Plate buckling

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Plate buckling

Postby Babita » 16 Jun 2009, 13:36

How can I make a plate buckling calculation for a 2Dmember model in Scia.
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Re: Plate buckling

Postby Cyril Heck » 16 Jun 2009, 17:01

Hi,

you need to make sure of the following:
- do you have the stability modules on your key ?
- the functionality "stability" must be activated
- create a stability combination
- in the "solver setup", specify the number of stability modes you want, and launch the calculation.

That's it !

Best regards,
Cyril Heck
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Re: Plate buckling

Postby memento » 17 Jun 2009, 15:45

Cyril Heck wrote:Hi,

you need to make sure of the following:
- do you have the stability modules on your key ?
- the functionality "stability" must be activated
- create a stability combination
- in the "solver setup", specify the number of stability modes you want, and launch the calculation.

That's it !

Best regards,


Hi Cyril Heck,

I have two questions about the buckling calculation in scia:
Is that possible (or necessary) to calculate stiffened plate buckling accroding to eurocode 3 EN 1993-1-5?
What does the critical load coefficients mean from the stability calculation results?

Thanks in advance!
regards,
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Re: Plate buckling

Postby Cyril Heck » 18 Jun 2009, 11:04

Hi,

the only part of the EN 1993-1-5 currently taken into account is the calculation of effective section properties for class 4 profiles.
The critical load coefficient is the multiplier that has to be applied to the considered load combination (stability combination) to reach Euler buckling (bifurcation load). This is valid for:
- shell buckling
- plate buckling
- global buckling of a frame structure
- lateral torsional buckling of a steel beam modeled with surface elements
- etc

It is also possible to investigate this behavior with a 2nd order analysis and a suitable initial imperfection, as well for frames as for plates & shells.

Hope this helps,
Cyril Heck
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SCIA
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Re: Plate buckling

Postby P. Van Tendeloo » 18 Jun 2009, 11:27

In case of 2D elements, Chapter 10 of EN 1993-1-5 can be followed.

This chapter specifies the use of the Reduced Stress Method wich has as advantage that no effective section properties are needed.

Using a stability calculation, Scia Engineer provides the Alfa,cr load amplifier.

The Von Mises stresses in the results menu (Sig E) can be used to calculate the Alfa,ult load amplifier.

=> Both can then be used to calculate the plate slenderness lambda,p

A manual calculation will then be required to determine the reduction factors according to Chapter 10 article (5) after which formula (10.1) can be applied for the verification.
Peter Van Tendeloo
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Re: Plate buckling

Postby memento » 19 Jun 2009, 13:50

Hi Cyril Heck and Tendeloo,
Thank you so much for your answers! :D
To be honest, I am still comfused. Here are more questions. Hope you can help. Thanks!
Cyril Heck,
1. If I got a critical load coefficient less than one from SCIA, could I say my stiffened plate structure would not fail in any kind of buckling mode according to the code with which I check?
2. How should I play with the "number of critical values"? It seems this value doesn't change the result.

Tendeloo,
1. If I did like what you said, and the result pass the check, could I say my stiffened plate structure would not fail in any kind of buckling mode according to the code with which I check?
2. When doing stability check, does SCIA "smear" the stiffener into the plate or it is depends on the model I build (are the stiffeners plate ribs or just composed of 2D plates)?

Finally, are you guys familar with plate buckling check of the stiffened plates? I am new to this problem, so althouth I have all the theorys, still don't know how to use it. How can I say a stiffened plate would not fail by buckling?

Have a nice weekend!
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Re: Plate buckling

Postby P. Van Tendeloo » 19 Jun 2009, 14:08

I'll try to answer all four questions:

a) The clearest to explain the critical load coefficient is in the following way:

Ncr = acr * NEd

You can thus multiply the actual axial force NEd with the critical load factor acr before buckling (Ncr) occurs.

Therefore, if the critical load factor is lower than one it means the structure is not OK since it already buckles for the given loading. For example acr = 0,8 implies that at 80% of the current loading buckling occurs.

Also, keep in mind that a stability analysis concerns the determination of the critical buckling load. This is not the same as the buckling resistance according to the codes since the codes also apply imperfection values (depending on buckling curves) to account for geometrical imperfections and residual stresses.

b) 'Playing with the critical values': each critical value corresponds to a buckling shape. You can augment the number of buckling shapes to be calculated in order to check which shape you should use.
Normally the shape with the lowest positive critical load factor is limiting.

c) Yes, if you follow the procedure outlined in EN 1993-1-5 chapter 10 then the design is fully according to the code. I refer also to point a) above: EN 1993-1-5 'uses' the acr to do the overall buckling analysis (instead of just checking that acr > 1).

d) Since you are working with 2D elements I would advise modelling the stiffeners also as 2D elements.

There is a good background document regarding EN 1993-1-5 which also includes examples. It can be freely downloaded using the following link:
http://eurocodes.jrc.ec.europa.eu/showp ... php?id=100
Peter Van Tendeloo
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Re: Plate buckling

Postby memento » 22 Jun 2009, 18:05

Hi Tendeloo,

Thanks again! Your example of the code helps a lot!
Sorry to bother you again, some parts of your answers still confuse me.
a). Sorry! I wrote the coefficient wrongly. It should be higher than one.
For geometrical imperfections and residual stresses, do you mean, in SCIA, the two factors are not taken into account no matter from Stability calculation or Nonlinear Stability calculation, i.e. SCIA will always see the 2D member as perfect flat.

b). Could this number means how many half valves the 2D element could have (the number of m when calculate buckling load coefficient k)?

c). and d). Very clear! Thanks!

From your example, I have some thoughts (please refer to chapter 16 the worked example):
Can I say the buckling of stiffened plate could be checked both "manually" by the effective width method(as the example illustrated) and by the effective stress method (the acr have to be got from finite element method); the two method would have equivalent results?

Thank you and have a nice day!

regards,

Memento
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Re: Plate buckling

Postby P. Van Tendeloo » 23 Jun 2009, 09:59

a) Consider for example a default column buckling check according to EN 1993-1-1

You calculate the Euler load Ncr however this is not used for verification i.e. the check is not NEd < Ncr.

The code uses Ncr to calculate the reduction factor Chi. In the calculation of the reduction factor you will see the value of Alpha which is dependent on the buckling curves. It is this value Alpha which accounts for imperfections and residual stresses. The final verification is done as NEd < Nb,Rd (and not using Ncr).

So, for 2D elements according to EN 1993-1-5 the same logic applies: You calculate the Ncr (i.e. acr) and then, in the same way as above, the reduction factors Rho need to be determined as outlined in article 10 (5).

b) A stability analysis is the same type of analysis like a dynamic eigenvalue analysis. In theory you will get as much buckling modes as the number of degrees of freedom you have.

Typically for a single member there will be a buckling shape where the whole member buckles with one half sinus wave. Higher modes will contain more sinus waves however they will (mostly) have higher critical load factors so they are not limiting.

So, the number of buckling shapes literally means how many modes Scia Engineer needs to calculate, not the number of half sinus waves the element has.

c) As for your final question:
Article 2.4 of EN 1993-1-5 specifies that both the effective width method or the reduced stress method can be applied. As I specified before, the effective width method requires you to calculate effective section properties which is quite a tedious analysis. The advantage however is that the procedure is clear and well documented.
The reduced stress method allows you to consider the sections as class 3 thus no reduced section has to be calculated which saves a lot of effort. But offcourse you still need to calculate the reduction factors Rho as specified above. This method is also less documented in literature.

Both methods indeed have the same aim however personally I do not expect them to give exactly the same results. In a way, the effective width method is more analytical, it approximates the plate buckling behaviour through formulas while the reduced stress method is more numerical, using the actual buckling shape of the model.

There exists an example which illustrates the use of Chapter 10 which can freely be downloaded. The example is in French but the procedure should be clear. It can be downloaded through the following link:

http://www.cticm.eu/spip.php?article97& ... che=combri

In the first part (Combri_1.pdf) chapter 4.2.3 page 229 illustrates the method including determination of acr (both using a manual calculation and a software calculation), detrmination of a,ult, determination of the Rho factors and finaly the unity check.
Peter Van Tendeloo
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Re: Plate buckling

Postby memento » 24 Jun 2009, 16:41

Thank you man!
Thank you for your patience and your suggestions. You really helps a lot!

Have a nice day!

Memento
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Re: Plate buckling

Postby anita » 10 Jul 2009, 09:48

Hello all,

I am currently writting a report on plate buckling and I am having lots of dificulties finding information about the Reduced Stress Method (Eurocode Part 1-5 (10)).
Could anyone tell me if there is any document or book on this subject other than those you mentioned above?

Thank you!

Ana
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Re: Plate buckling

Postby T_Hoekstra » 07 May 2015, 08:33

Ana,

This can be useful for you: http://www.stalforbund.com/Fagboker/EUR22898EN.pdf

Kind regards,

Tunis
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