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This may be a dumb question, but is a solid block 100% of the time stronger than a 'strategically hollow' part. Strategically hollow meaning, the internal structure is design in such a way to support the loading direction?

My thinking is originating from both 3D printing internal supports but also from bio-mimicry designs. Looking at the internal structure of bones, they aren't solid, would they definitely always be stronger if they were solid? Intuitively I would say 'obviously' however could an internal structure is designed in such a way that the surface energy, and load paths create a stronger structure than a single solid block?

If there is a way to design the internal structure, how would you go about designing/analyzing this, if not is there a way to prove that it's not?

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Seems like you might not be considering the importance of the strength to weight ratio.

A solid bar will be stronger, but it will also weigh more. This additional weight will put extra stress on other components, and also have it's own internal stresses associated with supporting itself (imagine a long solid cantilever beam vs. a hollow one).

Often times engineers will use hollow members or designs with webbing. The general purpose of the designs is to put material where it needs to be for strength, but try not to overdo it, or else then you have weight that makes it harder to move (or causes more internal stress between the members).

Often this involves making things hollow, especially since things like bending and torsion get the most resistance from the furthest out points. You maximize your material where you would see the most stresses (think of I beams for bending and hollow poles for twisting).

Nature has this down to a pretty good science. Bones and such have strength where they need them, but there are enough cavities and such to fit other vital parts of the anatomy instead of being full of bone without as much structural benefit.

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  • $\begingroup$ To continue: as you rightly point out, lightweighting any unsupported structure is a good way to reduce sag. OTOH, a solid boulder lying on the ground is much harder to crush than any lightweighted structure of the same shape (and lying on the ground). $\endgroup$ – Carl Witthoft Mar 28 '17 at 17:40
  • $\begingroup$ I wasn't considering weight specifically but for larger members (where mass plays a larger role) you're bang on. I was more thinking can an internal structure specifically designed for a loading case be stronger than a solid block. So if the loads were parallel to the ground (minimizing loading due to mass), could a complex structure be stronger than a solid structure? $\endgroup$ – Diesel Mar 28 '17 at 19:54
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    $\begingroup$ @Diesel I can't see how less material would be stronger unless you started to get some odd effects beyond most engineering analysis. As far as I know there isn't really anything that would do that. The thing is, regardless of the total size, you reduce the self loading if you decrease the mass. It won't make the structure stronger, but you can remove structure where you don't expect loading. This is all working with continous scales where this is a solid bulk of material. On very small scales this may fail due to secondary molecular effects. $\endgroup$ – JMac Mar 28 '17 at 20:31
  • $\begingroup$ Intuitively I'd say your right. But if you drill the fracture point of a crack stopping it from spreading, you've increased the strength. Is there absolutely no other case where removing material could increase strength. And to your point I'm more interested in macroscopic volumes not individual atoms. Reordering the molecular structure definitely can change material properties, eg. single crystal titanium, nano-structures etc. $\endgroup$ – Diesel Mar 28 '17 at 23:19
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    $\begingroup$ @Diesel Drilling a hole in a crack doesn't make it stronger really. Cracks are a fault that can propagate themselves easily in some conditions. Drilling a hole removes the fault so it can not spread, but you're still removing strength from the material, just the crack removes even more strength due to its failure mode. $\endgroup$ – JMac Mar 28 '17 at 23:22

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