Lateral buckling of beams - The Behaviour and Design of Steel Structures to EC3 - page 286

Elastic buckling (flange in compression).
Using Figure 6.27,

d f = 306.0 − ( 19.6 + 0 )/ 2 = 296.2 mm.
z = 0 + ( 306.0 − 19.6 − 0 ) × ( 306.0 − 0 ) × 11.9 / 2

( 229.0 × 19.6 ) + 0 + ( 306.0 − 19.6 − 0 ) × 11.9 = 66.0 mm.
I y = ( 229.0 × 19.6 × 66.0 2 ) + 0 + ( 306.0 − 19.6 − 0 ) 3 × 11.9 / 12
+ ( 306.0 − 19.6 − 0 ) × 11.9 ×{ 66.0 − ( 306.0 − 0 )/ 2 } 2
= 68.64 × 10 6 mm 4
1 − ρ m = 0
z 0 = 0 × 296.2 − 66.0 =− 66.0 mm.

(
)

( 296.2 − 66.0 ) × ( 0 + 0 ) − 66.0

× 229 3 × 19.6 / 12 + 229 × 19.6 × 66.0 2

1
68.64 × 10 6

+ ( 296.2 − 66.0 − 0 ) 4

β y =

*

− ( 66.0 − 19.6 / 2 ) 4 × 11.9 / 4

− 2 × ( − 66.0 ) = 215.6 mm.
I w = 0
π 2 EI z

L 2 = π 2 × 210000 × 1966 × 10 4

= 1.630 × 10 6 N

5000 2

β y
2

π 2 EI z
L 2

= 215.6
2

1.630 × 10 6 = 137.6 × 10 3 ×

Using equation 6.76,

( 1.633 × 10 6 ) ×

[ 81000 × 76.9 × 10 4 + 0 + ( 137.6 × 10 3 ) 2 ]
+ 137.6 × 10 3 = 540.1 kNm.

M cr =

(Using the computer program PRFELB [18] leads to M cr = 540.1 kNm.)
Member resistance (flange in compression).

Using equation 6.25, λ LT = √ ( 140.7 / 540.3 ) = 0.510
Using the EC3 simple general method with β = 1.0, λ LT ,0 = 0.2 (Clause
6.3.2.2) and α LT = 0.76 (Tables 6.4 and 6.3), and equation 6.27,

Φ LT = 0.5 { 1 + 0.76 ( 0.510 − 0.2 ) + 1.0 × 0.510 2 }= 0.748