ments showing the effect of variations in the thickness or size of bar in the case of cast iron on the sets under tensile and compressive strains; the effect of size on the set of bars loaded transversely we shall find (769) to be very considerable. (758.) The longitudinal set may be determined approximately from the observed set of a bar loaded transversely by the rules in (511), (642). Thus, by col. 5 of Table 106 a bar of wrought iron in which d= 1.515 inch, b 5.523 inches, l 13 feet, with a weight w = 952 lbs. in the centre, This by the rule (511) is equivalent to and lower edges of the section, or f or 162 inches, loaded has a set of 02 inch. a strain at the upper 3 x 162 x 952 = = = = 17920 lbs., or 8 tons per square inch; 2 x 5.523 × 1.515° and the longitudinal set with that strain by the rule in (642) 3 x 1.515 × ⚫02 2 x 812 becomes Ex = =000006927 of the length of a bar strained longitudinally with 8 tons per square inch. By direct experiment, the set by Table 11600000827, the -inch bar gave 00000678 with 18,672 lbs. by Table 95; and the bar in Table 94 gave 0000109 with 17,666 lbs. tensile strain. This method is useful in those numerous cases where we have no direct experimental information. PERMANENT SET UNDER TRANSVERSE STRAINS. (759.) "Cast Iron."-The elasticity of cast iron varies so much with different kinds of iron, and with the size of the casting, or rather with its thickness (744), that it is difficult to give rules for permanent set which will be even approximately correct for general cases. For rectangular bars of Cast iron we have the Rules: = = depth In which L = length between supports in feet: d breadth in inches: z = the ratios of strength as least dimension of the casting, and as given Permanent set in inches. = in inches: b = in Table 108 say we take Wp = 112 lbs. ; L 1.522 inch: b = 3.066 inches, &c. Then 8141 for 1 inch thick, the Rule (762) (1128141) x 000000002391 1.5225 × 3.0662 = .0196 inch Permanent set; col. 10: experiment gave 0192 inch, col. 8. The cols. 8 and 10 do not agree very well with heavy strains. (766.) The mean Transverse breaking weight of a cast-iron bar 1 inch square and 1 foot long is 2063 lbs. by col. 7 of Table 64, hence with Factor 3 we have 2063 ÷ 3 = 688 lbs. safe load, with which Rule (765) gives S = 1 × (6881)2 ×·000000002391 17 =001132 inch set with safe load, from which we obtain the Rules:— (767.) Permanent set with Safe Load = L2 d x 900 L2 dx 100 = In which Sg Permanent set with Safe Load, and S, with Breaking weight; the rest as before. These simple rules may be applied with approximate accuracy to cast-iron beams of all sections, the effect of size being eliminated (773): for example, a girder of any ordinary section, say 20 feet long, and 8 inches load, namely 3rd the breaking weight, &c. feet long: col. 3 Table 117 gives a summary of Mr. Hodgkinson's experiments on the mean set of 1-inch bars of cast iron, 4 has been calculated by Rule (765): thus with 4.5 × (561)2 ×·000000002391 S = 17 56 lbs. we obtain = 003075 inch, &c. TABLE 117.-Of the PERMANENT SET in CAST-IRON BEAMS, (769.) "Effect of Size of Casting."-This effect will be most clearly shown by reducing the experimental results from bars of different sizes to the equivalent loads, deflections, and permanent sets with a "Unit" beam, or a bar 1 foot long, 1 inch deep, and 1 inch wide, and in order to clear the investigation from the complications arising from the varying properties of different kinds of iron, it will be necessary to take one and the same iron for all the bars. We shall then obtain the deflection and set for bars of different sizes in a directly comparable form and shall show the effect of size alone cleared from all obscuring circumstances; see (746). For this purpose we will take Blaenavon No. 2 iron: bars about 1, 2, and 3 inches square, with lengths of 41, 9, and 131 feet respectively, were experimented upon by Mr. Hodgkinson: the loads, deflections, and permanent sets are given by cols. 1, 2, 3 of Table 118. To reduce the loads in col. 1 to equivalent ones of a “Unit” beam we have the Rule: (770.) W = w x L ÷ (ď2 × b). = In which Wreduced weight on Unit beam: w = the experimental weight on a bar whose depth d, and its breadth = b, both in inches while its length = L in feet. Thus a weight = 224 lbs. on a bar 13 feet long, 3.05 inches deep, and 3·095 inches wide is by the Rule equivalent to W = 224 x 13.5÷ (3.052 × 3·095) = 105 lbs. on a bar 1 × 1 inch × 1 foot long, as in col. 5 of Table 118. Then to reduce the experimental deflection with say the same 3-inch bar to that due to the equivalent load on Unit bar we have the Rule: (771.) In which = the deflection by experiment with a bar whose depth = d in inches, and its length L in feet: &, the deflection with equivalent weight on Unit beam: thus in our case d = 195 x 3.05 13.5 =·00326, as in col. 6 of Table 118. The reduced permanent set may be found by a modification of the same Rule: (772.) = = S1 = Sx d÷L2. In which S the permanent set by experiment with a bar whose depth d in inches and its length L in feet, and S1 = the equivalent set on a bar 1 inch square and 1 foot long:thus in our case 003 x 3.05 13.52 ·0000502 inch, as in col. 7, &c. = We thus find, that with a bar 13 feet long, 3·05 inches deep, 3.095 inches wide, a load of 224 lbs. produces by experiment a deflection of 195 inch, and a permanent set of 003 inch, and that this is equivalent to a load of 105 lbs. on a bar 1 foot long and 1 inch square, which load would give a deflection of ⚫00326 inch, and a permanent set of 0000502 inch, &c. Table 118 has been calculated in this way throughout. |
