the bottom is the weaker, and should govern the strength, but does not seem to do so. For example, with No. 9, the wrinkling strain by col. 11 is 17.32 tons per square inch, hence 17.32 x 437 x 15.75 = 119.2 tons, the resistance to compression: but for the bottom we have 21.6 x 272 × 15·75 = 92.54 tons only as the resistance to tension. In col. 9 the strength, as calculated from the resistance of the top plate, came out 33.74 tons, the error being only + 0.3 per cent. Now, if the strength were dominated by the bottom, as it should be by theory, we obtain 33.74 × 92.54 119·2 = 26·19 tons only, but little more satisfactory Let Fig. 108 be a section (412.) We should obtain a result by analysis, as in (320), (414), &c. of the same beam reduced to the case of a cantilever of half the length fixed at one end and loaded at the other, as in Fig. 92; the Neutral axis will not now be in the centre, but must occupy such a position as to give equality to the resistances of tension and compression. We must assume a position for it by judg ment, say as in the figure: for the top plate we found in (411) 119 2 tons, which with leverages of 11.182 and 180 inches gives at W, 119.2 × 11·182÷180 = 7·406 tons. Then, for the sides above the line N. A., the maximum strain at the top being 17 32 tons per square inch, that at o becomes 17.32 × 5.48211.4 = 8.329 tons. The area is 10.963 x 544 = 5.964 square inches; hence 5.964 x 8.329 × 5.482 × 180 = 2.016 tons at W, which added to the resistance of the top plate = 7.4062.016 9.422 tons at W due to compression alone. = Similarly, for the part of the section subjected to Tension, we have for the bottom plate 15 75 × ·272 × 21·6 × 12.214 ÷ 180 = 6.277 tons at W:-the strain at p is 21·6 × 6·039 ÷ 12.35 = 10.56 tons per square inch, which at W becomes 10.56 × 272 × 2 × 6·039 × 1·333 ÷ 180 = 3·106 tons, giving for the total resistance to tension 6·277 + 3·106 = 9∙380 tons at W, which is rather less than 9 422 tons due to compression, but is sufficiently near equality for our purpose. The sum of Tension and Compression is thus 9.380 + 9.422 = 18.802 tons at W, and is equivalent to 18.802 x 2 = 37·6 tons in the centre of the beam 30 feet long: experiment gave 33.64 tons, hence the error is 37 633.64 = 1.117, or 11.7 per cent. By rule (514) the error was + 0.3 per cent. only, as in col. 10. It is remarkable that the rule (514) by which col. 9 of Table 77 is calculated, which is based on the value of f, or the wrinkling strain, and is therefore not strictly applicable to beams which fail by tension, gives, notwithstanding, more correct results than any other rule:-the mean error of Nos. 7, 8, 9, and 10 being +2.7 per cent. only. (413.) "Lateral Strength of Beams."-In designing large Tubular beams for bridges, &c., it is necessary to consider and provide for the lateral strain due to the wind impinging on the side. The proportions of beams are usually fixed principally with a view to sustain a vertical load, and as a result they are frequently very weak in resisting horizontal pressure. With a view to obtain experimental information on this matter, Mr. Hodgkinson took the beam, No. 10 in Table 77, which in the ordinary vertical position broke with a load of 26 6 tons, and laid it on its side, as in Fig. 109, when it failed with 14·3 tons. In this abnormal position the whole of the conditions are so greatly changed, that the strength cannot be calculated correctly by the usual rule (408) for Tubular beams; thus, in our case, the wrinkling strain of the thin top plate by the rule in (308) becomes · 125 ÷ 24 × 104 = 7.505 tons per square inch, and the rule (514) then gives 7·505 × 2 × {163 × 24) – (15·753 × 23-384} 3 x 360 × 16 W = = 6.028 tons only, whereas experiment gave 14.3 tons. But this rule supposes that the beam is of the ordinary form and proportions, and that the strength is governed by the resistance to wrinkling or by f, but obviously the proportions might be such that the wrinkling strain would have little effect on the strength: for example, in Fig. 110 the great strength of the side plates a, b render them independent of the weak top plate c, and the rule which takes that plate as the exponent of the strength of the whole beam must necessarily fail to give correct results. (414.) The best method of calculating a beam of such an abnormal form is by analysis, as illustrated in (320) and (412). We shall assume that the top plate fails by wrinkling with = 7.5 tons per square inch, as calculated in (413), and the bottom plate with 21.6 tons, namely, the mean tensile strength of plate iron as given by Table 1. Then, for the sides, we shall take the maximum tensile strain at T and compressive strain at C = 21.6 tons per square inch also: the neutral axis N. A. must then be placed in such a position as to give equality to the tensile and compressive forces above and below that line respectively, and this position must be fixed tentatively by judgment. Reducing the case for the purposes of calculation to the equivalent one of a cantilever of half the length of the beam fixed at one end and loaded at the other, as in Fig. 92, we obtain the dimensions given by Fig. 109. For the top plate we obtain 24 × 125 × 7.5 22.5 tons, which with the leverages of 9·438 and 180 inches respectively, gives 22.5 x 9.4381801.18 ton at W. Then for that part of the side plates above N. A., the strain of 21.6 tons at C becomes 21.6 x 4.6889.5 = 10.66 tons per square inch at o, and their area being 9.375 × 616 = 5.775 square inches, with leverages of 4.688 and 180 inches gives at W, 5-775 × 3 × 4·688 ÷ 180 = 2.137 tons for the sides (495), making with that due to the top 1.182.137 3.317 tons from compression alone. Similarly, for the bottom plate, we obtain 24 × 125 × 21.6 6.48 tons, which at W becomes 6.48 × 6·438 ÷ 180 = 2.318 tons. The strain at p becomes 21.6 × 3·1886.5 = 10.59 tons per square inch. Then 6.375 × 616 × × 10·59 ÷ 180=982 ton at W due to the sides, making a total of 2.318982 = 3.3 tons due to Tension, or nearly the same as 3.317 tons due to compression. The sum of the two = 3.3 +3.317 6.617 tons at the end of the cantilever 15 feet long, equivalent to 6.617 × 2: 13.234 tons in the centre of the beam 30 feet long, as in our case: experiment gave 14.3 tons, hence 13.23414·3: = = = ⚫9255, showing an error of 1.09255 = ⚫0745, or 7.45 per cent. Considering the extremely abnormal form of the beam in this position, the error is perhaps not greater than might be expected. (415.) It will be observed that we have taken the full tensile strength of the iron, allowing nothing for rivet-holes, &c., on the supposition that there is no joint at or near the centre of the beam, where the strain is a maximum. With small and moderate-sized beams this condition is easily obtained, and with large structures we can secure practically the same condition by the adoption of chain-riveting (36). FORM FOR EQUAL STRENGTH THROUGHOUT. (416.) We have so far assumed that the beams were parallel, or of the same cross-section from end to end, and the load in the centre. In that case, the transverse strain is a maximum at the centre, and is progressively reduced towards the two supports, where it becomes nothing; to obtain throughout the length an equality between the strain and the strength it would be necessary to reduce the section of the beam toward each prop in proportion to the strain at each point, which can be effected by regulating the depths alone, while the breadths remain constant; or, on the other hand, by graduating the breadths, while the depths remain constant; or by a combination of the two methods. Let Fig. 111 be a beam 16 feet long, with a central weight of 10, producing of course a strain of 5 on each prop: this is obviously equivalent to a cantilever, Fig. 112, of half the length built into a wall at one end, and loaded with a weight of 5 at the other end. The ratio of the strains at each point along the beam is evidently proportional to the leverage with which the load of 5 acts, or to the distances 1, 2, 3 .. 8, &c., as in Fig. 112, and as by (324) the strength is proportional to D2 when the breadth is constant, it follows that D must be proportional to the square-roots of the respective strains, or √1 = 1; √2 = 1·41; √3 = 1.73, &c., as in the figure, these being of course proportional, and not real dimensions. We have thus obtained the depths in Fig. 112, which again give us the depths in Fig. 111. (417.) If we would obtain a uniform depth throughout, as in Fig. 113, which again is equivalent to the cantilever, Fig. 114, the breadths being simply proportional to the strains, the latter being 1, 2, 3, 4, &c., the former will be 1, 2, 3, 4, &c., also, and we thus obtain Fig. 115, giving a uniform taper from the wall to the end, and from this we obtain Fig. 116. To apply this to practice:-Let Fig. 117 be a beam of Beech, 14 inches deep, 10 inches wide, and 12 feet long, which by rule (324) with Mr = 585 by Table 66 breaks with 142 × 10 × 558 1291140 lbs. in the centre. The square of the depth at each point being proportional to the distance of that point from the nearest prop as shown by (416), and that at 6 feet being 14 inches, we have 1426= 3.27, a constant, which multiplied by the distance of each point, will give the square of the depth at that point, thus: At A we get 32.7 × 6 or 196·0/= 14·00 inches. We thus obtain the depths in Fig. 117, the breadth being 10 inches throughout. If, on the other hand, we maintain the depth at 14 inches throughout, the breadth would taper off uniformly from 10 inches at the centre to nothing at the props, as in Fig. 116. (418.) It may seem anomalous that the size at the end should be nothing, whereas a strain of 45,570 lbs., or half the central weight, has to be borne by it; but we have been considering the transverse strain only, which is really nil at G; the strain of 45,570 lbs. is a shearing or cross-strain, and must be provided for, but is a matter quite foreign to the proper subject of this chapter (123); see (403). (419.) "Load out of Centre."-Having thus found the forms of beams with a single central load, we may proceed to consider 1st, the effect of a single load out of the centre; and 2nd, of two or more loads variously distributed. In order to give precision to the investigation we will take a case, say that of a beam of Beech 12 inches square and 16 feet long; taking the value of Mr from col. 6 of Table 66 at 25 ton, the rule (324) gives for a central load 122 × 12 × 2516 27 tons breaking weight, as in Fig. 118. T |