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May 14, 2014
12:42 PM

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May 14, 2014
12:42 PM

Explanation of Buckling

Hello,

Someone he already made simulations of buckling.

Because I do not understand the result in buckling of a rod with several ( BLF) bending load factor , I still get

the same scale of displacement, with a maximum displacement of 1 [mm] .

I made several simulations of various shapes and load and I have always the same result, a displacement of 1 [mm] .

To view the buckling, we have to go through a static simulation model with all the stress conditions and charges

before proceeding to the simulation of buckling which includes the previous results of the static simulation.

Where I think the actual load must be important.

In the knowledge base PTC it is written :

For the stresses and displacements of the structure at the critical buckling load , the greater the BLF must be multiplied

by the initial compression load used in the static analysis . The initial load shall be changed

this value and static analysis restarted. Pro / SIMULATE not calculate the stresses due to buckling ,

because it is the field of nonlinear large deformation calculations .

If you have an idea , it will be welcome or a sample file .

Cordially.

Denis .

9 REPLIES 9

May 14, 2014
12:55 PM

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May 14, 2014
12:55 PM

Hi Denis,

A partial answer for you.

BLF= Buckling Load Factor

I believe the modal solver is used to calculate the buckling mode shapes. These mode shapes (eigen vectors) are normalised to amplitude 1.

Multiply the load applied in the initial static analysis by the BLF to get the load at which buckling will occur.

Regards

Charles

May 14, 2014
01:18 PM

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May 14, 2014
01:18 PM

Hello, Charles,

Thank you for the very quick response, I'll see what happens.

Cordially.

Denis.

May 14, 2014
01:58 PM

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May 14, 2014
01:58 PM

As Charles pointed out, the buckling analysis in Creo Simulate (Pro/Mechanica) is solving an eigenvalue problem (same as a modal analysis), where your eigenvalues are the Buckling Load Factors (the factor of the applied load "needed" to cause buckling) and the eigenvectors are the buckling mode shape. For an eigenvalue problem, Mechanica will always normalize the results (i.e. the largest "displacement" is 1).

Now, all that being said, it's very important to keep in mind that last sentence from the knowledge base: **"Pro / SIMULATE not calculate the stresses due to buckling, because it is the field of nonlinear large deformation calculations."**

The analysis you're doing is sometimes called a **linear bucking analysis**, because the solution is a linear perturbation of the intial change in the structures stiffness due to the base static analysis. There are two very important points about a linear buckling analysis that must be kept in mind:

- There is a unknown amount of error between the linear and non-linear buckling solution.
- The linear buckling solution is
**unconservative**(i.e. the calculated buckling load is higher than the true buckling load).

The implication of these two points (particularly the last one) means that you need to be very careful how you use the results from a linear buckling analysis. Realistically, a linear buckling analysis should only be used for preliminary comparisons (what type of buckling mode could exist, and how design changes effect the trend of the BLF). If the results are used for anything else, then a high factor of safety is needed. More accurate results can be obtained by running a non-linear analysis with large-displacement formulation and a non-linear material property. If you do this, be sure to include some eccentricity in your model (whether load or geometric).

May 14, 2014
02:28 PM

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May 14, 2014
02:28 PM

Hello, Shaun,

Thank you to you for this very interesting explanation.

I'll take it all.

Cordially.

Denis.

May 14, 2014
02:36 PM

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May 14, 2014
02:36 PM

Hi Shaun,

For interest and to broaden discsussion a little to perhaps pick up any other's input, hints, tips, literature references ...

I have found the linear solutions very limited in their application. It is possible to get the text book answers but it is rare that real life models are text book cases.

I have opted for more complex assemblies, contolling the force input by using enforced displacement contraints (to eliminate the instability) then running static large deformation studies. Plotting reactions vs displacement.

These studies need an initial eccentricity otherwise the model would simply crush. This initial eccentricity accounts for misalignments and material flaws etc. How does one judge this initial eccentricity?

Then there is the situation where the simpler Euler calculation predicts a buckling load that is too low, a calculation carried out on a product that has worked for years. The real situation being that as a structure buckles it encounters lateral support changing the stiffness and introducing new buckling modes at higher loads and these are the ones we would like to estimate. These assymmetrical models are difficult to constrain without adding further layers of stucture, contacts, springs to control instabilities. And then what stiffness the springs?

What if the structure carrying the load is excited and vibrating ...

etc.

Thanks

Charles

May 20, 2014
07:57 PM

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May 20, 2014
07:57 PM

Hello,

I go back to my problem of buckling for which I asked these questions.

I blade 8mm long 5mm wide and 0.25mm thick.

It would take on one of its ends a width of 5mm force 375N.

Its ends are rounded, there is no frame of the blade.

I did not find how to constrain the ends of the blade for it to burn freely.

I used the pivot constraint freely rotatable and laterally blocked on both ends, but the application of the force is also done on the pivot constraint.

The results of the static analysis are not already correct.

It is possible to give a slight curve to the blade to help blaze.

Now to make this simulation, should create an assembly with rounded surfaces recessed for the blade?.

Cordially.

Denis.

May 23, 2014
01:31 PM

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May 23, 2014
01:31 PM

Hello,

I return to my question, if someone an idea for my stress leaf spring buckling.

I enclose a picture of the blade, so that little help.

Cordially.

Denis

May 23, 2014
05:55 PM

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May 23, 2014
05:55 PM

Denis,

In reality as the blade is compressed it will buckle.

To buckle there has to be initial eccentricity or assymmetry. In reality nothing is perfect so these are present.

To make the model buckle you must have an initial eccentricity (otherwise it crushes) which means the model is difficult to constrain ... as you have found.

For a buckling hand calculation, the initial effective strut length is the distance between the axes of the cylindrical ends. Any bending from the initial straight position immediately means that the load's mechanical advantage changes and therefore other than the initial straight structure it does not behave as Euler suggested but as noted in an earlier post.

For this model you end up having to make extra parts, build an assembly, use contact, time-stepped enforced displacement constraints and large deformation.

This means you cannot use the buckling solver.

Create a component with shalllow cylindrical groove and assemble to each end to 'capture' the blade.

Fix one end and enforce displacements of the other.

Create reaction measures to plot against distance moved.

Remember, you must have an initial misalignment so do not apply your force exactly along the blade. This is a judgement.

Plot the graph of the reaction measure vs distance and look for the change (reduction) in gradient. This is where collapse is iminent.

time stepped enforced displacement constraints prevent run-away collapse.

Problem is that the large movements of the structure before collapse often means it encounters new boundary conditions (contacts neighbouring components) and maybe not actually collapsing at all ...

Hope this helps

Charles

May 23, 2014
07:36 PM

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May 23, 2014
07:36 PM

Hello, Charles,

Thank you for your explanations.

This confirms that I will have to travel with an assembly.

You did very well describes the problem and is outside the flanbage.

Cordially.

Denis.

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