Let н be the centre of gravity of the triangle ABE, through which draw KHI parallel to the slope face of the earth BE. Now the centre of gravity H may be accounted the place of the triangle ABE, or the point into which it is all collected. Draw HL parallel, and KP perpendicular to AE, also KL prep. to IK or BE. Then if HL represent the force of the triangle ABE in its natural direction HL, HK will denote its force in its direction HK, and PK the same force in the direction PK perpendicular to the lever EK, on which it acts. Now the three triangles EAB, HKL, HKP are all similar; therefore EB EA (HL HK) the weight of the triangle EAB · EA w, which will be the force of the triangle in the direcThen, to find the effect of this force in the direc

EB

tion HK.
tion PK, it will be, as HK: PK :: EB : AB ::

the force at K, in direction PK, perpendicularly on the lever EK, which is equal to AE. But AE. AB is the area of the triangle ABE; and if i be the specific gravity of the earth, then AE. AB. m is as its weight. Therefore

EA AB
B2

AE. AB =

EA2. AB2
2EB

m is the force acting at к in

direction PK. And the effect of this pressure to overturn the wall, is also as the length of the lever KE or AE*: con

*The principle now employed in the solution of this 45th prop. is a little different from that formerly used; viz. by considering the triangle of earth ABE as acting by lines IK, &c. parallel to the face of the slope BE, instead of acting in directions parallel to the horizon AB; an alteration which gives the length of the lever EK, only the half of what it was in the former way, viz. EKAE instead of AE: but every thing else remaining the same as before. Indeed this problem has formerly been treated on a variety of different hypotheses, by Mr. Muller, &c. in this country, and by many French and other authors in other countries. And this has been chiefly owing to the uncertain way in which loose earth may be supposed to act in such a case; which on account of its various circumstances of tenacity, friction, &c. will not perhaps admit of a strict mechanical certainty On these accounts it seems probable that it is to good experiments only, made on different kinds of earth and walls, that we may probably hope for a just and satisfactory solution of the problem.

The above solution is given only in the most simple case of the problem. But the same principle may easily be extended to any other case that may be required, either in theory or practice, either with walls or banks of earth of different figures, and in different situations.

VOL. II

24

sequently

against K, to overset the wall AEFG. Which must be balanced 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 w were suspended from the point N of the lever FN. Hence, 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 7. 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 therefore A. R FN

AF3 AB3 бEB 2

EA3. AB2
6E B2

m, in case of an equilibrium.

234. But now, both the breadth of the wall Fɛ, 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 ▲ = AE. FE; consequently the effort of the wall A. n. FN is = which must be =

FE2; AE n;

AE 3 AB3
6E B2

m, the effort of And the resolution of this equation gives the

the earth
breadth of the wall FE =

EB

AB. AE m
3n

perp to FB. So that the breadth of the wall is always proportional to the prep depth AQ of the triangle ABE But the 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 ɛ 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 to 2000; or they may be taken as equal; then ✔ ≈1 very

m

n

nearly; and hence FEAE, or AE nearly. That is, whenever a brick rectangular wall is made to support earth, its thickness must be at least or of its height. But if

the

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

m

m

then =

, and ✔

=

22

=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 INFE, and the area AFE. AE; consequently its effort a. n. FN is = FE2 AE n; which being put =

AE3. AB2 6BE3 АВ . ЛЕ

E. B

m, the equation gives FE =

m

An

BC

m
=AQ √ for the breadth

2n

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

AE And when this wall is of brick, then

m

n

But when it is of stone; then √ 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 support a given Arch.

BCDA, in the direction of gravity, this will resolve into κQ, the force acting against the pier perp to the Do the part of the force parallel to the same.

joint SR, and Now KR de

notes

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 DF. FG2 KQ. GP A an equation, by which will be determined the thickness of the pier FG; A denoting the area of the half arch BCDA*.

KL

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, DH GE= 72. Also by measurement are found ok = 50.3, KL = 40.6, LO = 29.7, TD = 30-87, ко 24, the A; and putting FG the breadth

area BCDA ==

of the pier. Then TE

750 =

=

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 Dr.

FG2 =

KQ. GP. A

KL

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 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.

7.26x=490-7; the root of which quadratic equation gives x= 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.

cle, whose centre is w, the radius wв = OB +OC2 40%+502

20B

80

= 514; hence ow = 511-40114; and the area of the semi-segment оBC 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 вDECB. Hence, by the method of balancing this space, and measuring the lines, there will be found, KC = 18, IK = 34.6, IX = 42, KX = 24, ox = 8, IQ 19.4, TE = 35 6, and тH = 35.6x, putting x EH, the breadth of the pier. Then IK: KX :: TH: HV = 24.7 +0.7x; hence GH 41.3-0.7 = GV, and IX: IK: GV: GP 34.02. These values being now substituted in the theorem EF. FG2 =

IQ. GP A

IK

gives 33x2 = 15431.47

- HV =

0.58x.

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.

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