Frames - The Behaviour and Design of Steel Structures to EC3 - page 377

For the lower-storey columns, using equation 8.4,

2210 × 10 4 / 5000 + 5259 × 10 4 / 5000
2210 × 10 4 / 5000 + 5259 × 10 4 / 5000 + 1.5 × 14136 × 10 4 / 12000

k T =

= 0.458

k B =∞ (theoretical value for a frictionless hinge).

k cr , l = 2.4(using Figure 3.21b).

N cr , l = π 2 × 210000 × 5259 × 10 4 /( 2.4 × 5000 ) 2 N = 756.9 kN

Using equation 8.7 and the axial forces of Figure 8.4f,

2 × 883.6 / 5.0
( 37.6 + 42.4 )/ 5.0 = 22.1

α cr , su =

2 × 756.9 / 5.0
( 103.3 + 136.7 )/ 5.0 = 6.31 < 22.1 = α cr , su

α cr , sl =

∴ α cr = 6.31 (compare with the storey deflection method value of α cr = 7.31
calculated above and the computer analysis value of α cr = 6.905 given in
Figure 8.4d).

8.5.4 Example 4 - buckling of unbraced single storey frame

Problem . Determine the effective length factor of the member 12 of the unbraced
frameshowninFigure8.13b(andforwhich π 2 EI / L 2 = 1),andtheelasticin-plane
buckling loads.

Solution .Theswaymember12isunrestrainedatend1andmoderatelyrestrained
at end 2 by the lightly loaded member 23. Its effective length factor is therefore
greater than 2, the value for a rigidly restrained sway member (see Figure 3.15e).
Accordingly, assume k cr ,12 = 2.50.
Then, by using equation 8.2, N cr ,12 = π 2 EI /( 2.50 L ) 2 = 0.16,
and Q cr = N cr ,12 / 5 = 0.032,
and N 23 = Q cr = 0.032.
Using Figure 3.19,

1 −

0.032
4 π 2 ( 2 EI )/( 2 L ) 2

α b 23 = { 6 × ( 2 EI )/( 2 L ) } = 5.904 EI / L