Steel Buildings in Europe

Title Worked Example: Web Resistance and Shear Resistance 2 of 2 8 - 28 k F = 3,41 < 6 ℓ e = y w 2 F w 2 f h k E t but ≤ s s + c EN 1993-1-5 Eq (6.13) ℓ e = 468 2 355 10,2 210000 3,41 2     = 224 ≤ 100 + 10 = 110 therefore ℓ e = 110 ℓ y1 = s s + 2 t f   2 1 1 m m   = 100 + 2  16   1 19,6 17,11   = 325 mm EN 1993-1-5 Eq (6.10) ℓ y2 = ℓ e + t f 2 2 f e 1 2 m t m          = 110 + 16 17,11 16 110 2 19,6 2         = 248 mm EN 1993-1-5 Eq (6.11) ℓ y3 = ℓ e + t f 2 1 m m  = 110 + 16 19,6 17,22  = 207 mm EN 1993-1-5 Eq (6.12) ℓ y = min ( ℓ y1 ; ℓ y2 ; ℓ y3 ) = min (325; 248; 207) = 207 mm F cr = 0,9 k F w 3 w h E t = 0.9  3,41  210000  468 10,2 3 = 1461406 N F  = cr y w y F t f  = 1461406 207 10,2 355   = 0,72 F  = 0, 72 > 0,5 Therefore the initial assumption was correct and the web resistance can be calculated based on this value of  F . Should the calculated value of  F be less than 0,5 then the calculation would need to be carried out again, using the appropriate expression for M 2 χ F = 0,72 0,5 0,5 F   = 0,69 χ F = 0,69 L eff = χ F ℓ y L eff = 0,69  207 = 143 mm F Rd = M1 eff w y  f L t = 1,0 143 10,2 355   = 518 kN EN 1993-1-5 § 6.2 (1)