Fig. 10. The truss on the plan of Fig. 6, may be lengthened by introducing additional pairs of inde pendent braces, as seen in Fig. 9. b = For the analysis of these trusses, using the same notation as before, as far as applicable, that is, making v = verticals ck, in 8 and 10, and equal to nn', pm', etc., in Fig. 9; hab, width of panel in each figure, = } whole chord; uniform weights at the four bearing points in each, and м = weight of material required to sustain a stress equal to w, with length equal to 1; then, making in truss 8, it is obvious that the two abutments at a and ƒ together sustain 4w, with the common centre of gravity of all the weights midway between abutments, whence each abutment sustains 2w, equal to weight sustained by al. The compression of al therefore equals 2. But al =✔12+4v2, which substituted in last preceding expression, gives §✔h2 + § v2w,=compression of al. Whence, multiply 2al ing by length, +, and changing w to м, we have (32+13v) м = material for al. 2 w The horizontal thrust of al, [xvi (4)] equals h zo 3h -w, tension of chord af. = The oblique member lk, sustains weight=w (through the vertical ck), and has a vertical reach = = }', Bik w, w = whence it suffers compression equal tow, 3 ✔h2 + žv2w, and requires material equal to v §o)м, while its horizontal thrust equals 1w, - 3h = v compression of ki, by which it is contracted. The ma = 3h2 terial required for ki, therefore, M. Material for g อ and gf, is the same as above found for al and lk, and. doubling those quantities, and adding amount just found for ki, we obtain 15 +3v) M, — — material in the whole arch. The tension of the chord af (Fig. 8), has been seen to be equal to 3w, whence, multiplying by the length, 3h 5h, and changing w to M, we have 15, — material for chord. = = The 4 verticals sustain each, weight w, and the aggregate length being 3v,... material 3vM. This, added to amount in chord, gives (15+ 3}v)м, = tension material required to support a full uniform load, as above assumed. But since any number of the points , c, d, e, are liable to be loaded while the others are unloaded, it is obvious that in such case, the arch will not be in equilibrio, the loaded points tending to be depressed, while the unloaded, tend to be thrust upward. Hence the arch requires the action of the obliques, or diagonals, in the three quadrangular panels, to counteract such tendency; and, as will appear further on, these members will require material equal to about one-third of the amount required in the chord, thus increasing the amount of tension material for the truss to about (20+ 43v)M. XXI. In truss Fig. 9, each brace obviously sustains a portion, x, of the weight w, which is to w, as the horizontal reach of its antagonist, as to horizontal action, or its fellow and assistant vertically, is to the whole length of chord; that is, the weight x, bearing at m, through mn', is to w, as ns to ms; or, x: w:: 4:5. Hence xw. This, multiplied by the horizontal reach, equal to h, and divided by v, gives the horizontal thrust of the brace, equal to w. In like manner, mm' sustains a weight x', which is to w, as ps to ms, i. e., x': w:: 3 : 5, whence x'= the horizontal thrust-2/w= 2. qw, and w; and in general, the horizontal thrust of a brace in this kind of truss, equals w, multiplied by the product of the number of sections of chord at the right and the left of the point of application (of the weight), and divided by the whole number of chord-sections, and, by the vertical reach (v) of the brace. = But the horizontal thrust of mn' equals that of n's, that of mt,and the horizontal thrust of mm', equals that of m's that of mu, whence, horizontal thrust of the 4 braces bearing at m, equals twice that of mn' and 20h mm', together, = 2 (; 4/12 + 8 1/2 ) u w= =w= 4. This бо h v being equal to the tension of the chord, multiplying by length 5h, and changing w to м, gives material for chord = M. Adding to this, 4vM, for 4 verticals 2012 v with зtress=w, and length = v, each, makes whole amount of tension material equal to (2022 +4v)M, being very nearly the same as for truss, Fig. 8. v XXII. As for braces in truss 9, we have already seen that each brace sustains weight equal to w, multitiplied by the number of panels crossed by its fellow, and divided by the number of panels in the whole truss. Hence mn' sustains w, with length equal to ✔h2+v2. 2. Therefore, stress+w, which mul v tiplied by length, and w changed to м, gives material for _mn' = ¿(22+v)m, ≈ (4)2 + 40 ) M. mm sustains = 4h2 3w with length = √4h2 + v2, whence material = mu sustains w with length =✔92+v2, and material equals(+2), while mt, = sustains w, with length ✓16+2 requiring (16%3 (16+). Then, adding and doubling material = = 2v. these amounts, we obtain (20+4v), against (15 + 3)M, for truss 8; a difference of about 30.6 per cent when h = v, and about 32.6 per cent when h Thus, truss Fig. 8 has over 30 per cent advantage over truss Fig. 9, in the economy of amount of action upon thrust material, with the advantage as to efficiency of action of this material, undoubtedly, also on the side of truss 8. Tension material is nearly the same in both. XXIII. It truss Fig. 9 be inverted, dropping the ob !ique members below the road-way of the bridge, thus reversing the action of thrust and tension members, the thrust material would act with nearly equal advantage in both plans, and with about the same amount of action. But the 30 per cent advantage as to amount of action upon tension material, would still be in favor truss 8. Besides, it is only in exceptional cases that this arrangement can be adopted, on account of interference with the necessary space below the bridge. XXIV. In truss Fig. 10, suppose the points b, c, d, e, to be loaded successively from left to right, with uniform weights equal to w each, and suppose the truss to be without weight, as we have hitherto done. When b alone is loaded, w must bear at f, [XVIII] which may be effected, either by tension of bl, thrust of le, tension of ck, thrust of kd, &c., by tension vertical and thrust diagonal alternately, till it reaches f; acting in its course upon 4 verticals, and 4 obliques, with a weight upon each, equal to w. Or, the weight may be transferred by tension of bk, ci and dg, and thrust of ke, id and gf. These alternatives are subject to the control of the builder, and he will form and connect the parts accordingly. Let it be assumed that the truss has tension diagonals, and thrust uprights at c and d, while l |
