of the cement of an arch, but only those which tend to the more uniform diffusion of the pressures through its mass, to enter into the conditions of its equilibrium, these equations embrace the entire theory of the cemented arch. The hypothesis here made probably includes all that can be relied upon in the properties of cement as applied to large struc

tures.

An arch may FALL either by the sinking or the rising of its crown. In the former case, the line of resistance passing through the top of the key-stone is made to cut the extrados beneath the points of rupture; in the latter, passing through the bottom of the key-stone, it is made to cut the extrados between the points of rupture and the crown.

In the first case the values of X, Y, and P, being determined as before and substituted in equation (454), and Р being assumed = (1+a), the value of 4, which corresponds to p=(1+a)r, will indicate the point at which the line of resistance cuts the extrados. If this value of be less than the angle of the semi-arch, the intersection of the line of resistance with the extrados will take place above the springing, and the arch will fall.

In the second case, in which the crown ascends, let the maximum value of p be determined from equation (454), p being assumed =r; if this value of p be greater than R, and the corresponding value of less than the angle of rupture, the line of resistance will cut the extrados, the arch will open at the intrados, and it will fall by the descent of the

crown.

If the load be collected over a single point of the arch, the intersection of the line of resistance with the extrados will take place between this point and the crown; it is that portion only of the line of resistance which lies between these points which enters therefore into the discussion. Now if we refer to Art. 336., it will be apparent that in respect to this portion of the line, the values of X and Y in equations (453) and (454) are to be neglected; the only influence of these quantities being found in the value of P.

Example 1.-Let a circular arch of equal voussoirs have the depth of each voussoir equal to

B

Y

་

th the diameter of its intrados, so that a=2, and let the load rest upon it by three points A, B, D of its extrados, of which A is at the crown and B D are each distant from it 45°; and let it be so distributed that ths of it may rest upon each of the points. B and D, and the remaining upon A; or let it be so distributed within 60° on either side of the crown as to produce the same effect as though it rested upon these points.

Then assigning one half of the load upon the crown to each semi-arch, and calling the horizontal distance of the centre of gravity of the load upon either semi-arch

from C, it may easily be calculated that sin. 45=

5303301. Hence it appears from equation (463) that no loading can cause the angle of rupture to exceed 65°. Assume it to equal 60°; the amount of the load necessary to produce this angle of rupture, when distributed as above, will then be determined by assuming in equation (460),

¥=60°, and substituting a for A, 2 for a, and 5303301 for 2.

Y

We thus obtain 0138. Substituting this value of

Y

29 and

also the given values of a and in equation (457), and observing that in this equation is to be taken =1+a and P

the equation (454), we have for the final equation to the line of resistance beneath the point B

2426 vers. +1493

p=.0138 sin. +·1183 cos. § + ·22 § sin. 8°

B

A

If the arc of the arch be a complete semicircle, the value of p in this

ཡ

equation corresponding to 8 = will 2

determine the point Q, where the line of resistance intersects the abutment; this value is p=1·09r.

If the arc of the arch be the third u of a circle, the value of p at the abutment is that corresponding to

8 =

; this will be found to be r, as

it manifestly ought to be, since the points of rupture are in this case at the springing.

In the first case the volume of the semi-arch and load is represented by the formula.

Thus, supposing the pier to be of the same material as the arch, the volume of its material, which would have a weight equal to the vertical pressure upon its summit, would in the first case be 3594r', and in the second case 2442, whilst the horizontal pressures P would in both cases be the same. viz. 11832; substituting these values of the vertical and horizontal pressures on the summit of the pier, in equation (377), and for k writing a-(p-r), we have in the first

where H is the greatest height to which a pier, whose width is a, can be built so as to support the arch.

If a-11832=0, or a=48647, then in either case the pier may be built to any height whatever, without being overthrown. In this case the breadth of the pier will be nearly equal to th of the span.

The height of the pier being given (as is commonly the case), its breadth, so that the arch may just stand firmly upon it, may readily be determined. As an example, let us suppose the height of the pier to equal the radius of the arch. Solving the above equations in respect to a, we shall then obtain in the first case a=2978r, and in the second a='3r.

If the span of each arch be the same, and r, and r, represent their radii respectively, then r=r, sin. 60°: supposing then the height of the pier in the second arch to be the same as that in the first, viz. r,, then in the second equation we must write for H, r, sin. 60°. We shall thus obtain for a the value 28,

The piers shown by the dark lines in the preceding figures are of such dimensions as just to be sufficient to sustain the arches which rest upon them, and their loads, both being of a height equal to the radius of the semicircular arch. It will be observed, that in both cases the load Y=0138, being that which corresponds to the supposed angle of rupture 60°, is exceedingly small.

Example 2.-Let us next take the example of a Gothic arch, and let us suppose, as in the last examples, that the angle of rupture is 60°, and that a=2; but let the load in this case be imagined to be collected wholly over the crown of the arch, so that sin. 30°. Substituting in equa

=

tion (459), 30° for e, and 60° for Y, and 2 for a, and sin. 30°

Y

we shall obtain the value 21015 for; whence by

for 2%

B

have thus all the data for determining the width of a pier of given height which will just support the arch. Let the height of the pier be supposed, as before, to equal the radius of the intrados; then, since the weight of the semi-arch and its load is 5556r', and the horizontal thrust 24057, the width a of the pier is found by equation (379) to be 4195r.

The preceding figure represents this arch; the square, formed by dotted lines over the crown, shows the dimensions of the load of the same materials as the arch which will cause the angle of the rupture to become 60°; the piers are of the required width 41957, such that when their height is equal to AB, as shown in the figure, and the arch bears this insistent pressure, they may be on the point of overturning.

TABLES OF THE THRUST OF ARCHES.

344. It is not possible, within the limits necessarily assigned to a work like this, to enter further upon the discussion of those questions whose solution is involved in the equations which have been given; these can, after all, become accessible to the general reader, only when tables shall be formed from them.

Such tables have been calculated with great accuracy by M. Garidel in respect to that case of a segmental arch* whose loading is of the same material as the voussoirs, and the extrados of each semi-arch a straight line inclined at any given angle to the horizon. These tables are printed in the Appendix (Tables 2, 3).

*The term segmental arch is used, here and elsewhere, to distinguish that form of the circular arch in which the intrados is a contiguous segment from that in which it is composed of two segments struck from different centres, as in the Gothic arch.