# Elasticity of a drilled plate

If I take a large thin plate and I uniformly drill a very large number of holes (diameter "d") very close together (minimum spacing "s"), what will be the stiffness of this new plate? When I say "stiffness" I mean the derivative of average macroscopic displacement with respect to stress. With no holes it would be Young's modulus.

Here is a way to visualize how the drills are to positioned, i.e. close packed circles with space in between.

This is the question I want to answer. Of course, the stiffness will depend on which direction the stress is applied - for example, if I am pulling along the "grain" it will be stiffer than if I turn the plate 45 degrees and pull.

I thought about doing some sort of analytic model (found a few papers but they are hard to understand). Or a finite element model. However, I am interested in how the parameters "d" and "s" (see above) change the stiffness for each type of circle packing pattern. I am stuck now, what would you recommend as the best way to approach this problem?

Also, if this problem has already been solved (wouldn't be surprised) and someone could point me to a reference that would work too.

• This site answers your question for the most common hole geometry. – Chris Mueller May 19 '15 at 2:23
• This is useful, thanks. What I am more interested in is building a model that will allow me to predict 50+ different variations. – Nic May 20 '15 at 18:06
• Then why aren't you doing a finite element model? What exactly are you stuck with? – null May 21 '15 at 15:32
• I would do a finite element model but don't know how. Is there a FEA program that allows you to write scripts to define the geometry? Also how do I approximate an infinite plane, just have a large number of holes and let the computer deal with it? – Nic May 22 '15 at 18:31
• LUSAS would meet the required spec of being an FEA program which you can run using scripts. Not just to define the geometry - you could write a script to define geometry, apply load, solve, output results to excel, and then loop through with different geometry. – AndyT May 26 '15 at 16:31

$$\eta = \frac{\frac{E_f}{E_m} -1}{\frac{E_f}{E_m}+1}$$
$$E_c = \frac{E_m(1+\eta f)}{(1-\eta f)}$$
Since $E_f = 0$, we have $\eta = -1$, so:
$$E_c = E_m(\frac{1-f}{1+f})$$