| Fig. 25. |
The latter relation shows that the bending moment varies as the area cut off by the ordinate in the graph of F. In the case of uniform load we have
F = −wx + A, M = 1⁄2wx2 − Ax + B,
(15)
where the arbitrary constants A,B are to be determined by the conditions of the special problem, e.g. the conditions at the ends of the beam. The graph of F is a straight line; that of M is a parabola with vertical axis. In all cases the graphs due to different distributions of load may be superposed. The figure shows the case of a uniform heavy beam supported at its ends.
| Fig. 26. |
§ 5. Graphical Statics.—A graphical method of reducing a plane system of forces was introduced by C. Culmann (1864). It involves the construction of two figures, a force-diagram and a funicular polygon. The force-diagram is constructed by placing end to end a series of vectors representing the given forces in magnitude and direction, and joining the vertices of the polygon thus formed to an arbitrary pole O. The funicular or link polygon has its vertices on the lines of action of the given forces, and its sides respectively parallel to the lines drawn from O in the force-diagram; in particular, the two sides meeting in any vertex are respectively parallel to the lines drawn from O to the ends of that side of the force-polygon which represents the corresponding force. The relations will be understood from the annexed diagram, where corresponding lines in the force-diagram (to the right) and the funicular (to the left) are numbered similarly. The sides of the force-polygon may in the first instance be arranged in any order; the force-diagram can then be completed in a doubly infinite number of ways, owing to the arbitrary position of O; and for each force-diagram a simply infinite number of funiculars can be drawn. The two diagrams being supposed constructed, it is seen that each of the given systems of forces can be replaced by two components acting in the sides of the funicular which meet at the corresponding vertex, and that the magnitudes of these components will be given by the corresponding triangle of forces in the force-diagram; thus the force 1 in the figure is equivalent to two forces represented by 01 and 12. When this process of replacement is complete, each terminated side of the funicular is the seat of two forces which neutralize one another, and there remain only two uncompensated forces, viz., those resident in the first and last sides of the funicular. If these sides intersect, the resultant acts through the intersection, and its magnitude and direction are given by the line joining the first and last sides of the force-polygon (see fig. 26, where the resultant of the four given forces is denoted by R). As a special case it may happen that the force-polygon is closed, i.e. its first and last points coincide; the first and last sides of the funicular will then be parallel (unless they coincide), and the two uncompensated forces form a couple. If, however, the first and last sides of the funicular coincide, the two outstanding forces neutralize one another, and we have equilibrium. Hence the necessary and sufficient conditions of equilibrium are that the force-polygon and the funicular should both be closed. This is illustrated by fig. 26 if we imagine the force R, reversed, to be included in the system of given forces.
It is evident that a system of jointed bars having the shape of the funicular polygon would be in equilibrium under the action of the given forces, supposed applied to the joints; moreover any bar in which the stress is of the nature of a tension (as distinguished from a thrust) might be replaced by a string. This is the origin of the names “link-polygon” and “funicular” (cf. § 2).
If funiculars be drawn for two positions O, O′ of the pole in the force-diagram, their corresponding sides will intersect on a straight line parallel to OO′. This is essentially a theorem of projective geometry, but the following statical proof is interesting. Let AB (fig. 27) be any side of the force-polygon, and construct the corresponding portions of the two diagrams, first with O and then with O′ as pole. The force corresponding to AB may be replaced by the two components marked x, y; and a force corresponding to BA may be represented by the two components marked x′, y′. Hence the forces x, y, x′, y′ are in equilibrium. Now x, x′ have a resultant through H, represented in magnitude and direction by OO′, whilst y, y′ have a resultant through K represented in magnitude and direction by O′O. Hence HK must be parallel to OO′. This theorem enables us, when one funicular has been drawn, to construct any other without further reference to the force-diagram.
| Fig. 27. |