(488.) "Effect of the Weight of the Beam itself.”—The laws for "Similar Beams" will enable us to show most clearly the effect of the weight of the beam itself on the total load which it can bear. It is important to remember that the breaking weight of a beam must always be composed of two different loads, one being the weight of the beam itself, and the other the extra load laid upon it; moreover, that the proportion between these two is very variable (490). In short beams, the weight bears a very small proportion to the load, and in most ordinary cases may be safely neglected, but with long beams, the proportion rises with the length, until the whole strength of the beam is required to carry its own weight, and it can bear no extra load whatever (492). In the large tube of the Conway Bridge the weight was 1080 tons (474), which is equivalent to 1080÷2 = 540 tons in the centre, and this is about th of the total strength of the beam, or 2145 tons, as calculated in (473).

(489.) If we take a series of "Similar" beams, with lengths and all other dimensions in the ratio 1, 2, 3 . . . 10, the 6th law in (461) shows that the breaking weights will be proportional to the squares of those numbers, and by the 10th law in (468) the weights of the beams will be as the cubes. Thus, with beams having all their dimensions, lengths, depths, breadths, and thicknesses in the ratio:

1 2 3 4 5 6 7 8 9

10

the total breaking weights taken as a distributed load, including the strain from the weight of the beam itself, would be in the ratio of the squares of those numbers (461), say

10

40 90 160 250 360 490 640 810 1000

Now, say that the weights of the beams between supports, being by (468) in the ratio of the cubes of the dimensions,

are:

1 8 27 64 125 216 343 572 729 1000 Then, the useful load borne, over and above the weights of the beams themselves, would obviously be found by deducting the

weight of the beam from the total breaking load, and become :

ΤΟ

9 32 63 96 125 144 147 128 81 0 Hence, with the smallest beam of the series, the weight of the beam itself is th, and the useful load ths of the total strength with the length, &c. : 5, the weight of the beam is, and the useful load the total strength. But with the largest beam of the series, the weight of the beam itself is equal to the total strength, and it can bear no useful load whatever.

(490.) We have seen (488) that with the large Conway tube the weight of the beam was 4th of the total strength. With the 75-foot model tube (473) it was 2.989·150325, orth of the total strength. With the little model tube No. 8 in Table 78, the weight between supports was 4.34 lbs., equivalent to 4·34 ÷ 2 = 2·17 lbs. in the centre, and the total breaking weight being 672 + 2 = 674 lbs., the weight of the beam is 2.1767400322, or 31th of the total strength.

(491.) "Effect with Large Beams."—With very large beams the effect of the strain due to the weight of the tube is to complicate considerably the application of the "Factor of Safety." The majority of our leading Engineers admit 6 as the value of that Factor for Railway Bridges, or that the beam may be strained to th of its breaking weight only, for a moving load such as a train. But we have seen that in the case of the Conway tube the weight of the tube alone strains the structure far beyond that limit, in fact to th of the breaking weight: hence the highest possible Factor = 4. Thus the calculated breaking weight of that tube by the most correct method of calculation (473) being 2145 tons, the safe load with Factor 6 is 2145 ÷ 6 = 358 tons, but as we have seen (474) the strain due to the weight of the tube itself is 540 tons.

It is shown, however, in (839) that in most cases the strain produced by a rapid train is very little greater than that due to a dead load, for which, as shown by (885), (886), the Factor for cast or wrought iron may be taken at 3. Then, with the Conway tube we obtain 21453 715 tons total safe load,

and as the strain due to the weight of the tube is 540 tons, we have 715 540 = 175 tons for the weight of the train.

(492.) It may easily be shown that if the length of the Conway tube were doubled, the span becoming 800 feet, and all the cross-sectional dimensions were retained, the beam would break with its own weight; for obviously the breaking load would be reduced to half or to 2145 ÷ 2 = 1073 tons, while the weight of the beam being doubled becomes 540 x 2 tons.

=

1080

If, on the other hand, all the dimensions were increased in the same ratio, the length with which the beam would break with its own weight would be 1600 feet, or 4 times its present span. In that case, the breaking weight increasing as the square of the dimensions becomes 2145 × 42 = 34320 tons, and the weight of the tube itself increasing as the cube of the dimensions becomes 510 x 43 = 34560 tons, &c. We have here, however, taken the theoretical ratio R2 instead of the more correct experimental ratio R (471).

CHAPTER XII.

THE CONNECTION OF THE TENSILE AND CRUSHING STRAINS WITH THE TRANSVERSE STRAIN,

(493.) When a beam is fixed at one end and loaded at the other, the transverse strain is resolved into a tensile one at the upper part of the section, and a crushing one at the lower part. There is therefore an intimate connection between the transverse strength of a material, and its strength in resisting tensile and crushing strains, and our present object is to investigate the nature of that connection, and to obtain rules that will enable us to calculate any one of those three strains, when the other two are known by experiment.

(494.) Let Fig. 165 be a beam or cantilever built into a wall and unloaded, the depth being 20 inches, the breadth 1 inch, and the length 100 inches. Fig. 166 is the same beam loaded

at the end with a weight W, the tendency of which is to cause the beam to rotate round the "neutral axis" N. A., the fibres above that line being stretched in a ratio increasing in simple arithmetical progression from the neutral axis, where it is nothing, to the upper edge of the section, where it is a maximum, and if we admit that the strains are simply proportional to the extensions (which is not strictly true (604)), those strains will obviously increase in arithmetical progression also, and it will follow that if the beam is loaded up to the point of rupture, it will give way at first by the fracture of the upper fibres, the rest following in succession. Obviously, the fibres below the neutral axis are compressed and subjected to a crushing strain increasing from the neutral axis to the lower edge of the section, where it becomes a maximum.

(495.) If the tensile and crushing strengths are equal to one another, the neutral axis will be in the centre of the section as in Fig. 166: assuming for illustration that they are thus equal, and that the maximum strength is 10 tons per square inch, we have in Fig. 164 a section of the beam to a larger scale, and can calculate the strains on each square inch of that section, also the leverage with which it acts; and its effect in sustaining a load at W, in Fig. 166.

Thus, the square inch B is strained with 10 tons per square inch at its upper edge, and 9 tons at the lower edge; the mean is 9.5 tons, which, acting with a leverage of 9.5 inches (namely, the distance from its centre of gravity to the neutral axis), will exert, in sustaining a load at W, a strain of 9.5 × 9.5 100 =9025 ton. Similarly, D gives 8.5 x 8.5 100 =·7225 ton: following out the calculation in this way throughout, the sum of the whole series of strains for that half of the section subjected to tension will come out 3.325 tons: if the series had been infinite the sum of the resistances would have been 3 tons.

We might have obtained the same result more easily by multiplying the whole area 10, by the mean strain 5, and by the mean leverage 5, and taking of that product; thus 10 × 5 × 5 × ÷ 100 3 tons, as before.

This strain is due to tension alone, and that due to com

pression being the same we obtain 63 at W. If we would convert this case of a cantilever loaded at one end, Fig. 166, into an equivalent beam supported at each end and loaded in the centre, Fig. 167, we have evidently a length of 200 inches, a strain of 63 tons on each prop, and a central load of 13 tons.

This mode of calculation may be expressed by general Rules which become:

(499.)

T=

W x L x 4.5

2

{d − √ W × L × 4 · 5 ÷ (T × b}" × b

4·5

W x L x 4.5

{d – √ W × L × 4·5 ÷ (C × b}" × b

In which T = the maximum tensile strain, or that at the extreme edge of the section, per square inch.

C = the maximum crushing strain, or that at the extreme edge of the section, per square inch.

d = the depth of rectangular beam, in inches.

L = length of beam, supported at both ends, in

Of course, T, C, and W must all be taken in the same terms, tons, lbs., &c.

(500.) Applying rule (496) to the case of the beam whose strength was investigated analytically in (494), the length