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 is represented by the following equation for the bending moment, \(M(x, t)\), along the length of the beam: \(M(x, t) = - EI \frac{d^2 w(x, t)}{d x^2}\) where \(w(x, t)\) is the deflection of the beam at position \(x\) and time \(t\), and \(d^2 w(x, t)/d x^2\) is the second derivative of the deflection with respect to the position.

Since the beam is simply supported on its ends, the boundary conditions are as follows: 1. \(w(0, t) = w(L, t) = 0\): the deflection at both ends of the beam is zero. 2. \(M(0, t) = M(L, t) = 0\): the bending moment at both ends of the beam is zero.

The distributed load on the beam varies sinusoidally with time, so we first write that down: \(p(x, t) = q_0 \sin \Omega t\) Taking this load into account, the equation for the deflection of the beam becomes: \(EI \frac{d^4 w(x, t)}{d x^4} = m \frac{d^2 w(x, t)}{d t^2} + p(x, t)\)

We now need to find the general solution for the deflection \(w(x, t)\), which satisfies both the boundary conditions and the differential equation. Using standard techniques for solving partial differential equations, we obtain the following expression for the deflection of the beam: \(w(x, t) = A \cos (\omega x) \sin (\Omega t)\) where \(\omega\) is a constant related to the frequency of oscillation and \(A\) is the amplitude of the deflection.

Now we need to find the second derivative of the deflection with respect to the position, \(d^2 w(x, t)/d x^2\), in order to determine the bending moment at the center of the span: \(\frac{d^2 w(x, t)}{d x^2} = -A \omega^2 \cos (\omega x) \sin (\Omega t)\)

We will now substitute the expression for the second derivative of the deflection into the equation for the bending moment, and then evaluate the bending moment at the center of the span, \(x = L/2\): \(M(L/2, t) = EI \cdot \left[-A \omega^2 \cos \left(\omega \frac{L}{2}\right) \sin (\Omega t)\right]\) Since we want the bending moment at the center, we plug in \(x = L/2\): \(M(L/2, t) = - EI A \omega^2 \cos \left (\frac{\omega L}{2} \right) \sin \Omega t\) Still, we don't know the exact value for \(A\omega^2\). To find that, we can use the given \(r_G/L=0.1\) and the fact that \(r_G=\frac{A\omega^2}{q_0}\). Therefore, \(A\omega^2=0.1q_0\). Plugging that back in to the expression for the bending moment at center: \(M(L/2, t) = -0.1 EIq_0 \cos \left (\frac{\omega L}{2} \right) \sin \Omega t\) This is the expression for the bending moment at the center of the span.