## 2 Computational Model
Considering that \(u_{0y}=0\) and \(\phi_{0z}=0\), Equation (4) is reduced to Equation (5):  
\[[K]\{U\}=\{F\} \tag{5}\]  
where [K] = [\(K_{0}\) (3:2n + 2, 3:2n + 2)] is the stiffness matrix after considering the boundary conditions and has a dimension of \(2n\times 2n\) (the first two rows and columns of the matrix \([K_{0}]\) are removed), \(\{U\}=\left\{u_{1y},\varphi_{1z}.....,u_{iy},\varphi_{iz}.......,u_{ny}, \varphi_{nz},\right\}^{T}\) is a vector of global coordinates after considering the boundary conditions and has a dimension of \(2n\times 1\), and \(\{F\}=\{F_{A1},0,F_{A2},0....F_{Ai},0..F_{An},0\}^{T}\) is a vector of node loads after considering the boundary conditions and has a dimension of \(2n\times 1\). \(F_{A0}\) and \(M_{A0}\) are determined using Equation (6):  
\[[K_{1}]\{U\}=\left\{ \begin{array}{l}F_{A0}\\ M_{A0}\end{array} \right\} \tag{6}\]  
where [\(K_{1}\)] = [\(K_{0}\) (1:2, 3:2n + 2)] is a \(2\times 2n\) matrix formed by the first two rows and columns from 3 to \(2n+2\) of the matrix [\(K\)]. Classical FEM analysis was used up to this step. In Equation (5) (unlike the classical problem), the displacements \(u_{\rm iy}\) are known, but the forces \(F_{\rm Ai}\) and rotations \(\varphi_{\rm iz}\) of the nodes are not known. To make the equations easier to solve (with MATLAB R2020b software), it is necessary to discretize this continuous mechanical system. To determine the unknowns, the system is transformed into the form given in Equation (7):  
\[[X]\{p\}=\{Z\}{\Rightarrow}\{p\}=\ [X]^{-1}\{Z\} \tag{7}\]  
In Equation (7), the term \(\{p\}\) has a dimension of \(2n\times 1\) and contains the unknowns \(F_{\rm Ai}\) and \(\varphi_{\rm iz}\). It is determined using Equation (8):  
\[\{p\}=\{F_{A1},F_{A2},...F_{An},\varphi_{1z},\varphi_{2z}......\varphi_{nz} \}^{T} \tag{8}\]  
The term \(\{Z\}\) in Equation (7) has a dimension of \(2n\times 1\) and contains the predetermined displacements \(u_{\rm iy}\) of the nodes (coordinates of \(\rm A_{i}\) along the \(y\)-axis direction) and elements of the matrix [\(K\)]. It is determined using Equation (9):  
\[\{Z\}=\left\{ \begin{array}{l}K_{1,1}u_{1y}+K_{1,3}u_{2y}......+K_{1,2n-1}u_{ my}\\ K_{3,1}u_{1y}+K_{3,3}u_{2y}......+K_{3,2n-1}u_{ny}\\...\\ K_{2n-1,1}u_{1y}+K_{2n-1,3}u_{2y}......+K_{2n-1,2n-1}u_{ny}\\ K_{2,1}u_{1y}+K_{2,3}u_{2y}......+K_{2,2n-1}u_{ny}\\ K_{4,1}u_{1y}+K_{4,3}u_{2y}......+K_{4,2n-1}u_{ny}\\...\\...\\ K_{2n,1}u_{1y}+K_{2n,3}u_{2y}......+K_{2n,2n-1}u_{ny}\end{array} \right\} \tag{9}\]The matrix \([X]\) has a dimension of \(2n\times 2n\), and contains the null matrices and the elements of matrix \([K]\). It is determined using Equation (10):  
\[[X]=\left[ \begin{array}{l}1,O_{1,n-1},-K_{1,2},-K_{1,A},-K_{1,6},........,-K_{1,2n }\\ 0,1,O_{1,n-2},-K_{3,2},-K_{3,4},-K_{3,6},........,-K_{3,2n}\\....\\...\\ 0_{1,n-1},1,-K_{2n-1,2},-K_{2n-1,A},-K_{2n-1,6}........,-K_{2n-1,2n}\\ O_{1,n},-K_{2,2},-K_{2,4},-K_{2,6}........,-K_{2,2n}\\ O_{1,n},-K_{4,2},-K_{4,4},-K_{4,6}........,-K_{4,2n}\\...\\....\\ O_{1,n},-K_{2n,2},-K_{2n,A},-K_{2n,6},........,-K_{2n,2n}\end{array} \right] \tag{10}\]  
In this way, the unknowns from Equation (8) can easily be solved with MATLAB software. After determining the unknowns \(F_{A1},F_{A2},...F_{An},\phi_{12},\phi_{22},.....\phi_{nz}\) (contained in \(\{p\}\)), \(M_{A0}\) and \(F_{A0}\) can be determined from Equation (6) or from the two equilibrium conditions of the entire beam.