The previous determinations have been confined to arches which are assumed to be acted on solely by vertical forces. For flat segmental arches, or even for elliptical arches where the arch is very much thickened at each end so that the virtual abutment of the arch is at a considerable distance above the nominal springing line, such a method is sufficiently accurate, and it has the advantage of simplicity of computation; but where the arch has a very considerable rise in comparison with its span, the pressure on the extrados, which is presumably perpendicular to the surface of the extrados, has such a large horizontal component that the horizontal forces cannot be ignored. The method of determining the amount and direction of the force acting on each voussoir, is illustrated in Fig. 225. The reduced load line, found as previously described, is indicated in the figure. A trapezoid represents the loading resting on the voussoir ac. The line df represents, at some scale, the amount of this vertical loading. Drawing the line de perpendicular to the extrados ac, we may complete the rectangle on the line df, and obtain the horizontal component, while the equivalent normal pressure on the voussoir is represented by de.

Drawing a vertical line through the center of gravity of the voussoir, and producing it (if necessary) until it intersects ed in the point v, we may lay off vw to represent, at the same scale, the weight of the voussoir. Making vs equal to de, we find vt as the resultant of the forces; and it therefore measures, according to the scale chosen, the amount and direction of the resultant of the forces acting on that voussoir. Although the figure apparently shows the line de as though it passed through the center of gravity of the voussoir, and although it generally will do so very nearly, it should be remembered that de does not necessarily pass through the center of gravity of the voussoir.

A practical graphical method of laying off the line vt to represents the actual resultant force is as follows: The reduced load line, drawn as previously described, gives the line for a loading of solid stone,-which would be the equivalent of the actual load line. If this loading has a unit-value of, say, 160 pounds per cubic foot, and if the horizontal distance ab is made 2 feet for the load over each voussoir, then each foot of height (at the same scale at which ab represents 2 feet) of the line gd represents 320 pounds of loading. If the voussoir were actually a rectangle, then its area would be equal to that of the dotted parallelogram vertically under ac, and its area would equal ab X dk; and in such a case, dk would represent the weight of that voussoir, and the force vw could be scaled directly equal to dk, without further compu-

3S5 tation. The accuracy of this method, of course, depends on the equality of the dotted triangle below c and that below a. For vous-soirs which are near the crown of the arch, the error involved by this method is probably within the general accuracy of other determinations of weight; but near the abutment of a full-centered arch, the inaccuracy would be too great to be tolerated, and the area of the voussoir should be actually computed. Dividing the area by 2 (or the width ah), we have the equivalent height in the same terms at which gd represents the external load, and its equivalent height would be laid off as vw.

418. Application To A Definite Problem

We shall assume for this case a full-centered circular arch whose intrados has a radius of 15 feet. The depth of the keystone computed according to the rule given in Equation 47, would be 1.57 feet, which is practically 19 inches. By drawing first the intrados of the arch as a full semicircle (see Fig. 226), and then laying off the crown thickness of 19 inches, we find bytrial that a radius of 20 feet for the extrados will make the arch increase to a thickness of about 2 1/2t at a point 45 degrees from the center, which is usually a critical point in such arches. We shall therefore draw the extrados with a radius of 20 feet, the center point being determined by measuring 20 feet down from the top of the keystone. We shall likewise assume that this arch is one of a series resting on piers which are 4 feet thick at the springing line.

By drawing a portion of the adjoining arch, we find that its extrados intersects the extrados of the arch considered at a point about 7 feet 6 inches above the pier. By drawing a line from this point toward the center for joints, which is about midway between the center for the extrados and the center for the intrados, we have the line for the joint which is virtually the skewback joint and the abutment of the arch.

Fig. 225. Resultant of Oblique Pressures.

Fig. 225. Resultant of Oblique Pressures.

The center of the pier is precisely 17 feet from the center of the arch. We shall assume that the arch is overlaid with a filling of earth or cinders which is 18 inches thick at the crown, and that it is level. Drawing a horizontal line to represent the top of this earth filling, we may divide this line into sections which are 2 feet wide, commencing at the vertical line through the center of the pier. Extending this similarly to the other side of the arch, we have eight sections of loading on each side of the keystone section. Drawing lines from the points where these verticals between the sections intersect the extrados, toward the center for joints, previously determined, we have the various joints of the voussoirs. Assuming, as in the previous numerical problem that the cinder fill weighs 100 pounds per cubic foot, and that the stone weighs 160 pounds per cubic foot, we determine the reduced load line for the top of the earth fill over the entire arch.

Fig. 226. Resultant Forces Acting on Voussoirs of a Full Centered Arch.