The complete figures obtained by drawing first the force-diagrams of a system of forces in equilibrium with two distinct poles O, O′, and secondly the corresponding funiculars, have various interesting relations. In the first place, each of these figures may be conceived as an orthogonal projection of a closed plane-faced polyhedron. As regards the former figure this is evident at once; viz. the polyhedron consists of two pyramids with vertices represented by O, O′, and a common base whose perimeter is represented by the force-polygon (only one of these is shown in fig. 28). As regards the funicular diagram, let LM be the line on which the pairs of corresponding sides of the two polygons meet, and through it draw any two planes ω, ω′. Through the vertices A, B, C, ... and A′, B′, C′, ... of the two funiculars draw normals to the plane of the diagram, to meet ω and ω′ respectively. The points thus obtained are evidently the vertices of a polyhedron with plane faces.
| Fig. 28. |
| Fig. 29. |
To every line in either of the original figures corresponds of course a parallel line in the other; moreover, it is seen that concurrent lines in either figure correspond to lines forming a closed polygon in the other. Two plane figures so related are called reciprocal, since the properties of the first figure in relation to the second are the same as those of the second with respect to the first. A still simpler instance of reciprocal figures is supplied by the case of concurrent forces in equilibrium (fig. 29). The theory of these reciprocal figures was first studied by J. Clerk Maxwell, who showed amongst other things that a reciprocal can always be drawn to any figure which is the orthogonal projection of a plane-faced polyhedron. If in fact we take the pole of each face of such a polyhedron with respect to a paraboloid of revolution, these poles will be the vertices of a second polyhedron whose edges are the “conjugate lines” of those of the former. If we project both polyhedra orthogonally on a plane perpendicular to the axis of the paraboloid, we obtain two figures which are reciprocal, except that corresponding lines are orthogonal instead of parallel. Another proof will be indicated later (§ 8) in connexion with the properties of the linear complex. It is convenient to have a notation which shall put in evidence the reciprocal character. For this purpose we may designate the points in one figure by letters A, B, C, ... and the corresponding polygons in the other figure by the same letters; a line joining two points A, B in one figure will then correspond to the side common to the two polygons A, B in the other. This notation was employed by R. H. Bow in connexion with the theory of frames (§ 6, and see also Applied Mechanics below) where reciprocal diagrams are frequently of use (cf. [Diagram]).
When the given forces are all parallel, the force-polygon consists of a series of segments of a straight line. This case has important practical applications; for instance we may use the method to find the pressures on the supports of a beam loaded in any given manner. Thus if AB, BC, CD represent the given loads, in the force-diagram, we construct the sides corresponding to OA, OB, OC, OD in the funicular; we then draw the closing line of the funicular polygon, and a parallel OE to it in the force diagram. The segments DE, EA then represent the upward pressures of the two supports on the beam, which pressures together with the given loads constitute a system of forces in equilibrium. The pressures of the beam on the supports are of course represented by ED, AE. The two diagrams are portions of reciprocal figures, so that Bow’s notation is applicable.
| Fig. 30. |
| Fig. 31. |
A graphical method can also be applied to find the moment of a force, or of a system of forces, about any assigned point P. Let F be a force represented by AB in the force-diagram. Draw a parallel through P to meet the sides of the funicular which correspond to OA, OB in the points H, K. If R be the intersection of these sides, the triangles OAB, RHK are similar, and if the perpendiculars OM, RN be drawn we have
HK·OM = AB·RN = F·RN,
which is the moment of F about P. If the given forces are all parallel (say vertical) OM is the same for all, and the moments of the several forces about P are represented on a certain scale by the lengths intercepted by the successive pairs of sides on the vertical through P. Moreover, the moments are compounded by adding (geometrically) the corresponding lengths HK. Hence if a system of vertical forces be in equilibrium, so that the funicular polygon is closed, the length which this polygon intercepts on the vertical through any point P gives the sum of the moments about P of all the forces on one side of this vertical. For instance, in the case of a beam in equilibrium under any given loads and the reactions at the supports, we get a graphical representation of the distribution of bending moment over the beam. The construction in fig. 30 can easily be adjusted so that the closing line shall be horizontal; and the figure then becomes identical with the bending-moment diagram of § 4. If we wish to study the effects of a movable load, or system of loads, in different positions on the beam, it is only necessary to shift the lines of action of the pressures of the supports relatively to the funicular, keeping them at the same, distance apart; the only change is then in the position of the closing line of the funicular. It may be remarked that since this line joins homologous points of two “similar” rows it will envelope a parabola.