= prop, 2133 x 1.875 × 2000 inch-lbs., being thus precisely the same in all the four cases. We have here neglected the Inertia due to the weight of the beam, the effect of which is to disturb perfect equality, as shown in (811). (809.) From this we find that while the safe load increases with the distance from the centre, the deflection with that increased load is reduced in exactly the same ratio, so that their product, or W × 8, is everywhere constant. The same fact is shown by col. 6 of Table 104. The result is, that the power of a beam to resist Impact is the same at whatever part of the length it is struck: that is to say, a given weight which falling on the centre from a certain height, will break the beam, or strain it to a given fraction of the breaking weight, will have the same effect at any other point in the length. This remarkable result has been confirmed by experiment, as shown by Table 126. Mr. Hodgkinson found that a bar of cast iron 1 inch square and 4 feet long, struck horizontally at the centre by a ball of 203 lbs., acting as a pendulum with a radius of 12 feet, required a fall through a chord of 6 feet to break it: and two similar bars struck at span, or midway between the centre and one support, broke with the same fall precisely. (810.) Let Fig. 205 represent the case: it is an axiom that the velocity at V acquired by a body falling by gravity say from Q, is the same whether the body falls through the arc Q, V, or vertically through the height X, V, &c. If we take chords 1, 2, 3, as in the figure, the vertical heights are in the ratio 1, 4, 9 as shown, and as the velocities are proportional to the square-roots of the heights we have velocities of 11, √42, and √93; that is to say, the velocities are simply as the chords, and the vertical heights as the square-roots of the chords: hence we have the Rule = = In which Rp the radius of the pendulum: Cp the chord fallen through, and hy the corresponding vertical fall, all in the same terms (feet or inches): thus in our case hy = 6 ÷ (12 × 2) = 1·5 foot, as in col. 6 of Table 126. = In calculating the effect of blows at the centre and elsewhere, perfect equality will be disturbed by the Inertia of the bar (781): with the 1-inch bar, the weight between supports being 11.2 lbs., the Inertia I: 11·2 ÷ 2 = 5.6 lbs., and by Rule 1 x 1 x 4 x 81 x (5.6+ 20.75) (782) we obtain h = 20.752 19.8 inches, or 1.65 feet, as in col. 8. I becomes 5.6 × 2 = 11.2 lbs., and h = = When struck at span, 1 x 1 x 4 x 81 x (11·2 + 20.75) or 2 feet, &c. 20.752 = 24 inches, = 93 lbs., and With the 3-inch bars, the weight of the bar between supports = 186 lbs., hence when struck at the centre I 3 × 3 × 6.75 × 81 × (93 + 603) 6032 = Rule (782) gives h = 9.42 inches, or 785 feet. When struck at span, I = 93 × 2 3 × 3 × 6.75 × 81 × (186 + 603) = 186 lbs., and h = 10.67 inches, or 8898 feet, &c. 6032 TABLE 126.-Of EXPERIMENTS on the ULTIMATE RESISTANCE of CAST IRON to IMPACT, when struck at the Centre and at Span. = (812.) "Resistance to Impact as Deflection Squared.”—With beams whose elasticity is perfect the statical or dead loads and the corresponding deflections are simply proportional to each other, thus, deflections 1, 2, 3, &c., require loads 1, 2, 3, &c., but the dynamic or falling weights are proportional to the deflection squared. Thus deflections 1, 2, 3 require with the same fall, weights in the ratio 12, 22, 33, or 1, 4, 9, &c., for obviously, as the power required to bend a beam, or R, is proportional to the deflection multiplied by the weight producing it, or to 8 x W, it follows that with statical weights, 1, 2, 3 and corresponding deflections in the ratio 1, 2, 3, 8 x W becomes 1 x 1 =1: 2 x 2 4: and 3 × 3 = 9, &c., or as the deflection squared. Conversely when the falling weight is constant, the height of fall is in the ratio of deflection squared; this is proved to be true experimentally by Table 120, deflections of 1.25 and 2.62 inches, which are in the ratio 1 to 2 nearly, required falls of 12 and 48 inches, which are in the ratio 1 to 22, or 1 to 4. (813.) In the case (775) and Fig. 186, in which the deflection is throughout 1 inch per pound for the statical weights 1 2 3 4 5 6 7 the power is respectively 2 4 8 12 18 241 inch-lbs., and for each successive inch 8 9 10 lbs. 32 401 50 1 11 21 31 34 51 61 7 81 91 inch-lbs. While therefore 1 lb. statically produces 1 inch deflection throughout, the first inch takes inch-lb. only dynamically, and the last, 9 inch-lbs. It is apparently an anomaly that the last inch deflection requires 9 inch-lbs., and yet that it is produced by the addition of 1 lb. falling 1 inch, and it is still more remarkable because the dynamic effect of the latter is only an inch-lb., for, as in the first inch so with the last, the mean weight 1+0 during the 1-inch fall is = inch-lb. But it should be 2 addition of the first pound had no other inch-lb. due to it, the addition of the effect of causing the 9 lbs. with which observed that while the effect than to yield the last pound had also the the beam was already loaded to descend 1 inch with the full uniform weight, thus yielding 9 inch-lbs., which added to the inch-lb. due to the weight itself, makes the total dynamic weight 9 inch-lbs. for the last inch, as before stated. = (814.) The power to deflect the beam from 0 to 9 inches, as shown by (801), being 40 inch-lbs., and from 0 to 5 inches 12 inch-lbs., it follows that to increase the deflection from 5 to 9 inches will be 401 - 121 28 inch-lbs., or the same result as in (813). = Table 121 gives the results of Mr. Hodgkinson's experiments on the resistance of wrought-iron bars to impact: comparing cols. 4 and 5 it will be seen that the height of fall is practically as the square of the deflection, as due by theory. (815.) The imperfect elasticity of cast iron causes a considerable divergence from the rule, as is shown by Table 122, which gives the result of similar experiments on cast-iron bars: comparing cols. 2 and 3, or 5 and 6, it will be seen that the experimental fall in col. 2 or the power in col. 5, is in a much lower ratio than 2, the difference being due to defect of elasticity (688). The ordinary rule supposes that the deflections are simply proportional to the statical loads, but col. 4 of Table 108 shows that this is not even nearly true of cast iron, the deflections increasing with successive 56 lbs. from 3754 inch with the first, to 1.157 inch with the last. Say for illustration, that we had a material in which defect of Elasticity was such, that to produce deflections 1, 2, 3, required loads 1, 1, and 2, respectively, instead of 1, 2, 3, as due with perfect elasticity. Then 8 x W becomes 1 x 1=1: 2 × 1 = 3: and 3 x 2 = 6 we thus obtain the ratio 1, 3, 6, instead of 1, 4, 9, which is that of the square of the deflection. = (816.) Comparing the first experiment in Table 122 with the last, the deflections being in the ratio 1 to 9, or 1 to 6, the vertical fall, or the work done by the falling ball, should be as 62 or 36 to 1, but experiment gave 1.19440638 18.8 to 1, which is in the ratio of the 1.64 power of the deflection, or 81.64 instead of 82. Thus in our case the ratio of the deflections was 1 to 6, and the log. of 6788: the ratio of the work done in producing the deflections 1 to 18.8, the log. of 1.638, say which 1.274: then we obtain 1·274 ÷ ·778 = 1.64 the power of the deflection to which the work done is proportional. Cols. 4 and 7 were calculated by that ratio, taking the experimental numbers for 1 inch deflection as a basis. It should be observed that the difference between 814 and 62 is so great as to double the work with deflections in the ratio 1 to 7, for 7164 243 and 7o 49 or about double. = (817.) "Effect of Inertia not Constant."-The inertia of the beam and its load is most influential when the falling weight is light in proportion thereto. If the beam were unloaded, and itself without weight, it would be quite immaterial what the falling weight might be so long as the fall multiplied by that weight was constant: thus 100 lbs. falling 10 feet would produce the same effect in straining the beam as 10 lb. falling 100 feet, &c. But this is not true when the inertia of the beam and its load is considered: say that we have a beam whose inertia, or half-weight between supports = 1000 lbs., and we require the heights from which weights of 1, 10, and 100 lbs. must fall to produce one and the same deflection. Of course if the inertia was nothing, those heights would be inversely proportional to the weights simply, or 100, 10, and 1. We require only proportional numbers, and may take from (782) h = (I + w)÷w2, which becomes in our cases (1000+ 1) ÷12 = 1001; (1000 + 10) ÷ 102 = 10·1; and (1000 + 100) 100211 respectively, or nearly in the ratio 100, 1, and th, instead of 100, 10, and 1, as with a beam whose inertia was nothing. 10 The Mechanical power R, required to bend a beam being by (780) equal to w × h, and h being proportional to (I + w) ÷ w2, we have the ratio of the power R = {I + w)÷w2) × w, or which in our three cases becomes (1000 + 1)÷1 = 1001; (1000+10) 10 = 101; and (1000+ 100) 100 = 11 ÷ respectively, which agrees with the preceding calculations of the heights of fall with weights of 1, 10, and 100 lbs. falling 1001, 10.1, and 0.11, the products of the weight by the height |