Part 14

391. Supposing we had a beam of 40" span, 2" broad, and 0"·5 deep, we can easily see that it is equivalent to two bars like that of No. 3 placed side by side; and we infer generally that the strength of a bar is proportional to its breadth; or to speak-more definitely, _if two beams have the same span and depth, the ratio of their breaking loads is the same as the ratio of their breadths_.

392. We next examine the effect of the _depth_ of a beam upon its strength. In experimenting upon a beam placed edgewise, a precaution must be observed, which would not be necessary if the same beam were to be broken flatwise. When the load is suspended, the beam, if merely laid edgewise on the supports, would almost certainly turn over; it is therefore necessary to place its extremities in recesses in the supports, which will obviate the possibility of this occurrence; at the same time the ends must not be prevented from bending upwards, for we are at present discussing a beam free at each end, and the case where the ends are not free will be subsequently considered.

393. Let us first compare together experiments Nos. 2 and 3; here we have two bars of the same dimensions, the section in each being 1"·0 × 0"·5, but the first bar is broken edgewise, and the second flatwise. The first breaks with 77 lbs., and the second with 38 lbs.; hence the same bar is twice as strong placed edgewise as flatwise when one dimension of the section is twice as great as the other. We may generalize this law, and assert that the strength of a rectangular beam broken edgewise is to _the strength of a beam of like span and section broken flatwise, as the greater dimension of the section is to the lesser dimension_.

394. The strength of a beam 40" × 0"·5 × 1" is four times as great as the strength of 40" × 0"·5 × 0"·5, though the quantity of wood is only twice as great in one as in the other. In general we may state that if a beam were bisected by a longitudinal cut, the strength of the beam would be halved when the cut was horizontal, and unaltered when the cut was vertical; thus, for example, two beams of experiment No. 4, placed one on the top of the other, would break with about 40 lbs., whereas if the same rods were in one piece, the breaking load would be nearly 80 lbs.

395. This may be illustrated in a different manner. I have here two beams of 40" × 1" × 0"·5 superposed; they form one beam, equivalent to that of No. 1 in bulk, but I find that they break with 80 lbs., thus showing that the two are only twice as strong as one.

396. I take two similar bars, and, instead of laying them loosely one on the other, I unite them tightly with iron clamps like those represented in Fig. 56. I now find that the bars thus fastened together require 104 lbs. for fracture. We can readily understand this increase of strength. As soon as the bars begin to bend under the action of the weight, the surfaces which are in contact move slightly one upon the other in order to accommodate themselves to the change of form. By clamping I greatly impede this motion hence the beams deflect less, and require a greater load before they collapse; the case is therefore to some extent approximated to the state of things when the two rods form one solid piece, in which case a load of 152 lbs. would be required to produce fracture.

397. We shall be able by a little consideration to understand the reason why a bar is stronger edgewise than flatwise. Suppose I try to break a bar across my knee by pulling the ends held one in each hand, what is it that resists the breaking? It is chiefly the tenacity of the fibres on the convex surface of the bar. If the bar be edgewise, these fibres are further away from my knee and therefore resist with a greater moment than when the bar is flatwise: nor is the case different when the bar is supported at each end, and the load placed in the centre; for then the reactions of the supports correspond to the forces with which I pulled the ends of the bar.

398. We can now calculate the strength of any rectangular beam of pine:

Let us suppose it to be 12' long, 5" broad, and 7" deep. This is five times as strong as a beam 1" broad and 7" deep for we may conceive the original beam to consist of 5 of these beams placed side by side (Art 391); the beam 1" broad and 7" deep, is 7 times as strong as a beam 7" broad, 1" deep (Art. 393). Hence the original beam must be 35 times as strong as a beam 7" broad, 1" deep; but the beam 7" broad and 1" deep is seven times stronger than a beam the section of which is 1" × 1", hence the original beam is 245 times as strong as a beam 12' long and 1" × 1" in section; of which we can calculate the strength, by Art. 388, from the proportion—

144" : 40" :: 152 : Answer.

The answer is 42·2 lbs., and thus the breaking load of the original beam is about 10,300 lbs.

399. It will be useful to deduce the _general_ expression for the breaking load of a beam _l"_ span, _b"_ broad, and _d"_ deep, supported freely at the ends and laden in the centre.

Let us suppose a bar _l"_ long, and 1" × 1" in section. The breaking load is found by the proportion—

_l_ : 40 :: 152 : Answer;

and the result obtained is 6080/_l_. A beam which is _d"_ broad, _l"_ span, and 1" deep, would be just as strong as _d_ of the beams _l"_ × 1" × 1" placed side by side; of which the collective strength would be—

6080 ———————— × _d_. _l_

If such a beam, instead of resting flatwise, were placed edgewise, its strength would be increased in the ratio of its depth to its breadth—that is, it would be increased _d_-fold—and would therefore amount to

6080 ————————— × _d_². _l_

We thus learn the strength of a beam 1" broad, _d"_ deep, and _l"_ span. The strength of _b_ of these beams placed side by side, would be the same as the strength of one beam _b"_ broad, _d"_ deep, and _l"_ span, and thus we finally obtain

6080 ———————— _d_² × _b_. _l_

Since _b d_ is the area of the section, we can express this result conveniently by saying that the breaking load in lbs. of a rectangular pine beam is equal to

area of section × depth; 6080 × ————————————————————————— span

the depth and span being expressed in inches linear measure, and the section in square inches.

400. In order to test this formula, we have calculated from it the breaking loads of all the ten beams given in Table XXIV. and the results are given in the sixth column. The difference between the amount calculated and the observed mean breaking weight is shown in the last column.

401. Thus, for example, in experiment No. 7 the span is 20", breadth, 1", depth 0"·5; the formula gives, since the area is 0"·5,

0·5 × 0·5 _P_ = 6080 —————————— = 76 20

This agrees sufficiently with 74 lbs., the mean of two observed values.

402. Except in experiments Nos. 5 and 10, the differences are very small, and even in these two cases the differences are not sufficient to make us doubt that we have discovered the correct expression for the load generally sufficient to produce fracture.

403. We have already pointed out that a beam begins to sustain permanent injury when it is subjected to a load greater than half that which would break it (Art. 368), and we may infer that it is not in general prudent to load a beam which is part of a permanent structure with more than about a third or a fourth of the breaking weight. Hence if we wanted to calculate a fair working load in lbs. for a beam of pine, we might obtain it from the formula.

area of section × depth. 1500 × ————————————————————————— span

Probably a smaller coefficient than 1500 would often be used by the cautious builder, especially when the beam was liable to sudden blows or shocks. The coefficient obtained from small selected rods such as we have used would also be greater than that found from large beams in which imperfections are inevitable.

404. Had we adopted any other kind of wood we should have found a similar formula for the breaking weight, but with a different numerical coefficient. For example, had the beams been made of oak the number 6080 must be replaced by a larger figure.

A BEAM UNIFORMLY LOADED.

405. We have up to the present only considered the case where the load is suspended from the centre of the beam. But in the actual employment of beams the load is not generally applied in this manner. See in the rafters which support a roof how every inch in the entire length has its burden of slates to bear. The beams which support a warehouse floor have to carry their load in whatever manner the goods are disposed: sometimes, as for example in a grain-store, the pressure will be tolerably uniform along the beams, while if the weights be irregularly scattered on the floor, there will be corresponding inequalities in the mode in which the loads are distributed over the beams. It will therefore be useful for us to examine the strength of a beam when its load is applied otherwise than at the centre.

406. We shall employ, in the first place, a beam 40" span, 0"·5 broad, and 1" deep; and we shall break it by applying a load simultaneously at two points, as may be most conveniently done by the contrivance shown in the diagram, Fig. 53. A B is the beam resting on two supports; C and D are the points of trisection of the span; from whence loops descend, which carry an iron bar P Q; at the centre R of which a weight W is suspended. The load is thus divided equally between the two points C and D, and we may regard A B as a beam loaded at its two points of trisection. The tray and weights are employed which we have used in the apparatus represented in Fig. 58.

407. We proceed to break this beam. Adding weights to the tray, we see that it yields with 117 lbs., and cracks across between C and D. On reference to Table XXIV. we find from experiment No. 2 that a similar bar was broken by 77 lbs. at the centre; now ³/₂ × 77 = 115·5; hence we may state with sufficient approximation that the bar is half as strong again when the load is suspended from the two points of trisection as it is when suspended from the centre. It is remarkable that in breaking the beam in this manner the fracture is equally likely to occur at any point between C and D.

408. A beam _uniformly_ loaded requires twice as much load to break it as would be sufficient if the load were merely suspended from the centre. The mode of applying a load uniformly is shown in Fig. 54.

In an experiment actually tried, a beam 40" × 0"·5 × 1" placed edgewise was found to support ten 14 lb. weights ranged as in the figure; one or two stone more would, however, doubtless produce fracture.

409. We infer from these considerations that beams loaded in the manner in which they are usually employed are considerably stronger than would be indicated by the results in Table XXIV.

EFFECT OF SECURING THE ENDS OF A BEAM UPON ITS STRENGTH.

410. It has been noticed during the experiments that when the weights are suspended from a beam and the beam begins to deflect, the ends curve upwards from the supports. This bending of the ends is for example shown in Fig. 54. If we restrain the ends of the beam from bending up in this manner, we shall add very considerably to its strength. This we can do by clamping them down to the supports.

411. Let us experiment upon a beam 40" × 1" × 1". We clamp each of the ends and then break the beam by a weight suspended from the centre. It requires 238 lbs. to accomplish fracture. This is a little more than half as much again as 152 lbs., which we find from Table XXIV. was the weight required to break this bar when its ends were free. Calculation shows that the strength of a beam may be even doubled when the ends are kept horizontal by more perfect methods than we have used.

412. When the beam gives way under these circumstances, there is not only a fracture in the centre, but each of the halves are also found to be broken across near the points of support; the necessity for three fractures instead of one explains the increase of strength obtained by restraining the ends to the horizontal direction.

413. In structures the beams are generally more or less secured at each end, and are therefore more capable of bearing resistance than would be indicated by Table XXIV. From the consideration of Arts. 408 and 411, we can infer that a beam secured at each end and uniformly loaded would require three or four times as much load to break it as would be sufficient if the ends were free and if the load were applied at the centre.

BEAMS SECURED AT ONE END AND LOADED AT THE OTHER.

414. A beam, one end of which is firmly imbedded in masonry or otherwise secured, is occasionally called upon to support a weight suspended from its extremity. Such a beam is shown in Fig. 55.

In the case we shall examine, A B is a pine beam of dimensions 20" × 0"·5 × 0"·5, and we find that, when W reaches 10 lbs., the beam breaks. In experiment No. 8, Table XXIV., a similar beam required 36 lbs.; hence we see that the beam is broken in the manner of Fig. 55, by about one-fourth of the load which would have been required if the beam had been supported at each end and laden in the centre.

We shall presently have occasion to apply some of the results obtained by the experiments made in the lecture now terminated.

LECTURE XIII. _THE PRINCIPLES OF FRAMEWORK._

Introduction.—Weight sustained by Tie and Strut.—Bridge with Two Struts.—Bridge with Four Struts.—Bridge with Two Ties.—Simple Form of Trussed Bridge.

INTRODUCTION.

415. In this lecture and the next we shall experiment upon some of the arts of construction. We shall employ slips of pine 0"·5 × 0"·5 in section for the purpose of making models of simple framework: these slips can be attached to each other by means of the small clamps about 3" long, shown in Fig. 56, and the general appearance of the models thus produced may be seen from Figs. 58 and 62.

416. The following experiment shows the tenacity with which these clamps hold. Two slips of pine, each 12" × 0"·5 × 0"·5, are clamped together, so that they overlap about 2", thus forming a length of 22": this composite rod is raised by a pulley-block as in Fig. 49, while a load of 2 cwt. is suspended from it. Thus the clamped rods bear a direct tension of 2 cwt. The efficiency of the clamps depends principally upon friction, aided doubtless by a slight crushing of the wood, which brings the surfaces into perfect contact.

417. These slips of pine united by the clamps are possessed of strength quite sufficient for the experiments now to be described. Models thus constructed have the great advantage of being erected, varied or pulled down, with the utmost facility.

We have learned that the compressive strength, and, still more, the tensile strength of timber, is much greater than its transverse strength. This principle is largely used in the arts of construction. We endeavour by means of suitable combinations to turn transverse forces into forces of tension or compression, and thus strengthen our constructions. We shall illustrate the mode of doing so by simple forms of framework.

WEIGHT SUSTAINED BY TIE AND STRUT.

418. We begin with the study of a very simple contrivance, represented in Fig. 57.

A B is a rod of pine 20" long. In the diagram it is represented, for simplicity, imbedded at the end A in the support. In reality, however, it is clamped to the support, and the same remark may be made about some other diagrams used in this lecture. Were A B unsupported except at its end A, it would of course break when a weight of 10 lbs. was suspended at B, as we have already found in Art. 414.

419. We must ascertain whether the transverse force on A B cannot be changed into forces of tension and compression. The tie B C is attached by means of clamps; A B is sustained by this tie; it cannot bend downwards under the action of the weight W, because we should then require to have on the same base and on the same side of it two triangles having their conterminous sides equal, but this we know from Euclid (I. 7) is impossible. Hence B is supported, and we find that 112 lbs. may be safely suspended, so that the strength is enormously increased. In fact the transverse force is changed into a compressive force or thrust down A B, and a tensile force on B C.

420. The actual magnitudes of these can be computed. Draw the parallelogram C D E B; if B D represent the weight W, it may be resolved into two forces,—one, B C, a force of extension on the tie; the other, B E, a compressive force on A B, which is therefore a strut. Hence the forces are proportional to the sides of the triangle, A B C. In the present case

A B = 20", A C = 18", B C = 27";

therefore, when W is 112 lbs., we calculate that the force on A B is 124 lbs., and on C B 168 lbs. A B would require about 300 lbs. to crush it, and C B about 2,000 lbs. to tear it asunder, consequently the tie and strut can support 1 cwt. with ease. If, however, W were increased to about 270 lbs., the force on A B would become too great, and fracture would arise from the collapse of this strut.

421. When a structure is loaded up to the breaking point of one part, it is proper for economy that all the other parts should be so designed that they shall be as near as possible to their breaking points. In fact, since nothing is stronger than its weakest part, any additional strength which the remaining parts may possess adds no strength to the whole, and is only so much material wasted. Hence our structure would be just as strong, and would be more properly designed if the section of B C were reduced to one-fifth, for the tie would then break when the tension upon it amounted to 400 lbs. When W is 270 lbs. the compression on A B is 300 lbs., and the tension on B C is 405 lbs., so that both tie and strut attain their breaking loads together. The principle of duly apportioning the strength of each piece to the load it has to carry, involves the essence of sound engineering. In that greatest of mechanical feats, the construction of a mighty railway bridge across a wide span, attention to this principle is of vital importance. Such a bridge has to bear the occasional load of a passing train, but it has always to support the far greater load of the bridge materials. There is thus every inducement to make the weight of each part of the bridge as light as may be consistent with safety.

A BRIDGE WITH TWO STRUTS.

422. We shall next examine the structure of a type of bridge, shown in Fig. 58.

It consists of two beams, A B, 4' long, placed parallel to each other at a distance of 3"·5, and supported at each end; they are firmly clamped to the supports, and a roadway of short pieces is laid upon them. At the points of trisection of the beams C, D, struts C F and D E are clamped, their lower ends being supported by the framework: these struts are 2' long, and there are two of them supporting each of the beams. The tray G is attached by a chain to a stout piece of wood, which rests upon the roadway at the centre of the bridge.

423. We shall first determine the strength of this bridge by actual experiment, and then we shall endeavour to explain the results in accordance with mechanical principles. We can observe the deflection of the bridge by the cathetometer in the manner already described (Art. 362). By this means we shall ascertain whether the load has permanently injured the elasticity of the structure (Art. 367). We begin by testing the deflection when a load is distributed uniformly, as the weights are disposed in the case of Fig. 62. A cross is marked upon one of the beams, and is viewed in the cathetometer. We arrange 11 stone weights along the bridge, and the cathetometer shows that the deflection is only 0"·09: the elasticity of the bridge remains unaltered, for when the weights are removed the cross on the beam returns to its original position; hence the bridge is well able to bear this load.

424. We remove the row of weights from the bridge and suspend the tray from the roadway. I take my place at the cathetometer to note the deflection, while my assistant places weights H H on the tray. 1 cwt. being the load, I see that the deflection amounts to 0"·2; with 2 cwt. the deflection reaches 0·43"; and the bridge breaks with 238 lbs.

425. Let us endeavour to calculate the additional strength which the struts have imparted to the bridge. By Table XXIV. we see that a rod 40" × 0"·5 × 0"·5 is broken by a load of 19 lbs.: hence the beams of the bridge would have been broken by a load of 38 lbs. if their ends had been free. As, however, the ends of the beams had been clamped down, we learn from Art. 411 that a double load would be necessary. We may, however, be confident that about 80 lbs. would have broken the unsupported bridge. The strength is, therefore, increased threefold by the struts, for a load of 238 lbs. was required to produce fracture.

426. We might have anticipated this result, because the points C and D being supported by the struts may be considered as almost fixed points; in fact, we see that C cannot descend, because the triangle A C F is unalterable, and for a similar reason D remains fixed: the beam breaks between C and D, and the force required must therefore be sufficient to break a beam supported at the points C and D, whose ends are secured. But C D is one-third of A B, and we have already seen that the strength of a beam is inversely as its length (Art. 388); hence the force required to break the beam when supported by the struts is three times as large as would have been necessary to break the unsupported beam. Thus the strength of the bridge is explained.

427. As a load of 238 lbs. applied near the centre is necessary to break this bridge, it follows from the principle of Art. 408 that a load of about double this amount must be placed uniformly on the roadway before it succumbs; we can, therefore, understand how a load of 11 stone was easily borne (Art 423) without permanent injury to the elasticity of the structure. If we take the factor of safety as 3, we see that a bridge of the form we have been considering may carry, as its ordinary working load, a far greater weight than would have crushed it if unsupported by the struts and with free ends.

428. The strength of the bridge in Fig. 58 is greater in some parts than in others. At the points C and D a maximum load could be borne; the weakest places on the bridge are in the middle points of the segments A C, D C, and D B. The load applied by the tray was principally borne at the middle of D C, but owing to the piece of wood which sustained the chain being about 18" long, the load was to some extent distributed.

The thrust upon the struts is not so easy to calculate accurately. That down C F for example must be less than if the part C D were removed, and half the load were suspended from C. The force in this case can be determined by principles already explained (Art. 420).

A BRIDGE WITH FOUR STRUTS.

429. The same principles that we have employed in the construction of the bridge of Fig. 58 may be extended further, as shown in the diagram of Fig. 59.