T. 18537, 18323, 18312, 18085, 19501, 17734, 20242. D. 23171, 22904, 22890, 20606, 24626, 22167, 25302. The ratios of D to T were found to average as 1 to 78, and those of Mr. Barlow as 1 to 81. That a similar thing occurs in the case of wrought iron was also observed, but that material is difficult to measure the compressive resistance of, because it does not crush like cast iron, but bends and becomes distorted before its ultimate strength is arrived at; but it is observed that, although there is a great difference between the ultimate tensile and compressive resistances of wrought iron, yet the force necessary to overcome the elastic resistance does not so vary.

Mr. Barlow referred this accession of strength to the molecular disturbances occasioned by the flexure of the beam, and termed the increment of strength the resistance of flexure.

The difference of action of the force is this: in direct longitudinal strain the layers of molecules during extension move together; one is not more extended than the next, but when a beam is bent, the outer layers are lengthened and shortened more than the inner, and therefore one layer moves upon another, and so gives rise to the resistance of flexure.

Applying our formula for transverse resistance to a cantilever 1 inch square and 1 inch long, we have, if we take the resistance from the experiments of direct strain, breaking weight at end=

8. b. d2
6

18000 × 1 × 12
6

= 3,000 lbs.,

but find that 7,000 to 8,000 lbs. practically represents the ultimate strength of the cantilever.

The average of a number of experiments on cast-iron bars 1 inch square, supported with a bearing of 36 inches, gave as the mean breaking weight at the centre 844 lbs.

Ꮃ l 8 b d2

4

=

6

=45,476 lbs. as the longitudinal resistance tensile and to flexure: the latter may be regarded as the elastic resistance to shearing stress.

The last figures give, for the resistance of a sectional square inch of the material to transverse breaking, 7,596 inch lbs.

The resistance of cast iron to shearing force has been found to be double that of its tensile resistance, and it would appear that the elastic resistance of flexure agrees with this, for the total movement in the bent beam of one layer upon another will average one-half of the total extension, and the resistance to flexure is shown to be roughly equal to the tensile resistance proper, which would correspond to a shearing resistance double that to tension.

With wrought iron, on the other hand, the shearing resistance is somewhat less than the tensile strength.

In flanged girders, where the flanges are thin compared with the depth, this resistance to flexure will diminish as the extension and compression of the fibres become sensibly uniform.

CHAPTER IV.

FRAMED STRUCTURES.

I AM now passing on to consider a class of structures wherein the strains act longitudinally upon the different component parts, the whole work consisting of bars framed together in such manner that they form, as it were, channels for the strains to pass along.

In calculating the strains recourse will be had to the principle of the parallelogram of forces already explained in Chapter II.

There are two characteristic dispositions of bars shown in Fig. 27, by the use or combination of which all braced works are built up. In the first, two bars, A B and A C, joined at A and resting on abutments at B and C, support between them a load, W. Each bar carries part of the load as such on to its abutment, the proportion carried in each direction depending upon the position of W in regard to B and C. If Bd be a horizontal line between the abutments, extending from centre to centre of the feet of the bars B and C, and this be intersected at e by the vertical line joining A and the load W, then the part of the load e d Be B d'

on B will be Wx and that on C will be Wx:

Bd

H, I, J represents another arrangement in which the whole load is carried on the abutment H by the bar IH; IJ being horizontal, and therefore carrying no load, its duty

is to keep the end I of the bar IH from falling towards J.

Returning to the first arrangement, on A e mark off the distance Ao (to any convenient scale) to equal the weight W; complete the parallelogram Ap oq, making op parallel to A B and o q parallel to A C. Then the thrusts on AB and AC will be represented by Aq and Ap respectively, and will pass away into

the abutments in the directions fand g; the vertical component on each abutment being the part of the load already assigned to it. From 9 and Ρ draw horizontal lines meeting A e in the points and t, then

Ar and Aq will be the vertical forces corresponding to the strains upon the bars A B and AC, for if Bf be made equal to Aq, and this oblique thrust on the abutment resolved into its vertical and horizontal components, by completing the parallelogram Buƒv, uf or Bv will be the vertical component, and be

cause Ar and Bv are both vertical they are therefore parallel, and meeting the straight line Af at the points A and B, make the angle rAB equal to the angle v Bf. Again, qr is drawn horizontal, and also fv; hence they are parallel, and meeting the line Af at q and f, make the angle r q A equal to the angle vƒB, and the remaining angles fv B, qr A are equal; hence the triangles

are similar, but the side Bf has been made equal to the side Ag; therefore the triangles Aqr, Bfv are equal, and their sides Ar, B v are equal-that is, Ar is equal to the vertical load carried on the abutment B. And in like manner it can be shown that At is equal to the vertical load on the abutment C.

But, again, by similar triangles, Aq is to Ar in the same ratio as A B is to Ae, because Arq, A e B are both right angles, and the angles A qr, A Be are equal; hence the strain on A B is to the weight it carries as the length A B is to the height A e.

Now the length A B is the length of the bar carrying a certain proportion of load and A e is the differ

(W xed).

ence in level between the top and bottom of the bar, or, in other words, Ae is the vertical height of the bar; hence we have a simple rule for the strain on an inclined bar when the vertical load carried by it is known, i.e. the strain on the bar is equal to its vertical load multiplied by the length of the bar and divided by its vertical height. All these dimensions must be measured on lines running along the centres of the elemental bars, as shown dotted in the figure.

In the present case, for example, let the load be 5 tons, the length of AB 7 feet, its vertical height 5.75 feet, the distance Bd 6.3 feet, and the distance e d 2.5 feet, then the

2.5

part of the load on A B will be W x = 5 X =

e d Bd

6.3

1.984 tons, and the strain upon the bar A B will be 1.984

The same rule can, of course, be shown to apply to A C or to any other inclined bar, the vertical load upon which is known, or can be ascertained.

All that remains is to determine the nature of the strain