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.

In Fig. 216, have been drawn the same forces A, B, and C, having the same relative positions as in Fig. 215. The lines of action of the two vertical forces R' and R" have also been drawn in the same relative position as in Fig. 215. The point n has also been located on the load line in the same position as in Fig. 215. Thus far the lines are a repetition of those already drawn in Fig. 215, the remainder of the figure being omitted for simplicity. Since the point n in Fig. 215 is the end of the line from the trial pole o, which is parallel to the closing line yz, and since the point n is a definitely fixed point and determines the abutment reactions regardless of the position of the trial pole o, we may draw from n an indefinite horizontal line, such as no', and we know that the pole of any force diagram must be on this line if the closing line of the corresponding equilibrium polygon is to be a horizontal line. For example, we shall select a point o' on this line at random. From o' we shall draw rays to the points p, s, r, and q. From the point y, we shall draw a line parallel to o'p. Where this line intersects the force A, draw a line parallel to the ray o's. Where this intersects the force B, draw a line parallel to the ray o'r. Where this intersects the force C, draw a line parallel to the ray o'q. This line must intersect the point z , which is on a horizontal line from y. The student should make some such drawing as here described, and should demonstrate for himself the accuracy of this law. This equilibrium polygon is merely one of an infinite number which, if acting as struts, would hold these forces in equilibrium, but it combines the special condition that it shall pass through the points y and z". There are also an infinite number of equilibrium polygons which will hold these forces in equilibrium and which will pass through the points y and zr.

Fig. 216. Equilibrium Polygon with Horizontal Closing Line.

We may also impose another condition, which is that the first line of the equilibrium polygon shall have some definite direction, such as yl. In this case the ray from the point p of the force diagram must be parallel to yl; and where this line intersects the horizontal line no' (produced in this case), is the required position for the pole o". Draw rays from o" to s, r, and q, continuing the equilibrium polygon by lines which are respectively parallel to these rays. As a check on the work, the last line of the equilibrium polygon which is parallel to o"q should intersect the point z'. The triangles ykh and o'pn have their sides respectively parallel to each other, and the triangles are therefore similar, and their corresponding sides are proportional, and we may therefore write the equation: o'n: yh:: pn: kh.

Also, from the triangles ylh and o"pn, we may write the proportion: o"n: yh:: pn: Ih.

From these two proportions we may derive the proportion: o'n: o"n:: lh: kh; but o'n and o"n are the pole distances of their respective force diagrams, while kh and Ih are intercepts by a vertical line through the corresponding equilibrium polygons. The proportion is therefore a proof) in at least a special case, of the general law that the perpendicular distances from the poles to the load lines of any two force diagrams are inversely proportional to any two intercepts in the corresponding equilibrium polygons. The above proportions prove the theorem for the intercepts hk and hl. A similar combination of proportions would prove it for any vertical intercept between y and h. The proof of this general theorem for intercepts which pass through other lines of the equilibrium polygon, is more complicated and tedious, but is equally conclusive. Therefore, if we draw any vertical intercept, such as tvw, we may write out the general proportion: o"n: o'n:: tw;: vw..(46)

In this proportion, if o"n were an unknown quantity, or the position of o" were unknown, it could be readily obtained by drawing two random lines as shown in diagram c, and laying off on one of them the distance no', and on the other line the distances vw and tw. By joining v and o' in diagram c, and drawing a line from t parallel to vo', it will intersect the line no' produced, in the point o". As a check, this distance to o" should equal the distance no" in diagram b. A practical application of this case, and one that is extensively employed in arch work, is the requirement that the equilibrium polygon shall be drawn so that it shall pass through three points, of which the abutments are two, and some other point (such as v) is the third. After obtaining a trial equilibrium polygon whose closing line passes through the points y and z', the proper position for the pole o" which shall give the equilibrium polygon that will pass through the point v, may be easily determined by the method described above.

The process of obtaining an equilibrium polygon for parallel forces which shall pass through two given abutment points and a third intermediate point, may be still further simplified by the application of another property, and without drawing two trial equilibrium polygons before we can draw the required equilibrium polygon. It may be demonstrated that if the pole distance from the pole to the load line is unchanged, all the vertical intercepts of any two equilibrium polygons drawn with these same pole distances are equal. For example, in Fig. 215, a line is drawn from o, vertically upward until it intersects the horizontal line drawn through n in the point o". This point is the pole of another equilibrium polygon whose closing line will be horizontal, because the pole lies on a horizontal line from the previously determined point n in the load line. Any vertical intercept of this equilibrium polygon will be equal to the corresponding intercept on the first trial equilibrium polygon; therefore, in order to draw a special equilibrium polygon for a given set of vertical loads, the polygon to pass through two horizontal abutment points and a definite third point between them, we need only draw first a trial equilibrium polygon, the rays in the force diagram being drawn through any point chosen as a pole. Then, if we draw a line from the trial pole which shall be parallel with the closing line of this trial equilibrium polygon, the line will intersect the load line in the point n. Drawing a horizontal line from the point n in the load line, we have the locus of the pole of the desired special equilibrium polygon. Then draw a vertical through the point through which the special equilibrium polygon is to pass. The vertical distance of this point above the line joining the abutments, is the required intercept of the true equilibrium polygon. The intersection of that vertical with the upper line and the closing line of the trial equilibrium polygon, is the intercept of the trial polygon. The pole distance of the true equilibrium polygon is then obtained by the application of Equation 46, by which the pole distances are declared inversely proportional to any two corresponding intercepts of the equilibrium polygons.

Another useful property, which will be utilized later, and which may be readily verified from Figs. 215 and 216, is that, no matter what equilibrium polygon may be drawn, the two extreme lines of the equilibrium polygon, if produced, intersect in the resultant R; therefore, when it is desired to draw an equilibrium polygon which shall pass through any two abutment points, such as yz or yz', we may draw from these two abutment points, two lines which shall intersect at any point on the resultant R. We may then draw two lines which will be respectively parallel to these lines from the extremities p and q of the load lines, their intersection giving the pole of the corresponding force diagram.

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