Beam-columns - The Behaviour and Design of Steel Structures to EC3 - page 338

Using α m of equation 6.5 for C 1 with β m = 0 leads to C 1 = 1.75

1 − N Ed
N cr , T

1 − N Ed
N cr , z

C 1 4 0.2

1 − 200
1636

√

1 − 200
584.4

1.75 × 4

= 0.2 ×

= 0.231 < 1.093 = λ 0 TA.1

a LT = 1 − I t / I y = 1 − 15.4 / 5537 = 0.997 > 0 TA.1

W el , y = 45 × 10 6 200 × 10 3 × 47.2 × 10 2

ε y = M y , Ed
N Ed

A

433 × 10 3 = 2.453

TA.1

Using N cr , y = 1417 kN from Section 7.7.2,

C my ,0 = 1 − 0.18 N Ed / N cr , y = 1 − 0.18 × 200 / 1417 = 0.975

TA.2

√ ε y a LT
1 + √ ε y a LT

C my = C my ,0 + ( 1 − C my ,0 )

√

2.453 × 0.997
1 + √ 2.453 × 0.997 = 0.990

= 0.975 + ( 1 − 0.975 ) × TA.1

a LT

C mLT = C my

1 − N Ed
N cr , z

1 − N Ed
N cr , T

0.997

= 0.990 2 ×

= 1.287 > 1

TA.1

1 − 200
1636

1 − 200
584.4

w y = W pl , y

W el , y = 483 433 = 1.115,

200 × 10 3
47.2 × 10 2 × 275 / 1.0 = 0.154

N Ed
N Rk /γ M 1 =

n pl =

TA.1

2 − 1.6

w y C my λ max − 1.6 w y C my λ 2

C yy = 1 + ( w y − 1 ) n pl − b LT

max

≥ W el , y
W pl , y