= x 353100) = (215000 + 264800) = 158200 lbs., as in the Table, which has been calculated throughout in this way. With four exceptions, the whole of the numbers required correction for incipient crushing. When the strength as a pillar is greater than the tensile strength in the second column, then the latter limits the strength of the whole :—for instance, a 3-inch rod, say 3 feet long, would bear as a pillar 251,500 lbs., but the tensile strength at the key-way, &c., is 203,200 lbs. only, and as the strain during the up and down strokes is usually equal, the strength both ways is limited by the lesser. It will be observed that in the Table the strengths of say a 5-inch rod 5, 6, and 7 feet long are all alike, being, in fact, equal to the tensile strength, or 564,600 lbs., &c. (216.) Table 49 gives the particulars of Piston-rods from cases in practice :—the strain in col. 5 was obtained by multiplying the area of the piston by the pressure of steam, plus the vacuum in condensing engines, where the total has been taken at 20 lbs. per square inch; with the High-pressure Engines, the pressure has been taken at 45 lbs. above atmosphere. In col. 6, we have given the caleulated breaking weight as found by Tables 48, 56, &c., and in col. 7, the “Factor of Safety,” (880) or the ratio of the breaking to the working strain, the mean value of which is about 12 for ordinary Condensing-engines, and 14 for Marine Engines; it will be observed that the whole of the latter are dominated by the tensile strength, in consequence of the shortness of the stroke, and are marked by a *. In col. 8 we have given the calculated diameters as found by Tables 48, 56; thus in the 80-Horse rod, No. 6, with Factor 12 we have 32520 x 12 = 390240 lbs. breaking strain, the nearest number to which in the column for 8 feet length in Table 48 is 391,100 lbs., opposite 4 inches diameter; the actual diameter was 4 inches. Again, in the 60-Horse Marine Engine, No. 9, with Factor 14 we obtain 29040 x 14 = 406560 lbs. breaking strain, the nearest number to which in the column for 5 feet long is 408,000 lbs., opposite 4 inches diameter, which was also the actual size ; being governed by the Tensile strain in col. 2. Wrought iron is not often used in modern times for first-class Engines; the superior character of steel, combined with its = TABLE 48.-Of the STRENGTH of WROUGHT-IRON PISTON-RODs to resist TENSILE and COMPRESSIVE STRAINS. :::: 33,430 15,620 11,030 7,9104 5,4907 52,220 28,840 18,600 16,070 12,100 36,500 28,280 22,180 17,400+ 28,970 65,280 53, 250 43,760 36,300 353,100 238,500* 238,500* 238,300 208,500 181,800 158, 200 138,100 120,700 361,400* 361,400* 357,300 | 319,100 284,000 . 252, 600 224,700 179,100 603,900 408,000* 408,000* 408,000* 374,200 335,200 300,000 268, 200 215,500 4 457,100 676,700 457,100* 457, 100* 434,400 391,100 351,600 316,100 256,000 564, 600 835,900 564,600* 564,600* 564,600* 515,900 468,700 425,000 319,300 5. 100 0 1,011,000 683, 100* 683, 100* 660, 700 605,000 552,900 460,800 812,700 1,203,000 812,700* 812,700* 812,700* 760,400 698,900 590,200 Tensile Breaking Crushing Weight Strain CP. inches. lbs. Ibs, 361,400 535,000 683, NOTE.-The strains marked * are limited by the tensile strength in col. 2; those marked † are due to flexure simply. The rest have been reduced for incipient crushing by the rule in (164), with the value of Cr given by col. 3. reduced cost, should commend it for universal adoption; the strength of Steel Piston-rods is considered in (271). Table 49.-Of the STRENGTH and PROPORTIONS of PISTON-RODS to STEAN. From Cases in Practice. ENGINES. Condensing 18 80 45. 20 32 43 50 60 90 ft. lbs. Ibs. 3.5 2,460 32,710 13.30 6 0 39, 270,536,500* 13.66 3:3 4,085 73,410 17.96 21 Iron 3.5 8,250 96,050* 11.64 1 Steel 3:3 5,521 115,400 20.85 4 5 5,985 77,530 12.95 4.5 6,435 77,530 12.05 41 16 11] 108 High-pressure. 27 11 2 23 23 3 41 (8) 22 (9) (217.) “Radius-rods of Steam-engines.”—In ordinary Beam Engines, the Radius-rods and “parallel bars” of the motion are subjected alternately' to equal tensile and compressive strains ; the latter, tending to cripple the rod as a pillar, is the most influential, and the diameter necessary to resist that strain will usually be sufficient for the tensile strain also. For Radiusrods we may give the following Empirical Rule : (218.) d = V (H x L = 150). = In which H = the reputed or Nominal Horse-power of the Engine ; L the length of the radius-rod between centres, in feet; and d = the diameter of the rod in inches. Table 50 has been calculated by this rule. The “parallel bars ” are usually made of the same diameter as the radius-rods; Table 51 gives the particulars of both from cases in practice, col. 4 having been calculated by the rule. Thus, for the 30-horse Engine with a TABLE 50.-Of the Sizes of RADIUS-RODS to STEAM-ENGINES. NOTE.-The parallel bars may be of the same diameter as the Radius-rods. TABLE 51.–Of RADIUS-RODS, &c., to STEAM-ENGINES. Cases in Practice. = rod 3? feet long, we have 30 x 12:25 • 150 = 2.45; then the logarithm of 2:45, or 0.389166 = 4 = 0.097291, the natural number due to which, or 1.251 inch, is the diameter required. Or we might have obtained the same result without logarithms by finding the square-root of the square-root of 2.45, thus 12.45 = 1.565, and 71.565 = 1.251, as before: see also, Table 85. (219.) “Hollow Cylindrical Pillars of Wrought Iron.”—For hollow cylindrical pillars, the rules in (196) require a simple modification and become: (220.) F = Mp X (D34 – ) -- L'. Mp = by flexure, dependent on Mp in Table 34. (223.) Thus, with No. 3, the logarithm of 6 187 or 0.791480 x 3.6 = 2.849328, the natural number due to which, or 706.9, is the 3.6 power of D:—then the logarithm of 6, or 0.778151 x 3.6 = 2.801344, the natural number due to which, or 622.9, is the 3.6 power of d. Hence 706.9 622.9 = 74, and the rule (220) becomes F = 299600 x 74 = 100 = 221700 lbs., the breaking weight by flexure, or F, as in col. 10. This requires correction for incipient crushing by the rule (164); taking the crushing strength at 19 tons, or 42,560 lbs. per square inch (201), and the area by col. 5 being 1.799 square inches, Cp becomes 1.799 x 42560 = 76570 lbs., as in col. 11, and {Cp=57420 lbs.; hence (221700 X 76570) = (221700 + 57420) = 60810 lbs., |