One of these lines the line of resistance-determining the point of application of the resultant of the pressures upon each of the surfaces of contact of the system, and the other—the line of pressure-the direction of that resultant, the determination of the two includes the whole theory of the equilibrium of the system.

In its application to the theory of the arch there belong to the line of resistance all those properties treated of in my former paper which have reference to the condition "that the voussoirs shall not turn upon the angles of one another."

It follows, therefore, on the principles established in that paper that this line touches the intrados of the arch at certain points equidistant from the crown, called points of rupture, and that the position of these points, and, consequently, that of the point of application of the resultant of the pressures upon the key-stone, are subject to the condition that this resultant is a minimum; and this condition being supposed, all the circumstances which connect themselves with the equilibrium of the circular arch, as a complete segment, and a broken or gothic arch, subjected to any variety of loading, are discussed and determined in the eleventh section of the following paper.

The condition, however, that the resultant pressure upon the key-stone is subjected in respect to the position of its point of application to the condition of a minimum, is dependant upon hypothetical qualities of the masonry. It supposes an unyielding material for the arch-stones, and a mathematical adjustment of their surfaces. These have no existence in practice. On the striking of the centers the arch invariably sinks at the crown, its voussoirs slightly opening there upon their lower edges, and thus pressing upon one another exclusively by their upper edges. Practically, the line of resistance then touches the extrados at the crown; whilst the condition of the minimum is satisfied by its contact with the intrados at the points of rupture in the haunches. This condition being assumed, all consideration of the yielding quality of the material of the arch or of its abutments is eliminated. It is thus discussed as a practical question in the twelfth section of this paper.

1. Let a continuous mass to which are applied certain forces of pressure, be supposed to be intersected by a plane whose equation is

≈ = Ax + By + C......(I.)

Let the sums of the forces impressed upon one of the parts and resolved in directions parallel to three rectangular axes, be respectively M1, M2, M3, and the sums of their moments N1, N2, N3.

Let, moreover, the position of the plane be such that these forces are reducible to a single resultant, a condition determined by the equation

If between the four preceding equations in which M1, M2, M3, N1, N2, N3 are functions of A, B, C, these three quantities A, B, C be eliminated, there will be obtained an equation in x, y, z, which is that to a surface of which this is the characteristic property; that it includes all the points of intersection of the resultant force with its corresponding intersecting plane in every position, which, according to the assumed conditions, this last may be made to take up.

This surface is the SURFACE OF RESISTANCE.

If to the preceding conditions there be added this, that in each two consecutive positions of the intersecting plane the corresponding resultants shall intersect, the surface of resistance will resolve itself into a line, which is the LINE OF RESISTANCE.

Differentiating on this hypothesis the equation III. in respect to A, B, C, we have

From the elimination of A, B, and C, between the five equations
I, II, III, V, will result the two equations to the LINE OF RESIST-
ANCE; and from the elimination of the same three quantities between
the five equations II, III, IV,* the two equations to the LINE OF
PRESSURE.

The inclination intersecting plane, in

of the resultant pressure to a perpendicular to the any of its positions, may be determined (see paper on Equation of Arch), independently of the line of pressure, from the equation

COS=

AM1 + BM2 + M3

2

{(A2 + B2 + 1) (M; + M2 + M})}''

2. Let the mass be a PRISM whose axis is horizontal, and the forces applied to which are, its weight and certain pressures, P, whose directions are in planes perpendicular to its axis and inclined at angles to axis of, and whose points of application are uniformly distributed along lines on the surface of the prism parallel to its axis; all those pressures which are applied in each line being equal to one another.

* It will be observed that the condition V. is included in these. VOL. VI. PART III.

30

The relation of the forces which compose the equilibrium of the whole Prism, will then be the same with that of the forces impressed on any one of its sections perpendicular to the axis.

Let CBD, (Fig. 1,) represent any one of these sections. Suppose the mass to be intersected in any direction parallel to its axis by a plane, and let N N be the intersection of this plane with the section CB of the mass.

And first, let this intersecting plane in altering its position be supposed to remain always parallel to itself.

Take A, the axis of %, perpendicular to N, N2, and let it make an angle with the vertical.

M2 =ΣP sin - sin → ƒ(y, −y2) dC,

M2 =ΣP cos Þ + cos → ƒ (yı — y2) dC,

N1 = cos → Л(y2 — y2) dC + sin → ƒC (y1 − y2) dC + Σ + Pk cos ✨,

This hypothesis with regard to the position of the axis of ≈, and these substitutions being made, all the equations of condition vanish except equation I, the second of equations III, and the second of equations IV. These resolve themselves into the following:

%=

y=

C........

{EP sin -sin f(y1−y1) dC} z + cos → [(y2-y2) d C+sin → [C(y,-y1) dC++ Pk cos EP cos + cos e f (yı — y2) dC

dΣP sin

y

dC

.(1),

.(2),

y=

The equation to the line of resistance is determined by eliminating C between the equations (1) and (2), and that to the line of pressure by eliminating it between (2) and (3).

If the first elimination be made, and it be observed, that

y

there will be
resistance,

obtained the following general equation to the line of

2

≈ΣP sin + cos → f(y2 — y1⁄23) dz — sin → [f(y1 — y ) dz2 + Σ± Pk cos o
ΣP cos +cos → ƒ(y, — y2) dz

(4).

The second elimination is
P, Þ, k are independent of C.

greatly simplified in the case in which Since in this case, equation (3) gives

0............(5).

y - (y1 + Y2) + (≈ C) tan → =
1

If the intersections be supposed to be made horizontally, (Fig. 2,)
If they be made vertically, (Fig. 3) → = 2*

→ must be assumed = 0.

In the latter case, equation (5) gives C = %.

π

The elimination of C between (3) and (2) is therefore the same as that between (1) and (2), and the line of pressure in this case, coincides with the line of resistance.

3. Let the mass be a trapezoidal form. (Fig. 4.)

Let AB and CD be inclined to the axis of ≈ at angles a, a,

assume CA = a; .. Yi

Hence y y3) dz

and

= a + z tan a1, Y2 = tan aọ.

= az (a + ≈ tan a1) + §≈3 (tan2 a,

Therefore by substitution in equation (4) we have for the equation

to the line of resistance

} z3 {tan a, – tan a ̧}{tan a, +tan a ̧ ¬tan } + a z3 {tana, -tan☺}+{secΣP sin +}a2} + sec ☺Σ±Pk cos ©

{tan a1- tan a,} + az + sec ΣP cos

·(7).