Steel Buildings in Europe

Title Worked Example: Axial compression and bending interaction (N-M Interaction) 2 of 5 8 - 19 1.2. Buckling verification The buckling checks due to the interaction of axial compression and bending moment are carried out using expressions 6.61 and 6.62 from EN 1993-1-1. 1, 0 M1 z,Rk z,Ed z,Ed yz M1 y,Rk LT y,Ed y,Ed yy M1 y Rk Ed             M M M k M M M k N N 1, 0 M1 z,Rk z,Ed z,Ed zz M1 y,Rk LT y,Ed y,Ed zy M1 z Rk Ed             M M M k M M M k N N Expressions (6.61) and (6.62) These expressions can be simplified as follows: y,Ed M  = 0 and z,Ed M  = 0 for Class 1, Class 2 and Class 3 sections. M z,Ed = 0 Therefore expressions (6.61) and (6.62) can be written as: 1,0 b,Rd y,Ed yy b,y,Rd Ed   M M k N N and 1,0 b,Rd y,Ed zy b,z,Rd Ed   M M k N N 1.3. Equation 6.61 (EN 1993-1-1) 1.3.1. Flexural buckling resistance about the major axis, N b,y,Rd b h  190 450  2,37 t f  14,6 mm buckling about y-y axis:  Curve a for hot rolled I sections   y  0,21 Table 6.1 Table 6.2  1 = y f E  = 355 210000  = 76,4 §6.3.1.3 y  = y 1 cr 1  i L = 76,4 1 185 1700  = 0,12  y =     2 y y y 0, 2 0,5 1        y =     2 0,5 1 0, 21 0,12 0, 2 0,12    = 0,50 §6.3.1.2  y = 2 y 2 y y 1      = 2 2 0,50 0,50 0,12 1   = 1,0