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therefrom to the abutments. When suspension chains are used, it may properly be called a suspension bridge. If braces be employed, it is usually termed a trussbridge.

HORIZONTAL ACTION OF OBLIQUE MEMBERS.

VII. Before advancing further, it will be proper to refer to an important principle or fact which has not yet been taken into account, though a fact by no means of secondary interest.

The sustaining of weight by oblique forces, gives rise to horizontal forces, for which it is necessary to provide counteraction and support, as well as for the weight of the structure and its load.

FIG. 2.

The two equal and equally inclined braces, ac and be, Fig. 2, in supporting the weight w at c, act in the directions of their respective lengths, each with a certain force, which is equivalent to the combined α action of a vertical and a horizontal force, [Elementary Mechanics-Statics] which may be called the vertical and horizontal constituents of the oblique force. These two constituent forces bear certain determinate relations to one another, and to the oblique force, depending upon the angle at which the oblique is inclined.

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Now, we know that the vertical constituent alone contributes to the sustaining of the weight, and consequently, must be just equal to the weight sustained, in in this case equal to w. We know moreover, from the principles of statics, that three forces in equilibrio, must have their lines of action in the same plane, and

meeting at one point; and must be respectively proportional to the sides of a triangle formed by lines. drawn parallel with the directions of the three forces; and that each of the three forces is equal and opposite to the resultant of the combined action of the other two. We have, then, at c, the weight w, the oblique force in the line ac, and a third force, equal and opposite to the horizontal constituent of the oblique force in the line ac. Then, letting fall the vertical de, and drawing the horizontal ad, the sides of the triangle acd, are respectively parallel with the three forces in equilibrio at the point c. Hence, representing the vertical cd, by v, the horizontal ad, by h, and the oblique by o; and calling the horizontal force x, and the oblique force, y, we have the following proportions:

(1).

(2).

zw:x::v:h, whence, x=

tw: y::v: o, whence, y=tw°

But equals the weight sustained by the oblique ac. Therefore, from the two equations above deduced, we may enunciate the following important rule:

The horizontal thrust of an oblique brace, equals the weight sustained, multiplied by the horizontal and divided by the vertical reach of the brace; and the direct thrust (in the direction of its length), equals the weight sustained multiplied by the length, and divided by the vertical reach of the brace.

VIII. Now, it is obvious that the brace exerts the same action, both vertically and horizontally, at the lower, as at the upper end, though in the opposite directions; the brace being simply a medium for transmitting the action of weight from the upper to the

lower end of the brace. Hence, the weight sustained by the brace ac, exerts the same vertical pressure at the point a, as it would do if resting at that point, while the brace requires a horizontal resistance to prevent its sliding to the left, as would be the case if its foot simply rested upon a smooth level surface. This horizontal resistance may be provided by abutments of such form, weight, and anchorage in the earth, as to enable them to resist horizontally as well as vertically, or by a horizontal tie, in the line ab, connecting the feet of opposite braces.

These two methods are both feasible to a certain extent, and in certain cases; and, both involve expense. Under particular circumstances, it may be a question whether the former should not be resorted to, wholly or partially. But for general practice, in the construction of bridges for heavy burthens, such as rail road bridges, and especially iron truss bridges, where expansion and contraction of materials produce considerable changes, it is undoubtedly best to provide means for withstanding the horizontal action of obliques, within the superstructure itself; and this principle will be adhered to in the discussions following.

The preceding remarks and illustrations as to the ac-. tion of braces, or thrust obliques, obviously apply in like manner to obliques acting by tension, with only the distinction, that in the latter case, the weight is applied at the lower, and its action transmitted to the upper end of the oblique, and the horizontal action (at the remote end), is inward, and toward the vertical through the weight, instead of outward; and consequently, must be counteracted by outward thrust, as by a rigid body between the points p p', Fig. 1, or by heavy towers, and anchorage capable of withstanding

the inward tendency. Hence, in applying the rule be fore given, to tension obliques, and their vertical and horizontal constituents, the word pull should be substituted for the word thrust, wherever the latter occurs in said rule.

TWO PANEL TRUSSES.

IX. There are three forms of truss adaptable to bridges with a single central beam or cross bearer

α

FIG. 8

B

C

(which may be called two panel trusses), the general characteristics of which, are respectively represented by Figures 3, 4 and 5. Fig. 3 represents a pair of rafter braces, with

feet connected by a horizontal tie and with a vertical tie by which the beam is suspended at or near the horizontal tie, or the chord, as usually designated.

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For convenience of comparison, let bd=v,-1,-vertical reach of oblique members in each figure. Also, let each chord equal 4v, 4, and the half chord = 2= h=horizontal reach of obliques in Figs. 3 and 4. Then ad, Fig. 3, equals ✔h2+2 5, and if the truss be loaded with a weight w, at the point b, bd will have a tension equal to w, and abc, [see rule at end of Sec. VII], a tension equal to w, weight sustained by ad), multiplied by the horizontal, and divided by the vertical reach of ad; that is, equal to w, w,= w; while ad suffers compression from end to end, equal to But ad=✔5, and v=1. Whence w

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Now, as the cross-section of a piece, or member, exposed to tension (or to thrust, when pieces are similar in figure), should be as the stress, it follows that the weight of each such member should respectively, be as the stress sustained, multiplied by the length, + an additional amount taken up in forming connections; which latter, for purposes of comparing the general economy of different plans, may be neglected.

X. Then, representing by M, the amount of material required to sustain a stress equal to w, with a length equal to bd, = 1, we have only to multiply the stress of a member in terms of w, by the length in terms of bd, or v, and change w to M, to obtain the amount of material required for the member in question, omitting the extra amount in the connections. Hence, the length of the vertical tie bd, being equal to 1, and having a stress equal to w, requires an amount of material equal to 1M.

For the horizontal tie, or chord, length = 4, and stress (as seen above), w, whence material - 4M. This added to 1м, required for the vertical, makes a total of 5м, for material exposed to tension in truss 3.

The two thrust braces, as already seen, sustain compression equal to w✔5, which multiplied by length, ✔5, and w changed to м, give material § M, for each, or 5 м, for the two.

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XI. I the case of truss Fig. 4, the obliques manifestly sustain a weight = w, by tension, giving stress = w5, which multiplied by length, ✔5, gives §M = material for each, and 5м, for the two. The compres sion of ki, equals w×h = {w×2 wx2=w, while the length= 4, whence, material 4м; and, each end post sustain

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