1.7 ratio for beams of cast iron is 1.5 to 1.0 (361), and we might suppose that the same ratio would apply to pillars also, but the experiments we have seem to show that the theoretical ratio is more correct. Admitting the theoretical ratio 1.7 to 1.0 we obtain for square and rectangular pillars the modified values of Mp in Table 34. Putting S for the side of the square and the rest as in (151), we have the following general rules for the resistance of square pillars of cast iron to flexure, irrespective of incipient crushing (163). For solid square pillars:(171.) F = Mp x 536 : L'? (172.) S = 3/(F L1? = Mp). (173.) L = '7 (Mp x 536 = F). In which S = the side of the square pillar, and the rest as in (155). For hollow pillars these rules become (174.) F = Mp (936 – 886)=L'. (175.) (936 – 9306) = F x L'7; Mp. (176.) L = ") {Mpx (836 – 394) = F}. In which S = the side of the square externally, s = the side of the square internally, and the rest as before, in (155). (177.) “ Rectangular Pillars of Cast Iron.”-A rectangular pillar, other than square, will fail by bending in the direction of its least dimension, and in that case, the strength will be simply proportional to the breadth or greatest dimension. For long pillars failing simply by flexure, the rules become :(178.) F = Mp X 126 x 6 : L'?. (179.) t = 2) {F x L'"=(Mp * b}. (180.) b = F x L: (Mp X t?). (181.) L = 'V (Mpx 2*6 x 6 = F). In which t = the thickness or least dimension of a rectangular pillar. b =- the breadth or greatest dimension of a rectangular pillar. P P Х And the rest as in (155). The value of 12*8 may be found from col. 3 of Table 35; of L17 from Table 36; and of Mp from Table 34. For hollow rectangular pillars the rules are modified, and become : (182.) F = Me * {f** * b) – (*,** x bo} = L*?. ) 1 (184.) As an example of the application of the rules in (177) say we take a pillar 1} x 7 inches, 14 feet long, with both ends flat: the value of Mp from Table 34 is 75 tons; of 1126 from Table 35 is 2.87, and of 1417 from Table 36 is 88:8. Then F = 75 x 2.87 x 7 = 88.8 = 16.96 tons = - F. This will not require correction for incipient crushing, for the area being 1.5 x 7 = 10:5 square inches, Cp becomes 10.5 x 19 = 199.5 tons, and Cp = 49.875 tons. F being less than that, the correction is not required. Again, say we require the thickness t, for a pillar 9 inches wide, 18 feet long, both ends flat, and 15 tons breaking weight. 6 = = Then Mp being as before = 75, and 1817 = 131.1 by Table 36, we have 15 x 131•1(75 x 9) = 2.913, the nearest number to which in col. 3 of Table 35 is 2.87, which is opposite 1.} inch, the thickness required, &c. (185.) “ Cast-iron Pillars of + Section." - This form of pillar is commonly used for the connecting-rods of large steam-engines, and not unfrequently for carrying the floors of warehouses. The strength of such pillars may be found by a modification of the rules for rectangular pillars in (177). Let Fig. 38 be the section of such a pillar 8 x 8 inches, { inch thick, 15 feet long, with both ends flat; assuming that the pillar will fail by flexure in the direction of the arrow B, we have virtually two pillars, one a, a, forced to bend in the direction of its larger dimension, flexure in the contrary direction being prevented by the ribs c,c. We have in that case t 8 inches, and b = 1 inch; the other pillar c, c, will bend in the normal direction, or that of its smaller dimension, t being inch, and b = 8 -3 = 7 inches. (186.) By col. 3 of Table 35, 82.6 = 223, and 726 •707; by Table 36, 1517 = 99.8, and by Table 34, Mp = 75. Then for a, a, we have 223 xj 195, and for c, c, .707 x 7.125. = 5:—the sum of the two is 195 + 5 = 200, and it will be observed that c, c, adds only 2 per cent. to the strength, being 5 on a total of 200 (243). Having thus found the combined value of 126 x b 200, the rule in (178) becomes F = 75 x 200 • 99.8 150 tons, the breaking weight by flexure. In order to ascertain whether correction for incipient crushing (163) is necessary, we find the area of the section to be 15.23 square inches, and the crushing strength of cast iron being 43 tons per square inch (132), Cp becomes 15.23 x 43 = 655 tons, hence } Cp 164 tons; F, or 150 tons, being less than 1 Cp, the correction for incipient crushing is not necessary in this case (169), the correct breaking weight of the whole pillar is that due by flexure simply, or 150 tons. (187.) We found in (186) that the cross ribs c, c, contributed only 2 per cent. to the strength of the pillar Fig. 38, hence there would be no appreciable error if we omit them altogether in calculating the strength, which then becomes simply that due = to the pillar a, a, forced to bend in the direction of its larger dimension. Table 40 has been calculated on that principle. This, however, is true for those cases only where the length of the pillar is so great that it fails by flexure only; with short pillars requiring correction for incipient crushing, the cross ribs c, c, yield their full share of resistance to the load. It should be observed, that the breaking weight of a rectangular pillar like a, a, breaking by flexure in the direction of its larger dimension, will be simply proportional to its thickness, other things being the same ; for instance, in Fig. 38, the breaking weight when inch thick being 150 tons, with finch thick, it would be 75 tons, and with 14 inch thick = 300 tons, &c. TABLE 40.–Of the STRENGTH of CAST-IRON PILLARS of + SECTION, the ends being Flat and well supported. 146 176 201 227 261 291 326 370 411 452 500 600 700 826 96+ 1101 243 270 309 313 378 210 257 293 9 10 x 10 592 543 651 760 889 1037 1185 710 829 956 1115 1275 422 506 591 715 834 953 332 398 465 578 674 771 12 x 12 14 17 square inches. (188.) “ Relative Strength.”—The fact that nearly half the — section of a + pillar goes for nothing, will prepare us to find that this form of section is very uneconomical as compared with a cylindrical pillar of the same diameter, and area of section. Thus a pillar 8 inches external and 65 inches internal diameter will have an area of 15.8 square inches, or practically the same as that of Fig. 38, which we found (186) to be 15.23 Then from Table 35, 88.6 = 1783, and 6986 = 904, and Mp being 44.19, the rule in (157) gives 44.19 X (1783-904) = 99.8 = 380 tons breaking weight by flexure. This, however, will require correction for incipient crushing by the rule (164); Cp 15.8 X 43 680 tons, hence Cp 510 tons, and the rule Po F x Cp = (F+Cp) becomes 380 x 680 = (380 + 510) = 290 tons, which is nearly double 150 tons, the breaking weight of a + pillar of the same weight of metal. This ratio, however, is not constant; with a great length, such that both pillars would fail by flexure simply, the ratio in our case would be as we have seen, 380 to 150, or about 21 to 1; with a length of 15 feet, 2 to 1 ; as the length is reduced the ratio approaches equality, and with very short pillars in which the strength is governed almost exclusively by resistance to crushing, the two kinds of pillar become practically equal. (189.) Mr. Hodgkinson made an experiment on a pillar of + section 3 x 3 x 48 inch thick, Fig. 36, the length being , 7.562 feet, and both ends pointed; the breaking weight was 17,578 lbs. Calculating as in (186), and taking Mp at 56,100 lbs. from Table 34, we have for a, a, 32*6 = 17.4 by Table 35, then to find L!?, we must use logarithms. The log. of 7.562 = 0.878637 x 1.7 1:4936829, the natural number due to which, or 31.17, is the 1.7 power of 7.562:-then 56100 x 17.4 x •48 - 31.17 = 15035 lbs. For c, c, we have to find • 482.8 ; the log. of .48 = 1.681241 x 2.6 = 7:1712266, the natural number due to which, or · 1483, is the 2:6 power of •48; then b = 3 - •48 = 2.52, and the rule becomes 56100 x 1483 x 2.52 - 31.17 = 673 lbs. The sum of the two, or 15035 + 673 = 15708 lbs., is the breaking load of the entire pillar, which is 10.6 per cent. less than 17,578 lbs., the experimental breaking weight. |