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}

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s