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                                                    DESIGN OF BUILDING MEMBERS


                                                                              DESIGN OF BUILDING MEMBERS  6.19

                                                   TABLE 6.3 Q n for Partial Composite Design (kips)
                                                   Location of PNA    Q n and concrete compression
                                                       (1)                    A x F y
                                                      (2)–(5)               A s F y –2∆A f F* y
                                                       (6)                0.5[C(5) + C(7)] †
                                                       (7)                   0.25A s F y
                                                     *∆A f = area of the segment of the steel flange above the plastic
                                                   neutral axis (PNA).
                                                     † C(n) = compressive force at location (n).


                                    When a concentrated load occurs on a beam, the number of shear connectors between the con-
                                  centrated load and the inflection point should be adequate to develop the maximum moment at the
                                  concentrated load.
                                    When the moment capacity of a fully composite beam is much greater than the applied moment,
                                  a partially composite beam may be utilized. It requires fewer shear connectors and thus has a lower
                                  construction cost. A partially composite design also may be used advantageously when the number
                                  of shear connectors required for a fully composite section cannot be provided because of limited
                                  flange width and length.
                                    Figure 6.6 shows seven possible locations of the plastic neutral axis (PNA) in a steel section. The
                                  horizontal shear between the steel section and the concrete slab, which is equal to the compressive
                                  force in the concrete C, can be determined as illustrated in Table 6.3.


                      6.13 LRFD FOR COMPOSITE BEAM WITH UNIFORM LOADS

                                  The typical floor construction of a multistory building is to have composite framing. The floor con-
                                        1
                                  sists of 3 / 4-in-thick lightweight concrete over a 2-in-deep steel deck. The concrete weighs 115 lb/ft 3
                                  and has a compressive strength of 3.0 ksi. An additional 30% of the dead load is assumed for equip-
                                  ment load during construction. The deck is to be supported on steel beams with stud shear connec-
                                  tors on the top flange for composite action (Art. 6.12). Unshored construction is assumed. Therefore,
                                  the beams must be capable of carrying their own weight, the weight of the concrete before it hard-
                                                                                                   1
                                                                                  3
                                  ens, deck weight, and construction loads. Shear connectors will be  / 4 in in diameter and 3 / 2 in long.
                                  The floor system should be investigated for vibration, assuming a damping ratio of 5%.
                                    A typical beam supporting the deck is 30 ft long. The distance to adjacent beams is 10 ft. Ribs
                                  of the deck are perpendicular to the beam. Uniform dead loads on the beam are, for construction,
                                  0.50 kip/ft, plus 30% for equipment loads, and for superimposed load, 0.25 kip/ft. Uniform live load
                                  is 0.50 kip/ft.
                                  Beam Selection.  Initially, a beam of A992 steel that can support the construction loads is selected.
                                  It is assumed to weigh 22 lb/ft. Thus, the beam is to be designed for a service dead load of 0.5 × 1.3 +
                                  0.022 = 0.672 kip/ft.
                                                       Factored load = 0.672 × 1.4 = 0.941 kip/ft
                                                   Factored moment = M u = 0.941 ×  30  2  = 105.8 kip⋅ft
                                                                             8
                                  The plastic section modulus required therefore is
                                                           Z =  M u  =  105 8 .  ×12  = 28 2 . in 3
                                                              φ F y  09 .  × 50

                                                                                4
                                                       3
                                  Use a W14 × 22 (Z = 33.2 in and moment of inertia I = 199 in ).

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