Idealy a beam subjected to an axial compression force would be compressed but since there are always imperfections, it is not compression what determines strength of such members:
Slender (slim) members in compression are subjected to buckling. The assessment is
$$ \begin{equation} N_{Ed} \leq N_{b,Rd}, \end{equation} $$where $N_{Ed}$ is design value of the compression force (from load) and $N_{b,Rd}$ is design buckling resistance of the compression member. The resistance is being found as a resistance in pure compression which is lowered by reduction factor $\chi$ (effect of buckling):
$$ \begin{equation} N_{b,Rd} = \boldsymbol{\chi} \frac{A f_{y}}{\gamma_{M1}} \label{eq:ass_buck}. \end{equation} $$For class 4 sections in $(\ref{eq:ass_buck})$ $A_{eff}$ instead of $A$ is used. Nonsymmetric class 4 members need allowance also for additional moment $\Delta M_{Ed}$ from eccentricity.
Note: holes for fasteners at the columns' ends need not to be considered to determine $A, \ A_{eff}$.
Reduction factor $\chi$ is found from the relevant buckling curve (according to cross-section) or from the formula:
$$ \begin{eqnarray} \chi &=& \frac{1}{\Phi + \sqrt{\Phi^2 - \overline{\lambda}^2}} \hskip2em \chi \leq 1.0 \label{eq:buck_chi} \\ \Phi &=& 0.5(1+\alpha(\overline{\lambda} -0.2) +\overline{\lambda}^2) \label{eq:buck_fi} \\ \overline{\lambda} &=& \sqrt{\frac{A f_y}{N_{cr}}}, \end{eqnarray} $$Symbol $\lambda$ is non-dimensional slenderness (for class 4: $A_{eff}$ is used instead of $A$), $\alpha$ is an imperfection factor (according to the profile taken from table 6.2 of the Eurocode), $N_{cr}$ is elastic critical force for the relevant buckling mode.
Buckling curve | $a_0$ | $a$ | $b$ | $c$ | $d$ |
$\alpha$ | 0.13 | 0.21 | 0.34 | 0.49 | 0.76 |
The elastic critical force comes from the Euler theory (1757):
$$ \begin{equation} N_{cr} = \pi^2 \frac{EI}{L^2_{cr}}, \label{eq:buckl_cL}\\ L_{cr} = \beta L, \end{equation} $$letter $\beta$ is coefficient of the length for buckling according to Euler and the basic cases are listed on the picture. In practice we determine non-dimensional slenderness as
$$ \begin{eqnarray} \overline\lambda &=& \frac{\lambda}{\lambda_1}, \\ \lambda_1 &=& \pi \sqrt{\frac{E}{f_y}} = 93.9 \sqrt\frac{235}{f_y} = 93.9 \epsilon,\\ \lambda &=& \frac{L_{cr}}{i}. \end{eqnarray} $$Above $i = \sqrt{{I}/{A}}$ is radius of gyration about relevant axis and for standard sections can be found in tables.
Then
$$ \begin{equation} \overline\lambda = \sqrt {A f_y / N_{cr}} \end{equation} $$is a comparisson between full strength with no buckling against the critical force. Sidesway movement is expected when critical force is reached. For slenderness $\overline{\lambda} \leq 0.2$ (or for $N_{Ed}/N_{cr} \leq 0.04$) the buckling effect can be ignored.
In steel design a beam is usually subjected to buckling against both axis $y,\ z$ at the same time. Therefore we need to know a way how to combine such cases.
Closed sections | Flexural buckling | $\bot y$ or |
$\bot z$ | ||
Doubly symmetrical sections | Flexural buckling | $\bot y$ or |
$\bot z$ or | ||
Torsional buckling | $\omega$ | |
Symmetrical sections | Flexural buckling | $\bot y$ or |
Flexural torsional buckling | $\bot z + \omega$ | |
Non-symmetrical sections | Flexural torsional buckling | $\bot y + \bot z + \omega$ |
Critical length $L_{cr,\omega}$ in $(\ref{eq:buck2})$ is established as an analogy to Euler's method:
rotation allowed: | no constraint |
rotation prevented: | simple support |
deplanation prevented: | fixed support |
deplanation allowed: | simple support |
$I_\omega$ is torsion warping constant, $I_t$ is torsion section constant. The polar moment of inertia $I_p$ is being found as
$$ \begin{equation} I_p = I_y + I_z + Aa^2, \end{equation} $$where $I_y$ and $I_z$ are moments of inertia to the main axes, $A$ is section area, $a$ is the distance of shear center $C_s$ from the centroid $C_g$.
We have to combine
Let us focus on the case $\bot z+ \omega$ only, since the other direction is an analogy: $\lambda_{z\omega} = f(\lambda_z,\ \lambda_\omega)$.
are slendernesses $\lambda_z$, $\lambda_\omega$ ordered from the greatest to the smallest.
$$ \begin{equation} \alpha = \left(\frac{a_z}{i_p}\right)^2\\ i_p = \sqrt{I_p/A}. \end{equation} $$Length $a_z$ is ordinate of $C_s$ to $C_g$ to axis $z$.
We have to combine $\lambda_y$, $\lambda_z$ and $\lambda_\omega $ into $\lambda_{yz\omega}$ or $\lambda_{yz\omega} = f(\lambda_y,\ \lambda_z,\ \lambda_\omega)$
are $\lambda_y$, $\lambda_z$, $\lambda_\omega$ are ordered from the greatest to the smallest. As $\alpha_1 \geq \alpha_2$ are either $\alpha_y$ or $\alpha_z$:
\begin{equation} \alpha_y=\left(\frac{\alpha_y}{i_p}\right)^2, \hskip2em \alpha_z=\left(\frac{\alpha_z}{i_p}\right)^2 \end{equation}