T

With 14 lbs., the deflections were ⚫032 inch, and ·025 inch.

Showing almost perfect equality up to 112 lbs., which is about 3rd of the breaking weight in position T (364 lbs.), but th only in position (1120 lbs.). The old Rule (670) will give the same deflection in either position: thus with Fig. 72, and W = say 112 lbs. we obtain 8 6 × 112 × ⚫00002886 ÷ {1.553× 5) - (1.253 x 4.64 periment gave 273 inch.

×

Ι

=

1853 inch deflection: ex

(672.) "Wrought-iron I Sections."-The Rules we have given for cast-iron, T, and I sections will apply equally to wroughtiron ones, with the proper value of C, which for lbs. = .00001565 Thus, with Fig. 154, by Rule (669)

by col. 4 of Table 105. D1 = 10000, d' or 83*

183 5832, and with W

=

=

5862, b

=

=

44, &c., L3 or say 30 cwt. or 33,880 lbs. we obtain

♪ = 5832 × 33880 ×·00001565÷10000×43)–(5862×4}}

8

10

=

1369 inch deflection: experiment gave 16 inch. Table 73 gives in col. 8 the experimental deflection of a series of rolled beams of ordinary equal-flanged sections: col. 9 has been calculated by Rule (669) and shows an error of 14 per cent. The old Rule (670), although not so correct in principle, will give results which agree better with experiment; col. 10 has been calculated by that rule; the mean error of the whole is +5 per cent. only.

66

Unequal Sections."-When the top and bottom flanges are unequal, as in Fig. 80, the most correct method of calculation will be to estimate from the bottom or the line N. A., as we found to be necessary in calculating the strength in (378): we then have the rule :

(673.)

♪ = L × W ×·00001565÷{D3 - ď] × B)

+ (ď− d'] × b) + (d' x C).

TABLE 109.-Of the DEFLECTION of ROLLED IRON I BEAMS.

The values of D, d, d1, B, b, and C, are given by Fig. 80, and the rest as in (664). Figs. 89, 90 are sections of beams experimented upon by Mr. Fairbairn, the deflections being given in cols. 2, 5 of Table 109. Thus in Fig. 89, with 885 lbs., the Rule gives

♪ = 113 × 885 × ·00001565 ÷ { 73

+(63 — ·383] × ·325) + (.38 x × 4}

= .04751 inch

Cols. 3, 6 of

deflection experiment gave 04 inch, &c. Table 109 have been calculated by this rule, and show a fair agreement with experiment: the calculated deflections with Fig. 89 show an error of +11.9 per cent., and of Fig. 90, +2.3 per cent.: the mean of the whole being + 6.4 per cent.

It is remarkable that this rule applied to sections with equal flanges does not give satisfactory results: col. 11 of Table 73 has been calculated by it and shows a mean error of +42 per cent.; while Rule (669) gave 14, and Rule (670), + 5 per cent. In order to render Table 73 directly available for practical purposes, we have given in col. 13 the experimental

deflection for each section by 1 cwt. in the centre of a beam 1 foot between supports, and as the deflections are simply proportional to the cube of the length and the weight, we have the Rule:

= deflection in inches.

In which To the Tabular number in col. 13; L = length in feet; W = weight in cwts.; and 8 Thus with No. 6, say L 20 feet; W = 35 cwt. then we obtain 00000239 × 8000 × 35 0.6692 inch deflection: experiment gave ğ inch, col. 8.

=

=

(675.)" Wrought-iron T Sections."-This form of section in wrought iron should always be loaded with the top flange uppermost in the ordinary case of a beam supported at both ends, for reasons given in (377), we must then calculate the deflections by measuring the depths from the bottom or from the line N. A. in Fig. 132, and we have the Rule:

(676.)

8 = L3× W ×·00001565÷{ D3 - ď3] × B) + (ď3 × b}. In which the values of D, d, B, b, are given by Fig. 132 and the rest as in (664). Thus, Fig. 87 is the section of a bar, which with a length of 10 feet, deflected inch with 2 cwt. in the centre: then the Rule gives 8 = 103 × 224 × ⚫00001565÷ = 0.2359 inch deflection: experiment gave 0.25 inch, &c. Table 71 gives in col. 11 the experimental deflections of a series of T iron bars: col. 12 has been calculated by the Rule.

(213 - 213] × 21) + (213 × }}

When the depth is equal to the breadth and the thickness is the same all over, the rule becomes

(677.) 8 = L3 × W × 00001565 ÷ (D1·

d').

In which the section A is regarded as composed of two bars B, C, as in Fig. 70, which is not strictly correct, as explained in (337), a more correct rule would be :—

(678.)

D5

♪ = L3 x W x 00001565 ÷ (Do13).

The effect of the two rules may be shown if we take the section Fig. 132, and calculate the deflection by both, for say 12 cwt., or 1344 lbs., with a length of 10 feet. Then Rule (677) becomes 109 × 1344 ×·00001565 ÷ (31 — 211) = 0·5008 inch By Rule (678) we obtain = 103 × 1344 × 35-215 3

deflection.

00001565

= 0.4341 inch deflection.

TABLE 110. Of the DEFLECTION of ROLLED WROUGHT-IRONT BEAMS, 1 foot long, with a weight of 1 cwt. in the Centre, the flange being uppermost.

Table 110 gives the deflection of Standard sizes of Tiron bars 1 foot long, with a load of 1 cwt. in the centre, calculated by the Rule :

(679.)

8 = .001773 ÷ (D1 — d1).

The deflection for any load and length may be easily found from Table 110 by the Rule (674): thus, a T bar 4 × 4 × inch thick, 20 feet long, with 10 cwt. in the centre, will deflect ⚫00001673 × 8000 × 10 = 1.34 inch, &c.

(680.) Deflection of Wrought-iron Lattice Girders."-The deflection of lattice girders may be calculated from elementary principles. To do this with scientific accuracy is a difficult mathematical problem, but we may obtain approximately correct results by ordinary reasoning and common arithmetic.

In a lattice beam the strength is almost entirely in the top and bottom members, and the deflection of the beam arises from the alteration in length which those members suffer by their respective strains :-in the case of a beam supported at the ends and loaded in the centre, the top suffers a crushing strain and becomes shorter, while the bottom bears a tensile strain and becomes longer. If we know the respective strains we may calculate the corresponding extension and compression -and knowing these we can calculate the deflection of the beam. Let A in Fig. 188 be a beam deflected by a transverse strain to the form shown :-for ordinary cases in practice in which the deflections are very small compared to the length of the beam, we may admit that the difference in length of the top and bottom members arising from the deflection is equal to the sum of B and C, and knowing one, say C, we may calculate G, F, which is greater than C in the ratio of D, G to D, E, or in other words the ratio of half the length of the beam to its depth, D, F being perpendicular to D, E and a tangent to the curve D, H at the point D. Having thus found G, F, we may easily calculate G, H, or the deflection required, for the curve of the beam approximates to a parabola, and it is a principle that the height of a parabola J, K in Fig. 189, is always half the distance J, L, L being the point in the axis where a tangent to the base of the parabola at M, cuts the axis. Returning to Fig. 188 we thus find that the deflection sought, G, H, is equal to half the distance G, F.

(681.) The application of all this will be best illustrated by a case in practice worked out in detail. We will take the case in (446) of a lattice girder, Fig. 155, 32 feet between supports, loaded in the centre with 4 tons, the section being as shown by Fig. 141, the top is formed by two angle-irons 4 × 2 × 1, whose united area = 6 square inches, the rail R gives 1.6 square inches more, making the total area of the top 7.6 square inches. The bottom, formed of two angle-irons 21 × 2 × 1, has an area of 4.5 square inches.

(682.) In calculating the deflection from these data, we have first to find the strains on the top and bottom members of the girder; these are equal to one another in all cases, but are not