Frames - The Behaviour and Design of Steel Structures to EC3 - page 379

For antisymmetric buckling,

R = ( 12080 × 10 4 / 4000 )/( 12080 × 10 4 / 12369 ) = 3.09 ( 8.13 )

ρ c = 3 /( 10 × 3.09 + 12 ) = 0.0699 (8.12a)

ρ r = 1 (8.12b)

N cr , c = π 2 × 210000 × 12080 × 10 4 / 4000 2 N = 15648 kN (8.10)

N cr , r = π 2 × 210000 × 12080 × 10 4 / 12369 2 N = 1637 kN (8.10)

1.2 − 1 / 1.2
= 13.0

1.2

60.3
0.0699 × 15648

50.5
1 × 1637

α cr , as = + (8.9)

For symmetric buckling,

R H = ( 12369 / 4000 ) × ( 3000 / 12369 ) = 0.75 (8.15)

4.8 + 12 × 3.09 × ( 1 + 0.75 )
2.4 + 12 × 3.09 × ( 1 + 0.75 ) + 7 × 3.09 2 × 0.75 2

ρ c =

= 0.664 (8.14a)

ρ r = 12 × 3.09 + 8.4 × 3.09 × 0.75
12 × 3.09 + 4

= 1.377 (8.14b)

1.2 − 1 / 1.2 1.2

60.3
0.664 × 15648

50.5
1.377 × 1637

α cr , s = + = 38.4

> 13.0 = α cr , as (8.9)

andsotheframebucklingfactoris α cr = 13.0.Thisisreasonablyclosetothevalue
of 13.9 predicted by a computer elastic frame buckling analysis program [18].

8.5.6 Example 6 - moment amplification
in a braced frame

Problem .Determinetheamplifieddesignmomentsforthebracedframeshownin
Figure 8.3. (Second-order effects are rarely important in braced members which
havesignificantmomentgradients.ItcanbeinferredfromClause5.2.2(3)bofEC3
that second-order effects need not be considered in braced frames. Nevertheless,