EA3AB2
6E B2

sequently its effect is m, for the perpendicular force Which must be baagainst K, to overset the wall AEFG. lanced by the counter resistance of the wall, in order that it may at least be supported.

Now, if м be the centre of gravity of the wall, into which its whole matter may be supposed to be collected, and acting in the direction MNW, its effect will be the same as if a weight Hence, w were suspended from the point N of the lever FN. if A be put for the area of the wall AEFG, and n its specific gravity; then A. n will be equal to the weight w, and A. n. FN its effect on the lever to prevent it from turning about the point F And as this effort must be equal to that of the triangle of earth, that it may just support it, which was before found equal to

EA3. AB2
6EB2

-m; therefore A

n FN

234. But now, both the breadth of the wall FE, and the lever FN, or place of the centre of gravity м, will depend on the figure of the wall. If the wall be rectangular, or as broad at top as bottom; then FN = FE, and the area a = AE. FE; consequently the effort of the wall A. n. FN is = FE'. AE. 7; which must be = -m, the effort of the earth. And the resolution of this equation gives the

AE3. AB2

ABAE m
EB

3n

m

drawing AQ

3n

But the

So that the breadth of the wall is always properp. to FB. portional to the prep. depth AQ of the triangle ABE. breadth must be made a little more than the above value of it, that it may be more than a bare balance to the earth.If the angle of the slope E be 45°, as it is nearly in most cases;

235. If the wall be of brick, its specific gravity is about 2000, and that of the earth about 1984; namely, m to n as 1984

m

to 2000; or they may be taken as equal; then == 1 very

nearly; and hence FEAE, or whenever a brick rectangular wall is its thickness must be at least or

AE nearly. That is, made to support earth, of its height. But if

the

the wall be of stone, whose specific, gravity is about 2520;

m

then =, and √

H

m

n

895; hence FE 358 AE

AE: that is, when the rectangular wall is of stone, the breadth must be at least of its height.

236. But if the figure of the wall be a triangle, the outer side tapering to a point at top. Then the lever FN=FE, and the area AFE. AE; consequently its effort A. n. FN is FE AE.n; which being put = m, the equation gives FE =

AE3. AB

6BE AB, AE

EB

1

F

E

=AQ √ for the breadth

2n

BC

of the wall at the bottom, for an equilibrium in this case also. -If the angle of the slope E be 45°; then will FE be = And when this wall is of brick, then

FEJAE nearly. But when it is of stone; then √

•447 = nearly; that is, the triangular stone wall must have its thickness at bottom equal to of its height. And in like manner, for other figures of the wall and also for other figures of the earth.

PROPOSITION-XLVI.

237. To determine the Thickness of a Pier, necessary to sufi port a given Arch.

BCDA, in the direction of gravity, this will resolve into KQ, the force acting against the pier perp. to the joint SR, and La the part of the force parallel to the same. Now KQ de

notes

notes the only force acting perp on the arm GP, of the crooked lever FGP, to turn the pier about the point G; conseq. KQ X GP will denote the efficacious force of the arch to overturn the pier.

Again, the weight of the pier is as the area

FG. FG, or DF FG2, is its effect on the lever FG, to prevent the pier from being overset; supposing 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 JDF. FG2 = an equation, by which will be determined the thickness of the pier FG; A denoting the area of the half

KQ. GP. A

KL

arch BCDA.

Example 1. Suppose the arc ABM to be a semicircle; and that CD or OA or OB 45. BC 7 feet, AF 20. Hence AD = 52 DF GE=72. Also by measurement are found OK = 50 3, KL =406, LO = 297, TD = 30 87, KQ = 24, the area BCDA = 750 = ; and putting FG the breadth

of the pier.

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

EV 22.58 +0.73x.

then GE-EV

GV 49 42-73x,

lastly OK: KL :: GV: GP

39 89-59x.

These values being now substituted in the theorem DF.

give 36x2 17665 261.5x, or x2 +

Note. As it is commonly a troublesome thing to calculate the place of the centre of gravity K of the half arch ADCB, it may be easily, and sufficiently near, found mechanically in the manner des cribed 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 fines, 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 x; 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.

7.262

80

radius wв =

51; hence ow

=

514-40111;

and the area of the semi-segment OBC is found to be 1491; which is taken from the rectangle ODEC=OD. OC-46 X 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, KC 18, IK = 34.6, IX = = 24, 0x = 8, 1Q 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 0.58.x. These values being now substituted in the theorem EF.

42, KX

FG

IQ. GP
IK

A

gives 33x2 = 15431.47

Then

HV =

263x, 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.

ON THE STRENGTH AND STRESS OF BEAMS OR BARS OF TIMBER AND METAL, &c,

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

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

PROPOSITION XLVU.

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. suf ficient 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 ali equal areas, whether round, or square, or oblong, or solid, or hollow, &c.

241. Corol. 2. Hence also, the absolute strengths of dif ferent bars, of the same meterials, 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 And top, where it may always be expected to part asunder. hence we see the reason why longer bars are, in this way more liable to break than shorter ones, or with less ap pended weights. Hence also we perceive that, by gradually Increasing these weights, till the bar separates and breaks,

then