The stresses in the first three bars are not reversed by the live load, and are always tensile. In the bar JK we shall see that the live load puts a compressive stress upon the bar, thus Since JK can only resist tensile stresses, we must counterbrace the panel by inserting a diagonal crossing JK. Since it is hardly possible to ensure that each member shall perform the duty assigned to it, it is better practice to make the counterbrace take all the live-load shear, so that JK would have to sustain 30-65, and the counterbrace 9.432 tons. In the central bay there is no stress in the diagonals due to the dead load; the stresses due to the live load are +1.572(1 + 2 + 3 + 4) = +15.72 tons Hence the central bay must be counterbraced by inserting two diagonals crossing each other, each designed for a tensile stress of 15.72 tons. These diagonal counterbraces are shown in dotted lines, both in Figs. 151 and 152. The counterbraces only come into action when the live load tends to produce distortion of the panel by putting a compressive stress on the main brace; when this occurs, the main brace bends and throws the tensile stress upon the counterbrace, so that the two braces are never in action simultaneously. Stresses in Vertical Members.-The stresses in the verticals are the vertical components of the stresses in the diagonals, and one may be found from the other, or we may proceed independently, thus Hence we see that the live load puts tensile stress upon JH and HG. The maximum stresses are written on Fig. 156. In designing these girders we must compare the maximum stresses, as written on Fig. 156, with those produced by the dead load only, and thus find the range of stress for those bars where the stress is not reversed by the live load. Where there is a reversal of stress, as in the two central verticals, the range of stress is shown on Fig. 156. Hence the intensity of working stress is found as explained in Chapter I. When there are more than one system of triangulation each system may be calculated separately, as in Fig. 157, as the loads upon the apices of a particular system are transmitted by that system to the abutment independently of the others. 18 FIG. 157. FIG 158. Fig. 157 represents a lattice girder or truss of an odd number of panels, which may be subdivided into four different systems, as shown in Figs. 158 to 161. The subdivision illustrated in Figs. 158 and 159 represents the method for partial loading, and Figs. 160 and 161 show a possible division for complete loading; but here there is an ambiguity, as we are not sure whether the system illustrated in Figs. 158 and 159 would not hold. If the stresses are calculated for each case, and the maximum 10 FIG. 159. FIG. 160. FIG. 161. stresses adopted in the design of the girder, this ambiguity will not cause a want of strength. The student may work out the stresses in this case by assuming a live load of 10 tons and a dead load of 5 tons at each bottom apex, 0 = 30°, span 50 feet. This method of subdivision and superposition is often very convenient, and may be used in the two following examples, Figs. 162 and 165. Here, however, there is no ambiguity, as the number of panels is even, and there are only two subdivisions possible. Figs. 162 and 165 are both modifications of the Pratt or Murphy-Whipple truss, but in these forms they are sometimes. called Linville trusses; the former is suitable for a deck bridge, the latter for a through bridge. Fig. 165 represents the outline of the bridge over the Ohio River, 415 feet span. Fig. 168 shows a form of timber truss which has been largely used for road bridges in Australia. The vertical tension members are constructed with two or more bolts, which may be screwed up as required, and the diagonal members of the web fit into iron pockets, and are provided with double wedges at the ends for putting the member in initial compression and taking up shrinkage. The stresses written on the various members of the truss have been determined in the manner sufficiently explained, and may be verified by the student. The dead load is 0.6 tons per foot run, and the live load 0:53 tons per foot run on each truss. The counterbraces in the four central bays, shown in dotted lines, are chiefly necessary in consequence of the wedges and the stresses, as the various members of the truss may vary from those written on Fig. 168, depending on tightness or slackness of the wedges. The counterbraces should be made about half the sectional area of the main diagonal in the same bay, and should be arranged with their largest lateral dimensions at right angles to the plane of the truss in order to give lateral stiffness. The cross-beams in this truss are spaced on each side of the panel points; the main tie is subjected to a certain amount of transverse stress. Fig. 169 shows a Howe truss as used in a highway bridge of 70 feet span, in which the vertical members consist of two or more bolts, but the diagonal members of the web and the ends are inclined at the same angle, viz. 45°. The cross-beams carrying the floor rest upon the bottom chord at the panel points, the vertical bolts being spaced longitudinally at a distance apart equal to the width of the floor beam. There are no wedges at the ends of the diagonal members, as the shrinkage of the timber may be taken up completely by screwing up the nuts on the vertical bolts, hence only the central bay is counterbraced. The stresses written on Fig. 169 have been determined for a dead load of 0.414 tons and a live load of 0.470 tons per foot run for each truss. This truss is obviously superior to the foregoing for timber bridges. The method of constructing the joints in the bottom chord of these trusses will be considered in Chapter XIII. The bottom chord in both these trusses may be constructed of steel, and the compression members only of timber. This composite truss has the advantage of allowing the various timber members to be removed without stopping the traffic over the bridge. The following example is given to illustrate the method of calculating the stresses in lattice girders subjected to a permanent load due to the weight of the structure, and a live load due to an engine and train. This girder is not given as an example of good proportion, as it is too deep, and is unsuitable for so small a span. Let the permanent load considered as concentrated at each apex be 2 tons, denoted by W1, and the live load concentrated in a similar manner be 5 tons, denoted by W. There are 12 bays and 4 systems of triangulation, the bars being inclined at 45°. = 5 x 1·414 = 7·070 tons, and W1 sec = 2·828 = 0.589 W sec 12 1 By referring to Fig. 170 and the subjoined table of stresses, it will be observed that the live load W1 produces stresses on bars 3, 7, and 11; that W2 produces stresses on bars 4, 8, and 12; W, on bars 1, 5, 9, and 13. W1s also produces stresses on bars 1, 5, 9, and 13, thus 3 13 |