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Brockenbrough_Ch06.qxd  9/29/05  5:15 PM  Page 6.9



                                                    DESIGN OF BUILDING MEMBERS


                                                                               DESIGN OF BUILDING MEMBERS  6.9

                                    The shear forces in the web of wide-flange sections should be calculated, especially if large con-
                                  centrated loads occur near the supports. For LRFD, the AISC Specification requires that the factored
                                  shear V v (kips) not exceed
                                                                        φ v = 0.90                    (6.13)
                                                              V v =φ v V n
                                  The nominal shear strength V n (kips) is calculated as follows:
                                                                                                      (6.14)
                                                                V n = 0.6F yw A w C v
                                  where  h = clear distance between flanges (less the fillet or cover radius for rolled shapes, in
                                        t w = web thickness, in
                                       F yw = yield strength of the web, ksi
                                        A w = web area, in 2
                                        C v = web shear coefficient
                                      /
                                                                 /
                                                        /
                                           .
                                                             .
                                                 /
                                  For ht ≤110  k E F ,  or ht ≤ 234  E F yw  for rolled I-shaped members (k v = 5),
                                               v
                                                  yw
                                       w
                                                         w
                                                                   C v = 1.0                          (6.15)
                                  For 110 kE /F yw  <  / h t ≤ 137 kE /F yw  ,
                                                       .
                                      .
                                                           v
                                                   w
                                          v
                                                                          /
                                                                    .
                                                                   110  kE F yw
                                                                        v
                                                               C =                                    (6.16)
                                                                v
                                                                       /
                                                                       ht w
                                  For ht >137  k E F ,
                                           .
                                                 /
                                      /
                                               v
                                       w
                                                  yw
                                                            .
                                                      C =  151 Ek v                                   (6.17)
                                                       v
                                                           ht )
                                                          (/  w  2  F yw
                                                       k =+    5
                                                          5
                                                               +
                                                        v
                                                             ( ah) 2
                                                                              [
                                                                    /
                                                                           /
                                                         = 5  when  ah > 3  or  ah > 260/( /  2
                                                                                  h t)]
                                                         = 5  if no stiffeners are used
                                  where a = distance between transverse stiffeners
                      6.9 EXAMPLE—LRFD FOR SIMPLE-SPAN FLOOR BEAM
                                  Floor framing for an office building is to consist of open-web steel joists with a standard corrugated
                                  metal deck and 3-in-thick normal-weight concrete fill. The joists are to be spaced 3 ft center to center.
                                  Steel beams spanning 30 ft between columns support the joists. A bay across the building floor is
                                  shown in Fig. 6.3.
                                    Floor beam AB in Fig. 6.3 will be designed for this example. The loads are listed in Table 6.1.
                                  The live load is reduced in Table 6.1, as permitted by the applicable building code. The reduction
                                  factor R is given by the smaller of
                                                              R = 0.0008(A − 150)                     (6.18)
                                                                        D 
                                                              R = 0 231 1 +                           (6.19)
                                                                  .
                                                                        L 
                                                              R = 0.4  for beams                      (6.20)
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