## Rigid plate on four supports

Consider a rigid rectangular planar plate of width $W$ and height $H$ lying horizontal on four supports at its corners in $S_1=\left[0 \hspace{1 mm} 0 \right], S_2=\left[W \hspace{1 mm} 0 \right], S_3=\left[W \hspace{1 mm} H \right], S_4=\left[0 \hspace{1 mm} H \right]$ and with a concentrated vertical load $F$ at $\left[x \hspace{1 mm} y \right]$, with $0 \leq x \leq W$, $0 \leq y \leq H$.

How can I determine the reactions $R_1, R_2, R_3, R_4$?

This is a hyperstatic problem (the object has redundant constraints) and as such the equilibrium equations make an underdetermined system.

Force equilibrium along Z: $R_1 + R_2 + R_3 + R_4+F=0$

Torque equilibrium around X at y=0: $H R_3 + H R_4 + y F = 0$

Torque equilibrium around Y at x=0: $W R_2 + W R_3 + x F = 0$

Any other equilibrium equation I could write wouldn’t increase the rank of the system.

Such problems are usually solved by considering the deformation of the object and requiring a null displacement at the constraints.

I said that the plate is rigid, so it will stay planar even when the force $F$ is applied. How can I use this information to add a fourth equation?

Suppose the supports are elastic with a large constant $k$, so we do have small displacements $z_1 = - R_1/k, z_2 = - R_2/k, z_3=-R_3/k, z_4=-R_4/k$. If the corners lie on a plane, and neglecting for simplicity the slant of the plane, which is small if the displacements are small, it must be

$\left| \begin{array}{cccc} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{array} \right| = 0$

that is

$\left| \begin{array}{cccc} 0 & 0 & - R_1/k & 1 \\ W & 0 & - R_2/k & 1 \\ W & H & - R_3/k & 1 \\ 0 & H & - R_4/k & 1 \end{array} \right| = 0$

that is

$R_1 - R_2 + R_3 - R_4=0$

This is my fourth equation, whence the solution

$R_1 = -F \dfrac{3 H W - 2 H x - 2 W y}{4 H W}$

$R_2 = -F \dfrac{H W + 2 H x - 2 W y}{4 H W}$

$R_3 = -F \dfrac{- H W + 2 H x + 2 W y}{4 H W}$

$R_4 = -F \dfrac{H W - 2 H x + 2 W y}{4 H W}$

The reverse problem is easier. If I know $R_1, R_2, R_3, R_4$, I can directly obtain $F$ from the force equilibrium, $x$ from the equilibrium around Y and $y$ from the equilibrium around X:

$F = -R_1-R_2-R_3-R_4$

$x = -W \dfrac{R_2+R_3}{F}$

$y=-H \dfrac{R_3+R_4}{F}$

Advertisements
This entry was posted in Uncategorized and tagged , , . Bookmark the permalink.