if we conceive each buttress to be extended laterally until it meets the adjacent buttress, its material at the same time so diminishing its specific gravity that its weight when thus spread along the face of the wall may remain the same as before. There will thus be obtained a compound wall whose external and internal portions are of different specific gravities; the conditions of whose equilibrium remain manifestly unchanged by the hypothesis which has been made in respect to it.

THE LINE OF RESISTANCE IN A PIER.

289. Let ABEF be taken to represent a column of uniform dimensions. Let PS be the direction of any pressure P sustained by it, intersecting its axis in O. Draw any horizontal section IK, and take ON to represent the weight of the portion AKIB of the column, and OS on the same scale to represent the pressure P, and complete the parallelogram ONRS; then will OR evidently represent, in magnitude and direction, the resultant of the pressures upon the portion AKIB of the mass (Art. 3.), and its point of intersection Q with IK will represent a point in the line of resistance.

Let PS intersect BA (produced if necessary) in G, and let GC=k, AB=a, AK=x, MQ=y, POC=a, μ=weight of each cubic foot of the material of the mass. Draw RL perpendicular to CD; then, by similar triangles,

But QM=y, OM=CM-CO=x-k cot. a, RL=RN sin. RNLP sin. a, OL-ON+NL=ON+RN cos. RNL =μax+P cos. a;

which is the general equation to the line of resistance of a pier or wall.

290. The conditions necessary that the stones of the pier may not slip on one another.

Since in the construction of the parallelogram ONRS, whose diagonal OR determines the direction of the resultant pressure upon any section IK, the side OS, representing the pressure P in magnitude and direction, remains always the same, whatever may be the position of IK; whilst the side ON, representing the weight of AKIB, increases as IK descends the angle ROM continually diminishes as IK descends. Now, this angle is evidently equal to that made by OR with the perpendicular to IK at Q; if, therefore, this angle be less than the limiting angle of resistance in the highest position of IK, then will it be less in every subjacent position. But in the highest position of IK, ON=0, so that in this position ROM=a. Now, so long as the inclination of OR to the perpendicular to IK is less than the limiting angle of resistance, the two portions of the pier separated by that section cannot slip upon one another (Art 141.). It is therefore necessary, and sufficient to the condition that no two parts of the structure should slip upon their common surface of contact, that the inclination of P to the vertical should be less than the limiting angle of resistance of the common surfaces of the stones. All the resultant pressures passing through the point O, it is evident that the line of pressure (Art. 284.) resolves itself into that point.

291. The greatest height of the pier.

At the point where the line of resistance intersects the external face or extrados of the pier, y=a; if, therefore, H represents the corresponding value of x, it will manifestly represent the greatest height to which the pier can be built, so as to stand under the given insistent pressure P. Substituting these values for x and y in equation (377), and solving in respect to H,

a=

If P sin. apa2, H= infinity; whence it follows that in this case the pier will stand under the given pressure P, however great may be the height to which it is raised.

292. The line of resistance is a rectangular hyperbola. Multiplying both sides of equation (377) by the denominator of the fraction in the second member,

y(μax +P cos. a)= Px sin. a— Pk cos. a ;

dividing by pa, transposing, and changing the signs of all the

на

y) (x

(x+

to both sides,

μαλ

P cos. a
μα

P cos. a

=

a (k+ Psin. a);

a

μα

на

Psin. a-y+Peos. a) Peos. (k+ Pain. a).

=

μα

на

μα

This is the equation to a rectangular hyperbola, whose

293. The thickness of the pier, so that when raised to a given height it may have a given stability.

Let m be taken to represent the nearest distance to which the line of resistance is intended to approach the extrados of the pier, which distance determines the degree of its stability, and has been called the modulus of stability (Art. 286.). It is evident from the last article that this least distance will present itself in the lowest section of the pier. At this lowest section, therefore, y=a-m. Substituting this value for y in equation (377), and also the height h of the pier for x, and solving the resulting quadratic equation in respect to a, we shall thus obtain

294. To vary the point of application of the pressure P, so that any required stability may be given to the pier.

It is evident, that if in equation (377) we substitute a-m for y and h for x, the modulus of stability m

P

may be made to assume any given value G for a given thickness a of the pier, by assigning a corresponding value to k; that is, by moving the point of application G to a certain distance from the axis of the pier, determined by the value of k in that equation. This may be done by various expedients, and among others by that shown in the figure. Solving equation (377) in respect to k, we have

It is necessary to the equilibrium of the pier, under these circumstances, that the line of resistance should nowhere intersect its intrados below the point D.

THE STABILITY OF A WALL SUPPORTED BY SHORES.

D H A

295. Let the weight of the portion of the wall supported by each shore or prop, and the pressure insistent upon it, be imagined to be collected in a single foot of the length of the wall; the conditions of the stability of the wall evidently remain unchanged by this hypothesis. Let ABCD represent one of the columns or piers into which the wall will thus be divided, EF the corresponding shore, P the pressure sustained upon the summit of the wall, Q the thrust upon the shore EF, 2w its weight, x the

C & K

B

E E