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being 200 inches, or 16.667 feet, T and C each = 10 tons, d = 20 inches, b 1 inch, we obtain N 10

x 20 ) x1 x 10 = (16.667 x 4.5) = 13.333 toris, W10+ 10

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x1

) ,

as before.

With a bar 1 inch square and 1 foot long of the same material, we obtain

w =(

(110

10

X1 w 10 + N 10

1)' x1 x 10 = (1 x 4-5) = · 555 ton

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breaking weight in the centre. This it should be observed is Isth of T or C, for • 555 X 18 = 10 tons :-this, however, is true only where the tensile and crushing strengths are equal to one another (638).

(501.) But the Rules in (496), &c., will apply to cases where T and C are unequal, which, as shown by Table 79, is the case with most materials. Thus, with cast iron the mean value of T = 7.142 tons and of C = 43 tons per square inch : then a bar 1 inch square and 1 foot long will break transversely with 43

Х x1 , W 43 + 7.142 or 1795 lbs. in the centre. The mean transverse strength by experiment is 2063 lbs. (335), hence 2063 = 1795 = 1.15, so that the calculation shows an error of + 15 per cent.

No (502.) The equation

x d gives the depth of that

✓C+T part of the section subjected to tension ; thus with cast iron 43

6.557

6.557 x1, or

=7104 inch; ✓ 43+ 7.142 6.557 +2.673'

9.23 hence the depth of the part subjected to compression must be 1:0 – •7104 = 2896 inch, and thus we obtain the position of the neutral axis, as in Fig. 168. Similarly, the equation NT

x d gives the depth of that part of the section NT + N

or

or

or

or

2.673 subjected to compression, or in our case

2.673 + 6.557? 2.673

=•2896 inch, as before.' 9.23 Again, with Ash, T = 16576 lbs., and C

9023 lbs. per sqnare inch, and a bar 1 inch square will have as the depth

N 9023

95 subjected to tension =

9023 + 16576' 95 + 128.7' 95

= .4247 inch, hence the depth subjected to compression 223.7 must be 1.0 - •4247 = •5753 inch, and we thus obtain the position of the neutral axis, as in Fig. 169.

Of all known materials, glass gives the greatest inequality of T and C, their ratio to one another being 1 to 11.78, as shown by col. 3 of Table 79. With T = 2560 lbs., and C

vo 30150 lbs. per square inch, the equation

becomes

NO + NT
N 30150

173.6

173.6

=7734 inch, 30150 + 2560' 173.6 + 50:6 224.2 which is the depth subjected to tension, hence 1.0 -7734 = • 2266 inch for the depth subjected to crushing, and we thus obtain the neatral axis, as in Fig. 170.

(503.) The rule (498) for finding C, is of special value as applied to malleable materials such as Wrought Iron, Steel, Gun-metal, and Brass, whose resistance to crushing cannot be determined with precision by experiments conducted in the usual manner, namely, by crushing small specimens by direct pressure, because the ductile and semi-fluid character of such metals enables them to flow or expand laterally under pressure to an almost unlimited extent, instead of crushing suddenly into fragments as cast iron and similar materials do (133).

(504.) “ Wrought Iron.”—In applying the Rules (496), &c.,. to such materials we obtain the apparent rather than the real tensile and crushing strengths; at least this is true with heavy strains approaching the ultimate or breaking loads. It is assumed in the analytical investigation in (494) that the exten

or

or

T

sions and compressions are proportional to the distance from the neutral axis, and that the tensile and crushing strains are simply proportional to the respective extensions and compressions. This last assumption is practically correct for light strains or even up to half the ultimate weight, with wrought iron and steel, but is very far from the truth with heavier strains as shown by Table 95: plotting col. 3 in a diagram and eliminating the anomalies of experiment by a mean curve we obtain for

14 15 16 17 18 19 20 21 22 23 24 tons per square inch, the approximate extensions in parts of the length, cleared as far as possible from the effect of time (621),

are

.002.0035 .0054 .0076 0102 .0132 .0164 0198 024 .03 038

The extensions for lower strains are given by col. 4 of Table 96.

(505.) Let Fig. 203 be the section of a bar of wrought iron below the neutral axis when strained as a beam up to the point of rupture, the maximum strain or that at B being say 24 tons per square inch. Now, if the strains were simply as the extensions, we should have the series given by col. A in that figure, proceeding with which as in (495) we obtain column X. Area. Lever. Strain. X.

Area Lever. Strain. 1 x 1 x

1.5

1 x 1 x 14.25 = 14.25 1 x 3 x

4.5 =

13.5 1 x 3 x 16.70 50.10 7.5 =

37.5 1 x 5 X 18.53 = 92.65 1 x 7 x 10.5 73.5 1 x 7 X 19.55 = 136.85 1 x 9 x 13.5 = 121.5 1 x 9 x 21.28 191.52 1 x 11 x 16.5 = 181.5 1 x 11 x 22.25 = 244.75 1 x 13 x 19.5 253.5 1 x 13 x 23.00 299.00 1 x 15 x 22.5 = 337.5 1 x 15 x 23.70 = 355.50

Z.

1.5 =

1 x

5 X

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Thus taking B C and assuming the area = 1:0, as we require proportional numbers only, the mean strain at D becomes 22.5, and the leverage = 15, hence we obtain 1 x 15 x 22.5 = 337.5, &c., &c., as in column X.

But by (504) the extension at B due to 24 tons = .038, therefore at E, or half the distance from the neutral axis, we obtain •038 • 2 = .019, which, by the series of strains and corresponding extensions in (504), is due to a strain somewhere between 20 and 21 tons: by interpolation we find the exact strain to be 20.76 tons, as in the figure. Similarly at F, which is įth of the distance of B from the neutral axis, we obtain - 038 = 4 = .0095, which is between 17 and 18 tons : by interpolation we obtain 17.73 tons, as in the figure. Calculating in this way, we have obtained the column N: then taking B, C as before, we have 1 x 15 x 23.7 = 355.5, &c., &c., as in col. Z.

(506.) The result is, that although the maximum strain at B is the same in both cases, namely 24 tons, the mean resistance by col. Z is greater than in col. X in the ratio 1 to 1384.62 • 1020 = 1.36 nearly, or 36 per cent. Therefore, if we would calculate wrought-iron bars by the rules in (496), &c., we must take a fictitious value for the maximum strain at B, which would become in effect 24 x 1.36 = 32:64 tons per square inch. Thus, while the real maximum tensile and crushing strain = 24, the apparent strain by the ordinary rules is = 32.64 tons per square inch : see (133).

The near approach to equality in the strains throughout the section, as shown by col. N in Fig. 203, is remarkable: thus at G, toth of the distance of B from the neutral axis, the strain instead of being 24 = 16 = 1.5, is 14.25 tons per square inch.

By (374) it is shown that a bar of wrought iron 1 inch square and 1 foot long between bearings, and loaded as a beam, will break down (so far as that point can be definitely fixed) with 4000 lbs. or 1.786 ton in the centre. By the rule in (638) this is equivalent to 1.786 x 18 32.15 tons per square inch maximum strain at the top and bottom edges of the section, which far surpasses 25.7 tons, the mean tensile strength of British bar-iron, as determined by direct experiment and given by Table 1. But 32·15 is the apparent, not the real, strength of the iron, which by the ratio in (506) is reduced to 32.15 • 1.36 = 24 tons, differing 24 - 25•7 = .934 or 1.0 – 934 = .066, namely 6.6 per cent. only from the tensile strength by direct experiment (520).

With lower strains, say half the ultimate strength or 24 = 2 = 12 tons per square inch, the elasticity of wrought iron is practically perfect, that is to say, the strains are simply proportional to the extensions or to the distances from the neutral axis N. A. We should then have the ratios given by col. X in (505), which gives 1020 with the full strain, therefore 1020 = 2 = 510 with half the ultimate strength, and by col. Z 1384.62 with the full ultimate strain: hence the ratio becomes 1384.62 x 510 = 2.72 to 1.0: while, therefore, we double the maximum strain (namely 12 to 24 tons), we increase the mean apparent strain not to double only or to 2, but to 2.72 or to 2.72 x 12 = 32.64 tons, being the same as in (506).

(507.) “Steel.- Applying the rules in (496), &c., to steel, we obtain results analogous to those we have found for wrought iron: for example, a bar 1 inch square and 1 foot long breaks down with 6720 lbs., or 3 tons in the centre, as shown in (376). By Table 1, the mean tensile strength of a steel bar = 47.8 tons per square inch: then the Rule (498) becomes

3x1 x 4.5 C=

= 61.48 tons per {1 – V3 x 1 x 4.5 = (47.8 x 1}' x1 square inch apparent crushing strength. What the real crushing strength may be we have no means of determining exactly, because the compression of steel by that strain is not accurately known; but by the experiments on steel pillars, the mean resistance to crushing seemed to be 52 tons per square inch (268). If we admit this to be the real strain, and 61:48 tons the apparent strength under transverse strains, we have the ratio 61.48 ; 52 = 1.18, or 18 per cent. difference, being half of that obtained for wrought iron under similar strains.

(508.) Table 79 gives the results of the rules in (496), &c., as applied to many different kinds of materials, compared with the transverse strength, &c., as found by direct experiment. The values of T, C, and W are taken from Tables 1, 31, 32, 66, &c.;—the ratios of T to C, as given by col. 3, vary from 1 to · 34 with Willow, to 1 to 11.78 with Glass, and although there are in col. 6 departures from uniformity, they are not greater than might be expected under such extremely variable condi

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