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 41 inches diameter; the actual diameter was 44 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 44 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 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 ENGINES. From Cases in Practice. (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. 60 4:57 1.688 2 همدمم به مادر دن 3.27 1.117 (6) rod 31 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 1.565 = 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 (D– d-) L?. (221.) 196 – :6 = F xL Mp (222.) L = /{Mpx (D** – dot) =- F}. In which F = the breaking weight on the pillar in lbs. or tons by flexure, dependent on Mp D = the external diameter in inches. d = the internal L = the length in feet. Mp = Constant Multiplier, the value of which is given in Table 34. Table 52 gives the result of thirty-six experiments on hollow pillars, made of thin plate-iron, by Mr. Hodgkinson ; col. 10 has been calculated by the rule, the value of My for pillars with both ends flat being taken at 299,600 lbs. from 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 Dthen 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., = |