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Brockenbrough_Ch03.qxd  9/29/05  5:05 PM  Page 3.98



                                                        CONNECTIONS


                   3.98  CHAPTER THREE

                               beam as the beam-to-column connection is, it is advantageous to resist the moment here. To accomplish
                               this, an H c ′ is calculated in a manner similar to the H c calculation in the uniform force method, as follows:

                                                              H c ′ =  e c  R                      (3.75)
                                                                  e b  + β
                               The beam reaction is then distributed between the beam-to-column connection and the gusset-to-
                               column connection so as to eliminate moments in the extended single-plate connection. This is done
                               as follows:
                                                                    β
                                                              V c ′ =  R                           (3.76)
                                                                  e b  + β
                                                              V b ′ =  e b  R                      (3.77)
                                                                  e b  + β

                               These forces can then be included in the equations used with the uniform force method so that
                                                               β
                                                            V =  P + ∆ V + ′                       (3.78)
                                                                        V
                                                             c
                                                                r     b  c
                                                           H =  e c  PH′                           (3.79)
                                                                   +
                                                             c
                                                                r    c
                                                            V =  e b  P − ∆ V + ′ b                (3.80)
                                                                         V
                                                             b
                                                                      b
                                                                r
                               The moments at the interfaces can be calculated as
                                                               R           R 
                                                      M = α  V −  2    −  eH −  e c            (3.81)
                                                                    b
                                                        b
                                                                       b
                                                             b
                                                                            2
                                                      M =  e V −  H β                              (3.82)
                                                                c
                                                           c c
                                                        c
                                                                            −
                                                                    −
                                                      M =−β  H + α( V V +  e V e H                 (3.83)
                                                                       )
                                                                      c
                                                              c
                                                        g
                                                                          c
                                                                              b
                               Example of Extended Single-Plate Connection for Vertical Brace.  The above formulation of the
                               uniform force method will be applied to the design of the connection shown in Figs. 3.54 and 3.55.
                                 Determine Interface Forces
                                            θ
                                                              .9
                                       H = sin( ) P = sin(59 .74 )89  = 76 kips
                                             θ
                                                              .8
                                       V = cos( ) P = cos(59 .74 )89  = 44 kips
                                       α  =  e  tan ( ) −  e + tan( ) ( .900 ) tan(5974 )  −108  + ( ) tan(5974 )  = 201 in
                                                         θ
                                               θ
                                                           =
                                                     β

                                                                                       .

                                                                                9
                                                                            .
                                                                      .
                                                                                             .
                                                  c
                                           b
                                                                            +
                                       r = (α +  e ) 2  + (β +  e ) 2  = (20 .1 +10 . )  2  + (9..00 900 ) =  357 in
                                                                                 2
                                                                                      .
                                                                    8
                                                                              .
                                               c
                                                       b
                                            β
                                                  . 900 
                                      V c  = P  =  89   =  22 .4 kips
                                             r   35 .7 
                                            e c      10 .8 
                                      H c  = P  =  89   =  26 .8 kips
                                            r      35 .7 
                                            e b      . 900 
                                      V b  = P  =  89   =  22 .4 kips
                                            r      35 .7 
                                            α
                                                   20 .1
                                                           .
                                      H b  = P    =  89   35 .7   = 50 1 kips
                                            r
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