This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

The above method is rendered especially simple, owing to the fact that the forces are all vertical. When the forces are not vertical, the method becomes more complicated. The principle will first be illustrated by the problem of drawing an equilibrium polygon which shall pass through the points y, z, and v in Fig. 217. We shall first draw the two non-vertical forces in the force diagram. The resultant R of the forces A and B is obtained as shown in Fig. 213. Utilizing the property referred to in the previous article, we may at once draw two lines through y and z which intersect at some assumed point e on the resultant R. Drawing lines from p and q parallel respectively to ez and ey, we determine the point o' as the trial pole for our force diagram. As a check on the drawing, the line joining the intersections b and c should be parallel to the ray o's, thus again verifying one of the laws of Statics. If the line be is produced until it intersects the line yz produced, and a line is drawn from the intersection x through the required point v, it will intersect the forces A and B in the points d and g. Then dg will be one of the lines of the required equilibrium polygon. By drawing lines from q and p parallel to yd and zg, we find their intersection o", which is the pole of the required force diagram. There are two checks on this result: (1) the line so" is parallel to dg; and (2) the line o'o" is horizontal.

If the line be is horizontal or nearly so, the intersection (x) of be and yz produced is at an infinite distance away, or is at least off the drawing. If be is actually horizontal, the line dg will also be a horizontal line passing through v. When be is not horizontal, but is so nearly so that it will not intersect yz at a convenient point, the line dg may be determined as is indicated by the dotted lines in the figure. Select any point on the line yz, such as the point o. Through the given point v, draw a vertical line which intersects the known line be in the point k. From some point in the line be (such as the point b), draw the horizontal line bh and the vertical line bn. The line from o through k intersects the horizontal line from b in the point h. From the point h, drop a vertical; this intersects the line ov produced, in the point m. From m, draw a horizontal line which intersects the vertical line from b. This intersection is at the point n. The line vn forms part of the required line dg. As a check on the work, the lines zg and yd should intersect at some point / on the force R. Another check on the work, which the student should make, both as a demonstration of the law and as a proof of the accuracy of his work, is to select some other point on the line yz than the point o, and likewise some other point on the line bc than the point b, and make another independent solution of the problem. It will be found that when the drawing is accurate, the new position for the point n will also be on the line dg.

Fig. 217. Equilibrium Polygon through Three Chosen Points.

In applying the above principle to the mechanics of an arch, the force A represents the resultant of all the forces acting on the arch on one side of the point v through which the desired equilibrium polygon is required to pass; and the force B is the resultant of all the forces on the other side of that point. A practical illustration of this method will be given later.

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