11.4 per number due to which=1767: from these we obtain 1864-1767 = 97 for the value of $3-6 – 58.6 in the rule in (230). Taking the value of Mp from Table 34, at 498,500 lbs., we obtain 498500 x 97 = 100 = 483600 lbs. F, as in col. 9. (256.) This will require correction for "Incipient Wrinkling”: we have first to find Ww by the rule (250), namely Ww = (vi: Vb) x 80, which in our case becomes (W.065 W8.1) x 80, or ( 2449 = 2.846) x 80 = 5.468 tons per square inch, as in col. 11. This is a very low result, and is due to the extreme thinness of the plate. To find Cw in lbs., the area of the whole section being 2.07 square inches by col. 4, we have 5.468 x 2240 x 2.07 = 25350 lbs., as in col 10, therefore i Cw = 19010 lbs., and the rule (252), or Pw = F x Cw; (F + Cw) becomes 483600 x 25350 - (483600 + 19010) 24390 lbs., the reduced breaking weight, as in col. 7. The experiment gave 27,545 lbs., hence we have 24390 = 27545 = .886, showing an error of 1.0 – 886 = .114, or cent., as in col. 8. We may now give an illustration of the error that would have occurred in this case if we had neglected “Wrinkling' and calculated for incipient crushing in the ordinary method for solid pillars. Thus C, or the crushing strength of wrought iron in pillars, being 19 tons, or 42,560 per square inch, and the area 2.07 square inches, Cp becomes 42560 x 2.07 = 88100 lbs., and Cp = 66070 lbs. ; then the rule Po F X Cp • (F + { Cp), becomes 483600 x 88100 - (483600 + 66070) = 77510 lbs., whereas experiment gave 27,545 lbs. only. The calculated strength is 77510 = 27545 = 2.814 times the experimental ; an error of 181.4 per cent.! (257.) “ Rectangular Pillars.”—As an example of the method of calculating the strength of rectangular pillars of thin plateiron by the rules in (248) and (252), &c., we will take No. 15 in Table 55, whose section is shown by Fig. 51. For t®, the logarithm of 4:1, or 0.612784 x 2.6 = 1·593238, the natural number due to which is 39.2, hence 2*8 x b, becomes 39.2 X 8.175 = 320.5; then for 4,2-6, the logarithm of 3.978, or 0.599665 x 2.6 = 1.559129, the natural number due to which is 36.235, and 1,26 x bo, "becomes 36.235 x 8.053 = 291.8. With these values, and a length of 2-333 feet, the rule in (248), namely F = Mpx {*** x 6) – (43** x bo} - L’, becomes in our case 498500 x (320-5 – 291.8) -- 5.444 = 2,628,000 lbs. . breaking weight by flexure = F, as in col. 9. (258.) Correcting this result for “Incipient Wrinkling " (249), we have to find W, by the rule (250), or W, = (NT; ) x 80, which in our case becomes (W.061 = 18.175) x 80, or (-247 - 2.859) x 80 = 6.911 tons per square inch, and the area of the whole section being 1.532 square inch by col. 4, we obtain 6.911 x 2240 x 1.532 = 23720 lbs. for the value of Cw, and 17790 lbs. for i Cy. Then the rule in (252), namely Pw= F x Cv = (F + { Cw) becomes 2628000 x 23720 :(2628000+17790) = 23560 lbs., the reduced breaking weight, as in col. 7. Experiment gave 24395 lbs., as in col. 6; hence 23560 • 24395 = .966, showing an error of 1.0 - 0.966 = .034, or 3.4 per cent., as in col. 8. (259.) If in this case we had neglected “Wrinkling," and corrected F for incipient crushing, the error would have been enormous. Thus the area being 1.532 square inch, Cp becomes 42560 x 1.532 = 65200 lbs., and Cp 48900 lbs. : then the rule (253) becomes 2628000 x 65200 = (2628000 + 48900) = 80585 lbs., whereas experiment gave 24,395 lbs. only. The calculated strength by this erroneous method is 80585 -- 24395 = 3.30 times the experimental, showing an error of 230 per cent.! (260.) “ Cellular Pillars."-Several of the rectangular pillars in Table 55 were cellular, having a plate in the centre by which they were divided into two compartments or cells: in calculating the strength of such pillars some modification in the method of applying the rules becomes necessary. We will take No. 17 as an illustration, the section of which is shown by Fig. 52. Obviously the pillar would fail by flexure in the direction of its least dimension, t will therefore be 4.1 inches, and b = 8.1 inches, as usual. To find bo, we may imagine the central plate a to be divided into two equal portions, one of which is added to each of the two side plates 2, 2, and we thus obtain the simple section Fig. 53: we can then proceed with the calculation in the usual way, as illustrated in (257). Thus to find t's, the logarithm of 4.1, or 0.612784 x 2.6 = 1.593238, the natural number due to which is 39.2; then for to*:6, the logarithm of 3.982, or 0.600101 x 2.6 = 1.560263, the natural number due to which is 36.33: the value of Mp 498500 lbs. as before, and the length or 7.6252 being 58.14, the rule in (248), namely F = Me * {tz** * b) – (t,** * bo} -I', becomes 498500 x {39.2 x 8.1) – (36-33 x 7.923} • 58.14 = 254700 lbs. = F, as in col. 9. (261.) In correcting for incipient wrinkling it must be observed that the effect of the central plate is to reduce the breadth of the wide plate, in our case to half or 4.05 inches; then the greatest breadth unsupported is the end plate or 4:1 inches, and the wrinkling strain must be calculated for that width. The rule in (250), namely W, = (17:VD) X 80, becomes in our case (W.059 • 14.1) x 80, or (-2429 = 2.035) x 80 = 9:596 tons per square inch, as in col. 11: then the area of the whole section being 1.885 square inches by col. 4, Cw becomes 9.596 x 2240 x 1.885 = 40520 lbs., as in col. 10, and i Cw = 30390 lbs. The rule in (252), namely Pw = F x Cw: (F + {Cw), becomes 254700 x 40520 = (254700 + 30390) = 36190 lbs., as in col. 7. Experiment gave 45,451 lbs., as in col. 6; hence 36190 - 45451 = .796, showing an error of 1:0 0.796 = .204, or - 20•4 per cent., as in col. 8. (262.) “ Four-celled Pillars.”—Four of the pillars in Table 55, Nos. 6 to 9, were divided by central plates into four compartments or cells, as shown by Fig. 54, the lengths varying from 10.1 feet to 2} feet. In calculating the strength of this pillar we may assume that it will bend, and fail by flexure, say in the direction of the arrow. In that case the cross-plate B will be in the neutral axis N, A, and will add nothing to the strength of the pillar in resisting flexure. The effective section therefore becomes as in Fig. 55, and the case is assimilated to (260), and may be calculated in the same way: adding the thickness of the plate e to f and g in equal portions, we finally reduce the case to Fig. 56, with this exception, that the “Wrinkling" strain will be that due to the breadth H, or 4.05 inches, Fig. 55, not 8.1 inches as at D in Fig. 56; again, although the cross-plate B adds nothing to the strength in resisting flexure, it has its full effect in resisting wrinkling, and so finally affects the strength of the pillar. (263.) Taking No. 6 as an example of the mode of calculating pillars of this form :-to find to, the logarithm of 8.1 or 0.908485 x 2.6 = 2.362061, the natural number due to which is 230.176. For 1,2-6, the logarithm of 7.79726, or 0.901599 x 2-6 = 2.344157, the natural number due to which is 220.9. The rule (248), namely F = Mpx {*** x b) – (*.** Xbo} -- L”, becomes 498500 x 230.176 x 8.1) { (220.9 x 7.9089 7.9089) ) • 102 = 573800 lbs. F, as in col. 9. Reducing for incipient wrinkling (249), we have to take H in Fig. 55, or half the side of the square for the breadth, as explained in (262), and the rule Ww = (vt vb) x 80, becomes (V.0637 = 14:05) x 80, or (-25239 = 2.012) x 80 = 10.04 tons per square inch, as in col. 11, and the area of the whole section being 3.551 square inches by col. 4, we obtain 10.04 x 2240 x 3.551 = 79830 lbs. for the value of Cw, therefore 59,870 lbs. for { Cw, and the rule (252), or Pw FxCw=(F + Cw) becomes 573800 x 79830 = (573800 + 59870) = 72280 lbs. breaking weight, as in col. 7. Experiment gave 70,070 lbs., as in col. 6; hence 72280 - 70070 showing an error of + 3.1 per cent., as in col. 8. (264.) Table 55 has been calculated throughout in the way we have thus explained and illustrated :—the mean average error on the 27 experiments in which the pillars were broken, is – 24 per cento, for the sum of all the minus errors in col. 8, is 185.5, and of all the plus errors 124:6, giving a difference of 185.5–124.6=60.9, which on 27 experiments gives an average of 60.9 ; 27 = 2.25 per cent. The range of the errors is pretty nearly equal, the greatest + error being 28.5 per cent., and the greatest error, 23 per cent. (959). It will be observed that the highest wrinkling strain in col. 11, is 18.36 tons per 1.031, square inch, which being less than 19 tons, the crushing strength of wrought iron in pillars, correction had to be made for “Incipient Wrinkling" rather than “Incipient Crushing," as explained in (252). Comparing cols. 7 and 9, it will be seen that correction was required in every case, without exception. In two cases, Nos. 1 and 11, the pillars were not strained up to the breaking point, but it is satisfactory to observe that the calculation shows they were not likely to break with the experimental strain, the calculated breaking weight in col. 7, being in both cases in excess of that strain in col. 6. (265.) “ Economic Value of Cells.”—We are now in a position to estimate the value of division-plates in cellular beams as a matter of economy. Taking a plain rectangular pillar with sides in the ratio of 2 to 1, and so short that the case is • practically governed by the resistance to wrinkling, irrespective of flexure; by the addition of a central plate, we increase Ww, or the wrinkling strain per square inch, in the ratio of 2 to N1 or 1.414 to 1.0, or 41.4 per cent. But the area of the whole section is also increased in the ratio of 7 to 6, or from 1.0 to 7 = 6 = 1.167 or 16.7 per cent., so that the total increase in strength is 1.414 x 1.167 = 1.65, or 65 per cent. As the strength is increased by the centre plate 65 per cent., and the weight 16.7 per cent., the net economic advantage from it is 65 - 16.7 = 48.3 per cent. But this will apply only to a pillar so short that the strength is governed exclusively by the resistance to wrinkling, irrespective of flexure, which in long pillarsówill so affect the case that no general ratio can be given, as it will vary with the length and general sizes of the pillar. (266.) The fairest comparison may be made by taking a concrete case : say we take the four-celled pillar No. 6 in Table 55 and Fig. 54, whose calculated strength as given in col. 7 was 72,280 lbs. Then removing the two centre-plates and adding the material in them to the four side-plates we obtain a plain square pillar as in Fig. 57, whose total area or weight is precisely the same as that of the cellular one Fig. 54, thus the 4 sides in the plain pillar, or ·09555 x 4 = .3822, and the 4 sides |