the effective width of the 5-inch plate would be 5 feet 57 inches, and the wrinkling strain by the rule (308) becomes ✓ •625 = 65•75 x 80 = 7.8 tons per square inch, which is 7.8 • 19 = .41, 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 f becomes 3 x 58.66 x 360 x 24 19.2 tons, 2 x {248 x 15.5) – (22.958 x 14.45} as in col. 12. (320.) It will be interesting and instructive to check this result by an analytical investigation. The beam being 30 feet long supported at each end, our first 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 x 23.475 • 24 = 18.78 tons per square inch, this being simply propor = 12.04 square tional to the relative distances from the neutral axis N. The area of say the top plate being 15.5 X •525 = 8.137 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 x 18.78 x 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. Following the same course with the sides, we have below N the area of the two half sides = 22.95 x •525 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 11.475 - 24 = 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 x 9.18 x 11.475 x ) = 180 = 9.4 tons at W, as the resistance of the sides, making a total of 19.93 + 9.4 = 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). (321.) The value of f 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, f 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 Table 77 will give the value of My 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 Mw 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. 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 B = W = beam supported at both ends and loaded with a dead weight in the centre: the effect of other conditions may be considered afterwards. For solid Rectangular sections we have the Rules :(324.) W = D2 x B x Mr : L. (325.) D = NW x L) = (M. X B). (326.) (W x L) = (D2 x Mr). (327.) M. (W x L) = (D2 x B). (328.) For solid square sections: W = D x M.: L. (329.) For hollow square sections: D* - di x M.; L. D (330.) For hollow rectangular sections :(D x B) - (d xb) x ML. D (331.) For solid cylindrical sections : W = D.3 x Mr : L. (332.) For hollow cylindrical sections D. - di X M.: L. D. (333.) For solid Elliptical sections : W = D X D X M.-L. (334.) For hollow Elliptical sections :(D X DB) - (dB x dB) x MrL. DD In which D = the external, and d = the internal depth, in inches. B = the external, and b = the internal breadth, in inches. W = W = W = TABLE 64.-Of the TRANSVERSE STRENGTH, &c., of British Cast Iron, reduced to bars 1 inch square, 1 foot long between supports, and loaded in the centre. inches. inch-Ibs. 17,873,100 . 00002422 2403 .06836 82.17 801 .01940 7.770 16,802,000 00002571 2358 .07998 94.29 786 .02021 7.948 7.051 |