If, moreover, the forces P be supposed to be resistances, the line of resistance will touch the extrados (Fig. 12);

Suppose that there is only one force P applied at each extremity; therefore by equations (28) and (29),

P sin &= ab, a2 = ±ak + 1⁄2 P sin ☀ tan 4;

This last equation gives that portion of the force P which is resolved in the direction of a horizontal line. If k=0 or the force P be applied at the angle A of the mass,

2a

tano = 20, P = b √ aa +‡b3, Pcos • =

b2......(33).

It is worthy of remark, that the last of these expressions is independent of a, the depth of the voussoirs.

Let a straight arch be supposed to be supported upon the edges of two vertical piers, (Fig. 12).

The point of rupture in the extrados of the pier, or its greatest height, so as to stand unsupported, will then be determined by the following equation derived from equation (23), by taking k = 0 and writing for its complement, since in the straight arch is measured from the horizontal axis of x, in the pier from the same axis in a vertical position, so that in the one case is the complement of in the other:

Eliminating therefore P sin , and P cos, between this equation and equations (33), and calling a, the width of the pier to distinguish it from the depth of the arch, we have

7. Let it now be supposed that the forces P are impressed upon all the points of the face BC (Fig. 2) of a mass, and that the plane of intersection is horizontal. Let moreover all these forces P be parallel to one another, and let them be represented respectively in magnitude by the values of a function P of ≈, continuously from Z to z;

.. ΣP cos ↓ = cos ↓ f*Pdz, ΣPsin = sin f* Pdz,

Σ + Pk cos ↓ = √ P(y, cos o – z sin 4) dz.

These substitutions being made, the equation (4) to the line of resistance gives the following:

or y =

4)

§ ƒˆ (yi — y1) d≈ + z sin ↳ Pdz + P(y, cos — ≈ sin ‹Þ) dz

0

.(35),

.(36).

8. DYKES AND EMBANKMENTS.

Let the forces P be the pressures of a fluid mass upon the face of an embankment, supposed a plane, inclined to the vertical at the angle a2,

(see Fig. 13); will then become + a, and P will vary as ; let it

equal μ≈.

y =

2

From this equation the line of resistance may be determined for any given inclination of the internal face or form of the external face of the embankment; or conversely, these circumstances may themselves be determined according to a given equation to the line of resistance.

If the section of the embankment be of the form of a trapezoid (Fig. 14), by the integration (6) we have

-

·

≈3 { } (tan3 a, — tan2 a) — μ (seca, — cosa)} +az1tana, +≈ (α2 − μ Za cosa;) + 3⁄4μ Zseca, **{tana, -tana, - sin a}+2az+μZ sin a

If Z = 0 or the fluid extend to the edge of the embankment, this becomes the equation to an hyperbola.

The point of rupture of the extrados, or the greatest possible height of the embankment, may be determined as before, by assuming

{ z3{(tan a, – tan a1)(2 tan a, + tan a2) —– μ [3 cos (α, − a) sec a1 − 2 sec a2]}+az2{2 tan a1−tan a ̧—μ sina,

...

.(38).

= 0...(39).

The supposition now about to be made, with regard to the line of resistance, is, that it traverses the center of the embankment. We have thence, by equation (37),

(Y1 + Y2) (Y1 Y2) dz = [[(y1 — Y2) dz. d (y, + y12) + Л(yi — y?) dz

√(y} −

= μ {(x2 - Z3) [≈ cos a2 + (y1 + y) sin a,] - (~33 - Z3) sec a,}......(40). 1 }

Whence substituting for y, its value ≈ tan a,, and differentiating twice in respect to ≈, a differential equation will be obtained, the solution of which will determine the required relation of y, and z. If a, = 0, or the intrados be a vertical plane, we obtain, by the first differentiation in respect to %,

if A be the breadth of the embankment on the level of the surface of the fluid;

The equation to an hyperbola, whose center is in the inner edge of the embankment A, the ratio of whose axes is, and whose semi-axis

The plane of intersection has hitherto been supposed, in its successive changes of position, to remain always parallel to itself. Let this hypothesis now be discarded, and as the simplest case of a variable inclination of the plane, let it be supposed to revolve about a given horizontal line or axis within itself. Let moreover the extrados and the intrados be supposed to be cylindrical surfaces, having this line for their common axis; and suppose this arch to have a load uniformly distributed along the extrados in a line parallel to the axis and at a horizontal distance from it equal

to x.

Let ABD (Fig. 15), represent any section of this mass perpendicular to the axis C, and X the corresponding load. Let the horizontal force P be applied in AD at a vertical distance p from C; and let CT be any position of the intersecting plane, intersected by the resultant of the forces P and X, and the weight of the mass ASTD in R, PCR 0, CT= CR = p.

PCA = 0,

=

0, CTR, CS = r,

R

Therefore by the condition of the equality of moments,

R

["fr2 sin @dodr + Xx+Pp=p{P cos 0+X sin @ + sin @ [*[¶rdedr} ....(42).

r

At the point of rupture the line of resistance meets the intrados; therefore at this point

Also, generally, sufficient dimensions of the arch being supposed, the line of resistance touches the intrados at this point;

Assuming to be the value of at the point of rupture, and substituting it for 0 in the five preceding equations, we may eliminate between them the four quantities p, R, r, p, or the four p, R, r, Ψ. There will result an equation involving the quantities P and in the one case, and P and p in the other.

Now if the force P be supplied by the pressure of another opposite and equal semi-arch, it has been shewn, (see Memoir on the Theory of the Arch, Vol. V. Part III.) that if the masonry be supposed perfect, P is a minimum in respect to the variable p; moreover if the masonry be supposed to possess those yielding properties which obtain in practice, and which shew themselves in the settlement of the arch, then p R; according to either of these conditions, P and may therefore be determined.

=

The values of P and p becoming thus known, they may be substituted in equation (42), and the equation to the line of resistance will thus be completely determined.