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inches, L = 11 feet, both ends flat, we have 223 × (43 ̈·o — 313 ́6) ÷ 112, or 223 × (147 – 90·9) ÷ 121 = 103.5 tons F. Correcting for incipient crushing (164), the area of the section being 4.75 square inches, Cp becomes 4.75 × 19 90.25 tons, Cp 67-69 tons, hence Po = (103.5 × 90.25) (103.5+ 67-69) 54.56 tons breaking weight.

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With thin plates of wrought iron correction is required for incipient" Wrinkling" (249) rather than incipient crushing: in our case, however, the wrinkling strain is (√25√4) x 80, or (52) × 80 20 tons per square inch, which being in excess of 19 tons, the crushing strength of wrought iron in pillars (201), the strength is governed by the latter.

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(233.) "Rectangular Pillars of Wrought Iron."-For rectangular sections, other than square, the rules for square pillars are modified, and we have:

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L = √(Mp × 6 × b÷F).

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P

F × L2 ÷ (ť2·6 × b).

In which the letters have the same signification as in (177), namely F the breaking weight by flexure in lbs., tons, &c., dependent on the value of Mp as given by Table 34, t = the thickness or least dimension of the rectangular pillar, the 2.6 power of which is given by col. 3 of Table 35; b = the greatest dimension, and L = length in feet, &c.

(239.) Table 53 gives the results of twenty-one experiments on solid rectangular pillars of wrought iron by Mr. Hodgkinson; col. 9 has been calculated by the rule (234), the value of Mp for rectangular pillars with both ends flat being taken at 498,500, or say 500,000 lbs., from Table 34. Thus taking No. 9 as an example, to find 26, the logarithm of 995 or 1.9978 × 2.6 = 1.99428, the natural number due to which, or 9869, is the 2-6 power of t, and 7.52 being 56.25, the rule becomes 500000

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× 9869 × 5·86 ÷ 56.25 = 51410 lbs. F, or the breaking weight by flexure, as in col. 9. We shall find that this does not require correction for incipient crushing; the area of the section is 5.86 x 995 5.8307 square inches, and the crushing strength being 19 tons or 42,560 lbs. per square inch (201), Cp becomes 42560 × 5·8307 = 248100 lbs., as in col. 8, therefore Cp 62025, and as F is less that, namely 51,410 lbs., the correction is not required (164). The experimental breaking weight was 54,114 lbs., as in col. 5, hence we have 5141054114 = 95, showing a difference or error of 5 per cent., as in col. 7.

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TABLE 53. Of the STRENGTH of RECTANGULAR, SOLID PILLARS of WROUGHT IRON, both ends Flat.

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Το vary the illustration we will take a pillar very similar in section to the last, but only 5 feet long, say No. 14 in the Table 53: the logarithm of 996 = 1.99826 x 2.6=1.995476, the natural number due to which, or ⚫9896 = 126; then 500000 X 9896 x 5.8425 = 115600 lbs. = F, as in col. 9.

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Then the area ·996 × 5·84 = 5.8166 square inches, hence Cp becomes 5.8166 × 42560 = 247600 lbs., as in col. 8, Cp = 185700 lbs., and the rule in (164), gives (115600 × 247600) (115600185700) 94980 lbs., reduced breaking weight, as in col. 6. The experimental result was 102,946 lbs., the difference being 94980102946923, or 1·0·923077, showing an error of 7.7 per cent.

(240.) Table 53 has been calculated in this way throughout:— the sum of all the errors in col. 7 is 108.9, and of all the - errors, 109 4, they are therefore practically equalized. The greatest error is 22.4 per cent., and the greatest error = 34.8 per cent. (959).

(241.) Messrs. Kennard have made some experiments on wrought-iron pillars of cruciform, I iron, and other sections, the results of which are given by Table 54; unfortunately there is some doubt as to the form at the ends, some were cut away there, so as to approximate the case to a pillar with both ends pointed, and others approximate to the form of pillars with ends flat. All we can do is to calculate for both conditions, and it will be found that the experimental results are between those extremes which do not differ nearly so much as 1 to 3, (149) being governed by incipient crushing.

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(242.)" Wrought-iron Pillars of Section."-The cruciform pillars were of the sizes shown by Fig. 44, and were 5 feet long; they would therefore fail by flexure in the direction of the least dimension (2 inches), and the strength may be calculated as in (239) by the rule (234). Thus 22 10.84 by col. 3 of Table 35; for a pillar with both ends pointed Mp = 162900 by Table 34, then, 162900 × 10 84 × 375 662200 lbs. due to the rib a as a pillar one foot long. For the resistance of the rib c, 3260781 by col. 3 of Table 35, and the breadth b in the rule being 3 2.625, we obtain 162900 x .0781 x 2.625 33400 lbs. The sum of the two is 662200 + 33400 =

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TABLE 54.-Of EXPERIMENTS on WROUGHT-IRON PILLARS of T, L, and other Sections.

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695600 lbs., which for a length of 5 feet becomes 695600 5o = 27824 lbs., the breaking weight by flexure F. Reducing this for incipient crushing (164); the area of the whole section is 1.88 square inch, hence 1.88 × 42560 = 80013 lbs., the value of CP, and 60,010 lbs. Cp; then the rule (164) becomes Pc = (27824 × 80013) (27824 + 60010) 25350 lbs., the reduced breaking weight with both ends pointed, as in col. 6 of Table 54:-experiment gave 34,944 lbs., col. 5, which seems to show that the form at the ends did not conform to that condition.

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With both ends flat, F would be greater, in the ratio of the respective Multipliers Mp given by Table 34, and becomes 27824 x 500000 162900 = 85400 lbs., which corrected for incipient crushing becomes (85400 × 80013) ÷ (85400+ 60010) = 47000 lbs., col. 3; experiment gave 38,304 lbs., col. 4, which seems to show that the condition of both ends flat, so favourable to strength, was not fully attained. The small difference between the results of the two experiments, namely 38304 34944 = 3360 lbs. only, or about 9 per cent., shows that the form at the ends were nearly alike, although intended to be very different. The calculated difference was 47000 - 25350 =

21650 lbs., and the experimental difference would have been. about the same if the conditions assumed had been fulfilled. But it will be observed that the two experimental results fall between the two calculated ones.

(243.) "Wrought-iron | Pillars.”-Let Fig. 45 be the section of airon pillar of wrought iron, say 10 feet long, both ends flat. Such a pillar may fail by flexure in any one of the three directions shown by the arrows A, B, C, and of course it will select the one in which the resistance is the least, or where 16 x b has the lowest value.

In the direction A, we have (426 x 0.5) + (0.526 x 3.5) or (36.7 x 0.5) + (165 x 3.5) = 18.9275.

It should be observed that the rib R contributes only 3 per cent. to the strength: it might therefore be neglected with impunity thus (165 x 3.5) 18.927503, or 3 per cent. (186). In the direction of B, reckoning for wrought iron. from the line N, as we found it necessary to do in calculating the transverse strength (378), we have (3.56 × 0·5) + {426-3-520) × 4) = 55.8, or nearly three times the resistance in the direction of A: the pillar will therefore not give way in that direction unless it is forced to do so by the mode of fixing.

In the direction of C we must calculate for wrought iron from the line P, and 6 × b then becomes (0.526 × 4) + {42 -0.50) × 0.5} = 18.9275, or the same as in the direction A; the pillar may therefore fail in either direction indifferently. Then by the rule in (234) we have 223 × 18.9275 100 = 42.24 tons = F. Then for incipient crushing (164), the area of the section being 3.75 square inches, Cp becomes 3.75 x 1971-25 tons, and Cp 53.54 tons, hence (42.24 x 71.25) (42.24 + 53·44) = 31.46 tons breaking weight.

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(244.) Experiments were made on pillars whose section is given by Fig. 46, the length being 5 feet, one of them having flat ends or approximately so, and the other pointed. Assuming that the pillar will bend in the direction of the arrow D, and calculating as in (243), we have (326 x 375) +

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