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The table shows that the diameter is more influential on the CONN

ength with wrought-iron pillars than with cast-iron ones, which is due in part to the lower value of C: this, as we have seen (132), is 43 tons per square inch with cast iron, and 19 tons with wrought iron in the form of a pillar (133).

(204.) “Wrought-iron Connecting-rods."—Wrought-iron rods are commonly used for steam-engines, pumps, and many other

purposes :-being jointed at both ends, they are assimilated to Fi pillars with both ends pointed (149). The connecting-rods for

double-acting pumps are subjected to heavy shocks from a mass of water in motion throughout the system of suction and delivery pipes, which are only partially obviated by airvessels, &c. :—they therefore require special rules, which are given in (207).

Table 45 has been calculated by the rule (197); for a pillar with both ends pointed this becomes F 95848 x D3•6 L, which has been corrected where necessary for incipient crushing by the rule (164), taking the crushing strain C at the reduced value of 33,600 lbs., or 15 tons per square inch, from which we have obtained Cp in col. 2.

(205.) It should be observed that the rod of a steam-engine is subjected to an alternating tensile and compressive strain which is exceedingly destructive to any material, necessitating the adoption of a high “factor of safety” (915). If we admit that for a statical or dead load the "factor" should be 3, that is to say the working or safe load should be įrd of the dead breaking load, then by col. 5 of Table 141 the equivalent alternating dynamic strain would be įth of the dead load, or } x 5 = lath of the statical breaking weight, the “ factor” being = 18.

Table 46 gives the sizes of a series of connecting-rods from cases in practice :—the breaking weight as calculated by the rules is given by col. 6, and the factor of safety by col. 7, its mean value being 15:52. Thus, taking No. 2 as an example: the 3.6 power of 31 is by Table 35 = 116; 7 feet 2 inches being =

= 7.167 feet, and 71672 51.36, and the rule becomes 95848 x 116 • 51.36 216500 lbs. F, or the breaking weight by flexure, as in col. 4. This requires correction for incipient crushing (163):-the area due to 3 inches diameter = 11.04 square inches, hence 11.04 x 42560 = 469800 lbs., the value of Cp, as in col. 5, and Cp = 352400 lbs., from which we obtain (216500 x 469800) = (216500 + 352400) = 178800 lbs., the reduced breaking weight Pc, as in col. 6. The value of the factor of safety is 178800 - 11000 16.25, as in col. 7.

The diameter of a connecting-rod might be calculated with sufficient precision for practice by an empirical rule as follows: (206.)

D = 88 (W x L = 4640). In which w = the strain on the rod in lbs., as found from the diameter of the cylinder and pressure of steam, &c.; L = the length between centres in feet, and D = the diameter at the centre in inches. Thus, for No. 3 in Table 46 we have 6411 x 25 = 4640 = 34.54 the nearest number to which in col. 4 of Table 35 is 32:3 opposite 2 inches diameter, &c. : col. 8 in Table 46 has been calculated by this rule.


STEAM-ENGINES. Cases in Practice.

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(207.) “Double-acting Pump-rods." -The ordinary rules for wrought-iron pillars do not apply satisfactorily to the rods of double-acting pumps, partly, perhaps, because those rules are adapted only to a statical load or dead weight, whereas pumprods have to sustain heavy shocks at every stroke from the alternate inertia and momentum of the water in the pump and mains, the amount of which depends on the frequency of the alternations or the speed of the pump. The following empirical Rule is derived from long and varied experience :

(208.) D = ŕ (W L R : 200000). In which w = the weight or strain on the rod in lbs. ; L = the length between centres of joints in feet; R = the revolutions per minute of the engine, or of the crank working the pump; and D = the diameter of the rod at the centre in inches.

(209.) Table 47 gives the proportions of double-acting pump-rods from cases in practice; col. 6 has been calculated by the Rule. Thus, with No. 9 we have 17:52 306, then 25125 X 306 x 24.5 = 200000 942, which is the 4th power of the diameter. The logarithm of 942 or 2.974051 - 4 = 0.743513, the natural number due to which, or 5.54 inches, is the diameter at the centre, as in col. 9. We should have obtained the same result without the use of logarithms by finding

the square-root of the square-root of 942; thus 942 = 30.69, TABLE 47.-Of the STRENGTH of WROUGHT-IRON CONNECTING-RODS for

DOUBLE-ACTING PUMPs, to resist a COMPRESSIVE STRAIN. From Cases in Practice,


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ft. in. 13 50 2,926 16 23 0 11,683 | 4.00 31

71 180 3,221 25 20 0 21,805 6.80 3.1 13 64 3,724 28 13 0 29,492 7.93 3 9 160 | 4,450 24 14 0 31,790 7.14 3.16 71 250 | 4,490 34 13 045,372 10.10 33

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9 203 5,533 22 16 9 35,259 6.37 35 12 203 9,000 22 17 1 53,286 | 5.92 4 10 279 9,503 36 12 077,210 8.13 33 13. 400 25,125 24 17 6 145,028 5.78 5. 20 180 24,500 20 22 0 218,240 8.90 (1) (2) (3)

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and ✓ 30.69 = 5.54 inches, as before. Table 85 would have given nearly the same result with less trouble.

The near agreement of the calculated sizes in col. 9 with the actual ones in col. 8 is due to the fact that the latter were in many cases fixed originally by the Rule (208); still, they all stand well in practice, and have done so for many years without one failure, confirming so far the accuracy of the Rule.

(210.) “ Wrought-iron Piston-rods." — The piston-rods of Steam-engines are subjected alternately to a tensile and compressive strain, and in calculating their strength both of those strains must be considered.

First, for the tensile strain :—a common mode of connecting the piston-rod to the cross-head is shown by Fig. 42, in which we have a 2}-inch rod with a conical end, terminated by a 21-inch square-thread screw, 14 inch diameter at the bottom of the thread. The area of 24 = 4.9, and of 13

2:4 square inches; the area at the screw is therefore only half the area of the rod.

In large piston-rods a key is commonly used, as in Fig. 43; we have shown in (123) that the shearing strain is equal to the tensile, and as the key is subjected to a double shear, shearing at two places, its double area must be equal to the area of the rod at the key-way, supposing of course that the key is of the same material as the rod. With the sizes shown in the figure, the key will have a shearing area of 4.75 x 1.25 x 2 = 11.87 square inches. The area of 4 diameter being 17.72, and the area cut away by the key-way, 4.75 X 1.25 = 5.93, we have 17.72 – 5.93 = 11.79 square inches, being practically the same as the area of the key this, again, is only half the area of the 5-inch rod, which = 23.75 square inches.

(211.) We may therefore admit that the minimum area at the screw or key-way is half the area of the rod, and this of course limits the strength of the piston-rod ; for obviously, whatever may be the strength as a pillar during the up-stroke, the rod cannot bear more than is due to its tensile strength during the down-stroke.

(212.) The mean strength of British wrought iron may be

taken at 57,500 lbs. per square inch from (4) and Table 1, hence the maximum strain admissible on the body of the rod is 57500 : 2 = 28750 lbs. per square inch, and the second column of Table 48 has been thus calculated. When a rod is so short that we are certain the strain will be limited by the tensile strength, we may find the area direct by the Rule :

(213.) Area = W X Mx • 28750. In which W = the strain on the rod due to the area of the piston, pressure of steam, &c., and Me = the Factor of Safety (880). (216).

For the strength to resist the compressive strain at the upstroke we may consider the rod as a pillar with one end flat and the other pointed, namely, flat at the piston, and pointed at the cross-head. Taking the value of Mp at 197,700 lbs. from col. 2 of Table 34, the Rule (197) then becomes :(214.)

F = 197700 x D96 : L'. In which F = the breaking weight by flexure in lbs., L = the greatest length unsupported in feet, or in most cases, the distance from the gland of the stuffing-box, to the centre of the cross-head at the top of the stroke, D = the diameter in inches.

(215.) This rule gives the breaking weight by flexure only :as shown in (202), rods of a length less than 58.37 times the diameter will require correction for incipient crushing. Taking C, or the crushing strength of wrought iron at 19 tons, or 42,560 lbs. (201), and the areas due to the respective diameters, we obtain the value of Cp in the third column of Table 48, which gives the sizes of Piston-rods based on a combination of the rules for flexure (214), tensile (212), and crushing strength (164)

Thus, for a 31-inch rod, 8 feet long, 3186 being 69.6 by Table 35, the breaking weight by flexure becomes 197700 X 69.6 = 64 = 215000 lbs. F. This requires correction for incipient crushing (163), the area of 31 = 8.296 square inches, and we obtain 8.296 x 42560 = 353100 lbs., the value of Cp; hence Cp=264800 lbs., and the rule (164) gives Po = (215000

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