Another completely different way of visualizing the forces developed in a truss is to use an arch and cable analogy. It is evident, for example, that truss A can be conceived of as a cable with supplementary members [see Figure 4-2(d)]. Truss B can be conceived of as a simple linear arch with supplementary members. Diagonals in truss A are consequently in tension while diagonals in truss B must be in compression. The forces in other members can be determined by analyzing their respective roles in relation to containing arch or cable thrusts or providing load transfer or reactive functions. As Figure 4-3 indicates, special forms of more complex trusses can also be visualized in this same general way.
   Obviously, both of the methods discussed above for visualizing forces in trusses become exceedingly difficult or even impossible to apply when more complex triangulation patterns are present. A method of visualizing forces in such trusses on the basis of joint equilibrium considerations is considered in the next section. In general, however, complex truss forms must be mathematically analyzed according to methods to be discussed for correct results to be obtained.
4-3  ANALYSIS OF TRUSSES 
4-3-1  Stability
The first step in the analysis of a truss is always to determine whether the truss studied is indeed a stable configuration of members. It is usually possible to tell by inspection whether or not a truss is stable under external loads by considering each joint in turn to determine if the joint will always maintain a fixed relation to other joints under any loading condition applied to the truss.
In general, any truss composed of an aggregation of basic triangular shapes will be a stable structure.
The appearance of non triangular shapes in a bar pattern is an obvious sign that the truss should be looked at carefully, The truss indicated in Figure 4-4(a) is unstable and would collapse under load in the manner illustrated. This truss clearly does not have a sufficient number of bars to maintain a fixed geometrical relationship between joints. Assuming that the remaining truss members are adequately designed for the loads they carry, the addition of a member from B to E, as indicated in Figure 4-4(b), would make this into a stable configuration.
It is possible for a truss to include one or more figures that are not triangles and still be stable. Study Figure 4-4(c), which illustrates a truss of this type. It is made up of groups of rigid
FIGURE 4-4   Internal stability. Nontriangulated forms may collapse
triangulated bar patterns connected to form nontriangular, but still stable, figures. The group of triangles between A and C form a rigid shape, as do those between B and C. Joint C is thus held fixed in relation to joints A and B in much the same way as occurs in a simpler triangular figure.
The group of triangles between A and C can be thought of as a "member," as can those between B and C.
In a single truss it is quite possible to have more than the minimum number of bars necessary to achieve a stable structure. One of the two middle diagonal members of the truss shown in Figure 4-4(d) could be considered redundant. Either member BE or member CF could be removed and the resultant configuration would still be stable. Obviously, removing both would cause the structure to become a collapse mechanism.
The importance of determining whether a configuration of bars is stable or unstable cannot be overestimated, since there are few errors more dangerous. Total collapse occurs immediately after an unstable configuration is loaded. As an aid to ascertaining the stability of configurations, expressions have been developed which relate the number of joints present in a truss to the number of bars necessary for stability. A simple basic planar triangle has three bars (n = 3) and three joints (j = 3). Adding a single new connection beyond the original three requires the addition of two new bars or members. Hence the total number of members in a truss having a number of connections is given by following : n = 3 + 2(j - 3). [Three members are originally present and for each new joint (j-3) two] new members are added.] This same expression can bar networks. A simple tetrahedron is known to be a stable structure. Three bars are required to form each additional new node added to this basic form. There are six bars to begin with and new bars are added at the rate of three for each node beyond  the original four. Thus, n = 6 + 3 (j - 4) = 3j - 6. 建筑结构构件桁架英文文献和中文翻译(4):http://www.751com.cn/fanyi/lunwen_25653.html