## 7 Conclusions
Referring to a Cartesian system (\(R\)\(\equiv\) (O,XYZ)), which is fixed with respect to the ground, if a forced displacement (\(d_{b}\)) is imposed on \(B\) along the direction K, then, the \(\mathrm{I}_{n,S}\) nodes will all move by \(\mathrm{d}_{b}\) in the same direction. In mathematical terms, if N is the number of active degrees of freedom of the \(\mathrm{I}_{n,S}\) nodes, the displacement \(\underline{d}\) of the \(\mathrm{I}_{S}\)nodes will be equal to \(d_{b}\), \(\underline{e}\), where \(\underline{e}\) is the (N \(\times\) 1) vector that is characterized by unitary components in terms of the positions associated with the K direction, and by zero components in all the other positions To clarify this with an example, let us suppose that \(\mathrm{I}_{S}\) is characterized by eight nodes, and that each of the nodes has two active degrees of freedom that are related to the translations along the \(X\) and \(Y\) axis of a given reference frame (N = 16). In this case, if the displacement (db) is imposed along the X direction, then the vector (e) would have the following expression:  
\[(\underline{e})^{T}=\left[ \begin{array}{cccccccccccccccc}1&0&1&0&1&0&1&0&1&0& 1&0&1&0&1&0\end{array} \right]\]  
In the absence of external forces and dissipation, the dynamic response of \(S\) to an acceleration that is imposed at its base is governed by the following equation:  
\[\left[ \begin{array}{cc}m_{e}&\underline{0}\\ &\underline{0}&\underline{M}\\ (N\times 1)&\underline{M}&(N\times N)\end{array} \right]\left[ \begin{array}{c} \overset{..}{d}_{b}\\ \overset{..}{d}\end{array} \right]+\left[ \begin{array}{cc}\frac{\varepsilon ^{T}\underline{K}\,\underline{e}}{-\underline{K}\,\underline{e}}&\frac{- \varepsilon^{T}\underline{K}}{(N\times N)}\end{array} \right]\left[ \begin{array} []{c}d_{b}\\ \overset{..}{d}\end{array} \right]=\left[ \begin{array}{c}0\\ \underset{(N\times 1)}{0}\end{array} \right], \tag{1}\]  
where:  
\(m_{e}\) is the inertia of the base of S along the displacement direction that is individuated by the vector, \(\underline{e}\):  
\(\underline{M}\) is the mass matrix of S related to the \(\mathrm{In}\),S nodes only;  
\(\underline{\underline{K}}\) is the stiffness matrix of S constrained at its base;  
\(\underline{d}\) is the displacement vector of the \(\mathrm{In}\),S nodes. Let \(\underline{\Delta d}\) be the displacement vector of the \(\mathrm{I}_{S}\) nodes with respect to a reference system (\(R^{\prime}\)\(\equiv\) (O',X'Y'Z')), moving together with the base of S and having the axes X', Y', and Z', which are oriented as X, Y, and Z, respectively. Since it results in \(\underline{d}=d_{b}\)\(\underline{e}+\underline{\Delta d}\), the vector, \(\left[ \begin{array}{c}d_{b}\\ \overset{..}{d}\end{array} \right]\), can be expressed in the following way:  
\[\left[ \begin{array}{c}d_{b}\\ \overset{..}{d}\end{array} \right]=\left[ \begin{array}{cc}1&\underline{0}\\ &(1\times N)\\ \underline{e}&\underline{\underline{I}}\\ (N\times N)\end{array} \right]\left[ \begin{array}{c}d_{b}\\ \overset{..}{\Delta d}\end{array} \right] \tag{2}\]  
where \(\underline{\underline{I}}\) is the identity matrix of the size (N). By substituting Equation (2) into Equation (1), we obtain:  
\[\left[ \begin{array}{cc}m_{e}&\underline{0}\\ (1\times N)\\ \underline{0}&\underline{M}\\ \end{array} \right]\left[ \begin{array}{cc}1&\underline{0}\\ (1\times N)\\ \underline{\xi}&\underline{I}\\ \end{array} \right]\left[ \begin{array}{cc}\overset{..}{d}_{b}\\ \overset{..}{d}\end{array} \right]+\left[ \begin{array}{cc}\varepsilon^{T} \underline{K}\,\underline{e}&-\varepsilon^{T}\underline{K}\\ -\underline{K}\,\underline{e}&\underline{K}\end{array} \right]\left[ \begin{array} []{cc}1&\underline{0}\\ \overset{..}{(1\times N)}\end{array} \right]\left[ \begin{array}{c}d_{b}\\ \overset{..}{d}\end{array} \right]=\left[ \begin{array}{c}0\\ \underset{(N\times 1)}{0}\end{array} \right]\]or, in a more compact form:  
\[\left[