# Vibration and Reliability Analysis of Non-Uniform Composite Beam under Random Load
## 2 Theoretical Model
### Vibration Model of Non-Uniform Composite Beam under Random Load  
After the natures of the beam structure are solved by free vibration analysis and ADM, the forced vibration response can be solved using the modal analysis and Duhamel integral. The forced vibration governing equation of non-uniform composite beam considering damping under action of random load \(F(x,t)\) is given below,  
\[ \begin{split}\rho(x)A(x)\frac{\partial^{2}w(x,t)}{\partial t^{2}} +\frac{\partial^{2}}{\partial x^{2}}\Big{[}E(x)I(x)\frac{\partial^{2}w(x,t)}{ \partial x^{2}}+C(x)I(x)\frac{\partial^{3}w(x,t)}{\partial x^{2}\partial t} \Big{]}=F(x,t)=\\ -\rho(x)A(x)\partial^{2}Y\sin(\phi t),\end{split}  \tag{14}\]where \(C(x)\) is the strain rate damping coefficient. \(-\rho(x)A(x)\varphi^{2}Y\sin(\varphi t)\) is the distribution force generated by base motion. The displacement and acceleration of base motion are \(y=Y\sin(\varphi t)\) and \(a=-\varphi^{2}Y\sin(\varphi t)\), and \(A_{a}=\varphi^{2}Y\) is the amplitude of acceleration. \(Y\) and \(\varphi\) are amplitude and angular frequency of base motion. Substituting Equation (6) into Equation (14), the following equation can be obtained,  
\[ \begin{split}\rho(x)A(x)\sum\limits_{i=1}^{\infty}W_{i}(x)\frac {d^{2}q_{i}(t)}{dt^{2}}\\ +\frac{d^{2}}{dx^{2}}\Big{[}E(x)I(x)\sum\limits_{i=1}^{\infty}q_ {i}(t)\frac{d^{2}W_{i}(x)}{dx^{2}}+C(x)I(x)\sum\limits_{i=1}^{\infty}\frac{dq_ {i}(t)}{dt}\frac{d^{2}W_{i}(x)}{dx^{2}}\Big{]}\\ =-\rho(x)A(x)\varphi^{2}Y\sin(\varphi t).\end{split}  \tag{15}\]  
Based on the understanding from the free vibration governing equation, for the \(i\)-th mode of free vibration, we have,  
\[\frac{d^{2}}{dx^{2}}\Big{[}E(x)I(x)\frac{d^{2}W_{i}(x)}{dx^{2}}\Big{]}=\rho(x )A(x)\omega_{i}^{2}W_{i}(x). \tag{16}\]  
With the understanding from the relationship given in the above equation, both sides of Equation (15) multiplied by \(W_{j}(x)\) and integrated from 0 to \(L\) in space domain leads to,  
\[ \begin{split}\int_{0}^{L}\rho(x)A(x)\sum\limits_{i=1}^{\infty}W_ {i}(x)\frac{d^{2}q_{i}(t)}{dt^{2}}W_{j}(x)dx\\ +\int_{0}^{L}\frac{d^{2}}{dx^{2}}\bigg{[}E(x)I(x)\sum\limits_{i=1} ^{\infty}q_{i}(t)\frac{d^{2}W_{i}(x)}{dx^{2}}+C(x)I(x)\sum\limits_{i=1}^{ \infty}\frac{dq_{i}(t)}{dt}\frac{d^{2}W_{i}(x)}{dx^{2}}\bigg{]}W_{j}(x)dx\\ =-Y\sin(\varphi t)\int_{0}^{L}\rho(x)\varphi^{2}A(x)W_{j}(x)dx. \end{split}  \tag{17}\]  
Since in general, the damping function does not possess the orthogonality property, it is assumed that the structural damping is in the form of \(C(x)=\alpha E(x)\) where \(\alpha\) is a constant. According to the orthogonality of normal vibration modes, Equation (18) for the \(i\)-th mode can be obtained, while \(\alpha\omega_{i}^{2}=2\xi_{i}\omega_{i}\) is defined,  
\[\frac{d^{2}q_{i}(t)}{dt^{2}}+2\xi_{i}\omega_{i}\frac{dq_{i}(t)}{dt}+\omega_{i }^{2}q_{i}(t)=F_{i}(t)=\frac{-\varphi^{2}Y\sin(\varphi t)\int_{0}^{L}\rho(x)A( x)W_{i}(x)dx}{\int_{0}^{L}\rho(x)A(x)W_{i}^{2}(x)dx}, \tag{18}\]  
where \(\xi_{i}\) is the modal damping ratio of the corresponding \(i\)-th order natural mode. Equation (18) can be solved by the Duhamel integral, and the final time domain solution for \(i\)-th mode is,  
\[ \begin{split}& q_{i}(t)=\frac{1}{\omega_{d_{i}}}\int_{0}^{t}F_{i}( \tau)\cdot e^{-\xi_{i}\omega_{i}(t-\tau)}\cdot\sin(\omega_{d_{i}}(t-\tau))d \tau\\ =-\bigg{(}\Big{(}\int_{0}^{L}\rho(x)A(x)W_{i}(x)dx\Big{)}\big{(} -(-2\omega_{d_{i}}\xi_{i}\omega_{i}\cos(\omega_{d_{i}}t)+\sin(\omega_{d_{i}}t) \big{(}-\xi_{i}^{2}\omega_{i}^{2}-\varphi^{2} \tag{19}\] \[+\omega_{d_{i}}^{2})\big{)}\varphi e^{-\xi_{i}\omega_{i}t}+\big{(} (\xi_{i}^{2}\omega_{i}^{2}-\varphi^{2}+\omega_{d_{i}}^{2})\sin(\varphi t)-2 \xi_{i}\varphi\omega_{i}\cos(\varphi t)\big{)}\omega_{d_{i}}\big{)}\varphi^{2}Y \bigg{)}\\ &/\big{(}((\omega_{d_{i}}^{5}+\big{(}2\xi_{i}^{2}\omega_{i}^{2}-2 \varphi^{2}\big{)}\omega_{d_{i}}^{3}+\big{(}\xi_{i}^{2}\omega_{i}^{2}+\varphi ^{2}\big{)}^{2}\big{(}\xi_{i}^{2}\omega_{i}^{2}+\varphi^{2}\big{)}^{2}\omega_{ d_{i}})\\ &\Big{(}\int_{0}^{L}\rho(x)A(x)W_{i}^{2}(x)dx\Big{)}\Big{)},\end{split}