which needs no counterbrace; and we can then be sure that none of the outer bays will need it either. The foregoing of course only applies to the case in which all the W's are equal, and the original bracing as in Fig. 248; for other cases the work should be gone through as we did in the first instance. The effect of the passage of a long distributed load, as a train, is rather more complex, and we must refer for it to the larger treatise. EXAMPLES. 1. A lattice girder 20 feet span, 3 feet deep, in four divisions, is loaded at the top of the left hand quarter span with 5 tons, and supported at the ends of the bottom boom. The braces are stout flat bars. Find the stresses in the bars. Ans. Counting from the left-uprights, – 1, 2, 0, 0, - § tons; diagonals, ± 3.64, ± 1.21, ± 1.21, ± 1.21; top boom, 3.12, 5.21, 3.12, 1.04 compression; bottom boom, same as top, but tension. 2. How would the preceding results be modified if the braces were slighter, so that the compression brace only took half as much load as the tension brace? Ans. Uprights, - 21, - 31, 0, 0, - §; diagonals, +4.84, +1.61, +1.61, +1.61, − 2.42, - .81, - .81,-.81; top boom, -4, -5%, - 337, -1; bottom boom, +212, +438, +255, +38. 25 3. Find the stresses produced when a bridge platform, which rests on the bottom booms of two girders, similar to that of (1), is loaded with ton per foot run. Ans. Taking the bars in the same order as (1), − 18, + §, 0, +18; 1.82, ±1.21, ±1.21, ±1.82; 1.56, 3.66, 3.66, 1.56 compression, and the same tension for bottom boom. 4. Find the greatest stress in each bar of the bridge of question (3) when a travelling load of 10 tons crosses it. Ans. - 218, +3},±§,+31, − 218; ±5.46,±6.1,±6.1,±5.46; F5.2, II.5, 11.5, 5.2, for top and bottom booms respectively as before. 5. What would be the effect on the results of the preceding questions, if the trusses were made 4 feet deep instead of 3 ft. ? Ans. The stresses in the bars of the boom would be of those found before. In the diagonals the stresses would be reduced in the ratio of 59: 64 nearly. 6. An N girder in 10 divisions is loaded with 1 ton at each joint. A load of 16 tons travels along it. Find which bays will require a counterbrace. Ans. The middle six. 7. In the preceding find the stresses in all the bars, assuming the counterbraces equally efficient against tension and compression when the load is at the centre. CHAPTER XVIII BENDING OF SOLID BEAMS WE have seen that in a girder the bending is resisted by the booms, and the shearing by the diagonals, and this justifies our original separation of the effect of a load into these two actions, since they are resisted by distinct pieces, and H is independent of F while S is independent of M. We can from the preceding obtain a preliminary idea of how one kind at least of solid beam resists bending, for the above conclusions are true, no matter how long or short the bays may be; thus they are equally true for girders either as (a) or (b), Fig. 250. And they would still hold even if we reduced the lengths of the bays so much that the uprights and diagonals actually came into contact with each other. This being so, we have only now to change simple contact into connection to obtain a continuous solid beam consisting of two booms or flanges, united by a solid sheet of metal called the web, as Fig. 251. If then we assume that the connection of the bars where M now will be the moment at the section at which we wish to find the value of H. Since the boom is now supposed to be made up of an indefinitely large number of short rods, for each of which the value of H is different, there will be a continuous variation of H from point to point, and the opposite joint is now in the same plane as its indefinitely short bar, hence M is taken at the section as above stated. In a girder as Fig. 251 the top flange would be compressed and the bottom extended; and at any given section KK we should have the forces H, H, and H, H as drawn. These forces we represent by arrows drawn in the opposite directions to those we used in the framework girders, because there we put the arrows on the bar and they represented the action of the bar on the rest of the structure; here the bar is indefinitely short, and we cannot put arrows on it, so we put them as shown; hence they represent actions exerted on the bar at K, or since this is simply a point, they represent the mutual actions of the two parts of the flange on each other (compare chap. xvi. page 313). We then have Hh=MK, H being the total stress at K in either flange, so that if A sectional area of top flange, bottom flange. Intensity of stress is, on top flange, H/A,; on bottom flange, H/A. Then, by our assumption, the shear FK is distributed over the sectional area of the web only, and hence, assuming it to be uniformly distributed, Intensity of shear in web= FK where Aw sectional area of web. The assumptions here made give results for a beam of I section, useful for rough calculation; but we can see at once that they will not be even approximate for all beams. For example, some beams have no flanges at all, yet we know they can resist bending, while the preceding formula would give them no resistance to bending. We must then proceed to a closer investigation for solid beams, and as usual we will consider one action only at a time. We will commence then with bending, and investigate the stresses produced in a solid beam when subjected to bending only, or Pure Bending.—If a beam be subject to pure bending, the shear at every point must be zero, and we must first then see what kind of loading is necessary to produce this result. symmetrically at A and B. Then the support ing forces at A and B will be each W. Fig. 252. Taking now any point K between A and B we have FK W-W= 0, |