Question

For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).
\begin{center} \includegraphics[width=10cm]{50.png} \end{center}

Hint:

For double integration method, ensure you consider the reactions and moment distribution correctly before applying the equations for deflection calculation.

Answer 1

Correct Answer: 3

Let \( V_A \) and \( V_B \) be the vertical reactions at \( A \) and \( B \) respectively. The moment equation gives: \[ \Sigma M_A = 0 \quad \Rightarrow \quad (- V_B \times 2L) + M = 0 \] Thus, \[ V_B = \frac{M_0}{2L}, \quad V_A = -\frac{M_0}{2L}. \] By the double integration method, consider a section at \( x \), where the bending moment is: \[ M_0 - \frac{M_0}{2L}x + \left[ -M_x \right] = 0 \] At the mid-section \( x = \frac{L}{2} \), we have: \[ M_0 - \frac{M_0}{2L} \times \frac{L}{2} = (2EI) \frac{d^2y}{dx^2} \] Solving this equation gives: \[ \frac{d^2y}{dx^2} = \frac{3M_0}{8EI} \] Therefore, the value of \( \alpha = 3 \).