I am trying to find the buckling load of following column:

column with buckling load

I started from the system of equations $P = (K + K_g(N))U$, where $K$ is the stiffness matrix of a single 2D frame element and $K_g$ is the geometric stiffness matrix of the same element.

The boundary conditions give zero displacements and rotation in the node at the bottom of the column. So I tried to solve the equations for the 2nd part of the column, where the only unknowns are the displacements and the rotation in node 2, with $N = P$. But this gives me the same result as for a standard cantilever beam.

Here's an example on how to calculate the buckling load for a cantilever beam: cantilever beam example cantilever beam example cantilever beam example

I also tried solving the complete system of equations with unknown displacements in node 1 and 2, but without success. And since I need to be able to calculate it by hand, this doesn't seem to be the most efficient way to do it.

I don't know what I'm doing worng. Could someone help me to start in the right way?

  • $\begingroup$ you need to work at clarifying what you are asking. I don't get it. Whats this to do with finite elements? $\endgroup$
    – agentp
    Jan 20, 2018 at 17:46
  • $\begingroup$ I'm sorry for the confusion in my question. I added an example of a simple cantilever beam and tried to explain my question more clearly. I need to solve it for my courses about the finite element method, that's why I added the tag. $\endgroup$
    – user14605
    Jan 20, 2018 at 20:22

1 Answer 1


Obviously you solve for a column with prescribed displacements and rotations in node 2, and free conditions in the other node. Length must be "L" to apply the formulas depicted in your paper. The results (first critical load) is accurate. The value 2.49 corresponds to (nπ2) where the term "n" is the amount of critical load above or below the Euler critical load for columns. In your case n = 0.2522, and the real value is 0.25 for this particular case. The first value corresponds to the minimum autovalue or "eigenvalue" and you can use the "eigenvector" associated with the first "eigenvalue" to draw the final position or the deformed shape of the column due to this first critical load. The second eigenvalue corresponds to the second critical load. It's very large compared whith the first one.


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