## 4 General RC-Beam-Type Element--FEM Formulation  
In this paragraph, the prerequisite mathematical relations for the FEM formulation of the proposed model's algorithm are presented. The model algorithm is described in the next paragraphs.  
Figure 6 shows the forces, \(\overline{\mathbf{Q}}\), and the corresponding displacement degrees of freedom, \(\overline{\mathbf{q}}\), of the member's end nodes in its local coordinate system \(xyz\), considering rigid body mode (RBM). The torsional moment is neglected as it is considered constant and uncoupled with the rest of the degrees of freedom. Thus, the degrees of freedom of each element  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Parameter** & **Proposed Value** \\ \hline \(b\) & 0.02 \\ \(R_{o}\) & 10 to 20 \\ \(cR_{1}\) & 0.925 \\ \(cR_{2}\) & 0.15 \\ \(A_{1}\) & 0.00 \\ \(A_{2}\) & 1.00 \\ \(A_{3}\) & 0.00 \\ \(A_{4}\) & 1.00 \\ \hline \hline \end{tabular}
\end{table}
Table 1: Parameter values of steel constitutive model.  
consist of three forces and two bending moments corresponding to three displacements and two rotations per end node, respectively.  
In the present mathematical formulation, in addition to the complete kinematic state, the relative position of the member in the \(XYZ\) space is also adopted due to its transverse displacements. This relative position results from eliminating the transverse degrees of freedom to the axis of the element. The complete kinematic state can be recovered later based on the geometric relationship that relates the two states. Thus, there are five degrees of freedom left in the member, which correspond to the axial force \(Q_{5}\) and two bending moments at each node (\(Q_{1}\), \(Q_{3}\), and \(Q_{2}\), \(Q_{4}\), respectively), as shown in Figure 7. These forces correspond to an axial displacement \(q_{5}\) and to two rotations per end node (\(q_{1}\), \(q_{3}\), and \(q_{2}\), \(q_{4}\), respectively).  
Figure 6: Forces and displacements at element’s end nodes considering RBM in the local coordinate system _xyz_ ([PERSON] et al. [5]).  
Figure 7: Forces and displacements at the element’s ends without considering RBM ([PERSON] et al. [5]).  
Element displacement and force vectors, \(\overline{\mathbf{q}}\) and \(\overline{\mathbf{Q}}\), (considering RBM) are related to their corresponding vectors, \(\mathbf{q}\) and \(\mathbf{Q}\), (without RBM) via the following transformation matrix \(\mathbf{L}_{RBM}\):  
\[\overline{\mathbf{Q}}=\left[\begin{array}{ccccccccc}\overline{Q}_{1}&Q_{2}& \overline{Q}_{3}&\overline{Q}_{4}&\overline{Q}_{5}&\shortmid&\overline{Q}_{6}& \overline{Q}_{7}&\overline{Q}_{8}&\overline{Q}_{9}&\overline{Q}_{10}\end{array} \right]^{  
while the corresponding cross-sectional vectors are:  
\[\text{the force vector}\ \mathbf{D}(x)=\begin{bmatrix}M_{z}(x)\\ M_{y}(x)\\ N(x)\end{bmatrix} \tag{52}\]  
\[\text{and the displacement vector}\ \mathbf{d}(x)=\begin{bmatrix}\varphi_{z}(x)\\ \varphi_{y}(x)\\ \varepsilon_{\text{o}}(x)\end{bmatrix}. \tag{53}\]  
According to the classic FEM formulation, matrix \(\mathbf{b}(x)\) relates the element-nodal load vector \(\mathbf{Q}\) with the corresponding cross-sectional load vector \(\mathbf{D}(x)\) at a specific location \(x\), using the relation  
\[\mathbf{D}(x)=\mathbf{b}(x)\cdot\mathbf{Q} \tag{54}\]  
where  
\[\mathbf{b}(x)=\begin{array}{cccccc}Q_{1}&Q_{2}&Q_{3}&Q_{4}&Q_{5}\\ &&M_{z}(x)\\ \left[\begin{array}{cccccc}\frac{x}{L}-1&\frac{x}{L}&0&0&0\\ 0&0&\frac{x}{L}-1&\frac{x}{L}&0\\ 0&0&0&0&1\end{array}\right]&M_{z}(x)\\ &&N(x)\end{array}. \tag{55}\]  
In the above equation, it is assumed that the axial force, \(N(x)\), remains constant along the length of the member, while the distributions of the bending moments \(M_{z}(x)\) and \(M_{y}(x)\) are linear. Therefore, the interpolation functions described in matrix \(\mathbf{b}(x)\) do not cover the case of intermediate loads on a member. To be able to support this case, it is necessary to appropriately modify the interpolation functions of this matrix.