from which we have the simple law for a bar of those sizes, &c., that the maximum longitudinal strain is 18 times the transverse load: hence we have the Rule: We may now apply this Rule to Timber whose tensile and crushing strength is known by direct experiment: for the former we shall take Mr. Barlow's results in Table 1, and for the latter, Mr. Hodgkinson's in Table 32: the Transverse strengths will be given by Table 65. (640.) A bar of ash 1 inch square and 1 foot long, breaks with a load of 681 lbs. in the centre, the maximum longitudinal strain ƒ = 681 × 18 = 12258 lbs. per square inch; by direct experiments, the tensile strength T 17077 lbs.; and the crushing strength C = 9023 lbs. per square inch: the mean of the two is 13,050 lbs., or nearly as given by the Rule. = English oak breaks transversely with 509 lbs. ; hence ƒ = per square inch: by direct experiments 509 × 18 = 9162 lbs. 10389 lbs., and C of the two = 9330 lbs. Beech breaks transversely with 558 lbs.: hence f= 558 x 18 10044 lbs. per square inch; by direct experiments T = 11467 lbs., and C 8548 lbs.; the mean being 10,007 lbs. per square inch. = = Teak breaks transversely with 724 lbs.: hence ƒ 724 x 18 = 13032 lbs. per square inch; by direct experiments T = 15090 and C 10706, the mean being 12,898 lbs. per square inch, &c. The application of the rule to cast and wrought iron is given in (520). = (641.) Having found the value of f, or the maximum strain at the upper and lower edges of the section, we have now to find the extension and compression there from the deflection of the beam with a given load in the centre, for which we have the Rule: = Ex = in which D depth of rectangular bar in inches, = length between supports in inches; & deflection at the centre in inches and E2 = extension in parts of the length: it will be observed that the central load, and the breadth have nothing to do with the question in this case. The rule also supposes that the compression C is equal to Ex, which perhaps is nearly true for light strains (617) with most materials. (643.) For a bar 1 inch square, and 12 inches long, l÷2=6, 3 x 1 x 8 3 x 8 8 and the rule becomes E = or or 2 × 62 72 24' that is to say, the maximum extension and compression is 4th of the transverse deflection. Table 105 gives in col. 4 the mean deflection of bars of many materials, 1 inch square, 1 foot or 12 inches long, by 1 lb. in the centre, which by (638) is equivalent to 18 lbs. longitudinally. Thus with cast iron the rule becomes 00002886 ÷ 24 = б 24 ⚫0000012 for 18 lbs., therefore ⚫000001218 =⚫0000000667 for 1 lb., or 0000000667 × 22400001494 per ton, the length strained being 1.0. giving in this case the extension for 18 lbs. per square inch. For a standard bar 1 inch square, and 1 foot long, we have the Rules: (644.) (645.) the extension or compression in parts of the length per lb. = the extension or com = the deflection in inches by 1 lb. in centre: thus for cast iron d=00002886 by col. 4 of Table 105; hence, Rule (645) becomes 00002886 × 5.18 =0001494 as before (643). (646.) Calculating in this way we have obtained col. 8 of Table 105. Comparing these results for cast and wrought iron, with those obtained by direct and exact experiment in Table 91, we have for cast iron 0001494 by Table 105, while Table 91 gives for extension 00016556, and for compression ⚫0001695 parts of the length, by 1 ton. For wrought iron, Table 105 gives 00008106, while Table 91 gives for extension ⚫000080121, and for compression, 00010138 parts of the length, by 1 ton. These results agree fairly well together; hence we may have confidence in the method we have followed, and by which col. 8 of Table 105 has been obtained. We may now illustrate the application of these rules to practice say we have a wrought-iron bar 50 feet, or 600 inches long, with a tensile strain of 8 tons per inch: then the extension=00008106 × 600 × 8 = 0.389 inch. Direct experiment gives by col. 5 of Table 91, 000638774 × 600 = .383 inch. Again, a bar of Riga fir, with say 2 tons per inch, and a length of 30 feet, or 360 inches, will stretch ⚫00197 × 360 × 2 = 1.42 inch. Again, a pillar of English oak 12 feet or 144 inches long, with a compressive strain of ton per square inch will be shortened by the pressure 002042 × 144 × 3 = 0·22 inch: this being the mean, col. 8. CHAPTER XVI. ON THE DEFLECTION OF BEAMS. (647.) "Form of Curve of Flexure."-With a parallel beam, or one having uniform depth and breadth throughout the length, resting on bearings at each end, and loaded with a central weight, the strain on the material is a maximum at the centre, and is progressively reduced toward the ends, where it becomes nothing. In that case the elastic curve has its shortest radius at the centre, the curve becoming progressively flatter toward the ends, where it is a straight line. When the strength at every point is proportional to the strain there, for example when the depth is uniform, and the breadth is reduced toward the ends in arithmetical ratio as in Fig. 116, the elastic curve is uniform in its radius from end-toend, that is to say, it is a simple spherical curve. "Curve with Central Load."-Let Fig. 174 be a parallel beam resting on two bearings and loaded in the centre: then we have the Rule: ((3 L) × 202 2 In which L = length of the beam between bearings. 8 = deflection in centre by central strain. x = distance from centre to a point whose deflection is required. y = co-ordinate of the curve at that point. Of course all the dimensions must be in the same terms, feet, inches, &c. (649.) Thus, with a beam 10 feet long, say &= 0.25 foot; x= 3 x .25 3 feet: then by the Rule y = 5 x 9 27 6 = = 0.108 foot, or 1.296 inch: hence the deflection at that point is 31.296 1.704 inch. Calculating in this way we may obtain any number of points through which the entire curve of flexure may be drawn. Table 103 has been calculated by the rule, the half-length of the beam being divided into 10 parts, and the central deflection = 1·0, from which we may easily find the deflection at any point in a beam whose central deflection is known:-thus in our case, the point x being 3÷÷5 = 0·6 of the half-length distant from the centre, we find the deflection at that point 0.568 × 3 = 1.704 inch as before. The curve B in Diagram, Fig. 213, has been obtained from Table 103. = "Load out of Centre."-Let Fig. 175 be a parallel beam with a load W out of the centre: knowing the deflection which any TABLE 103.-Of the FORM of the CURVE of FLEXURE in a PARALLEL BEAM DEFLECTED by a CENTRAL WEIGHT, Fig. 212. weight would produce at the centre, we may find the deflection at any other point by the same weight applied at that point, by the Rule: : d = (L) X In which l = the distance from the weight to the nearest bearing in the same terms as L; d= the deflection at the point of application of the weight W, and the rest as before. Thus taking the beam in (649) whose central deflection 3 inches, and say we require the deflection at a point 3 feet from the end, therefore 2 feet from the centre:-then the Rule gives d = = = 0.1764 foot, or 2.1168 inches. × 3) — 3'}' = {10 x = Table 104 has been calculated by the Rule, taking the central deflection 1.0, and dividing the half-length into 20 parts, col. 2 gives the deflection at each point in the length by that same constant weight. (651.) For example, in our case, the point is 25 = 0.40 from the centre, hence the deflection there = 7056 × 3 = 2.1168 inches as before. The curve A in Diagram, Fig. 213, has been obtained from col. 2 of Table 104, and gives the deflection from centre to end for a rolling constant load. For example, say we have a beam 24 feet long, and we require the deflection at 6 feet from the centre by a certain weight applied at that point. We have first to find what the central deflection |
