TABLE 83.-For RooFs: SAFE TENSILE STRAIN on ROUND BARS: 7 Tons per Square Inch. only; and double-shear when supported on both sides. It is shown in (123) that the tensile and shearing strains are equal to one another; therefore Table 83 may be used for rivets and keys, as well as for tie-rods; for instance, a g-inch rivet gives 2.15 tons for single, and 4.3 tons for doubleshear, &c. (537.) "Practical Application."-We may now illustrate the application of the Figures and Tables to practice: say we have a truss like Fig. 184, 60 feet span, with 12 feet between each truss; then the area of roof on plan 60 × 12 720 square feet and the total load being 56 lbs., or ton per square foot, = 40 = as in (529), we have 720 ÷ 40 = 18 tons = on the truss. Then 20.5 tons, requiring C, 99 × 18 ÷ 100 = 18.8 tons, or 4 × 4 × : for B, 83 × 18 100 = 15 tons, or 3 × 3 × 1. = and for A, 66 × 18 ÷ 100 be in one piece from end to end, and in that case the maximum sizes, or those for D, would be used throughout. 100 = Then, for the struts; the strain on G = 23 × 18 4.14 tons, and the length being in our case about 12 feet, we require by Table 82 a T iron say 33 × 3 × 3. To show how serious an error would ensue if we had disregarded the length, and provided for the crushing strain simply, Table 81 would have given for 4·14 tons a T iron 1 × 1 ×. But Table 82 shows that the safe strain on a bar nearly of those sizes, and 12 feet long, would be 0.25 ton only, or less than th of 4.14 tons, the actual strain. (538.) It has been sometimes proposed to substitute for Tiron struts two flat bars of equal area, but in the case of long struts this is a most unsafe proceeding: for example, by exact calculation the angle-iron we have proposed for G, namely 32 x 32 x 3 gives by Rule (534), W = 50 x 33 x 3144 = 4.05 tons, which is near enough to 4.14 tons, the actual strain, for our purpose. But a flat bar left to itself would of course fail by bending in the direction of its least dimension (177): hence t 33, and the 2.6 power of 3, being 0781 by Table 35, the same rule gives W 50 x 0781 x 32144 0.1017 ton for one bar, or 0.2034 ton for a pair having the same area as the T iron, but th! only of the strength as a pillar. = = 3, and b = 20 To resume for the strut F we obtain 19 × 18100 = 3.42 tons, and the length being about 10 feet, we require a T iron say 3 × 3 × 3: for E we have 16 × 18 ÷ 100 = 2.88 tons, and the length being about 8 feet, we require a T iron, say 21 × 21 × 16. = = Then for the tie-rods: the strain on L 105 x 18 100 = 18.9 tons, or 1 inch diameter by Table 83: for M, 89 x 18100 = 16 tons, or 13 inch; for N, 74 × 18 100 13.3 tons, or 1 inch: for K, 46 × 18÷ 100 8.28 tons, or 1 inch for J, 12.5 × 18÷ 100 = 2.25 tons, or 1 inch: and for H, 6·5 × 18 ÷ 100 1.17 ton, or inch diameter, &c. Any of the different forms of truss shown by Figs. 180 to 184 may be used in the manner we have illustrated: it will be found that in order to obtain convenient sections for rafters and struts, Fig. 181 should be restricted to say 30 to 40 feet span: Fig. 182 from 40 to 50 feet: Fig. 183 from 50 to 60 feet and Fig. 184 from 60 to 70 feet, &c. (539.) "Curved Roofs."-For large spans, such as Railway Stations, &c., Curved Roofs are now very extensively used: Fig. 185 gives an outline of such a roof with the strains throughout as due to a total weight = 100, or 50 on each support, in order to assimilate the case to the other examples we have given, and to facilitate calculation of such roofs for various spans. Say we have a Roof 80 feet span, with 16 feet between the principals: then measured on plan, we have 80 × 16 = 1280 square feet on each principal: taking the load at 56 lbs. or 4 ton per square foot as in (529), we have 1280 ÷ 40 = 32 tons total load. The maximum compressive strain on the main rafter or upper ribs A, B, C, will then be 137 × 32 ÷ 100 = 43.8 tons, for which Table 81 gives a T section 7 × 7 x 11 inches, and in most cases that same section would be used throughout, although the strains on D, E are somewhat less. Then for F, we have 32 × 137 ÷ 100 = 43·8, requiring 24. Table 83. The struts M, O, Q, are pillars of the approximate lengths, in our case, of 8, 8, and 5 feet respectively: then (540.) "Wooden Roofs."—Iron is now used almost exclusively for large and important Roofs; but for ordinary purposes wood is still extensively employed, and is likely to be so for reasons of convenience and economy. The form of truss commonly used for small spans, say up to 30 feet, is shown by Fig. 179, another convenient form is shown by Fig. 181. The strains may be found by the methods already explained and illustrated with Iron roofs. The proportions of the different parts and the details of construction, are essentially practical questions, which are fully considered in most of the works on Carpentry, such as that of Tredgold and others. CHAPTER XIV. ON THE TORSIONAL STRAIN. (541.) It will be expedient to consider the Torsional strain under two different heads, namely "Torsional Strength," and "Torsional Elasticity." It is the more necessary to follow that course because the laws governing the Strength, differ entirely from those dominating the Stiffness. We shall deal with the Torsional Strength in this chapter and with the Torsional Elasticity in Chapter XVII. "Torsional Strength."-The fundamental laws for Torsional Strength may be expressed by the Rules: For Circular sections, (545.) or, W x L = QX S3 x 2357. For Rectangular sections, (546.) W x L = Qx d2 × b2 ÷ (√ dˇ × b2 × 3). In which R = radius of circular sections in inches: S = side of square in inches: d and b, depth and breadth of Rectangular sections in inches: L= the leverage in inches with which the twisting weight W acts: W weight, say in lbs.: Q = constant Multiplier, which has the same value for all the three forms of section, and is found from Experiment by the Rules: = W X L X 2 3.1416 × R$ W x L 1.5708 × R3 = (552.) "Ratio of Round to Square."-The Ratio of the Torsional Strength of round and square sections may be found by Rules (543) and (545), for while round bars 1.5708 × R3, square ones = 2357 × S3. Say we take 1-inch bars, then R = 1, and 12= ¦ or ·125, hence the ratio of strength will be (2357 × 13) (1.5708 x 125) 1.20 to 1.0, showing that square bars are 20 per cent. stronger than round ones. = But, when we calculate the values of Q for round and square bars from experiment, as in col. 4 of Table 84, we find that the latter are 34 per cent. greater than the former; thus 3590726800 = 1.34. So great a difference shows that there is some error in the Rules, although they are based on laws given by the highest authorities: admitting the experiments to be correct, the Rules require modification, and become :For Square sections, (553.) (554.) W x L = Q× S3 ×·3158. Q W x L Calculating Q by Rule (554) we obtain col. 5 of Table 84: the mean = 26,800, or nearly the same as for the round bars which gave by Rule (542), 26,709. From this it appears that the experimental ratio of the strength of square and round bars is 1.6 to 1.0, the theoretical ratio being 1.2 to 1.0. (555.) "Practical Rules.”—The theoretical rules are inconvenient, although they are fundamental, giving the laws governing the Torsional Strength, and have the advantage of a |