2.15 x 19 = 40.85 tons, and i Cp = 30.64 tons, hence (30.23 x 40.85) = (30.23 + 30.64) = 20.28 tons, col. 3. + Experiment gave 17:1 tons, col. 4. For the same pillar with two pointed ends we obtain 72.74 x 3.3857 = 25 = 9.85 tons = F, which being less than 1 Cp, correction for incipient crushing is not required (169), and the breaking weight is 9.85 tons: experiment gave 14 tons, col. 5. The experiments gave 17.1 and 14 tons respectively; the difference is only 3.1 tons, whereas the calculated difference is 20.28 – 9.85 = 10:43 tons, which shows that in the experiments the assumed form at the ends was not complied with perfectly in either case (241). Here again the experimental results fall between the calculated ones, as in (242). Table 54 gives a collected statement of the results of experiment and calculation on these pillars of unusual sections (242) to (247). “ Hollow Rectangular Pillars of Wrought Iron."-For rectangular pillars other than square the rules in (230) become : (248.) F F = Me* {120 x 6) – (1,59 x bo} = L'. In which t = the least dimension of the pillar externally, and to internally; b = the breadth or largest dimension externally, and bo, internally, all in inches, and the rest as in (233), namely L = length in feet, Mp = multiplier, whose value is given by Table 34, and F = the breaking weight by flexure, in lbs. or tons dependent on Mp. Thus for a pillar 3 x 4 externally, and 2} x 3} internally, in which 3; 6 4; to = 2}; bo = 3.2; L = 9 feet, both ends being flat, we have 223 x {328 x 4) – (2%** * 3}} =- 81, or 223 x{17.4 x 4) – (10-84 x 3.5}=- 81 = 87.24 tons = F. = = Correcting for incipient crushing, the area being 3.25 square inches, Cp becomes 3.25 x 19 = 61.75 tons, i Cp = 46:31 tons, therefore by the rule in (164), (87• 24 x 61.75) = (87• 24+ 46:31) = 40.34 tons breaking weight. In this case, as in (232), correction for incipient “Wrinkling” is not required. 2.6 2.6 P = = : (249.) “Incipient Wrinkling.”—It is shown in (306) that thin wrought-iron plates, where the breadth is considerable, will fail by wrinkling or corrugating under a compressive load, with a strain very much less than the absolute crushing strength of wrought iron. For plates forming the sides of a pillar, we have the rule : (250) Ww = (NE: 10) x 80. In which t = the thickness of plate, and b = the breadth unsupported, both in inches, Ww = the compressive wrinkling strain in tons per square inch. Thus for a plate } inch thick, and 9 inches wide, forming one side of a pillar 9 inches square, we have (11=79) 80, or (-3535 = 3) x 80 = 9.427 tons per square inch, being less than half the crushing strength of wrought iron in pillars (201), which is 19 tons per square inch. When, however, the plate is thick in proportion to the breadth, the wrinkling strain may become greater than the crushing strength, and in that case the strength of the pillar is governed by the latter. An example of this is given in (409). Table 63 gives the wrinkling strain for wrought-iron plates of various thicknesses and breadth when forming part of a pillar particularly, for, as shown in (321) and Table 62, the resistance of plate-iron in beams is greater than in pillars in the ratio of 104 to 80. (251.) It should be observed that the strength of a pillar is governed by that of the weakest plate :—for instance, in a rectangular pillar whose sides are in the ratio of 2 to 1, with equal thickness all over, the narrow end plates are stronger than the sides in the ratio of N 2 to w1, or 1.414 to 1.0, but when the wide plate fails by wrinkling under the strain, the whole of the load is thrown upon the end plates, and they fail under that increased load in spite of their superior strength, which in such a case goes for nothing. Evidently, the most judicious course is to make the wide plates thicker than the narrow ones, so as to produce equality of strength all over; and as the resistance to wrinkling is proportioned to vi: b by the rule in (250), it follows that the thickness of plate should be simply propor = tionate to the breadth unsupported, so that for breadths in the ratio of 1, 2, 3, the thicknesses should be in the ratio 1, 2, 3 also. When it is thus found that the Wrinkling strain is less than the crushing, the correction of F must be made by the rule : (252.) P = F x Cw (F + { Ow). But when the crushing strain is less than the wrinkling, the rule becomes : (253.) Po = F x Cp : (F + Cp). In which F = the breaking weight of the pillar due to flexure by the rule in (248), &c.; C = the specific crushing strength of wrought iron per square inch, namely 19 tons, or 42,560 lbs.; Cp = resistance of the pillar to crushing, due to the area of the section, and the value of C; Ww= the wrinkling strain in lbs., tons, &c., per square inch, which varies with the thickness and breadth of the plate, and may be calculated by the rule in (250) or (308). Cw = the wrinkling strain on the whole pillar, due ) to the area of the section, and the value of Ww; P. breaking weight of the pillar reduced for “ Incipient Crushing” in tons, lbs., &c., dependent on the terms of F and C; Py = the breaking weight of the pillar reduced for “Incipient Wrinkling” in tons, lbs., &c. Of course it will be understood that the whole must be taken in the same terms; for instance, if F be taken in Tons, all the rest must be in Tons also. (254.) “ Experimental Results." -Table 55 gives the results of 29 experiments by Mr. Hodgkinson on square and rectangular pillars of thin wrought-iron plate; col. 9 gives the calculated breaking weights by flexure, or F; the square pillars by the rules in (230), and the rectangular ones by those in (248). (255.) " Square Pillars."—As an example of the former we may take No. 4, whose section is shown by Fig. 50; to find 93:6, we have the logarithm of 8.1 = 0.908485 x 3•6 = 3.270546, the natural number due to which 1864. Then for 9868, the logarithm of 7.98 = 0.902003 x 3.6 = 3• 247211, the natural = the = |