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Table 98.-Of the MEAN COMPRESSION of WROUGHT IRON by
CRUSHING STRAINS : from Direct Experiment.
TABLE 99.-Of the COMPARATIVE EXTENSION of Cast and WROUGHT
IRON under the same TENSILE STRAINS.
2.066 2:189 2.276 2.421 2.567 2.807 3:008
and as B must also be stretched to that same extent, the question is, What will be the tensile strain due to that extension with wrought iron ? By col. 4 of Table 96, the strain is some
where between 9 and 10 tons per square inch, and by interpolation we shall find the exact strain to be 9.555 tons; the weight W in Fig. 123 is therefore 4 + 9.555 13.555 tons. Instead, therefore, of the two bars A and B dividing the load equally between themselves, we find that it is very unequally divided : Table 100 has been calculated in this way; col. 4 shows that the ratio of the strains is variable, attaining a maximum with 4 tons per square
inch. TABLE 100.-Of the COMPARATIVE RESISTANCE of Cast and
WROUGHT IRON to the same amount of EXTENSION by TENSILE STRAINS.
(631.) We found in (617) that the working safe tensile strain on cast iron was about 2} tons, with which the extension was .000413; now with that extension the resistance of wrought iron would, by col. 5 of Table 91, be somewhere between 5 and 6 tons per square inch, and by interpolation we find the exact strain to be 5.31 tons, from which we find the weight W
2.33 + 5.31 = 7.64 tons.
(632.) We have so far supposed that the bars A, B, were of the same area for the sake of illustration ; obviously the areas might be adjusted so that the weight borne by the two bars would be equalized, but the strains per square inch could not be altered without a violation of correct principles. Thus, if the cross-sectional area of A were 1 square inch, then by making that of B 2.33 : 5.31 = .44 square inch, the weight borne by B would be the same as that borne by A, namely, .44 x 5.31
= 2.33 tons, and we thus obtain W = 4.66 tons, the strain per
the wrought-iron bar being still 5:31 tons.
(633.) It will be evident from this that with any ordinary combination of cast and wrought iron the full value of the working strength of the latter, namely, 8 tons, cannot be realised; for this and other reasons such combinations are not expedient, at least in cases where considerable strains have to be borne. With 8 tons per square inch on the wrought iron the extension by col. 4 of Table 96 would be .000638774, and with that extension the strain on the cast iron by col. 4 of Table 88 would be nearly 3.1 tons, or about half the breaking weight, which would not be safe.
(634.) "Comparative Compression of Cast and Wrought Iron.”— The relative ultimate cohesive strength of cast and wrought iron being 7.142 and 25.7 tons per square inch, we found in (629) that the strength of a combination of the two metals in resisting a tensile strain was governed potentially by that of the weak cast iron. But in resisting compressive strains this order is reversed, the ultimate strength of cast iron being 43 tons per square inch, while that of wrought iron does not practically exceed 13 or 14 tons (627) for even a temporary load, the metal yielding so much that it becomes valueless : for permanent loads the strain should not exceed 8 or 9 tons, as we have seen (624), while that for cast iron may be as much as 14 tons (618). Table 101 gives the compressions of cast and wrought iron under the same crushing strains.
(635.) In Fig. 124 let C be a cast-iron and D a wroughtiron pillar one square inch in area, so close to one another, &c., that they must of necessity be shortened equally by the weight W, and let the strain on C be 5 tons per square inch, with which by col. 4 of Table 89, the compression of the cast iron will be .000899 : with that compression the resistance of wrought iron by col. 2 of Table 98 would be between 9 and 10 tons per square inch; by interpolation we find the exact strain to be 9.226 tons per square inch, the weight W will therefore be 5 + 9.226 = 14.226 tons. Calculating in this way we obtain Table 102.
Beyond 14 tons in col. 3, the TABLE 101.-Of the COMPRESSION of Cast and WROUGHT IRON
under the same CRUSHING STRAINS.
resistance of the wrought iron has been taken as constant, which is practically true even with strains of a very temporary character (634), to which alone, in fact, this part of the table strictly applies. With strains not exceeding 5 tons on the castiron and 9.226 tons per square inch on the wrought-iron bars, the Table may be regarded as accurate for permanent loads.
(636.) It appears from this investigation that in combinations of cast and wrought iron under compressive strains, the working load on the cast iron must not exceed 5 tons per square inch, although the ordinary safe strain is 14 tons, and the ultimate, or crushing strain 43 tons per square inch. This shows clearly the inexpediency of such combinations in ordinary cases and for heavy loads.
(637.) “ Extension and Compression of Timber, &c."-We have unfortunately no direct experiments on the extension, &c., of most materials, and shall be obliged to refer to experiments on the transverse strength and stiffness, and to calculate the longitudinal strains and elasticities from the transverse ones. There
TABLE 102.-Of the COMPARATIVE RESISTANCE of Cast and
Wrought Iron to the same AMOUNT of COMPRESSION by CRUSHING
is considerable uncertainty in this method, which has also the further disadvantage of giving inseparably the result of the compressive and tensile strains combined, so that we cannot determine the value of either alone. It will, therefore, be well to see how far this method agrees with direct experiments on such materials as cast and wrought iron, whose strength and elasticity are known with certainty (600).
(638.) We require 1st from a known weight in the centre of a rectangular beam of given dimensions, to find the maximum longitudinal strains, or those at the upper and lower edges of the section. Obviously the strain varies from nothing at the neutral axis a maximum at the upper and lower edges (494). Then 2nd, from the observed deflection we have to determine the extension and compression produced by those longitudinal strains.
To find the maximum longitudinal strains from the transverse load, we have the Rule (513), or f =
3 x W x 1
For a bar 2 x D x B