## 4 Frequency Response Function (FRF)
### FRF of the Beam Bridge Model with Three DOF  
Consider a beam computational model of a bridge with three DOF (Figure 7). Assume that the right beam support (point \(b\)) moves in a vertical direction because of harmonic motion:  
\[v_{bk}(t)=v_{bs}\cdot\sin(\omega\cdot t)=v_{bs}\cdot\sin(2\pi\cdot f\cdot t). \tag{26}\]  
Point 1 at 1/4 of the span from the support a then performs a vertical harmonic motion:  
\[v_{1k}(t)=v_{1s}\cdot\sin(\omega\cdot t)=v_{1s}\cdot\sin(2\pi\cdot f\cdot t), \text{ where }v_{1s}=\frac{v_{bs}}{l}\cdot\frac{1}{4}=v_{bs}/4. \tag{27}\]  
Point 2 at 2/4 of the span from the support \(a\) then performs a vertical harmonic motion:  
\[v_{2k}(t)=v_{2s}\cdot\sin(\omega\cdot t)=v_{2s}\cdot\sin(2\pi\cdot f\cdot t), \text{ where }v_{2s}=\frac{v_{bs}}{l}\cdot\frac{2l}{4}=\frac{v_{bs}}{l}\cdot\frac{1}{2}=v_{bs }/2. \tag{28}\]  
Point 3 at 3/4 of the span from the support \(a\) then performs a vertical harmonic motion:  
Figure 6: FRF in complex form. Figure 7: Kinematic excitation at point \(b\), the computational model with 3 DOF. \[v_{3k}(t)=v_{3s}\cdot\sin(\omega\cdot t)=v_{3s}\cdot\sin(2\pi\cdot f\cdot t),\text{ where }v_{3s}=\frac{v_{bs}}{l}\cdot\frac{3l}{4}=3\cdot v_{bs}/4. \tag{29}\]  
If \(v_{bs}=1\), then \(v_{1s}=v_{bs}/4=1/4\),  
\[v_{2s}=2\cdot v_{bs}/4=2/4,\] \[v_{3s}=3\cdot v_{bs}/4=3/4.\]  
The equations of motion have the form:  
\[ \begin{split}& m_{1}\cdot\ddot{v}_{1}(t)+k_{11}\cdot(v_{1}(t)-v_{ 1k}(t))+k_{12}\cdot(v_{2}(t)-v_{2k}(t))+k_{13}\cdot(v_{3}(t)-v_{3k}(t))=0,\\ & m_{2}\cdot\ddot{v}_{2}(t)+k_{21}\cdot(v_{1}(t)-v_{1k}(t))+k_{22 }\cdot(v_{2}(t)-v_{2k}(t))+k_{23}\cdot(v_{3}(t)-v_{3k}(t))=0,\\ & m_{3}\cdot\ddot{v}_{3}(t)+k_{31}\cdot(v_{1}(t)-v_{1k}(t))+k_{32 }\cdot(v_{2}(t)-v_{2k}(t))+k_{33}\cdot(v_{3}(t)-v_{3k}(t))=0.\end{split}  \tag{30}\]  
After substituting for \(v_{1k}\)(\(t\)), \(v_{2k}\)(\(t\)), \(v_{3k}\)(\(t\)) and adjusting we obtain:  
\[ \begin{split}& m_{1}\cdot\ddot{v}_{1}(t)+k_{11}\cdot v_{1}(t)+k_{12 }\cdot v_{2}(t)+k_{13}\cdot v_{3}(t)=+k_{11}\cdot v_{1k}(t)+k_{12}\cdot v_{2k }(t)+k_{13}\cdot v_{3k}(t),\\ & m_{2}\cdot\ddot{v}_{2}(t)+k_{21}\cdot v_{1}(t)+k_{22}\cdot v_{ 2}(t)+k_{23}\cdot v_{3}(t)=+k_{21}\cdot v_{1k}(t)+k_{22}\cdot v_{2k}(t)+k_{23 }\cdot v_{3k}(t),\\ & m_{3}\cdot\ddot{v}_{3}(t)+k_{31}\cdot v_{1}(t)+k_{32}\cdot v_{ 2}(t)+k_{33}\cdot v_{3}(t)=+k_{31}\cdot v_{1k}(t)+k_{32}\cdot v_{2k}(t)+k_{33 }\cdot v_{3k}(t).\end{split}  \tag{31}\]  
Respectively:  
\[ \begin{split}& m_{1}\cdot\ddot{v}_{1}(t)+k_{11}\cdot v_{1}(t)+k_{12 }\cdot v_{2}(t)+k_{13}\cdot v_{3}(t)=+k_{11}\cdot\frac{1}{4}v_{lk}(t)+k_{12} \cdot\frac{2}{4}v_{lk}(t)+k_{13}\cdot\frac{3}{4}v_{bk}(t)\\ & m_{2}\cdot\ddot{v}_{2}(t)+k_{21}\cdot v_{1}(t)+k_{22}\cdot v_{ 2}(t)+k_{23}\cdot v_{3}(t)=+k_{21}\cdot\frac{1}{4}v_{lk}(t)+k_{22}\cdot\frac{ 2}{4}v_{lk}(t)+k_{23}\cdot\frac{2}{4}v_{bk}(t)\\ & m_{3}\cdot\ddot{v}_{3}(t)+k_{31}\cdot v_{1}(t)+k_{32}\cdot v_{ 2}(t)+k_{33}\cdot v_{3}(t)=+k_{31}\cdot\frac{1}{4}v_{bk}(t)+k_{32}\cdot\frac{ 2}{4}v_{bk}(t)+k_{33}\cdot\frac{3}{4}v_{bk}(t).\end{split}  \tag{32}\]  
Based on the Fourier transform, the equations of motion will be transformed from the time to the frequency domain, as follows:  
\[ \begin{split}&-q^{2}\cdot m_{1}\cdot V_{1}(q)+k_{11}\cdot V_{1}(q)+k_{ 12}\cdot V_{2}(q)+k_{13}\cdot V_{3}(q)=+k_{11}\cdot\frac{1}{4}V_{bk}(q)+k_{12} \cdot\frac{2}{4}V_{bk}(q)+k_{13}\cdot\frac{3}{4}V_{bk}(q)\\ &-q^{2}\cdot m_{2}\cdot V_{1}(q)+k_{21}\cdot V_{1}(q)+k_{22} \cdot V_{2}(q)+k_{23}\cdot V_{3}(q)=+k_{21}\cdot\frac{1}{4}V_{bk}(q)+k_{22} \cdot\frac{2}{4}V_{bk}(q)+k_{23}\cdot\frac{3}{4}V_{bk}(q)\\ &-q^{2}\cdot m_{3}\cdot V_{3}(q)+k_{31}\cdot V_{1}(q)+k_{32} \cdot V_{2}(q)+k_{33}\cdot V_{3}(q)=+k_{31}\cdot\frac{1}{4}V_{bk}(q)+k_{32} \cdot\frac{2}{4}V_{bk}(q)+k_{33}\cdot\frac{3}{4}V_{bk}(q).\end{split}  \tag{33}\]  
Let us define FRF as follows:  
\[