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The central breaking weight being thus found, we require 1st, the extent to which that same weight would strain the parallel beam if placed at a given point out of the centre, and 2nd, the breaking weight at any other point in the same parallel beam.

(420.) A weight placed anywhere on a beam divides the whole length into two imaginary lengths—equal if in the centre, unequal if elsewhere—and when the weight is constant the strain produced by it at each point is proportional to the product of those two imaginary lengths, and will be found to be a maximum when the load is in the centre. Thus in Fig. 118, with 27 tons in the centre the beam is divided into two equal 8-foot lengths, the product of which is 8 x 8 = 64; now if that load of 27 tons is removed to 4 feet from one prop, therefore 12 feet from the other, as in Fig. 119, the product becomes 4 x 12 = 48, and the beam is strained at A to 4fths only of the breaking weight at that point. Hence the breaking weight there would be 27 x 64 = 48 = 36 tons. Calculating in this way we have obtained the ratios in col. 4 of Table 104 and in curve of Fig. 213, which give the equivalent load out of the centre corresponding to a central load of 1.000.

(421.) For general purposes perhaps a simpler course is to find a reduced or imaginary length to be used in the ordinary rules in (323), &c., instead of the actual length as for central loads. For this purpose :- divide the product of the two lengths into which the load divides the beam by 4th the actual length, and the quotient is the reduced length to be used in calculation. Thus in Fig. 119 we have 4 x 12 = (16 ; 4) = 12 feet :-—then with this reduced length, the rule W = d x 6 x M.:L becomes in our case 122 x 12 x .25 12 = 36 tons, as before. The same result would have been found by using the ratios 64 to 48 as in (420); for 16 x 48 = 64 = 12 feet, the reduced length, &c.

(422.) The contour of the beam in elevation may now be found as in (417), the breadth being 12 inches throughout, by making the square of the depth proportional to the distance from the props, but so that the depth at A shall be 12 inches. Thus from A to E we have four divisions :—then 12 ; 4 = 36 is a constant which multiplied by the distance of each point E, D, C, B, A from A in Fig. 120 will give the square of the depth at that point.

Thus at E we have 36 x 4 or 144N = 12 inches.
D
36 x 3

108N

= 10.4
36 x 2 728 8.5
B
36 x 1

367

6.0 А

36 x 0 ON 0:0

99

Similarly from E to Q we have 12 divisions, hence 122 = 12 = 12 is a constant which multiplied by the distances of the points F, G, &c., &c., from Q will give the square of the depth at each point from E to Q.

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Thus at E we have 12 x 12 or 144x = 12 inches.
F
12 x 11 132

= 11.5
G

12 x 10 120 W = 11:0
H

12 x 9 108 N = 10.4
I
12 x 8

96 N = 9.8
12 X 7 84N 9.2
K
12 x 6

72N 8.5
L
12 x 5

60N 7.7
M

12 x 4 487 = 6.9
N
12 x 3

36N

6.0
0
12 x 2 247

4.9
P

12 x 1 12N = 3.5 Q

12 x 0 ON = 0.0

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We have thus found the depth at every foot in the length of the beam; for ordinary purposes a smaller number of points would have sufficed, at least between E and Q, but we have a special purpose in view presently for which we require the depth at numerous points.

(423.) “ Effect of Two or more Loads."—We may now proceed to consider the effect on the form of a beam, of two or more loads variously distributed. We will take the case of Fig. 122, where we have two weights each of 36 tons, both being 4 feet from one prop and 12 feet from the other, and we will take it first as composed of two similar beams, as in Figs. 120, 121.

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Obviously Fig. 121 is the counterpart of Fig. 120, and if we imagine those two beams placed side by side, as in Fig. 122, we should have a compound beam that would fulfil the given conditions. Now in the beam Fig. 122, the depth squared multiplied by the breadth, must at each and every point be equal to the sum of the depths squared multiplied by the breadths of the corresponding points in the two beams Fig. 122, which are given by Figs. 120, 121. The breadths of all the beams we are considering being alike, or 12 inches, we may eliminate b and shall deal only with d'. Then, the square of the depth at b in Fig. 122 must be equal to the sum of the squares of the depths at B in Fig. 120 and at P in Fig. 121, or 6+ 3.52

48.25, and 48.25 = 6.95 inches, the depth at b in Fig. 122.

(424.) Calculating in this way we obtain the depths at each foot of length as follows :

Inches.
(B? + P) or ( 6.0+ 3.5%)V : 6.95 at b in Fig. 122.
(C~ + 0?) , ( 8:52 + 4.92)

= 9.81
(D2 + N) (10•4+ 6:02) N = 12:00 d
(E2 + Mo) (12:02 + 6.92)N = 13.85
(F2 + L?) „ (11.52 + 7.72N = 13.85 f
(G? + K%) , (11.0% + 8.5?)X = 13.85

9
(HR + J?) „ (10.42 + 9.2%) = 13.85 h
(I + IP ) , ( 9.82 + 9.8%) = 13.85 i
(J+ H) ( 9.22 + 10•4) = 13.85 j
(K? + GP) , ( 8.5% + 11:02) = 13.85 k
(L + FP) ( 7.72 + 11.5%) = 13.85 1
(M? + E) ( 6.92 + 12:02)n = 13.85
(N + Do) , ( 602 + 10:4°) = 12:00
(O2 + Con ( 4.92 + 8.5%) N = 9.81
(P2 + B* ) , ( 3.5% + 6:02) = 6.95

P (Q? + A') , (0.02 + 0.0%)N = 0.00 9 It will be observed that between the points e and m the beam is a parallel one, and it is curious to see how this is brought about by the fact that the sum of the squares of the depths at the corresponding points in the two parabolas in Figs. 120, 121 is constant between those points.

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(425.) The analytical method of investigation and calculation adopted in the above may be laborious, but it has two great advantages: 1st, it covers and includes all possible cases, of any number of weights, and any method or order of distribution, for obviously it is only necessary to calculate the form and dimensions of beam for each weight independently, and then combine them in the manner we have illustrated; 2nd, this method has the great advantage of allowing every step in the process of calculation to be seen by the operator, and the rationale to be understood.

We have applied the method of finding the theoretical forms to timber beams for convenience of illustration : for that material the parallel form would almost always be adopted in practice from motives of economy of labour, &c. ; the saving of material which would accrue from the adoption of the theoretical form would not compensate for the labour and trouble required to obtain it. With cast- and wrought-iron beams the case is different.

(426.) Equally Distributed Load."—When a load is spread equally all over the length of a beam, it may be considered as divided into any given number of equal weights equidistant from each other, and then the effect of each of these imaginary weights may be separately calculated. Let Fig. 125 be a beam 16 feet long supported at each end, with its load divided into 16 parts :this is equivalent to a cantilever, Fig. 126, of half the length with its load divided into 8 parts, and we require 1st the depth at the points a, b, c, &c., the breadth being constant, and 2nd we require to find the breadth at those points when the depth is constant. For this purpose we must suppose each weight to rest on the beam at two points, each giving half the strain due to the whole weight. Thus the weight W being 1:0 will give a pressure of į at a and } at b; then the weight o being also 1.0 will give } at b, and } at c, &c.; the combined effect of W, and w at b is therefore } + } 1.0.

(427.) The transverse strain at every point may now be calculated; thus the strain at b is that due to the weight of acting with a leverage of 1.0, or x1 = 0.5, which is the ratio of the breadth at that point when the depth is constant ;

e

but if the breadth is constant, we have 10.5 = .707, which is the ratio of the depth at that point. Again at c we have a weight of } with leverage of 2, plus the weight B or 1.0 with a leverage of 1.0 or (1 x 2) +(1 x 1) = 2, which is the ratio of the breadth at that point when the depth is constant; but if the breadth is constant, then 12 = 1.414 is the ratio of the depth at c: again at d, we have (1 x 3) + (1 X 2) + (1 x 1) = 4:5, the ratio of breadth when the depth is constant, and 4.5 = 2.12, the ratio of the depth when the breadth is constant, &c. Calculating in this way we find that at b d

f g h k where the depth is constant, the breadths should be in the ratios Fig. 129, or

0.0 0.5 20 4.5 8.0 12.5 18.0 24.5 32 but when the breadth of the beam is constant throughout its length, then the depths should be in the ratios Fig. 127, or

0.0 0.707 1.414 2.12 2.83 3.53 4.24 4.95 5.65

It will be observed that when the depth of the beam is constant, the breadths are in the ratio 1, 4, 9, 16, &c., or as the squares of the distances from the end of the beam, as in Fig. 128. But when the breadth is constant, the depths follow the simple arithmetical ratio 1, 2, 3, 4, &c., and the profile of the beam is then a triangle, as in Fig. 127. Applying this to a beam supported at each end and with the load equally distributed, then, when the depth is constant, the breadths are as given by Fig. 129, but when the breadth is constant, the profile is that of two triangles united at the base, as at x x in Fig. 127. The proportions of depths to breadths may be varied to any extent so long as d? x b follow the ratio of the middle line above, or 0.5, 2.0, 4.5, &c.

(428.) “ Form, as governed by Taste, &c.”—The forms of beams which we have thus obtained, although theoretically correct for the transverse strains, are not such as to satisfy the requirements of taste, moreover, they do not provide for the shearing strain at the ends (403) nor for a fair area of bearings at the supports so as to spread over a large surface the insistent

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