Bridgman Paradox?! Masters thesis

[Iteklf]

Hello!

Im working on my masters thesis and might have found an unexpected answer in inventor.

The setup is very simple:

Step 1: Faced with the seemingly trivial physical problem with a pressurized chamber with a bar protruding the chamber walls through sealing glands, most people opts for solution 1; no resulting axial force; T =0.

Step 2: By reversing the pressure, there is 100% agreement that solution 2 gives the right answer. A compression force corresponding to the overall pistion effect.

Step 3: Most people agree that lowering external pressure to negative gives a suction force and corresponding tension in the piston of T =∆pA.

Then, the question is: Is not step 3 identical to step 1?

Here are my screenshots from Inventor:

Files are attached:

So, Why is there no axial force in the rod when pressure is applied from inside?

[graemev]

There is no axial area on which the pressure may act, therefore no axial force is produced.  If the rod were to be necked down between the ports in the box, the necked down portion would see tension when the internal pressure exceeds the external pressure and, of course, compression when the reverse.  The only time the pressure differential will create tension in the rod is if the chamber pressure is sufficiently high to compress the diameter of the rod a significant amount, thus generating a necked effect.

Thought experiment:

Use the same setup, but sever the rod in the middle.  A pressurized chamber will push each rod out, as expected.  Now put a sleeve across the break in the rod and seal the break from the interior pressure of the chamber.  You now have effectively three pressure zones: external, chamber, and gap.  Changes in external pressure will cause motion in the rods and a corresponding change in pressure in the gap, but not in the chamber.  Assuming frictionless seals, the pressure in the gap will change to match the external pressure by acting on the rods until forces balance.

[JDMather]

Post the source of your reference information (I did a search myself and think what I found answers your question).

Also, I would have Split Faces rather than Split Part into separate solid bodies.  I am not sure of your reasoning for splitting the part.

Can you explain that reasoning?

And what are the limitations of Inventor FEA regarding linear static stress beyond yield?

[Iteklf]

Appreciate your reply! 🙂

The source is my supervisor. I dont think I have a direct web-address. Still, there was a thesis written last year starting on this theme [link]. My supervisor think that the rod is subjected to tension as soon as the internal pressure get higher than the external.

I put up this problem as an experiment. It looks like the rod will fail when internal pressure [750bar] reach the ultimate limit in tension of the rod [70 MPa]. Rod = acrylic/pmma[source]. I also tested with a POMC rod. Looking at the fracture surface in SEM, the surface looks like it was subjected to tension.

Why I used split part? Well, I just started learning Inventor. This was the only way I knew how to draw this up, - until now. Would it make a difference in the result?

[JDMather]

I had trouble with your hyperlink - I think this is the document it referrenced -

http://brage.bibsys.no/xmlui/bitstream/handle/11250/182928/Reve,%20Morten.pdf?sequence=1

[Iteklf]

Yep, your link is correct!

I can see my link is missing something. But it should end up at the same document. I will fix

[rhinterhoeller]

I concur that the internal pressure has no way of inducing an axial force on the rod (it will, due to poisson's ration cause the length of the rod to change.

The only axial force will be compression due to atmospheric pressure on the rod ends.  The force on each end is equal so there is no net force on the rod.  Increasing the pressure as in scenario 2 will increase the axial compression.  Changing the internal pressure is irrelevant.

Richard

[graemev]

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@rhinterhoeller wrote:

I concur that the internal pressure has no way of inducing an axial force on the rod (it will, due to poisson's ration cause the length of the rod to change.

The only axial force will be compression due to atmospheric pressure on the rod ends.  The force on each end is equal so there is no net force on the rod.  Increasing the pressure as in scenario 2 will increase the axial compression.  Changing the internal pressure is irrelevant.

Richard

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Yes.  My bad, however, for not looking up just what Bridgman's paradox is before replying.  *sigh*

One wonders if the source of load in the paradox isn't the sum total axial area found in the surface finish of the rod.  No material has a perfectly smooth texture, thus no permeation of the fluid into the rod should be necessary to generate localized axial stress.  The fluid pressures involved would be startlingly high, though lower for rougher surface finishes.  (?)

I doubt Inventor is suitably sophisticated to generate these loads based on trivially simple geometry.

[Iteklf]

Okay,

But then, what would you say make the rod break?

Von Mises stress is high, but that is just saying the energy is enough for a break, not meaning that a break should appear without any disturbances?

[JDMather]

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@graemev wrote:

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The fluid pressures involved would be startlingly high, though lower for rougher surface finishes.  (?)

I doubt Inventor is suitably sophisticated to generate these loads based on trivially simple geometry.

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...which is why I suggested the OP think about the yield stress for the material and the limitations of the Inventor linear static stress analysis solver.

One of the sources I found did refer to necking of the part, where poisson's ratio will come into play.  That sounds to me like a boundary condition outside of Inventor FEA solver guidelines.

I suspect though, that something else is missing in the digital set-up of this problem that could make it more like the real-world set-up.

[graemev]

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@Iteklf wrote:

Okay,

But then, what would you say make the rod break?

Von Mises stress is high, but that is just saying the energy is enough for a break, not meaning that a break should appear without any disturbances?

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I'm not quite following your terminology.  What do you mean by "disturbances?"

[Iteklf]

Some small force that will disturb the equilibrium of forces and make the rod fracture.

Might sound a bit on the edge, not reality...

My point, can anything fracture because of Von Mises stress only?

And/or/if the poisson effect is the thing that will make the rod fracture, how?

Any ideas?

[graemev]

I would guess that subjecting the rod to sufficient hydrostatic pressure would cause the core to flow to the ends, much like squeezing a garden hose.  Under those conditions I would expect the rod to fail near the ports of the chamber, but I'm no metalurgist and have a weak understanding of the nature of metal grain strengths, particularly under conditions of plastic and near-plastic strain.

The failure Bridgman observed may be a matter of scale propagation of microscopic flaws.  If the rod is assumed to have a surface that deviates from a geometric ideal, hydraulic forces will act locally in the groove.  (Consider a mountain valley full of water: the mountains on either side, acting as dams, are being thrust apart by the hydraulic force of the fluid.)  This localized force will induce tensile stress at the root of the groove.  This force is highly localized in the material, with the attendant strain falling off towards the interior of the rod.  It is quite conceivable to exceed the yield point of the material locally but not globally by this type of load.  (In short, a hydraulic wedge is being forced into the groove.)  If the material develops a fracture - most likely along a grain boundary - the fluid will have a greater axial area on which to act, thus increasing the axial load, thus increasing the stress, and 'round we go again unless the particular direction of flaw propagation fails to increase the axial area.  This would propagate the scale of the flaw, quite probably in a fairly dramatic fashion.  I am as yet at a loss to explain why rod failures occur near the center of the rod's length, but even something as small as gravitational loading may be sufficiently significant in such a case.

[QuasiMojo]

I took the bait.  I don't see any paradox. I see an abrupt change in compression stress in a material with a non-zero Poisson's ratio.  The reference says: "...and the pressure inside the vessel is in the numerical region of the maximum tensile strength of
the solid’s materiel, the bar will rupture as if in a tensile test."  This is enormous pressure and will undoubtedly cause radial compression.  Since the free length of the bar has no compression (abruptly) there is a gradient between a stress state (already near the tensile limit) and zero.  There will be radial compression in the pressurized zone and not in the free length: a strain gradient.  At this pressure your bar is being squirted out like so much toothpaste (just like graemev said).  Below are some axi-symmetric FEA plots showing this.  I used the material properties and geometry from the reference, with 0.7 GPa pressure in  a linear analysis.  This would result in 20% radial strain, which is way past failure for anything but polymers.  I don't see any mystery here.  What am I missing that makes this paradoxical?

[Iteklf]

Well, at 0,7GPa, do doubt..

The rods(PMMA) broke at 0,07GPa... The more flexible rods(POM-C) broke at 0,1GPa

[QuasiMojo]

Sure, that's an extreme pressure, but that seemed like the point; it also makes it easier to visualize.  Of course, the linear analysis is quantitatively garbage at this extreme...but it makes a good picture of what's going on