44 Replies Latest reply on Jun 15, 2015 11:13 AM by Alan Auger

    What´s wrong here?

    Alan Auger

      Hi I'm Alan, from Argentina... I'm having a little issue when I try to simulate a simple design. I will post some pics to be more clear.

       

      Here is how a fixed the piece and where is the force acting (12000N), it's a simple traction problem.

      Sujecion.pngvonMises.png

      Tensiones normales.png

      As you can see the max stress is about 2000 MPa in both pictures. The tensile stress to break AISI1045 is about 785 MPa (so it would explote haha), but this piece is real, have the same dimensions, same material, and it break's with a force about 17000N with the same fixture and point of aplication of the force.In the simulation it seems that the piece will break with 12000N... but in the real world this isn't true.

      So, what am I doing wrong?

       

      Thanks!

        • Re: What´s wrong here?
          Alex Henry

          Hi Alan, Interesting issue. The part seems simple - if you can provide dimensions and forces, I can run the same here and compare. Also you can go to your reseller...

          • Re: What´s wrong here?
            Mikael Martinsson

            Hi there.

            A couple of things.

             

            First of all you use a linear static simulation, with a linear material model, so all stresses above the material yield level is incorrect. You can't use this kind of simulation to predict failure. If you go for a non-linear simulation, with a correct material curve, you'll find stress above yield, but probably also below UTS with 12kN.

             

            Secondly, your restraint is not logical. The area where you added the force are allowed to deform, but the area where you added the fixtures will be completely rigid. You could go for a quarter symmetry solution, or perhaps a 3-2-1 minimum restraint with balanced loading on the complete part. You will then see a symmetric pattern for the stresses.

            • Re: What´s wrong here?
              James Riddell

              Alan,

              Your restraints are not realistic if I understand the application (i.e. buckle for web strap or some such).  I'd suggest trying equal and opposite forces on the two surfaces you are currently using, put a slider/roller on either the 'top' or 'bottom' (largest) surface and a linear restraint on one of the smaller ends, maybe even only a couple points, just to keep it from running away.

               

              IDK if that will give you the answer you are looking for but at least it will deflect more realistically.

              • Re: What´s wrong here?
                Mikael Martinsson

                Here you see the solution with 3-2-1 restraint and balanced loading.

                White areas are above yield strength of 530MPa (according to SW database, not sure if this is correct)

                 

                VonMises.PNG

                • Re: What´s wrong here?
                  Alan Auger

                  Wow guys, thank you very much, it's true I should balance the load for a simple action and reaction fact. In a couple of hours I'll do all the modification you told me and see if I can get a confirmation to the tensile stress to break of 1045, but I'm pretty sure it's around 785 MPa.

                  Thanks again!!!

                  • Re: What´s wrong here?
                    Alan Auger

                    I'm having problems to do the 3-2-1 restraint, how did u do it guys?

                      • Re: What´s wrong here?
                        Mikael Martinsson

                        Hi.

                        Balanced loading with minimum restraint (3-2-1) you can see in my picture below:

                        I lock 3 points in the z-direction, 2 points in y-direction and 1 point in x-direction

                         

                        This setup is useful in some situation, where the part should be able to deform freely.

                        For this example though, the best solution is a 2d Q-symmetry setup as Joe suggested.

                         

                        3-2-1.PNG

                          • Re: What´s wrong here?
                            Joe Galliera

                            @Mikael Martinsson, thanks for the vote of confidence. 

                             

                            I completely agree that you want to have just enough restraints to be able to run (to create a unique solution), and not so much to overconstrain the model.  It's a tenuous balance to be able to master, so anyone who says they know, they don't.  But you can start with a lot of great examples.  And this is a very good one.

                             

                            If working with a geometry like yours, then it naturally lends itself to exploit symmetry wherever possible.  These symmetry conditions create natural boundary conditions that don't over restrain the body.  If you've taken SW Sim training, you may know the first example of the hole in the plate (find the training model files online).  Take a look at this problem again and ask yourself, how could I run this problem using symmetry?  (Hint, the restraints provide an equal and opposite reaction force to balance out the applied load.  Same as in your original setup.)  I mentioned the 2D as an added bonus, which again you pick up when you're old school like me and the computers were not as sophisticated back when I got started.  But you know what, we still use it all the time for Nonlinear problems because these generally take much longer to solve than simple Static linear studies.

                             

                            [Notes on symmetry:  Symmetry is not defined by geometry alone.  It also depends on the loading and (to a far lesser frequency) material properties.  Think to what goes into the equation: {f} = [K]{x}.  When confronting a nonlinear problem, especially in CFD / fluid dynamics, you can have all the pre-run clues that the result may be symmetric and it will not be, e.g. flag in the wind.  So remember, symmetry of results is always king.]

                             

                            The key to it all is that you know what the answer will be beforehand, and then you can set up the problem to give you that answer.  Doesn't make sense, does it?!  What the software does for you though is to provide you the detailed numbers, which you cannot do in your head.  So in conclusion, experience is very helpful, so keep on practicing.

                            • Re: What´s wrong here?
                              Jaja Jojo

                              Hello sir im new in solidworks simulation how did you set up this 3-2-1 restraint im trying but its fixed all the direction

                          • Re: What´s wrong here?
                            Alan Auger

                            Thanks Joe and Mikael, learning a lot here...

                            But now that I could get the right stress distribution, I find myself in the first problem again haha.

                             

                            The part seems to fail (break) in the white area, where the stress is above 785 MPa (with 12kN), but the real part can take a 17kN load... I still don't understand why in the simulation fails...

                            IMG_20150611_124513840_HDR.jpg...

                              • Re: What´s wrong here?
                                Mikael Martinsson

                                Well, as I mentioned in my first post in this thread, you have to consider non-linear material in a non-linear study when you want to investigate stress/strain above yield for a ductile material. Linear static simulation results are only valid up to the yield strength of the material.

                                 

                                I don't have the non-linear package, and I also don't know the exact stress strain curve of the material on the real part that you're showing. This type of simulation raises the bar regarding your knowledge in simulation and in materials. But if you have the non-linear module, and have a stress / strain curve from the test that you're referring to, you could perhaps give it a try. Remember to recalculate the stress from the test curve because it will probably show engineering stress/strain but the input for a plasticity von mises material in SW is engineering strain and true stress. Formula for true stress is: σtr = σe (1+ϵe)

                                 

                                If the part that you tested is AISI1045, and the yield strength is 530 MPa, as set in my scale, then we now already that this part will go inte plasticity, and it will have a permanent deformation and residual stress when you release the 12kN load. We don't know if it will break due to overload, but it will probably fail with repeated loading of 12 kN due to fatigue. For many applications the yield limit is the design limit and you also want to add some safety factor to that limit. So in some ways I agree with your concern about the results. Is the part really designed for that load?

                                 

                                More information on the product, material and application for the product will probably be necessary to understand the results.

                                  • Re: What´s wrong here?
                                    Alan Auger

                                    We don't manufacture this part with 1045 anymore, today we are using 15B30 steel... but we used to produce the part and several other parts like this one with 1045, and the behavior is practically the same with the two different materials. That's why I simulate it with 1045, because the material is in the SW's library.

                                     

                                    The thing is that wen you do the math in the paper it doesn't break at all... not even enter into the plasticity period, with just 12kN, so I don't think this is happening because a non-linear behavior if I'm not even close to the yield strength. Now when we load it with +17kN we get the plasticity, the break, the part fails... in the reality and in the papers.

                                     

                                    I post a pic of the calculation answering James down here.

                                      • Re: What´s wrong here?
                                        Mikael Martinsson

                                        But this isn't exactly a simple rod in tension (F/A). If you try to simulate a simple rod in tension with the dimensions we have, you'll find that the simulated stress would be very close to the calculated stress. You mentioned that by 17 kN the part brakes. With the simple F/A caculation the part is still below yield strength at 17kN so it wouldn't break nor yield if that was the case.

                                         

                                        Steel is steel, with a modulus of Elasticity of between 200-210 GPa so in static linear simulation you can more or less choose what ever steel you want. Or for that matter, if you'r only interested in stress, not displacement, and the level is below yield you could use aluminium or some other linear material in the database. The stress would be the same.

                                         

                                        If you heat treat steel, the yield and UTS limit will raise, so the material can handle higher loads without yielding or breaking. So therefore my question about material to understand the result from simulation. You now mention that the material is tempered to 40 Hrc and that corresponds to approximately 1250 MPa UTS. This will also raise the yield limit from the 530 MPa for a cold drawn material up to about 1000 MPa for the tempered, so that is the answer to your original question.

                                          • Re: What´s wrong here?
                                            Alan Auger

                                            Ok, it might be not as simple as a F/A problem... and supouse the part is tempered and quenched to 40 HRc... that give us a 1250 MPa UTS... but the part still breaks with just 12kN

                                            break.png

                                            Even if it's tempered and quenched at 46 HRc that give us 1530 MPa UTS, so it will be break at the 2000 MPa point.

                                             

                                            So this is happening because of a non-linear behavior of the steel? Solidworks doesn't realize that the steel doesn't act in a linear way up to the yield limit?

                                            How can I design a part if I can't see if it's really going to break?

                                             

                                            Thanks again Mikael, you are very kind taking the time to explain me these things.

                                              • Re: What´s wrong here?
                                                J. Mather

                                                Alan Auger wrote:

                                                 

                                                ....... but the part still breaks with just 12kN.......

                                                Just out of curiosity - how and where is SolidWorks static FEA indicating the part will break (at any load)?

                                                  • Re: What´s wrong here?
                                                    Alan Auger

                                                    If the ultimate tensile strength is about 1250 MPa, and I have a +2000 MPa in the red zones that makes me think it will break...

                                                      • Re: What´s wrong here?
                                                        J. Mather

                                                        You must have missed this response by Mikael early in the discussion.

                                                        "First of all you use a linear static simulation, with a linear material model, so all stresses above the material yield level is incorrect. You can't use this kind of simulation to predict failure."

                                                         

                                                        I don't think your results have any meaning.  SolidWorks certainly isn't indicating "fracture" (because it can't give that result with this analysis).

                                                        You might need to change your type of analysis and define your (software) criteria of "break" or "failure".

                                                          • Re: What´s wrong here?
                                                            Alan Auger

                                                            Definitely, Mikael was right in the first place and he told me this some days ago, but since I didn't understand how the software was working I couldn't introduce that concepts in my head I was going in the wrong way, as you say the results I'm having here are useless because I must design to the fracture, not to the plasticity period. The part can warp, but it must not break, so I have, as you and Mikael said, change my criteria because I'm not going to design up to the yield limit hahahaha.

                                                            Thanks again to all you people, and excuse for my slow learning... but I'm just doing that, learning.

                                                             

                                                            Now I'm using the non linear static simulation, but I need the material data... I need the Stress-Strain curve, I've tried to get it from the matereality web that solidworks offers you but when I choose Non linear study - Plasticity Von Mises in material model and I choose Metal it tells me that there not results found with all the database available. It founds many plastics, but no metals... any idea on how can I get the material data?

                                                             

                                                             

                                                              • Re: What´s wrong here?
                                                                J. Mather

                                                                The first question I would ask is, "What do you want to know and why do you want to know it?"  And that isn't a trivial question.  A lot of times when I ask someone that question after a bit of discussion they could not have answered the question prior to the discussion.  (although they thought they did know the what and why)

                                                                  • Re: What´s wrong here?
                                                                    Alan Auger

                                                                    What I want is to understand how to create a simulation that represents the real behavior of the part in the real world designing up to fracture, many times parts can have deformations, but they musn't break. Why I want to know this? Because if I can create that scenario I will have a trustable know-how that shows me what is gonna happen in the real life in other design working above the yield limit or even up to it.

                                                                      • Re: What´s wrong here?
                                                                        J. Mather

                                                                        We can work this through step-by-step

                                                                         

                                                                        Alan Auger wrote:

                                                                        ....design working above the yield limit or even up to it.

                                                                         

                                                                        This could be a matter of language barriers, but this phrase doesn't make sense. 

                                                                        Yield occurs when the part goes from elastic deformation into plastic deformation. 

                                                                        So I would expect the phrase to read, "Design up to yield limit or even above yield limit."

                                                                         

                                                                        Of course we often work above yield limit, work hardening parts for example.
                                                                        The ultimate strength of a part might be well above yield strength.

                                                                          • Re: What´s wrong here?
                                                                            Alan Auger

                                                                            Haha that's right, please excuse, my english isn't that good, and talking about yield limits, ultimate tensile strength, etc. It just makes it more difficult, because I know those names in spanish, know I know them in English too haha

                                                                             

                                                                            You are right I need a trustable source to design up to the yield limit or even above the yield limit.

                                                                             

                                                                            We should continue the post at the bottom of the topic because some will loose the topic thread...

                                                      • Re: What´s wrong here?
                                                        James Riddell

                                                        OK, I don't get it, what's the big deal here?  You say you have AISI 1045 steel.  The load it can handle is 17 kN (3822 lbs) - (I'm from the only country that still uses English units and is still the only one that has landed a man on the moon.)  The cross-section area is 0.054 sq. in.  Simple math shows that to be 70,778 psi.  Yield strength for 'quenched and tempered to 595 HB' is 183 ksi.  Cold drawn is 77 ksi.  That should be the end of the story or am I missing something?

                                                          • Re: What´s wrong here?
                                                            Alan Auger

                                                            Hahaha, I hate conversions, but you are right, in paper all seems to be beatiful and works fine...

                                                            IMG_20150611_170204346.jpg

                                                             

                                                            BUT there are two big diferences:

                                                            1) The part is quenched and tempered to 40HRc, that is about 369HB, so the yield strenght isn't 183ksi = 1260MPa

                                                            2) We don't manufacture this part with 1045 anymore, today we are using 15B30 steel... but we used to produce the part and several other parts like this one with 1045, and the behavior is practically the same with the two different materials. That's why I simulate it with 1045, because the material is in the SW's library.

                                                             

                                                            PS: Do you have any table that tells you the yield strength vs the hardness of material?? That would be beatifull, the ones I have tells the yield strength to a particular NQT temperature.

                                                              • Re: What´s wrong here?
                                                                Bill McEachern

                                                                I didn't read this whole thread so forgive me if this has all been covered......The failure mode needs to be considered - when you harden up the material the loss of ductility could lead to a change in mode from ductile failure to brittle fracture which could happen at a lot less than the bulk average yield stress of the whole cross section. It could crack and then rip like glass.  Failure is tricky and SWX Sim at best can give you an indication but you need to know what you are looking for. If you really want to model fracture you typically need (and there are other approaches)  an explicit code with a Johnson-Cook plasticity and failure/fracture model with element deletion and even this sort of elaboration requires extensive testing to define the material parameters and it is still considered approximate.........If it's brittle fracture consider  when P1 hits the UTS stress in say a notch and away it goes.

                                                                  • Re: What´s wrong here?
                                                                    Alan Auger

                                                                    Thank for the appreciation,I know what you are trying to say, but lets be honest, is just 40HRc, not 50-54 HRc, I know it's wrong to say, but the curves of the material might be something like this...

                                                                    IMG_20150613_243106803_HDR.jpg

                                                                    But I think this isn't the real issue here, the failure can be more brittle, yes that's right. But both have a non-linear behavior as Mikael said before.

                                                                    I'm a newbie with FEA software, but it's hard for me to believe that the software can't give me something a little bit more "real". I mean, I know it can do it, but I didn't understand how yet.

                                                                • Re: What´s wrong here?
                                                                  Mikael Martinsson

                                                                  Well, at least you drive on the right side in your country. Conversions are easier to handle than left lane traffic. But are you sure about the moon thing? Some peoples with tin foil hats would argue that this was a Hollywood production.

                                                              • Re: What´s wrong here?
                                                                Roland Schwarz

                                                                You're using a fixed faced constraint on a face that covers a significant percentage of your part. This exists nowhere in nature or lab. Same for your application of force.

                                                                 

                                                                If your model does not reflect reality, your model is simply wrong.

                                                                • Re: What´s wrong here?
                                                                  Alan Auger

                                                                  Ok I'm making some progress here . I' m trying to make a nonlinear analisys with 2D simplification.

                                                                   

                                                                  Please tell me if you see something wrong in my analisys. First the material, the type is plasticity Von Mises and I have the stress-strain curve of this material, it's not 1045, but it's one of the few materials that has the stress-strain curve in Solidworks database. It will work at least for now until I understand how to make a right simulation.

                                                                  material.png

                                                                  Now the restraints. Based on the 3-2-1 restraint Mikael told me before I made something like a 2-1 restraint since I don't have Z direction. A point has X and Y restraint and the other one just in X direction. Here is where I have my doubts, is this restraint right?

                                                                  2-1 restraint.png

                                                                  And here are the results...

                                                                  stress.png

                                                                  Please guys if you see something wrong tell me...

                                                                   

                                                                  PS: you won't believe me but I just was talking with my dad and he told that the part can take a 44kN load in the laboratory tests... and guess

                                                                  what... F/A=44000N/34,76mm2=1265MPa and the the real material has around 1250-1290MPa UTS depending of the tempered and quenched. Please sorry for the huge mistake...

                                                                    • Re: What´s wrong here?
                                                                      James Riddell

                                                                      "Father Knows Best"  Sometimes it's a good idea to talk to the old guys.

                                                                       

                                                                      That being said, the set-up is not all that realistic.  I'm sure that the force is not applied completely uniform and strictly along the flat surface, the restraints likewise.  Until you have the exact conditions you won't get exact answers.  It has been my experience that static linear analysis only indicates where failure might occur and is better if you use it to demonstrate that you have an 'X' factor of safety that the load will won't produce any stress levels near to the YS of the material.

                                                                       

                                                                      Funny thing, decades ago we designed many things that are till holding together with F/A analysis.

                                                                       

                                                                      There is another thread that points to the lack of exactness of the material properties in the SW database.  While I & nu aren't much changed (unless you believe a certain "peer-reviewed" paper from Nigeria).

                                                                        • Re: What´s wrong here?
                                                                          Alan Auger

                                                                          James, thanks for the answer... The loads are applied along all the surface, because the test is made with devices that holds all the width of the two holes you see in the part. Respect to the restraints I was hopping if somebody has a better idea, since the restraints in the test are made by those devices that locks the parts in X and Y directions, and pull in Z direction.

                                                                          The material I choose is just for practise until I get the real stress-strain curve of the material we use here, otherwise I can make some tensile specimen, (I don't if it's right to write that, haha) run some test in the university and get the curve I need and edit a customized material in my SW database.

                                                                           

                                                                          However I can design this part with a F/A analisys I really want to know how to simulate this right, because once I get it I have to design more complex parts that won't be as easy as a F/A problem.

                                                                            • Re: What´s wrong here?
                                                                              James Riddell

                                                                              So your test set up is sort of like a metal hook in each side connected to your tensile test rig?  In that case the constraint could be the 1-2-3 but should work equally well if you merely restrain one side with roller/slider and a no-lateral-movement on one of the 'edges' (thickness of the link on either side). 

                                                                               

                                                                              In the end, if you are trying to 'make' it fail, the linear static is not going to work at all.