Page 137 - Structural Steel Designers Handbook AISC, AASHTO, AISI, ASTM, and ASCE-07 Design Standards
P. 137

Brockenbrough_Ch03.qxd  9/29/05  5:05 PM  Page 3.69



                                                           CONNECTIONS


                                                                                          CONNECTIONS  3.69

































                                              FIGURE 3.42 Bolted column splice for biaxial bending. (Source: A. R. Tamboli,
                                              Handbook of Structural Steel Connection Design and Details,  McGraw-Hill,
                                              1999, with permission.)


                                  The flange force due to major-axis bending is

                                                          020φ M   0 20  × 649 ×12
                                                           .
                                                                    .
                                                      F =       px  =          = 110 kips
                                                             d         14 2 .
                                                       f x
                                  The flange force due to minor-axis bending is calculated as follows:
                                                       020φ M           ×  ×
                                                        .
                                                                 .
                                                   M =       py  =  0 20  × 0 9 50 836.  = 376 in ⋅kips
                                                    f
                                                          2            2
                                  Check that

                                                   M = 376  ≤  3  × 9 50 83 6 14100.  ×  ×  .  =   in ⋅kips  OK
                                                    f
                                                            8
                                  Calculate

                                                 
                                        ε=  1  ×14 6 1 − 1 −  376     = 0 523.   in
                                                 
                                               .
                                           4           1410  
                                        T  = 0 75 1 8 50.( .  ×  ) × 0 780 0 523 27 5.  × .  =  .  kips  and  F  = × 27 5 55 0.  =  .  kips
                                                                                      2
                                                                                   f y
                              Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
                                         Copyright © 2004 The McGraw-Hill Companies. All rights reserved.
                                          Any use is subject to the Terms of Use as given at the website.
   132   133   134   135   136   137   138   139   140   141   142