not necessarily in the same precise ratio. XIV. Applying this rule to thrust members in plan Fig. 3, being the braces, the compressive force equals w✓5, and square of length = 5. Hence the quotient Tw✓5=0.2236w. The piece ki Fig. 4, has length =4 and compressionw, whence, force divided by square of length gives w =0.0625w. This shows the material to be capable of sustaining much more to the square inch in the former, than in the latter case, though it does not give the true ratio. On the other hand, ek and gi, with length -1, and stressw, give quotient=0.5w. Hence, with similar cross-sections, these parts have greater power to the inch than either of the former, but not enough to balance the inferiority of ki, as compared with ad and de, in Fig. 3. = = = With regard to truss Fig. 5, ql and pn, suffer each compression equal to w✔2, with square of length = 2, giving quotient w√2, 0.371w, while pq, has compression w, and square of length 4, and quotient w=0.25w. Hence it appears that this plan not only possesses a decided advantage in the less amount of action upon materials, but also, a considerable advantage as to ability of compression, or thrust members, to withstand the forces to which they are exposed. = XV. Still another modification for a truss to support a single beam, is formed by reversing Fig. 3, thus converting tension members into thrust members, and vice versa; the oblique members falling below, instead of rising above the grade, or road-way of the bridge. In this case, the long horizontal thrust member ac, is divided and supported in the centre, and its economy of action becomes the same as that of pq, in Fig. 5; and the truss gives the same exponents for both thrust and tension material as when in the position of Fig. 3. This arrangement affords no side protection, and is not always admissible, on account of interference with the necessary open space beneath. XVI. We seem to learn from what precedes, that: (1). Since all heavy bodies not in motion toward, or not approaching the centre of the earth (or receding from it under the influence of previous impulse), exert a pressure equal to their respective weights [VIII], *By the expression, amount of action, is meant, the sum of products of stresses into lengths of parts, or members. either directly or indirectly upon the earth; and, since, a body crossing a bridge, having (as bridges are always supposed to have), a void space underneath, preventing a direct pressure, it follows, that every such body exerts an indirect pressure at some point or points at greater or less horizontal distance from the body. (2). That the pressure of a body at a point or points not directly below it, can only take place through one or more intermediate bodies, or members, capable of exerting (by tension or thrust), one or more oblique forces upon the first named body, and it is the office of a bridge to furnish the medium of such horizontal transfer of pressure [IV]. (3. That a single oblique force can not alone prevent a heavy body from falling toward the earth (since two forces can only be in equilibrio when acting oppositely in the same line), and that each oblique force is equal to the combined action of a vertical and a horizontal constituent, of which the first alone is equal to the weight sustained and transferred by the oblique member, while the horizontal constituent, acting at both extremities of the oblique medium, must be counteracted by means outside of the oblique and the weight sustained by it; which means are usually to be supplied by other members of the structure [VIII]. (4). The direct force exerted by an oblique member (in the direction of its length), is equal to the weight sustained, multiplied by the length, and divided by the vertical reach of the oblique, while the horizonta. constituent equals the weight sustained multiplied by the horizontal, and divided by the vertical reach of the oblique [VII]. (5). The amount of material required in a tension member, is as the stress multiplied by the length of |