Steel Buildings in Europe

Title APPENDIX D Worked Example: Design of portal frame using elastic analysis 6 of 44 4 - 87  cr,s,est =                      NHF max R,cr R,Ed 200 1 0,8 1  h N N =                    1,6 6000 200 1 772 0,8 1 130 = 12,5 Appendix B of this document Thus  cr,est =  cr,s,est = 12,5 > 10 First order elastic analysis may be used and second order effects do not need to be allowed for. Section 2.2 of this document 6. Frame imperfections The global initial sway imperfection may be determined from  =  0  h  m  0 = 1/200  h = 0,82 6,0 2 2   h  m = 0,5(1 1 ) 0,87   m = ) 2 0,5(1 1  = 0,87 m = 2 (number of columns)  = 3 0,82 0,87 3,56 10 200 1      EN 1993-1-1 §5.3.2 Initial sway imperfections may be considered in two ways:  By modeling the frame out of plum  By applying equivalent horizontal forces (EHF). Applying equivalent horizontal forces is the preferred option and the method that is used in this worked example. The equivalent horizontal forces are calculated as: H EHF =  V Ed However sway imperfections may be disregarded where H Ed  0,15 V Ed. EN 1993-1-1 §5.3.2(4) Table 1 shows the total reactions for the structure to determine H Ed and V Ed . Table 1 Vertical and horizontal reactions Left column (kN) Right column (kN) Total reaction (kN) 0,15 VEd (kN) H Ed V Ed H Ed V Ed H Ed V Ed Reactions 116 168 –116 168 0 336 50 H Ed = 0  0,15 V Ed Therefore the initial sway imperfections have to be taken into account.