## 6 Example  
The interpolation and the usefulness in the optimization of this method are well documented in [20; 21]. The example focuses on the construction of the Bernoulli beam model from FEM data.  
We choose the simplest example in order to be able to verify the result using analytic calculations. Consider the massless cantilever beam that has a mass at the end (see Figure 6). This example is actually the building block of the lumped-mass FEM model of a Bernoulli beam therefore it can be easily extended to more complex cases.  
The beam is an aluminum rod with a rectangular cross-section of 20 mm \(\times\) 40 mm, with the following parameters:  
* \(L=1\) m
* \(I=26\),666 mm4
* \(E=70\) GPa
* \(m=1\) kg  
The applied load is a 0.1 Nm pure bending moment along the entire span. The unloaded nodes are projected onto the projection line (see Figure 7), then the corresponding displacement is carried to the next step.  
Figure 6: Cantilever beam with end mass.  
In the next step, a polynomial is fitted to the projected data. Since the cross-section of the beam and the Young's modulus of the material remain constant throughout, the solution to the Bernoulli equation is a quadratic polynomial [32]; therefore, a quadratic function is selected for the fitting process. However, in more general cases, the order of the polynomial must be selected carefully to obtain an acceptable result while avoiding overfitting. A higher-order polynomial may provide a closer fit to the data points, but it could also introduce artificial oscillations that do not reflect the true behavior of the system. Conversely, a lower-order polynomial might fail to capture important features.  
In Figure 8, the node data and the fitted polynomial are nearly indistinguishable due to the accuracy of the polynomial fit. To provide insight into the deviation between the fitted polynomial and the original data, an additional plot displaying the error between these two datasets is included.  
Since the fitted polynomial is a quadratic polynomial, its second derivative is a constant value given by \(v^{\prime\prime}=-5.3327\times 10^{-5}\). Using Equation (5), the stiffness can be calculated as  
Figure 8: Polynomial fit for projected node data.  
Figure 7: Loaded beam (The deformation is not to scale).  
\(IE_{FEM}=1875.2\) N/m. The theoretical calculation gives IE = 26,666 \(\times\)\(10^{-12}\times 70\times 10^{9}=1866.6\) N/m, resulting in an error of less than one percent.  
We now derive the transfer function of the beam, where the input is a force applied perpendicular to the tip, and the output is the resulting deflection of the beam.  
The cantilever beam is modeled as a spring-mass-damper system with Rayleigh damping. The stiffness of a massless cantilever beam with modulus \(E\), moment of inertia \(I\), and length \(L\) is [28]:  
\[k=\frac{3 EI}{L^{3}}. \tag{46}\]  
Rayleigh damping is defined as follows [37]:  
\[c=\alpha m+\beta k \tag{47}\]  
where \(\alpha\) is the mass-proportional damping coefficient, and \(\beta\) is the stiffness-proportional damping coefficient.  
The equation of motion for the mass \(m\) at the tip is the following:  
\[m\ddot{x}(t)+c\dot{x}(t)+kx(t)=F(t). \tag{48}\]  
Substituting \(c=\alpha m+\beta k\),  
\[m\ddot{x}(t)+(\alpha m+\beta k)\dot{x}(t)+kx(t)=F(t). \tag{49}\]  
Taking the Laplace transform (assuming zero initial conditions),  
\[ms^{2}X(s)+(\alpha m+\beta k)sX(s)+kX(s)=F(s). \tag{50}\]  
Factoring \(X(s)\),  
\[X(s)\Big{[}ms^{2}+(\alpha m+\beta k)s+k\Big{]}=F(s). \tag{51}\]  
The transfer function is  
\[G(s)=\frac{X(s)}{F(s)}=\frac{1}{ms^{2}+(\alpha m+\beta k)s+k}. \tag{52}\]  
After this point, the result expressed in [20,21] applies.