As described above, the arrows representing the forces in Fig. 226 are drawn at a scale such that each of an inch represents 2 cubic feet of masonry, or 320 pounds; therefore every inch will represent the quotient of 320 divided by s, or S53 pounds per linear inch. The practical method of making a scale for this use is illustrated in the diagram in the upper right-hand corner of Fig. 226. We may draw a horizontal line as a scale line, and lay off on it, with a decimal scale, a length ca which represents, at some convenient scale, a length of SOO. Drawing the line ab at any convenient angle, we lay off from the point c the length cli to represent 853 at the same scale as that used for ca. The line cd is then laid off to represent 7,000 units at the scale of SOO units per inch. By drawing a line from d parallel to ha, we have the distance cc, which represents 7,000 units at the scale of 853 units per inch. By trial, a pair of dividers may be so spaced that it steps off precisely seven equal parts for the distance cc; or the line ce may also be divided into equal parts by laying off on cd to the decimal scale, the seven equal parts of 1,000 each which are at the scale of SOO units per inch; and then lines may be drawn from these points parallel to ha and dc. The last division may be similarly divided into 10 equal parts, which will represent 100 pounds each.
Using dividers, the resultant force on each voussoir from No. 1 to No. 17 may be scaled off as follows: The student should note the three dotted curves in the lower part of the figure, which have been drawn through the extremities of the forces. The only object in drawing these three curves is merely to note the uniformity with which the ends of these arrows form a regular curve. If it was found that one of the forces did not pass through this curve, it would probably imply a blunder in the method of determining that particular force. Even if such curves are not actually drawn in, it is well to observe that the points do come on a regular curve, as this constitutes one of the checks on the graphi cal solution of problems.
Fig. 226 is merely the beginning of the problem of determining the stresses in the arch. In order to save the complication of the figure, the arch itself and the resultant forces (1 to 17) are repeated in Fig. 227, the direction, intensity, and point of application of these forces being copied from one figure to the other.
Forces Nos. 1 to 17 are drawn in the force diagram of Fig. 227 at the scale of 4,000 pounds per inch. Forces 1 to S, inclusive, have a resultant whose direction is given by the line marked R," which joins the extremities of forces 1 to S. Similarly, the direction of the re sultant or of forces 9 to 17, inclusive, is given by the line which joins the extremities of this group. The direction of the resul tant of all the forces Nos. 1 to 17, is given by the line joining the ex tremities of these forces in the force diagram, this resultant being marked By choosing a pole at random (the point oz in the force diagram), drawing rays to the forces, and beginning at the left-hand abutment, we may draw the trial equilibrium polygon, which passes through the point a on force No. 17. The line through a parallel
to the last ray, has the direction oh. Producing the section of the polygon which is between forces S and f) (and which is parallel to the ray which reaches the load line between forces S and 0), it inter sects the first and last lines of the trial equilibrium polygon at the points h and d. The point b is therefore a point on the resultant of forces Nos. 0 to 17 inclusive; and by drawing a line parallel to the force R,' in the force diagram, we have the actual line of action of the resultant.
Similarly, the line of action of 'the force 1?," is determined by drawing from the point d a line parallel to in the force diagram. Their intersection at the point c gives a point in the line of action of the resultant of the whole system of forces, and by drawing from the point e a line parallel to I?, of the force diagram, we have the line of action of R,. We select a point (f) at random on the resultant R,, and join the point f with the center of each abutment. By drawing lines from the extremities of the load line parallel to these two lines from f, they intersect at the point A horizontal line through 0," is there fore the locus' of the pole of the true equilibrium polygon passing through the center of both abutments. The line fn intersects 1?; in the point g, and the line intersects the force in the point h. The intersection of gh with the vertical through the center (the point i) is the rrial point which innst be raised up to the point r, which is done by the method illustrated in Article 401. The application of this method gives the line k/, passing through e; and the line in is there fore the first line of the special equilibrium polygon for the complete system of forces from No. 1 to No. 17; and the line kin is similarly the last line of that polygon. By drawing lines from the extremities of the load line, parallel to In and kin, we find tliat they intersect at the point which is the pole of the special equilibrium polygon passing through it, c, and nt, for the complete system of forces Nos. 1 to 17.
As a check on the work, the intersection of these lines from the ends of the load line, parallel to In and kin, must be on the horizontal line passing through By drawing rays from the new pole a,'" to the load line, and completing the special equilibrium polygon, we should find as a double check on the work, that both of these partial polygons starting from in and n should pass through the point c; and also that the section of the polygon between forces Nos. S and 9 lies on the line kl. This gives the special equilibrium polygon for the system of forces Nos. 1 to 17, which corresponds with the second condition of loading, as specified above.