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Lets assume we have a mechanical element from uniform material that is supposed to work in a static load conditions. Moreover lets assume there is no gravity or gravity is negligible. Are there any examples when reducing the element mass/volume actually gives an element more strength? I mean pure strength, no strength to weight ratio, the element cost is also irrelevant in this case.

I have mechanics lectures some time ago and if I remember correctly, the lecturer have provided an example of that. Unfortunately I cannot remind myself any details about that. Maybe the "trick" was that he reduced the mass simultaneously changing the shape a little bit in a way that some pressure concentration was dispersed?

Unfortunately I cannot produce or find any example of that now and I am not even sure if I remember this all correctly.

I will appreciate any help.

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    $\begingroup$ Why aren’t bridges solid? $\endgroup$
    – Solar Mike
    Sep 6 '21 at 9:25
  • $\begingroup$ @SolarMike because of gravity and cost I suppose, which are both irrelevant here as I mentioned. $\endgroup$ Sep 6 '21 at 11:10
  • $\begingroup$ I think for other reasons, but you can work on them - as if I suggest them you will say likely "irrelevant". $\endgroup$
    – Solar Mike
    Sep 6 '21 at 11:14
  • $\begingroup$ Regarding items such a metal beam, smaller items made from the same metal, using the same fabrication process, can higher strength properties simply because they contain fewer atom dislocations within their crystals & similar imperfections. The number of such strength reducing imperfections increases with the size of an item. $\endgroup$
    – Fred
    Sep 6 '21 at 11:49
  • $\begingroup$ @Fred I completely agree with your comment. My only qualm is that with smaller objects the absolute strength (not strength to weight) reduces (although as I am writing this comment I realise that the term strength here is probably ill defined as I mean it more as load, where traditionally strength is load over area). $\endgroup$
    – NMech
    Sep 6 '21 at 15:55
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I just found an answear, the keyword is "stress concentration". Sometimes, the material is removed to make a stress relief in an element. Some examples: Stress relief examples

The origin source I found it: https://www.youtube.com/watch?v=QtSki5nfO2g

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    $\begingroup$ The adaptations in the top right panel don't actually increase the radius of curvature at the crack tip, so I guess the way they reduce the stress concentration falls into the category of "subtle nuance". In that case, it might be important that "fatigue" is not "static load". $\endgroup$ Sep 14 '21 at 12:14
  • $\begingroup$ Maybe I was not precise. By static load I mean that there are no moving elements; I want to exclude any inertia as it will cause the question to be trivial. $\endgroup$ Sep 15 '21 at 22:32
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I think that if you apply tensile loading, and a uniform cross-section is used, then if you remove material you will end up with a lower strength.

The only class of examples I can think of (although not for static loading) is removing mass from a rotating element like a gear or a pulley. The reason is that the added mass increases the inertial forces (when rotation is involved), which in turn results in the component to be subjected to higher dynamic loads during operation. When you remove, masses that don't contribute as much to the structural integrity you end up with smaller masses, and smaller forces.

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  • $\begingroup$ I suggest looking at the ‚area moment of inertia‘ of square tubes vs square bars. $\endgroup$ Sep 7 '21 at 12:16
  • $\begingroup$ The second moment or area makes sense in bending and torsional problems, and its basically a rearrangement of the cross-section (so its better utilisation of material). However if you start from a square cross-section and you end up with any hollow cross-section the absolute force will be less (the force per weight is a different matter). I interpreted the question's "pure strength, no strength to weight ratio, " in that light. $\endgroup$
    – NMech
    Sep 7 '21 at 12:35
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    $\begingroup$ Yes, I'd agree. according to their description it seems to me, they're talking about the area of moment...but the correct answer would require the correct question :) $\endgroup$ Sep 7 '21 at 12:51
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If the load is applied in flexure (i.e. bending), then a small amount of material a long way from the neutral plane can contribute more to the load-bearing capacity than a large amount of material near the neutral plane (that's what the second moment of area is all about). Similarly, if the load is applied in torsion then a small amount of material a long way from the axis about which the torsion is applied can contribute more to the load-bearing capacity than a large amount of material near the axis about which the torsion is applied (that's what the polar moment of area is all about).

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one example would be a very long tower under self weight.

  • if you remove mass from the columns on the higher levels where it is not needed, the structure Will be stronger, because you have reduced the dead load.

  • a cantilever heavy beam will be stronger if you taper the section and remove some of the self weight.

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  • $\begingroup$ There is no gravity or gravity is negligible in this question. $\endgroup$ Sep 13 '21 at 14:03
  • $\begingroup$ @radoslav006, thank you, my mistake. shouldn't expect more when you answer a question when you taxi behind several noisy jets. $\endgroup$
    – kamran
    Sep 13 '21 at 16:50
  • $\begingroup$ There is no gravity in the question, bit I wouldn't be surprised if the tapered cantilever (reduced self weight means higher load can be supported at the end) was the example OP can't remember from the lecture... This was the first thing I thought of. $\endgroup$ Sep 14 '21 at 12:37

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