Experiments on the tranverse strength and other properties of malleable iron: with reference to its uses for railway bars

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B. Fellowes, 1835 - Iron - 97 pages
 

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Page 28 - ... 7. New bar, by Messrs. Gordon 10 tons. We may consider, therefore, that the elastic power of good iron is equal to about ten tons per inch, and that this force varies from ten to eight tons in indifferent and bad iron. It appears, also, (considering -000096 as representing in round numbers j^-0th) that a bar of iron is extended one ten-thousandth part of its length...
Page 17 - As far as relates to ultimate strength, there can be no doubt Mr. Stephenson's rail is equal to that of an elliptic rail, and, consequently, to that of a rectangular rail of the same depth; but there is still an important defect in all elliptical bars, viz: that although this form gives a uniform strength throughout, it is by no means so stiff as a rectangular bar of a uniform depth, equal to that of the middle of the curved bar, and it is the stiffness rather than the strength that is of importance;...
Page 57 - B : which is the whole effect. " For those who understand the integral calculus, this solution is sufficient; but as the article will probably be consulted principally by practical men, it will be more convenient to give a specific solution for a rail, embracing under one general figure all the usual forms, the only variations being in the depth, breadth, and thickness of the parts.
Page 28 - We may consider, therefore, that the elastic power of good medium iron is equal to about ten tons per inch, and that this force varies from ten to eight tons in indifferent and bad iron. It appears, also, (considering -000096 as representing in round numbers iooootn,) that a bar of iron is extended one ten-thousandth part of its length...
Page 40 - ... tons,) had their restoring power just preserved with a transverse strain of two and a half tons on a bearing length of thirty-three inches. Hence, in the formula: 4dat we have / = 33, w = 2%, d = 2, a = 2, / = 9.5, and d' = 1.62 inches, depth of tension. Consequently, d...
Page 15 - BI — v (r2+d2— 2rd cos x). And by this formula the ordinates of the curves have been computed for two different fish-bellied rails ; the extreme depth in both being five inches, but the lesser depth in one three inches, and in the other three and three-quarter inches, the latter being that proposed by Mr. Stephenson for the London and Birmingham Railway. The ordinates are taken for each 10°, or for every inch of the half-length, and in the last column are given the ordinates of the true ellipse.
Page 46 - ... the deflection. That is, all rectangular bars having the same bearing length, and loaded in their centre to the full extent of their elastic power, will be so deflected, that their deflection (8) being multiplied by their depth (d) the product will be a constant quantity, whatever may be their breadths or other dimensions, provided their lengths are the same. "Let us see how nearly our several results agree with this condition. "In the several bars, Nos. 8, 9, 10, 11, 12, 13, multiplying the...
Page 57 - ... which is precisely the expression for the centre of oscillation of a disc of the same figure. " We have hence the following general rule for finding the resistance or the weight which any given bar or rail will support at its middle point, within the limits of its elastic power, that is, Calling the integral of formula...
Page 68 - Stevenson, by a judicious and scientific distribution of the metal, has avoided, and no doubt such fractures would be with his rail less common; but the objection I offer above applies not merely to the fish-bellied rail, but to the truly elliptical form itself, if it were possible to arrive at it. Thirdly. — The parallel rail is the best, because it enables the engineer to keep the blocks and chairs of the two rails directly opposite to each other, so that the wheels of the carriage shall pass...
Page 63 - Then the former product, divided by the latter, will be the resistance in tons due to the head, not including the continuation of the middle rib. Resistance of the Centre Rib. Multiply the whole depth of the rail by the whole depth, minus half an inch, and that product by 10 times the thickness of the rib; and the last product divided by 3, will be the resistance in tons of the middle rib continued through the whole depth, ie, through the upper and lower tables.

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