Reduced. TABLE 37.-Of the STRENGTH of SOLID and HOLLOW PILLARS Length of the 8 Breaking Weight, Outside. Inside. By Flexure. Reduced. By Flexure. Reduced. By Flexure. of CAST IRON, the ends being Flat and well supported by Iron cast on both ends. 55.0 55.0 33.6 33.6 33.8 33.8 ::: ::: ::: : ::: 46.0 46.0 117 103 89.0 85.8 71.8 71.8, 264 212 202 178 162 153 ::: : :::::: ::::::::: :::::::::: : : : : ::: :::::::::: ::: 1932 823 1421 738 1087 659 871 593 711 532 593 478 2777 993 1660 885 1270 782 1018 709 831 633 693 568 2578 835 1892 783 1448 715 1160 651 948 598 792 547 3250 1106 2390 1014 1826 922 1464 843 1195 768 997 701 3841 1337 2825 1223 2160 1110 1730 1013 1412 923 1178 841 4357 1556 3205 1418 2448 1287 1963 1171 1602 1062 1337 969 : ::: ::: have a cast-iron pillar 6 inches diameter externally, and 5 inches internally, therefore inch thick, and 14 feet long, with both ends flat. x = From Table 34 the value of Mp = 44.19, say 44 tons; from Table 35, col. 4, the value of 636 = 633, and of 53-6 = 328; from Table 36 we obtain 88.8, say 89, for the value of Lor 14. Then the breaking weight by flexure by rule (157) becomes 44 × (633328)89 151 tons, which being due to flexure only, will require correction for incipient crushing as shown by (168). Table 37 gives the breaking weight of solid and hollow pillars of cast iron from 1 to 12 inches diameter, and from 5 to 20 feet long, calculated in the way we have illustrated, the result being there entered as due to flexure, which is corrected for incipient crushing in the next column when necessary. The breaking weight due to flexure is thus given separately in order to adapt the table to conditions other than those where the pillar is flat at both ends: thus, the pillar which we have found to have a strength of 151 tons when both ends were flat, would bear only 151 ÷ 3 = 50 tons with both ends rounded, and 50 x 2 = 100 tons when one end is flat, and the other rounded, &c., correction being made for incipient crushing in all cases where necessary (163). (161.) Table 38 gives a selection of all the more important experiments of Mr. Hodgkinson on solid and hollow pillars of cast iron, and in order to show the correctness of the rules in (151), col. 9 has been calculated by them, the value of Mp being taken from Table 34. In col. 7 these results are corrected where necessary for incipient crushing by the method explained in (163), the value of C, or the crushing strain being taken at 49 tons, or 109,760 lbs. per square inch, this being the strength of the particular iron used by Mr. Hodgkinson, as found by him from direct experiment. The mean crushing strength of British cast-iron is 43 tons, as shown in (132), and this value should be used in ordinary cases. In col. 8 we have given the error or difference between the calculated and experimental results:-the sum of all the + errors is 163, and of all the errors, 141·7; hence we have as a general average result of the forty experiments (163 — 141·7) 40 = +0.532, or per cent. only. It will also be observed that the range of the error is nearly equal, the greatest + error being 19.8, and the greatest - error is 22.1 per cent. (959). = (162.) As an example of the application of the rules in (157) to cases where exact results are required, as in Table 38, we will take No. 35, in which D 2.01 inches, d 1.415 inch, and L 7.395 feet. We require logarithms for working these rules; then for D36 we have, log. of 2.01 = 0.303196 x 3.6 = 1.091505, the natural number due to which, or 12 35, is the 3.6 power of 2.01. Similarly, for ds, we have log. of 1.415 0.150756 x 3.6 542721, the natural number due to which, or 3.49, is the 3-6 power of 1·415. For the 1.7 power of L, we have log. of 7.395 0.868938 x 1·7 = 1.477194, the natural number due to which is 30. The value of M, from Table 34, for pillars with both ends flat, as in our case, is 99,000 with these data, the rule in (157) gives 99,000 × (12·35 – 3·49) ÷ 30 = 29230 lbs. as in cols. 9 and 7; cor rection for incipient crushing not being necessary in this case. = · = INCIPIENT CRUSHING. (163.) If we calculate the strength of a series of pillars with a progressively diminishing length, the calculated strain increases as the length is reduced until it eventually becomes greater than the absolute crushing strength of the material. Obviously, the pillar cannot sustain a load greater than the crushing strain due to the area of the section:-there is therefore a limit to the shortness of pillars, beyond which the rules in (151) do not apply without correction. It might be supposed, that down to a certain length, the pillar would break simply by flexure with the strain given by the rules, and that with any length less than that, the breaking weight would be simply the crushing strain due to the area of the section and the specific strength of the material, irrespective, therefore, of any further reduction of length. But Mr. Hodgkinson found that long before that length was reached there was a falling off in the strength of long pillars, and he was led to the following |
