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(0.37526 x 2.625) or (17.4 x 375) + (0781 x 2.625) = 6.73, which is 126 × b in the rule (234), hence 162900 × 6·73

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2543850 lbs. = F, or the breaking weight by flexure. The area of the section = 2.1 square inches, hence Cp 2.1 × 42560 = 89380 lbs., and Cp = 67030 lbs. :-then the rule in (164) becomes (43850 x 89380) (43850 + 67030) = 35350 lbs. breaking weight of a pillar with both ends pointed, col. 6 of Table 54: experiment gave 42,000 lbs., col. 5.

The same pillar with both ends flat, gives 500000 × 6·73÷ 25 134600 lbs. for the value of F, hence (134600 × 89380) ÷ (13460067030) = 59670 lbs. breaking weight, col. 3: experiment gave 62,989 lbs., col. 4.

(245.) "Wrought-iron ↳ Pillars."-This form of pillar frequently occurs in the struts of roofs and other structures, the determination of the strength is therefore a matter of considerable practical importance. We have first to find the direction in which such a pillar will fail by flexure, which of course will be the one in which it is the weakest.

Let Fig. 48 be the section of a 3-inch angle-iron, inch thick, which as a pillar may fail by flexure in one of three directions indicated by the arrows A, B, C. To find 126 × b in the direction A we must calculate from the line N for wrought iron (378), and we have (326 × 3) + {326 – 32·6) ×

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6.73. In the other

direction B, we must calculate from the line P, and we obtain

{326 - 252) × 3} + (256 × 3) or {17·4 — 12·28) × 3} +

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(12.28 × 375) 19.965, or nearly three times the strength in the direction A. For the direction C, t must be measured angle-wise, and in our case becomes 24 inches as in the figure, the breadth b becoming the sum of the thicknesses of the two ribs, or inch. We thus obtain 2126 × 3 or 8.23 ×·75 6.1725, which being less than either of the others, shows that the pillar will fail in the direction C.

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(246.) Experiments were made on pillars whose section is given by Fig. 47; there were four, flat at both ends, the lengths

L

being 11, 3, 4, and 5 feet respectively; one was 5 feet long with both ends pointed or approximately so. Admitting that the pillar will fail by flexure in the direction of the arrow as in (245), t will be 21 inches, and b = inch, and taking Mp from Table 34 at 223 tons, and 216 at 8.23 from Table 35, we obtain for 1 feet long 223 x 8.23 × 625÷2·25 = 510.2 tons for the value of F. This requires correction for incipient crushing (164): the area being 1.78 inch, and the crushing strength of wrought iron in pillars 19 tons per square inch (201), Cp becomes 1.78 x 19 = 33.82 tons, and Cp: = 25 36 tons; then (510 2 x 33.82) (510-225-36) = 32.22 tons, col. 3: experiment gave 31.3 tons, col. 4. Table 54 has been calculated in this way for the several lengths, from 1 to 5 feet.

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For the pillar with both ends pointed, Mp = 72.7 from Table 34, and the rule becomes 72.7 x 8.23 ×·625 ÷ 25 14.96 tons F: then (14.96 × 33·82) (14.96 + 25·36) · 12.55 tons, col. 6, the reduced breaking weight: experiment gave 12.25 tons, col. 5

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(247.) "Wrought-iron Pillars."-Experiments were made on two pillars whose section is given by Fig. 49. Such a pillar might fail by flexure in one of three directions as indicated by the arrows A, B, C. In the direction A, the thin ribs R, R would be subjected to compression, therefore (378) we must calculate the strength from the line N, and 6 × b by which the strength is governed, becomes (36 × 3) + {13a°— 3oo) × }}

or (0781 × 3)+{4.28

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32-6

0781) 75) = 3.3857. In the direction B, we have (1326 × 3) - (136 × 24) or (4·28 × 3) – (2.28 × 2·25) = 7.71, or about double that in the direction A. In the third direction C, we obtain (326 × 13) – (212·6 × 13) or (17.4 x 1.75) (8.23 x 1.375) = 19 134, or nearly six times the strength in the direction A. We find from this that the strength of the pillar must be calculated for A; then for both ends flat Mp 223 × 3.3857 ÷ 25

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223, and for a length of 5 feet we have = 30·23 tons = F. Reducing for incipient crushing, the area being 2.15 square inches, Cp becomes

2.15 x 19 40.85 tons, and Cp = 30.64 tons, hence (30-23 x 40-85) (30.23 + 30.64)= 20.28 tons, col. 3. Experiment gave 17.1 tons, col. 4.

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For the same pillar with two pointed ends we obtain 72.74 x 3.385725 9.85 tons = F, which being less than Cp, correction for incipient crushing is not required (169), and the breaking weight is 9.85 tons: experiment gave 14 tons, col. 5.

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The experiments gave 17.1 and 14 tons respectively; the difference is only 3.1 tons, whereas the calculated difference is 20.289.85 10.43 tons, which shows that in the experiments the assumed form at the ends was not complied with perfectly in either case (241). Here again the experimental results fall between the calculated ones, as in (242).

Table 54 gives a collected statement of the results of experiment and calculation on these pillars of unusual sections (242) to (247).

"Hollow Rectangular Pillars of Wrought Iron."-For rectangular pillars other than square the rules in (230) become :

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(248.) F = Mp × {t2 × b ) − (126 × b} ÷ L2.

In which the least dimension of the pillar externally, and to internally; b = the breadth or largest dimension externally, and b, internally, all in inches, and the rest as in (233), namely L = length in feet, Mp multiplier, whose value is given by Table 34, and F = the breaking weight by flexure, in lbs. or

tons dependent on Mp.

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Thus for a pillar 3 × 4 externally, and 2 × 3 internally, in which t = 3; b

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4; to

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21; b = 31; L

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9 feet, both ends being flat, we have 223 × (326 × 4) - (2126 × 31)÷81,

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or 223 × {17·4 × 4) – (10·84 × 3·5)÷81 = 87·24 tons = F.

Correcting for incipient crushing, the area being 3.25 square inches, Cp becomes 3.25 x 19 = 61.75 tons, Cp = 46.31 tons, therefore by the rule in (164), (87.24 x 61.75) (87.24+ 46.31) 40.34 tons breaking weight. In this case, as in (232), correction for incipient "Wrinkling" is not required.

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(249.) "Incipient Wrinkling."-It is shown in (306) that thin wrought-iron plates, where the breadth is considerable, will fail by wrinkling or corrugating under a compressive load, with a strain very much less than the absolute crushing strength of wrought iron. For plates forming the sides of a pillar, we have the rule:

(250.)

Ww = (√t÷√b) × 80.

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In which t = the thickness of plate, and b = the breadth unsupported, both in inches, Ww the compressive wrinkling strain in tons per square inch. Thus for a plate inch thick, and 9 inches wide, forming one side of a pillar 9 inches square, we have (√ √9) × 80, or (35353) × 80 = 9.427 tons per square inch, being less than half the crushing strength of wrought iron in pillars (201), which is 19 tons per square inch. When, however, the plate is thick in proportion to the breadth, the wrinkling strain may become greater than the crushing strength, and in that case the strength of the pillar is governed by the latter. An example of this is given in (409). Table 63 gives the wrinkling strain for wrought-iron plates of various thicknesses and breadth when forming part of a pillar particularly, for, as shown in (321) and Table 62, the resistance of plate-iron in beams is greater than in pillars in the ratio of 104 to 80.

(251.) It should be observed that the strength of a pillar is governed by that of the weakest plate:—for instance, in a rectangular pillar whose sides are in the ratio of 2 to 1, with equal thickness all over, the narrow end plates are stronger than the sides in the ratio of √2 to 1, or 1.414 to 1.0, but when the wide plate fails by wrinkling under the strain, the whole of the load is thrown upon the end plates, and they fail under that increased load in spite of their superior strength, which in such a case goes for nothing. Evidently, the most judicious course is to make the wide plates thicker than the narrow ones, so as to produce equality of strength all over; and as the resistance to wrinkling is proportioned to √√b by the rule in (250), it follows that the thickness of plate should be simply propor

tionate to the breadth unsupported, so that for breadths in the ratio of 1, 2, 3, the thicknesses should be in the ratio 1, 2, 3 also.

When it is thus found that the Wrinkling strain is less than the crushing, the correction of F must be made by the rule:

(252.)

P = F x C÷÷ (F + ₫ Cw).

W

W

But when the crushing strain is less than the wrinkling, the rule becomes :—

(253.)

Po FX Cp÷ (F + 3 C2).

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In which F = the breaking weight of the pillar due to flexure by the rule in (248), &c.; C = the specific crushing strength of wrought iron per square inch, namely 19 tons, or 42,560 lbs. ; Cp = resistance of the pillar to crushing, due to the area of the section, and the value of C; Ww the wrinkling strain in lbs., tons, &c., per square inch, which varies with the thickness and breadth of the plate, and may be calculated by the rule in (250) or (308). Cw the wrinkling strain on the whole pillar, due to the area of the section, and the value of Ww; Po the breaking weight of the pillar reduced for "Incipient Crushing' in tons, lbs., &c., dependent on the terms of F and C; Pw breaking weight of the pillar reduced for "Incipient Wrinkling" in tons, lbs., &c.

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Of course it will be understood that the whole must be taken in the same terms; for instance, if F be taken in Tons, all the rest must be in Tons also.

(254.) "Experimental Results."-Table 55 gives the results of 29 experiments by Mr. Hodgkinson on square and rectangular pillars of thin wrought-iron plate; col. 9 gives the calculated breaking weights by flexure, or F; the square pillars by the rules in (230), and the rectangular ones by those in (248).

(255.) "Square Pillars."-As an example of the former we may take No. 4, whose section is shown by Fig. 50; to find 83, we have the logarithm of 8.1 = 0.908485 x 3.6 = 3.270546, the natural number due to which = 1864. Then for 836, the logarithm of 7.98 = 0·902003 × 3.6 3.247211, the natural

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