= = breadth required for that length and weight being greater than any other, governs the case (448). These rectangular pillars are under peculiar conditions, being forced to yield by flexure in the direction of their larger dimension; of course an ordinary rectangular pillar of unequal dimensions, left to itself, would fail by bending in the direction of its least dimension : in a lattice girder this is prevented by the diagonals. By col. 4 of Table 34, Mp = 162900 lbs., which with Factor of Safety 3 becomes 162900 - (112 x 3) 485 cwt.: then the rule (235) may be modified into:(447.) B24 = (W x L) - (480 x t). In which W = the safe working dead load in cwts.; L = the length of the pillar in feet; B = the breadth or greatest dimension of a rectangular pillar, and t = the thickness or least dimension of the same, both in inches. For 24 feet and 5 tons, or 100 cwt., as in our case, and assuming t } inch, we obtain B2-6 = (100 x 576) = (480 x }) 240, which is the 2.6 power of B. By logarithms log. 240 = 2:38 = 2.6 = .915, 2 which is the log. of 8.22 inches, the breadth B, required. To show that the breadth is a maximum with the length we have taken, we may calculate it for the next greater and next less lengths: thus, with 20 feet, or d, d, in Fig. 155, the strain by (146) is 7 tons, or 140 cwt., and we have (140 x 400) = (480 Х * }'= = 8.142 inches. For 28 feet, or e, e, in Fig. 155, and 3 tons, or 60 cwt., we obtain (60 x 784) -- (480 x 72 7.615 inches; both being less than 8.22 inches as found for 24 feet. The actual breadth of the top flange in Fig. 141 given by two angle-irons and the 3-inch diagonals was 8.1 inches. (448.) In an ordinary plate-iron girder with a continuous vertical web the strains would be found to increase, not by steps of 2 tons, as in Fig. 155, but in regular arithmetical progression from the ends where it is nothing to the centre where it attains a maximum, and it will be found that the tendency to yield by flexure at the centre is a maximum with one-third of the maximum or central strain, which by the conditions of the 1 1 = case is exerted on an acting length of two-thirds of the extreme length of the girder. Thus, let Fig. 142 be a girder 9 feet long and 1 foot deep loaded with 4 tons in the centre, giving 2 tons on each prop, which is equivalent to a cantilever, Fig. 143, 4.1 feet long, fixed in a wall at one end, and loaded with 2 tons at the other. The strain on the top and bottom members at the wall is therefore 2 x 4 = 1 9 tons, equivalent to 9 tons at the centre of the girder, Fig. 142. Then we have the series of longitudinal strains given in the figure, increasing in arithmetical progression, and we thus obtain a series of imaginary pillars with lengths of : 0 1 2 3 4 5 6 7 8 9 feet, the corresponding longitudinal compressive strains being 9 8 7 6 5 4 3 2 1 0 tons. Then the tendency to break by flexure at the centre being in the ratio W X L', becomes ’ 0 8 28 54 100 108 98 64 0 80 which is a maximum with 6 feet, or two-thirds of the extreme length, and the corresponding strain of 3 tons, or one-third of the maximum or central strain. The practical application of these laws to plate-iron girders is illustrated by (394). (449.) “Effective, and Extreme Depth."-It should be observed that in the diagrams, Figs. 136–151, the strains are taken as acting on mathematical lines, or centres of strain, and that in dealing with practical cases, the depth taken as a basis for calculation should not be the extreme depth of the girder, but rather the distance between centres of the resisting forces of cohesion and crushing, which may be taken as coincident with the centres of gravity of the sections of the top and bottom members of the girder (385), (684). The difference between the effective depth thus measured and the extreme depth is sometimes considerable, but in practical rules it is allowed for in fixing the value of the constant (390). 9 B = (450.) Beams are termed “Mathematically Similar” when ) all the dimensions of one bear a given proportion to all the corresponding dimensions of the others. Thus, in Fig 152, A, B, C are three similar tubular beams of plate-iron, or other material, the length, depth, breadth, and thickness of B, being double the corresponding dimensions of A, and in C, triple those of A. Again, in Fig. 153, D, E, F are three similar cast-iron girders, all the corresponding dimensions being in the ratio 1, 2, 3, as before. We may now complete the illustrations by calculating the breaking weight of these beams. For the wrought-iron tubular beams A, B, C, we may take the value of Mī at 3200 lbs. (375), then rule (330) becomes with (28 x 1) - (1.83 x 8) A = x 3200 = 4 = 1333.76 lbs., 2 (48 x 2) – (3:68 x 1.6) x 3200 = 8 5335.04 lbs., 4 (69 X 3) - (5.4 x 2.4) x 3200 = 12 = 12003.84 lbs. 6 Now, it will be observed that the linear dimensions being in the ratio 1, 2, 3, the loads are in the ratio of the squares of those numbers, or 1, 4, 9. We should obtain the same Ratios by the special rules (377), or even by the incorrect old rule (337), although the calculated loads would differ in amount. (451.) For the cast-iron girders D, E, F, we may take the value of My at 92 ton (335); then calculating by the special rule in (378), we obtain :D = {}" x 2) + (33 – }”] +) + (– 3-] x 4} * *92 + 5 = 6.049 х { 12 * * 4)+(62 – 19J x 1) +(88 – 6*] 8} * •92 + 10 = 24 · 196 x F = {1}+x6) + (92–1}*]x1})+(122–9*] x 12} * •92 + 15 = 54 : 441 C = Tons E= Here again, the linear dimensions being in the ratio 1, 2, 3, the loads are in the ratio of the squares of those numbers, or 1, 4, 9, &c. . (452.) “ General Laws for Similar Beams.”—Mr. Tate and others have shown that there are some general laws governing the relations of similar beams, which are very useful, by enabling us to reason from one whose strength is known by experiment, to another of very different sizes, but similar proportions, whose strength we desire to know. Thus, as an extreme case, it is shown in (475) that the strength and sizes of the great tubular bridge at Conway, 400 feet span, and weighing 1080 tons, might be calculated with approximate accuracy from the experimental strength of a little model tube 31 feet span, weighing only 4} lbs. Figs. 152, 153, will enable us to illustrate the principal laws for similar beams. (453.) 1st. The breaking loads of similar beams are to each other as their sectional areas :- thus, the areas of A, B, C are 0:56, 2.24, and 5:04, and of D, E, F, 6•25, 25.0, and 56.25 square inches respectively, which in both cases are in the ratios of the breaking weights 1, 4, 9. (454.) 2nd. In similar beams, the cubes of the breaking loads are to each other as the squares of the weights of the beams between supports. Thus, the breaking loads in our figures being in the ratio 1, 4, 9, we have 1', 49, 99, or 1:0, 64, and 729, which are the ratios of the squares of the weights of the beams, for 1%, 8, and 27% are = 1, 64, and 729. (455.) 3rd. The breaking weights of beams of similar sections, but of varying lengths, are equal to the continued product of the whole cross-sectional area, the depth, and a constant derived from experiment for the particular form of beam, material, &c., divided by the length, or distance between supports, hence we have the rules :(456.) W = a x d x M: 1. (457.) (W x 2) = (d x M). (458.) M (W x1) = (a x d). In which W = the weight or load on the beam in lbs., tons, &c., dependent on the terms of M; a = the whole cross-sectional = a = = area in square inches; d = the depth in inches ; l = the length or distance between supports in inches; and M = a constant from experiment, &c. Thus taking the girder E, Fig. 153, we may find the value of M: here W 24.196 tons, as found in (451); a = 25 square inches; d 8 inches, and I = 120 inches. Then M = (24.196 x 120) = (25 x 8) = 14.5176. Now . applying this, for example, to the girder F, where a = 56.25, d = 12, and 1 180, we obtain W = 56.25 x 12 x 14.5176 : 180 = 54:441 tons, as before, &c. (451). (459.) 4th. The breaking weights of beams of similar sections, but all of the same length, are as the cubes of the linear ratios of the sections. Thus, if the beams A, B, C had all been say 4 feet long, or the length of A, the breaking weights would evidently have been in the ratio 1, 8, 27, instead of 1, 4, 9:—then the linear ratio of the sections being 1, 2, 3, we have 13, 23, 33, or 1, 8, 27, which is also the ratio of the breaking weights. (460.) 5th. In beams having similar sections but different lengths, L and 1, corresponding to breaking weights W and to, the relations are expressed by the equation w = R3 x W x L X XL • 1, in which R = the linear ratio of the two sections. Thus, with the cast-iron girder D in Fig. 153, we have W 1, and L = 5; then comparing with the girder F, we have R = 3, from which we obtain the relative breaking weight of F 33 x1 x 5 = 15 = 9; that is to say, if the breaking weight of D = 1, that of F will be 9. Similarly, reasoning from D to E, we have R 2, and w comes out 23 x 1 x 5 + 10 = 4, the relative breaking weight of E, &c. (461.) 6th. The breaking weights of similar beams are to each other as the squares of their linear dimensions. Thus, with Figs. 152, 153, in A, B, C, or in D, E, F, the ratio of the linear dimensions is 1, 2, 3, and 1°, 2, 3, or 1, 4, 9, is the ratio of the breaking weights, as given by the figures and shown by (450), (451). (462.) 7th. In beams of similar sections the cube-root of the breaking weights multiplied by the respective lengths, is in the simple ratio of the linear dimensions, or V W X L = R. Thus, in the tubular beams A, B, C, we have breaking weights in the ratio 1, 4, 9, corresponding to lengths in the ratio 4, 8, = |