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its adoption; and it will not be considered in the discussions and comparisons in regard to trusses of greater span, to which we may now proceed.

TRUSSES WITH SEVEN PANELS.

THE ARCH TRUSS.

XXVII. In Fig. 11, let md = v, and h➡ ab=ai. If each of the points b, c, d, etc., be loaded with a weight equal to w, then, in order that the arch may be in equilibrio under the effects of these weights, without any action of the diagonals, it is necessary that each section of the arch have the same horizontal thrust, since, if one section have a greater horizontal thrust than the one opposed to it at either end, the diagonals alone can sustain the surplus. And, that the sections may have the same horizontal thrust each must have a vertical reach (the horizontal reach being the same for all), proportional to the weight (W) sustained by each. For illustration, horizontal thrust being equal to W, in order that this expression may represent a constant quantity, h remaining constant, v must be as W.

Now, ml being horizontal, can sustain none of the weight acting at the point m, through the vertical md; hence mn must sustain a weight equal to w. This is transferred to no [VIII], and in addition to the weight at c, makes 2w sustained by no. The latter weight is in turn transferred to oa, and, in addition to the weight at b, makes 8w, to be sustained by ao. The vertical reaches, therefore, beginning with ao, should be as 3, 2 and 1; whence, ob should equal tv, and nc should equal v.

The thrust of ao, then, equals 3w

6w✔2++; whence, amount of action

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Adding these amounts, and repeating the first three,

we have (4212 + 43v) ×м, equal to amount of action upon the arch when fully loaded.

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The stress of the chord obviously equals the horizontal thrust of ao, equal to 3w, 64; and is the same throughout, when the truss is fully loaded throughout. Hence, for the whole chord, we have, stress = multiplied by length (=7h), and w changed to M, = 421×M, representing the material required for the

chord.

ha

6w1

The above are assumed, for the present, to be the greatest stresses that any part of the chord or arch can

*By amount of action upon material, is meant the stress of a member multiplied by its length.

be subjected to, in any condition of the load; w, being the maximum weight for any one of the sustaining points, b, c, d, &c. This is a point we shall be better enabled to verify after considering the

STRESSES OF VERTICALS AND DIAGONALS.

XXVIII. As the diagonals do not act under a full load of the truss, the verticals must each sustain a tension equal to w, when the weights are applied at the chord; and, the diagonals acting by tension, serve, when in action, to diminish the tension of verticals, or to subject them to compression, but can never increase their action of tension. Hence, the maximum tension stress of each vertical equals w.

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In order to bring the diagonals into action, the truss must obviously be unequally loaded; and, to determine the maximum stresses of the several diagonals respectively, we may begin by removing the weight at g; (see Fig. 11A), and, to facilitate the process, let w" represent w divided by the number of panels in the truss (= 7 in this case), i. e., w" 4w. Then, the full load of the truss bearing with a weight of 3w, =21w", at i, ... with load removed from g, the bearing at i, equals 21" — 6w", = 15w"; and produces a thrust upon j equal to 15w+. Then, taking ją by any convenient scale, on ij produced, to represent the thrust of ÿj (reduced to w" with a numerical coefficient, according to the proportions of the truss), and drawing qr parallel with ƒj, and meeting jk produced in r, it is obvious that the three forces acting at j, namely, the thrust of ij and jk, and the tension of fj, will be represented respectively by the sides of the triangle jar, parallel respectively with the directions of those forces; and may be mea

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sured by scale and dividers, or calculated trigonometrically.

Now, it will be seen. that the greater the pressure at i, the greater the thrust of ij, represented by jq, and consequently, the greater the line qr, representing the tension of fj. But the pressure at i, is manifestly the greatest possible in the case here supposed, except wher the weight at gis wholly or partially restored, in which case the tension of fj would be wholly or partially relieved. Hence, it follows that the maximum stress of fj, occurs when all the supporting points except g, have their full load, and the point g is without load.

Taking, then, fs = qr, and drawing st parallel with gf, st will represent the horizontal, and ft, thevertical effect (equal to 3w") of the action of fj; the former

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effect being, in addition to the tension of fi, resisted by the tension of ef, while the latter is counteracted by the weight at f, and the tension of fk is thereby diminished, but not exhausted*. But if the weights were applied at the arch instead of the chord, then ƒk, in this condition, would suffer compression represented by ft.

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= 10w"

XXIX. If the points g and ƒ be unloaded, the pressure at i is reduced to 10w", and the thrust of ÿj ✔2+¦2. Then, taking ja' by the scale, to represent this quantity, and drawing q'r' parallel with fï, q'r', compared with the same scale, will give the stress of fj; from which, as in the preceding case, we obtain ft' (= 2w"†), to represent the vertical effect of the action of fj; and, there being no weight at f, this force

*Since the vertical reach of jk is that of ij, and their vertical actions (horizontal thrust, and horizontal reach being the same), as their respective vertical reaches, jk must react downward at j, with of the lifting force exerted by ij. Then, if a weight be suspended at g, by the vertical jg, equal to of the weight bearing at i, the forces acting at j, through ij, jk, and jg, are in equilibrio. But if the weight at g be less than of the pressure at i, the tendency of the point j is upward, and exerts a lifting force upon fj,. But the action of fj, brings into play horizontal reaction in jk, equal to that of fj, which gives jk a depressing action at j, equal to of the lift of fj. This depressing power of jk, depends on forces acting directly or indirectly at k, and which go to make up part of the pressure at i. Hence, jk supports at the upper end, at k, first, of the weight bearing at i, in virtue of the horizontal thrust received through ij, and second, of the other (when there is no weight at g), in virtue of the horizontal thrust communicated through fj. Now, g being alone unloaded, the bearing at i is 15w", of which, 10w" is received through jk, in virtue of horizontal thrust counteracted by ij. and of the other 5", in virtue of horizontal thrust counteracted by fj, in consequence of the latter sustaining of said 5". Hence, fj sustains or, while jk sustains, or of all the weight bearing at i, when g alone is unloaded; and 30", therefore, is the maximum weight sustained by fj.

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+Since jq represents the action of ij, due to a pressure of 15w" at i, and jq', the action of ij, due to a pressure of 10w", it follows that jq' =jq; whence q'r' obviously equals & qr, and fs' =fs. Consequently, ft' ft. But ft represents the lift of fj,= 3w", whence ft' repre sents 210".

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