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gravity 21, that of earth being nearly 2, that is, m = 2, and

m

n=21; then ✅✔✅ ✓895, of which is 298, and the

n

§

breadth x=•298a=a nearly. That is, the thickness of the stone wall must be of its height.

10

PROBLEM III.

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

In this case, the area of the wall AEF C B

is only half of what it was before, or
only | AE X EF ax, and the weight
waxn. But now, the centre of gra-
vity is at only of FE from the line AE,
3 FE x. Consequently FM D
Xw=2xX1αxn= ax2n. This, as be-
fore, being put the pressure of the

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or FN

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a2 e3m,

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for the

3n

6n

earth, gives the equation ax3n = ¦a3e2m, or x2n = 2

and the root x, or thickness EF = de✓✓ slope of 45°.

Now when the wall is of brick, or mn nearly, this becomes x = a✓ } = · 408α = &a, or of the height nearly. But when the wall is of stone, or m to n as 2 to 21, then

m

n

10

✓=, and the thickness x or a✓

a nearly, or nearly of the height.

PROBLEM IV.

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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.

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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 prep. to EF; and conceive the two weights w, w, to be suspended from the centres of gravity of the rectangle AH and the triangle GAF, 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 prep. to AE.

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Now put the required thickness AG or EHx, 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

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that is FH = AE or Ja, the lever FM will be of ja
and the lever FN = FH + EH = a + 1x. But the area
of GHF = GH X 1HF =α × √ α =
a Τσα a2=w, and the area
AH = AE X AG = ax=w; these two drawn into the re-

spective levers FM, FN, give the two momenta, aw =

×

ax

a2 =α3, and (a + x) x ax = ax+ax2; theref,
이?,

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the sum of the two, (}ax2+}a3x+a3)n must be = }} a3m

m

or dividing by fan, x2 + zax +75a2 = fa2 × =

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; now adding to both sides to complete the square, the equation becomes xa+jax +z a3 =}α2·· a2, the root of which

m

5

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n

m

is x+}a=α√ (12'5 +), and hence x = α √ (13+ ~~) — zα.

In

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And the base EF = α √ (3} + — ).

In

Now, for a brick wall, m = n nearly, and then the breadth
x = α √(√ + }) — ja = }} α √34
a (+) · 75

Ja in brick walls. But the stone walls,

ja

189a, or almost

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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

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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 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

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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.

PROBLEM V.

To Determine the Quantity of Pressure sustained by a Dam or
Sluice, made to pen up a Body of Water.

By art. 813 Hydrostatics, (in this volume) the pressuré 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 20×6=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 120×3= 360 cubic feet is the content of the column of water. But each cubic foot of water weighs 1000 ounces, or 621 pounds; therefore 360 X 1000 = 360000 ounces, 22500 pounds, or 10 tons and 100lb. 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

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Here the gate is in form of a trapezoid, AH-E
having the two parallel sides AB, CD, VIZ. AB
=24, and CD=16, and depth 6 feet. Now,
by mensuration problem 3, volume 1, 1
(AB+CD) × 6, 20 X 6 = 120 the area of
the sluice, the same as before in the 1st ex-
ample 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.

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Now, to determine the centre of gravity
K of the trapezoids AD, produce the two sides
AC, BD, till they meet in &; also draw GKE and
CH perp. to Aв then AH: CHAE GE, that is, 4: 6:: 12:
18 GE; and EF being 6, theref. FG= =12. Now, by Statics
art. 229, EF=6=EG gives F the centre of gravity of the tri-
angle ABG, and F1=4=1FG gives 1 the centre of gravity of the
triangle CDG. Then assuming x to denote the centre of AD,
it will be, by art. 212 this vol. as the trap. AD: ▲ CDG :: IF :
FK, or A ABC Δ ▲ CDG: A CDG :: IF: FK, or by theor. 88
Geom. GE2 - GF2 : GF2 :: IF FK, that is 183-122 to 122 or
32 -2a to 2a or 5: 4 :: IF=4: 16=31=PK; and hence EK
=6-33 = 24 = 4 14 is the distance of the centre x 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 80lb. 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 bot-
tom, and 8 feet depth of water ?

PROBLEM VI.

To determine the Strongest Angle of Position of a Pair of Gates
for the Lock on a Canal or River..

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A

E

D

B

Let AC, BC be the two gates, meet-
ing 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
Ac3. Therefore the resistance or the strength of the gate
must be as the reciprocal of this ac2.

Now

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1

Now produce ac to meet BD, prep. to it, in D; and draw ce to bisect AB perpendicularly in E; then, by similar triangles, as AC AE :: AB: AD; where, AE and AB being given lengths, AD is reciprocally as Ac, or AD2 reciprocally as Ac2; that is, AD2 is as the resistance of the gate AC. But the resistance of ac is increased by the pressure of the other gate in the direction BC*. Now the force in BC is resolved into the two BD, DC; the latter of which, Dc, being parallel to ac, has no effect upon it; but the former, BD, acts perpendicularly on it. Therefore the whole effective strength or resistance of the gate is as the product AD2 X bd.

2

If now there be put ABα, and BD = x, then Ad2 = AB2 -BD2 = α2 — x2; conseq. AD2 XBD=(A2 — x2) Xx—a3 x − x 3 for the resistance of either gate. And, if we would have this to be the greatest, or the resistance a maximum, its fluxion must vanish, or be equal to nothing: that is, a2 x − 3x2 x=0; hence a2=3x2, and xa3a/357735a, the natural sine of 35o 16': that is, the strongest position for the lock gates, is when they make the angle a or в= 35° 16', or the complemental angle ACE or BCE =54° 44′, or the whole salient angle ACB = 109° 28′.

Scholium.

Allied to this problem, are several other cases in mechanics, such as, the action of the water on the rudder of a ship, in sailing, to turn the ship about, to alter her course; and the action of the wind on a ship's sails, to impel her forward; also the action of water on the wheels of water-mills, and of the air on the sails of wind-mills, to cause them to turn round. Thus, for instance, let

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ABC be the rudder of a ship ABDE, sailing in the direction BD, the rudder placed in the oblique position BC, and consequently striking the water in the

direction CF, parallel to BD. Draw EF prep. to BC, and BG prep. to cr. Then the sine of the angle of incidence, of the direction of the stroke of the rudder against the water, will be BF, to the radius cF; therefore the force of the water against the rudder will be as BF2, by art. 3, Mot. of bod. in Flui. this vol. But the force BF resolves into the two BG, CF, of which the latter is parallel to the ship's motion, and there

*The reasoning in this and in two following sentences does not appear to me to be founded on mechanical principles.

fore

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