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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 STRAINS.
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 to 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 3 × W × l load, we have the Rule (513), or ƒ = For a bar
2 x D x B
1 inch square and 1 foot, rule evidently becomes f=
or 12 inches, between supports, the
from which we have the simple law for a bar of those sizes, &c., that the maximum longitudinal strain is 18 times the transverse load: hence we have the Rule:
We may now apply this Rule to Timber whose tensile and crushing strength is known by direct experiment: for the former we shall take Mr. Barlow's results in Table 1, and for the latter, Mr. Hodgkinson's in Table 32: the Transverse strengths will be given by Table 65.
(640.) A bar of ash 1 inch square and 1 foot long, breaks with a load of 681 lbs. in the centre, the maximum longitudinal strain ƒ = 681 × 18 12258 lbs. per square inch; by direct experiments, the tensile strength T = 17077 lbs.; and the crushing strength C = 9023 lbs. per square inch: the mean of the two is 13,050 lbs., or nearly as given by the Rule.
English oak breaks transversely with 509 lbs. ; hence ƒ = 509 x 18 9162 lbs. per square inch: by direct experiments 10389 lbs., and C = 8271 lbs. per square inch; the mean of the two = 9330 lbs.
Beech breaks transversely with 558 lbs.: hence ƒ= 558 × 18 = 10044 lbs. per square inch; by direct experiments T = 11467 lbs., and C 8548 lbs.; the mean being 10,007 lbs. per square inch.
Teak breaks transversely with 724 lbs.: hence ƒ = 724 × 18 = 13032 lbs. per square inch; by direct experiments T 15090 and C = 10706, the mean being 12,898 lbs. per square inch, &c. The application of the rule to cast and wrought iron is given in (520).
(641.) Having found the value of f, or the maximum strain at the upper and lower edges of the section, we have now to find the extension and compression there from the deflection of the beam with a given load in the centre, for which we have the Rule:
in which D depth of rectangular bar in inches, 7 = length between supports in inches; 8 = deflection at the centre in inches and Ex : = extension in parts of the length: it will be observed that the central load, and the breadth have nothing to do with the question in this case. The rule also supposes that the compression C is equal to Ez, which perhaps is nearly true for light strains (617) with most materials.
(643.) For a bar 1 inch square, and 12 inches long, l÷2=6, and the rule becomes Ex
3 × 1 × 8 3 x 8
2 × 62
th of the
is to say, the maximum extension and compression is transverse deflection.
Table 105 gives in col. 4 the mean deflection of bars of many materials, 1 inch square, 1 foot or 12 inches long, by 1 lb. in the centre, which by (638) is equivalent to 18 lbs. longitudinally. Thus with cast iron the rule becomes 00002886 ÷ 24 =
⚫0000012 for 18 lbs., therefore 000001218·0000000667 for 1 lb., or 0000000667 × 2240·0001494 per ton, the length strained being 1.0.
The rule giving in this case the extension for 18 lbs. per
For a standard bar 1 inch square, and 1 foot long, we have the Rules:
of Table 105; hence, Rule (645) becomes 00002886 × 5.18. =0001494 as before (643).
(646.) Calculating in this way we have obtained col. 8 of Table 105. Comparing these results for cast and wrought iron, with those obtained by direct and exact experiment in Table 91, we have for cast iron 0001494 by Table 105, while Table 91 gives for extension 00016556, and for compression ⚫0001695 parts of the length, by 1 ton. For wrought iron, Table 105 gives ⚫00008106, while Table 91 gives for extension 000080121, and for compression, ·00010138 parts of the length, by 1 ton. These results agree fairly well together; hence we may have confidence in the method we have followed, and by which col. 8 of Table 105 has been obtained.
We may now illustrate the application of these rules to practice: say we have a wrought-iron bar 50 feet, or 600 inches long, with a tensile strain of 8 tons per inch: then the extension=00008106 × 600 × 8 = 0·389 inch. Direct experiment gives by col. 5 of Table 91, 000638774 × 600 =
Again, a bar of Riga fir, with say 2 tons per inch, and a length of 30 feet, or 360 inches, will stretch 00197 × 360 × 2 = 1.42 inch.
Again, a pillar of English oak 12 feet or 144 inches long, with a compressive strain of ton per square inch will be shortened by the pressure 002042 × 144 × 3 = 0.22 inch : this being the mean, col. 8.
ON THE DEFLECTION OF BEAMS.
(647.) "Form of Curve of Flexure."-With a parallel beam, or one having uniform depth and breadth throughout the length, resting on bearings at each end, and loaded with a central weight, the strain on the material is a maximum at the centre, and is progressively reduced toward the ends, where it becomes