2
$\begingroup$

I am designing something with a sandwich construction that needs to squeeze the internal layer uniformly. The lower plate is to be made from cast iron and the upper plate needs to be made from copper. The squeezing will be done with four threaded rods, nuts and some spring washers.

enter image description here

The squeezing force I need to apply is 30720N, so I guess 7680N per nut (or is it not so simple?). And what I really don't know is, how do I determine a thickness of material required, given its properties and my design details? My requirements are

  1. Plates will not deform like playdough between 20˚-150˚C
  2. Plates will not buckle like a spring very much, so that the sandwich pressure is not applied unevenly.

kamran's answer describes the red line below as the lever distance in his model.

Kamran points out I can model my fulcrum distance as this length

I have tried my best to read up on Young's Modulus and Shear Modulus, etc., but I do not grok how to correctly apply these to real life.

Question: Can someone kindly point me to a formula or a workflow so that I can calculate the answer myself?

$\endgroup$
  • $\begingroup$ how stiff is the stuff in the middle and how uniform do you need the pressure to be? $\endgroup$ – agentp Feb 1 '18 at 23:53
  • $\begingroup$ Middle layer is going to be things that are not going to change size/shape when squished with the design pressure. I would like the pressure to be as uniform as possible given flat surfaces (I do not want the complexity of elastically deforming a curved surface into a flat one) $\endgroup$ – Douglas Held Feb 2 '18 at 0:19
  • $\begingroup$ I forgot to add that the construction is static. I don't anticipate any dynamic forces (until/unless a bolt breaks :( $\endgroup$ – Douglas Held Feb 2 '18 at 0:53
  • 1
    $\begingroup$ First point, pulling the corners together is a terrible way to go about this if you are concerned about applying the pressure evenly. You want to push the centers together. This has been understood for some time now. Care to supply some dimensions for the sandwhich? $\endgroup$ – Phil Sweet Feb 3 '18 at 2:24
  • 1
    $\begingroup$ Next question, the iron and copper are heat sinks. What is the copper doing with the heat? The copper has to be thick enough to conduct heat away to the ultimate sink. You also don't want heat transfer back down through the bolts. Have you seen this document? Sorry about the pressure remark, I figured a 2 x 2, not 4 x 4 array. But the pressure still seems high by a factor of 2 or so. $\endgroup$ – Phil Sweet Feb 3 '18 at 15:25
3
$\begingroup$

First off you want to make sure if you should divide the required squeezing force by four. Assuming that is correct we can proceed.

Because of symmetry you can assume the flange overhang will have to resist a bending moment that would want to bend it down like a cantilever slab.

There are actually many steel column base-plates that are loaded very similar to this and by checking a column base-plate design manual you'll find an identical configuration and use the equations and recommendation there.

An alternative as a reasonable simplification is to assume the plate's bending moments, once on x-x direction and then on y-y direction, so we disregard the stress concentration at the corner and assume the flange is under a concentrated load by one bolt acting at the distance measured perpendicular to the edge of the plate from center of the bolt to inside edge (0.707 x your call out distance), let's call it $l$, and let's call your bolt tension $p$.

So your moment is $ M= pl$

your slab bending moment $I=bh^3/12$

Where $b$ is your plate width and $h$ is its thickness.

Now your stress $\sigma =My/I= Mh/2I$

And you can test if your plate is strong enough to support the bolt load if you have its allowable stress. The only catch is you apply only one bolt tension per edge.

| improve this answer | |
$\endgroup$
  • $\begingroup$ kamran, I cannot visualize this measurement $l$ thought I have re-read several times. Do you mean to measure diagonally from the bolt center to the fulcrum point of the bend? Similar to my red line? And can you explain where you get 0.707? And the $y$ in the stress calculation -- is this the Young's Modulus of the material? $\endgroup$ – Douglas Held Feb 2 '18 at 0:49
  • $\begingroup$ My apologies for using my phone and not having my reading glasses. Basically we are measuring straight from the center of the bolt to a dash line continueation of edge of your specimen. The y is not the young modules, it is the heigth of the point in cross section of your plate under consideration for stress and because this is bigest at the top and bottom surfaces of your plate, we use height/2, h/2. I is called second moment of inertia. $\endgroup$ – kamran Feb 2 '18 at 1:01
  • $\begingroup$ @DouglasHeld, let's just plug in the numbers using English units. Copper yield tensile strength =4830 psi, say we set factor of safety at 40% thus our limit is say 2000 psi allowable. Then say the thickness of your plate, h is 1/8" so you get 2000= 7680x l (say the distance from bolt cen to edge of specimen in inches).1/16"/I. and you calculate I as mentioned in my answer. If your answer is bigger than 2000 you need thicker copper, if smaller you consider thinner plate and check it again! $\endgroup$ – kamran Feb 2 '18 at 3:32
  • $\begingroup$ And you get 7680 how? And what does x represent? $\endgroup$ – Douglas Held Feb 2 '18 at 23:32
  • $\begingroup$ I picked your number but I did not convert N to lbs, you'd need to do it. X is multiplication sign and lower case l means the vertical distance of the bolt center to the edge of specimen. That is the moment your copper plate needs to support. $\endgroup$ – kamran Feb 2 '18 at 23:48
2
$\begingroup$

From the information you've given it is difficult to be too specific, but looking at the loads and design you might get some ideas by considering elastomeric bridge bearing plates or sole plate design standards.

A couple of useful references may be found for bearing plates and sole plates.

There is also another post that considers stiffness versus thickness of steel plate here. Essentially it comes down to stiffness, which is a function of Moment of Inertia and Modulus of Elasticity.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, I will check those references. What further information could I give that would be helpful? $\endgroup$ – Douglas Held Feb 2 '18 at 0:50
  • $\begingroup$ @DouglasHeld Well the dimensions (in plan) of the plate and the internal object and their relative position could be important as there may be a need for more than 4 threaded rods in order to apply pressure evenly. This was my first thought - one of the more difficult tasks will be the uniform application of pressure - depending upon the material contained (elastic/plastic/homogeneity response) there will be an interaction that may require a greater number of bolts (threaded rods) to achieve uniformity. $\endgroup$ – AsymLabs Feb 2 '18 at 8:49
1
$\begingroup$

Okay, you're not going to like this answer, but you need to space the individual devices in the array to accommodate intermediate fasteners. The assembly as you drew it would cost a couple hundred dollars in materials and fabrication, and would be very heavy. So leave 3 6mm gutters between devices perpendicular to the wire gutters, drill a 6mm hole between each unit, drop in a 4x4mm nylon shoulder washer, and install a 4M machine screw. This gives you 12 intermediate fasteners. Then run 16 more around the perimeter. These last could be 3M in a 3x4mm shoulder washer if you wanted to get fussy. This gets the thickness of copper down to about 6mm for good results. Going with fancier spring isolation washers of at least 10mm dia can get you down to 5mm copper. A happier solution would be to find a Peltier form factor that has fastener notches so you can close-pack them.

The fact that you have a thermal gradient across the copper makes it want to distort. So you don't really want too thick of a copper plate, the thermal distortion forces would soon overwhelm the clamping forces.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I like this answer very much! Although all the holes do complicate my design, it will be safer as well as lighter. Well I would love to make the whole thing out of diamond, but I don't know anybody who can CNC it, let alone to find a billet large enough. Thank you for this advice, which is also what I heard from friends today. $\endgroup$ – Douglas Held Feb 3 '18 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.