Local buckling of thin-plate elements - The Behaviour and Design of Steel Structures to EC3 - page 148

Using the lower of the web and plate yield stresses of f yw = 355 N/mm 2 and
ε = 0.814,

A eff , s = 2 × ( 16 × 100 ) + ( 2 × 15 × 0.814 × 10 + 16 ) × 10
= 5801 mm 2 EC3-1-5 F9.1

I eff , s = ( 2 × 100 + 10 ) 3 × 16 / 12 = 12.35 × 10 6 mm 4 EC3-1-5 F9.1

i eff , s = √ 12.35 × 10 6 / 5801 = 46.1 mm.
λ 1 = 93.9 × 0.814 = 76.4

6.3.1.3(1)

L cr = 1.0 × 1500 = 1500 mm EC3-1-5 9.4(2)

λ = 1500 /( 46.1 × 76.4 ) = 0.426 6.3.1.3(1)

For buckling curve c, α = 0.49 EC3-1-5 9.4(2),T6.1

Φ = 0.5 ×[ 1 + 0.49 × ( 0.426 − 0.2 ) + 0.426 2 ]= 0.646

6.3.1.2(1)

χ = 1 / [ 0.646 + √ ( 0.646 2 − 0.426 2 ) ]= 0.884 6.3.1.2(1)

N b , Rd = 0.884 × 5801 × 355 / 1.0 N 6.3.1.1(3)

= 1820 kN > 706.4 kN = N Ed , s

OK.

a / h w = 1800 / 1500 = 1.20 < √ 2 EC3-1-5 9.3.3(3)

1.5 h w t w / a 2 = 1.5 × 1500 3 × 10 3 / 1800 2 EC3-1-5 9.3.3(3)

= 1.563 × 10 6 mm 4 < 12.35 × 10 6 mm 4 = I s OK.

4.9.9 Example 9 - load-bearing stiffener

Problem. Checktheadequacyofapairofload-bearingstiffeners100 × 16ofS460
steel which are above the support of the plate girder of Section 4.9.6.The flanges
of the girder are not restrained by other structural elements against rotation. The
girder is supported on a stiff bearing s s = 300mm long, the end panel width is
1000mm, and the design reaction is 1400kN.

Stiffener yield resistance.

t s = 16 mm, f y = 460 N / mm 2 EN10025-2

ε = √ ( 235 / 460 ) = 0.715

T5.2

c s /( t s ε) = 100 /( 16 × 0.715 ) T5.1

= 8.74 < 9 and so the stiffeners are fully effective. T5.1

F s , Rd = 2 × ( 100 × 16 ) × 460 / 1.0 N = 1472 kN.