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6. Demonstrate 1st, That the surface produced by a plane line or curve by revolving about an axis in the plane of that curve, is equal to the product of the generating line or curve into the path described by the centre of gravity.

And 2dly. That the solid produced by the revolution of a plane figure about an axis posited in the plane of that figure, is equal to the product of the generating surface into the circumstance described by the centre of gravity.

ON THE EQUILIBRIUM OF ARCHES.

116. A very interesting department of the science of Statics, is that which relates to the stability of arches, as introduced in the construction of bridges, powder-magazines, &c. Every such structure is a system of forces, and the examination of its firmness, therefore, requires the application of the general principles of equilibrium. We shall here present a few useful propositions in elucidation of the more received theories.

117. PROP. The force of a voussoir depending on the magnitude of the angle formed by its sides, the impelling force, and the resistance to be overcome, is on the first account directly as the radius of curvature of the arch at that point, on the second as the square of the sine of the angle included between the tangent of the curve at the given point and the vertical passing through that point, and on the third, as the sine of the same angle.

1. Let EABF, eabf, be two similar concentric curves, and AB, ab, two voussoirs similarly situated, whose sides perpendicular to the curve converge to the centre C. The forces of these vous. soirs considered as portions of wedges, are inversely as the sines of the half vertical angles (schol. art. 82.) or,

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because each wedge occupies an equal portion of its respective arch, directly as the radii of curvature.

2ndly, Let нh be the invariable breadth of the voussoirs on the arch cabf, GgHk the incumbent weight, which, since GH is supposed given, is as the breadth hk, or as the sine of the angle huk: by the resolution of the force gн into two hн, HK, the latter is the force impelling the voussoir to split the

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arch, which, since gн is given, varies as the sine of нg, or hнk: wherefore, the force impelling the voussoir is as the square of the sine of hнk.

3dly, The wedge impelled in a direction perpendicular to the curve endeavours to split the arch, and therefore to move one segment about the fulcrum e, the other about the fulcrum f. Hence the force of the voussoir acting on the levers нf, He, being as either of the perpendiculars fr, ea, is as the sine of the angle fcr or hнk.

We have supposed the centre of curvature of the arches at the points a, a, h, H, to be at c: but this is merely to prevent the figure from being too complex, and makes no alteration in the nature of the demonstration.

Corol. Hence, if the height of the wall incumbent on any point н of the intrados is inversely as the cube of the sine of hнk into radius of curvature at that point, or directly as cube of the secant of the angle formed by hн and the horizon, and inversely as the radius of curvature, all the voussoirs will endeavour to split the arch with equal forces, and will be in perfect equilibrium with each other.

The general expression, therefore, for the thickness GH over any point of an arch, is

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where rrad. of curvature at the vertex

a thickness of material there

R = rad. of curvature at H.

The radii of curvature for the different curves are deter. minable by the method of fluxions, or by other means: they are here supposed known.

1. Suppose, for example, it were required to find the requisite thickness over any point of a circular arc, to ensure equilibration, the thickness a= DK, at the crown of the arch being given.

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Hence we have a convenient logarithmic expression for computation; viz.

Log. DK +3 log. sec. DH = log. GH.

In this example, the curve of equilibration, GKS, runs up to an infinite height over B, the springing of a semicircular arch. But over a portion of 30° or 35° on each side the vertex, as DH, the curve KC of the extrados accords very well with what would be required for a roadway.

Ex. 2. Determine the requisite thickness for equilibration over any point of a parabola.

Here if DR = 7, RH = y, R = (4x+p)3

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which, at the vertex, where x vanishes becomes r=tp:

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T

n

D

R

RT = 2x, RH = y = √px, тH = √/HR2 + RT2 = √px+4x2

TH

sec. THR

HR

‚px + 4x2 p+ 4x

=√

P

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a = α = KD.

So that the extrados is a parabola equal to the intrados, and every where vertically equidistant from it.

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R (parallel to sc) HO= 2sc = 2✓d2-dx; r=2cD=2d:

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By computing the value of GH for several corresponding values of DR, and CR, and thence constructing the extrados by points, it will, as in the figure, appear analogous to that for the circle, but rather flatter till it approach the extremities of the arch, where the curve runs off to infinity, as in the case for the circle.

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на

=QR.

Also, sec. THR=sec. HQR====÷ (c-x) = f(c — x) *

QR

Radius of curvature at = R =

t2(c

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πρ

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r (rad. curv. at D) =

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as before, a convenient expression for logarithmic operation. Here, again, computing values of GH for several assumed values of CR, the curve of the extrados may thence be constructed, and, like that for the cycloid, it will be found rather flatter than that for the circle, but still analogous to it.

EXAM. 5. For the Catenary. (See the fig. to Exam. 2.) Here, put DR = x, GR = y, DG=z, t = tension at the vertex D when the chain hangs from A and B. Then, by the nature of the curve z2 = 2x + x2, subtang. TR = 12+z2

Rad. curv. at G =

zy t

= R, and therefore at D where z

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Corol. If at, or the thickness at the crown equal to a line whose weight expresses the tension,

then GH = a + x = KD + DR.

Corol. 2. If a t, the exterior curve will proceed

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Corol. 3. If DK, the thickness at the crown, be very small compared with t, then will the thickness over H be nearly

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