# Vibration and Reliability Analysis of Non-Uniform Composite Beam under Random Load
## 2 Theoretical Model
### Iteration Process Considering Random Load  
The premise of the iterative method is that at \(t=0\) s, the initial condition of vibration is known, and there is no force applied to the beam. At this time, the beam is considered to be at rest, and the deflection and velocity at any position of the beam are,  
\[\left\{ \begin{array}{c}w(x,t)=0\\ \frac{dw(x,t)}{dt}=0\end{array} \right.. \tag{21}\]  
Because some variables, like deflection, corresponding generalized coordinate, excitation, in each iteration step are different in the process of iteration of the system, they are defined by iteration step subscripts. The subscript 1 represents these variables in the first time period/iteration step in period \(t_{1}\)-\(t_{2}\). The subscript 2 represents these variables in the second time period/iteration step in period \(t_{2}\)-\(t_{3}\) and so on. The subscript \(n\) represents these variables in the \(n\)-th time period/iteration step in period \(t_{n}\)-\(t_{n+1}\). According to the engineering experience, the step length of each iteration is determined by base motion excitation frequency, \(\Delta t\)=\(t_{n+1}\) - \(t_{n}\)=\(\frac{1}{10\times f(n)}\), \(1\leq n\leq\infty\), and \(n\) is the number of iteration steps, \(f(n)\) is the base motion excitation frequency during time period \(t_{n}\)-\(t_{n+1}\). The initial condition for time \(t_{1}\) (\(t_{1}=0\) s) is,  
\[\left\{ \begin{array}{c}w(x,t_{1})=0\\ \frac{dw(x,t)}{dt}\Big{|}_{t=t_{1}}=0\end{array} \right.. \tag{22}\]  
When the time is in \(t_{1}\)-\(t_{2}\), there is no free vibration and the response of the forced vibration is,  
\[ \begin{array}{c}w_{1}(x,t)=\sum_{i=1}^{\infty}W_{i}(x)q_{1_{i}}(t)=\sum_{i=1 }^{\infty}W_{i}(x)\frac{1}{\omega_{d_{i}}}\int_{0}^{t-t_{1}}F_{i_{i}}(\tau+t_{ 1})\cdot e^{-\xi_{i}\omega_{i}(t-t_{1}-\tau)}.\\ \sin\left(\omega_{d_{i}}\cdot(t-t_{1}-\tau)\right)d\tau\;(t_{1}\leq t\leq t_{ 2}).\end{array}  \tag{23}\]  
To clarify the iteration progress, we start the derivation from the second time step, \(t_{2}\)-\(t_{3}\), while the first step vibration solutions is only from the Duhamel integral assuming the structure is at rest before excitation as described in Equation (23). The free vibration response from the initial condition at \(t_{2}\) is,  
\[ \begin{array}{c}w_{free_{2}}(x,t)=\sum\limits_{i=1}^{\infty}W_{i}(x)\cdot q _{free_{2_{i}}}(t)\\ =\sum\limits_{i=1}^{\infty}W_{i}(x)\cdot e^{-\xi_{i}\omega_{i}t}(A_{2_{i}} \cos\omega_{d_{i}}t+B_{2_{i}}\sin\omega_{d_{i}}t)\;(t_{2}\leq t\leq t_{3}), \end{array}  \tag{24}\]  
\(A_{2_{i}}\) and \(B_{2_{i}}\) are determined by \(q_{1i}(t_{2})\) and \(\frac{dq_{1i}(t)}{dt}\) at \(t=t_{2}\). \[ \begin{array}{c}e^{-\xi_{i}\omega_{i}t_{2}}\big{(}A_{2_{i}}\cos\omega_{d_{i }}t_{2}+B_{2_{i}}\sin\omega_{d_{i}}t_{2}\big{)}=q_{1i}(t_{2}),\end{array}  \tag{25}\]  
\[ \begin{array}{c}\big{(}\omega_{d_{i}}\cos\omega_{d_{i}}t_{2}-\zeta_{i}\omega _{i}\sin\omega_{d_{i}}t_{2}\big{)}e^{-\xi_{i}\omega_{i}t_{2}}B_{2_{i}}\\ -\big{(}\omega_{d_{i}}\sin\omega_{d_{i}}t_{2}+\zeta_{i}\omega_{i}\cos\omega_{d _{i}}t_{2}\big{)}e^{-\xi_{i}\omega_{i}t_{2}}A_{2_{i}}=\left.\frac{dq_{1i}(t)}{ dt}\right|_{t=t_{2}}.\end{array}  \tag{26}\]  
\(A_{2_{i}}\) and \(B_{2_{i}}\) can be obtained by the above formula,  
\[ \begin{array}{c}B_{2_{i}}=\frac{e^{\xi_{i}\omega_{d_{i}}t_{2}}\cdot\left( \left.\left(\zeta_{i}\omega_{i}q_{1i}(t_{2})+\left.\frac{dq_{1i}(t)}{dt}\right| _{t=t_{2}}\right)\cos(\omega_{d_{i}}t_{2})+\omega_{d_{i}}q_{1i}(t_{2})\sin( \omega_{d_{i}}t_{2})\right)}{\omega_{d_{i}}},\end{array}  \tag{27}\]  
\[ \begin{array}{c}A_{2_{i}}=\frac{q_{1i}(t_{2})\cdot e^{\xi_{i}\omega_{i}t_{2} }-B_{2_{i}}\sin(\omega_{d_{i}}t_{2})}{\cos(\omega_{d_{i}}t_{2})}.\end{array}  \tag{28}\]During the period, \(t_{2}\)-\(t_{3}\), the total vibration response is,  
\[