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Different reaction forces

Question asked by Stefan Tenev on Sep 26, 2019
Latest reply on Oct 29, 2019 by Jay Seaglar

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Hi, all. I have tried to simulate load of a shaft. I use 2d simplification and set two forces 51250 N per item. I have three fixtures. The shaft is d=300mm. You can see that i have R1=17400 N, R2=67700 N and R3=17400. When i set d=10mm, the reaction forces are changing like R1= 16000 N, R2=70500 N and R3=16000 N. I have calculate by hand the reaction forces of beam like that shaft i received reaction forces near to R1= 16000 N, R2=70500 N and R3=16000 N.

Why i have different reaction forces when i change to d=300mm?

Stefan Tenev Oct 29, 2019 9:33 AM

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Hi to all. Thank you to everyone for replays. I have found answer of my problem (with little help of prof. Akin). The mathematics equations of the model in SolidWorks was calculations of stiffness matrix with counting of shear deflection. A variation of reaction force of model in SolidWork and mathematics calculations is 0.3 %. That is perfect. Once again thanks to everyone.

P.S. If someone want to see calculations, i can upload it on pdf.

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Steven McCallion Sep 26, 2019 5:24 AM
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Hi Stefan,
What is one of the first and most important assumptions you make when looking at beam theory?
Hint, what answer do you get if you scale the length of the beam by how much you increased the diameter i.e. 300/10.
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Ananda Ganesh Madheswaran Sep 26, 2019 7:14 AM
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hi
whatever may be the size the reaction forces will not vary..
pl look at the input and restraints
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Stefan Tenev Sep 26, 2019 7:59 AM
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Thanks to all for reply. Ananda is right. If in three moment equation rise length with 300/10 it is the same. My input and restrains are ok, but i still get difference. I have tried to get reaction forces in AutoCad. AutoCad give me exactly same results that i have made by hand. My input is length=1500 mm, diameter=300 mm, first F=51250 N is set to Z=-375mm(1500/4), second F=51250 N is set to Z=-1125 mm. Some other suggestions?
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- Steven McCallion @ Stefan Tenev on Sep 26, 2019 8:15 AM
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  At 300mm it is not a beam, your hand calcs are wrong for the 300mm case because your hand calcs assume that it is a beam, which it is not.
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- Tim Boerschig @ Stefan Tenev on Sep 26, 2019 8:32 AM
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  I think what Steven is suggesting is to consider whether assuming simple beam theory applies to the short, large-diameter shaft (e.g. consider the difference between Euler-Bernoulli and Timoshenko beam theories).
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  - Steven McCallion @ Tim Boerschig on Sep 26, 2019 8:54 AM
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    But then he's getting different answers in autocad so I don't know what's going on. Depends on how he's doing this in autocad.
    Stefan, what type of elements are you using in autocad?
    I'm not sure Timoshenko will still handle cases like this any better, it's not developed to handle local deformations any more than EB right?
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    - Tim Boerschig @ Steven McCallion on Sep 26, 2019 9:38 AM
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      I didn't mean to suggest Timoshenko was appropriate as we are really focused on reaction forces at the moment -- only that the geometry of the large, short diameter shaft may not really act like a simple beam in certain situations. Perhaps the difference may be in the contraints(?) but I'm not sure.
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Ryan Navarro Sep 26, 2019 11:03 AM
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I agree with others that the beam bending hand calcs likely aren't valid at the larger diameter size you have pictured. I assume you are using Axisymmetric 2D symmetry since it is a circular shaft

It makes sense that it would roughly line up with the hand calcs at D=10 because your geometry is closer to the assumptions in which you would typically consider beam theory acceptable.

Both D=10 and D=300 problems sum up to the same total Y reaction force so it passes that check.

I think what you're seeing is your D=300 mm problem is likely more accurate in FEA for your given boundary conditions

AutoCAD is also likely using 1D beam bending theory for it's calculations so that's why the results line up with your hand calc. If you want to do the same thing in SOLIDWORKS Simulation you could setup a problem using 1D beam elements.

You will likely get a notice that the member is too short to be considered a beam but I'm guessing the results would line up closer to your hand calcs.

There are some good validation examples built into the software under the pull down menus, Help -> SOLIDWORKS Simulation -> Validation -> Verification Problems and NAFEMS benchmarks

Verifying the results is a great step to take and I'd encourage you to continue to do that. But in this case the only way I would validate your D=300 case would be against another FEA code.

If you can set it up you could run a 3D Solid Mesh simulation to rule out any problems with the 2D simplification.

There could be a problem in your boundary conditions too, that becomes more prominent at the larger shaft sizes.
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- Bill McEachern @ Ryan Navarro on Sep 26, 2019 10:35 AM
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  The issue it probably in the linear FEA. Turn on the large deflection option and the equilibrium iterations will probably make them the line up assuming the problem isn't in the next sentence. The other thing if you want good reactions you need to pick single nodes and you need to have it kinematically constrained to match the beam calc. Otherwise it is over constrained the individual reaction will be proportional to the stiffness to the restraint. I take it hey all sum up to the applied load in each case? I did not look to carefully so please ignore if I have completely mis understood the issue.
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- Tim Boerschig @ Ryan Navarro on Sep 26, 2019 3:16 PM
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  Just a minor point, as I don't think it whould explain why Stefan is getting different reaction forces, but I don't think he is using simplified 2D Axisymmetric linear study type for his simulation but perhaps he is using a simplified 2D Planar study (at a constant depth/thickness?) -- otherwise I don't know how he would get a reaction force at all. I agree though that running a 3D Solid Mesh simulation may help rule out any problems.
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Stefan Tenev Sep 30, 2019 5:14 AM
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Hi to all. Sorry for my lite reply. I will answer to the questions.
Steven: I am using a AutoCad Mechanical 2020. I did two different simulations - first is with d=10mm and second is d=300mm. You can see that we have a equal reaction forces and bending moment. Difference is stress, deflections and angle of elastic curve of the shaft.
My hand calculations have a same results for reaction forces, bending moment and deflections (on both).

Now the SolidWorks calculations for d=10mm are same.
But when i change to d=300, i get:
We have difference over 1400 N.
If we see the beam theory, when we have 17400 N reaction force, that means a bending moment in middle point is drop down (bending moment is minus). Why is drop down?
When i change the length with 300/10 (Steven you are right), i get:
In short shaft with big diameter, we have different deflection of elastic curve ( w=wm+wq+w0), but in long shaft (w=wm).
wm-deflection of bending moment
wq-deflection of shear forces
But I think the reaction forces have to be the same. (like calculation in AutoCad and by hand).
Even i simulate 3d model in solidworks d=300mm L=1500mm - reaction force is 18767 N (?).
Ryan: No, i am using Plane stress, but with Axissymmetric it is 19057 N (?).
Bill McEachern: On large deflection option is even more differnet 18767 N.
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- Steven McCallion @ Stefan Tenev on Sep 30, 2019 8:17 AM
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  Hi Stefan,
  I can't directly see what elements you're using in AutoCad, but as Ryan mentioned it's most likely beam elements, and considering the images you've posted they most certainly are beam elements. What this means is that the underlying math used by the autocad simulation is making the same assumptions as you are when you're doing it by hand. Hence the same answers.
  
  This comes back to what assumptions are being made when one uses beam theory to calculate the reaction forces. SolidWorks' own help files state this pretty clearly "For the beam formulation to produce acceptable results, the length of the beam should be 10 times larger than the largest dimension of its cross section."
  
  Page 211 of the following pdf gives the best summary of what's going that I've found so far:
  http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_I/BookSM_Part_I/07_ElasticityApplications/…
  
  Your 300 to 1500 ratio is right on the edge of what is acceptable to be using beam theory. It's very much a rule of thumb, but I have seen 1:4, 1:5 and 1:10 kicked around.
  
  On a more intuative note (I hate that word in the context of engineering, everything is relatively intuative...), imagine if your beam was 1,500 long, but 100,000,000 tall, thus no longer a beam, but rather a tall column sitting on those three points. In this extreme case, if you pushed down on the top of it, would you expect your beam calculations to still hold? Personally I'd expect the force to smear out as it moved down the column and distribute itself much more evenly, nearly 1:2:1, over those three supports.
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Stefan Tenev Oct 29, 2019 9:33 AM
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Hi to all. Thank you to everyone for replays. I have found answer of my problem (with little help of prof. Akin). The mathematics equations of the model in SolidWorks was calculations of stiffness matrix with counting of shear deflection. A variation of reaction force of model in SolidWork and mathematics calculations is 0.3 %. That is perfect. Once again thanks to everyone.

P.S. If someone want to see calculations, i can upload it on pdf.
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- Jay Seaglar @ Stefan Tenev on Oct 29, 2019 3:57 PM
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  In case anyone else has a similar concern in the future, please also take a look at the following SOLIDWORKS Knowledge Base Solution:
  
  S-071982: "In the SOLIDWORKS® Simulation software, why is the deflection obtained for my beam not equal to the result from my manual calculation?"
  
  The solution explains that in SOLIDWORKS Simulation, Total deflection = deflection in bending + deflection in shear + deflection in torsion, whereas hand calculations typically neglect shear deflection. The solution also includes sample models and spreadsheets as file attachments.
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