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for Ww 19 tons, bw = (√1 × 10419) 29.96, say 30 inches for the top flange of a girder, supported at one edge only, and measured as in Fig. 60, the breadth would be 30 ÷ 4 = 7 inches.

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Calculating in this way we may find the breadths of wroughtiron plates under different conditions, such that the Wrinkling Strain shall be equal to the Crushing strain, or 19 tons per square inch in all cases: for thicknesses of:

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the breadth of plate supported at both edges and forming part

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4}} 62 9 111 13

15

18 inches.

21 The same plates forming the top of a Tubular beam would have breadths of:

3

18 221 26

30 inches.

7 111 15 The breadths for plates supported at one edge only in pillars

become :

1% 1층 18 21

21 3 31 4 inches,

and in plates forming the top flange of a plate-iron girder, Fig. 93, we have :

1

4

5

6

11 27 84 7 inches. With breadths greater than those given above the Wrinkling strain would be less than 19 tons per square inch: with less breadths the wrinkling strain by calculation would come out more than 19 tons, but this would not be realised; in that case the strength of the plate would be limited by the crushing strength.

(318.) Let Fig. 62 be the section of a short pier for a bridge, &c., 6 feet square, of wrought-iron plate inch thick, strengthened with T ribs A, B, &c., giving 11 inches between their edges, we should then have 19 tons per square inch wrinkling strain, and should thus have obtained the utmost possible effect from the material. If in this case we dispense with the ribs,

TABLE 63.—Of the RESISTANCE of WROUGHT-IRON PLATES to WRINKLING in Tons per Square Inch.

Breadth of Plate, supported at Both Edges: in Inches.

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the effective width of the g-inch plate would be 5 feet 5 inches, and the wrinkling strain by the rule (308) becomes √625 ÷ 65·75 x 80 7.8 tons per square inch, which is 7.81941, or 41 per cent. only of the maximum strength, and thus 59 per cent. would be wasted in that case.

The vertical ribs would not only increase the strength of the thin plate, but would also yield their full quota of strength to the pier in the simple proportion to their own area or weight. The additional strength to the plates is thus a clear net advantage.

If the pillar is of considerable length it would be expedient to make it cellular as in Fig. 54, rather than a simple square as in Fig. 62. See (266).

(319.)" Wrinkling Strain in Beams."-In an ordinary rectangular tubular beam, supported at both ends, a transverse load causes a compressive strain on the top plates. With very thick plates the limit to that strain is the crushing strength of the material, or 19 tons per square inch, but with ordinary thicknesses and breadths the plate will fail by Wrinkling with a much lower strain, as shown by col. 12 of Table 77.

We have first to ascertain the longitudinal compressive strain from the transverse load, &c., which will be given by rule (514). Thus, taking No. 5 in Table 77 as an example, whose section is shown by Fig. 63, the rule (514) for finding ƒ becomes

3 × 58.66 × 360 × 24

2 × {243 × 15·5) — (22·953 × 14·45

as in col. 12.

= 19.2 tons,

(320.) It will be interesting and instructive to check this result by an analytical investigation. long supported at each end, our first

The beam being 30 feet step will be to reduce it

to the equivalent case of a cantilever of half the length, say built into a wall, and loaded at the end with half the central load. See (385) and Figs. 91, 92.

Assuming the value of f, or 19-2 tons per square inch, as the maximum strain at A and B in Fig. 63, this is reduced at C and D, or the centres of the top and bottom plates to 19.2 × 23.475

24 18.78 tons per square inch, this being simply propor

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tional to the relative distances from the neutral axis N. The area of say the top plate being 15.5 x 5258.137 square inches, and the leverage by which the strain acts in supporting the load at W, 23.475 and 180 inches, we obtain 8.137 × 18.78 × 23.475 180 = 19.93 tons at W, and as we have taken for the leverage the whole depth C, D, not the distance from the neutral axis N to C or D, this will be the sum of the resistances of tension at C and compression at D, which in that case occupy the positions of fulcra and resistances to each other reciprocally.

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Following the same course with the sides, we have below N the area of the two half sides = 22.95 × 525 12.04 square inches. The mean resistance at o and p will be proportional to the distances from N: hence 19.2 tons at A is reduced to 19.2 x 11.47524 9.18 tons per square inch at o and p. By (495) it is shown that the true mean is not found by multiplying the area by the mean tensile strain and the mean leverage simply, but by of that product. Then we obtain (12.04 × 9.18 × 11·475 × 3) ÷ 180 9.4 tons at W, as the resistance of the sides, making a total of 19·939.4 29.33 tons at the end of a cantilever 15 feet long, which is equivalent to 29.33 x 2 = 58.66 tons in the centre of the girder 30 feet long, agreeing precisely with the experiment; col. 8 of Table 77. See (412) and (414).

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(321.) The value of ƒ in col. 12 of Table 77 has been calculated by the rule (514); it represents the maximum strain at the edge of the section due to the transverse load, but does not determine whether failure takes place by wrinkling or by crushing. When, however, f is much less than 19 tons per square inch, the plate must have failed by wrinkling; in two cases, Nos. 13 and 15, ƒ was greater than 19 tons, namely, 23.13 and 24 56 tons respectively, which must be regarded as exceptional and anomalous. They may be accounted for by the variableness common to all materials under all kinds of strain, as shown by Table 147, which gives for Boiler-plate under tensile strain 29 per cent. in excess of the mean strength, col. 1. In our two cases the excess was 22 and 29 per cent. respectively under wrinkling or crushing strains.

(322.) "Value of Mw for Beams." - The experiments in

W

Table 77 will give the value of Mw for a plate forming part of a tubular beam and subjected to compression, which is usually the top plate. Selecting cases where the top and bottom plates were of one and the same thickness, and where the results are likely to be more correct than under other conditions, omitting also the anomalous cases Nos. 13 and 15, we obtain Table 62; the mean of the whole is 104, which is 30 per cent. higher than 80, the mean value of Mw for pillars where the plate is subjected to direct compression (312). We found the same remarkable difference to prevail in the crushing strength of Wrought-iron and Steel, the former giving 26 and the latter 18 per cent. greater resistance in Beams than in Pillars (133). The variations in the value of My for beams in Table 62 are 117.1 104 = 1.126 or 12.6 per cent., and 85.1 104 = .818 or 18.2 per cent., which are not greater than the variableness of plate-iron under tensile strains, namely +29 and 33 per cent., as shown by Table 149. This is the more satisfactory when it is remembered that the thicknesses ranged from 03 to 75 inch, or 1 to 25, and the breadths from 1.9 to 24 inches, or 1 to 12.6. These relative numbers, however, fail to give an adequate idea of the great differences of the dimensions: Fig. 100, where the beams are drawn to the same scale, will convey a clearer conception. It should also be observed that the largest and the smallest beams give nearly the same value for Mw, namely, 100·4 and 106 4 respectively, and that both differ but little from 104, the mean value of the whole.

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The application of the laws of Wrinkling to rectangular pillars is shown by (249), and to Tubular Beams by (406), &c.

CHAPTER X.

ON THE TRANSVERSE STRAIN.

(323.) The general investigation of the Transverse Strength of Materials is complicated very much by the variable conditions in the mode of fixing and loading. It will therefore be expedient to take first a standard case, say that of a horizontal

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