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the pier is as the area

DF XFG; therefore DF. FG. FG, or 1DF. FG2, is its effect on
the levere, to prevent the pier from being overset; sup-
posing the length of the pier, from point to point, to be no
more than the thickness of the arch.


But that the pier and the arch may be in equilibrio, these
two efforts must be equal. Therefore we have 1DF. FGa
, an equation, by which will be determined the
thickness of the pier FG; A denoting the area of the half arch



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Example 1. Suppose the arc ABM to be a semicircle; and
that co or oa or OB 45, BC 7 feet, Ar=20. Hence AD =
52, df=ge=72. Also by measurement are found ox=50·3,
KL=40·6, LO=29.7, TD=30.87, KQ-24, the area BCDA 750
A and putting FG x the breadth of the pier.

Then TE TD+ DE = 30.87+x, and KL: LO: TE: ÉV

then GE EV=Gv=49.42 73x,

lastly OK KL:: GV: gr=39·89-59x.
These values being now substituted in the theorem DF.
give 36x217665 -261.5x, or x2 +



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*Note. As it is commonly a troublesome thing to calculate the place of the centre of gravity K of the half arc ADCB, it may be easily, and sufficiently near, found mechanically in the manner described in art. 211, thus: Construct that space ADCB accurately by a scale to the given dimensions, on a plate of any uniform flat substance, or even card paper; then cut it nicely out by the extreme lines, and balance it over any edge or the sides of a table in two positions, and the intersection of the two places will give the situation of the point K; then the distances or lines may be measured by the scale, except those depending on the breadth of the pier FG, viz. the lines as mentioned in the examples.



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7.26x = 490·7 ; the root of which quadratic equation gives 18.8 feet = DE or FG, the thickness of the pier sought. Example 2. Suppose the span to be 100 feet, the height 40 feet, the thickness at the top 6 feet, and the height of the pier to the springer 20 feet, as before.

Here the fig. may be considered as a circular segment, having the versed sine OB 40, and the right sine OA or oc= 50; also BD = 6,

CF 20, and EF =

66. Now, by the nature of the cir

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cle, whose centre is w, the radius wв = OB2 +OC2 402 +502


511; hence ow 511-40=114;

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34.6, IX


2оB and the area of the semi-segment оBC is found to be 1491; which is taken from the rectangle ODEC OD .oc = 46 × 50 2300, there remains 809 = A, the area of the space BDECB. Hence, by the method of balancing this space, and measuring the lines, there will be found, «c = 18, IK 42, KX = 24, ox = 8, IQ 19.4, TE 35-6, and TH 35.6 + x, putting x EH, the breadth of the pier. IK: KX :: TH : HV = 24·7 + 0·7x; hence GH 41·3 — 0·7 — GV, and IX : IK :: GV : GP = 34·02′ 34·02 - 0·58x. These values being now substituted in the theorem EF. gives 33x2 = 15431·47 15431·47 — 253x, or x2 + 8x=467·62, the root of which quadratic equation gives x = 18 = EH or FG, the breadth of the pier, and which is probably very near the truth.




HV =


238. Another use of the centre of gravity, which may be here considered, is in determining the strength and the stress of beams and bars of timber and metal, &c. in different positions; that is, the force or resistance which a beam or bar makes, to oppose any exertion or endeavour made to break it and the force or exertion tending to break it;




both of which will be different according to the place and position of the centres of gravity.


239. The Absolute Strength of any Bar in the Direction of its Length, is Directly Proportional to the Area of its Transverse Section.

SUPPOSE the bar to be suspended by one end, and hanging freely in the manner of a pendulum; and suppose it to be strained in direction of its length, by any force, or weight acting at the lower part, in the direction of that length, sufficient to break the bar, or to separate all its particles. Now, as the straining force acts in the direction of the length all the particles in the transverse section of the body, where it breaks, are equally strained at the same time; and they must all separate or break together, as the bar is supposed to be of uniform texture. Thus then, the particles all adhering and resisting with equal force, the united strength of the whole, will be proportional to the number of them, or as the transverse section at the fracture.

240. Corol. 1. Hence the various shapes of bars make no difference in their absolute strength this depending only on the area of the section, and must be the same in all equal areas, whether round, or square, or oblong, or solid, or hollow, &c.

241. Corel. 2. Hence also, the absolute strengths of different, bars, of the same materials, are to each other as their transverse sections, whatever their shape or form may be.

242. Corol. 3. The bar is of equal strength in every part of it, when of any uniform thickness, or prismatic shape, and is equally liable to be drawn asunder at any part of its length, whatever that length may be, by a weight acting at the bottom, independent of the weight of the bar itself; but when considered with its own weight, it is the more disposed to break, and with the less additional appended weight, the longer the bar is on account of its own weight increasing with its length. And, for the same reason, it will be more and more liable to be broken at every point of its length, all the way in ascending or counting from the bottom to the top, where it may always be expected to part asunder. And hence we see the reason why longer bars are, in this way more liable to break than shorter ones, or with less appended weights. Hence also we perceive that, by gradually increasing these weights, till the bar separates and breaks,



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then the last or greatest weight, is the proper measure of the absolute strength of the bar. And the same is the case with a rope, or cord, &c. So much then for the longitudinal strength and stress of bodies. Proceed we now to consider those of their transverse actions.


243. The Strength of a Beam or Bar, of Wood or Metal, &c. in a Lateral or Transverse Direction, to resist a Force acting Laterally, is Proportional to the Area or Section of the Beam in that Place, Drawn into the Distance of its Centre of Gravity from the Place where the Force acts, or where the Fracture will end.

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Let AB represent the beam or bar, supported at its two ends, and on which is laid a weight w, to cause a transverse fracture abee. The force w acting downwards there, the fracture will commence or open across the fibres, in the opposite or lowest line ab; from

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thence, as the weight presses down the upper line ee, the fracture will open more and more below, and extend gradually upwards, successively to the parallel lines of fibres cc, dd, &c. till it arrive at, and finally open in the last line of fibres ee, where it ends; when the whole fracture is in the form of a wedge widest at the bottom, and ending in an edge or line ee at top. Now the area de contains and denotes the sum of all the fibres to be broken or torn asunder; and as they are supposed to be all equal to one another, in absolute strength, that area will denote the aggregrate or whole strength of all the fibres in the longitudinal direction, as in the foregoing proposition. But, with regard to lateral strength, each fibre must be considered as acting at the extremity of a lever whose centre of motion is in the line ce: thus, each fibre in the line ab, will resist the fracture, by a force proportional to the product of its individual strength into its distance de from the centre of motion, consequently the resistance of all the fibres in ab will be expressed by ab Xae. In like manner, the aggregate resistance of another course of fibres, parallel to ab, as cc, will be denoted by ceXce; and a third, as dd by dd Xde; and so on throughout the whole fracture. So that the sum of all these products will express the total strength or resistance of

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all the fibres or of the beam in that part. But, by art. 222, the sum of all these products is equal to the product of the area deeb, into the distance of its centre of gravity from ee. Hence the proposition is manifest.

244. Corol. 1. Hence it is evident that the lateral strength of a bar, must be considerable less than the absolute longitudinal strength considered in the former proposition, and will be broken by a much less force, than was there necessary to draw the bar asunder lengthways. Because, in the one case the fibres must be all separated at once, in an instant; but in the other, they are overcome and broken successively, one after another, and in some portion of time. For instance, take a walking stick, and stretching it lengthways, it will bear a very great force before it can be drawn asunder; but again taking such a stick, apply the middle of it to the bended knee, and with the two hands drawing the end towards you, the stick is broken across by a small force.


245. Corol. 2. In square beams, the lateral strengths are as the cubes of the breadths or depths.

246. Corol. 3. And in general, the lateral strengths of any bars, whose sections are similar figures, are as the cubes of the similar sides of the sections.

247. Corol. 4. In cylindrical beams, the lateral strengths are as the cubes of the diameters.

248. Corol. 5. In rectangular beams, the lateral strengths are to each other, as the breadths and square of the depths.

249. Corol. 6. Therefore a joist laid on its narrow edge, is stronger than when laid on its flat side horizontal, in proportion as the breadth exceeds the thickness. Thus if a joist be 10 inches broad, by 24 thick, then it will bear 4 times more when laid on edge, than when laid flat. Which shows the propriety of the modern method of flooring with very thin, but deep joists.

250. Corol. 7. If a beam be fixed firmly by one end into a wall, in a horizontal position, and the fracture be caused by a eight suspended at the other end, the process would be the same, only that the fracture would commence above, and terminate at the lower side; and the prop. and all the corollaries would still hold good.

251. Corol. 8. When a cylinder or prism is made hollow, it is stronger than when solid, with an equal quantity of mate


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