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 any 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.
| Fig. 32. |
In structural applications a frame must be stiff, 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 2n 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 − 3 independent relations to determine the 2n − 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 case, 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 new bars meeting there, is called a simple frame; it is obviously just rigid. The stresses produced by extraneous forces in a simple frame can be found by considering the equilibrium of the various joints in a proper succession; and if the graphical method be employed the various polygons of force can be combined into a single force-diagram. This procedure was introduced by W. J. M. Rankine and J. Clerk Maxwell (1864). It may be noticed that if we take an arbitrary pole in the force-diagram, and draw a corresponding funicular in the skeleton diagram which represents the frame together with the lines of action of the extraneous forces, we obtain two complete reciprocal figures, in Maxwell’s sense. It is accordingly convenient to use Bow’s notation (§ 5), and to distinguish the several compartments of the frame-diagram by letters. See fig. 33, where the successive triangles in the diagram of forces may be constructed in the order XYZ, ZXA, AZB. The class of “simple” frames includes many of the frameworks used in the construction of roofs, lattice girders and suspension bridges; a number of examples will be found in the article [Bridges]. By examining the senses in which the respective forces act at each joint we can ascertain which members are in tension and which are in thrust; in fig. 33 this is indicated by the directions of the arrowheads.
| Fig. 33. |
| Fig. 34. |
When a frame, though just rigid, is not “simple” in the above sense, the preceding method must be replaced, or supplemented, by one or other of various artifices. In some cases the method of sections is sufficient for the purpose. If an ideal section be drawn across the frame, the extraneous forces on either side must be in equilibrium with the forces in the bars cut across; and if the section can be drawn so as to cut only three bars, the forces in these can be found, since the problem reduces to that of resolving a given force into three components acting in three given lines (§ 4). The “critical case” where the directions of the three bars are concurrent is of course excluded. Another method, always available, will be explained under “Work” (§ 9).
When extraneous forces act on the bars themselves the stress in each bar no longer consists of a simple longitudinal tension or thrust. To find the reactions at the joints we may proceed as follows. Each extraneous force W acting on a bar may be replaced (in an infinite number of ways) by two components P, Q in lines through the centres of the pins at the extremities. In practice the forces W are usually vertical, and the components P, Q are then conveniently taken to be vertical also. We first alter the problem by transferring the forces P, Q to the pins. The stresses in the bars, in the problem as thus modified, may be supposed found by the preceding methods; it remains to infer from the results thus obtained the reactions in the original form of the problem. To find the pressure exerted by a bar AB on the pin A we compound with the force in AB given by the diagram a force equal to P. Conversely, to find the pressure of the pin A on the bar AB we must compound with the force given by the diagram a force equal and opposite to P. This question arises in practice in the theory of “three-jointed” structures; for the purpose in hand such a structure is sufficiently represented by two bars AB, BC. The right-hand figure represents a portion of the force-diagram; in particular ZX> represents the pressure of AB on B in the modified problem where the loads W1 and W2 on the two bars are replaced by loads P1, Q1, and P2, Q2 respectively, acting on the pins. Compounding with this XV>, which represents Q1, we get the actual pressure ZV> exerted by AB on B. The directions and magnitudes of the reactions at A and C are then easily ascertained. On account of its practical importance several other graphical solutions of this problem have been devised.