Page 71 - Handbook of Structural Steel Connection Design and Details
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Design of Connections for Axial, Moment, and Shear Forces

                    56    Chapter Two

                    section. Note that this can be done because M of Fig. 2.4, although shown
                    at the centroid of the section, is actually a free vector that can be applied
                    anywhere on the section or indeed anywhere on the free-body diagram.
                    This being the case, there is no reason to assume that the bending stress
                    distribution is symmetrical about the center of the section. Considering
                    the distribution shown in Fig. 2.6, because the stress from A to the center
                    is too high, the zero point of the distribution can be allowed to move
                    down the amount e toward B. Equating the couple M of Fig. 2.4 to the
                    statically equivalent stress distribution of Fig. 2.6 and taking moments
                    about point D,
                                           t         2           2
                                      M 5 [f sa 1 ed 1 f sa 2 ed ]
                                                          2
                                              1
                                           2
                    where t is the gusset thickness. Also, from equilibrium
                                          f (a   e) t   f (a   e)t
                                           1           2
                    The above two equations permit a solution for f and f as
                                                                1     2
                                                      M
                                              f 5
                                               1
                                                   atsa 1 ed
                                                      M
                                              f 5
                                               2
                                                   atsa 2 ed
                    For a uniform distribution of normal stress,
                                              f   f   f   f
                                              1    a   2   a
                    from which e can be obtained as

                                         1      M    2          M
                                                           2
                                      e 5  c  a     b 1 4a 2       d
                                         2 B at f a            at f a
                    Substituting numerical values,

                         1          7280       2       2        7280
                     e 5  c   a               b 1 4s21d 2                 d 5 8.10 in
                         2 B s21ds0.75ds9.97d              s21ds0.75ds9.97d

                    Thus,
                                                7280
                                   f 5                      5 15.9 ksi
                                    1
                                        s21ds0.75ds21 1 8.10d
                                                7280
                                   f 5                      5 35.8 ksi
                                    2
                                        s21ds0.75ds21 2 8.10d



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