# Dynamic Response of Slope Inertia-Based Timoshenko Beam under a Moving Load
## 4 Finite Element Solution
### Numerical Results  
The case of a simply supported beam reported in [17; 34; 35] is considered. The beam is discretized into an even number of elements with equal length and using 100 equal incremental time steps [38]. The numerical results of the normalized mid-span displacement and moment of the beam with \(b=0.03\) and \(c=0.11\) discretized into different numbers of elements are plotted in Figure 4. It can be seen that when the number of elements is 20, the numerical results of displacement and bending moment are in good agreement with the analytical results. Therefore, in the following finite element analysis, the beam is discretized into an even number of elements and at least 20. Figure 5 shows the numerical results of the normalized mid-span deflections and bending moments for various moving speeds of the load. As stated earlier, the critical speed is used as a reference [7; 17; 34; 35; 39; 40; 41]. Note that for the SIBT model, the fundamental frequency \(\omega_{1}\) is defined by Equation (17), which is different from \(\omega_{1}^{*}=\left(\pi/L\right)^{2}\sqrt{EI/\rho A}\). With \(b=0.03\), one can easily obtain the critical speed parameter \(c^{*}=0.998\). This means that the velocity of the moving load is just equal to \(\omega_{1}L/\pi\) and that the ratio between the moving time of the load on the beam and the fundamental period \(\tau/T_{1}\) is 0.5. Five moving speeds are considered here, namely, \(c=0.125c^{*}\), \(0.25c^{*}\), \(0.5c^{*}\), \(c^{*}\) and \(1.5c^{*}\). For \(c<c^{*}\), as the c increases, the beam displacement response deviates further from the static one. The maximum normalized mid-span displacements are 1.139, 1.269 and 1.710 for \(c=0.125c^{*}\), \(0.25c^{*}\) and \(0.5c^{*}\) respectively. For \(c=c^{*}\), the beam displacement response reaches the maximum when the moving load leaves the beam (\(t/\tau=1\)), and the corresponding maximum displacement amplification factor is 1.554. For \(c=1.5c^{*}\), the mid-span displacement increases monotonously with the marching time, and the maximum displacement amplification factor is 1.008. In this case, the maximum response occurs after the moving load has passed the beam, similar to an impact problem [7]. This phenomenon is related to the speed of the waves in the beam. As for the bending moment, the maximum normalized mid-span moment is less than the maximum normalized mid-span displacement. However, the bending moment histories are more irregular. Figure 4: Numerical solutions with different element numbers of the beam with \(b=0.03\) and \(c=0.11\). To further visualize the dynamic character of the beam under the moving load with different speeds, the dynamic magnification factors are carefully presented for the whole range \(0.01\leq c\leq 1.0\) and different cross-sections \(b=0.03\), \(0.15\) and \(0.30\), in Figure 6. The dynamic magnification factors of mid-span displacement and mid-span moment are defined as \(D_{1}=\max[y(L/2,t)/y_{st}(L/2)]\) and \(D_{2}=\max[M(L/2,t)/M_{st}(L/2)]\), respectively. We can clearly see that with the increase in the moving speed of the load, the displacement amplification factor changes nonlinearly, which is associated with the oscillations discussed in Figure 5. In addition, with the increase in the cross-section size of the beam, the displacement amplification factor has a tendency of constant amplification. This phenomenon can be attributed to the increasing influence of the shear effect of the SIBT beam. Compared to the displacement magnification factors \(D_{1}\), we can note that the moment factors \(D_{2}\) are more irregular; \(D_{2}\) is smaller than \(D_{1}\) for all \(c\); and \(D_{2}\) is even less than unity for certain \(c\) values. Taking \(b=0.03\) for an example, the main increases in the factors \(D_{1}\) and \(D_{2}\) occur only in the intervals \(0.2\leq c\leq 0.62\) and \(0.2\leq c\leq 0.38\), respectively. The max (\(D_{1}\)) = 1.738 for \(c=0.62\) and max (\(D_{2}\)) = 1.437 for \(c=0.38\).