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them alone would; but using two bars is equivalent to doubling the width of one, hence it is clear that the strength of the bar varies as its breadth.

Fig 11.

a

Let A B (Fig. 11) represent a solid bar supposed to consist of a number of layers, and a b one arm of the assumed bent lever, referred to in previous demonstrations, in the arm b d may be found a point c at which the resisting efforts of all the fibres may be supposed to be centred, and this will always bear a certain relation to the B whole depth of the beam; hence, if the bar is increased by layers, shown by the dotted lines (but the whole being solid, not in separate layers), the length to this centre of resistances will increase from

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A

be to be, varying as the depth of the bar. Again, by increasing the depth of the bar, we also, in like proportion, increase the number of layers of resisting fibres; hence thus also increase the strength. Thus, by increasing the depth of the bar, we increase the number of resisting fibres as the depth, and increase their mean leverage as the depth; wherefore it is found that the strength of the bar varies as the square of its depth, and thus the accuracy of the law is proved. There is another law to be mentioned in connection with rectangular bars; it is this:

LAW. All other things being unaltered, the strength of the bar varies inversely as its length.

This is evident, for as the length of the bar varies so does the length of the arm a b of the assumed lever

upon which the load acts vary, and the greater the leverage of the load the less will be the amount of load which the bar is capable of sustaining.

Having thus disposed of the strains upon straight beams, we must now proceed to treat of the effects of distributed load upon arches and chains. The intensity of the load will in both cases be determined by the same rules, the difference between the two being that the upright arch is in compression, but the inverted arch or chain is in tension.

Let A B, Fig. 12, represent an arch carrying an uniformly distributed load upon the platform over it. A B is the span, and de the "rise or "versine" of the arch. It is required

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to find the thrust at the centre or crown of the arch.

D

f

Fig. 12.

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B

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e

k

RULE (17). Multiply the load per lineal foot by the square of the span of the arch in feet, and divide the product by eight times the versine of the arch in feet.

Example:-Let the arch be 240 feet span, its versine 30 feet, and the load per lineal foot 5 tons :—

240 feet span,

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00

1200 tons' thrust at crown

of arch.

It is now requisite to determine the thrust at any other point, e, lying vertically under the point f. RULE (18). To find the thrust at any other point than the crown, multiply the load per lineal foot by the distance of such point from the crown, square the product, and add to it the square of the thrust at the crown, then the square root of the sum will be the thrust at the point referred.

(Note. This rule will of course give the thrust at the abutment by taking the point at a distance of half the span from the crown of the arch.)

Example:-Let the thrust be required in the case of the above arch at a distance of 40 feet from the crown or centre:

1200 tons' thrust at

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1440000 square of thrust at crown, 1)1480000(1216 tons' thrust at point (nearly).

1

22) 48

44

241) 400
241

2426)15900
14556

1344

After what has already been shown in previous pages, the demonstration of these rules will be comparatively simple.

Σ

Let CD (Fig. 12) represent the outline of an arch having a span C D = = l, and a versine kg = v; let the load per lineal unit = w, the thrust at centre = T, and the thrust at any point distant x from the centre = S: -As the arch is uniformly loaded throughout its length, the actual weight or vertical reaction on each pier will be half the total load, or

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and regarding C kg as an imaginary bent lever, this force will act with the leverage Ck, producing a thrust at the end g of the arm k g; hence the thrust on the crown of the arch will be the same as that at the centre of the top flange of a flanged beam having a span equal to that of the arch and a depth equal to its versine, therefore

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The thrust at any point e is the resultant of two forces acting at right angles to each other, the one being the horizontal thrust above given, the other the vertical weight of the load between h and g. Let h m and he represent these two forces, then, according to the parallelogram of forces, m e will be the resultant strain. But because e h m is a right-angled triangle (Euc., Bk. 1, prop. 47)

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hm =

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em = S

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From equation (6) we have the rule for the strain at the centre:

RULE. Multiply the load per lineal foot by the square of the span of the arch in feet, and divide the product by eight times the versine of the arch in feet.

From equation (c) the rule is found for the strain at any other point.

mul

RULE. To find the thrust at any other point than the crown, tiply the load per lineal foot by the distance of such point from the crown, square the product, and add to it the square of the thrust at the crown, then the square root of the sum will be the thrust at the point referred to.

CHAPTER III.

COMBINATIONS OF ELEMENTS AND DISTRIBUTION OF LOADS.

In the preceding chapter we have shown how to calculate the strains on various kinds of elements when the loads or forces by which such strains are caused are known; in the present we intend treating of the distribution of loads due to certain combinations of elements, as preliminary to explaining the modes of proportioning complex structures according to the weights they are designed to sustain. In order to make clear the mode

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