Theory of Plates and Shells Lecture

Date: 09-03-2016

Example Problem:

Determine the deflection and stress in a very long and narrow rectangular plate if it is simply supported at edges

y = 0 and y = b

. a) The plate is carrying a uniform loading expression \begin{equation*} p(y) = P_o Sin\frac{\pi y}{b} \end{equation*} where the constant $P_o$ represents the intensity along the line passing thru $y = \frac{b}{2}$ parallel to x=axis. $\nu = \frac{1}{3}$ , $t = 10mm$, $E = 200GPa$, $P_o = 10kPa$ and $b = 0.4m$.

Solution:

Since the deflection is a function of 'y' only, the equation reduces to, \begin{equation*} \frac{P}{D} = \frac{\partial^4\omega}{\partial y^4} or \frac{\partial^4\omega}{\partial y^4} = \frac{P}{D} Beam Equation \end{equation*}

Substitute:
\begin{equation*}P = P(y) = P_o sin \frac{\pi}{b} y \end{equation*}
\begin{equation*} \frac{d^4\omega}{dy^4} = \frac{P_o sin\frac{\pi}{b}y}{D} \end{equation*}
\begin{equation*} \frac{d^3\omega}{dy^3} = \frac{P}{D}\frac{b}{\pi}[-\cos \frac{\pi}{b}y] + C_1 \end{equation*}
\begin{equation*} \frac{d^2\omega}{dy^2} = \frac{P}{D}\frac{b^2}{\pi^2}[-\sin \frac{\pi}{b}y] + C_1 y + C_2 \end{equation*}
\begin{equation*} \frac{d\omega}{dy} = \frac{P}{D}\frac{b^3}{\pi^3}[\cos \frac{\pi}{b}y] + C_1*\frac{y^2}{2} + C_2 y+ C3 \end{equation*}
$$ \omega = \frac{P}{D}\frac{b^4}{\pi^4}[\sin \frac{\pi}{b}y] + C_1 \frac{y^3}{3} + C_2 \frac{y^2}{2} + C_3 y + C_4 $$

Apply boundary conditions:
$ \omega = 0 $ at $y = 0$ and $y = b$

$$ \omega = \frac{P}{D} \frac{b^4}{\pi^4} sin \frac{\pi}{b}y $$

Moment Values:

$$ M_x = -d(\nu \frac{\partial^2\omega}{\partial^2}) \quad since \frac{\partial^22\omega}{\partial x^2} = 0 \quad because \frac{\partial \omega}{\partial x} = 0\quad\quad \text{Slope along x-axis}$$

$$ M_x = -D \nu [\frac{P}{D}\frac{b^2}{\pi^2}(-\sin \frac{\pi}{b}y)] $$ $$ M_x = \nu P_o \frac{b^2}{\pi^2} \sin \frac{\pi}{b}y $$ $$ M_x = \frac{1}{3 \pi^2} P_o b^2 $$
$$ \sigma_x = \frac{12 M_x z}{t^3}\quad = \quad \frac{12 M_x (\frac{t}{2})}{t^3} $$

$$ M_y = -D \frac{\partial^2\omega}{\partial y^2} = -D [\frac{P b^2}{D \pi^2}(- sin \frac{\pi}{b}y)] $$$$ M_y = \frac{P b^2}{\pi^2} sin \frac{\pi}{b}y \quad \text substitute \frac{b}{2} = y$$$$ M_y = \frac{P b^2}{\pi^2} sin \frac{\pi}{2} $$$$ \sigma_y = \frac{12 M_y z}{t^3} \quad z = \frac{t}{2} \text for max stress$$