Buckling resistance of members (6.3)

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$
Firstly the cross-section is categorized by means of table 6.2 within the Eurocode. From the table buckling curve is found and then from the above table (Table 6.1 within the code) imperfection factor $\alpha$ is taken to evaluate $(\ref{eq:buck_fi})$, $(\ref{eq:buck_chi})$. The other approach is to use curves from the below chart.

Buckling reduction factor $\chi$ can be read for given section (ie. for buckling curve) and for given slenderness

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.

The factor $\beta$ for basic configurations to determine critical length in $(\ref{eq:buckl_cL})$ (by Euler, 1757).
Example from the picture: the second case is less stable compared to the third one. Thus the critical force $N_{cr}$ for the second case needs to be four times smaller (to show buckling)


$$ \begin{equation} \overline\lambda = \sqrt {A f_y / N_{cr}} \end{equation} $$

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.

a) Double symmetric sections are subjected either to flexural buckling (against axis $y$ or $z$) or to torsional buckling $\omega$
b) Symmetric sections are subjected to flexural buckling against soft axis or to a combination of lateral and torsional buckling
c) Non-symmetric sections are subjected to lateral torsional buckling (combination of $y$, $z$, $\omega$)
Note: the red dot above is shear center
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 sectionsFlexural torsional buckling $\bot y + \bot z + \omega$

Allowance of combinations $\boldsymbol{\bot y+ \omega}$ and $\boldsymbol{\bot y+\bot z+\omega}$

$$ \begin{equation} \lambda_y = \frac{L_{cr,y}}{i_y}, \hskip2em \lambda_z = \frac{L_{cr,z}}{i_z}, \hskip2em \lambda_{\omega} = \sqrt{\frac{I_p}{\frac{I_\omega}{L^2_{cr,\omega}} + \frac{I_t}{25}}} \label{eq:buck2} \end{equation} $$

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 $C_g$.

Combination $\boldsymbol{\bot y+ \omega}$

We have to combine

Let us focus on the case $\bot z+ \omega$ only, since the other direction is an analogy: $\lambda_{z\omega} = f(\lambda_z,\ \lambda_\omega)$.

A. Slendernesses differ much

$$ \begin{equation} \lambda_{z\omega}=\sqrt{\lambda_1^2 + \alpha \lambda_2^2}, \hskip2em \text{where} \lambda_1 \geq \lambda_2 \end{equation} $$

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 $z$.

B. Slendernesses do not differ much

$$ \begin{eqnarray} \lambda_{z\omega} &=& \kappa_z \sqrt{\lambda_\omega^2 + \lambda_z^2}, \\ \kappa_z &=& \sqrt{\frac{1+a_z/i_p}{2}} \end{eqnarray} $$

Combination $\boldsymbol{\bot y+ \bot z + \omega}$

We have to combine $\lambda_y$, $\lambda_z$ and $\lambda_\omega $ into $\lambda_{yz\omega}$ or $\lambda_{yz\omega} = f(\lambda_y,\ \lambda_z,\ \lambda_\omega)$

A. Some of the slendernesses prevails

$$ \begin{equation} \lambda_{yz\omega}=\sqrt{\lambda_1^2 + \alpha_1 \lambda_2^2 + \alpha_2 \lambda_3^2}, \hskip2em \text{where}\ \ \lambda_1 \geq \lambda_2\geq \lambda_3 \end{equation} $$

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}

B. Slendernesses are in a range

\begin{eqnarray} \lambda_{yz\omega} &=& \kappa_z\sqrt{\lambda^2_\omega + \lambda^2_y + \lambda_z^2}\\ \kappa_z &=& \sqrt{\frac{1+a/i_p}{3}} \end{eqnarray}

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