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Here again, the linear dimensions being in the ratio 1, 2, 3, the loads are in the ratio of the squares of those numbers, or 1, 4, 9, &c.

(452.) “General Laws for Similar Beams."-Mr. Tate and others have shown that there are some general laws governing the relations of similar beams, which are very useful, by enabling us to reason from one whose strength is known by experiment, to another of very different sizes, but similar proportions, whose strength we desire to know. Thus, as an extreme case, it is shown in (475) that the strength and sizes of the great tubular bridge at Conway, 400 feet span, and weighing 1080 tons, might be calculated with approximate accuracy from the experimental strength of a little model tube 33 feet span, weighing only 4 lbs. Figs. 152, 153, will enable us to illustrate the principal laws for similar beams.

(453.) 1st. The breaking loads of similar beams are to each other as their sectional areas:- thus, the areas of A, B, C are 0.56, 2.24, and 5.04, and of D, E, F, 6·25, 25·0, and 56.25 square inches respectively, which in both cases are in the ratios of the breaking weights 1, 4, 9.

(454.) 2nd. In similar beams, the cubes of the breaking loads. are to each other as the squares of the weights of the beams between supports. Thus, the breaking loads in our figures being in the ratio 1, 4, 9, we have 13, 43, 93, or 1·0, 64, and 729, which are the ratios of the squares of the weights of the beams, for 12, 89, and 272 are = 1, 64, and 729.

(455.) 3rd. The breaking weights of beams of similar sections, but of varying lengths, are equal to the continued product of the whole cross-sectional area, the depth, and a constant derived from experiment for the particular form of beam, material, &c., divided by the length, or distance between supports, hence we have the rules :—

(456.)

(457.)

(458.)

In which W

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the weight or load on the beam in lbs., tons, &c., dependent on the terms of M; a = the whole cross-sectional

area in square inches; d = the depth in inches; l = the length or distance between supports in inches; and M = a constant from experiment, &c. Thus taking the girder E, Fig. 153, we may find the value of M: here W = 24.196 tons, as found in (451); a = 25 square inches; d= 8 inches, and 7 = 120 inches. Then M (24.196 × 120) ÷ (25 × 8) = 14.5176. Now applying this, for example, to the girder F, where a = 56·25, d = 12, and l = 180, we obtain W = 56.25 × 12 × 14·5176 18054 441 tons, as before, &c. (451).

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(459.) 4th. The breaking weights of beams of similar sections, but all of the same length, are as the cubes of the linear ratios of the sections. Thus, if the beams A, B, C had all been say 4 feet long, or the length of A, the breaking weights would evidently have been in the ratio 1, 8, 27, instead of 1, 4, 9:—then the linear ratio of the sections being 1, 2, 3, we have 13, 23, 33, or 1, 8, 27, which is also the ratio of the breaking weights.

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(460.) 5th. In beams having similar sections but different lengths, L and 7, corresponding to breaking weights W and w, the relations are expressed by the equation w = R3 x W x L 1, in which R the linear ratio of the two sections. Thus, with the cast-iron girder D in Fig. 153, we have W = 1, and L = 5; then comparing with the girder F, we have R = 3, from which we obtain the relative breaking weight of F x 1 x 515 9; that is to say, if the breaking weight of D = 1, that of F will be 9. Similarly, reasoning from D to E, we have R 2, and w comes out 23 × 1 × 5 ÷ 10 relative breaking weight of E, &c.

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(461.) 6th. The breaking weights of similar beams are to each other as the squares of their linear dimensions. Thus, with Figs. 152, 153, in A, B, C, or in D, E, F, the ratio of the linear dimensions is 1, 2, 3, and 1o, 2o, 3o, or 1, 4, 9, is the ratio of the breaking weights, as given by the figures and shown by (450), (451).

(462.) 7th. In beams of similar sections the cube-root of the breaking weights multiplied by the respective lengths, is in the simple ratio of the linear dimensions, or W × L = R. Thus, in the tubular beams A, B, C, we have breaking weights in the ratio 1, 4, 9, corresponding to lengths in the ratio 4, 8,

12; then we

have

and 3/9 × 12

=

=

3.174;

1 x 4 = 1.587; 3/4 x 8 4.762, which are in the linear ratio of the sections, or 1, 2, 3. Similarly with the cast-iron girders D, E, F, in Fig. 153, we have 1 x 5 = 1·71; 4 × 10 = 3·42; and 9 × 15 = 5.13; which are still in the linear ratio of the sections, or 1, 2, 3. This rule will be found useful in giving the sizes for "Unit" girders, or girders for a load = 1.0 with a length = 1.0, from which the sizes necessary for any load and length are easily obtained from direct experiment on beams of any size or form (485).

(463.) 8th. The breaking weights of similar cast-iron girders are to each other as the areas of their bottom flanges. Thus, in the girders D, E, F, Fig. 153, the areas of the bottom flanges are 4, 16, and 36 square inches respectively, or in the ratio 1, 4, 9, which is also the ratio of the breaking weights.

(464.) 9th. The breaking weights of girders of similar sections are to each other as the areas of their bottom flanges, multiplied by the respective depths, and by a constant adapted to the particular form of section, divided by the length between supports, or

(465.)

(466.)

(467.)

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the area of the bottom flange in square inches, and the rest as in (458). The value of M will vary with the form of section; for the section given by Fig. 153 it is 22·68 for tons-thus with F we have (54 441 x 180) (36 x 12) = 22.68 M.

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For the form of section recommended by Mr. Hodgkinson (351) and Fig. 79, with top and bottom flanges having areas in the ratio 1 to 6, he gives the value of M at 26, for finding the breaking weight in tons. With that value for M, we obtain for the girder D, 4 × 4 × 26 ÷ 60 6.933 tons; for E, 16 x 8 × 26 120 27.73 tons; and for F, 36 × 12 × 26 ÷ 180

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= 62.4 tons, and those breaking weights are in the ratio 1, 4, 9, as in Fig. 153.

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TABLE 78.-Of the RELATIVE STRENGTH of "MATHEMATICALLY SIMILAR" TUBULAR BEAMS.

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(468.) 10th. In similar beams the weights of the beams between supports are proportional to the cubes of the linear ratio of the dimensions. Thus, in Figs. 152, 153, the linear ratio being 1, 2, 3, we have 13, 23, 33, or 1, 8, 27, which is also the ratio of the weights of the beams.

We may now give some illustrations of the application of these laws to cases in practice.

(469.) The 6th law in (461) states that the breaking weights of similar beams are to each other as the squares of their linear dimensions. Mr. Hodgkinson made some experiments with the special object of testing the accuracy of this law as applied to tubular beams. These experiments had particular reference to the great Britannia and Conway bridges, in which tubular beams were used for the first time on a large scale. Little or nothing was at that time known of the strength of beams of the tubular form, and it was highly necessary that the theoretical laws should be checked by experiment. In Table 78 the beams are grouped together in pairs, selected so as to be "similar" in all respects as nearly as possible, but also very different in dimensions, the special object being to show how far the strength of a large Tubular beam could be calculated with accuracy from that of its small fellow. Of course it was impossible in practice to preserve precisely any given ratio between all the dimensions; for instance, in Nos. 9 and 10 the ratio intended was 1 to 12, as in col. 6, but the actual ratios of the depths, breadths, and thicknesses of plate were 11.92, 12.31, and 12.29 respectively, as shown by cols. 7, 8, and 9. The best that can be done under these circumstances is to take the mean ratio of the four dimensions as the general ratio or value of R, and this is given by col. 10 = 12∙13.

(470.) Now, the experimental breaking weight of the small beam of this pair, as given by col. 2, is 1.127 ton, and 12·132 being 147 1, we obtain 1.127 × 147·1 = 165.8 tons as the strength of the large beam, col. 14. But experiment gave 118 02 tons only, hence we have 165.8 118.02 = 1·405, or an error of +40.5 per cent., as in col. 15. Calculating in this way throughout, the mean error of the whole series of experi17.58 per cent., col. 15, showing that the theoretical

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