Basic equations
The basic equations which shows the relationship between the deflection $w$ and resultant forces can be shown below.
Flexural rigidity of a plate
\begin{equation*}
D=\cfrac{E h^3}{12 (1-\nu^2)}
\end{equation*}
where, $E$ is Elastic modulus, $\nu$ is Poisson's ratio and $h$ is the plate thickness.
Moments
\begin{align*}
&M_x=-D \left(\cfrac{\partial^2 w}{\partial x^2} + \nu\cfrac{\partial^2 w}{\partial y^2}\right)\\
&M_y=-D \left(\cfrac{\partial^2 w}{\partial y^2} + \nu\cfrac{\partial^2 w}{\partial x^2}\right)\\
&M_{xy}=-M_{yx}=D (1-\mu) \cfrac{\partial^2 w}{\partial x \partial y}
\end{align*}
Shearing forces
\begin{align*}
&Q_x=-D \cfrac{\partial}{\partial x}\left(\cfrac{\partial^2 w}{\partial x^2} + \cfrac{\partial^2 w}{\partial y^2}\right)\\
&Q_y=-D \cfrac{\partial}{\partial y}\left(\cfrac{\partial^2 w}{\partial x^2} + \cfrac{\partial^2 w}{\partial y^2}\right)
\end{align*}
Reactions along the edges of a plate
\begin{align*}
&V_x=\left(Q_x - \cfrac{\partial M_{xy}}{\partial y}\right)_{x=0}\\
&V_y=\left(Q_y - \cfrac{\partial M_{xy}}{\partial y}\right)_{y=0}
\end{align*}
Reaction at the corner of a plate
\begin{equation*}
R=2 (M_{xy})_{x=a, y=b}
\end{equation*}
Solution for the simply supported rectangular plates (Navier solution)
Deflection
A rectangular plate with the length of $a$ in $x$ direction and the length of $b$ in $y$ direction is considered. In this method, distribution of the deflection $w(x,y)$ is expressed in the form of a double trigonometric series shown below.
\begin{equation*}
w(x,y)=\sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \sin\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b}
\end{equation*}
where, $a_{mn}$ is acoefficient including the load factor. $a_{mn}$ and $q_{mn}$ can be expressed below.
\begin{gather*}
a_{mn}=\cfrac{q_{mn}}{D \pi^4 \left\{\left(\cfrac{m}{a}\right)^2 + \left(\cfrac{n}{b}\right)^2\right\}^2} \\
q_{mn}=\cfrac{4}{a b}\int_0^a \int_0^b q(x,y) \sin\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b} dx dy
\end{gather*}
where, $q(x,y)$ is load distribution. If uniform load with intensity of $q_0$, $q_{mn}$ can be expressed below.
\begin{equation*}
q_{mn}=\cfrac{16 q_0}{m n \pi^2} \qquad (m, n = 1,3,5, \cdots)
\end{equation*}
Resultant forces in tge form of double trigorometric series
Derivatives of $w$ which are required to express resultant forces are shown below.
\begin{align*}
&w = \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \sin\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b} \\
&\cfrac{\partial w}{\partial x} = \pi \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \cfrac{m}{a} \cos\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
&\cfrac{\partial w}{\partial y} = \pi \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \cfrac{n}{b} \sin\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b} \\
&\cfrac{\partial^2 w}{\partial x^2} = - \pi^2 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \cfrac{m^2}{a^2} \sin\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
&\cfrac{\partial^2 w}{\partial y^2} = - \pi^2 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \cfrac{n^2}{b^2} \sin\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
&\cfrac{\partial^2 w}{\partial x \partial y} = \pi^2 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \cfrac{m n}{a b} \cos\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b} \\
&\cfrac{\partial}{\partial x}\left(\cfrac{\partial^2 w}{\partial x^2} + \cfrac{\partial^2 w}{\partial y^2}\right)
= - \pi^3 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m^2}{a^2}+\cfrac{n^2}{b^2}\right) \cfrac{m}{a} \cos\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
&\cfrac{\partial}{\partial y}\left(\cfrac{\partial^2 w}{\partial x^2} + \cfrac{\partial^2 w}{\partial y^2}\right)
= - \pi^3 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m^2}{a^2}+\cfrac{n^2}{b^2}\right) \cfrac{n}{b} \sin\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b}
\end{align*}
Using above, moments and shear forces can be obteined as shown below:
\begin{align*}
M_x = D \pi^2 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m^2}{a^2} + \nu \cfrac{n^2}{b^2}\right) \sin\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
M_y = D \pi^2 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\nu \cfrac{m^2}{a^2} + \cfrac{n^2}{b^2}\right) \sin\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
M_{xy} = D (1-\nu) \pi^2 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m n}{a b}\right) \cos\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b} \\
Q_x = D \pi^3 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m^2}{a^2} + \cfrac{n^2}{b^2}\right) \cfrac{m}{a} \cos\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b} \\
Q_y = D \pi^3 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m^2}{a^2} + \cfrac{n^2}{b^2}\right) \cfrac{n}{b} \sin\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b}
\end{align*}
And using the following relationships, reactions $V_x$ and $V_y$ can be obtained.
\begin{align*}
\cfrac{\partial M_{xy}}{\partial x} = - D (1-\nu ) \pi^3 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m n}{a b}\right) \cfrac{m}{a} \sin\cfrac{m \pi x}{a} \cos\cfrac{n \pi y}{b} \\
\cfrac{\partial M_{xy}}{\partial y} = - D (1-\nu ) \pi^3 \sum_{m=1}^\infty \sum_{n=1}^\infty a_{mn} \left(\cfrac{m n}{a b}\right) \cfrac{n}{b} \cos\cfrac{m \pi x}{a} \sin\cfrac{n \pi y}{b}
\end{align*}
Program
| Filename | Description |
|---|---|
| a_f90.txt | Shell script for execution |
| f90_navier.f90 | Coefficients by Navier method under uniform distributed load (1) |
| f90_navier_m.f90 | Coefficients by Navier method under uniform distributed load (2) |
Program (1) is for making below table, and program (2) is for getting only one case of model size by input of Elastic modulus, Poisson's ratio, plate size a, b and t.
Example of an analysis
In the following table, numerical factors α, β, γ, δ and n for uniformly loaded and simply supported rectangular plate are show.
| b/a | (w) | (Mx) | (My) | (Qx) | (Qy) | (Vx) | (Vy) | (R) |
|---|---|---|---|---|---|---|---|---|
| (b>=a) | α | β | β1 | γ | γ1 | δ | δ1 | n |
| 1.000 | 0.00406 | 0.0479 | 0.0479 | 0.337 | 0.337 | 0.420 | 0.420 | 0.065 |
| 1.100 | 0.00487 | 0.0555 | 0.0493 | 0.360 | 0.346 | 0.440 | 0.438 | 0.071 |
| 1.200 | 0.00565 | 0.0627 | 0.0501 | 0.379 | 0.353 | 0.456 | 0.453 | 0.076 |
| 1.300 | 0.00639 | 0.0694 | 0.0503 | 0.396 | 0.358 | 0.468 | 0.464 | 0.080 |
| 1.400 | 0.00708 | 0.0755 | 0.0502 | 0.411 | 0.361 | 0.478 | 0.473 | 0.083 |
| 1.500 | 0.00772 | 0.0812 | 0.0498 | 0.424 | 0.364 | 0.485 | 0.479 | 0.086 |
| 1.600 | 0.00831 | 0.0862 | 0.0493 | 0.435 | 0.366 | 0.491 | 0.485 | 0.088 |
| 1.700 | 0.00884 | 0.0908 | 0.0486 | 0.444 | 0.367 | 0.496 | 0.489 | 0.090 |
| 1.800 | 0.00932 | 0.0948 | 0.0479 | 0.452 | 0.368 | 0.499 | 0.492 | 0.091 |
| 1.900 | 0.00974 | 0.0985 | 0.0471 | 0.459 | 0.369 | 0.501 | 0.494 | 0.092 |
| 2.000 | 0.01013 | 0.1017 | 0.0464 | 0.465 | 0.370 | 0.503 | 0.496 | 0.093 |
| 3.000 | 0.01223 | 0.1189 | 0.0406 | 0.493 | 0.371 | 0.505 | 0.501 | 0.095 |
| 4.000 | 0.01282 | 0.1235 | 0.0384 | 0.498 | 0.371 | 0.502 | 0.501 | 0.095 |
| 5.000 | 0.01297 | 0.1246 | 0.0377 | 0.499 | 0.371 | 0.500 | 0.501 | 0.095 |
| 6.000 | 0.01301 | 0.1249 | 0.0376 | 0.500 | 0.371 | 0.500 | 0.501 | 0.095 |
| 7.000 | 0.01302 | 0.1250 | 0.0375 | 0.500 | 0.371 | 0.500 | 0.501 | 0.095 |
| 8.000 | 0.01302 | 0.1250 | 0.0375 | 0.500 | 0.371 | 0.500 | 0.501 | 0.095 |
| 9.000 | 0.01302 | 0.1250 | 0.0375 | 0.500 | 0.371 | 0.500 | 0.501 | 0.095 |
| 10.000 | 0.01302 | 0.1250 | 0.0375 | 0.500 | 0.371 | 0.500 | 0.501 | 0.095 |
| Poisson's ratio $\nu=0.3$ | ||||||||
The meanings of the factors in above table are shown below.
\begin{gather*}
w_{max}=\alpha \cdot \cfrac{q a^4}{D} \\
(M_x)_{max}=\beta \cdot q a^2 \qquad
(M_y)_{max}=\beta_1 \cdot q a^2 \\
(Q_x)_{max}=\gamma \cdot q a \qquad
(Q_y)_{max}=\gamma_1 \cdot q a \\
(V_x)_{max}=\delta \cdot q a \qquad
(V_y)_{max}=\delta_1 \cdot q a \qquad
R= n \cdot q a^2
\end{gather*}