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May 18, 2017 at 18:42 comment added Bill Greene Thanks, but unfortunately I don't have access to the journal containing the Bailey paper and I couldn't find a pdf of it elsewhere.
May 18, 2017 at 18:41 answer added Bill Greene timeline score: 3
May 18, 2017 at 17:25 comment added nicoguaro @BillGreene, it seems that the solution is from: Bailey, R., and Hicks, R., 1960, Behavior of perforated plates under plane stress, J. Mech. Eng. Sci., Vol. 2, p. 143.
May 18, 2017 at 16:03 comment added Moonwalker @BillGreene it is from Peterson's Stress Concentration Factors eu.wiley.com/WileyCDA/WileyTitle/productCd-0470048247.html page 324
May 18, 2017 at 16:02 history edited Moonwalker CC BY-SA 3.0
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May 17, 2017 at 20:18 answer added nicoguaro timeline score: 1
May 17, 2017 at 19:45 comment added nicoguaro I would insist in the reference for the problem, as @BillGreene asked.
May 15, 2017 at 11:54 comment added Bill Greene Sorry, I do not know how to apply boundary conditions like that for a FE model. If I were solving this problem, I would create several quarter models with symmetry BC applied at $x=0$ and $y=0$. The first model would have two holes, the second three, etc. As you increase the number of holes, the stresses at the center hole will converge relatively rapidly due to St Venant's principle. Not an elegant approach, but a straightforward one.
May 15, 2017 at 11:13 comment added Moonwalker @BillGreene yes, you are right! The displacement itself is not a periodic, it was mine mistake. But, the displacement could be represented as $u = A\tilde{u} + u^*$ where $u^*$ is a periodic part. So, if I can somehow remove linear part $A\tilde{u}$ I would get a periodic problem. Not sure though, how do I remove the linear part first. Would like to get any help.
May 14, 2017 at 21:40 comment added Bill Greene What is the reference for the original elasticity solution to this problem? It is unclear to me that periodic boundary conditions are appropriate for this problem.
May 14, 2017 at 16:01 history asked Moonwalker CC BY-SA 3.0