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weight which otherwise would crush the material, -stone, brick, &c., on which the beam rests. To meet these requirements, the theoretical forms may be modified at pleasure, care being taken, however, that the sizes demanded by theory are not curtailed by the lines required by taste, convenience, or other considerations. Thus Fig. 117 might be modified to Fig. 130, in which the length is increased by the supplementary pieces m, n, the amount of which must be fixed by judgment so as to give a good bearing. The two semi-parabolas o, p, &c., are the same as in Fig. 117, and the curve r, s, t is an ellipsis which is perhaps the most beautiful of simple curves, and may be. easily described by taking a piece of paper with a perfectly straight edge P, making the distance a, b, equal to S, V, and b, c, equal to U, V; then passing it over the latter so that b and c are always in contact with the major and minor axes of the ellipse respectively, and making dots with a pencil or needle point at b, b, &c., a sufficient number of guide-dots is obtained, through which the perfect ellipse may be drawn by a French curve, &c.

(429.) The same principles may be applied to find the section at different points in the length of I girders, whose profile has been determined by taste or convenience. Let Fig. 131 be a girder 16 feet between bearings, resting 18 inches on the wall at each end, 30 inches deep in the centre, whose section there is given at A, and the load being a central one, or by the Pillar E. The strength at the centre (350), or the reduced value of ď xb, is for the top flange 22 x 8 = 32; for the vertical web (289 – 22) x 1} = 1170: and for the bottom flange (302 – 282) x 18 = 2088 Saltogether 32 + 1170 + 2088 = 3290. Dividing the extreme half-length of the girder into four parts, we can now determine the strength, or do x b at each point B, C, D, that at A being 3290 on the principles explained in (423) and by Fig. 122. Thus, at D it will be 3290 - 4 = 822; at C, 3290 = 2 1645; and at B, 3290 x 3; 4 = 2467. The depth of the girder at those points having been predetermined by Fig. 131, we have now to find the breadths of the bottom flange necessary to give the required strength, that of the top flange being maintained at 8 inches throughout. Thus, at B, the depth being 25 inches, we have: top flange, 2 x 8 = 32; vertical web (23 – 2*) x 1} = 787; or 32 + 787 819 together; and as we require 2467 at that point, the bottom flange must yield 2467 – 819 = 1648, and as the depth there squared is 252 – 23= 96, the breadth of the bottom flange must be 1648 - 96 = 17.1 inches. Similarly, at C we have: top flange, 22 x 8 = 32; vertical web (192 – 24) x 1} = 535; or 32 + 535 = 567 together; hence the bottom flange must yield 1645 – 567 = 1078, and the depth squared being 21 – 192 80, the width must be 1078 · 80 = 13} inches. Finally, at D we have for the top flange and vertical web (2 x 8) + (16' – 2'] x 1}) = 410, and as we require 822, the bottom fange must yield 822 – 410 = 412, and the depth squared being 18 - 16' = 68, the width must be 412 = 68 6.06 inches, &c.

In many cases the form of the bottom flange as thus found would need modification to meet the requirements of taste, care of course being taken that the calculated sizes are not curtailed (428)

(430.) Effect of Modes of Fixing and Loading.”—There are three principal methods of fixing beams :—1st, when supported at the two ends; 2nd, when fixed, or built into walls at the two ends; and, 3rd, when fixed or built into a wall at one end only, the other end being free, and the beam then becomes a cantilever.

With each of these modes of fixing beams there are two principal methods of arranging the load, namely, 1st, a single weight in the centre of beams that are fixed or supported at the two ends; and, 2nd, when the load is distributed equally all over the length. Similarly, with a cantilever, the load may be, 1st, a single one at the remote end, and, 2nd, it may be equally distributed all over the length.

(431.) The ratios of the loads in these various cases are as follow:


of Loads. Supported at two ends, and loaded in the centre 1:0 Supported at two ends, load spread equally all over 2.0 Fixed at two ends, and loaded in the centre ..

1.5 Fixed at two ends, load spread equally all over 3.0 Fixed at one end, and loaded at the other

0.25 Fixed at one end, load spread equally all over 0.50


Of course the distribution of the load may be varied endlessly; in (419), &c., the whole matter is fully investigated, and the effect on the sizes and forms of beams is considered in detail on a principle that admits of universal application.

The ratios given above are easily applied in practice :- Say that we require the depth of a cantilever of Riga Fir, projecting 5 feet from the wall, to carry safely a load of 1900 lbs. distributed all over, the thickness being 3 inches. We find first from the ratio 0·50 given above that 1900 lbs. equally distributed over a cantilever is equivalent to 1900 = -5 = 3800 lbs. in the centre of a similar beam of the same length supported at the two ends as in the rules in (323), and the value of M for safe load being for Riga Fir, 78 lbs. by col. 3 of Table 67, the rule d = w{Wx L) = (M,x } becomes in our case (3800 x 5) = (78 x 3}v = 9 inches, the depth required, &c.

LATTICE GIRDERS. (432.) The investigation of the strains on the several parts of lattice girders is an interesting study on its own account, and is also instructive as illustrating the internal strains in girders of other kinds, where the phenomena are often very obscure. In lattice girders the tensile and compressive strains are confined to certain definite lines formed by the different members of the structure, and this fact enables us to estimate the force, direction, and resultants of those strains with a facility and precision not attainable with girders of other kinds.

The forms of lattice girders are so very variable, that in most cases the strains on the various parts must be found by analysis and reasoning rather than by set rules; but for the main question of the Load which can be borne, the Rules for Plate-iron Girders in (386), &c., will equally apply to Lattice Girders.

(433.) Let A, B, C, Fig. 134, be a triangular frame loaded with a weight W of 1 ton, the angle of the strut B being 45°. The weight W may be resolved into two forces on A and B respectively by the well-known parallelogram of forces; making the diagonal a equal to the weight W by a scale of equal parts, the strain on B will be equal to the length d or e, and will be in our case 1.414, or say 1•4 ton by the same scale. The strain on A will be equal to the length for g, namely = 1.0. Then the strain on B, or 1.4 ton, may be resolved by another parallelogram into the two strains on C and h, by making the diagonal R = 1.4, when the length of m or n will give the strain on C = 1:0, and o or p the strain on h = 1:0 also. The strain on B is a compressive one, and those on A and C are tensile ones: the former are represented by full lines, the latter by dotted lines.

Let Fig. 135 be a system of framed rods, in which A, B, с are obviously under precisely the same conditions as in Fig. 134, and bear the same strains. We may now find the other strains by reasoning, thus :—the strain on A or 1.0 must evidently be transmitted to D, and be borne by that member of the system; but D has also to bear the extra strain from the thrust of E, and as the strain of 1.4 on B caused a strain of 1:0 on A, so the strain of 1.4 on E will cause a strain of 1.0 on D in addition to that transmitted from A; hence the total strain on D = 2.0. The strain on E is known to be 1.4, because, as in Fig. 134, the weight W = 1:0 produced a strain of 1.4 on B, and 1.0 on C, so in Fig. 135 will the strain of 1:0 on C produce a strain of 1.4 on E and 1.0 o on r. Similarly, as B gave a strain of 1.0 on C, so will E cause a strain of 1:0 on F, while G simply taking the place of h in Fig. 134 bears the same strain, mamely 1.0, but the point q having received 1:0 from E, has a total strain on it = 2.0, namely, 1:0 from G and 1.0 from E.

(434.) Fig. 136 is a long girder or cantilever, composed of a series of frames, as in Fig. 135, and the strains throughout may be found by pursuing the same reasoning :-thus we have seen that every diagonal, B, E, &c., causes an extra strain of 1.0 on those members of the top and bottom which receive it, the total strains on the top become 1, 2, 3, 4, 5, and 6; and those on the bottom 1, 2, 3, 4, and 5. After H has passed the foot of the last diagonal K and received its thrust, the strain at J becomes 6 also. The strain at the wall, or at J and L, might be found direct from the leverage with which the weight W acts, for as the length L, M is six times the depth J, L, the weight of 1.0 at M will evidently give a strain of 6 at J and L, as found by the preceding analysis. It should also be observed that, whatever the length of the girder may be, the strain on the verticals and diagonals is constant from end to end. In all the figures solid lines represent compressive strains, and dotted lines tensile ones.

In Fig. 137 we have a similar girder, but in a reversed position, the principal effect being that the strains on the diagonals which were compressive in Fig. 136 become tensile in Fig. 137, while the tensile strains on the verticals become compressive ones.

(435.) If we now combine Fig. 136 with Fig. 137 by superimposing one upon the other we obtain Fig. 138; the strains on the top and bottom members will evidently be the sum of those in Figs. 136 and 137; those on the diagonals will remain unchanged in character and amount, but those on the verticals, being tensile in one case and compressive in the other, will neutralise each other, showing that there will be no strain upon them. They may therefore be omitted altogether as useless, as is done in Fig. 138.

The strain on top and bottom members is of course a maximum at the wall, and is equal to 12 tons, as shown by Fig. 138; we may check this result in another way. Thus, the depth of the girder being taken as = 1:0, the length in our case = 6:0, therefore the load of 2 tons becomes a strain of 2 x 6 = 12 tons at the wall, as found before.

It will be observed that when, as in Figs. 136, 137, there is only one set of diagonals, the strains on the top and bottom members increase towards the wall in the simple arithmetical ratio 1, 2, 3, &c.; but where there are two sets at similar angles, as in Fig. 138, those strains increase in the order of the odd numbers 1, 3, 5, 7, &c. Another important fact is, that the strain on the diagonals is constant throughout, whatever may be the length of the girder. But these statements are true only where the strain is taken as a single load at the end, the weight of the girder itself being neglected. The effect of the latter, being equivalent to an equally distributed load, is considered in (444)

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