Steel Buildings in Europe

Title Worked Example: Axial compression and bending interaction (N-M Interaction) 4 of 5 8 - 21 M b,Rd = M1 pl,y y LT   W f = 6 3 10 1, 0 355 10 1702 0,961      = 581 kNm M Ed = 356 kNm < 581 kNm OK 1.3.3. Interaction of axial force and bending moment The interaction factor, k yy , is calculated as follows: k yy =                              b, y,Rd Ed my b, y,Rd Ed y my 1 0,8 ; 0, 2 1 min N N C N N C  The expression for C my depends on the values of  h and  .  = 1,0. Therefore C my is calculated as: C my = 0,6 + 0,4 ψ = 0,4 + 0,4  1,0 = 1,0 Annex B Table B.3 k yy =                        3507 ; 1 1,0 0,8 127 3507 min 1,0 1 0,12 0,2 127 = min [0,997; 1,029] = 0,997 Annex B Table B.2 b,Rd y,Ed yy b, y,Rd Ed M M k N N  = 581 0,997 356 3507 127  = 0,647 < 1,0 OK The member satisfies the in-plane buckling check. 1.4. Expression 6.62 (EN 1993-1-1) 1.4.1. Flexural buckling resistance about minor axis bending, N b,z,Rd b h  190 450  2,37 t f  14,6 mm buckling about z-z axis  Curve b for hot rolled I sections   z  0,34 Table 6.1 Table 6.2  1 = y f E  = 355 210000  = 76,4 §6.3.1.3 z  = z 1 cr 1  i L = 76, 4 1 41, 2 1700  = 0,540  z =     2 z z z 0, 2 0,5 1        z =     2 0,5 1 0,34 0,540 0, 2 0,540    = 0,704 §6.3.1.2