Lateral buckling of beams - The Behaviour and Design of Steel Structures to EC3 - page 277

Using Figure 6.7, α m = 1.35

= 0.4 × 1.35 × ( − 612.2 / 2 ) × 1449 × 10 3
569 × 10 6

0.4 α m z Q N cr , z
M zx0

=− 0.421

Adapting equation 6.11,
M cr = 1.35 × 569 { √ [ 1 + ( − 0.421 ) 2 ]+ ( − 0.421 ) } Nmm = 510kNm.

(UsingthecomputerprogramPRFELB[18]with L = 7806mmleadsto M cr =
522kNm.)

Member resistance.
Using equation 6.25, λ LT = √ ( 974 / 510 ) = 1.382

h / b = 612.2 / 229.0 = 2.67 > 2

Using the EC3 simple general method with β = 1.0, λ LT ,0 = 0.2 (Clause 6.3.2.2)
and α LT = 0.34 (Tables 6.4 and 6.3), and equation 6.27,

Φ LT = 0.5 { 1 + 0.34 ( 1.382 − 0.2 ) + 1.0 × 1.382 2 }= 1.656

Using equation 6.26,
M b , Rd = 974 / { 1.656 + √ ( 1.656 2 − 1.382 2 ) } kNm
= 379kNm < 433kNm = M Ed

and the beam appears to be inadequate.
Using the EC3 less conservative method with β = 0.75, λ LT ,0 = 0.4 (Clause
6.3.2.3) and α LT = 0.49 (Tables 6.5 and 6.3), and equation 6.27,

Φ LT = 0.5 { 1 + 0.49 ( 1.382 − 0.4 ) + 0.75 × 1.382 2 }= 1.457

Using equation 6.26,
M b , Rd = 974 / { 1.457 + √ ( 1.457 2 − 0.75 × 1.382 2 ) } kNm
= 426kNm < 433kNm = M Ed

and the design moment resistance still appears to be inadequate.
Thedesignmomentresistanceisfurtherincreasedbyusingequations6.29,6.30,
and 6.28 to find
f = 1 − 0.5 × ( 1 − 1 / √ 1.35 ) { 1 − 2 × ( 1.382 − 0.8 ) 2 }= 0.978, and
M b , Rd , mod = 426 / 0.978 = 436kNm > 433kNm = M Ed

and the design moment resistance is adequate after all.