2

I need some simple and clear explanation of what is called linear buckling analysis and why it is also called eigenvalue buckling analysis?

In other words how natural vibration frequency or eigenfrequency refers to the static stability of mechanical system?

To be more specific I need some basic understanding of this subject because I'm currently studying a problem of topology (or structural) optimization of some mechanical construction using finite element analysis software. And as I know only linear buckling criteria can be used for topology optimization today. Correct me if I'm wrong.

Below is the most understandable illustration of this item I've found on the internet.

Picture

Here is the link. Though it explains the sense and purpose of described items, but actually gives no full and clear explanation of what is actually going on. So here I hope to get at least a good link.

Bzhenko
  • 21

1 Answers1

2

The eigenvalue problem has in this case not much to do with the system vibration but an analogy can be drawn.

Defining the problem and clarifying vibrations

In a rather general case, a mechanics problem finite element approximation ends up in the form $$ [M]\{\ddot u\} + \{\psi\}=\{\phi\}$$ where $[M]$ is the mass matrix, $\{\psi\}$ the vector of internal forces and $\{\phi\}$ the vector of external forces.

Let us assume the linear case where $\{\psi\}=[K]\{u\}$. Solving the eigenvalue problem: $$ (\omega^2[M]+[K])\{u\}=\{0\} $$ will provide us with eigen-frequencies associated with modes, i.e. the corresponding structural shapes.

Linear buckling analysis

Now what we consider is the quasi-static problem, with no inertia (mass) effect : $$\{\psi(u)\}=\{\phi\}$$ that we linearize around a reference state supposing small displacements and negligible evolution of external forces, ending up with the form $$[K_T]\{u\}=\{0\}$$ $[K_T]$ can be split into the sum of a material contribution $[K^m]$ and a geometrical contribution $[K^g]$.

  • if $\det [K_T]\neq0$ then no displacement is admissible and the problem is stable.
  • if $\det [K_T]=0$ then there exists a non zero displacement solution that requires no additional force : this is buckling.

$[K^m]$ is usually positive definite while $[K^g]$ may not be; it is therefore of interest in linear buckling analysis. The eigenvalue problem we consider in linear buckling analysis is: $$([K_T]+\lambda[K^g])\{u\}=0$$ i.e. we seek a loading factor $\lambda$ and the associated eigen-modes that will bring the system's determinant to zero.

neerby
  • 179
  • 1
  • 9