Steel Buildings in Europe

Title Worked Example: Axial compression and bending interaction (N-M Interaction) 5 of 5 8 - 22  z = 2 z 2 z z 1      = 2 2 0,540 0, 704 0, 704 1   = 0,865 N b,z,Rd = M1 z y   Af = 3 10 1, 0 355 9880 0,865     = 3034 kN N Ed = 127 kN < 3034 kN OK 1.4.2. Interaction of axial force and bending moment §6.3.3(4) The interaction factor, k zy is calculated as follows: For z   0,4 : k zy =                             b,z,Rd Ed mLT b,z,Rd Ed mLT 0, 25 0,1 ; 1 0, 25 0,1 max 1 N N C N N C z  The bending moment is linear and constant. Therefore C mLT is 1,0. Annex B Table B.3 k zy =                            3034 127 1 0, 25 ; 1 0,1 3034 127 1 0, 25 max 1 0,1 0,540 = max (0,997, 0,994) = 0,997 Annex B Table B.2 b,Rd y,Ed zy b,z,Rd Ed M M k N N  = 581 0,997 356 3034 127  = 0,653 < 1,0 OK