Rational

Prolong the side of the trial equilibrium polygon through b' to g where it intersects the line of the crown thrust prolonged. Accord ing to the principles of graphic statics, g is a point on the resultant of the forces F„ F„ F,, F5, and The section of the arch from the crown joint to joint 6 is at rest under the action of the crown thrust T, the resultant of the external forces, and the reaction of joint 6. Since the first two intersect at g, and since it has been assumed that the center of pressure for joint 6 is at b—the inner extremity of the middle third, —a line bg must represent the direction of the resultant reaction of joint 6; and hence the line R„ in the force diagram drawn from the upper extremity of parallel to bg, to an intersection with P'0, represents, to the scale of the load line, the amount of the reaction of joint 6. Then P0, to the same scale, represents the crown thrust corresponding to the line of resistance passing through U and b; and a line—not shown in Fig. 197—from the upper extremity of to the lower extremity of F,, would represent, in both direction and amount, the resultant of F,, F„ F5, and Having found the thrust at the crown, complete the force dia gram by drawing the lines R,, R„ R„ etc.; and then construct a new equilibrium polygon exactly as was described above for the trial equilibrium polygon. The equilibrium polygon shown in Fig. 197 by a solid line was obtained in this way.

The points in which the sides of the new equilibrium polygon cut the corresponding joints are the centers of pressure on the respec tive joints. The stability olthe arch may be discussed as in 4 1233 35.

Unsymmetrical Load.

The design for an arch ring should not be considered perfect until it is found that the criteria of safety (1 1197) are satisfied for the dead load and also for every possible position of the live load. A direct determination of the line of re sistance for an arch under an unsymmetrical load is impossible. To find the line of resistance for an arch under a symmetrical load, it was necessary to make some assumption concerning (1) the amount of the thrust, (2) its point of application, and (3) its direction; but when the load is unsymmetrical, we neither know any of these items nor can make any reasonable hypothesis by which they can be determined. For an unsymmetrical load we know nothing concern ing the position of the joint of rupture, and know that the thrust at the crown is neither horizontal nor applied at one third of the depth of that joint from the crown; and hence the preceding methods can not be employed. When the load is not symmetrical, the following method may be employed to find a line of resistance; but it gives no indication as to which of the many possible lines of resistance is the true one.

Let it be required to test the stability of a symmetrical arch having a uniform live load covering only half the span. The problem could

be solved by determining the external forces as in § 1219-20; but for variety and to explain a method of determining the loads that is frequently used, in one form or another, in discussions of the stability of voussoir arches, a different method of determining the vertical forces will be employed. This method consists in reducing the actual load upon an arch (including the weight of the arch ring itself) to an equivalent homogeneous load of the same density as the arch ring. The upper limit of this imaginary loading is called the reduced-load contour.

To find the Reduced-Load Contour. Assume that it is required to find the reduced-load contour for the dead load on the arch in Fig. 198. Assume that the weight of the arch ring is 160 pounds per cubic foot, that of the rubble backing 140; and that of the earth 100. Then the ordinate at a to the load contour of an equivalent load of the density of the arch ring is equal to ab + be 160 cd say, g f. The value of g f is laid off in Fig. 199. Com puting the ordinates for other points gives the line EF, Fig. 199, which is the reduced-load contour for the load shown in Fig. 198. The area between the intrados and the reduced-load contour is proportional to the dead load on the arch.

In a similar manner, a live load (as, for example, a train) can be reduced to an equivalent load of masonry,—in which case the re duced-load contour would consist of a line GH above and parallel to EJ for that part of the span covered by the train; while for the remainder of the span, the line JF is the reduced-load contour.

To utilize the reduced-load contour, draw the arch and its re duced-load contour upon thick paper or card-board, to a large scale; and divide the arch into any number of imaginary voussoirs, and erect verticals from the upper ends of the imaginary joints. Then measure, with a planimeter or otherwise, the area of each voussoir and its load, from which the weight of each voussoir and its load can be easily determined. Next, with a sharp knife, carefully cut out the area representing the load on each arch stone. The center of gravity of each piece, as ijklmn, Fig. 199, can be found by bal ancing it on a knife-edge successively in two positions at right angles to each other; and then the position of the center of gravity is to be transferred to the drawing of the arch.

To find the Line of Resistance. Assume that it is desired to find the line of resistance for the arch shown in Fig. 200, page, 636, whose reduced-load contour is as shown. Assume that the vertical forces have been determined as explained in the preceding section; and also assume that the horizontal component of the earth pressure is one third of the vertical intensity, and that it has been computed as in f 1208.