# How would I do a buckling analysis with an additional fixed known lateral force?

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!

• Cross posted here: stackoverflow.com/q/59578115/4961700 Jan 3 '20 at 13:36
• Picture = 1000 words, diagram = 10,000. Jan 3 '20 at 14:27
• @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. Jan 3 '20 at 17:13

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.