Let us now apply the method to a more complex two rod system in Figure 1.3. We introduce a preliminary step - idealization of the problem. For complex problems we are rarely if ever able to analyze the base structure directly. We need to simplify it first.
The axial deformation behavior of a rod subjected to load is much like
that of a spring. 
where F is the applied force, A is the cross section area,E
is the elastic modulus, L is the length, and d is the deflection.
If we identify 
as a spring constant then it is indeed spring like behavior. We must point
out that this is a great simplification since we assume that lateral deformations
are negligible, area of cross section and elastic modulus are constant
etc.
Thus, we simplify the 2 rod structure to an equivalent 2 spring system as in Fig 1.4.
Figure: 2 spring elements and 3 displacement degrees of freedom
after FE discretization.
We decompose the structure into 2 elements each comprised of one spring. Thus we end up with 2 elements and 3 nodes.
On each element we can obtain the local approximation by looking at the sum of forces at each node.
Figure 1.5: General spring with stiffnes kp and under
loads fip and fjp
For node i and node j in figure 1.5
Rewriting in matrix form and moving the forces to the right:
or in compact notation
where [k] is called the element stiffness matrix, 
is the element displacement vector and 
is the element force vector.
Equations (1.3) repesents the behavior of any generic element. Let us now specialize it for the two elements in our structure.
Element 1
The specialization is simply an identification of the symbols p with element number, i and j with the appropriate nodes and kp with the correct stiffness coefficient. Thus for element 1
and correspondingly (1.3) is now
Element 2
For element 2
and correspondingly (1.3) is now
We need to put together (1.4) and (1.5) to get the total system behavior. Examination of Fig. (1.4) shows that the forces at the nodes add up (we took care in setting up the system that they were all in the same direction). Fig. 1.6 shows all the forces and their locations and directions.
Writing down the sum of forces at each node as shown in Fig. 1.6 in matrix-vector form yields:
or in compact notation
Now take a careful look at (1.6).
We notice that this equation is simply an overlap of (1.4)
and (1.5)
with addition of terms corresponding to the shared node 
. Thus, the same equation could have been obtained without reference to
Fig. 1.6
and writing sums of forces by simply overlapping [k] , and
from (1.4)
and (1.5)
appropriately.
