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6667, and we obtain 6667 +1243 791 ton:-similarly two of those parts, or 1243 × 2 = 2486, deducted from 1.0397 gives 1.0397 -·2486 = ·791, as before; the ratio of which to the breaking weight, as found by special rule and experiment, is 2791 2.528 to 1.0 instead of 3 to 1, as by the ordinary method of procedure. The effect of this is to add ·791 ÷·6667 =1.1865, or 18.65 per cent. to the working load:-taking a beam whose breaking weight = 100 tons, then, when strained to 3rd, the load will not be 100 ÷ 3 = 33.33 tons, but 118.65 ÷3 39.55 tons.

(358.) The application of these principles to practice is very simple; for instance, the breaking weight of the girder, Fig. 79, was found in (351) to be 40.5 tons, as calculated by the special rule; then if we adopt 4 as the factor of safety, the working load would be 40.54 = 10.12 tons by the ordinary course, but by (356) we obtain 10 12 × 1.28 13 tons nearly. If we adopt 3 as the Factor, we have 40·5 ÷ 3 = 13.5 tons in the usual way, which by (357) becomes 13.5 × 1.1865 16 tons, &c.

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(359.) "Ratio of Square to Round Bars."-It is shown in (519) that the theoretical ratio of the strength of square to round bars of the same dimensions is 1.7 to 1.0; but this is strictly true for those cases only where the tensile and compressive strengths and the corresponding extensions and compressions are equal to one another, and this, as we have seen, is not realised perfectly with any materials, except with very light strains (617). Under these circumstances, experiment alone can determine the real ratio for the Breaking weight, &c.

(360.) For cast iron we have the experiments of Mr. W. H. Barlow, which were made with direct reference to this question: the results are given in Table 69. It will be observed that the round bars were not made of the same linear dimensions as the square ones, but rather of the same sectional area. The object of this was possibly to avoid the complications due to the size of the casting, which, as shown by (932), is very influential on the transverse strength of cast iron. Thus, a bar 2 inches square would have the same area as another 24 inches diameter, and presumably there would be equality of strength so far as that is affected by the size of casting.

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In the first set of experiments the bars were about one square inch in area: the mean value of Mr from five square bars = 2530 lbs., and from five round ones 1697 lbs., the ratio being 25301697 = 1.491 to 1.0. In the second set the bars were about 4 square inches in area: four square bars gave Mr 2173 lbs., and nine round ones = 1399 lbs., the ratio being 21731399 = 1.553 to 1.0.

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TABLE 69.-Of EXPERIMENTS on the RELATIVE TRANSVERSE STRENGTH of SQUARE and ROUND BARS of CAST IRON, all 5 feet long.

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2173÷1399 = 1·553 to 1; Ratio of square to round.

(361.) The mean ratio from the whole of these experiments,

23 in number, is 1.522 to 1.0, of 1.7 to 1.0, as due by theory. as the real ratio for cast iron.

or nearly 1.5 to 1.0, instead This may be taken, therefore,

For timber we have the experiments of Mr. P. Barlow on 2-inch round and square bars cut from the same plank of Christiania deal 4 feet long:-there were three round bars which broke with 740, 780, and 796 lbs. respectively, the mean being 772 lbs. Two square bars gave 1125 and 1110 lbs. respectively, the mean being 1117 lbs., and the ratio 1117÷772 = 1.447 to 1.0. Mr. Couch's experiments on round beams in the form of spars 3 inches diameter give as the value of Mr for

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T

English Oak

337 lbs.

By Table 66, the mean value of Mr for square bars of the same materials is

389

491

446

509 lbs.

hence the ratio of the transverse strengths of square to round

bars comes out

1.314

1.584

1.64

1.51

The mean = 1.512 to 1.0: Mr. Barlow's experiments, as we have seen, gave 1.447 to 1.0.

(362.) We may therefore take 1.5 to 1.0 as the ratio of square to round beams of Timber, being the same as found for Cast iron; both being for the breaking weights (354).

For wrought iron, and still more certainly for steel, whose elasticity is nearly perfect, especially for the moderate strains usually adopted as the working load in practice, we may admit the theoretical ratio 1.7 to 1.0.

(363.) "Bottoms of Round Vessels: Cast Iron."-The bottoms of air-vessels and the sides of valve-boxes for pumps, &c., are frequently made flat for reasons of necessity or convenience, although that form is not well adapted to bear the heavy internal pressure to which they are usually subjected; the calculation of the strength is therefore a matter of some importance. We may find the strength of the whole bottom by taking a portion of it, say a strip in the direction of a diameter, 1 inch wide, and having found the strength of that by the ordinary rules for beams, then multiplying the result by the

number of such strips in the whole circumference will give the strength of the whole.

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(364.) Let Fig. 82 be a vessel 3 feet internal diameter with a plain flat bottom 1 inch thick:- -we have then a bar A, B, 1 inch deep, 1 inch wide, 3 feet long, fixed at both ends and loaded equally all over by the pressure of the water. By column 5 of Table 66, the value of MT 2063 lbs. in the centre when the bar is merely supported at the ends, but when fixed at the ends and loaded equally all over, as in our case, the ratio given by (431) is 3, hence MT becomes 2063 x 3 : = 6189 lbs., and rule (324) gives 12 × 1 × 61893 = 2063 lbs. spread equally all over. Now, a circle 36 inches diameter has a circumference of 113 inches, and as each of our imaginary beams occupies two inches, namely an inch at each end, we have 113 256.5 such beams in the whole bottom, and the breaking weight of the whole number will be 2063 × 56.5 = 116560 lbs. This load being distributed all over an area of 1018 square inches gives 1165601018 114 lbs. pressure per square inch, or 114 × 2.3 262 feet, the height of the column of water which would burst the vessel. An air-vessel is exposed to more or less violent shocks from the alternating action of the pump, &c.; it is therefore expedient to adopt a high Factor of safety (924) for such cases, say 10, hence the safe working pressure would be 262 ÷ 10 =26.2 feet of water.

(365.) It may be objected to this method of calculation, that we have reckoned on the strength of the material at the centre, or where the bars cross each other, twice over or more; but this is not the fact. Taking, for instance, the two bars A, B, and C, D, in Fig. 82, at right angles to each other, it will be observed that the longitudinal filaments which form the strength of A, B, are not at all strained by loading C, D, because the filaments of the latter form another scries at right angles to those of the former; and the same reasoning applies to all the imaginary beams of which the bottom is composed.

(366.) "Bottoms of Square Vessels."-In applying this method of calculation to the bottoms of square vessels some modifications are necessary. Let Fig. 83 be a square vessel with a bottom 1 inch thick, &c. :-we have as before an imaginary

bar 1 inch square and 3 feet long, with a breaking load of 2063 lbs. distributed all over, and if all the imaginary beams of which the bottom is composed were allowed to deflect alike, and therefore to be equally strained, we should have 2063 × 36 × 2 = 148536 lbs. as the total breaking load. But it will be observed that from the conditions of the case, this equality of strain is not realised, for while the beams E, F, and G, H, have equal and great deflections, the beam I, J, will deflect very little, being prevented from doing so by the beam E, F, the fibres of both beams being interlaced with one another. It will be thus seen that between the edges where the deflection is nothing, to the centre where it is a maximum, we have a series of beams with a progressively increasing deflection. The mean deflection of the whole series of beams is 3rds of the maximum central deflection, hence the bottom will bear 3rds only of the strain we calculated before, or 148536 × 2 ÷ 3 = 99024 lbs. distributed over an area of 36 × 36 = 1296 square inches, or 99024

1296 = 76 lbs. per square inch. Hence the ratio of the strength of round to square bottoms is 114÷76 = 1.5 to 1.0. For plain unribbed flat bottoms of round and square vessels, we have the rules:

(367.) For Round Vessels p

(368.)

=

t2 x 148390÷d2.

h = 12 x 342780 ÷ d2.

(369.) For Square Vessels p× 99010 S2.

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In which t = the thickness of the plate in inches; d = inside diameter in inches; S = side of square, inside, in inches; p bursting pressure in pounds per square inch; and h = head of water, bursting pressure, in feet. Thus, for a vessel 40 inches diameter 14 inch thick, rule (368) gives 2.25 × 342780 ÷ 1600 = 482 feet of water bursting pressure, or with Factor 10, 48.2 feet working pressure, &c. See (961). (371.) "Ribbed Bottoms."-The same reasoning may be applied to a flat-bottomed vessel, with strengthening ribs inside or outside; with cast iron, and an internal pressure, the ribs

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