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Brockenbrough_Ch03.qxd  9/29/05  5:05 PM  Page 3.100



                                                        CONNECTIONS


                   3.100  CHAPTER THREE

                                               =
                                                                      ⋅
                                   M =  e V −  H β 108 347) − 900 414)  = 0 in kips
                                                      .
                                                  .(
                                                            (
                                                               .
                                                          .
                                       c c
                                    c
                                             c
                                                        −
                                                −
                                  M =−β  H + α( V V +  e V e H =−9 00 41 4) + 9 00 10 2)  +10 75 44 8) − 8 995 76 9)
                                                                         .
                                                                            (
                                                                              .
                                                                     .
                                                   )
                                                                .
                                                                  (
                                                                                        .
                                                                                            .
                                                                                               (
                                                                                     (
                                                                                                  .
                                                                                   .
                                                          b
                                                  c
                                                      c
                                          c
                                    g
                                           ⋅
                                     = 491 in kips
                                 Determine Interface Forces for Bottom Bracket
                                            θ
                                      H = sin( ) P = sin(59 .74 )140  = 121 kips
                                            θ
                                                               .5
                                      V = cos( ) P = cos(59 .74 )140  = 70 kips
                                                        θ
                                                    β
                                              θ
                                                          =
                                      α  =  e tan( ) − e + tan( ) ( .900 )tan(5974 )  −108  + ( . )tan(5974 )  = 227 in
                                                                     .
                                                                                             .
                                                                                       .
                                                                              1
                                                                               05
                                                                           .
                                          b
                                                  c
                                       r = (α  +  e ) 2  + (β  +  e ) 2  = (22 .7 +10 . ) 2  + ( .5 9900.  ) =  387 in
                                                                           +
                                                                                2
                                                                   8
                                                                        10
                                                                                     .
                                                      b
                                               c
                                            β
                                                 10 5 . 
                                      V c  = P  = 140   =  38 0 . kips
                                             r   38 7 . 
                                            e c     10 8 . 
                                     H c  = P    = 140   =  39 1 . kips
                                            r      38 7 . 
                                                     .
                                            e b      900 
                                      V b  = P  =  140  =  32 6 . kips
                                            r     38 7 . 
                                            α
                                                  22 7 . 
                                                           .
                                     H b  = P    =  140  38 7 .    = 82 1 kips
                                            r
                                 Distribute Half of Beam Reaction to Top Gusset
                                                . ( . ) 8
                                       R )
                                                         .6
                                  H c ′ =  (/2  e c  =  24 5 10  = 13 kips
                                       e b  + β  . 900  +105
                                                    .
                                       R )β
                                               . ( . ) 5
                                  V c ′ =  (/2  =  24 5 10  = 13 .2 kips
                                      e b  + β  . 900  +105
                                                    .
                                                . ( . 900
                                       R )
                                  V b ′ =  (/2  e b  =  245  )  = 11 kips
                                                         .3
                                                    .
                                       e  + β  . 900  +105
                                       b
                                         ∆
                                            + V
                                  V b  = V b  − V + ′ = 32 6 .  − +11 3 .  = 43 9 . kips
                                                      0
                                               b
                                           b
                                         ∆
                                                      0
                                              V
                                  V c  = V c  − V b  + ′ = 38 0 .  + +13 2 .  = 51 2 . kips
                                               c
                                                 +
                                               .
                                  H c  = H c  + H ′ = 39 1 13 6 .  = 52 7 . kips
                                           c
                                                        R
                                                                   −
                                                                                                 ⋅
                                                                             (
                                                               (
                                                                          .
                                                                 .
                                                                                     8
                                                            .
                                                                                        5
                                  M b  = α   V b  −  R    − eH b  − e c    = 9 00 43 9 24 5 . ) − 9 00 68.. )5 − 10 . (24 . ) =  707 in kips
                                                       
                                                b
                                            2
                                                        2
                                                                     ⋅
                                                       −
                                  M c  = e V  − H β  =  10 .(51 . ) 10 .(52 . ) =  0 in kips
                                                  8
                                                      2
                                                           5
                                                              7
                                       c c
                                            c
                                                                               +
                                                       −
                                                                      +
                                                                                        −
                                       β
                                  M g  =− H c  + α ( − V c ) + e V e H  =− 10 . (52 . ) 9 .00 ( . ) 10 . (70 . ) 8 .995 (121 )
                                                                           19
                                                                     7
                                                                                   8
                                                                                      5
                                                                              3
                                              V
                                                                 5
                                                     c
                                                          b
                                     =  707 in kips
                                           ⋅
                               Nonorthogonal Trusses.  The uniform force method as originally formulated can be applied to
                               trusses as well as to bracing connections. After all, a vertical bracing system is just a truss, as can be
                               seen in Fig. 3.55, which shows various arrangements. However, bracing systems generally involve
                               orthogonal members whereas trusses, especially roof trusses, often have a sloping top chord. To han-
                               dle this situation, the uniform force method has been generalized as shown in Fig. 3.55 to include
                               nonorthogonal members. As before, α and β locate the centroids of the gusset-edge connections
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