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Let G be now assumed = * , the sections will then be vertical (Fig. 3),

the line of resistance will coincide with the line of pressure (Art. 2), and the equation common to both will be

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If a = a, (Fig. 10), the equation to the line of resistance and to the line of pressure becomes

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The equation to a parabola whose axis is vertical, whose parameter is as Peoso, and the co-ordinates of whose vertex are XP sin q) ax+ Pk cos (p + 3 (S P sin (b) it to to and & a > PCOS q)

The elements of this parabola are thus independent of the position of the mass.

Let us suppose the impressed forces P to be placed symmetrically (Fig. 11), then will the vertex of the parabola manifestly be situated midway between the extremities of the mass. Let the length of it be 2b; > Psin op - - —z-


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

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

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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,

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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 q its complement, since in the straight arch p is measured from the horizontal axis of x, in the pier from the same axis in a vertical position, so that p in the one case is the complement of p in the other:

Vol. VI. PART III. 3 P

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Eliminating therefore P sin P, and P cos b, 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 x, continuously from Z to x: ... SP cost =cos 1 s Pdx. SPsin q =sin 4's Pd,

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

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

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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

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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

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

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

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

is z 14). AM


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 ar.

Let A B D (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 A STD in R,

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