EA3. AB2 sequently its effect is 6EB2 -m, for the perpendicular force 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 M 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 . 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 EA3. AB2 before found equal to AE3. AB2 6EB2 therefore A n FN m, in case of an equilibrium. 234. But now, both the breadth of the wall FE, and the lever FN, or place of the centre of gravity M, will depend on the figure of the wall. If the wall be rectangular, or as broad at top as bottom; then FNFE, and the area A = AE. FE; consequently the effort of the wall A n. FN is FE. AE.n; which must be = the earth. And the resolution of this equation gives the breadth of the wall FE = AE3. AB2 '6LB2 AB. AE m EB 3n m, the effort of m N =AQ V. drawing AQ perp. to EB. So that the breadth of the wall is always proportional to the perp. 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 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 earth about 1984; namely, m to n as 1984 to 2000; or they may be taken as equal; then ✔ nearly; and hence FEAE, or AE nearly. That is, wherever a brick rectangular wall is made to support earth, its thickness must be at least 3 oref its height. But if the the wall be of stone, whose specific gravity is about 2520; 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 FNFE, and the area AFE. AE; consequently its effort a. n. FN is = FE. AE. n; which being put = AE3. AB2 6BE2 -m, the equation gives FE = BC F NE D 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√ m. FEAE nearly. But when it is of stone; then m n 447 nearly: that is, the triangular stone wall must have its thickness at bottom equal to 4 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. LET ABCD be half the arch, and DEFG the pier. From the centre of gravity K of the half arch draw KL perp. OA; also OKR, and TKQP perp. to it; also draw LQ and GP perp. to TP, or parallel to OKR. Then if KL represent the weight of the arch BCDA, in the direction of gravity, this will resolve into KQ, the force acting against the pier perp. to the joint SR, and LQ the part of the force parallel to the same. Now de KQ 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. KQ. GP. A But that the pier and the arch may be in equilibrio, these two efforts must be equal. Therefore we have DF. FG2 = 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 simicircle; 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 = 503, KL 40'6, Lo 29.7, TD = 30.87, KQ = 24, the area BCDA = 750 A; and putting FG = the breadth of the pier. Then TE Ev then GE TD + DE 30.87+x, and KL : LO :: TÈ: 22.58 +0.73x, lastly OK KL :: GV GP 39.89 These values being now substituted in the theorem DF. FG2 = KQ. GP. A , give 36x 17665 261.5x, or *2+ *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 side 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. 7.26% 4907; 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 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 F G cle, whose centre is w, the radius WB = OB2 + оe2 402 +502 514; hence ow1-40=11'; = and the area of the semi-segment OBC is found to be 1491; which 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 194, TE 356, and TH= 35.6+x, putting x = EH, the breadth of the pier. Then IK: KX :: TH: HV 247 +0.7%; hence GH 41.3 - 0.7x GV, and IX: IK:: GV: GP 34.02 - 0.58x. These values being now substituted in the theorem EF. IQ. GP. A IK HV = FG2 = gives 33x15431:47-263x, or x2 + 8x=467-62, the root of which quadratic equation gives * = 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; bot! both of which will be different, according to the place and position of the centres of gravity. PROPOSITION XLVII. 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 a 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. Corol. 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, then |