VOUSSOIR ARCHES SUBJECTED TO OBLIQUE FORCES 417. Determination of Load on a Voussoir. The previous determinations have been confined to arches which are assumed to be acted on solely by vertical forces. For flat segmental arc1ies, 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 suffi ciently accurate, and it has the advantage of simplicity of computa tion; 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 hori zontal 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 redu6ed load line, found as previously described, is indicated in the figure. A trapezoid represents the loading resting on the voussoir ac. The line df repre sents, at some scale, the amount of this vertical loading. Drawing the line de perpendicular to the extrados ac, we may complete the rec tangle 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 vous soir, 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 represent. 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 actual ly a rectangle, then its area would be equal to that of the dotted parallelogram vertically under cc, and its area would equal ab X dk; and in such a case, c/k would represent the weight of that voussoir, and the force vw could be scaled directly equal to dk,without further compu 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 arc near the crown of the arch, the error involved by this method is probably within the general accuracy of other determina eons 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 aS), we have the equivalent height in the same terms at which yd represents the external load, and its equivalent height would be laid off as pw.
418. Application to a Definite Prob= lem. 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 semi circle (see Fig. 226), and then laying off the crown thickness of 19 inches, we find by trial that a radius of 20 feet for the extrados will make the arch increase to a thickness of about 21 feet 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.