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COMPARISON OF DIFFERENT PLANS OF IRON TRUSS BRIDGES.

CXXXII. It is the purpose of this chapter to canvass the relative merits of most of the several systems of IRON BRIDGE TRUSSING, which have claimed and received more or less of public notice and approval during the last few years; and of which the distinctive principles have already been discussed in preceding pages; though not in the precise combinations here about to be presented.

We may take the number, lengths and stresses (the latter governing principally the required cross-sections), of the several long pieces or members of the truss, in the manner employed in the fore part of this work, as affording a near criterion of the comparative cost and economy of the bridges respectively. Then, after reference to such peculiarities as may seem advantageous or otherwise, leave the reader to his own conclusions in regard to the relative merits.

THE BOLLMAN TRUSS, FIG. 47,

Is founded upon the general principle discussed in sections XXII and XXIII, with oblique tension rods, and a thrust upper chord, in place of the thrust braces and tension lower chord as represented in Fig. 9.

Let Fig. 47, represent a truss 15′ high, and 100' long; or, in the proportion of 1 to 63. Also, let w represent the maximum variable load for each of the points c, d, e, etc., and w' (say, w), the permanent weight of one panel of superstructure, supposed to be constantly bearing at each of said points. Then making W wxw', we have W = weight sustained by ac.

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Now, we have seen [VII], that the stress upon an oblique in such case, equals the weight sustained, multiplied by the length, and divided by the vertical reach of the oblique; and, assuming that the member requires a cross-section proportional to the stress, it follows that (making ab= 1), the amount of material required in ac, will be as the weight it sustains, multiplied by the square of its length. Hence, the material required in ac, must be as W × ac2. Then, diminishing be until ac coincides with ab, W x ab2 becomes W, which is still proportional to the material required in ac (which has now become ab, = 1), and, being replaced by M, representing the actual material required to sustain the weight W, with a length equal to ab (our unit of length), in a vertical position, we have only to substitute M × ac for W x ac2, to know the actual material necessary to sustain the weight W (at a given stress per square inch of cross-section), with any length and position, retaining the same vertical reach, equal to unity.

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It must be obvious, therefore, that M, with the coefficient used before W, to express the weights respectively sustained by the several oblique rods ia truss 47, will, when multiplied by the squares of the respective lengths of those obliques, show the amount,of material required in their construction, under the conditions above expressed.

Let mM, and h M, and h = bc. Then, we manifestly have, for material in the 14 obliques of the truss in question 7m (h2+1)+ 6m (4h2+1) + 5m (9h2+1)+ 4m (16h2+1)+ 3m (25/2+1)+2m (36h2+1) + 1m (49h2+1) = (336h2+ 28)m, for those meeting at a, and a like amount for those meeting at l; making a total of (672h2+56)m. But h-0.694, which substituted in the last expression, gives 522.368m, = 65.296M.

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The thrust of the chord al, equals the horizonta. action of the 7 obliques connected with either end. Making then x = W, and h = bc, bk, it is obvious that each oblique carries weight equal to x × the number of panels not crossed by it, while its horizontal reach equals h the number of panels it does cross. Hence, the horizontal action of each oblique, equals hx x the product of the numbers of panels at the right and loft respectively, of the lower end of the oblique.

The compressive force acting from end to end, upon al, then, must be equal to hx (7, + 2×6, + 3×5, + 42+ 5×3, + 6×2, +7), = 84hx, = 101 W×0.833, = 8.75W. Multiplying stress by length, and substituting M, we have 8.75 × 6.66M583M material required in al, at a given stress per square inch of cross-section; M being the amount required for a unit of length (ab), to sustain the unit of weight (W), at the same rate of stress.

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Add 7M for two end posts, with length equal to 1 and bearing weight equal to 7W, and we obtain 654M as a total for thrust material in long pieces, not including 7 intermediate uprights, not properly to be classfied with other parts, as their action is merely incidental, except that of supporting the weight of upper chord.

The parts above considered, mainly determine the character of the truss as to economy of material.

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. Other parts, such as short bolts, nuts, connecting pins, &c., although just as essential, are comparatively, of small amount and cost, except the intermediate uprights, which will be referred to hereafter.

If the truss be used in a deck bridge, and the end posts be replaced with masonry, the intermediates will sustain the same weight as the ends sustain in a through bridge, thus giving the same representative of material as above found.

THE FINCK TRUSS, FIG. 48,

CXXXIII. Possesses several of the characteristics which distinguish the Bollman plan. Both dispense with the bottom chord, which is common to most, if not all other plans of truss, for both iron and wooden bridges. Both also employ a pair of tension obliques acting in horizontal antagonism to each other, at each of the supporting points c, d, e, &c. But while in the one, the members of each pair of obliques are of equal length and tension, in the other, the pairs consist of unequal members (except at the centre), as the diagrams will sufficiently illustrate.

It will readily be seen that Fig. 48 exhibits three classes of obliques, consisting respectively of 2, 4, and 8 members to the class. Supposing a truss of the same dimensions and proportions, and subjected to the same load, as in case of Fig. 47, and using the same notation, as far as applicable; it is manifest that each of the 8 short obliques, sustains W. The 4 next longer sustain upon each, a weight equal to W-one half directly, and the other, through the short obliques and uprights. The two long obliques sustain 2W each, being the half of 1W, received directly at f, and 1 and 2 respectively

through the upright, from members of the other classes, meeting at the point p.

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The material required for all the obliques, then, (ab being 1, and be = h), is 8 × † (h2 + 1) + 4 × 1 (4h2 + 1) + 2 × 2 (16h2 + 1)M, being the number of pieces in each class multiplied by co-efficients of W in weights sustained, and by squares of length respectively, and the sum of products multiplied by M.

Subsituting in the above expression the value of h3, (0.694), and, reducing and adding terms, we derive material in obliques = 70.296 M.

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The compression upon the chord a l, is equal to the horizontal action of one member of each class of obliques, communicated at each end; that is, equal to (3 h + 2h+8h) W, 103 h W; and, multiplying by length (= 6.66), and substituting 0.833 for h, and M for W, we have (10.5 x .833 x 6.66)м 58. M, to represent the material required in al; - the same as in case of Fig. 47.

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The uprights of the Finck truss obviously sustain 12W, namely, 3 at each end, 3 in the middle, and 1 at each of the quarterings, r and n. But, in comparing this with the Bollman truss, it seems fair to offset 6 uprights, not including the end and centre ones, in the Finck, against 7 in the Bollman truss not estimated; hus leaving 10M for uprights in the former, making

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