Compression members - The Behaviour and Design of Steel Structures to EC3 - page 95

3.12.6 Example 6 - effective length factor in a braced frame

Problem. Determinetheeffectivelengthfactorofmember1-2oftheframeshown
in Figure 3.29a if bracing is provided which prevents sway buckling.
Solution.
Using equation 3.44,
k 1 = 2 EI / L

0.5 × 0 + 2 EI / L = 1.0
k 2 = 2 EI / L

0.5 × 2 ( 2 EI )/( 2 L ) + 2 EI / L = 0.667
Using Figure 3.21a, L cr / L = 0.87.

3.12.7 Example 7 - checking a non-uniform member

Problem. The 5.0 m long simply supported 203 × 203 UC 60 member shown in
Figure 3.29b has two steel plates 250 mm × 12 mm × 2 m long welded to it,
one to the central length of each flange. This increases the elastic buckling load
about the minor axis by 70%. If the yield strengths of the plates and the UC are
355 N/mm 2 , determine the compression resistance.
Solution.
ε = ( 235 / 355 ) 0.5 = 0.814

T5.2

c f /( t f ε) =[ ( 205.8 − 9.3 − 2 × 10.2 )/ 2 ] /( 14.2 × 0.814 ) = 7.62 < 14

T5.2
c w /( t w ε) = ( 209.6 − 2 × 14.2 − 2 × 10.2 )/( 9.3 × 0.814 ) = 21.24 < 42
T5.2

and so the cross-section is fully effective.
N cr , z = 1.7 × π 2 × 210000 × 2065 × 10 4 / 5000 2 = 2910 kN.

Af y
N cr , z =

76.4 × 10 2 × 355
2910 × 10 3

λ z = = 0.965

For a rolled UC section with welded flanges ( t f ≤ 40 mm), buckling about the
z -axis, use buckling curve (c) with α = 0.49

T6.2,T6.1

Φ z = 0.5 [ 1 + 0.49 ( 0.965 − 0.2 ) + 0.965 2 ]= 1.153 6.3.1.2(1)

1

χ z = √

1.153 2 − 0.965 2 = 0.560

6.3.1.2(1)

1.153 +

γ M 1 = 0.560 × 76.4 × 10 2 × 355

N b , z , Rd = χ Af y

= 1520 kN 6.3.1.1(3)

1.0