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To 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 = 42-6; then 500000 x •9896 x 5.84 • 25 = 115600 lbs.

F, as in col. 9. Then the area = .996 x 5.84 = 5.8166 square inches, hence Cp becomes 5.8166 X 42560 247600 lbs., as in col. 8, i Cp = 185700 lbs., and the rule in (164), gives (115600 x 247600) • (115600 + 185700) = 94980 lbs., reduced breaking weight, as in col. 6. The experimental result was 102,946 lbs., the difference being 94980 = 102946 = .923, or 1.0 – 923 = .077, 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 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,

(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 2126 = 10.84 by col. 3 of Table 35; for a pillar with both ends pointed Mp = 162900 by Table 34, then, 162900 x 10.84 x .375 = 662200 lbs. due to the rib a as a pillar one foot long. For the resistance of the rib c, 32:6 = .0781 by col. 3 of Table 35, and the breadth b in the rule being 3 – 0.375 = 2:625, we obtain 162900 X •0781 x 2.625 = 33400 lbs. The sum of the two is 662200 + 33400

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

and other Sections.

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47,000 lbs. 38,304 lbs. 34,944 lbs. 25,350 lbs.
21.0 tons 17:1 tons 15.6 tons 11.32 tons
59,670 lbs. 62,989 lbs. 42,000 lbs. 35,350 lbs.
26.65 tons 28.12 tons 18.75 tons 15.78 tons
21.79 18.75
24.99 23.43
28.21 28.12
32.22 31.30

12:25

12.55 20.28 17.10

14.00

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695600 lbs., which for a length of 5 feet becomes 695600 = 52 = 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 X 42560 = 80013 lbs., the value of Cp, and 60,010 lbs.= { Cp; then the rule (164) becomes Po= (27824 x 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.

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 x 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 1 Pillars.”—Let Fig. 45 be the section of a I iron 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 196 x 6 has the lowest value.

In the direction A, we have (496 x 0.5) + (0.5" 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) - 18.9275 = .03, 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 (3.526 x 0.5) + {4** – 3.589) * 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 x b then becomes (0·526 x 4) +

0.5*9) * 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 x 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 19 = 71.25 tons, and Cp = 53.54 tons, hence (42.24 x 71.25) = (42.24 + 53.44) = 31.46 tons breaking weight.

(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 (38-6 X •375) +

(0.37526 x 2.625) or (17.4 x .375) + (0781 x 2.625) = 6.73, which is f-x b in the rule (234), hence 162900 x 6.73 • 25 = 43850 lbs. = F, or the breaking weight by flexure. The area of the section = 2•1 square inches, hence Cp = 2:1 x 42560 = 89380 lbs., and i 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 x 6.73 • 25 = 134600 lbs. for the value of F, hence (134600 x 89380) • (134600 + 67030) = 59670 lbs. breaking weight, col. 3: experiment gave 62,989 lbs., col. 4.

(245.) “ Wrought-iron L 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 t28 x b in the direction A we must calculate from the line N for wrought iron (378), and we have (3** * 3) + {3** – J*) * 3} (•0781 x 3)+{17•4 - .0781) * •375} = 6·73. In the other direction B, we must calculate from the line P, and we obtain {3*** – 25+4) R 3} + (29** * ?) or (17•4 – 12-28) 3} + (12-28 X375) = 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 27 inches as in the figure, the breadth b becoming the sum of the thicknesses of the two ribs, or 4 inch. We thus obtain 2126 x 1 or 8.23 x .75 = 6.1725, which being less than either of the others, shows that the pillar will fail in the direction C.

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

or

L

being 13, 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 = sinch, and taking Mp from Table 34 at 223 tons, and 272-8 at 8.23 from Table 35, we obtain for 14 feet long 223 x 8.23 x.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 i Cp = 25.36 tons; then (510.2 x 33.82) = (510.2 + 25.36) = 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.

For the pillar with both ends pointed, Mp = 72.7 from Table 34, and the rule becomes 72.7 X 8.23 x .625 = 25 = 14.96 tons = F: then (14.96 X 33.82) - (14.96 + 25•36) = 12.55 tons, col. 6, the reduced breaking weight: experiment gave 12:25 tons, col. 5

(247.) “ Wrought-iron uPillars.”—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 t26 x b by which the strength is governed, becomes (3** 3) + (1.346 – 3*)* }} or (-0781 x 3) + {4:28 – • 0781) * •75} = 3.3857. In the direction B, we have (1726 x 3) - (1325 x 21) or (4.28 x 3) (2.28 x 2.25) = 7.71, or about double that in the direction A. In the third direction C, we obtain (32-6 x 17) - (2126 x 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, and for a length of 5 feet we have 223 x 3.3857 = 25 = 30.23 tons F. Reducing for incipient crushing, the area being 2.15 square inches, Cp becomes

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