inch, giving 254 x 65 2240 7.4 tons, requiring 7.4÷ ÷ = 2.87 2.6 square inches area at the key-way, equivalent by (210) to 2.6 x 2 = 5.2 square inches of the body of the rod, or 2 inches diameter where it works through the gland of the pump. With a deep-well pump, there would be long rods to the surface with welds here and there: by Table 1 the strength of welded joints = 47266 lbs., or 21 tons per square inch Breaking weight, which by the "Ratio" in col. 3 of Table 141 is reduced to 21 x 7 tons Dynamic strain, and with Factor 3, to 7 ÷ 3 2.33 tons Working strain. Hence we obtain 7·42·33 = 3·2 square inches, or 2 inches diameter: Table 28 gives practically the same result: the Working load for 2-inch rods being 15,708 lbs., or 7 tons, being nearly 7.4 tons, the strain in our case. = = = If this same pump had been double-acting, the strain being alternately Tensile and compressive, col. 5 of Table 141 gives 3.2 tons per square inch Breaking weight, or 3.23= 1.07 ton working load, requiring 7·4÷1·07 = 7 square inches area, or 3 inches diameter of the body of the rod. Here the key-way question is eliminated, being covered by the large diameter due with an alternating strain (210). = (930.) Again: say that we have a short pillar of English Oak subjected to an intermittent load of 10 tons acting without shock: then, col. 2 of Table 141 gives 1.85 tons per square inch Breaking weight, Table 137 gives the Factor 5, hence we obtain 1.85537 ton per square inch working load, and require 10 ÷ ·37 = 27 square inches area, say 5 inches square. If this Oaken rod had worked a Double-acting pump, the strain being both Alternating and Dynamic, col. 5 of Table 141 gives 0.46 ton Breaking weight, or 46 5·092 ton working load, which is 5th of 3.7 tons, the Crushing dead load by col. 1: we then require 10 ÷ 092 = 109 square inches area, or 10 inches square, &c. = Again: say we have a rocking-beam working a double-acting pump, 12 inches diameter, 100 feet head of water = 1002.3 = 44 lbs. per square inch: the area of 12 113 square inches, hence 113 x 44 2240 2.2 tons, which being a Dynamic and Alternating load is by the "Ratio" in col. 5 of Table 141, equivalent to 2.2 x 6 13.2 tons dead load: wrought-iron. = TABLE 141.-Of the STRENGTH of MATERIALS with DIFFERENT KINDS of STRAIN being the Ultimate or Breaking Loads. NOTE. In alternating strains, the loads marked are limited by the strength in the opposite direction. TABLE 141.-Of the STRENGTH OF MATERIALS with DIFFERENT KINDS of STRAIN, &c.-continued. = Say that the beam is a cantilever 5 feet long, equal by (431) to a beam 5 × 4 = 20 feet long, supported at both ends: then with Mr = 4000 lbs., or 1.8 ton, and assuming the thickness or B 1 inch, we may find the depth D by Rule (325) or D = √(13·2 × 20) ÷ (1·8 × 1) = 9 inches deep. case the working load ÷ 3 = 1 of the dead load. the value of MT = 66722403 ton, from col. 5 of Table 141, the Rule (324) gives W 972 × 1 × 320 = 2.2 tons breaking dead load, as before. = = In this Taking APPENDIX. EFFECT OF SIZE OF CASTING ON THE STRENGTH OF CAST IRON. (931.) The transverse strength of cast iron has usually been determined by experiments on bars 1 inch square, and it was supposed that the data thus obtained were applicable without correction to bars of all sizes and to girders of all forms of section. More recent observations have shown that these conclusions were not correct, and that, 1st, bars of large sizes are specifically weaker than small ones: 2nd, that in bars of rectangular section the strength is governed by the thickness or least dimension rather than by the greater. (932.) "Effect of Thickness on Transverse Strength.”—All the properties of cast iron seem to be more or less affected by the size or rather by the least thickness of the casting; so far as we have experimental knowledge, that is to say between 1 inch and 3 inches square, the tensile, crushing, and transverse strengths, also the Modulus of Elasticity (738) are reduced as the size of the casting is increased. Table 142 gives reduced results of experiments by Mr. Hodg kinson and Captain James, R.E.; the latter are more numerous and more consistent among themselves than those of Mr. Hodgkinson, which are anomalous, giving the same specific strength for 2-inch as for 3-inch bars; they give, however, one important fact, that in a rectangular bar of unequal dimensions, namely 3 × 1, the transverse strength is practically the same as that of a bar 1 inch square, as shown by the Table, where the value of MT, or the specific transverse strength, is nearly an arithmetical mean between the strength of 1-inch and 2-inch square bars. This experiment seems to show that the larger dimension of rectangular bars, has no sensible influence on the |