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linear buckling / strand7
2

linear buckling / strand7

linear buckling / strand7

(OP)
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!!

RE: linear buckling / strand7

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".

RE: linear buckling / strand7

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/

RE: linear buckling / strand7

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/

RE: linear buckling / strand7

(OP)
thanks :)

I will try to make what you say..

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

THX!

RE: linear buckling / strand7

(OP)
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!!

RE: linear buckling / strand7

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

RE: linear buckling / strand7

Quote:

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/

RE: linear buckling / strand7

(OP)
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! :)

RE: linear buckling / strand7

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

RE: linear buckling / strand7

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

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: linear buckling / strand7

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.

  • If you are applying loads as node loads the corner nodes should have half the load to get a uniform distribution.
  • It is easier to apply the load as an plate edge pressure, especially when you have mid-side nodes.
  • 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.
  • Results are slightly improved by using 8 noded or 9 noded plate elements.
  • 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.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: linear buckling / strand7

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

RE: linear buckling / strand7

(OP)
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!

RE: linear buckling / strand7

Quote:

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.


Quote:

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/

RE: linear buckling / strand7

(OP)
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.

RE: linear buckling / strand7

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/

RE: linear buckling / strand7

(OP)
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.

RE: linear buckling / strand7

jazzmine - I really think you'd get more from doing some more experimentation on your own. The stresses you have generated are for a uniform compressive stress of 4500 combined with a flexural stress of +-500. Most likely there was an existing stress and you clicked the "add" button, rather than "apply".

Note that the stress is applied as a series of stepped pressures, so the stress on the outer plates will be +-475 (i.e. the stress on the mid point of the plate), not +-500. Also all the top and bottom edge nodes are restrained in the X and Y directions in your model, which is very different to the conditions in Roark Table 15.2.3.

The buckling stress is just the maximum applied stress (projected to the edge of the plate) times the buckling factor found by the analysis. You can calculate the axial load and bending moment using the formulae given in your previous post (assuming that W is the elastic section modulus).

By the way, be careful with abbreviations, it looks like you are from a non-English speaking country, so use of abbreviations may cause confusion where we have different conventions.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: linear buckling / strand7

(OP)
Ok. Thanks for your advices and help!

RE: linear buckling / strand7

Quote:

Ok. Thanks for your advices and help!

Good luck! Feel free to keep the discussion going if you have any more questions, and I would be interested to see how your final results compare with the Roark values, and also what effect the central hole has.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: linear buckling / strand7

(OP)
I apply bending as you said. I guess that I should multiply bucklig load factor with 0,5 MPa ? If I calculate stress ratio with formula which is given in Roark`s table (15.2.3), I get α=0,5 for bending. So I shoud compare my results with the first line in that table. Can you give me your opinion?
Thanks

RE: linear buckling / strand7

Quote (jazzmine)

I apply bending as you said. I guess that I should multiply bucklig load factor with 0,5 MPa ? If I calculate stress ratio with formula which is given in Roark`s table (15.2.3), I get α=0,5 for bending. So I shoud compare my results with the first line in that table. Can you give me your opinion?

Yes, I agree with both points. I said previously that the Roarke table did not cover the pure bending case, but that was wrong. α is the maximum stress divided by the stress difference, so α = 0.5 is pure bending and α = infinity is pure compression.

I'm getting a k value about 6-7% lower than the Roark value for the pure bending case.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: linear buckling / strand7

(OP)
I need your advice about combination of compression and bending. What is your opinion about restraints, should I leave them like for bending? I made two models, first one with the restraints like for compression and second one like for bending. For first model, Mode 1: buckling load factor is 0 and Mode2: 17,3741. And for the second model: Mode 1: 17,3741. If I multiply buckling load factor with 1,5 MPa (I add compression load to the bending load) and compare results with the Roarke table, the difference is about 2%.
Thanks.

RE: linear buckling / strand7

I'd suggest keeping restraints to the minimum that models the restraints in the actual structure and provides sufficient restraint to allow the analysis to solve. In the case of getting a buckling factor of zero (or very close to it) have a look at the mode shape (set the deflection magnification to 10%) and confirm that this corresponds to a deformation that can't happen in practice. In this case it shows rotation of the plate in the XY plane about the centre point, which is clearly not a real buckling mode. Having done that you may choose to do another run with added restraints, as you did. This gives identical results, and also good agreement with the Roarke results, so you can be confident that the added restraints are not restraining a buckling mode.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: linear buckling / strand7

(OP)
Thank you.

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