Automatic Buckling lenght RSA calculation

[seifer92]

Hi all,

I have some question regarding how RSA calculate buckling length with automatic procedure. I know that it assumes the hypothesis of EN 1993, annex E, however I do get some problems when I perform my analysis. In the picture attached I extrapolated a part of my structure, where you can see, basically, my constraint. The beams are fixed to the column (that is continuous in its length) too in both direction. Now here's my first questione. When I use the automatic procedure with the column fixed at the base, I get similar results with automatic one and buckling coefficient with 6 members. Now, is it correct to not assign entry n°3 and 6 (the one regarding the column above and below) if my column is continuous over the storeys?

When, instead, I choose to use an hinge at the bottom of my column, with the automatic procedure my column is unstable along z-axis. Using the buckling coefficient with 6 members, the column is stable again. Is it a problem regarding the fact RSA wants a fixed column when using automatic procedure or maybe is it linked to the fact my column is continuous? (Indeed I also tried splitting my column where the beams join and the result is correct both with fixed and hinged condition.

Thanks in advance for any answer 🙂

[Simau]

@seifer92 

Please share your model and specify the bars number you want to design

[seifer92]

Hi Simau,

thanks for the answer. Here's the model, the column is n° 7. I created two members definition, "colonna_new" and "colonna_new2". The first one uses automatic definition, the second one 6 members definition for buckling length calculation. You can see that, with the same hinge constraint at the bottom, using the first defnition will result in an unstable column, while the second one will give a stable result.

[Simau]

@seifer92 

I splitted the column in 2 parts and created 2 kinds of member: Lower column and upper column.

Keep i, mind that it's a non sway structure.

Attached the model

[seifer92]

Thank you. I do have, however, some questions regarding what you've done. In the image 1, you should have put member n° 8 as column below in member 6 instead of member 3, like you did in image 2 for instability along z-axis,

right?

 

I can't understand at all why automatic procedure doesn't give correct result when the column is continuous. Why does such a thing occur? Indeed in another structure I was designing, my columns were under the level of the steel structure. In that case, switching from fixed to hinged at the bottom, gave me the same problem. Maybe it's because, in that case, I have to split the bar above the column, even if my beam in reality is continuous?

Regarding the non sway problem, yes you're right, but it's an extrapolated part with that constraint. My entire structure, however, doesn't have any bracing system against horizontal forces, so it's a sway structure. This example, with that particular constraint and forces, is a non sway one.

[Simau]

@seifer92 

...you should have put member n° 8 as column below...

Yes you are right

I can't understand at all why automatic procedure doesn't give correct result when the column is continuous. Why does such a thing occur?

If i understand you correctly and in my opinion, it's difficult to set buckling parameters in a continuous columns. Each column level has it's own parameters.

Share some pictures of what you put in parameters for that continuous column

[seifer92]

@Simau 

What a noticed after some checks is that the automatic procedure fails when a column is continuous. This is, probably, due to the fact the buckling length is based on Annex E of Eurocode, where it clearly states each column between storeys must be analyzed with the corresponding stiffness of beams and columns that join together. However two things I found strange in this process: 1) changing the constraint, from fixed to hinged, at the base of the column will always result in a instability condition (i.e. an automatic buckling procedure with a continuous column is stable with a fixed constraint and unstable with hinged constraint... the same column splitted will be stable in both cases); 2) when a column is not constrained by beams in a particular direction, the automatic procedure will alway give an instability condition.

In your opinion, to overcome the second point, should I choose a more traditional buckling coefficient according to end constraint condition (hinged-hinged, hinged-free, fixed-free, etc)?

The first point makes me think that when I have a structure with more than one storeys I should always split the column, since the automatic calculation would give the correct result, the same I would have if I manually choose the members adjoining that particular column (I think this is a convenient option, in order to avoid to choose the members on every column).

Second point, instead, states that when I have one storey or a column that would be jointed to a continuous beam, I should use a mixed method: automatic procedure in both directions if I have beams in both directions, automatic procedure and traditional method where I don't have a beam in a particular direction.

Do you agree with me?

(sorry for the long post but I like to have a fruitful discussion and a different point of view 🙂 )

[Simau]

@seifer92 

It's more easy if you could share your different models with continuous columns and instabilities.
Please don't use groups in your models to be able to use object inspector

[seifer92]

Here's another simple example of what I mean. The problem seems to occur, when I have a continuous beam and a column jointed below it, when using sway definition. However the structure I'm analyzing has no bracings and some horizontal deformations due to particular constraint definition.

In the example attached, I tried automatic length calculation and classical hinged-hinged condition for both cases. When I choose non sway, I obtain the results expected: using automatic definition rather than hinged-hinged, results in a reduced slenderness in both ways, since I have beams constraining my column.

Instead, when I choose sway, with automatic lenght the column is unstable again, while with hinged-hinged definition, it is stable.

I really don't get the point here. Maybe I'm missing something about the behaviour of the stiffness of the beams adjoining my column when the structure can sways.

[Simau]

@seifer92 

I never used the automatic procedure, because I don't know what is really hidden there.

[seifer92]

The automatic procedure does exactly the same thing you would do manually with 3 or 6 members... of course without specifying members for every column of your structure (I checked it with a simple model like the one attached).

After some investigations I came up with the fact the really big problem is to know if your structure is really sway or non sway.

The classical behaviour (hinged-hinged, fixed-free, etc) are valid for non sway structure only... I don't know why RSA gives the opportunity to choose between sway and non sway, since in these cases the results are the same. Maybe they can fix it in the next version, in order to avoid confusion.

[Simau]

@seifer92 

Please look at this model in which:

- the left structure is Sway in X and non Sway in Y

- the right structure is Sway in both directions

[seifer92]

@Simau 

I just checked the structure, I don't get a thing. Why for instablity along y-axis of the column you always choose the third end member condition? What does it stand for?

For instability along z-axis, instead, you chose  pinned end condition but I can't see any release on the members.

Does this manual choice have to be done for every column in the structure? When I edit member parameters I see only one column to have the correspondant member, however the other columns seem to dont' have the definition of  their adjacent members.

However from the results it's clear that without bracings and with pinned support at the bottom a column classified as sway will always be unstable.

Is there a way to declare if a particular structure is sway or non sway? I read a topic in the forum where it states that a buckling analysis should be done and check if the elastic critical buckling has a safety factor > 10 versus vertical global loads, in order to classify a structure as non sway (that is the classification of Eurocode 3 too).