Page:EB1911 - Volume 17.djvu/981

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962
[sTAtics
MECHANICS


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 golygons A, B in the other. This notation was employed by R. H. ow 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.

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, o M

N,

H

-F

P .

FIG. 31.

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. The “ centre” (§ 4) of a system of parallel forces of given magnitudes, acting at given points, is easily determined graphically. We have only to construct the line of action of the resultant for each of two arbitrary directions of the forces; the intersection of the two lines gives the point required. The construction is neatest if the two arbitrary directions are taken at right angles to one another.

§ 6. Theory of Frames.-A frame is a structure made up of pieces, or members, each of which has two joints connecting it with other members. In a two-dimensional frame, each joint may be conceived as consisting of a small cylindrical pin fitting accurately and smoothly into holes drilled through the members which it connects. This supposition is a somewhat ideal one, and is often only roughly approximated to in practice. We shall suppose, in the first instance, that extraneous forces act on the frame at the joints only, i.e. on the pins. On this assumption, the reactions on arry member at its two joints must be equal and opposite. This combination of equal and opposite forces is called the stress in the member; it may be a tension or a thrust. For diagrammatic purposes each member is sufficiently represented by a straight line terminating at the two joints; these lines will be referred to as the bars of the frame. F IG. 32.

In structural applications a frame must be stij, or rigid, i.e. it must be incapable of deformation without alteration of length in at least one of its bars. It is said to be just rigid if it ceases to be rigid when any one of its bars is removed. A frame which has more bars than are essential for rigidity may be called over-rigid; such a frame is in general self-stressed, i.e. it is in a state of stress independently of the action of extraneous forces. A plane frame of n joints which is just rigid (as regards deformation in its own plane) has 2n-3 bars, for if one bar be held fixed the 2(n-2) co-ordinates of the remaining n-2 joints must just be determined by the lengths of the remaining bars. The total number of bars is therefore 2(n-2) -{- 1. When a plane frame which is just rigid is subject to a given system of equilibrating extraneous forces (in its own plane) acting on the joints, the stresses in the bars are in general uniquely determinate. For the conditions of equilibrium of the forces on each pin furnish zn equations, viz. two for each point, which are linear in respect of the stresses and the extraneous forces. This system of equations must involve the three conditions of equilibrium of the extraneous forces which are already identically satisfied, by hypothesis; there remain therefore 2n - 5 independent relations to determine the zn- 3 unknown stresses. A frame of n joints and 2n- 3 bars may of course fail to be rigid owing to some parts being over-stiff whilst others are deformable; in such a case it will be found that the statical equations, apart from the three identical relations imposed by the equilibrium of the extraneous forces, are not all independent but are equivalent to less than 2n- 3 relations. Another exceptional case, known as the critical ease, will be noticed later (§ 9). A plane frame which can be built up from a single bar by successive steps, at each of which a new joint is introduced by two