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the general Ratios of loads, and actual safe loads of a bar 4 x 4 x }, 12 feet long :

Ratio of

Load. Load.

cwts. Normal position T, dead load in centre 1:0 13

T, distributed dead load 2.0 26
T, Rolling load

6.5 Reversed position I, dead load in centre 3

5.2 I, distributed dead load .. $ 10.4 1, Rolling load

}

2.6 (381.) “ Wrought-iron I Beams.”—The strength of rolled beams with double flanges may be calculated on the same principles as those of I and T section. Figs. 89, 90 are sections of two wrought-iron beams experimented upon by Mr. Fairbairn. Fig. 89 sunk with 12,955 lbs. in the centre; the length between bearings being 11 feet. Calculating from the bottom as before (378), the reduced value of Dx B becomes for the top flange (7-62) x 2 = 32.5; for the vertical web (62 – .38%) x .325 = 11:7; and for the bottom flange, .382 x 4 = 0.576. The sum of the whole = 44776, which with MI = 3200 lbs. (375) and rule (324) becomes 44.776 x 3200 = 11 = 13,026 lbs., which is į per cent. only in excess of 12,955 lbs., the experimental weight.

Fig. 90 sunk with 18,962 lbs. in the centre, the length between bearings being 10 feet. Calculating from the bottom as before, the bottom flange gives · 442 x 4.3 = .83; for the vertical web ( 7.44%) *.35 = 17.08; and for the top flange (8° - 74) x 24 = 41.25. The sum of the whole 59.16, with which the rule (324) becomes 59.16 x 3200 -- 10 = 18931 lbs., which is only 0.17 per cent. less than 18,962 lbs., the experimental weight. See (673) and Table 109 for the Deflections.

(382.) These results show not only the accuracy of the method of calculation, but also the correctness of Mr 3200 lbs. derived from T bars (375) as applied to I sections. Both beams are described as yielding by lateral flexure of the top flanges under compression, although their areas were much greater than those of the respective bottom flanges. As shown in (445), with

wrought-iron beams, the flanges subjected to compression should have considerable width, to enable them to resist flexure laterally, and this is especially necessary where the length of the beam is great.

(383.) These rolled beams are now made of large sizes, and are deservedly used extensively. In Table 73 are given the sizes of the most useful sections supplied by various makers; col. 14 gives the safe working loads for beams 1 foot long between bearings, from which the load for any length may be easily found. Thus with No. 7, which is 8f inches deep, the safe dead load, with a length of 12 feet, would be 706 - 12 = 59cwt.; or fora moving load 29 cwt.(923). Again: say we require a beam to carry a dead load of 20 cwt. with a length of 14 feet, which is equivalent to 20 x 14 = 280 cwt. with a length of 1 foot, for which No. 4, 6} inches deep, 279 cwt. would be used.

The deflections may be found by (674) and by col. 13 in the same Table:—thus with 61 cwt. in the centre, No. 7, 12 feet long would deflect •000001486 x 123 x 61 = •1514 inch. With 20 cwt. in the centre, No. 4, 15 feet long deflects .000006351 x 153 x 20 = .34 inch, &c.

(384.) Plate-iron Girders." — The investigation of the strength of plate-iron girders may be effected most easily by analysis from elementary principles with the known tensile and crushing strengths of the material, in the manner illustrated for plain rectangular sections in (494). By Table 1, the mean tensile strength of plate-iron is 21.6 tons per square inch; and by (201) the mean crushing strength = 19 tons, the ratio being as 8 to 7, which will be the ratios of the areas of the top and bottom flanges also.

(385.) Let Fig. 93 be the section and Fig. 91 the elevation of a girder, say 20 feet or 240 inches long between bearings, loaded in the centre, and 30 inches deep, but the effective depth, or that at the centres of gravity of the top and bottom sections may be taken at 29 inches (449). This will evidently be equivalent to & cantilever of half the length or 120 inches, built into a wall at one end and loaded at the other with half the central load, as in Fig. 92:—but it is necessary to remember here that in a beam supported at both ends and loaded in the centre, the lower part

TABLE 73.-Of the STRENGTH and STIFFNESS of ROLLED IRON I BEAMS.

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• 2481 .2857 4228 10.7
· 0948 1092 1616 13.7
• 3645 .4770 6032 9.4
• 4454 5828 •7371 8.6
•5309 .6948 8785 7.9
1.5680 2.0520'2.5950 4.5
. 4612 .5258 .7115 13:0
•4887 5572! •7539 12.5
.6526 7442 1.0070 10.9
.19212084 2878 28.0
.8745 •94841.3100 17.4
.5506 5972 8246 16.4
• 6031 6541 9031 16.4
•5352 6183 8589 20:1
. 6723 7767 1:0790 18:0
• 8070 9323 1.2950 16.2
• 4422 •5063 -6951 29.5
• 6296 7214 •9903 25:1
. 94861.0870'1.4920 20:1
6250 7823 1:0650 32:1
7727 .96731.3170 29.4
•9659 1.20901.6460 29.4
1.1590'1:45101.9750 29.4
1.3520 1.6920 2.3010 29.4
• 1369 -1643 2504 53.0
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of the section is subjected to tension and the upper part to compression. In a cantilever loaded in the usual manner this is just reversed :-to prevent confusion, therefore, we have in Fig. 92 taken the weight as acting upwards by means of a pulley, thus eliminating that difficulty, and restoring the normal conditions with a beam supported at both ends and loaded in the centre.

The top and bottom flanges in Fig. 93 occupy the position of fulcrum and resistance to one another reciprocally, the strength may therefore be found from one of them, say the bottom. We have first to find the reduced or net area of the bottom by deducting the metal cut away by the rivets, or by taking the area through the line of rivet-holes :—the rivets being inch diameter, the area of the bottom plate (12 - 14) x = 8.06 square inches. If we admit that the part of the middle web between the two angle-irons compensates for one pair of rivet-holes (which is very nearly the fact), we shall have for the area of the two angle-irons (3+24) – } } x 2 = 4.875 square inches. The sum of the two areas = 8:06 + 4.875 = 12.935, say 13 square

inches. Then, the tensile strength of plate-iron being 21.6 tons per square inch by Table 1, the resistance of the bottom becomes 13 x 21.6 = 280.8 tons, and the leverages being 29 and 120 inches respectively, as in Fig. 92, we have at the end of the 10-foot cantilever 280.8 x 29 - 120 = 67.86 tons breaking weight, equivalent to 67.86 x 2 = 13572 tons in the centre of our beam, 20 feet long.

(386.) It is shown in (464) that the breaking weights of girders of similar or nearly similar sections are in the ratio of the respective areas of their bottom flanges, multiplied by the depth, and divided by the length: hence we have the rules :

(387.) W = A XD X C;L. (388.) A = (W x L) = (D X C). (389.) D = (W x L) = (A x C). (390.) C = (W x L) = (A x D).

In which W = the central load on the beam in lbs., tons, &c.,

dependent on the terms of C. A = the area of the bottom plate and angle-irons, in

square inches.

D = the depth in inches or feet.
L = the length between supports in the same terms

as D.

C = a constant adapted to the particular materials,

&c.; for plate-iron Girders 75 tons. It will be expedient to take for A, the gross area of the bottom (making no deduction for rivet-holes), and for D, the total depth (449). The value of C, as adapted to those conditions, may then be found by rule (390), taking was found by analysis or as given by direct experiment.

(391.) Thus, the gross area of the bottom plate in Fig. 93, = 12 x = 9, and of the two angle-irons, (3 + 2) } x 2 = 5.5; the sum being 9 + 5.5 = 14:5 square inches. Then, the breaking weight W as found in (385) being 13572 tons, rule (390) gives C = (135.72 x 240) + (14.5 x 30) = 74.88,

say 75.

The rule (387), namely W = A xDXC : L, is the wellknown one given by Mr. Fairbairn, the value of C as given by him being 80 for tubular beams, and 75 for ordinary flanged girders, the latter having precisely the value that we found for it by an altogether independent method.

(392.) We must now consider the top flange, in connection with which there are three points requiring attention :—1st, to see that the area is sufficient to bear the crushing strain :-2nd, that the breadth is sufficient to prevent failure by lateral flexure; and 3rd, that the thickness is sufficient to prevent “Wrinkling."

The area which has to sustain the crushing strain is really the gross area, for the material lost at the rivet-holes is replaced by the rivets, which being put in and riveted while hot, effectually fill the holes and restore the section to its normal condition so far as compressive strains are concerned :—but of course this does not apply to tensile strains, for which the net area at the rivet-holes must be taken.

(393.) The gross area of the top plate = 12 x } = 9 square

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