How thick can I cut a beam from a thin card without L-T-Buckling?

[NuclearNerd]

Here's an interesting problem for you. I'm trying to make a mechanism consisting of a deck of laser-cut thin plastic sheets. Each sheet will have a beam cut into one edge, as shown in the attached sketch. The mechanism works by bending each of these beams in the deck independently.

Ideally, I would like the beam section depth (b) to be as large as possible so that the shape isn't too weak. However I need to keep the thickness (t) to a minimum (probably < 2*b), which is the opposite aspect ratio you would normally want for a flexing beam.

So my question to you is: How deep (b) can I make the beam of thickness t, and length L, so that it flexes by (d) without buckling out of the plane? I'm not familiar with this kind of stability calculation, so your help is greatly appreciated. If it helps, the cards will probably be made from Delrin, and are 5" x 8" x 0.03" before cutting.

Thanks in advance

 

[paddingtongreen]

When you say beam, do you mean flat strip? With force "P" over the length "L" you want to close the gap "d" without buckling?
There are four unknowns there, I don't know if a dimensionless solution is possible.

Normally, we start with real dimensions, an estimated section, and then test it.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[NuclearNerd]

Yes, the flat strip, thanks.

I took your advice and stopped looking for a dimensionless solution, and started playing with real numbers. I used the LTB formula presented in the American Wood Council's design guides:

http://www.awc.org/pdf/tr14.pdf

to find the critical moment Mcr where buckling occurs:

Mcr = 1.3*Cb*Ce*E*Iy / L
(where Cb, Ce are loading factors defined in the paper)

I then calculated the end load and resulting moment needed to deflect the beam by d and compared to Mcr. Iterating with some numbers, I was able to find a reasonable relationship between b and L for *just critical* beams. L tends to grow exponentially with b, so it seems I should make the beam as thin (b) and short (L) as possible to maximize stiffness out of the page.

I hope this approach makes sense. One follow up: Is this the appropriate formula to be using for Mcr?

 

[ishvaaag]

Quite likely not. For once, it is for wood, whereas you have a plastic of yet unstated properties. Then, you seem to use the formula at p. 14, that seems to be for segments of beams braced at Lb distances.

Your problem differs from that in some aspects; since you seem to say that the cards will be one besides the others, the inner ones will have scarce ability for proportionally moderate forces to buckle out of plane, at least if the deformation is not such that brings the cantilever out of lateral restriction. Hence every cantilever except the ones at extremes (if not themselves boxed by some limit surface of enough strength) will be restrained laterally by all the other cards at the sides.

Of course to design each card for LTB as if no restriction was available is a safe and conservative assumption.

But in any case you need to state boundary conditions for your cantilever; how is it to be restrained in your model that it may represent the actual situation in practice?

Quite likely you may do even for the whole card set and pushing any cantilever in SAP 2000, Autodesk Simulation Multiphysics or other even more complete package like Ansys or Abaqus. You will enter the restrictions and fixities, and the mutual restraints, then press any end of cantilever.

It is important to acknowledge that the root of your cantilevers (if the cards were by themselves) may provide less than true fixity and so you will likely would have to count on a lower critical moment than a closed form formula for the critical moment of a cantilever with full fixity were then providing. Of course given the mutual lateral restrictions in your cards that for many of them will mean almost complete prevention of lateral torsional buckling may be irrelevant. For the extreme cards if laterally unrestrained (and others where the lateral restriction is not stiff and strong enough), it will be significant, since there may be LTB.

For a single card alone you may model it in SAP2000 and make at least two kid of analyses, one for the elastic buckling load (from whee if the buckling was in the elastic realm you would get the limit load), and one or two considering nonlinearities (geometrical and material etc).

Then it may be that your device needs this thing for imposed impact and very repetitive tasks. SAP wouldn't be ideally suited then for the analysis, and quite likely you should resource to some FEM fatigue analysis program. Without resourcint to them, impact itself may require factoring the load for static equivalence analyses, or otherwise use dynamic analyses where the loading regime (as true as possible to the actual loads) is used as input.

 

[prex]

The answer is at page 24 of the reference you cited, that gives the otherwise known (theoretical) formula for the critical load in a cantilever loaded at the tip:
P_cr=4.013[&radic;](EI_yEI)/L²
By rearranging this expression with your symbols and other known relationships (valid for steel but possibly also for your plastic sheets) and also assuming b/t=10, it turns out that
P_cr=4.016Et³b/L²
Now the load to obtain the deflection d is
P=3EI_x/L³
and rearranging it turns out that the critical value of b is (if I'm not mistaken)
b_cr[&cong;]4t[&radic;](L/d)
You see that for e.g. L/d=16 you should have
b<16t
a quite comfortable value.
Note however that all the above is quite theoretical, to stay on earth at least the following considerations should be taken into account:
- P_cr above is theoretical, a safety factor of 2 or 3 should be taken wrt it
- P_cr above is valid if the load is applied at mid beam height, it will be lower if it is applied above the tip or even farther (as in your sketch), but not by a substantial quantity (some 10% reduction)
- the assumption of b/t=10 is quite safe, as with an even much higher value the reduction in P_cr would be of a few percent
- all the above assumes that your 'beam' does not go into plastic deformation
- the phenomenon pointed out by ishvaag, the lateral restraint in the card deck, depending on how your deck is formed and restrained, could completely suppress the possibility of buckling

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[TLHS]

Is this for fun, or is it some sort of mechanism for business? If it's the latter and you're making more than a couple of them it's probably more efficient to make a few test items of different shapes and either test until you get the load you want or test until you have enough samples to come up with a mathematical relationship you can use to come to a solution. Personally, unless I was incredibly conservative I probably wouldn't trust a one off of such an extreme edge case like this using unfamiliar materials and machining tolerances without some load testing.

If it's for fun then it's a whole different story!

 

[NuclearNerd]

It's for fun. The mechanism is part of a large toy I want to build. I will probably do some experimentation, but I'd like to understand the structural equations so that I can minimize the number of wasted attempts.

prex: thanks for the analysis. I need a bit of time to go through your algebra. b/t = 10 seems very thin from my experience - I would expect a beam that narrow to rotate easily.