linear buckling / strand7 2

[jazzmine]

Hi guys!
I have problem. I am new in this program.
I have to do a linear buckling analyses of rectangular simple supported plates with circular and rectangular holes in various positions subjected to axial compression, so i can calculate buckling coefficient k.

what i do with restraint ? In y direction I mark Dy Dz and Rx Ry , in x direction I mark Dx Dz and Rx Ry for simple supported plate.

the load was applied directly to the nodes.

my question is : are this restraint ok? and what do I have to do with FREEDOM CASES?

I first do linear analysis and then buckling analysis. How to calculate buckling coefficient k with buckling load factor?

THX!!

 

[Denial]

You will only need a single "freedom case". (Multiple cases are required only when you have a problem involving imposed displacements.)

Use this single freedom case to globally remove those degrees of freedom that are irrelevant to your problem. If your plate lies in the XY plane, and your problem contains no elements other than coplanar plate elements, the only dof that you can eliminate this way is RZ, the so-called "drilling freedom".

 

[IDS]

The restraints applied in the Freedom Cases are applied every node. For plate shell elements in the XY plane you could restrain the RZ rotation, but I suggest leaving it unrestrained because it doesn't make any difference to the results, and if you want to add some beam elements to the model it must be unrestrained to allow the beams to bend.

For simply supported restraints you should not restrain RX and RY, the buckling load will be substantially reduced when these restraints are removed.

The buckling load is given by the applied load (in the linear static analysis) x buckling load factor.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

Also I suggest reading the theoretical manual for more detailed background information, and try generating your own models for some examples from the verification manual to make sure you get the same results.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jazzmine]

thanks

I will try to make what you say..

if I had any other uncertainties i will send you questions.

THX!

 

[jazzmine]

Hi!
More questions. What does it means if I have negative buckling factor ( 1st eigenvalue is negative)? I read that
a negative buckling factor means that the structure will buckle when the directions of the applied loads are all reversed and if loads cannot be reversed,the negative load factors can be ignored. Should I ignore that eigenvalue and use the second one?
Thanks!!

 

[jazzmine]

Does anyone know how to apply bending moment directly to the nodes of plate in Strand7?
Thx!

 

[IDS]

What does it means if I have negative buckling factor ( 1st eigenvalue is negative)? I read that
a negative buckling factor means that the structure will buckle when the directions of the applied loads are all reversed and if loads cannot be reversed,the negative load factors can be ignored. Should I ignore that eigenvalue and use the second one?

Yes, but make sure the loads really can't be reversed! You might want to run the analysis with the loads reversed, even if this can't happen in practice, just to see what happens. Also have a look at the deformed shapes you get for each mode and make sure they make sense (e.g. you haven't included any restraints that don't exist in the actual structure).

You can apply point moments anywhere on a plate with Attributes-Plate-Point Load-Moment, or directly to a node with Attributes-Node-Moment.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jazzmine]

Helo!
I need help with the model. I can not get the buckling coefficient k=4 for a simply supported plate. If I multiply applied load with buckling load factor I get k=7,7.
I made model in the following way :
- I open the dialog box MODEL UNITS and select units ( m, kg, J, kPa, kN, K)
- I create a SNAP GRID ( length of model is 2 m, width is 1 m )
- From the CREATE menu i select Element and use Quad 4
- I SUBDIVIDE element ( Targets I left by default, for plate- Quad 4 , for Brick Quad8)
- I applied the load ( uniform axial compression ) directly to the nodes as a system of conservative forces ( 1 kN in each of 21 nodes)
- In LOAD AND FREEDOM CASES, as you said I left all unrestrained
- RESTRAINTS: I restrain in y direction I mark Dy Dz , in x direction I mark Dx Dz
- In PROPERTY for plate , for the type I use plate/shell , for material Steel (structural) , Membrane thickness same as bending thickness 0,005 m
- And I use Linear static solver, then Linear buckling solver.

For the perforated plate I create a circular and rectangular holes with GRADE PLATES AND BRICKS, and after subdividing elements, I CLEAN MESH ( left all by default, with Zip tolenace 1,0 x 10^-4 ,click apply and after that again apply and close the box).

So my question is do you see mistake in modeling , did I do something wrong or I skipped something?
I am very grateful for your help!

 

[jazzmine]

This is the buckling load factor that I get for the first mode (the plate without hole) : 8.02018644E+00 .

 

[IDS]

jazzmine - If you post your model I'll have a look at it.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jazzmine]

This is a plate without hole and I will post you a perforated plate to see. I am not good in this program and really don't know what is the problem, so thank you for helping!

 

[jazzmine]

Here's second model.

 

[IDS]

jazzmine - The main problem with your model is that the edges running in the X direction should not be restrained in the X direction, since that is the direction you are applying the force. The attached spreadsheet shows that if the X restraints are removed the buckling load is found to be within 2.5% of the theoretical value. The comments below may also be helpful, but the X restraint was the main problem.

[ul]
[li]If you are applying loads as node loads the corner nodes should have half the load to get a uniform distribution.[/li]
[li]It is easier to apply the load as an plate edge pressure, especially when you have mid-side nodes.[/li]
[li]Be careful with units. To get the results to be compatible with theoretical equations you need to use consistent units. In the spreadsheet I have used MPa and millimetres, and I have applied a 1 MPa edge pressure, but you can do everything in kPa and metres if you prefer.[/li]
[li]Results are slightly improved by using 8 noded or 9 noded plate elements.[/li]
[li]You can remove all the restraints other than in the Z direction. You get a warning that the model may be unstable, and the first buckling factor is very close to zero, but it solves without problem (because the nett force in both directions is zero) and the results for the second buckling mode are very close to the theoretical values.[/li]
[/ul]


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jazzmine]

I did what you said. Thank you very much for help!

 

[jazzmine]

Can you give me some suggestion how to apply, to the same model, bending moment (pure bending) and combination of axial compression and bending moment? And how to calculate buckling coefficient in that cases?
I supposed I should compare the results with the results in table 15.2.3.
Thanks!

 

[IDS]

Can you give me some suggestion how to apply, to the same model, bending moment (pure bending) and combination of axial compression and bending moment? And how to calculate buckling coefficient in that cases?

Just apply the bending moment and axial load in the static analysis then re-run the buckling analysis. You can't apply an edge moment to the plates so you will have to apply it as node loads or plate point loads. You will need to calculate the bending moment/node that gives the ratio of bending moment/axial load that you want. Note that if there is no axial load you will get a warning that the matrix diagonal terms are all zero, and the buckling results will not be valid.

I supposed I should compare the results with the results in table 15.2.3.

If table 15.2.3 is a table of buckling factors for a plate under the same restraints and loads then yes, that would be a good idea.

(Just had a look in Roark; I guess that's the table 15.2.3 you meant, and yes that looks like the appropriate table. Note that all the tabulated values include some axial load).

I'll be interested to see your results.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jazzmine]

The problem is that I have to apply just bending moment like on a picture that I post. The stress ratio ψ must be equal -1 for pure bending. Do you have other idea how to apply just bending moment? Is it possible to apply rigged load in Strand7 and how?
I assumed that the buckling stress I should calculate as σ=M/W for pure bending, and as σ=N/A + M/W for combination?
Thanks.

 

[IDS]

OK, I was looking at a moment about the Y axis (assuming the plate is in the XY plane), and you want a moment about the Z axis.

You need to apply a varying edge pressure to the plates, which you can do with an equation. Instead of entering a value for the edge pressure enter:
1000*(Y-0.5), which will automatically generate a pressure varying from 500 kPa at the top to 500 kPa at the bottom.
With this loading it works better if you restrain one node on the longitudinal centre line in the XYZ directions and the node the other end in YZ directions.

I'm not sure what you mean by a "rigged load", but I'm guessing you mean applying a moment to one node and transferring it along the edge with rigid links. This is also a possibility. You might want to try both ways and check they give the same answer (and the same as Roark).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jazzmine]

I am afraid that I did something wrong again beacause, when I apply the load, I get at the top -4,025*10^3 edge pressure and at the bottom -4,975*10^3, as you can see in the model for bending . And not as you said. And I have not found how to calculate buckling stress in this case. Can you help me with that to?
Thanks.