# Dynamic Response of Slope Inertia-Based Timoshenko Beam under a Moving Load
## 3 Results  
To solve the problem analytically, we apply a coupled finite integral transformation method. Firstly, for the transverse displacement \(y\), one may employ the Fourier sine transformation. For the bending rotation \(\varphi\), the Fourier cosine transformation is used. \[Y(j,t)=\int\limits_{0}^{L}y(x,t)\sin\frac{j\pi x}{L}dx,\hskip 14.226378 ptj=1,2,3,\ldots \tag{9}\]  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]  
\[\psi(j,t)=\int\limits_{0}^{L}\varphi(x,t)\cos\frac{j\pi x}{L}dx,\hskip 14.226378 pt j=1,2,3,\ldots \tag{11}\]  
\[\varphi(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}\psi(j,t)\cos\frac{j\pi x} {L} \tag{12}\]  
where \(Y(j,t)\) and \(\psi(j,t)\) are image functions of \(y(x,t)\) and \(\varphi(x,t)\), respectively. Multiply Equations (4) and (5) with \(\sin\frac{j\pi x}{L}\) and \(\cos\frac{j\pi x}{L}\), respectively, and integrate them with respect to \(x\) over a finite interval \([0,L]\). The following is obtained considering the boundary conditions stated in Equation (7):  
\[\rho A\bar{Y}(j,t)-kGA\left(-\frac{j^{2}\pi^{2}}{L^{2}}Y(j,t)+\frac{j\pi}{L} \psi(j,t)\right)+\rho I\frac{j^{2}\pi^{2}}{L^{2}}\bar{Y}(j,t)=P\sin\frac{j\pi vt }{L} \tag{13}\]  
\[-EI\frac{j^{2}\pi^{2}}{L^{2}}\psi(j,t)+kGA\left(\frac{j\pi}{L}Y(j,t)-\psi(j,t )\right)=0 \tag{14}\]  
Figure 1: Simply supported beam under a moving load \(P\) at speed \(v\). Beam length \(L\), cross-sectional area \(A\), moment of inertia \(I\), elastic modulus \(E\), shear modulus \(G\), shear correction coefficient \(K\) and mass density \(\rho\). Substituting Equation (14) into Equation (13), one obtains  
\[\left(\rho A+\rho Ij^{2}\pi^{2}/L^{2}\right)\ddot{Y}(j,t)+\frac{EIkGA\dot{J}^{4} \pi^{4}/L^{4}}{kGA+EIj^{2}\pi^{2}/L^{2}}Y(j,t)=P\sin\frac{j\pi vt}{L} \tag{15}\]  
which can be rewritten as  
\[\ddot{Y}(j,t)+\omega_{j}^{2}Y(j,t)=P_{j}\sin\overline{\omega}_{j}t \tag{16}\]  
where  
\[\omega_{j}^{2}=\frac{EIkGAj^{4}\pi^{4}/L^{4}}{(kGA+EIj^{2}\pi^{2}/L^{2})(\rho A +\rho Ij^{2}\pi^{2}/L^{2})} \tag{17}\]  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]0  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]1  
Equation (16) depicts a harmonic vibration without damping and can be solved according to [PERSON] [32]. Herein, the Laplace-Carson transformation method [2] is applied, which leads to  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]2  
Transforming Equation (16) in accordance with Equation (20), and in view of the initial condition as Equation (8), one obtains  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]3  
By rearrangement, Equation (21) can be written as  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]4  
According to the inverse Laplace-Carson transformation [2], the image function \(Y(j,t)\) can be obtained as  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]5  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]6  
where \(\beta_{j}\) is the frequency ratio, defined as  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]7  
Substituting Equations (23) and (24) into Equation (14), \(\psi(j,t)\) can be expressed as  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]8  
\[y(x,t)=\frac{2}{L}\!\sum\limits_{j=1}^{\infty}Y(j,t)\sin\frac{j\pi x}{L} \tag{10}\]9At last, substituting Equations (23), (24), (26) and (27) into Equations (10) and (12), one obtains the series solutions of the original functions \(y(x,t)\) and \(\varphi(x,t)\)  
\[\begin{array}{rl}y(x,t)&=\frac{2}{L}\sum\limits_{ \begin{array}{c}j=1\\ j\
eq\alpha\end{array}