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Let’s say we have a thin band (made out of PLA, 3D printing material) with a rectangular cross section. In drawing, it would look something like this: enter image description here

Let’s say a force of 15 N is to be applied from the top down to the band, how am I supposed to figure out the minimum thickness (and or height) that could endure the force and do not undergo deformation or failure?

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  • $\begingroup$ Deformation always occurs. How much deformation can you tolerate? $\endgroup$ – Paul Oct 26 '18 at 4:49
  • $\begingroup$ @Paul Probably less than 1cm. Would you be so kind to talk me though the process of how to go about this for this problem? $\endgroup$ – Kwun Oct 26 '18 at 4:53
  • $\begingroup$ A person with more experience can probably give you a more elegant answer, but i would model it using finite elements for linear elasticity (axisymmetically), applying a total load of 15N and then setup a optimization loop to determine the configuration that produces minimal displacement on the top surface. This is essentially what an experimentalist would do, except i’d do it all on computer simulations. $\endgroup$ – Paul Oct 26 '18 at 4:56
  • $\begingroup$ It almost seems like you would want to do buckling analysis for the case of very thin rings. $\endgroup$ – Paul Oct 26 '18 at 4:59
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As a very rough first estimate we would consider this as a $20\pi = 62.8\text{ cm}$ polygon made of approximately 63 equal sides of 1 cm wide by 3 cm high. After multiplying the load by a factor of safety of 2 we get 30 N.

And we ignore the slenderness concerns now because these columns are laterally supported and considered short, $L< 10$ radius of gyration.

So now we can calculate the required thickness.

$$ t= \frac{30\text{ N}}{63\text{ cm} \times \sigma} $$

where $\sigma$ is the allowable compression stress of PLA.

Edit

Answering some comments considering the arbitrary selection of a polygon to substitute the circle:

Yes I agree that if we add the number of sides of polygon more and more, at the limit we reach the perimeter of the circle, but that is not the point as the difference is less than 3%. I chose 20 as a simple linear approximation to circle, to give OP the kind of answer he can immediately envision.

Short columns with a length smaller than 20 times radius of gyration crush ultimately rather than buckle under stresses more than yield stress . In this case with radius of gyration of 10cm, lengths of more than 200cm would be cause for buckling concerns.

And local buckling happens only under concentrated loads which is not the case here.
For deeper review here is the link to NASA research on buckling of thin wall cylinders.

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  • $\begingroup$ You can ignore bucking if you want, but that doesn't stop it happening: youtube.com/watch?v=NbJcTLiy2Ao $\endgroup$ – alephzero Oct 26 '18 at 14:31
  • $\begingroup$ Approximating this as a polygon with arbitrary number of sides is a pointless complication - why choose 20 sides, not 13 or 97? If you ignore buckling, the ring will behave exactly the same as a rod with the same cross-section area. $\endgroup$ – alephzero Oct 26 '18 at 14:45
  • $\begingroup$ @alephzero, at this scale and configuration I would hardly worry about buckling. In fact if I get a chance I will make a model out of card-board to show it can support the 15N. If it was 30 meter height then I would use Euler's formula, Pcr = Pi^2EI/L^2. But as it is its a short column. L < 10 GR, radius of gyration. $\endgroup$ – kamran Oct 26 '18 at 16:01
  • $\begingroup$ My concern would be local buckling, with the bottom (or top) slipping out from underneath the ring. The radial stiffness would be rather low given the thin cross-section. $\endgroup$ – Wasabi Oct 26 '18 at 18:55
  • $\begingroup$ @kamran thanks for the answer! I took a different approach, and I am not sure if my approach will eventually lead to your answer. However, I am not fully getting the concept yet. Since it does not require a great accuracy, I though I would do a beam analysis and multiply it with some correction factors to account for its shape. so, I ended up getting an equation k = (4Ebh^3)/(L^3), where k is stiffness, E is modulus, b is width, h is thickness, L is the length. and the pressure force being applied to the top of the band is 178 Pascal. elastic modulus of the material im using is 3.5 Gpa. $\endgroup$ – Kwun Nov 1 '18 at 1:42

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