aproksymacje Fouriera obciążenia i rozwiązania (ugięcia) $$ q(x)=\sum_{k=1}^{K} q_{k} \sin \left(\frac{k \pi x}{L}\right), \quad y(x)=\sum_{k=1}^{K} y_{k} \sin \left(\frac{k \pi x}{L}\right) $$ wyznaczenie wspólczynników obciążenia $$ q_{k}=\frac{2}{L} \int_{\frac{L}{4}}^{\frac{3}{4} L} q_{0} \sin \left(\frac{k \pi x}{L}\right) d x=-\left.\frac{2 q_{0}}{L} \frac{L}{k \pi} \cos \left(\frac{k \pi x}{L}\right)\right|_{\frac{L}{4}} ^{\frac{3}{4} L}=-\frac{2 q_{0}}{k \pi}\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right) $$ wyznaczenie wspólczynników ugięcia - podstawienie aproksymacji Fouriera dla $q(x)$ i $y(x)$ do równania ugięcia belki $$ \begin{aligned} &E I \cdot \frac{d^{4} y}{d x^{4}}+\kappa \cdot y(x)=q(x) \\ &E I \frac{k^{4} \pi^{4}}{L^{4}} \cdot \sum_{k=1}^{K} y_{k} \sin \left(\frac{k \pi x}{L}\right)+\kappa \cdot \sum_{k=1}^{K} y_{k} \sin \left(\frac{k \pi x}{L}\right)=\sum_{k=1}^{K} q_{k} \sin \left(\frac{k \pi x}{L}\right) \end{aligned} $$ czyli $$ E I \frac{k^{4} \pi^{4}}{L^{4}} \cdot y_{k}+\kappa \cdot y_{k}=q_{k}, \quad k=1,2, \ldots, K $$ stąd $$ \begin{aligned} &y_{k}=\frac{q_{k}}{E I \frac{k^{4} \pi^{4}}{L^{4}}+\kappa}=\frac{-\frac{2 q_{0}}{k \pi}\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right)}{E I \frac{k^{4} \pi^{4}}{L^{4}}+\kappa}=\frac{-\frac{2 q_{0}}{k \pi}\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right)}{\frac{E I k^{4} \pi^{4}+\kappa L^{4}}{L^{4}}}= \\ &=\frac{-2 q_{0} L^{4}\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right)}{k \pi\left(E I k^{4} \pi^{4}+\kappa L^{4}\right)}, k=1,2, \ldots, K \end{aligned} $$ wyznaczenie funkcji ciaglych $$ \begin{aligned} &y(x)=\sum_{k=1}^{K} y_{k} \sin \left(\frac{k \pi x}{L}\right)=-\frac{2 q_{0} L^{4}}{\pi} \sum_{k=1}^{K} \frac{\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right)}{k\left(E I k^{4} \pi^{4}+\kappa L^{4}\right)} \sin \left(\frac{k \pi x}{L}\right) \\ &M(x)=E I \frac{\pi^{2}}{L^{2}} \sum_{k=1}^{K} k^{2} y_{k} \sin \left(\frac{k \pi x}{L}\right)=-2 E I \pi q_{0} L^{2} \sum_{k=1}^{K} \frac{k\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right)}{E I k^{4} \pi^{4}+\kappa L^{4}} \sin \left(\frac{k \pi x}{L}\right) \\ &T(x)=-E I \frac{\pi^{3}}{L^{3}} \sum_{k=1}^{K} k^{3} y_{k} \cos \left(\frac{k \pi x}{L}\right)=2 E I \pi^{2} q_{0} L \sum_{k=1}^{K} \frac{k^{2}\left(\cos \left(\frac{3 k \pi}{4}\right)-\cos \left(\frac{k \pi}{4}\right)\right)}{E I k^{4} \pi^{4}+\kappa L^{4}} \cos \left(\frac{k \pi x}{L}\right) \end{aligned} $$ wyznaczenie wartości da $x=L / 2 i k=1$. $$ \begin{aligned} &y\left(x=\frac{L}{2}\right)=\frac{2 \sqrt{2} q_{0} L^{4}}{\pi\left(E I \pi^{4}+\kappa L^{4}\right)} \\ &M\left(x=\frac{L}{2}\right)=2 \sqrt{2} \frac{E I \pi q_{0} L^{2}}{E I \pi^{4}+\kappa L^{4}} \\ &T\left(x=\frac{L}{2}\right)=0 \end{aligned} $$