=

in Fig. 71; E broke with 1008 lbs.: and F with 270 lbs.; the ratio being 1008270 = 3.73 to 1. The length between supports was 4.25 feet. Calculating from the line N, A, in both cases we have with E, for the vertical web 1·12 × 302, and for the bottom flange (1·35o 1·1o) × 4 = 2.45: the sum of the two is 302 +2.45 = 2.752, which is the reduced value of D2 x B in rule (324), and taking MT = 2063 lbs., we obtain 2.752 x 20634.25 = 1336 lbs.

In the position F we have for the top flange 252 × 4 = ·25, and for the vertical web (1.352 — ·252) × ·25 = ·44:—the sum of the two = · 25 ·44 =69, which is the reduced value of D2 × B, and rule (324) becomes 69 × 2063 ÷ 4.25 335 lbs. The ratio of the strength in the two positions is 1336 335 3.988 to 1: experiment gave 3.73 to 1.

=

(344.) In Fig. 72 we have sections of similar beams experimented upon by Mr. Hodgkinson, the length between supports being 6 feet; G broke with 1120 lbs., and H with 364 lbs., the ratio being 1120364 3.08 to 1. Calculating from the line N, A as before, with G we have for the vertical web 1.26 x 365 = 58; for the bottom flange (1.562 1.262) x 54.23. The sum of the two 58 + 4·23 = 4.81, with which rule (324) becomes 4·81 × 2063 ÷ 6·5 = 1526 lbs. In the position H, we have for the top flange 32 × 5 = ·45; for the vertical web (1.552·32) × ·36 = ·8325:-the sum of the two = ·45 + ·8325 1.2825, with which rule (324) becomes 1.2825 × 2063 ÷ 6·5 = 407 lbs., giving as the ratio of strengths in the two positions, 1526407 3.75 to 1: experiment gave 3.08 to 1.

=

Considering the extreme disproportion between the flange and web in these experimental beams, whose forms were not such as would usually be found in practice, but were designed for the special purposes of research, the calculated results are perhaps as correct as could be expected.

(345.) "Hollow Rectangular Beams."-Let Fig. 73 be the section of a hollow beam 4 inches square externally, 3 inches internally, 6 feet long, MT 92 ton. Calculating from the neutral axis N, A, D2 x B becomes for the top plate a, 52 × 4 = 10; for the two sides b b, (312 — 12) × 1 = 12.0; and for

the bottom c (42

1 + 12 + 15 =

- 312) × 4 = 15.0. The sum of the whole is 28, with which the rule (324), or W = D2 × B × MTL, becomes 28 x 9264·293 tons breaking weight in the centre.

By the old rule (337) we should have had (43 - 33) ×·92

65.67 tons. By rule (329) we obtain 66.71 tons.

(4-3)

4

(346.) We have thus obtained three very different results, and in the absence of experiment should not know which was correct; fortunately, we have Mr. E, Clark's experiments on hollow beams of various sections by which the various rules may be tested.

8

8

Fig. 74 is the section of a hollow square beam, the mean breaking weight of which with a length of 6 feet was 2.152 tons. There were three experiments which gave 2·0, 2·05, and 2.405 tons respectively. Calculating from the line N, A, as in the last example, we have for the top 32 × 3 = 439; for the two sides (23232) x = 5.666; and for the bottom (312 3 232) × 31 = 6.885. The sum of the three is 12.99: then rule (324) becomes W = 12.99 x 9262 tons nearly, agreeing precisely with one of the experiments, but differing from the mean of the three 2.0 2.152 = 929, giving an error by Special Rule of 1·0 — ·929 = .071 or 7.1 per cent.

If we calculate the same beam by rule (329), we obtain 31234

W =

31

X.926 = 3.119 tons: hence 3·1192.152

= 1·45, or an error of +45 per cent. by the ordinary rule. (347.) Fig. 75 gives the section of hollow rectangular beams, the mean breaking weight of which by Mr. Clark's experiments was 2.3 tons, the length being 6 feet. There were four experiments, the maximum 2.45, and the minimum 2.2 tons. Calculating from the line N, A, we have for the top, 32 x 2.21 3107; for the two sides (3.6652 - 32) × for the bottom (4.042 - 3.6652) × 2.21

=

=

=

8

9.975; and = 6.033. The sum

of the whole is 16.319; then the rule becomes 16 319 x ⚫92 62.502 tons; hence 2.502 ÷ 2·3 = 1·088, or an error of 8.8 per cent.

By rule (330) we obtain

×·926

=

3.558 tons:-hence 3.558 ÷ 2·3 = 1.547, or

an error of + 54.7 per cent. by the ordinary rule.

=

=

7.14 tons per

(348.) "Circular Sections."-For circular sections, or cylindrical beams, we must still calculate from the top edge of the section, but shall have to resort to analysis in the manner explained and illustrated for rectangular sections in (494). Taking first a solid round bar 1 inch diameter, and 1 foot long, we know (335) that a square bar of those dimensions would break with 92 ton, and admitting for the breaking weight 1.0 to 1.5 as the experimental ratio of the strength of round to square bars (361), a round bar would break with 92 ÷ 1.5 6133 ton. Now by analysis, we have to reckon from the line N, A in Fig. 76; the maximum tension at C square inch (4), therefore at B = 7·14 ÷ 2 area 7854 square inch, hence 7854 x but it is shown in (495) that to obtain the take of that product or 2.8 × 4÷3 ducing the case to a cantilever of half the length of the beam, as in Fig. 92, we find that the leverages by which the strain of 3.733 tons acts are inch and 6 inches, then the strain at the end of the cantilever becomes 3.733 × 6·3111 ton, which is equivalent to 3111 x 2 = 6222 ton in the centre of the beam 1 foot long, differing little from 6133 ton, as we found before.

=

3.57 tons. The

3.57

= 2.8 tons:

true mean, we must = 3.733 tons. Re

(349.) Applying this method of calculation to hollow cylindrical beams; Fig. 77 gives the section of three which by Mr. Clark's experiments broke with 2.0875, 2.358, and 2·416 tons respectively, the mean being 2.287 tons, with a length of 6 feet. The area of the section = 4.12 square inches, hence 4.12 x 3.57 × 1 19.6 tons, which with leverages of 2 and 36 inches, gives 19.6 x 2÷36 1.09 ton at the end of a cantilever 3 feet long, equivalent to 1.09 x 22.18 tons in the centre of a beam 6 feet long. Experiment gave as the mean 2.287 tons; hence 2.182.287 = 9532, or 1.0 — ·9532 0468, giving an error of 4.68 per cent.

=

=

=

By rule (332) taking the value of Mr for the breaking weight

T

(335) of circular sections at ·92 ÷ 1·5 = ·6133, we obtain 3.8751 3.1251

3.875

×·61336 = 3.432 tons: hence 3.432 ÷ 2.287 = 1.50, or an error of + 50 per cent. by the ordinary rule.

(350.) "Girder-sections."-The same method of calculation will apply to girders of all sections with equal or unequal flanges. Fig. 78 is the section of one which by Mr. Hodgkinson's experiment broke with 6678 lbs. in the centre; the weight of the beam itself between supports 4 feet asunder was 32 lbs., equivalent to 16 lbs. in the centre, giving a total of 6678 + 16 = 6694 lbs. Calculating from the line N, A, and taking Mr = 2063 lbs., we have for the top flange 422 × 1·76 = ·31; for the vertical web (4.7352 •422) × 29 6.45; and for the bottom flange (5·1252 - 4.7352) × 1·76 = 6.758: the sum of the whole is 31 +6.45 + 6·758 = 13.518 which is the reduced value of D2 × B in rule (324), which then becomes 13.518 × 20634.5 6197 lbs.: experiment gave 6694 lbs., hence 61976694 ·9258, showing an error of 1·0 –·9258 = .0742, or - 7.42 per cent.

=

=

By rule (330) we should have had

W =

(5.125 x 1.76) (4.3153 x 1.47)

5.125

× 2063 ÷ 4.5 = 10630 lbs.: hence 10630 ÷ 6694 = 1·588, or an error of +58.8 per cent. by the ordinary rule.

(351.) As an example of the application of the special rules to girders with unequal flanges, we may take Fig. 79, which gives the section of girders of the proportions recommended by Mr. Hodgkinson, the ratio of the areas of flanges being exactly 1 to 6. Mr. Owen, the Government Inspector of Metals, made a series of 13 experiments on these girders, with very various kinds of British cast iron, some pure and others mixed; this fact, together with the large scale of the experiments, enhances their value very much. The general results are given by Table 68 the maximum = 471 tons, the minimum 30 tons, and the mean of the whole = 38.3 tons in the centre of the girder, 16 feet between supports.

=

Calculating as before from the line N, A, we have for the top flange 12 × 3 = 3.5; for the vertical web (1212 – 12) × 1 = 149.0; and for the bottom flange (142 - 1212) × 12 = 552. The sum of the whole = 3.5+ 149.0 + 552·0 = 704·5, which is the reduced value of D2 x B in rule (324), which then becomes 704.5 x 92 16 = 40.5 tons: experiment gave a mean = 38.3 tons; hence 40.5 38.3 = 1.0575, or an error of +5.75 per cent. by Special Rule.

To calculate this girder in the ordinary way, we shall have to modify rule (330), which then becomes in our case 11 × 2 × 92÷16

143 x 12 1213 × 8

=

W { 149

14

14

14

}

=56.46 tons, giving an error of 56.4638.3 = +47.5 per cent.

1.475, or

(352.) "General Results.”—The whole of these experiments on square, rectangular, circular, and ordinary girder sections with equal and unequal flanges, show that the method of calculating from the top edge gives in all cases the most correct results, the errors in the five different kinds of section by the special rules being - 7·1, +8.8, 4.68, 7.42, and +5.75 per cent. respectively; whereas by the ordinary rules they were + 45, + 54·7, + 50·0, + 58.8, and + 47.5 per cent. respectively. It should also be observed that in the special rules the sum of the three errors is 19.2, and the sum of the two +errors is 14.55, so that we have as a general average result (19.2 14.55) — 5 = 0.93, or less than 1 per cent. (353.) It is shown by all these experiments that the ordinary rules in (323), &c., give always much higher results than the special rules, the difference being 47.5 - 5.75 41.75 per cent. in the girder with unequal flanges (351), and 4·68 + 50·0 54.68 per cent. with the circular section in (349). Now it is admitted in (510) that for small strains, say up to 3rd of the breaking weight, the ordinary rules in (512), &c., which are based on the supposed equality of the tensile and compressive strengths of the material, are nearly correct.

=

(354.) According to that, we might calculate the safe load by the ordinary rule, while the breaking weight must be found by the special rule. This method would conduct us to some curious