increased by loading the beam. The cols. 6, 8, 10, and 12 of Table 124 have been thus calculated, and they show that resistance to impact, or height of fall, is a maximum when the load plus the falling weight, or I+w, is 3, the breaking weight being 9.

(803.) Engineers have been guided by experience to fix on 3rd of the statical breaking weight, as the safe statical load for ordinary beams, neglecting usually the weight of the beam itself, and it is remarkable that the power of a beam to resist impact is a maximum with that load. But this requires explanation, and must not be understood to mean that a beam loaded with 3rd of the breaking load, is at the maximum of its power to resist a blow from an extra falling weight because the dead load already strains it to the safe limit assigned. Take the case of a beam whose breaking weight is 9, as in Table 124, which would have 3 assigned as the safe load, but say that when loaded with 2, the rest, or 1 falls on the beam, which often happens by the accidental failure of a rope, &c. Then, if we admit the weight of the beam between supports to be 1, its inertia would be, hence I + w becomes (2 + 1) + 1 3 when by col. 6 of Table 124 the resistance is a maximum, h being 73.94 inches. Again: say that when the dead load is 1, the rest, or 2 falls, then I becomes (1+) + 2 3, when by col. 10 the height h = 24.61 inches and is still a maximum. (804.) It will also be found that a factor for safety less or greater than 3rd (or 3) will give a less resistance. Say we adopt instead of; then if when loaded with or 4 in the Table, the rest, or 1 falls; I + w becomes (4 + 1) + 1 = and by col. 6 the height of fall is 55.7 inches, whereas the same weight of 1 fell 73.94 inches to break the beam when the factor was or 3. Again; when the dead load is 3 and the rest, or 2 falls, we get I + w = (3 + 1) + 2 = 5, and by col. 10 the height of fall is 20.8 inches instead of 24 61 inches with the same falling weight of 2 when the factor was rd.

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1

5,

If we adopt a lower factor, say instead of 3, then, if when loaded with 1 the rest, or 1 falls, we get (1 + 1) + 1 21 as the value of I+w, and by col. 6, the height of fall is 70.31 inches instead of 73.94 inches as with a factor of 3rd.

(805.) In Table 125 are given the results of Mr. Hodgkinson's experiments on the resistance to impact of bars of Cast iron 3 inches square, 13 feet long between supports. To obtain exact data for calculation, precisely similar bars of the same iron were subjected to experiments with a dead load, the bars being strained horizontally so as to eliminate the complications arising from the weight of the bar itself. The mean breaking

weight was 2865 lbs., and the ultimate deflection 4.939 inches: to obtain the deflection with smaller weights before defect of elasticity (688) became considerable, it was observed that with 672 lbs. the deflection was 633 inch or 633672 = ⚫000942 inch per lb. in the centre, which is equivalent to ·000942 × 00059 inch per lb. distributed all over the beam.

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(806.) Taking the 3rd experiment in the Table, the deflection due to the weight of the beam is 376 ×·00059 = 222 inch : that due to the central weight of 28 lbs. is .000942 × 28 026 inch, the sum is 222 +⚫026 248 inch, col. 6, so that the flexure required from impact is 4.939 ..248 4.691 inch, col. 7. The inertia of the bar is 3762 188 lbs., which added to 28 lbs., the central load, gives I = 188 + 28 = 216 lbs., col. 4, and the falling weight being 303 lbs., the Rule (789) 4.6912 × 2865 × (216 + 303)

4.939 × 3032 × 2

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36.08 inches,

becomes h = col. 11: experiment gave 42 inches, col. 9, but the last observed fall which did not break the bar was 39 inches. Obviously the fall which really broke the beam was somewhere between 39 and 42 inches, taking it at a mean of the two we obtain 40 inches, col. 10: hence we have 36.08÷40·5·891, showing a difference of 1.0·891·109, or 10.9 per cent., col. 12. The mean difference or error of all the experiments was 30.3 per cent., - 4 per cent., ranging from + 24.4 to showing a nearly equal divergence in both directions.

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(807.) "Impact out of the Centre of Beams."-We have so far considered only the case of a beam struck in the centre, but the same reasoning will apply to other cases; we have only to find the load out of the centre, and the deflection at the point of application, and the resistance to Impact R will be the product, or R = W x 8 x, as before (777).

Say we take the case of a bar of wrought iron 1 inch deep, 4 inches wide, and 16 feet long between end-bearings: the load in the centre, straining the bar to the "limit of Elasticity," with MT 2000 lbs. (374) by Rule (324) becomes W = 12 × 4 × 2000 ÷ 16 500 lbs., with which the deflection by 163 × 500 × ⚫00001565 = 8 inches, A,

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Rule (659) becomes ♪ =

Fig. 190.

13 x 4

By the Rule (650) the same weight of 500 lbs. at 1, 1, and span, or 4 feet, 2 feet, and 1 foot from one of the props, would give with L = 8 feet, and 8 = 8 inches

(808.) But if the loads were all proportional to the strength at each point, we should have had at B in Fig. 190, or at 4 feet from the prop, by the Rule in (420), 500 × (8 × 8) ÷ (12 × 4) = 666.7 lbs., and as 500 lbs. gave 4.5 inch, so by proportion 666.7 lbs. would give 4·5 × 666·7 ÷ 500 = 6 inches deflection at B. Similarly, at C, or 2 feet from a prop, we should have 500 × (8 x 8)÷ (14 × 2) = 1143 lbs., and a deflection of 1.531 × 1143500 3.5 inches: and finally at D, or 1 foot from a prop, we obtain 500 × (8 × 8) ÷ (15 × 1) 2133 lbs., giving a deflection of ·4395 × 2133 ÷ 500 =

1.875 inch.

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We can now find the resistance to Impact by a blow at A, B, C, or D in the figures, by simply multiplying the load at each point by the corresponding deflection, and taking the product, as explained in (776). At the centre A, we obtain 500 × 8 × 2000 inch-lbs.: at B, or 4 feet from one prop, we have 666.7 × 6 × 2000 inch-lbs: at C, or 2 feet from a prop, 1143 × 3.5 × = 2000 inch-lbs. and at D, or 1 foot from a

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