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4 minutes read
15 August 2017

Symmetry in FEA may be dangerous

4 minutes read

Symmetry in FEA is great, as it reduces the size of the task. This, in turn, shortens the computing time. Computing time is still a big issue, especially when you run complex models. Without a doubt, this is why using symmetry in FEA is so popular. There is a catch to this, however, as you make some assumptions while defining symmetry. Those assumptions may lead to incomplete outcomes of your analysis.

What is symmetry in FEA

Simply put symmetry is a set of boundary conditions that make part of your model work as if you have modeled the whole model.

This is almost always explained in a classical example of a beam loaded with 2 forces:

Symmetry in FEA - a classical example

You can clearly see here, that the model is symmetric, which means that if you split it in half (as shown with the axis) you get 2 mirror images. Such cases allow for use of symmetry, and the definition of it can be derived simply by observation.

You can notice that in the middle of the deformed shape there will be no rotation at all, but it moved downward along the symmetry line/plane. This is enough to see, that symmetry boundary conditions will look like this:

Symmetry in FEA - classical example with BC

With such a definition, you can simply solve half of your model, and the outcomes will be correct! This is a great benefit of symmetry, as it can greatly reduce model size.

Where is the catch?

Of course, there must be something, as the title suggested right? What I was silent about before is, that in the example above not only the model and loads were symmetric. Also, the outcome was!

This is an issue that can be easily overlooked. Symmetry in FEA will always produce a symmetric outcome, simply because well… it’s symmetry!

At first sight it may seem that this is not an issue. After all, if I have a symmetric model, with symmetric loads and boundary conditions the outcome must be symmetric right? Unfortunately, that is not always the case! The fact that in a lot of cases it is true, makes this even easier to forget!

If you analyze vibrations, buckling, or any other analysis that may produce an unsymmetrical outcome, you should especially remember about this!

Small example

I will show you how this problem works on another well-known problem: Euler’s case!

We will use Linear Bifurcation Analysis (LBA) to calculate the eigenvalues of the problem. After some simple modeling, I was able to obtain the first 5 eigenvalues of the beam:

You can easily notice, that the task is actually symmetric. So now let’s reduce it, to see what will happen:

Notice, that forms 2 and 4 from the full model is “gone”. The values for the forms I got are correct, and correspond with the forms I got in the first model. However, some of the forms are not there at all!

Take a closer look at Forms 2 and 4 from the first model. They are not symmetric! In fact, they are antisymmetric which is possible to get in symmetric models.

This is why symmetry in FEA may be dangerous:

It will show the correct answers, but will hide from you some forms of buckling or vibrations etc. Biggest problems appear if you miss something that is important, like that vibration frequency that you were seeking or buckling form you were not able to counteract.

Last few words

As you can see, from time to time, symmetry can work against you. But this doesn’t change the fact that it is an awesome tool, that should be used whenever possible.

Just be aware of its limitations, and you will do just fine!

If you enjoyed this post, definitely check out my FREE online FEA course. You can sign up below.

Author: Łukasz Skotny Ph.D.

I have over 10 years of practical FEA experience (I'm running my own Engineering Consultancy), and I've been an academic teacher for a decade. Here, I gladly share my engineering knowledge through courses, and on the blog!

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Comments (9)

Mahdi - 2020-07-11 09:48:07

( for static analysis )

Mahdi - 2020-07-11 09:38:54

Hi, great post again :)

I was wondering : is reduction of a model using its symmetry still work when this model is nonlinear ?

In another words : is the proof uses linearity assumption ?

Thanks in advance !

Łukasz Skotny Ph.D. - 2020-07-12 19:03:55

Hey Mahdi!

No, you don't have to assume linear conditions - I use symmetry in some nonlinear problems as well :)

All the best!

M Asghar Bhatti - 2017-08-25 19:29:16

The title is little misleading. It should be "Do not use symmetry with modal analysis". This is what I always stress when I teach structural dynamics.

Łukasz Skotny Ph.D. - 2017-08-25 21:51:18


I think you are partially right. I came from stability background and there you can have issues just as easily. I wanted others to read it as well, hence the title. Still, this title might be considered misleading - I will try to do a better job from now on. Thank you for pointing this out!


Rob Campbell - 2017-08-15 11:32:25

I understand that symmetry can be used with asymmetric loads in some cases. I'd be interested in learning more if this is true.


Łukasz Skotny Ph.D. - 2017-08-15 18:51:02

Hey, Rob!

I'm not sure about that but maybe this is a language issue. Do you mean antisymmetric (I'm not sure what asymmetric means)...

If you mean loads that are not symmetric I would say that if the differences are small symmetry might be acceptable, but to be honest I haven't seen symmetry used in not symmetric loading. Maybe if you have a big model and symmetry is in the middle while loads are at the edges (so they do not influence the model in the middle much). In such case, I would say this is reasonable.

Also if you have loads that can go from 2 directions (like wind) you can say that you will implement "one direction" and ignore the other since the structure is symmetric - that would make sense, as long as you are careful in post processing.

Antisymmetry is also possible of course.

I hope this helps a bit :)

Rob Campbell - 2017-08-15 11:31:17

Great post, as always. The limitation on using symmetry to calculate vibration modes applies by extension to dynamic analyses (shock and vibration), since calculating these modes is fundamental step in their solution.

Łukasz Skotny Ph.D. - 2017-08-15 17:29:05

Hey, Rob!

You are of course right :)

All the best


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