Steel Buildings in Europe

Title 5.4 Worked Example – Column Splice 4 of 8 5 – 96 Block tearing resistance For concentrically loaded bolt group: N bt,Rd = V eff,1,Rd § 3.10.2(3) 2 e 2 = 2  55 = 110 mm p 2 = 150 ≤ 2 e 2 Hence A fp,nt = t p ( 2 e 2 – d 0 ) = 12 (2  55 – 22) = 1056 mm 2 A fp,nv = 2 t p ( e 1 + ( n 1 – 1) p 1 – ( n 1 – 0,5) d 0 ) = 2  12 (40 + (2 – 1)  160 – (2 – 0,5)  22) = 4008 mm 2 V eff,1,Rd = 3 10 3 1, 0 4008 275 1, 25 1056 430               = 1000 kN N bt,Rd = 1000 kN N t,Rd = min(858; 802; 1000) = 802 kN F Ed = 43 kN ≤ 802 kN, OK Check for the suitability of ordinary bolts. (It is sufficiently accurate to base this calculation on the gross area of the flange) Ref [4] y,uc f,uc f,uc Ed f t b F = 355 12,5 260 43 10 3    = 0,04 < 0,1 There is no significant net tension in the column flange and the use of ordinary bolts in clearance holes is satisfactory. 5.2.1.3. Bolt group resistance = 40 1 1 2 e = 160 p 2 p e =150 = 55 1 1 Flange cover plate Shear and bearing resistance of the flange cover plate Basic requirement: F Ed ≤ F Rd