-m, becomes zaem, for the perpendicular pressure of the earth against the wall. And if that angle be 45°, as is usually the case in common earth, then is e2 = 1⁄2, and the pressure becomes a3m. PROBLEM II. I To determine the Thickness of Wall to support the Earth. AG In the first place suppose the section of the wall to be a rectangle, or equally thick at top and bottom, and of the same height as the rampart of earth, like AEFG in the annexed figure. Conceive the weight w, proportional to the area GE, to be appended to the base directly be CB ENF low the centre of gravity of the figure. Now the pressure of the earth determined in the first problem, being in a direction parallel to AG, to cause the wall to overset and turn back about the point F, the effort of the wall to oppose that effect, will be the weight w drawn into FN the length of the lever by which it acts, that is w X FN, or AEFG X FN in general, whatever be the figure of the wall. But now in case of the rectangular figure, the area GE AT XEF=ax, putting a=AE the altitude as before, and x = EF the required thickness; also in this case FN = EF =r, the centre of gravity being in the middle of the rectangle. Hence then ax xx = ar2, or rather tax'n is the effort of the wall to prevent its being overturned, n denoting the specific gravity of the wall. Now to make this effort a due balance to the pressure of the earth, we put the two opposing forces equal, that is ax'n = a3em, or rn = fatem, an equation which gives x=ae, for the requisite thickness of the wall, just to sustain it in equilibrio. 2m 2m n Corol. 1. The factor de, in this expression, is = the line AQ drawn perp. to the slope of earth BE: theref. the breadth x becomes = AQ, which conseq. is directly proportional to the perp. AQ.-When the angle at E is = 45°, or half a right angle, as is commonly the case, its sine e is = √, and the breadth of the wall x = a. Further, when m 12 L the wall is of brick, its specific gravity is nearly the same as the 1 the earth, or m = n, and then its thickness x = a, or onethird of its height. But when the wall is of stone, of the specific gravity 24, that of earth being nearly 2, that is, m m = 2, and n = 24; then ✓ = = √ = 895, + of which n a nearly. That is, is 298, and the breadth x = 298a PROBLEM III.. To determine the Thickness of the Wall at the Bottoni, when its Section is a Triangle, or coming to an Edge at Top. CB In this case, the area of the wall AEF is only half of what it was before, or only AE X EF = tax, and the weight w=axn. But now, the centre of gravity is at only of FE from the line AE, or FN = FE = r. Consequently FN X W = x xaxn = ax'n. This, as before, being put = the pressure of the earth, gives the equation fax'n=a3em, and the root x, or thickness EF = ae the slope of 45°. or x2n=a2cm, n = av en m 3n m for Now when the wall is of brick, or m = n nearly, this becomes x = a = 408a = a, or of the height nearly. But when the wall is of stone, or m to n as 2 to 21⁄2, then •365a = a nearly, or nearly of the height. PROBLEM IV. To determine the Thickness of the Wall at the Top, when the Face is not Perpendicular, but Inclined as the Front of a Fortification Wall usually is. Here GF represents the outer face of a fort, AEFG the profile of the wall, having AG the thickness at top, and EF that at the bottom. Draw GH perp. to EF; and conceive the two weights w, w, to be suspended from the centres of gravity of the rectangle AH and the tri angle GHF, and to be proportional to their areas respectively. Then the two momenta of the weights w, w, acting by the levers FN, FM, must be made equal to the pressure of the earth in the direction perp. to AE. Now $2 Now put the required thickness AG or EH = x, and the altitude AE or GH = a as before. And because in such cases the slope of the wall is usually made equal to + of its altitude, that is FH = AE or fa, the lever FM will be of a = rsa, and the lever FN = FH + H = a + . But the area of GHF = GH X HF = a × a = a2 = w, and the area AH = AE X AG=ax = w; these two drawn into the respect ive levers FM, FN, give the two momenta, ζαω = τα x Toa = a3, and (a + x) & ax = ax + ar2; theref. the sum of the two, (ax2+zu2x ++a3)n must be = m 72 2 a3m, or dividing by an, x2 + fax + a2 = a2 × ; now adding 73 to both sides to complete the square, the equation becomes x2 + ax + 222 + a2, the root of which m n 9n Now, for a brick wall, m = n nearly, and then the breadth x = a√(3 + 1) - za = a/34-ła = •189a, or almost I m Ja in brick walls. But in stone walls, = 4, and x = 4 n a(+)- a = 29-a=159a = a nearly, for the thickness AG at top, in stone walls. In the same manner we may proceed when the slope is supposed to be any other part of the altitude, instead of as used above. Or a general solution might be given, by assuming the thickness = part of the altitude... 1 C REMARK. Thus then we have given all the calculations that may be necessary in determining the thickness of a wall, proper to support the rampart or body of earth, in any work. If it should be objected, that our determination gives only such a thickness of wall, as makes it an exact mechanical balance to the pressure or push of the earth, instead of giving the former a decided preponderance over the latter, as a security against any failure or accidents. To this we answer, that what has been done is sufficient to insure stability, for the following reasons and circumstances. First, it is usual to build several counterforts of masonry, behind and against the wall, at certain distances or intervals from one another; which contribute very much to strengthen the wall, and to resist the pressure of the rampart. 2dly. We have omitted to include the effect of the parapet raised above the wall; which must add somewhat, by its weight, to the force or resistance of the wall. < 1 wall. It is true we could have brought these two auxiliaries to exact calculation, as easily as we have done for the wall itself: but we have thought it as well to leave these two appendages, thrown in as indeterminate additions, above the exact balance of the wall as before determined, to give it an assured stability. Besides these advantages in the wall itself. certain contrivances are also usually employed to diminish the pressure of the earth against it: such as land-ties and branches, laid in the earth, to diminish its force and push against the wall. For all these reasons then, we think the practice of making the wall of the thickness as assigned by our theory, may be safely depended on, and profitably adopted; as the additional circumstances, just mentioned, will sufficiently insure stability; and its expense will be less than is incurred by any former theory. PROBLEM V. To determine the Quantity of Pressure sustained by a Dam or Sluice, made to pen up a Body of Water. By art. 313 Hydrostatics, vol. 2, 6th edit. the pressure of a fluid against any upright surface, as the gate of a sluice or canal, is equal to half the weight of a column of the fluid, whose base is equal to the surface pressed, and its altitude the same as that of the surface. Or, by art. 314 of the same, the pressure is equal to the weight of a column of the fluid, whose base is equal to the surface pressed, and its altitude equal to the depth of the centre of gravity below the top or surface of the water; which comes to the same thing as the former article, when the surface pressed is a rectangle, because its centre of gravity is at half the depth. Ex. 1. Suppose the dam or sluice be a rectangle, whose length, or breadth of the canal, is 20 feet, and the depth of water 6 feet. Here 20 x 6 = 120 feet, is the area of the surface pressed; and the depth of the centre of gravity being 3 feet, viz, at the middle of the rectangle; therefore 120 x 3360 cubic feet is the content of the column of water. But each cubic foot of water weighs 1000 ounces, or 62 pounds; therefore 360 x 1000 = 360000 ounces, or 22500 pounds, or 10 tons and 100 lb, is the weight of the column of water, or the quantity of pressure on the gate or dam. Ex. 2. Suppose the breadth of a canal at the top, or surface of the water, to be 24 feet, but at the bottom only 16 feet, the depth of water being 6 feet, as in the last example: required the pressure on a gate which, standing across the canal, dams the water up? Here Here the gate is in form of a trapezoid, having the two parallel sides AB, CD, viz, AB = 24, and CD = 16, and depth 6 feet. Now, by mensuration, problem 3 vol. 2, (AB+CD) x 6 = 20×6= 120 the area of the sluice, the same as before in the 1st example: but the centre of gravity cannot be so low down as before, because the figure is wider above and narrower below, the whole depth being the same. AHE B Now, to determine the centre of gravity K of the trapezoid AD, produce the two sides AC, BD, till they meet in G; also draw GKE and CH perp. to AB: then AH: CH :: AE: GE, that is, 4:6:: 12: 18 = GE; and EF being = 6, theref. FG = 12. Now, by Statics art. 229 vol. 2, EF = 6 = EG gives F the centre of gravity of the triangle ABG, and FI = 4 = FG gives I the centre of gravity of the triangle CDG. Then assuming K to denote the centre of AD, it will be, by art. 212 vol. 2, as the trap. AD: A CDG :: IF: FK, or ABC - ACDG: A CDG :: IF: FK, or by theor. 88 Geom. GE - GF : GF :: IF : FK, that is 182- 122 to 122 or 32-22 to 22 or 5:4:: IF = 4: =3=FK; and hence EK = 6 - 3 = 24= is the distance of the centre k below the surface of the water. This drawn into 120 the area of the dam-gate, gives 336 cubic feet of water = the pressure, = 336000 ounces = 21000 pounds = 9 tons 80 lb, the quantity of pressure against the gate, as required, being a 15th part less than in the first case. Ex. 3. Find the quantity of pressure against a dam or sluice, across a canal, which is 20 feet wide at top, 14 at bottom, and 8 feet depth of water? PROBLEM VI. 1 To determine the Strongest Angle of Position of a Pair of Gates for the Lock on a Canal or River. Let AC, BC be the two gates, meeting in the angle c, projecting out against the pressure of the water, AB being the breadth of the canal or river. Now the pressure of the water on a gate Ac, is as the quantity, or as the extent or length of it, AC. And the mechanical effect of that pressure, is as the length of lever to the middle of AC, or as AC itself. On both these accounts then the pressure is as AC2. |