The need to solve two equations simultaneously can be obviated by using a rotated reference axis system (m'n') rather than the traditional vertical and horizontal (xy) system. Thus, with respect to m'n':
An easier method of analysis for this kind of problem is discussed in a subsequent example.   
 Simplifying Conditions.   Occasionally, truss analysis by the method of joints can be facilitated by paying attention to some special conditions that frequently arise. One of these is the appearance of zero-force members. Consider the truss shown in Figure 4-11 and refer directly to joint B. By inspecting the equilibrium of this joint in the vertical direction, it is possible to determine by observation that there is no internal force at all developed in member BI. There is no applied external load and members BC and BA are horizontal, having no component in the vertical direction. If there were any force in member BI, the joint would not be in equilibrium; hence,there can be no force present in member BI. It is a zero-force member. The same is true for the force in member HC. By isolating joint J and considering equilibrium in both the horizontal and vertical directions, it can be seen that both members framing into it (JA and JI) are also zero-force members. The same is true of the members framing into joint F.
    This approach of picking out selected members and joints for quick determination of forces also can be applied to some joints with external loads applied. Isolating joint D (see Figure 4-11), inspection of equilibrium in the vertical direction indicates that the force developed in member DG must be equal and opposite to the applied load. Thus, member DG carries a force of 2P and is in tension.
As a further example of simplifying conditions, consider the truss shown in Figure 4-10 that was analyzed previously. Inspection of joint B indicates that member BH must be a zero-force member. This implies that member HC is also a zero-force member. This can most easily be seen by thinking in terms of the rotated axis system shown in Figure 4-10 and considering equilibrium in the m direction. Member HC cannot have any force in it if joint H is to be in equilibrium in this direction (as it must be in any direction). Inspection of joint D indicates that DF must carry a tension force of P. This implies that member FC carries a compressive force of P. Again, this is most easily seen by thinking in terms of a rotated axis system, as indicated in Figure 4-10. Since   is shown to be equal to P in tension, the component of this force in the m' direction must be balanced by a similar component in member FC. The members make a similar angle to the m' axis, so it follows that the forces must be numerically the same if the components are the same; thus,  = P.
Simplifying conditions of the type described facilitates calculations enormously. But the technique is most valuable in the development of a more intuitive understanding of the force distributions present in a truss.
4-3-4 Equilibrium of Sections
In our discussion of the equilibrium of joints, the elemental portions of the truss defined for equilibrium considerations were the joints themselves but any portion of a structure must be in a state of equilibrium. It follows that the limits of the portion considered can be extended to a complete subassembly consisting of several joints and members. Consideration of the equilibrium of a larger subassembly can then be the vehicle used for finding unknown bar forces. This concept of considering the equilibrium of a portion of a structure larger than a point is a very powerful one and will form the basis for the analysis and design of many structures other than the truss forms currently under discussion.
Solution of bar forces by this approach is best illustrated by example. Again consider the truss shown in Figures 4-6 and 4-12. Assume that you wish to know the forces in members ED,BD, and BC. The truss is first considered to consist of two subassemblies. Free-body diagrams for these subassemblies are shown in Figure 4-12. The internal set of forces shown is developed as a consequence of the external loading of the whole structure and serves to maintain the equilibrium of the elemental subassemblies of the truss. The complete force system acting on a sub-assembly, consisting of internal forces and any external forces present, must form a system whose net result is zero. Both translational equilibrium and rotational equilibrium must be considered, since the forces acting on the body form a coplanar, but nonconcurrent and nonparallel,force system (see Section 2-3). 土木工程英文文献和中文翻译(5):http://www.youerw.com/fanyi/lunwen_34445.html