supposing, however, that the wear, by the action of the wheels amount to one tenth of a pound, per yard, per annum; if the top, or bearing part, of the rail, be made an inch, in depth, it will be sufficient, for all the purposes required. Any increased depth, and weight, which would not be required, for above eighty years; would, at compound interest, at the end of that period, amount to a greater sum, than it would be expedient to expend, for such a purpose, considering the remote period, at which it becomes useful. Having, therefore, determined the area of section of the head, or wearing surface, of the rail; it then becomes a question, in what manner, the other part of the material is to be distributed, so as to present the strongest form of section. It is necessary, however, to have, a projecting bead, or flanch, at the base of the rail, for the purpose of securing it to the chair. In cast-iron rails, this may be done; near the ends of the rail, leaving it open, to mould the other part of the rail into that form, which presents the strongest form of section. In wroughtiron rails, it is different, whatever breadth of base is required, to secure the rail to the chair, must, from the mode of manufacture, be given, throughout its whole length; it, therefore, becomes of paramount importance, to determine what width of base, can be given, without impairing the strength of the rail. Professor Barlow, in his "Second Report, to the London and Birmingham Railway Company," has given a solution of this question (see note B. Appendix), by which it is shewn, that a maximum strength -e2 a is obtained, when " (a+eb), x2 = 46 -; a being "the whole, of the sectional area, below the neutral "axis; b, the breadth of the middle rib, pq; and 66 e, the depth of the lower flanch; a being any variable depth of the rail. "From this, x may be determined, for any given "values of a b, and e. Thus, let A B C D represent "the section of a rail; ABEF A being the head, and GCHD the "lower rib, or flanch. Suppose the "middle rib is 78 inch, or b = 78, E "to find what lower flanch must be given, and the corresponding depth of rail, to produce the maximum strength. 66 66 G C P B H m m עי “The rail, being four and a "half inches below the neutral "axis nn, and its breadth pq 78, "its area is 78 x 43.51-a, and it "is required to distribute this area, "so as to produce a rail of maximum strength, the depth of the proposed flanch being 1 inch. Substituting a 3.51, b=78, ce=1, the foregoing equa"tion becomes a3-4·11 x2-1.12; whence, x=404, "the depth of the rail required. 66 66 Now, 4.04 × 78=3·15, area middle rib; a-b x= “3·51—3·15=•36=b′; or, ·36, the area of the lower flanch, which is, also, its breadth, its depth being 1. 66 "The strongest rail, therefore, of this weight, whose "breadth is 78, is that whose depth is 4.04 inches, "and the breadth of the lower flanch, including the "middle rib, is 78+36=1·14 inch.”—(Barlow's Second Report, p. 95.) In this solution, the strength, or resistance, of the head, which is very little, has been neglected; the resistance of the tensile portion of the rail, from the neutral axis, being taken; and this, it must be observed, is, perhaps, the most correct way of calculating; inasmuch as we should take the strength of the rail, when the head, or upper portion, is nearly worn down, and not when the rail is first put into use. It is true, as the head wears down, the neutral axis is changed; but still it is better, that the solution should shew the strength below, rather than above, the correct result. Comparing this result, with some of the rails now in use, viz., Figs. 3. 12, 13. Plate III., it will be seen, that most of them exhibit, too great a quantity of material, in the base of the rail. In a general way, suppose the depth of the head, or bearing part, of the rail is equal to one inch, and the neutral axis, nn, half an inch from the upper surface; we can, from the preceding formula, calculate that form of section of the base of the rail, which exhibits a maximum strength. § 3.-Rigidity of different Kinds of Rails. Having the best form of section, we then come to the degree of strength, or rigidity required; the two requisites are, that the rail present such a depth of wearing surface, as may be sufficient for all the economical purposes of durability; and that the deflection be such as not to present any resistance, to the carriages passing along them. We have already determined the requisite section of the head, or upper table, of the rail, the formula, page 70, determines that form of the remaining part of the rail, which presents the strongest section; we shall now, therefore, give some experiments, made with a view of ascertaining the strength, and stiffness of different sections of rails; and that amount of deflection, which presented an obstacle, to the wheels of the car. riages. We cannot, however, give experiments, upon all the forms of sections, required for the different rail ways; and, therefore, it becomes necessary, having first ascertained, by experiment, the degree of rigidity of a given form of section, to give formulæ, for calculating the rigidity of other rails, of different forms of sections. The law of resistance, of bars of cast and malleable iron, subjected to pressure, has been so well illustrated by Tredgold, in his work on cast iron; and by Professor Barlow in his report already alluded to; and in his recent publication on the strength of materials; that, without entering into all the details of their calculations and experiments, on the subject, we shall here give only their formulæ, for the calculation of the strength, and rigidity of the various forms of railway bars, referring the reader to those works, for complete information on this important subject. Mr. Tredgold, having found, from experiment, that the tensile, or cohesive power of cast iron, was equal to 15,300 per square inch, gives the following formula for calculating the strength of any section of rail, with a given weight of carriage. Let the marginal figure, page 70, represent a section of rail, and w=the utmost strain, to which the rail is subjected, in lbs. l=the distance between the supports, in feet. b=the extreme breadth of section A B C D, in inches. d the extreme depth of section A B C D. b the difference between the breadth in the middle, and the extreme breadth. Ρ d the depth of the narrow part, or rib p q. Then wl 850 = Iron, art. 148.) b d2(1—q p3).—(See Tredgold on Cast Consequently d=√(1—9p3), and 850 850bd2 W= (1-qp3). This calculation is on the supposition that the thickness of the head, middle rib, and lower flanch are all the same; but as these are varied in practice, we shall therefore also give Professor Barlow's solution, which includes every possible shape; and being, therefore, more particularly adapted to malleable-iron rails, will be more generally applicable, for this description of rails. Having previously ascertained, by experiment, that the ratio of the resistance of wrought iron, to tension and compression, was as 4: 1. Let A B C D, page 70, represent the section of a rail, nn the neutral axis. c=the centre of compression, c n being two thirds of hn, and the point m which is in the centre of rs. The breadths, n n and m m, are also known. t=the tension of iron per square inch, just within the limits of elasticity. Then the resistance of the whole section, referred to the centre of compression, c, may be considered to be made up of the three resistances. 1. Of the middle rib, continued through the head and foot tables, v tzw hs.ns.pq. t. 2. Of the head A EF B, minus the breadth of the centre ribh x.nx. n x ns t. 3. Of the lower web, G C DH, also minus the con tinuation of the centre rib-nm.rs. (mm-pq) (See Barlow, p. 58.) [Note C. Appendix.] t. These three resistances being computed, let their sum be called s, and the clear bearing 1; then 4 s =W, the load the bar ought to sustain, at its middle point, for an indefinite time, without injury to its elasticity. |