Compression members - The Behaviour and Design of Steel Structures to EC3 - page 93

Design axial load. N Ed = 561 kN, as in Section 3.12.1.
Solution.
Guess χ = 0.3

A ≥ 561 × 10 3 /( 0.3 × 355 ) = 5268 mm 2

Try a 250 × 150 × 8 RHS, with A = 60.8 cm 2 , i y = 9.17 cm, i z = 6.15 cm,
t = 8.0 mm.
For S355 steel with t = 8 mm, f y = 355 N/mm 2 EN 10025-2

ε = ( 235 / 355 ) 0.5 = 0.814
c w /( t ε) = ( 250.0 − 2 × 8.0 − 2 × 4.0 )/( 8.0 × 0.814 ) = 34.7 < 42 T5.2

and so the cross-section is fully effective.

N cr , y = L cr , y

Af y

1
λ 1 =

12000
( 9.18 × 10 )

1
93.9 × 0.814 = 1.710 6.3.1.3(1)

λ y =

i y

N cr , z = L cr , z

Af y 1

λ 1 =

6000
( 6.15 × 10 )

1
93.9 × 0.814 = 1.276 < 1.710
6.3.1.3(1)

λ z =

i z

Bucklingwilloccuraboutthemajor ( y ) axis.Forahot-finishedRHS,usebuckling
curve (a) with α = 0.21 T6.2,T6.1

Φ y = 0.5 [ 1 + 0.21 ( 1.710 − 0.2 ) + 1.710 2 ]= 2.121

6.3.1.2(1)

1
2.121 + √ 2.121 2 − 1.710 2 = 0.296

χ y = 6.3.1.2(1)

γ M 1 = 0.296 × 60.8 × 10 2 × 355

N b , y , Rd = χ Af y

= 640 kN > 561 kN = N Ed
6.3.1.1(3)

1.0

and so the 250 × 150 × 8 RHS is satisfactory.

3.12.4 Example 4 - buckling of double angles

Problem. Two steel 125 × 75 × 10 UAare connected together at 1.5 m intervals
toformthelongcompressionmemberwhosepropertiesaregiveninFigure3.28c.
The minimum second moment of area of each angle is 49.9 cm 4 . The member is
simply supported about its major axis at 4.5 m intervals and about its minor axis
at 1.5 m intervals. Determine the elastic buckling load of the member.