Announcement

Collapse
No announcement yet.

Metal compression

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • #16
    Originally posted by Bob Engelhardt View Post
    It's not a matter of elastic or plastic, your example was just plain wrong. Wrong, wrong. Not because it was too simple, but because you added the dimensional changes & called it a volume change.
    Actually, since the length changes are very small, the change in volume is equal to the sum of the length changes in the (very accurate) first order approximation. (All changes expressed as fractional changes, of course.)

    Comment


    • #17
      Originally posted by tomato coupe View Post
      Actually, since the length changes are very small, the change in volume is equal to the sum of the length changes in the (very accurate) first order approximation. (All changes expressed as fractional changes, of course.)
      That's interesting!! Using relative changes makes things much easier, but still not obvious:

      Vd/Vo = Ld/Lo x Wd/Wo x Hd/Ho ("d" & "o" for the deformed & original values)

      = (Lo - s)/Lo x (Wo + v x s)/Wo x (Ho + v x s)/Ho with s<< L, W, H

      I don't see the clever algebra that gets that to:

      = -s + (v x s) + (v x s)

      But checking it using an example of a 1" steel cube squished 0.001", the relative changes are -0.0010, +0.0003, +0.0003 & the relative volume change is

      -0.0010 + 0.0003 + 0.0003 = -0.0004000 where the actual product gives a volume change of -0.0004005. Very close, indeed!

      Interesting is that the volume change is negative! Which means that the Poisson ratio is good for dimensional changes, but not so good for volume changes?

      Comment


      • #18
        Originally posted by Bob Engelhardt View Post

        That's interesting!! Using relative changes makes things much easier, but still not obvious:

        Vd/Vo = Ld/Lo x Wd/Wo x Hd/Ho ("d" & "o" for the deformed & original values)

        = (Lo - s)/Lo x (Wo + v x s)/Wo x (Ho + v x s)/Ho with s<< L, W, H

        I don't see the clever algebra that gets that to:

        = -s + (v x s) + (v x s)

        But checking it using an example of a 1" steel cube squished 0.001", the relative changes are -0.0010, +0.0003, +0.0003 & the relative volume change is

        -0.0010 + 0.0003 + 0.0003 = -0.0004000 where the actual product gives a volume change of -0.0004005. Very close, indeed!

        Interesting is that the volume change is negative! Which means that the Poisson ratio is good for dimensional changes, but not so good for volume changes?
        It would be quite strange if the volume increased while it was being squished.

        It’s hard to type out equations, but calculate the volume using terms like X(1 + x) where X is an initial dimension and x is a (small) relative change in dimension. You end up with XYZ(1 + x + y + z + xy +xz + yz + xyz). For small x,y,z you can throw out the last four terms.

        Comment


        • #19
          Originally posted by Bob Engelhardt View Post


          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 &amp; its VOLUME doesn't change.
          If steel VOLUME wouldn't change(infinite bulk modulus) it would have 0.5 poisson's ratio.
          Poisson's ratio "sort of" defines compressibility. More exactly it defines the ratio of elastic modulus/bulk modulus. http://silver.neep.wisc.edu/~lakes/PoissonIntro.html

          v=0.5-E/6B
          v= poissons ratio
          E=elastic modulus
          B=bulk modulus

          Even metals vary quite a bit in compressibility. Potassium has about same level of bulk modulus as water and water will compress considerably at pressures like Mariana's trench or waterjet cutter. Potassium compresses nearly 10% at 50kPSI
          Location: Helsinki, Finland, Europe

          Comment


          • #20
            Originally posted by strokersix View Post
            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+0.3-1=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+0.5-1=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.
            The volume doesn't change. The material is displaced in the XY directions and whether or not the change is permanent depends on where you are on the stress-strain curve and whether or not you went to the plastic region. The volume of material will be the same. You are not changing the density of the metal
            -paul

            Comment


            • #21
              Originally posted by psomero View Post

              The volume doesn't change. The material is displaced in the XY directions and whether or not the change is permanent depends on where you are on the stress-strain curve and whether or not you went to the plastic region. The volume of material will be the same. You are not changing the density of the metal
              The volume changes if the Poisson ratio is < 0.5.

              Comment


              • #22
                The Poisson effect compression vs compression defined by the bulk modulus has me confused. The 0.001" strain on my 1" steel cube example produced a Poisson effect volume reduction of 0.0004 cu-in. That 0.001 strain would require (30 x 10^6) x 0.001/1.0 pounds stress (elastic modulus) = 30,000 lbf. To achieve a compression of 0.0004 cu-in with pressure on all sides, would require 25 x 10^6 psi (bulk modulus) x 0.0004/1.0 = 10,000 lbf.

                Well, I guess that pushing on all sides takes less force than pushing on one side is not that confusing. What is still confusing is how the Poisson effect causes a reduction in volume at all. The material is elastic and unconstrained in 2 axes - why doesn't it just spread out without compression? The forces resisting deformation are greater than the forces resisting compression?

                Comment


                • #23
                  The math I am seeing on this subject definitely involves approximations. They are at least partially justified by the fact that more than one physical process is taking place and even the more correct volume calculations that you are urging will still not produce exact results.



                  Originally posted by Bob Engelhardt View Post

                  The fundamental problem here is that you are summing linear values and using the result as a cubic. Length PLUS width PLUS height does not equal volume.

                  If L,W,H are the initial dimensions, "v" the P-R, and "s" the strain in the length axis, then the correct equation for the change in volume is:

                  L x W x H - (L-s) 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.
                  Paul A.
                  SE Texas

                  And if you look REAL close at an analog signal,
                  You will find that it has discrete steps.

                  Comment


                  • #24
                    Molecules can rearrange themselves. And different arrangements can have different volumes. Take water as an example. Water's volume decreases (it's density increases) as the temperature is lowered until you reach 4 degrees C. Then, the molecules rearrange themselves and they suddenly start to take up MORE volume. This if while it is still a liquid. The density change has undergone a reversal and just before freezing at 0 degrees C it is back to about the same density it had at about 8 or 9 degrees C. But then when it reaches the freezing point there is another rearrangement of those molecules and the density drops by almost 10%. This not only explains why ice floats on the top of water, but it also explains why only the top of a lake or pond freezes while the water below that layer of ice stays in the liquid state. Then, as the temperature drops even more, the density, once again increases. All of this is at a pressure of about 1 atmosphere and pressure does have an effect on this.

                    Perhaps carbon is a better example in that it can exist in different molecular arrangements at room temperature. Amorphous carbon has a density of about 1.8–2.1 g/cm3 while graphite is 2.267 g/cm3. But wait, diamond is way up there at 3.515 g/cm3, And pressure is one way to create diamond from the other forms of carbon. When the other forms of carbon become diamond, then the density increases and the volume DECREASES. So, YES, the volume can definitely decrease or increase depending on the molecular arrangement of the atoms and/or molecules.

                    When you are talking about changes in density under different circumstances, there is more than one physical process going on. Thus a change in volume across the three dimensions is not a simple thing to explain or to write an equation for. And a first order approximation may be very useful for many purposes even if it is not the ultimate explanation.

                    In contrast to what many people think, science is NEVER exact. There is always room for additional refinements. This is something that is often lost on many people and, strangely enough, that includes many scientists.



                    Originally posted by Bob Engelhardt View Post
                    The Poisson effect compression vs compression defined by the bulk modulus has me confused. The 0.001" strain on my 1" steel cube example produced a Poisson effect volume reduction of 0.0004 cu-in. That 0.001 strain would require (30 x 10^6) x 0.001/1.0 pounds stress (elastic modulus) = 30,000 lbf. To achieve a compression of 0.0004 cu-in with pressure on all sides, would require 25 x 10^6 psi (bulk modulus) x 0.0004/1.0 = 10,000 lbf.

                    Well, I guess that pushing on all sides takes less force than pushing on one side is not that confusing. What is still confusing is how the Poisson effect causes a reduction in volume at all. The material is elastic and unconstrained in 2 axes - why doesn't it just spread out without compression? The forces resisting deformation are greater than the forces resisting compression?
                    Paul A.
                    SE Texas

                    And if you look REAL close at an analog signal,
                    You will find that it has discrete steps.

                    Comment


                    • #25
                      As far as I know, there is no material that does not compress, i.e. reduce in volume/increase in density if pressure is applied to it on all sides. People would be very interested in any material which truly did not compress at all, under any circumstance

                      If it compresses at all, it compresses to some degree at essentially any pressure (although it may be impossible to measure at low pressures). If that is true, then when pressure is applied to the top and bottom of a solid sample, with the sides unconstrained, there must be a mixed effect with some displacement, and some volume reduction.
                      Last edited by J Tiers; 05-31-2022, 10:44 AM.
                      CNC machines only go through the motions.

                      Ideas expressed may be mine, or from anyone else in the universe.
                      Not responsible for clerical errors. Or those made by lay people either.
                      Number formats and units may be chosen at random depending on what day it is.
                      I reserve the right to use a number system with any integer base without prior notice.
                      Generalizations are understood to be "often" true, but not true in every case.

                      Comment


                      • #26
                        Molecules can rearrange themselves. And different arrangements can have different volumes.
                        Friendly reminder: there are no molecules in a pure metal, only atoms. Steel will have carbides and inclusions, but the iron atoms are in a crystal lattice, not a molecule.

                        Comment


                        • #27
                          What if your micrometer were sucked into a black hole, I reckon it would shrink, I jest btw
                          mark

                          Comment


                          • #28
                            Originally posted by boslab View Post
                            What if your micrometer were sucked into a black hole, I reckon it would shrink, I jest btw
                            mark
                            It won't shrink. It will get stretched out and pulled apart by the gravitational gradient.

                            Comment


                            • #29
                              Originally posted by boslab View Post
                              What if your micrometer were sucked into a black hole, I reckon it would shrink, I jest btw
                              mark
                              That happens every time I clean the shop.

                              Comment


                              • #30
                                Originally posted by tomato coupe View Post

                                It won't shrink. It will get stretched out and pulled apart by the gravitational gradient.
                                From what I have read about "black holes", that's true with regard to the process of entering one. But the bits and pieces would seem to have to end up occupying less space and being denser........... At least "space" as measured out here, which does not seem to translate well.........

                                That was going to be the argument..... that since density is huge in a black hole, everything is compressible, and so there would seem to be a mix of compression and deformation in the original question.

                                The counter argument is that "structure" is changed on the way in, so no material makes it in as what it originally was.
                                Last edited by J Tiers; 05-31-2022, 11:55 PM.
                                CNC machines only go through the motions.

                                Ideas expressed may be mine, or from anyone else in the universe.
                                Not responsible for clerical errors. Or those made by lay people either.
                                Number formats and units may be chosen at random depending on what day it is.
                                I reserve the right to use a number system with any integer base without prior notice.
                                Generalizations are understood to be "often" true, but not true in every case.

                                Comment

                                Working...
                                X