I made a bushing 25 x 49 x 40 mm long to mount a timing belt sprocket to a treadmill motor. I made the ID a bit too small, about .001" too small. It took a lot if force to press it on. I go to screw it on the shaft and it gets tight. Before pressing it screwed on by hand. To relieve tension, I drilled a series of holes in the bushing, 4 threaded for a puller and 4 plain drilled holes. Now it screws on by hand about 3/4 of the way to shoulder. I blued the shaft to prove the interference. So now the question, is there a way to calculate metal compression? Alternately how much tension do I relieve by drilling? image widget
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Originally posted by Lee Cordochorea View PostMetal doesn't actually "compress." The atoms in the lattice maintain a set distance from each other. Total volume remains the same. Simple geometry therefore applies. The hard part is predicting how much of the material will flow plastically in more than one direction where force is not applied.
A Poisson's ratio of 0.5 defines an incompressible material. Rubber is one material that is nearly incompressible. For the math to work out for rubber, Poisson's ratio is usually taken as 0.49 because the math crashes if you use 0.5. Divide by zero if I recall. This characteristic makes rubber useful for metal forming work.
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Thanks striker six for the link. I see my use of terms is incorrect. Strain should be substituted for compression. I have measured the results. Now I can use the numbers to fit the equation. If two rings are pressed together, the id of the inner ring gets smaller and the od of the outer ring get larger relative to materials, dia., etc.
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Originally posted by Lee Cordochorea View PostMetal doesn't actually "compress." The atoms in the lattice maintain a set distance from each other. Total volume remains the same. Simple geometry therefore applies. The hard part is predicting how much of the material will flow plastically in more than one direction where force is not applied.Originally posted by strokersix View Post
Actually, this is not true. Poisson's ratio defines how much a material will change in volume when load is applied. In other words, if you squeeze a block of material in one direction, how much will it pooch out in the other two directions. For steel, a Poisson's ratio of 0.3 is typical.
A Poisson's ratio of 0.5 defines an incompressible material. Rubber is one material that is nearly incompressible. For the math to work out for rubber, Poisson's ratio is usually taken as 0.49 because the math crashes if you use 0.5. Divide by zero if I recall. This characteristic makes rubber useful for metal forming work.
https://www.engineeringtoolbox.com/p...iod_1224.html
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Originally posted by strokersix View Post
Actually, this is not true. Poisson's ratio defines how much a material will change in volume when load is applied. In other words, if you squeeze a block of material in one direction, how much will it pooch out in the other two directions. For steel, a Poisson's ratio of 0.3 is typical.
A Poisson's ratio of 0.5 defines an incompressible material. Rubber is one material that is nearly incompressible. For the math to work out for rubber, Poisson's ratio is usually taken as 0.49 because the math crashes if you use 0.5. Divide by zero if I recall. This characteristic makes rubber useful for metal forming work.
The process of making rubber is archaic at best and has not changed in 100 years. It is full of inclusions and voids and is compressible to that extent. It is nice for metal forming in that is acts like a fluid exerting equal pressure in all directions, not that it is “incompressible”.
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As I understand:
Example: A cube of material placed on your milling machine table. Press down on it with a force to deflect it to a Z direction strain value of 1 for easy math. The resulting strains in X and Y directions are the Poisson ratio. A material with Poisson's ratio of 0.3 such as steel will strain 0.3 in X direction and 0.3 in Y direction with a resulting VOLUME CHANGE of (0.3+0.31=0.4).
A material with Poisson of 0.5 such as rubber will strain 0.5 in the X and 0.5 in the Y (0.5+0.51=0) which equals NO loss of volume or an incompressible material.
I'll happily concede if I'm wrong about this but I don't think so.
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Edit: In further reading I discovered some errors in my original answer. I have gone over and revised it. It should be more accurate now, but I will accept any honest corrections.
Actually this is dead on. By definition, the Poisson effect is the deformation (expansion or contraction) of a material in directions perpendicular to the direction of loading. That means sideways. It is not defined in terms of volume. It is defined in terms of the linear change.
BOTH the Poisson effect and actual compression in a metal, like steel, represent changes that are reversible, at least for the most part. But they differ in that actual compression is a reduction in volume which is not accompanied by any decrease in any of the actual dimensions (X, Y, or Z) while the Poisson effect is not that neat and while one dimension is getting smaller the others may be getting larger.
Deformation from the Poisson effect is what happens when a more or less normal force (one we can easily achieve in our shops) is applied to an object, like a piece of metal. It gets noticeably wider and longer at the expense of the height. And that change is, for the most part, reversible.If we remove the force, the dimensions return to their original values or at lease very close to them.
The actual compression of metal, while also possible, is a much more difficult thing to achieve. You can not just apply force between two opposite faces. You must restrain that metal object from ALL sides to prevent the Poisson effect from taking place with the displaced metal just exiting in one or more sideways directions. The compression of a metal (uranium 235) is one of the mechanisms for triggering an atomic explosion. To achieve that actual compression explosive charges completely surround a spherical, uranium core and a large number of detonators are simultaneously triggered on the outside of these explosive charges to produce a very symmetric shock wave toward that inner uranium core. That compression of the U235 raises the number of neutrons per unit volume above the critical point and fission proceeds. The number of neutrons flying around inside the U235 remains the same but the volume is reduced. BTW, those charges and the trigger circuitry are the hard parts of building an Abomb after the U235 isotope is produced.
I do not completely understand the exact geometry of the OP's bushing but since there is no means of restraining the sideways deformation of that part while he pressed it on, I feel confident that we are talking about the Poisson effect, along with some actual PERMANENT deformation and not the actual, volumetric compression of steel. If I assume that the bushing is the part shown on the (motor?) shaft in the provided photo, then, it is also reasonable to assume (there's that word again) that he applied the force at the exposed end of the threaded part of that bushing. Such an application of force would cause the threaded part of the bushing, which has the smallest effective diameter of that part (the minor diameter of the thread) to expand. Part of that expansion would be due to the reversible Poisson effect, but another part would be a permanent distortion of the metal which is not reversible. So the threaded portion of the bushing has expanded outwards in all RADIAL directions making the pitch diameter of that thread larger. Again, this is due to a permanent distortion of the metal when the Poisson effect exceeded the elastic limit of the metal.
He has attempted to relieve that permanent deformation by drilling holes in the part think that would relieve internal stresses and the dimensions would return to their prior values. That may work to a small extent, but, unlike actual compression of the metal, much of the deformation above and beyond the Poisson effect will be permanent. Heating the part may have had a similar effect of partially removing some of that deformation but I doubt that either heating nor drilling holes will eliminate all of it and bring the thread back to it's dimensions before the pressing operation. BTW, I do not recommend heating it at this point as that may loosen the press fit.
The best solution at this point would be to use a die to chase the threads. That should cut off the expanded metal, leaving a correct thread which should fit. If you do not have the correct die, another possibility would be the use of a thread file to remove that expanded metal. That filing should be done evenly on all sides of course.
If starting from scratch with a new bushing, I would not attempt to compensate for the expansion with altered dimensions for the thread. Instead I would make a tool from some round steel that is about three times the OD of the thread. Drill and tap a deep hole in the end to fit the thread and screw all the way on it. Use that tool, screwed against the shoulder of the threaded section of the bushing, to press the new bushing on. By pressing against the shoulder on the bushing, which is past the threads, that tool should prevent any permanent deformation of the threaded area. But it should be large enough in diameter to do so without expanding very much itself. Twice the OD of the thread will not cut it for this and four or more times would be best. This is definitely a case where bigger is better.
Originally posted by Bob Engelhardt View Post
To be precise: "compress" mean a reduction in volume. Everything can compress, but infinitesimally for steel, etc. Compression is a matter of its "bulk modulus", not Poisson's ratio. Poisson's ratio applies to deformation under load, not its compression.Last edited by Paul Alciatore; 05282022, 12:10 AM.Paul A.
SE Texas
And if you look REAL close at an analog signal,
You will find that it has discrete steps.
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Originally posted by strokersix View PostAs I understand:
Example: A cube of material placed on your milling machine table. Press down on it with a force to deflect it to a Z direction strain value of 1 for easy math. The resulting strains in X and Y directions are the Poisson ratio. A material with Poisson's ratio of 0.3 such as steel will strain 0.3 in X direction and 0.3 in Y direction with a resulting VOLUME CHANGE of (0.3+0.31=0.4).
A material with Poisson of 0.5 such as rubber will strain 0.5 in the X and 0.5 in the Y (0.5+0.51=0) which equals NO loss of volume or an incompressible material.
I'll happily concede if I'm wrong about this but I don't think so.
If L,W,H are the initial dimensions, "v" the PR, and "s" the strain in the length axis, then the correct equation for the change in volume is:
L x W x H  (Ls) x (W + v x s) x (H + v x s)
which is really messy and will include terms with L x H, W x H, L x W, & L, W, H. Not just the 3 strains.Last edited by Bob Engelhardt; 05282022, 01:15 PM.
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More food for thought: The simplified examples in the articles cited might confuse a person into thinking steel cylinders remain cylinders when force is applied to the ends. Under compressive axial force, the steel cylinder becomes a barrelshape. Under constant axial force, it becomes waspwasted. The greatest change in diameter is used to calculate Poisson's ratio.
Quick clarification, Poisson's ratio for steel varies with composition. I've seen numbers as low as 0.26 and as high as 0.31. I have a vague recollection that temperature can play a part as well. (I'm not sufficiently interested to go look it up right now.)
Just to keep the universe interesting, some materials (not steel) have a negative Poisson's ratio.
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Originally posted by Lee Cordochorea View PostMore food for thought: The simplified examples in the articles cited might confuse a person into thinking steel cylinders remain cylinders when force is applied to the ends. Under compressive axial force, the steel cylinder becomes a barrelshape. Under constant axial force, it becomes waspwasted. The greatest change in diameter is used to calculate Poisson's ratio.
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Sure, real, full size examples are more complex. I offered the most simple illustration possible in hope of facilitating understanding. My illustration applies to elastic deformations. When you get into plastic deformation that's a different story.
If you break the problem into small pieces (finite element analysis and the like) I think each little piece (often a cube but doesn't have to be) will behave exactly as I've described within the range of ELASTIC deformations.
Is steel compressible? Yes, defined by Poisson's ratio in the elastic range. For practical purposes steel can be considered incompressible for larger, plastic deformation.
Like a lot of things, the question is too broad for one simple answer...
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Originally posted by strokersix View PostSure, real, full size examples are more complex. I offered the most simple illustration possible in hope of facilitating understanding. My illustration applies to elastic deformations. When you get into plastic deformation that's a different story.
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Is steel compressible? Yes, defined by Poisson's ratio in the elastic range. For practical purposes steel can be considered incompressible for larger, plastic deformation.
Like a lot of things, the question is too broad for one simple answer...
There is never volume change (compression), elastic or plastic. The Poisson ratio does NOT define compressibility, it defines deformation. The bulk modulus for steel is 25 MILLION psi. Elastic or plastic, steel can be deformed to any extent & its VOLUME doesn't change.
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