III. A force on a rigid body can be re placed by any set of its components without dis turbing eqtulibrium. In an elastic body the state of internal stress would be altered by such a substitution. By means of III we can replace the weight of members, and roof, snow and wind loads by their equivalent components acting at the pins of a frame.
IV. The equilibrium theorems as expressed by the force and funicular polygons and the reciprocal diagrams explained below.
V. Theorems based on theprinciple of work and virtual displacements. These cannot be dealt with in a brief exposition. Consult Hudson, 'Deflections and Statically Indeter minate Stresses,' (1911) ; Johnson, Bryan and Turneaure, The Theory and Practice of Modern Framed Structures' (9th ed.); Grimm, 'Secondary Stresses in Bridge Trusses.' Reciprocal diagrams.— A frame acted on by pin loads or by intermediate forces replaced by their equivalent pin components, consists statically of as many sets of concurrent forces as there are pins. The resulting tensile and compressive forces in the members, which are two-point nieces, can be obtained in several ways, but in many problems in structural en gineering the graphical solution can be carried systematically as shown in connection with Fig. 8. i shows a simple roof truss carrying a load W and resting on supports; the reactions are vertical. They are to be found in the simplest way; in this case, from the symmetry, each is W/2. Now put letters in the spaces be tween the forces, and numbers in the panels. The load W is then called ah, the reactions being be and ca. The force in the vertical member is either 12 or 21 according to the pin on which it acts; in the member between A and 1 it is al or la. This is Bow's notation, mentioned above. Draw the external force polygon abca, taking the forces in alphabetical order. At the left pin there are three forces ca, al, lc which must be named and used around the pin in the order the letters have around the truss : clock wise in this case. Since ca, ar, 1c are in equilibrium they will form a triangle which is to be attached to the f. :e polygon as in ii. In the triangle, al (not la) acts from a towards 1 and therefore pushes on the left pin of the truss; al is thus in compression. 1c is in tension be cause lc (not cl) acts from 1 to c and represents a pull. The quadrilateral for the top pin forces la, ab, b2, 21 is drawn in ii; similarly for the other pins. No pin can be used at which more
than two unknowns act. Consider now the free body cut out by the line ss. Correspond ing to the forces on it — la, ab, b2, 2c, cl there is a polygon lab2c1; the same is true of any part of the truss as a free body. Con versely, for every force polygon in ii there is a free body in equilibrium in i. This is the most general form of the reciprocal relation already seen in the funicular and force polygons. The diagrams in Fig. 8 are called Maxwell-Cremona diagrams; they were used independently how ever by Rankine, 'Applied Mechanics' (1857) and shortly afterwards by W. P. Taylor, a practical draughtsman.
Henneberg's method.— The Maxwell-Cre mona construction cannot be employed in all cases. In Fig. 7, i and iv, there is no pin at which there are only two unknowns, but by using the theorem of four forces on the parts cut out by the section lines ss, some of the internal forces can be found; the method- then applies. Fig. 7, vi. cannot he solved by any method ex plained above. An ingenious graphical solution which has not yet become current in textbooks was devised by Henneberg, der starren Systeme' (1886). The frame in Fig. 9 will serve to illustrate the procedure. F. is given and F., F. are found in any convenient way. In order to make the structure indeterminate intro duce a link L anywhere. Being indeterminate it has an infinity of solutions two of which are to be got thus: (a) Select any pin at which there are not more than three rods, e.g., that at F.; this will always be possible — see the discussion of determinateness above. Draw any polygon ii for it and find the remaining forces (not shown). Observe that P', Q', R' are not the required forces in the links of the given frame; they are merely those forces in the members of the redundant frame which can equilibrate F& (b) The redundant frame may have initial stress, i.e., if L or any other link is too long or too short it will produce stress in all the mem bers without the aid of external forces. Find any set of initial forces Po as in iii for the pin at Fl and then find the rest Lo, Sm .... (not shown).
If P, Q, R, S, .... are the actual forces be fore L is introduced we have for the pin F. the vector sum