Compression members - The Behaviour and Design of Steel Structures to EC3 - page 94

Member buckling about the major axis.

N cr , y = π 2 × 210000 × 1495 × 10 4 / 4500 2 N = 1530 kN.

Member buckling about the minor axis.

N cr , z = π 2 × 210000 × 164.2 × 10 4 / 1500 2 N = 1513 kN.

Single angle buckling.

N cr , min = π 2 × 210000 × 49.9 × 10 4 / 1500 2 N = 459.7 kN

and so for both angles 2 N cr , min = 2 × 459.7 = 919 kN < 1520 kN.
It can be seen that the lowest buckling load of 919 kN corresponds to the case
where each unequal angle buckles about its own minimum axis.

3.12.5Example - effective length factor in
an unbraced frame

Problem. Determine the effective length factor of member 1-2 of the unbraced
frame shown in Figure 3.29a.
Solution. Using equation 3.48,

6 EI / L
1.5 × 0 + 6 EI / L = 1.0

k 1 =

6 EI / L
1.5 × 6 ( 2 EI )/( 2 L ) + 6 EI / L = 0.4

k 2 =

Using Figure 3.21b, L cr / L = 2.3

205.8

14.2

209.6 y

9.4

5 N 5 N

z

203 × 203 UC 60
I y = 6125 cm 4
I z = 2065 cm 4
A = 76.4 cm 2
i z = 5.20 cm
r = 10.2 mm
(b) Example 7

2

2 EI

3

L

EI EI

1

2 L

(a) Example 5

Figure 3.29 Examples 5and 7.