value for Q, which is, or should be, constant for all sections. The following Practical Rules are based on the theoretical ones, corrected by experiment (552): (562.) W = M1 × ď2 × b2 × 2·264 ÷ (√ ď2 + b2 × L). (563.) Mt = W × L × √ď2 + b2 ÷ (ď2 × b2 × 2·264). In which Mt = a constant, having the same value for all the sections: D diameter: S = side of square: d = = = depth, and b = breadth of rectangular sections: L leverage, all in inches: W = weight in lbs. acting with the length of lever, L. (564.) The values of Mt in col. 6 of Table 84 have been calculated by Rules (558) and (561): thus, the bar 4 inches diameter by Rule (558) gives Mt = 1938 × 17064 = 5148. Again, the bar 1 inch square, which broke with 231 lbs. and 36 inches leverage, gives by Rule (561), Mt = (18 x 1.6) 5197, &c. = 231 x 36÷ The mean for 9 bars varying from 2 inches to 4 inches diameter 5290: and the mean for 6 square bars = 5290 also, which may therefore be taken as the mean value of Mt for cast iron. The experimental ratio of square to round bars, namely, 1.6 to 1.0, has been adopted in these Rules, instead of the Theoretical Ratio 1.2 to 1.0 (552). = (565.) "Rectangular Sections."-The Rules for Rectangular Sections may be used for square ones, which of course are rectangular with equal sides: say we take a 3-inch square bar with 60 inch leverage: then taking Me 5290, Rule (559) gives W 5290 x 33 x 1.660 3809 lbs.: taking it as a rectangular bar, Rule (562) becomes W = 5290 x 32 x 32 x 2.264÷(√√3 +32 × 60) = 3809 lbs. also. = = When the two dimensions of a bar, rectangular in section, are very unequal, or the breadth very great in proportion to the thickness, the torsional strength is practically as the breadth simply, the thickness being constant. Thus, for cast iron, with 10, and a thickness of 1 inch and breadths L = 1 2 4 8 16 inches, the torsional strength, or W, becomes by Rule (562) = 847 2142 4648 9508 19,127 lbs. It will be observed that, while the breadths are in the ratio of 2 to 1 throughout, the strength for the greater breadths follow nearly the same ratio. (566.) "Hollow Sections."-We have seen by Rules (556) and (559) that with square and round bars the torsional strength is directly proportional to D3 or S3, so that for diameters in the ratio : the Torsional strength would be in the ratio: 5 125 By the laws of Torsional elasticity (716) the Stiffness with constant strain is inversely proportional to D' or S1, hence in our case D1 being in the Ratio: 1 16 81 256 625 the stiffness with constant strain will be in the inverse ratio, But when the strain is proportional to the strength, the Ratio of the torsional angles become : which is in inverse proportion to the arithmetical ratio of the diameters simply. (567.) "Old Rule."-The old rule commonly used by pracD3 - d3, D being the external, and d tical men is W = = the internal diameter of a hollow shaft, which is regarded as composed of two shafts having the respective diameters, and it is assumed that, deducting the strength of one from that of the other, we should obtain the strength of the annulus. But a shaft will not yield the full strength due to it, except it be allowed to twist proportionately to its diameter. Say we take a section 2 inches diameter externally and 1 inch internally; then by the old rule the strength will be 23 - 13 = 7: but we have seen by the preceding investigation that the diameters, being 2 and 1, the twist should be to 1, that is to say, the internal section should twist twice as much as the external one, which of course is impossible in our case, for obviously both must be twisted to the same extent. As the internal section is allowed to twist to half only of the extent due to its full strength, it will yield half only of that strength; hence we obtain 23 (13 × 1) 7.5 for the hollow section, instead of 23 — 13 = 7, as by the old Rule. = (568.) Hence the rule for hollow circular sections becomes W = (D1 — d1) ÷D, or in our case (21 — 1') ÷ 2 = 7·5, as before. Taking another case of a hollow shaft with diameters 4 and 3 then by the old rule we should have 43 33 = 37: but by the preceding analysis, the ultimate twist for diameters in the ratio 3 and 4 = and respectively; hence, instead of the internal diameter yielding 32 27, it would give 27 x , or 27 × 34 20-25 only, and the strength of the hollow shaft becomes 43 43.75, instead of 37. The rule (D' - d')÷D gives in our case (43) 443.75, as before. These principles apply to the transverse strain also (337). = = 20.25, or 64 - 20.25 = We have therefore for hollow shafts the rules: (569.) For Circular Sections, = the internal diameter in In which D = the external, and d inches: S = the external, and s = the internal dimensions of square sections in inches: W = the straining weight in lbs. acting with a leverage of L, in inches: Mt = a constant whose value is given by col. 6 of Table 84, and by (564). To illustrate the application of these rules: say we have a hollow cast-iron shaft in which D = 6 inches, d = 4 inches, 64 44 L = 84 inches: then Rule (569) becomes W = 5290 X 6 8410900 lbs. breaking weight. Again, with a hollow square shaft in which S = 5 inches, s = inches then by Table 85, 5 = 625, 3 inches, and L = 96 and 31 = 150, with 625-150 and 3* 625, which Rule (570) gives W = 5290 × = 8376 lbs. 5 × 1.696 TABLE 85.-Of the FOURTH POWER of NUMBERS. (571.) "Wrought Iron and Steel."-According to W. and D. Rankine's experiments, the torsional strengths of Cast iron, |