Biharmonic equation, which is a fourth order PDE is used for deflection of plates
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The plate's deflection \( w(x,y) \) depends on the applied load and the plate's material and geometric properties. Show more…
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The deflection of the plate can be noticed by going to a co- rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance $\xi$ from the axis is $a=\xi \beta$ and the moment of the forces exerted by the section between $x$ and $l$ is $$ N=\rho l h \beta \int_{x}^{l} \xi^{2} d \xi=\frac{1}{3} \rho \ln \beta\left(l^{3}-x^{3}\right) $$ From the fundamental equation $$ E I \frac{d^{2} y}{d x^{2}}=\frac{1}{3} \rho l h \beta\left(l^{3}-x^{3}\right) $$ The moment of inertia $I=\int_{-h / 2}^{+h / 2} z^{2} l d z=\frac{l h^{3}}{12 .}$ Note that the neutral surface (ie. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and $z$ is perpendicular to it. $$ \begin{gathered} \frac{d^{2} y}{d x^{2}}=\frac{4 \rho \beta}{E h^{2}}\left(l^{3}-x^{3}\right) . \text { Integrating } \\ \frac{d y}{d x}=\frac{4 \rho \beta}{E h^{2}}\left(l^{3} x-\frac{x^{4}}{4}\right)+c_{1} \end{gathered} $$ Since $\frac{d y}{d x}=0$, for $x=0, c_{1}=0 .$ Integrating again, $$ \begin{gathered} y=\frac{4 \rho \beta}{E h^{2}}\left(\frac{l^{3} x^{2}}{2}-\frac{x^{5}}{20}\right)+c_{2} \\ c_{2}=0 \text { because } y=0 \text { for } x=0 \end{gathered} $$ Thus $\lambda=y(x=l)=\frac{9 \rho \beta l^{5}}{5 E h^{2}}$
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