
Rigidity of GAP elements in contact
GAP element rigidity will depend on the material of the parts in contact... and also on the mesh size! Learn how to calculate it!
12 December 2022Using deformations from linear buckling as imperfections in nonlinear buckling cases is the most common approach. I will use a simple example to prove that this is not always the best idea. Proper selection of imperfections is a very complicated process – I hope to shed some light on this matter.
As you know, I like to show complex problems on simple models. Today we will deal with a small shell with some stiffening. Such solutions can be often observed in silos structures in their support zones. Below you can see a model I will use (simplified for the task), and one of the silos I have designed with similar support (marked in red).
Such a shell is usually loaded with friction (from the material inside), but at the bottom part, we can almost assume that the load is vertical from the top. I used such an approach in my simplified model as you can see above. Model is made from 1mm thick steel plate with stiffeners and rings 10mm thick. Total height and diameter are equal to 1000mm. Support is 10 degrees wide (around 87mm).
I want to measure how imperfections influence the capacity of the model. This means I need a reference to compare to. The easiest is to calculate the capacity of the model without any imperfections and use this value. Below are the outcomes from linear (LBA) and nonlinear (GNA) buckling without any imperfections (follow the links to read more about the methods).
As you can see above, the linear buckling gave the multiplier of 0.6813 while nonlinear buckling is 0.4725. This means that the critical load is 200 x 0.6813 = 136.3 kN/m for linear buckling (200 kN/m is the load I have applied in my model). Note that the capacity drops in nonlinear buckling (without any imperfections) by 44% which is a big difference to start with!
To distinguish somehow the outcomes I will compare imperfection influence on linear buckling and nonlinear buckling separately on the same charts. You will see what I mean in just a moment 🙂
At first, let’s do a “classical” approach. Below is the deformation from LBA, which I have used as an imperfection “shape”. Firstly I have chosen the maximal imperfection amplitude to be equal to shell thickness (in this case 1mm).
I love this part! Note that we have applied imperfections, but capacity in nonlinear buckling… is actually 5% higher! This is the main reason, why using imperfections from LBA is not always the best idea. Look at the outcomes from models without imperfections, compare the geometries. Nonlinear buckling often produces deformations that are much smoother than those from LBA. Imposing linear buckling imperfections in nonlinear buckling case, sometimes strengthen the model, as the shell want to buckle in a different way and have to “fight against” imperfections from linear buckling case!
Imperfections in linear buckling reduced the capacity… but note also that the “shape” of stability failure changed as well!
Since in nonlinear buckling shell want to fail in a certain way (with certain deformations), let’s help 🙂
Here I use deformations from nonlinear buckling (I have chosen one of the increments). They are scaled just as in the case of linear buckling: maximal imperfection amplitude is 1mm. This leads to the following results:
This time both capacities in linear and nonlinear buckling dropped. What is more interesting is that in linear buckling it dropped even more than when imperfections from linear buckling were used. This is of course just a coincidence, you can never be sure which imperfections are “correct” in your case!
I hope I have managed to convince you to use different sets of imperfections. Please remember that deformation shape from linear buckling is not the optimal and only choice… I know a lot of people use this as the only case, but this is not the best approach. Do not forget that by implementing those “linear” imperfections, you can actually strengthen your model, just as I have shown here.
Let’s take a look at the typical procedure here:
If I would perform linear buckling (r = 0.6813), and then implemented the imperfections and perform nonlinear buckling, I would got r = 0.4939.
Value is smaller as expected, so I would be happy.
While in fact, the capacity is much smaller. The simplest approach shows capacity reduced by 11%, but I haven’t searched for the minimal outcome. It can still be smaller!
This is why code EN 1993-1-6 when you want to perform nonlinear buckling demands, that you use the worst imperfections. You will never know which set is the worst one until you use several different sets and compare them. It is also required to check smaller and higher amplitudes, as bigger imperfections do not have to be worse… but this is a topic for another post 🙂
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