FEA Course Sample: Rigidity of Supports
This is definitely a topic we have to start with. It is so easy to simply implement support in the model. But in reality, such action means that you just have made a lot of assumptions!
I will discuss those assumptions briefly, and you will notice that those basic engineering concepts will come really handy now!
Support is… really, really rigid!
First of all, every support in your model is treated as infinitely rigid. Basically, you can put as much load on it as you want… and it won’t even move! Of course, nothing is “infinitely” rigid in reality. This means that this assumption is never met in real cases.
In most situations, this is not a big issue. A lot of elements are “rigid enough” that such assumptions don’t really influence the outcome. Sure when I will put a book on the table, the table will be a tiny bit shorter. This is discussed more in-depth in the basic engineering concepts module. In short, this happens because the table legs took additional compressive force from the weight of the book.
Also, the tabletop will deflect a little. This means that deformations (from the book’s point of view) would be even higher.
But does it really affect anything? No, not really.
Problems start when few supports in your model are not as rigid as others in reality. Depending on what you want to do this may not be important, or might be critical. Do you remember the gummy bears? There, when one leg was less rigid it took far less load.
This is not an issue in a typical model, in which you implemented all 3 legs (the solver will figure it out). Problems start when you want to “shrink” your model and use only the tabletop, treating legs as supports.
If you model supports instead of “full legs” things get tricky. Your model does not know that one support is from gummy bears! Solver will simply assume that all supports are equally rigid. This means that load in your model will be distributed in a different way than in reality!
This may be very bad for your model. I will show you this on a simple model.
We are getting serious: stones and rubber balls!
To show you why support rigidity is important I will use yet another simple model. Think about a board supported on 2 stones at the ends and a rubber ball in the middle:
First, let’s make few observations. If you would press the stone with a really big force it won’t deform at all. Well… maybe it will deform a tiny little bit, but we will ignore that here.
The rubber ball reacts to load a bit differently. It will shorten and that means that it won’t take as much load as it “should”. You can also calculate the rigidity of the “rubber ball support”. If you apply a force of F to the ball, and as a reaction, the ball will have a vertical deformation of D, supports rigidity is F/D. This is in [Unit of force] / [Unit of length].
So now think a bit about what can go wrong here. The simplest approach to this problem would be to simply model supports in all 3 points. In such a case you would get something like this:
This looks really reasonable, right? If I would attach the board to the stones it won’t slide. The ball can roll left and right so there is no horizontal support in the middle. But something is missing here, and that is the ball “rigidity”:
Stone deformation will be so small, that we can simply ignore it in the model. This is a common practice. It is important to take into account supports that are significantly less or more rigid than “normal” for our task.
Real example – no balls this time 🙂
It would be easy to finish here, but I want to drive the point home. Below you will see outcomes for a board from our example. I used a simple 100 x 20mm board and implemented several models with a different stiffness of the support in the middle. Board was 4m long (with support dividing it into equal halves).
Firstly let’s take a look at the deformations of a selected case:
Notice, that in the middle the board deflected, but not as much as if there was no support there at all. This is how a typical “elastic” support works. It is very easy to imagine that such action will have a significant influence on the design.
To see how support stiffness changes the outcomes, let’s compare the deformation in the middle (where the support is) for different values of K:
You can clearly see above that for K=1000kN/m rigidity is more or less irrelevant (since it is “close” to infinity). Please don’t draw any conclusions from this!
There is no “right” value of support rigidity that can be ignored. Sometimes 100kN/m will be “almost infinitely rigid” and sometimes 100000kN/m won’t be enough to treat it as a rigid support. Everything depends on your model and how much it will deform!
Of course, the deformation is not everything. Stresses also change in the model. Since in this case, this is a pure bending (with some shear that we will ignore) it is best to show the changes in the bending moment chart:
Notice how bending moment changes from a positive value (no support) to a negative value (rigid support). Also, bending moment distribution changed along the length of the beam. This is how much influence on outcomes support rigidity has!
Support rigidity and structural failure
It’s fun to analyze funny examples. But let’s strike a bit more serious note here.
As you could see above support rigidity can greatly influence the distribution of load in the model. We already discussed this with gummy bears examples in basic engineering concepts.
In reality, such a phenomenon can have drastic consequences. A few years back in my country there was a spectacular silos failure. What happened was, that someone supported a silo wrongly. Since the structure was supposed to stand over the railway track a frame had to be built that looked like this:
It is seemingly obvious that silo is supported on 4 points, as I marked in the top view. Unfortunately, the orange secondary beams were “weaker” and had higher deflections than the main beams. This meant they were made from “gummy bears” and didn’t take any significant load. Instead of 4 support points, the silo had only two… which obviously were overloaded. Support zones of the silo over the main beams failed, and the silo collapsed.
What is even more surprising is the fact that the difference in deformations of the primary and secondary beams wasn’t very high. But since the silo shell was almost undeformable in the vertical direction even a small difference was very significant.
What you can learn from this example is:
The higher the stiffness of the model the more rigid a support must be to be treated as “infinitely rigid”. For the board from the first example 1000kN/m meant that support can be treated as rigid. For a silo such rigidity (if onther supports are rigid) would meant that more or less there is no support at all!
There is not an easy way around it, but with experience you will learn to recognize correct rigidity values. Just start with precise analysis of few typical problems you will do – and you will gain the experience you will need later on!
How to deal with this?
If you have “elastic” supports in your model there are 2 ways you can follow:
- Firstly, you could include the rigidity of the support to your model. Simply input the value of support rigidity and do the analysis. Sometimes this will require a second model of the support to calculate the proper value of rigidity. You can also take it from tests or simply estimate it.
- Secondly, you could include “the ball” in the model. You can always model additional elements in the model that will deform. While modeling the “the ball” from the first example would be too much work, adding a beam that will deflect is only a few clicks away. It is not always quicker to estimate rigidity – sometimes implementing the elements itself is simply an easier choice.
As I mentioned above calculation of the “elastic” support rigidity is usually simple. You just need to know how much it will deform if you press with a certain force. Both deformation and force should be measured/applied in the direction of the support of course. If you don’t want to go into nonlinear rigidity (you don’t want to go there I think) you should measure it with the force that will be close to the final load in the model going to that support. Then divide the force by the deflection… and that is the stiffness of the support 🙂 To make it more visual I will once more use the drawing from above:
The second approach requires adding additional elements to the model – this may be useful (if you would like to check if the toy ball can even carry so much load) but sometimes it is uncalled for. Making a bigger than necessary model leads to problems as we discussed previously. Still sometimes adding one element is much less work than calculating the support rigidity. This approach has its uses without a doubt!
To be fair it doesn’t really matter which approach you will use… as long as you will use one of them!
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