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 1117772 = 1.447 to 1.0. Mr. Couch's experiments on round beams in the form of spars 31 inches diameter give as the value of MT for Riga Fir Red Pine Yellow Pine English Oak T 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. = (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 Mr = 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 × 3 = 6189 lbs., and rule (324) gives 12 x 1 x 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 116560 1018 = 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 = t x 148390 : . (369.) For Square Vessels p = x 99010 S2. = In which 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 1 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 P should always be inside for the same reason that a section girder should be broken flange downwards (342), the metal of the bottom plate being then brought into tension. With external pressure, the ribs should be external for the same reason. = Ꭲ Say, we have a vessel 3 feet diameter with ribs cast inside as in Fig. 84; we may take the central part, or the rib a, and its proper share of the bottom plate, namely, from b to c, as the index of the strength of the whole bottom. We have then in effect, a girder of the section given by Fig. 85, 3 feet long, fixed at the ends, and loaded all over. The value of MT for this case we found in (364) to be 6189 lbs., and by the method of calculation explained in (342), we have to calculate from the line N, A, and we have for the vertical web 4 × 1 = 16·0; for the bottom flange (52 42) × 9 81, the sum of the two = 97, and rule (324) gives 97 × 6189 ÷ 3 = 200000 lbs. spread all over. The circle 36 inches diameter has a circumference of 113 inches, and as each beam, like Fig. 85, occupies 18 inches, namely, 9 inches at each end, we have 113÷÷18 = 6.28 of such beams in the whole circumference, which will bear collectively 200000 x 6.28 1256000 lbs., distributed over an area of 1018 square inches, or 12560001018 = 1233 lbs. per square inch bursting pressure. With 10 for the Factor of safety, we obtain 123310 = 123 lbs., or 123 × 2.3 283 feet of water, safe working pressure. = = = (372.) As an illustration of the effect of placing the ribs injudiciously outside instead of inside, we have in Fig. 86 the same girder as in Fig. 85, but in a reversed position. Then calculating as before, but from the line N. A. we have for the top flange 12 × 9 9; for the vertical web (52- 12) × 1 = 24; the sum of the two = 9 +24 33, whereas in the other position we had 97. The strengths in the two positions will be in the ratio of those numbers, and without going through the whole calculation again, we obtain for the safe working pressure 283 x 3397- = 96 feet of water, instead of 283 feet as with internal ribs the ratio being 9733 2.94, or nearly 3 to 1 (343). = (373.) "Wrought Iron and Steel."-With very tough and ductile materials, such as wrought iron, there is great difficulty and uncertainty in determining the ultimate or breaking weight of a bar loaded transversely. A beam of cast iron or timber breaks more or less suddenly, and the breaking point is thus usually well defined; but it is impossible literally to break a beam of really good tough wrought iron, and it is difficult to say with what load the bar breaks down completely. But if the deflections of a bar of wrought iron or steel under a series of progressively increasing loads be very accurately observed, and the results are plotted in a diagram, as in Figs. 210, 211, it will be found, as with the tensile and compressive strains in the diagram, Fig. 215, that up to a certain point the elasticity is almost perfect, that is to say, the deflections are almost exactly proportionate to the loads, and the permanent set almost inappreciable until that point is reached, when the deflections and sets (757) begin to increase very rapidly, showing that the material is beginning to be crippled or over-strained. It will also be observed that beyond that point the deflections and sets increase rapidly with time, even with the same load. Now, the Diagram, Fig. 215, shows that both for the tensile and compressive strains the "limit of Elasticity" is about 12 or 13 tons per square inch, and as, by Table 1, the mean ultimate strength of wrought iron is 25.7 tons, it would appear that the "limit of Elasticity" is half the ultimate strength for 25.72 12.85 tons per square inch. Applying that ratio to the Diagram for the transverse strain, Fig. 210, we find that the "limit of Elasticity," or the point beyond which the bar would begin to be crippled, is about 1720 lbs. as indicated by a *, with which Rule (327) gives (1720 × 6·75) ÷ (1·0272 × 5·51) = 2000 lbs. as the value of Mr for the limit of Elasticity: then, admitting that to be half the ultimate or breaking-down strain, the value of Mr for the latter 4000 lbs. = = = = T (374.) This result is confirmed by the experiments of Mr. E. Clark on three bars of wrought iron, 14 inch square and 3 feet long, in Table 70. With MT 2000 lbs., Rule (328) gives W = 113 × 2000 (3 × 112) 20 1 cwt. as the "limit of Elasticity," and col. 4 shows that up to that point the deflections. are nearly in simple proportion to the weights, as due with perfect elasticity. But as the load is increased, the Ratio of |
