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The Transformation Matrix
Let's begin with the general equation transforming the forces in global coordinates into axial forces.

$$\begin{bmatrix}

P^{(e)}_{1}\\ P^{(e)}_{2} \end{bmatrix} = T^{(e)}

\begin{bmatrix}

f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{bmatrix}$$ Here P is the axial force. Here, the transformation matrix must be a 2 by 4 matrix in order for the math to be correct. Recall the elemental axial force displacement relationship: $$\hat{k}^{(e)}_{2x2}q^{(e)}_{2x1} = p^{(e)}_{2x1}$$ This translates to $$\hat{k}^{(e)}(T^{(e)}d^{(e)}) = (T^{(e)}f^{(e)})$$ The goal here is get the elemental force displacement relationship, so, in order for that to happen, one must "move" the transformation matrix from the right hand side to the left hand side by multiplying by the inverse of the transformation matrix. However, it is unfortunate that the transformation matrix is a rectangular matrix (2 by 4), so it cannot be inverted. How can we obtain the elemental force displacement matrix then? The answer is the following equation. $$[T^{(e)T}_{4x2}\hat{k}^{(e)}_{2x2}(T^{(e)}_{2x4}]d^{(e)}_{4x1} = f^{(e)}_{4x1}$$

After deriving this, it is necessary to indicate that the principle of virtual work is responsible for the elimination of columns 1, 2, 5, and 6 in the stiffness matrix so that $$\displaystyle k_{6x6}d_{6x1}=F_{6x1}$$ becomes $$\displaystyle k_{2x2}d_{2x1}=F_{2x1}$$

Why not solve the global force displacement relationship by multiplying both sides by the inverse of the stiffness matrix K -1 ? This is because the inverse of the stiffness matrix could not be computed and there is a singularity (i.e. the determinant is zero and thus K is not invertible). Recall, you need to compute 1 over the determinant of the matrix to find the inverse. Why? For an unconstrained system, there are three possible rigid body motions in 2 dimensions (2 translation and 1 rotation). Thus, we calculate the eigenvalues of the stiffness matrix to give us information of the motions of this constrained system.

The Finite Element Method eigenvalues can be compared to a dynamics eigenvalue problem such as $$ Kv = \lambda MV$$ where K is a stiffness matrix, $$\lambda$$ is an eigenvalue, and M is a mass matrix. In both the dynamics approach and the Finite Element approach, a zero eigenvalue represents zero stored energy.

Because eigenvalues are used in a broad range of applications, the terminology associated with them varies across the mathematical disciplines, and at times even within a single discipline. Eigenvalues are often referred to as modes, or mode shapes. These modes refer to rigid body modes and correspond to elastic energy for the problems covered in this course of Finite Element Analysis.