Linearized Buckling

2 min read • 288 words

Corotational frame elements are used to approximate Euler's buckling load.

Problem  

Corotational frame elements are used to approximate Euler’s buckling load which is given by:

$$P_{\mathrm{euler}} = \frac{\pi^2 EI}{L^2}$$

This example is adapted from https://github.com/denavit/OpenSees-Examples  . The files for the problem are buckling.py for Python, and buckling.tcl for Tcl.

Theory  

Buckling Analysis  

Loosely speaking, buckling happens when there are multiple shapes that a structure can deform into that will be in equilibrium with it’s applied loads. This implies that at the point of buckling, there are multiple independent displacement increments $\bm{u}$ which will be mapped to the same resisting load by the tangent $\bm{K}$ . In otherwords, The buckling load is the point at which $\bm{K}$ becomes singular. If we consider $\bm{K}$ as a function of the load factor $\bm{\lambda}$ , this condition can be expressed as the nonlinear root-finding problem:

$$\operatorname{det}\bm{K}(\lambda) = 0$$

For many classical models, the dependence of $\bm{K}$ on $\lambda$ is linear, and in this case the problem is equivalent to a generalized eigenvalue problem which is computationally much more tractable. However, even if $\bm{K}$ is nonlinear in $\lambda$ , one may still investigate the linearized buckling problem, where an eigenvalue problem is obtained by learizing $\bm{K}(\lambda)$ :

$$\bm{K}(\lambda) \approx \bm{K}(0) + \bm{K}^{\prime}(0) \lambda$$

where $\bm{K}^{\prime}$ is the derivative of $\bm{K}$ with respect to $\lambda$ .

Timoshenko Column Buckling  

$$\begin{gathered} \lambda=\sqrt{\frac{P L^2}{E I\left[1-P /\left(k_{\mathrm{s}} G A\right)\right]}}=\sqrt{\frac{P L^2}{\chi E I}} \\ \chi=1-P /\left(k_{\mathrm{s}} G A\right) \\ P=\chi \lambda^2 E I / L^2 . \end{gathered}$$ $$\begin{gathered} \chi=\frac{1}{1+\lambda^2 E I /\left(k_{\mathrm{s}} G A L^2\right)}=\frac{1}{1+\lambda^2 \varphi / 12} \\ P=\frac{\lambda^2 E I / L^2}{1+\lambda^2 \varphi / 12} . \end{gathered}$$ $$\tan \lambda_{\text {cr }}=\chi \lambda_{\text {cr }}=\frac{\lambda_{\text {cr }}}{1+\lambda_{\text {cr }} 2 \varphi / 12}$$