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!!
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
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
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
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: linear buckling / strand7
I will try to make what you say..
if I had any other uncertainties i will send you questions.
THX!
RE: linear buckling / strand7
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
Thx!
RE: linear buckling / strand7
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
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
RE: linear buckling / strand7
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: linear buckling / strand7
RE: linear buckling / strand7
RE: linear buckling / strand7
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: linear buckling / strand7
RE: linear buckling / strand7
I supposed I should compare the results with the results in table 15.2.3.
Thanks!
RE: linear buckling / strand7
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.
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
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
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
Thanks.
RE: linear buckling / strand7
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
RE: linear buckling / strand7
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
Thanks
RE: linear buckling / strand7
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
Thanks.
RE: linear buckling / strand7
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: linear buckling / strand7