As an example, consider the mesh of 6 elements (N = 6) and 7 nodes, shown in Fig. 6.4.

Figure 6.4 Mesh with 6 elements

The connectivity matrix for this mesh can be written as:

.

(6.42)

Let

,

(6.43)

and

(6.44)

be the element stiffness matrix and the element force vector of the elements =1,2,3,4,5,6.

Consider the element 1, i.e. . Note that

(6.45)

Then as per equation (6.36), components of the stiffness matrix of the element 1, i.e. of , occupy the following locations in the expanded matrix :

(6.46)

Similarly, as per equation (6.37), components of the force vector of the element 1, i.e. of , occupy the following locations in the expanded vector :

(6.47)

The remaining components of the expanded matrix and the expanded vector are zero. Thus, the matrix becomes:

.

(6.48)

and the vector becomes:

(6.49)

Similarly, we obtain the expanded versions of the element stiffness matrix and the element force vector for the remaining elements, i.e. for = 2,3,4,5,6. It can easily be verified that, for the 3 rd element (i.e. for ), the expanded matrix and the expanded vector are:

(6.50a)

(6.50b)

This completes the first step.

In the 2^nd step, we add all the expanded matrices and vectors. Thus, equation (6.38) gives the following expression for the global stiffness matrix:

(6.51)

Similarly, the sum of the expanded force vector becomes:

(6.52)

However, before we get the global force vector , we need to add the vector to the above expression. Since (no. of elements) = 6, the -th component, i.e. the 7-th component of the vector will be . The remaining components will be zero as per equation (6.41). Thus, becomes:

(6.53)

Substituting the expressions (6.52) and (6.53) in equation (6.39), we get the following expression for the global force vector :

(6.54)

Now, as in section 6.3, assume that and (distributed force) are constant for the entire bar. Further, assume that the length of each element is constant. Let us denote it by h . Then

(6.55)

Then, equation (6.32) implies that the element stiffness matrix is identical for each element and is given by

(6.56)

Similarly, equation (6.33) implies that the element force vector is identical for each element and is given by

(6.57)

Substituting the expression (6.56) in equation (6.51), we get

Further, substituting the expression (6.57) in equation (6.54), we get

(6.59)

In actual calculations, the assembly procedure is appropriately modified to reduce the computational time and storage requirements. When, the number of elements is large, storing of the expanded matrices and vectors for each element needs a lot of storage requirement. Therefore, the process is modified as follows:

• Once the expanded version of the element stiffness matrix of the first element is obtained, the element stiffness matrices of other elements are not expanded.
  
  
• Instead, the locations of the components of the stiffness matrix of the element two are determined using equation (6.36).
  
  
• From the connectivity (6.34), it is easy to see that

(6.60)