: Determine the element stiffness matrix for each element.

Before we can proceed with our FEM analysis, we must determine the stiffness matrix for each element. Since we are relating 4 nodal forces and moments, to 4 nodal displacements and rotations, the element stiffness matrix must be of order 4 x 4, which relates each nodal force and moment to the corresponding displacements and rotations at the nodes.

During our formulation of the Element Stiffness Matrix we found our elemental equation to be:

where the element quantities are:

We will assume our usual sign conventions where nodal forces and displacements acting upward are positive and moments and rotations acting counterclockwise are positive in accordance to the right-hand rule.

Element 1:

The local (I, J) nodes of each element are related to the global nodes in the element mesh, which we will number (the element mesh can be viewed at any time, by clicking on mesh in the lower left-hand corner of your screen). The nodes that are between elements are shared by them, and provide element compatibility (the absence of element gaps and overlaps). For element 1 local node I is global node 3 and local node J is global node 1.

The geometric and material properties for element 1, using consistent units are as follows:

L₁ = 96 in, I₁ = 23.1 in⁴ and E₁ = 29 x 10⁶ psi

where the subscript denotes that these properties are for element number 1. When these values are substituted into the element stiffness matrix, ,

the stiffness matrix for element 1 becomes:

where the nodal forces and moments, , are related to the nodal displacements and rotations, by the element stiffness matrix, , where the notation denotes a matrix and the E subscript for , and denotes elemental quantities.

Element 2

For element 2 local node I is global node 1 and local node J is global node 2, where the global node locations can be seen in the element mesh, shown in the lower left-hand frame.

The geometric and material properties for element 2, using consistent units are as follows:

L₂ = 116 in, I₂ = 83.2 in⁴ and E₂ = 29 x 10⁶ psi

When these values are substituted, the stiffness matrix for element 2 becomes:

The individual element stiffness matrixes should be checked at this point to see if they are show the proper characteristics.

: Determine the stiffness matrix for the entire mesh.

The FEM mesh (shown in the lower left-hand frame), contains three nodes, each having one vertical translational degree of freedom and one rotational degree of freedom. Therefore, the stiffness matrix must relate six nodal forces and moments to six translational and rotational degrees of freedom, this requires the stiffness matrix for the entire structure to be of order 6 x 6. We will use the direct assembly method to combine the stiffness matrixes for individual elements, to form the global stiffness matrix.

Our first step is to assign global names to each node in the stiffness matrix. Since element one incorporates local nodes I and J as global nodes 3 and 1, column 1 relates all the nodal forces and moments to the vertical displacement at global node 3 and column 2 relates all the nodal forces and moments to the rotation at global node 3. Similarly columns 3 and 4 relate all the nodal forces and moments to the vertical displacement and rotation at global node 1. The stiffness matrix for element 2 relates all the nodal forces and moments to displacements and rotations at global nodes 1 and 2. Therefore we label the columns as follows:

and

We must also be careful that the rows be correctly aligned in our global stiffness matrix, so we label the rows to indicate the degree of freedom from the displacement and rotation column matrix, that is associated with that row. In order to simplify our matrix assembly process, we partition our matrix (divide it up into sub-matrixes) where each sub-matrix contains the terms that relate the vertical displacement and rotation of a single node to the nodal force and moment for a single node. For example the terms in the upper right sub-matrix for element 1 (shown below) relate the displacement and rotation at node 1 to the nodal force and moment at node 3.

and

We begin assembling the global stiffness matrix, by assembling each sub-matrix of the stiffness matrix for element 1 into the position with the same designation in the global stiffness matrix. For example the sub-matrix from rows , and columns , of the element stiffness matrix is placed in rows , and columns , of the global stiffness matrix.

We now complete our global stiffness matrix, by assembling the sub-matrixes of element 2 into the global matrix and assigning a zero value to the empty spaces. In the positions indicated by column -row , column -row , column -row and column -row ( positions in the upper left-hand sub-matrix of the global stiffness matrix shown below), the values for element 2 are added to the element 1 values already in these positions.

Our completed global stiffness matrix is as follows:

where the global stiffness matrix is symmetric, just as the element stiffness matrix was. It should also be noted that all the stiffness terms in the main diagonal (upper left-hand corner to lower right-hand corner) must always be positive and that the stiffness matrix is singular, meaning that it's determinate is equal to zero. In order to obtain a unique FEM solution we need to constrain the structure. Having a known degree of freedom will eliminate the singularity by altering the stiffness matrix when we expand the matrix equation. The known degrees of freedom will be found from the problem's boundary conditions.

: Prescribe known boundary conditions and determine unknown nodal displacements and rotations.

Before we can find a solution for the vertical displacements and rotations, we must apply the known boundary conditions at the nodes. There are several things that need to be considered when applying boundary conditions, as follows:

• We must not be under-constrained, meaning we need to have as many equations as unknowns when our matrix is expanded. This will require that we have as many known boundary conditions as rows in our global stiffness matrix. Each node has four quantities that can be unknowns ( F, M, v and ) we need to know one of these quantities for each degree of freedom of our FEM equation.
• We should also avoid over-constraining the problem, which is where we have more equations than unknowns. A problem that is over-constrained will have some row(s) where both an applied force or moment and a degree of freedom (vertical displacement or rotation) will be specified. This situation can be dealt with by assuming one of these quantities to be an unknown (usually the vertical displacement or rotation, since they are usually easier to solve for), the solution for this assumed unknown quantity can then be compared to the known quantity as a check of the solution procedure.
• In order to avoid having rigid body motion, we must have at least two degrees of freedom specified for the structure, either two displacements or a displacement and a rotation.
• We get our boundary conditions from the element mesh. From the figure below we can see that our known boundary conditions for this example are , where the forces and moments are externally applied to the structure and the forces are negative in this case, by the positive = upward sign convention we are using.

We will use these boundary conditions along with our global stiffness matrix to formulate our FEM equation, which we will then solve for the unknown nodal displacements and rotations. Since the degrees of freedom which columns 3 and 4 relate to the nodal forces and moments have zero values in this example, we can eliminate these columns without effecting our solution. We can also eliminate rows 3 and 4, since the unknown force and moment at global node 2 will not assist us in finding the displacements or rotations at the other nodes.

We end up with the following reduced force-displacement equation:

where the above equation is in the form of a system of linear algebraic equations. In expanded form our matrix equation results in 4 equations with 4 unknowns, which we can now solve for the unknown nodal displacements and rotations . These simultaneous equations can be solved by hand, using techniques such as Gaussian Elimination or by computer, using a matrix solving program or an iterative solver. It is good to use two methods, in order to check your results, as it can be quit easy to make an error in calculation or data entry, that will make your entire FEM solution invalid.

The solution of these equations result in:

When we add the known displacement and rotation from global node 2, our solution is:

Solutions should always be checked to see if they make physical sense and verified by another method that can at least approximate the magnitude and direction of the FEM solution. In the above example, we can input the nodal displacements and rotations just obtained, back into our reduced force-displacement equation and solve for the forces and moments, which should match the applied forces and moments that we started with.

: Graphical interpretation of the displacement and rotation for the entire structure.

The solution characteristics of the beam element require that the vertical displacement and rotation vary continuously through the element and the entire mesh. The vertical displacement varies within each element according to a cubic polynomial and since the rotation is the derivative of the vertical displacement function it varies according to a quadratic polynomial within each element.