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AE3. AB2


m, becomes falem, 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, and the pressure becomes a3m.


To determine the Thickness of Wall to support the Earth.



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 below 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 XFN in general, whatever be the figure of the wall.

But now in case of the rectangular figure, the area GE=AF XEFax, putting a=AE the altitude as before, and r = EF the required thickness; also in this case FN EF = r, the centre of gravity being in the middle of the rectangle. Hence then axxxar, or rather arn 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 ar2n = a3e2m, or a2n = ža2e2m, an equation which gives for the requisite thickness of the wall, just

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to sustain it in equilibrio.

Coról. 1. The factor de, in this expression, is the line AQ drawn perp. to the slope of earth BE: theref. the breadth r becomes


40, which conseq. is directly propor


tional 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


a. Further, when

and the breadth of the wall r = }α√.


the wall is of brick, its specific gravity is nearly the same as


the earth, or mn, and then its thickness ra, or onethird of its height.-But when the wall is of stone, of the specific gravity 2, that of earth being nearly 2, that is, m = 2, and n = 21; then /== '895, of which 13



is 298, and the breadth = 298aa nearly. That is, the thickness of the stone wall must be of its height.


To determine the Thickness of the Wall at the Bottom, when its Section is a Triangle, or coming to an Edge at Top.

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In this case, the area of the wall AEF CB is only half of what it was before, or only AEX EF = ax, and the weight waxn. But now, the centre of gravity is at only of FE from the line AE, or FN=FE3r. Consequently FN X w = 3xx axnax'n. This, as before, being put the pressure of the earth, gives the equation ar1n={a3e2m, or x2n={a2e2m, and the root, or thickness EF = aeNT = a√√ T the slope of 45°.


m 3n



Now when the wall is of brick, or m = n nearly, this becomes rα=408a3a, or of the height nearly. But when the wall is of stone, or m to n as 2 to 24, then


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√, and the thickness r or a =α√73= 365a a nearly, or nearly of the height.


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



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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. $ 2


of a = a,

But the area

a2 = w, and the area

Now put the required thickness AG or EH = ', 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 a, the lever FM will be and the lever FN FH + EH = a + x. of GHF GH X HF = a × √α = AH = AE × AG=ax=w; these two drawn into the respective levers FM, FN, give the two momenta, aw Στα χ aa3, and (a + 4x) × ax = za2x + {ax*; theref. the sum of the two, ({ax2+}u2x+75a3)n must be =75a3m, or dividing by lan, x2 + }ax + za2 = ža2 × ing to both sides to complete the square, the equation becomes x2+zax +1⁄2za2±ža2. — + a2, the root of which



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-; now add


is x + ‡a = a√(13+), and hence x=a√(23+)− za.


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Now, for a brick wall, m = n nearly, and then the breadth x = a√ (23 + 1) — zα = 1}α/34-ja·189a, or almost

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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.


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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 thè 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


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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.


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 3 360 cubic feet is the content of the column of water. But each cubic foot of water weighs 1000 ounces, or 624 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 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 x 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.

Now, to determine the centre of gravity k of the trapezoid AD, produce the two


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sides AC, BD, till they meet in G; also draw CKE 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 F1 = 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 A CDG: A CDG :: IF FK, or by theor. 88 Geom. GE2 - GF2: GF2:: IF : FK, that is 182 122 to 122 or 32-22 to 22 or 5: 4 :: IF = 4: 10 = 35 = FK; and hence EK = 6 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.


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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?


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.

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