Steel Buildings in Europe

Title APPENDIX D Worked Example: Design of portal frame using elastic analysis 18 of 44 4 - 99  LT =     2 0,51 0,49 0,364 0,4 0,75 0,364     = 0,541  LT = 2 LT 2 LT LT 1        LT = 2 2 0,541 0, 75 0,364 0,541 1    = 1,02 EN 1993-1-1 §6.3.2.3  LT cannot be greater than 1.0, therefore:  LT = 1,0 M b,Rd = M1 pl,y y LT   W f = 6 3 10 1, 0 355 10 1, 0 2194      = 779 kNm M Ed = 616 kNm < 779 kNm OK Interaction of axial force and bending moment – out-of-plane buckling Out-of-plane buckling due to the interaction of axial force and bending moment is verified by satisfying the following expression: 1, 0 b,Rd y,Ed zy b,z,Rd Ed   M M k N N EN 1993-1-1 §6.3.3(4) For z   0.4, the interaction factor, k zy is calculated as: k zy =                             b,Rd,z Ed mLT b,Rd,z Ed mLT 0, 25 0,1 ; 1 0, 25 0,1 max 1 N N C N N C z  EN 1993-1-1 Annex B Table B.2 C mLT =  0, 6 0, 4   = 616 444 = 0,721 C mLT = 0, 6 0, 4 0, 721   = 0,888  0,4  C mLT = 0,888 EN 1993-1-1 Annex B Table B.3 k zy =                            3731 168 0,888 0, 25 0,1 ; 1 3731 168 0,888 0, 25 max 1 0,1 0, 448 k zy = max (0,996; 0,993) = 0,996 b,Rd y,Ed zy b,z,Rd Ed M M k N N  = 779 0,996 616 3731 168  = 0,832 < 1,0 OK 7.5.4. Lower segment (3800 mm) As previously the flexural buckling resistance and the lateral-torsional buckling resistance are checked individually and then the interaction between the two is verified by using interaction Expression 6.62.