3. Flexural buckling¶

The idea¶

A slender column loaded in compression can bend sideways and fail below its squash load \(N_{pl}\). The ratio between the buckling resistance and the squash load is the reduction factor \(\chi\). It depends only on the relative slenderness \(\bar{\lambda}\) and on the imperfections of the member.

Equations¶

Squash load and relative slenderness:

\[ N_{pl} = A\,f_y \qquad \bar{\lambda} = \frac{L_{cr}/i}{\lambda_1} \qquad \lambda_1 = \pi\sqrt{\frac{E}{f_y}} \]

Reduction factor:

\[ \Phi = 0.5\left[1 + \alpha\,(\bar{\lambda} - 0.2) + \bar{\lambda}^{2}\right] \qquad \chi = \frac{1}{\Phi + \sqrt{\Phi^{2} - \bar{\lambda}^{2}}} \le 1.0 \]

Design buckling resistance:

\[ N_{b,Rd} = \frac{\chi\,A\,f_y}{\gamma_{M1}} \]

The imperfection factor \(\alpha\) depends on the buckling curve:

Curve	a0	a	b	c	d
\(\alpha\)	0.13	0.21	0.34	0.49	0.76

The curve to use for a rolled I-section depends on the ratio \(h/b\), the flange thickness and the axis. The selection table is in the design standard.

Worked example: IPE 330, S235¶

Data: \(A = 62.6\) cm², \(i_y = 137\) mm, \(i_z = 35.5\) mm, \(f_y = 235\) N/mm², \(\gamma_{M1} = 1.0\). The buckling length is 6 m about the y-axis and 3 m about the z-axis (an intermediate support about the weak axis). Here \(h/b = 2.06\) and \(t_f = 11.5\) mm, so curve a is used about the y-axis and curve b about the z-axis.

First, the values that do not depend on the axis:

\[ N_{pl} = 6260 \times 235 / 1000 = 1471.1\ \text{kN} \qquad \lambda_1 = \pi\sqrt{\frac{210\,000}{235}} = 93.91 \]

Quantity	y-axis	z-axis
\(L_{cr}\) [mm]	6000	3000
\(L_{cr}/i\)	43.80	84.51
\(\bar{\lambda}\)	0.466	0.900
Curve, \(\alpha\)	a, 0.21	b, 0.34
\(\Phi\)	0.637	1.024
\(\chi\)	0.934	0.661
\(N_{b,Rd}\) [kN]	1374.6	972.8

The weak axis governs: \(N_{b,Rd} = 972.8\) kN. For a design force of \(N_{Ed} = 500\) kN the utilization is \(500 / 972.8 = 0.51\).

PythonResult

import numpy as np

def flexural_buckling(A_cm2, i_mm, Lcr_mm, alpha, fy=235.0, E=210000.0, gamma_M1=1.0):
    lam1 = np.pi * np.sqrt(E / fy)
    lam = (Lcr_mm / i_mm) / lam1                      # relative slenderness
    phi = 0.5 * (1 + alpha * (lam - 0.2) + lam**2)
    chi = min(1.0, 1.0 / (phi + np.sqrt(phi**2 - lam**2)))
    Nb_Rd = chi * A_cm2 * 100 * fy / gamma_M1 / 1000   # kN
    return lam, chi, Nb_Rd

print(flexural_buckling(62.6, 137.0, 6000, alpha=0.21))   # y-axis
print(flexural_buckling(62.6, 35.5, 3000, alpha=0.34))    # z-axis

(0.466, 0.934, 1374.6)    y-axis
(0.900, 0.661,  972.8)    z-axis

The same example, with the figure of the buckling curves, is in the article Flexural buckling of an IPE 330 column.