Solution
Fourier approximation of the load and the response (deflection)
$$ q(x)=\sum_{k=1}^{K} q_{k} \sin\!\left(\frac{k \pi x}{L}\right),\qquad y(x)=\sum_{k=1}^{K} y_{k} \sin\!\left(\frac{k \pi x}{L}\right) $$Load coefficients:
$$ q_{k}=\frac{2}{L} \int_{L/4}^{3L/4} q_{0} \sin\!\left(\frac{k \pi x}{L}\right)\,dx = -\frac{2 q_{0}}{k \pi}\Big[\cos\!\left(\frac{3 k \pi}{4}\right)-\cos\!\left(\frac{k \pi}{4}\right)\Big] $$Deflection coefficients: substitute the Fourier expansions of \(q(x)\) and \(y(x)\) into the governing beam equation.
$$ EI\left(\frac{k^{4} \pi^{4}}{L^{4}}\right) y_{k} + \kappa\, y_{k} = q_{k},\qquad k=1,2,\ldots,K $$Thus:
$$ y_{k} = \frac{q_{k}}{EI\left(\frac{k^{4} \pi^{4}}{L^{4}}\right) + \kappa} $$Continuous function:
$$ y(x) = \sum_{k=1}^{K} y_{k} \sin\!\left(\frac{k \pi x}{L}\right) $$Evaluate at \(x = L/2\) and \(k=1\).