A simply supported uniform beam of length \(L\), bending stiffness \(E I\) and mass per unit length \(m\) is subjected to the uniform distributed load \(p(x, t)=q_{0} \sin \Omega t\). Determine the bending moment at the center of the span if the behavior of the beam is predicted using Rayleigh Beam Theory with \(r_{G} / L=0.1\).

Rayleigh beam theory provides an approximate expression for the deflection, \(w(x,t)\), of a beam as: $$ w(x,t) = \sum_{n=1}^{\infty} \frac{1}{n \pi} \left[ \frac{1-\cos(n \pi r_{G})}{1-\cos(n \pi)} \right] \sin \frac{n \pi x}{L} \int_{0}^{t} \int_{0}^{L} p(x',t')\sin\frac{n\pi x'}{L}\,dx'\,dt' $$

Now, we need to substitute the given values of \(p(x, t)=q_{0} \sin \Omega t\) and \(r_{G}/L = 0.1\) into the Rayleigh Beam Theory equation for deflection, and solve for \(w(x,t)\). $$ w(x,t) = \sum_{n=1}^{\infty} \frac{1}{n \pi} \left[ \frac{1-\cos(n \pi (0.1))}{1-\cos(n \pi)} \right] \sin \frac{n \pi x}{L} \int_{0}^{t} \int_{0}^{L} (q_{0}\sin\Omega t')\sin\frac{n\pi x'}{L}\,dx'\,dt' $$ Now, let's integrate the expression with respect to \(x'\) and \(t'\): $$ w(x,t) = \sum_{n=1}^{\infty} \frac{1}{n \pi} \left[ \frac{1-\cos(n \pi (0.1))}{1-\cos(n \pi)} \right] \sin \frac{n \pi x}{L} (q_{0}\int_{0}^{t} \sin\Omega t'\,dt'\int_{0}^{L} \sin\frac{n\pi x'}{L}\,dx') $$ Compute the \(x'\) integral: $$ \int_{0}^{L} \sin\frac{n\pi x'}{L}\,dx' = \frac{L}{n\pi}(\cos(n\pi)-1) $$ Compute the \(t'\) integral: $$ \int_{0}^{t} \sin\Omega t'\,dt' = -\frac{1}{\Omega}+ \frac{1}{\Omega}\cos(\Omega t) $$ Put these two integrals back into the expression for \(w(x,t)\): $$ w(x,t) = -q_{0} \sum_{n=1}^{\infty} \frac{1}{(n \pi)^{2}} \left[\frac{1-\cos\left(n \pi (0.1)\right)}{1-\cos\left(n \pi\right)}\right] \sin \frac{n \pi x}{L}\left[\frac{L}{n\pi}(\cos(n\pi)-1)\right]\left[-\frac{1}{\Omega}+ \frac{1}{\Omega}\cos(\Omega t)\right] $$

Now that we have the expression for deflection, \(w(x,t)\). We can determine the bending moment \(M(x,t)\) at the center of the span by using the following relationship between deflection and bending moment: $$ M(x, t) = -E I \frac{d^2 w(x,t)}{dx^2} $$ To calculate the second derivative of the deflection with respect to x, we need to differentiate the deflection expression \(w(x,t)\) twice: $$ \frac{d^2 w(x,t)}{dx^2} = -q_{0} \sum_{n=1}^{\infty} \frac{n^{3}\pi}{L^{2}} \left[\frac{1-\cos\left(n \pi (0.1)\right)}{1-\cos\left(n \pi\right)}\right] \cos \frac{n \pi x}{L}\left[\frac{L}{n\pi}(\cos(n\pi)-1)\right]\left[-\frac{1}{\Omega}+ \frac{1}{\Omega}\cos(\Omega t)\right] $$ Now substitute the expression for the second derivative into the expression for bending moment and use x = L/2 for the center of the span: $$ M\left(\frac{L}{2}, t\right) = E I \left. \frac{d^2 w(x,t)}{dx^2}\right|_{x=\frac{L}{2}} =q_{0} E I \sum_{n=1}^{\infty} \frac{n^{3}\pi}{L^{2}} \left[\frac{1-\cos\left(n \pi (0.1)\right)}{1-\cos\left(n \pi\right)}\right] \cos \frac{n \pi}{2}\left[\frac{L}{n\pi}(\cos(n\pi)-1)\right]\left[-\frac{1}{\Omega}+ \frac{1}{\Omega}\cos(\Omega t)\right] $$ This expression represents the bending moment at the center of the span as a function of time.