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).
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For double integration method, ensure you consider the reactions and moment distribution correctly before applying the equations for deflection calculation.
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 \( \alpha = 3 \).
