In-plane bending of beams - The Behaviour and Design of Steel Structures to EC3 - page 220

Checking for lateral bracing .

229 3 × 19.6 / 12
229 × 19.6 + ( 612.2 − 2 × 19.6 ) × 11.9 / 6 = 59.1 mm

i f , z = 6.3.2.4

k c L c /( i f , z λ 1 ) = 1.0 × 1500 /( 59.1 × 93.9 × 0.942 ) = 0.287 6.3.2.4

λ c 0 M c , Rd / M Ed = 0.4 × 974 / 837 = 0.465 > 0.287

6.3.2.4

and the bracing is satisfactory.
Checking for shear .

V Ed = 186 × 6 / 2 = 558kN
A v = 159 × 10 2 − 2 × 229 × 19.6 + ( 11.9 + 2 × 12.7 ) × 19.6
= 7654mm 2 6.2.6(3)

V c , Rd = 7654 × ( 265 / √ 3 )/ 1.0 N = 1171 kN > 558 kN = V Ed 6.2.6(2)

which is satisfactory.
Checking for bending and shear . The maximum M Ed occurs at mid-span where
V Ed = 0,andthemaximum V Ed occursatthesupportwhere M Ed = 0,andsothere
is no need to check for combined bending and shear. (Note that in any case, 0.5
V c , Rd = 0.5 × 1171 = 585.5 kN > 558kN = V Ed and so the combined bending
and shear condition does not operate.)

Checking for bearing.

R Ed = 186 × 6 / 2 = 558 kN
t w = 11.9, f y , w = 275 N / mm 2

EN10025-2

h w = 612.2 − 2 × 19.6 = 573.0 mm.

Assume ( s s + c ) = 100 mm, which is not difficult to achieve.

k F = 2 + 6 × 100 / 573.0 = 3.047 EC3-1-5, 6.1(4)(c)
F cr = 0.9 × 3.047 × 210000 × 11.9 3 / 573.0 N = 1694 kN

EC3-1-5, 6.4(1)

m 1 = ( 265 × 229 )/( 275 × 11.9 ) = 18.54 EC3-1-5, 6.5(1)

m 2 = 0.02 × ( 572.7 / 19.6 ) 2 = 17.09 if λ F > 0.5

EC3-1-5, 6.5(1)