Nodal forces and moments are related to nodal displacements and rotations by the element stiffness matrix. We will formulate the element stiffness matrix using the direct approach, which uses principles of superposition and equilibrium equations to find the terms in the stiffness matrix. During our formulation of the element stiffness matrix for the one-dimensional beam element, we assume that there are no axial displacements or forces. There can be a vertical force and a moment applied at each node, which results in a vertical displacement and/or a rotation at each node. Since we are relating 4 nodal forces and moments, to 4 nodal displacements and rotations, the element stiffness matrix must be a 4 x 4 matrix, that relates each force and moment to the displacements and rotations.
In order to determine the coefficient values kij in the stiffness matrix, we will individually set each displacement and rotation equal to one, while keeping the others equal to zero. This is referred to as the unit displacement method. Each time we do this, we will be able to determine another column of the stiffness matrix. Each column of the element stiffness matrix is associated with an equilibrium problem.
Column 1:
We start by setting the vertical displacement at node I
equal to positive one, while keeping the vertical displacement at
node J and the rotations at both nodes equal to
zero. The element and physical problems associated with column 1
are shown below.
Our equation now becomes:
and in expanded form we can derive the following relationships:
Where the k values equal the forces and moments
necessary, to cause the deformation state of the beam element.
For a linear system, we can use the superposition principal
to find these forces and moments. The principle of superposition
sums the displacements and rotations of two or more cases in
order to model a more complicated situation. In the figure below
the sum of the rotations 
and 
in the two elements to
the right of the "=" sign, are equal to the rotation 
of
node I in the element problem at the left.
Similarly the sum of the vertical displacements 
and 
of the
two elements on the right are equal the total vertical
displacement vI at node I
of the element on the left.
Element problem using superposition.
Using beam deflection and slope equations, (which can be found in most strength of materials texts) we can derive the following equations:
Where positive element sign convention has been used for nodal displacements and rotations as in the right-hand side above. Solving the two equations simultaneously for the force and moment at node I, results in the following equations:
Next we use equilibrium equations and the usual sign conventions, setting the sum of the vertical forces and the sum of the moments each equal to zero, in order to find the last two terms in column one:
The terms for the first column of the element stiffness matrix are shown below :
Column 2:
In order to obtain the second column we assume a rotation at node
I that is equal to positive one, while keeping both
vertical displacements and the rotation at node J
equal to zero.
Using the same procedure as column 1, we find that the terms of the second column are:
These ar the nodal forces and moments necessary to sustain the
deformation state ( = 1, vI = vJ
= 
= 0). Column 2 can be derived in the same manner as column 1
using superposition.
Columns 3 and 4:
We can find the terms in the third and forth columns, by
individually setting the vertical displacement and rotation at
node J equal to one, as we did for node
I. The deformation states for these problems are shown
below:
Deformation state for column 3
Deformation state for column 4
When the common terms are factored out, we end up with the following stiffness matrix:
Our FEM equation for the one-dimensional beam element becomes:
where:
It should be noted that the stiffness matrix is always symmetric (i.e., kij = kji) and this characteristic can be used as a check. If the matrix is not symmetric, it is an indication of a mistake in the element formulation.