гра of tension. do zy dy = the direct resistance to compression, z dy is the total area which is subjected to tension, we = z dy the resistance to longitudinal shearing on the com EXAMPLE.-Let the beam be rectangular; b the breadth, and d the or the neutral axis coincides with the centre of the sections. If the horizontal plane which is passed through the middle of the depth, divides the beam into two equal and symmetrical parts, then will the neutral axis lie in this plane, for the longitudinal shearing will be the same in the two parts, and hence the second term of (99) will be zero, and the equation reduces to equation (87), which shows that the neutral axis passes through the centre of the sections, which, in this case, is at half the depth. Hence the neutral axis passes through the centre of beams having circular, elliptical, or regular polyginal sections, double T or H sections with equal flanges, and tubular girders in which the hollow is similar to the full section. But it does not pass through the centre of triangular or single T beams, or other non-symmetrical forms. EXAMPLE. To further illustrate the use of equation (99), I will apply it to a single T, Fig. 35. Using the notation as given in the figure, and equation (99) will give for the part above the neutral axis To find it for the lower part we suppose that the vertical web passes through the flanges to thebot tom; hence we have for the lower part of the web, is b(d—y)2+q b(d—y) FIC 35 d' According to the 4th principle, as stated in the preceding article, page 310, the longitudinal shearing stress in the flanges is The width of both flanges is b'-b. Hence, for the sum of all the forces in the flanges we readily find d' d-y 38(b'—b) [ (d—y)'—(d'—y)3]+ ¿2′ ̧‚o [(d—y)—(d"'—y)](b'—6), or 38 (b'—b) (d'd+d'd"—2d'y) + (b'—b). d/2 Day d-y · (102). Substitute (100), (101), and (102) in the first of (61), and it gives 38 by2+ by—18 b(d—y)2—q b(d—y)—žs (b'—b)(d'd+d'd"—2d'y)— 2. Let the moduli of elasticity be unequal. In this case, the direct strain, y 8, brings into play the elastic resistance; hence, for this strain. we proceed as in the preceding theory; but the expression for the longitudinal shearing will be exactly of the same form as in the pre 3. To find the position of the neutral axis so as to give a minimum strength. SI d1 This is equivalent to making the second member of equation (84), or +SS y dy dx, a minimum. As I have not succeeded in reduc ing this to a more definite form, dependent upon the form of section, I am obliged to reduce it for each special case. To show how to treat it, I will apply it to rectangular beams. As before, let b be the I= breadth, and d the depth, and 1=ffy dy da. Then we have (d3—3d3y+3dy3)+oo ( y2+(d—y)2) = minimum, (103) from which y may be found. If ys, as it does very nearly for cast iron beams having solid sections, we find y=0.54958d, or 0.04958d from the centre of the section. Comparing this result with that found under the former hypothesis, (see Example 1, which follows equation (94),) and we see that the neutral axis is nearer the centre of the beam at the instant of rupture, according to this hypothesis, than according to the former. If be zero, equation (103) will give the same result as (94). If y be very large compared with s, or s=0, equation (103) gives y=d; hence, we infer that all positive values of makes the neutral axis nearer the centre of the beam than if it were zero. The same general principles are applicable to beams of any form. B. Suppose that the deflecting forces are inclined to the axis of the beam. a. Adopt the common theory, and let the moduli be unequal. Let Ф be the angle which the direction of force makes with the axis The similarity of the triangles o L N and Nek, Fig. 31, and equation (89) The exact reduction of (106) is somewhat difficult, but as we have before observed Et nearly equals E, and practically may be considered exactly equal, and as this hypothesis simplifies the equation, we 2. Suppose that the moduli are equal. Then equation (106) becomes, Ythe distance from the neutral axis (the origin of co-ordinates) to the centre of gravity of the sections. Then, from mechanics we have dt de =SS,„, y dy dx=ƒƒy dy dx-ffy dy dx; KY= hence, (107) becomes ρ Since and are variable when the beam is curved, we see that the neutral axis is not parallel to the axis of the beam. If = 90°; Y=0. If 0=0; p is infinite and =x. 22°. Let the direction of the force coincide with the axis of the bar. We shall then have ẞ=7=90°; ∞ = 0° or 180° according as the bar resists tension or compression; b=c=90°, a = 180° or 0° according as the force is tensive or compressive; and equations (4) and (5), vol. xlviii, page 231, gives 2 ΣΕΞ Σ Ρ. The resistance F may be expressed in terms of the elastic resistances, or the ultimate resistance. For the former we have the well known expression Σ F = E K ; in which E is the modulus of elasticity, K the section, 2 the elongation, and i the length. For the latter F = s K or C K, in which K is the transverse section, and s the modulus of strength, and c the modulus of resistance to crushing. We have thus come out at the equation which forms the starting point of most writers on the resistance of materials. My aim has not been to exhaust any part of the subject, but rather to show the general relation and dependence of the equations used in computing the strains on the several parts of a bridge, and the strains on the fibres of beams, by deducing all of them from the fundamental equations of statics. In some cases I have reduced the equations so as to make them applicable to special cases, while in others they would require consid erable transformation. Of the former I might mention the parabolic arched truss; and of the latter, normally pressed arcs and some conditions for maximum shearing; but inasmuch as they are all statical problems, the equations which are applicable to them must be deducible from the equations which I have given. I might also have considered some problems in which we should determine the distribution. of the parts so as to fulfil certain conditions which might be assigned to the strains. Such problems may properly form independent arti- ^ cles. Some errors have crept into the work, most of which will be readily detected by the reader. I have noticed the following: ERRATA. Vol. xlviii, page 232, line 6, for ΣP(cos b-y cos a) read ΣP(x cos b―y cos a). On the Size of Pins for connecting Flat Links in the Chains of Suspension Bridges. By Sir CHARLES FOX. From the London Artizan, May, 1865. In the construction of chains of this kind, it is of the highest importance that the pins which pass through and connect together the links of which the chains are composed should be of the right size, inasmuch as their being too small, as compared with the links through which they pass, renders ineffective a portion of the iron contained in the latter, which then exists only as a useless load to be carried by such links; while, at the same time, if the pins and heads of the links be too large, they become uselessly cumbersome and expensive. Čareful examination and experiments made upon a large scale (which will be explained hereafter) have brought out facts by which a simple rule has been arrived at a rule that may safely be adopted as a guide in deciding upon the relative sizes of these two parts. On this rule mainly depends the economical use of iron in the construction of such chains. In this paper the term chains for suspension bridges implies such as are usually employed, and are composed of several flat bars of equal thickness throughout, placed side by side, but having their ends swelled edgeways, so as to form what are technically termed heads, and which are coupled together by pins passing through holes in such heads, as shown in Figs. 5 and 6 of the accompanying drawing. In deciding upon the size of the pins it has often been assumed as a close approximation, that, as about the same force is required for |