TABLE 66.-Of the TRANSVERSE STRENGTH of Bars 1 inch square, 1 foot long : value of M, for Breaking load. = working load of 1900 lbs. in the centre :—we may take Mr 78 lbs. from col. 3 of Table 67, and the rule (325) becomes W 1900 x 10) = (78 x 3) = 9 inches, the depth required. Again: to find the breadth for a beam of Pitch-pine, 12 feet long, 10 inches deep, to break with 24,000 lbs. in the centre :taking My from col. 5 of Table 66, at 577 lbs., rule (326) becomes (24000 x 12) = (100 x 577) = 5.16 inches, the breadth required, &c. TABLE 67.–Of the TRANSVERSE STRENGTH and STIFFNESS of BEAMS, 1 inch square and 1 foot long. "Limit of Elasticity": Wrought-iron = , and Steel = the Breaking-down loads. I of Breaking-down load. “Old Rule.”—The rule (329) shows that in hollow square D4 – d'. sections the strength is proportional to The old rule D very commonly used by practical men is : (337.) W = (D – d) M.: L. > or In this rule a hollow beam is considered as composed of two solid beams, one having the external, and the other the internal dimensions; then the strength of the smaller one subtracted from that of the larger was supposed to give the strength of the hollow one. But it is overlooked that the ultimate deflection is inversely proportional to the depth (694), and that the full strength will not be realised if that deflection is not permitted. For instance, in Fig. 64, the hollow beam A is supposed to be composed of the two solid beams B, Fig. 65, and C, Fig. 66, but the ultimate deflection of the two latter will be in the ratio 1 to 2. Now, when combined as in A, and when breaking with the deflection due to B, the deflection of C is half only of that due with its own breaking weight, therefore half only of its strength is to be subtracted from the full strength of B in order to find the real strength of A. By the old rule the strength of A would be 48 – 29, or 64 - 8 56, but allowing half of C only, we obtain 64 – 4 = 60 as the actual strength. We should D4 – dll obtain the same result by the rule (329), namely D 4* 24 256 16 240 60, as before, 4 4 4 which is 60 = 56 = 1:07, or 7 per cent. in excess of the old rule. (338.) Again : in Fig. 67, the internal dimensions are of the internal ; for the former we have 4 = 64, and of the latter becomes 3 x = 20.25; hence the strength of the hollow 44 – 34 beam is 64 20.25 = 43.75. The rule (329) gives 4 256 81 175 = 43.75 also. The old rule gives 4 3.19 37, the correct rule being 43.75 = 37 = 1.18, or 18 per cent. in excess of the old one. Again : in Fig. 68, the internal sizes are { of the external; hence, instead of deducting 3.18 = 42.875, we have to deduct 42.875 x 7:8 = 37.5, and we obtain 64 – 37.5 = 26.5 for 4' - 3.5" 256 150 the hollow beam. By rule (329) 4 4 in our case or or or or or or 106 = 26.5 as before. By the old rule 48 – 318 = 21. 125; 4 the correct rule being 26.5 - 21.125 = 1.254, or 25.4 per cent, in excess of the old rule. From all this we find that the old rule is entirely incorrect; and, further, we have the result that the error increases with the relative thinness of the metal. Thus with metal 1, 7, and 1 inch thick, the error in our case was 7, 18, and 25.4 per cent. respectively. (339.) The same erroneous method of calculation (337) is very commonly applied to girders of the ordinary flanged type. Thus in Fig. 69, A is the section of a girder, which is frequently considered as composed of two plain sections B and C:then for B by the rule (324) we have 92 x 3 = 241, and for C, 78 x 21 = 110.25, from which A becomes 241 110.25 = 130:75. But by the correct rule (330), we obtain (98 x 3) - (78 x 21) = 157.3; a difference of 157.3 : 130.75 9 = 1.20, or 20 per cent. SPECIAL RULES FOR CAST IRON. (340.) The Rules in (323) are perfectly correct for all materials so far as solid sections are concerned, but for hollow tubular, and ordinary I and I sections in cast and wrought iron they are not correct except for very light strains. The rules are absolutely correct for those cases only where the Tensile and Crushing strength of the material and the corresponding extensions and compressions are equal to one another, and this, as shown by Table 79, is not the case with any known material when strained nearly to the breaking point. With cast iron the ratio of those strains is 1 to 6, and special rules become necessary when we would calculate the breaking weight. With wrought iron there is much greater equality between T and C, and the ordinary rules would be nearly correct but for the fact that in thin plates of wrought iron there is a tendency to wrinkle or become undulated under a compressive load with a strain much less than is necessary to crush the material. The great strength of cast iron in resisting compression, and the great weakness of wrought iron, necessitate special rules for both materials, differing from one another, and differing also from the ordinary rules for other materials whose tensile and crushing strengths are more nearly equal. The necessity for special rules will be made apparent by comparing calculation with experiment. It is probable that with the working loads commonly adopted in practice, say krd of the breaking weight, the ordinary rules will be sufficiently correct for practice (353); the usual course, however, is to calculate the breaking weight, and then to find the working load by the use of the “Factor of Safety” (880). (341.) The best rule we can give for Cast-iron beams is an Empirical one, and is based on the assumption that the resistance to compression is infinite, and as a result, that the neutral axis coincides with the edge of that part of the section subjected to compression. This assumption is manifestly not absolutely true; nevertheless, we shall obtain with cast iron more correct results on that hypothesis than with any other. (342.) “ Beams of I and T Section.”—The best evidence of the necessity for special rules for cast-iron beams is given by sections of this form. Let A, in Fig. 70, be such a section. By the old method of calculation (339) this would be regarded as two solid beams, B and C; for B we have 82 x 6 = 384 ; 72 x 5 245; hence A becomes 384 – 245 = 139, which is the reduced value of D2 x B. Now, if we reverse the position as at D, we should by this mode of calculation obtain precisely the same result, whereas it is well known by experiment that with cast iron there is a great difference of strength in the two positions, A being much stronger than D. By the new method of calculation we must calculate in both cases from the line N, A; then with A we have for the vertical web 72 x1 = 49, and for the bottom flange (82 – 7) x 6 90; the sum of the two = 139, or the same as by the method. But with D we have for the top flange 12 x 6 6, and for the vertical web (8? – 1o) x 1 63; the sum of the two is 6 + 63 = 69, which is very nearly half the strength in the other position, namely 139. (343.) Mr. Hodgkinson made experiments on beams E and F for C, |