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 Euler-Bernoulli Theory.

For the given simply supported beam under a sinusoidal distributed load \(p(x, t) = q_{0} \sin \Omega t\), the Euler-Bernoulli equation can be written as: $$\frac{\partial^2}{\partial x^2}\left(EI\frac{\partial^2 w}{\partial x^2}\right) = -m \frac{\partial^2 w}{\partial t^2} - q_0\sin\Omega t$$ where \(w(x,t)\) represents the deflection of the beam along its length \(x\) and time \(t\), \(E\) denotes the Young's modulus, \(I\) is the second moment of area of the cross-section, and \(m\) is the mass per unit length.

For a simply supported beam, the deflection is zero at the supports (left end and right end), and the slope is zero at the midpoint. To represent these boundary conditions, we need to set the value of the deflection equation w(x, t) and its first and second derivatives: 1. At the left support (x = 0): $$w(0,t) = 0$$ 2. At the right support (x = L): $$w(L,t) = 0$$ 3. At the midpoint (x = L/2), the slope is zero: $$\frac{\partial w}{\partial x}(L/2,t) = 0$$

Now, we will solve the Euler-Bernoulli equation with the given boundary conditions to find the deflection equation \(w(x,t)\). The general solution for this equation will be a sum of steady and transient deflections, consisting of a time-dependent term \(T(t)\) and a space-dependent term \(X(x)\): $$w(x, t) = X(x)T(t)$$ First, we need to find the \(X(x)\) part. We rewrite the Euler-Bernoulli equation to focus on \(X(x)\): $$X''''(x)=\frac{q_0}{EI}\sin\Omega t$$ After integrating the above equation, we get: $$X(x) =A_0 + A_1x + A_2x^2 + A_3x^3$$ Now applying the boundary conditions given in Step 2 to the equation obtained above: 1. At the left support (x = 0): $$A_0 = 0$$ 2. At the right support (x = L): $$A_1L + A_2L^2 + A_3L^3 = 0$$ 3. At the midpoint (x = L/2), the slope is zero: $$A_1 + A_2L + 4A_3L^2/2 = 0$$ Solving the above set of equations, we can find the values of \(A_1, A_2, A_3\). After finding the values of \(A_1, A_2, A_3\), we can determine the spatial part of the deflection equation, \(X(x)\).

To calculate the bending moment at the center of the span, we need to use the relation between the bending moment and the curvature of the beam. The curvature is given by the second derivative of the deflection equation, \(\frac{\partial^2 w}{\partial x^2}\). The bending moment \(M(x,t)\) can be calculated using the following equation: $$M(x,t)=-EI\frac{\partial^2 w}{\partial x^2}(x,t)$$ Now, we substitute the deflection equation, \(w(x,t)=X(x)T(t)\), obtained in Step 3: $$M(x,t) = -EIA_2 - 6EIA_3x$$ Finally, we calculate the bending moment at the center of the span (x = L/2): $$M(L/2,t) = -EIA_2 - 3EIA_3L$$