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I am using Nastran for buckling analysis, and I'm new to all of this so please bear with me.

For a linear buckling analysis, I understand that you must first apply any compressive load in the static subcase. Then solve the eigenvalue buckling problem in the second subcase using the first static subcase. The buckling load is then the amount of compressive load applied multiplied by the resulting eigenvalue.

However, I now want to apply a second fixed force of exactly 50N laterally and see what the buckling load would be with this additional 50N force. Now if I apply both the compressive load and the lateral load to the static subcase, and then solve the eigenvalue buckling using that subcase, the resulting buckling load varies a lot depending on the compressive load applied in the static subcase.

Example 1:

  • Apply 1N compressive force, eigenvalue is 100. So buckling load is 1N*100 = 100N
  • Apply 10N compressive force, eigenvalue is 10. So buckling load is 10N*10 = 100N This makes sense, however:

Example 2:

  • Apply 1N compressive force and 1N lateral force, eigenvalue is 90. So buckling load is 1N*90= 90N.
  • Apply 10N compressive force and 1N lateral force, eigenvalue is 9.5. So buckling load is 10N*9.5=95N, which is higher than 90N.

This just proves that I am doing something wrong, and I am hoping someone can clarify what it is. Any help would be appreciated, thank you!

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  • $\begingroup$ Cross posted here: stackoverflow.com/q/59578115/4961700 $\endgroup$
    – Solar Mike
    Jan 3 '20 at 13:36
  • $\begingroup$ Picture = 1000 words, diagram = 10,000. $\endgroup$ Jan 3 '20 at 14:27
  • $\begingroup$ @JonathanRSwift In this particular case, I don't see how a picture would add anything to the question. Either you know Nastran well enough to answer it from the clear description given, or you don't. $\endgroup$
    – alephzero
    Jan 3 '20 at 17:13
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In your example 2, Nastran is scaling both loads as part of the buckling load.

In your first case, you have buckling with $90\times 1 = 90$N axially plus $90\times 1=90$N laterally.

In the second, you have buckling with $9.5\times 10 = 95$N axially plus $9.5\times 1 = 9.5$N laterally.

From example 1 you have a buckling load of 100N axially plus zero laterally.

You could plot a graph with those three points, and estimate the axial load when the lateral load is 50N. Then set up the Nastran input for those values and repeat if necessary.

Alternatively, since you have now bracketed the answer between 90N and 95N, you could run analyses with 91+50, 92+50, 93+50 and 94+50, and estimate when the buckling factor is exactly 1.0.

In a general situation, this sort of loading is likely to be incompatible with the assumptions of linear buckling. You could model it as a nonlinear analysis. First ramp up the side load from 0 to 50N, then increase the axial load and see what happens. You may find the structure doesn't "buckle" in the sense of a catastrophic instability, but just gets more flexible as you increase the axial load and lateral deflection increases.

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