Imágenes de páginas

(436.) The strain on the diagonals will vary with their angle; an equilateral triangle, as in Fig. 139, is commonly used, and the strains on A, B, &c., may be found by the parallelogram of forces as before. We have to determine first, the ratio between a, b, c, &c.: let M in Fig. 140 be an equilateral triangle whose sides are all = 1.0; we want to know the height of the vertical line f. Dividing the triangle into two equal parts, we obtain the right-angled triangle N, having its two sides = 1:0 and

respectively; then by the well-known rule for right-angled triangles, namely, that the square of the hypothenuse is equal to the sum of the squares of the other two sides, we have f, = (a” – c), which in our case becomes (1– }?) = .866; therefore when, as with O, f = 1:0 ton, c will be 1 ; .866 = 1.155 ton, and as e is half of c, we have e = 1.155 = 2 = •5775 ton. Transferring these numbers to Fig. 139, we obtain the strains on A and B respectively.

The strains on the diagonals will be constant, as we found in (435) and Fig. 136, &c., and will be 1.155 ton throughout. The strain on G in Fig. 139 will be the resultant of the thrust of B and the pull of C, and may be found by the parallelogram P in Fig. 140, where the compressive strain of B is converted into an equivalent tensile one B' at the same angle. Making B' and C each 1.155, G is in our case 1.155 also, from which we find that each pair of diagonals adds 1.155 ton to the top and bottom members, which receive their combined strains; hence H, Fig. 139, becomes 1.155 + 1.155 2.31 tons. After F has been passed, and the thrust of one more diagonal has been received, the strain becomes 2.31 + •5775 = 2.8875 tons. Again; in the top member, the strain on A =-5775 ton, as found in (435); on J it becomes • 5775 + 1.155 = 1.7325 ton; and on K, having received the thrust of F, and the pull of E, it becomes 1.7325 + 1.155 = 2.8875 tons, or the same as we found for the maximum strain on the bottom members.

(437.) “ Beams with Load in Centre, &c.”—The case of a cantilever, as in Fig. 139, may be easily converted into that of a girder of double length, supported at the ends and loaded in the centre with 2 tons. Obviously, in order to obtain a

strain of 1 ton at each end, the central load must be 2 tons, and the cantilever must be inverted. See Figs. 91, 92.

(438.) “Rolling Load.—The effect of change of position in the load, such as occurs where it rolls slowly without shock from end to end of a girder, may be illustrated by Figs. 144 to 148 inclusive, the strains throughout being calculated as in the preceding examples, starting in each case with the load on each prop as a datum. From these figures we may obtain some useful general facts.

(439.) First,-it will be observed that the strains on the diagonals change with the varying position of the load, not only in amount, but also in character, or from tensile to compressive, and vice-versa. Taking G for example, the strain changes from a tensile one of 5.77 tons in Fig. 146, to a compressive one of 3•46 tons in Fig. 147. A strain thus alternating or acting in opposite directions, is known (915) to be very trying to any material which in fact suffers from fatigue; the effect is equivalent to the sum of the two strains, or in our case to 5•77 + 3:46 = 9.23 tons acting in one direction only, but being alternately laid on and relieved ; this, again, is more trying than a steady and constant load. These facts should be remembered in fixing the “Factor of safety " (880) for the particular case; it will also be seen that in a girder for carrying a rolling load, all the diagonals must be adapted to sustain both tensile and compressive strains, and in many cases should be of I iron, or some such form of section, rather than simple thin bars.

(440.) Secondly,--the figures show that while the strains on the top and bottom members never change their character with a rolling load, the amount of the strain varies considerably; for instance, from 5.2 tons tensile strain on M in Fig. 144, to 5755 ton, also tensile, in Fig. 148; and again, from 8:08 tons on N in Fig. 145, to 1.155 ton in Fig. 148, both compressive strains. This partial relief of the strain every time the load passes is much more trying to the material than even the maximum load would be if it were a dead and constant strain (912), and the “Factor of safety" should be higher than for an equal statical load.

[ocr errors]

It should be observed that the strains on the diagonals being alternate, or in both directions, while those on the top and bottom are simply intermittent, should in strictness lead to the use of a higher Factor of Safety for the former than for the latter, in the ratio of 2 to 1 (915).

(441.) “Effect of Distribution of Load.—The effect of two or more equal loads may be found by adding together the respective strains in two or more of the figures ; thus, in Fig. 149 we have the strains with 10 tons on the apex of every triangle, which were obtained by adding together all the corresponding strains in the figures from 144 to 148 inclusive. Thus, on A A, &c., they are all compressive strains, and their sum is 28.86 tons; but on D, two are compressive, with a sum of 1.155 + 3.46 = 4.615 tons, and three are tensile, with a sum of 5.77 + 3.46 + 1.155 = 10.385 tons, so that the tensile, being 10.385 4.615 = 5.77 tons in excess of the compressive, is the final strain on D, as in Fig. 149.

(442.) It is remarkable that with an equally distributed load, as in Fig. 149, while the strain on the top and bottom members is a maximum at the centre, where it is 37•48 tons, and is reduced progressively toward the ends, where it becomes 14:42 tons; it is just the reverse with the diagonals, being a minimum, or 5.77 tons, in the centre, increasing progressively to a maximum of 28.86 tons at the ends, &c.

(443.) In Fig. 150 we have the strains with two weights each of 10 tons, placed at equal distances on each side of the centre, which were obtained by combining those in Figs. 145, 147. It will be thus found that the strains on the diagonals between the two weights neutralise one another, being tensile in one case and compressive in the other; they are therefore useless, and are omitted altogether in Fig. 150.

The same principles will apply to any other equal or unequal loads on the several points by simple proportion. Thus, taking for illustration the diagonal D in all the figures, say we have 5 at Z in Fig. 149, then Fig. 144 gives 1.155 X5 = 0·578 at D; say 2 at Y gives by Fig. 145, 3•46 x 2 = 0.692 at D, both being crushing strains, and their sum •578 +692 = 1.27. Then all the rest are tensile strains on D; say 8 at X by Fig. 146 becomes 577 x 8 = 4.616 at D; say 12 at V becomes by Fig. 147, 3:46 x 1.2 = 4.152 at D; and say 6 at U gives by Fig. 148, 1.155 x 6 = 0.693 at D. The sum of the three tensile strains = 4.616 + 4.152 +0.693 = 9.461, and the sum of the two crushing strains being 1.27, we have as a final result 9.461 – 1.270 = 8.191 tensile strain on D. Applying the same process to the other parts of the girder, we may obtain the strains throughout for loads of 5, 2, 8, 12, and 6, at Z, Y, X, V, and U respectively.

(444.) “ Effect of the Weight of the Girder itself.—We have so far considered only those strains which a given load would produce in the different parts of a lattice girder, irrespective of those due to the weight of the girder itself. In short girders the latter is usually very small in proportion to the load, and may in such cases be neglected ; but in long girders it becomes important, as shown in (488), &c., and must not be overlooked.

The weight of the girder itself is in effect equivalent to a load equally distributed all over the length; it will not, however, suffice to take it as borne on the top oply, as in Fig. 149, but rather by the top and bottom equally, as in Fig. 151. Calculating, as before, successively for the loads at R, S, T, Q, we obtain another series of strains which, added to those given by Figs. 144 to 148, give the strains in Fig. 151, which may be taken as representing the ratios of the strains on a girder arising from its own weight. It should be observed that we have not only 10 tons at the apex of every triangle, but also 5 tons at each end; but as these latter impinge direct on the bearings, they cause no strain on the members of the girder. The total weight on the girder thus becomes 100, or 50 on each bearing, but the straining weight is 45 tons only.

We have here assumed that the weight of the girder per foot run is equal from end to end, which it usually is in ordinary cases; where, however, the strength and corresponding weight of the several parts are graduated from the centre to the ends, as would or should be the case in large and important structures, the method of calculating must be modified accordingly.

The general effect of the weight of a beam on the load which it will bear is shown most clearly and fully in the case of “Similar” beams (488).

(445.) Top Flange.Lateral Stiffness.”—The form of section at the top of a lattice girder, or rather at that part subjected to compression, is not arbitrary; there must not only be area sufficient to bear the crushing strain, but also a flange of considerable breadth, so as to give lateral stiffness sufficient to prevent flexure sideways. The top flange of a girder is virtually a pillar, and while yielding by flexure vertically is prevented by the diagonals, there is nothing to resist flexure sideways except the resistance of the top flange. With long girders particularly, considerable breadth is necessary to prevent the top yielding by flexure with a strain much less than that necessary to crush the material.

(446.) For the more perfect investigation of the matter let us take the case of the long girder in (681) and Fig. 155: the total length being 32 feet with 16 bays, each bay is 2 feet. Now, obviously, the part of the top flange between the points a, a, is a pillar 4 feet long compressed with 15 tons; between b, b, a pillar 8 feet long loaded with 13 tons, &c., &c.; hence we have a series of pillars with lengths varying by increments of 4 feet, namely :

32 28 24 20 16 12 8 4 feet long, the corresponding compressive strains being

[blocks in formation]

tons, and the question is, with which of those lengths and corresponding strains the lateral strength required to resist flexure at the centre becomes a maximum.

By (147) it is shown that the tendency of a pillar of wrought iron to break by flexure is proportional to W X L', which in our case becomes :

[blocks in formation]

which being a maximum with 5 tons and a length of 24 feet, or c, c, in Fig. 155, shows that with a flange of the same size throughout, this is the weakest pillar of the set, and that the

« AnteriorContinuar »