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                                                    DESIGN OF BUILDING MEMBERS


                                                                              DESIGN OF BUILDING MEMBERS  6.23

                                                . / ) 2
                                              (13 7  − . 3 53 0
                                                         + .335
                                            ′′ =y             + . 720
                                                      2
                                           y ′′ = . 870  in =  distance from PNA to C web
                                                          y
                                          M n  = C c  × + C flange  × ′ + C web  ×    ′′ y
                                                  e
                                                                                        − .355 8
                                             = 81 .1 ×15 .60  + 50 0  × . 5 0  ×10 .36 0 23 13 7  − . 3 53 0  ]( .70 )}
                                                                        + . [( . / ) 2
                                                          { .335
                                          M n  = , 2 431  kip ft ⋅
                                               .
                                         φ φM =  0 90  × 2331
                                            n
                                                  2
                                                     ⋅
                                                 .
                                         φM = 182 3 kip ft
                                            n
                                    From the partial composite values 2 to 7, value 6 is greater than M u = 183.4 kip⋅ft.
                                    The AISC “Manual of Steel Construction” includes design tables for composite beams that greatly
                                  simplify the calculations. For example, the table for a W14 × 22, A992, composite beam gives φM n
                                  for the seven positions of the PNA and for several values of the distance Y 2 (in) from the concrete
                                  compressive force C to the top of the steel beam. For the preceding example,
                                                                 Y =  Y  −  a                         (6.29)
                                                                      con
                                                                  2
                                                                         2
                                  where Y con = total thickness of floor slab, in
                                         a = depth of the concrete compressive-stress block, in
                                  From the table for case 6, ∑Q n = 119 kips.
                                                           a =    119    = 0 519.  in
                                                               .
                                                              085 × 30 90.  ×
                                  Substitution of a and Y con = 5.25 in in Eq. (6.29) gives
                                                                      .
                                                            Y =  525−.  0 519  =  499 in
                                                                            .
                                                             2
                                                                      2
                                  The manual table gives the corresponding moment capacity for case 6 and Y 2 = 4.99 in as
                                                                             ⋅
                                                        φM = 195 kip ⋅ >183 9 kip ft  OK
                                                                   ft
                                                                        .
                                                          n
                                    The number of shear studs is based on C = 119.1 kips. The nominal strength Q n of one stud is
                                                                                   2
                                                    3
                                  given by Eq. (6.28). For a  / 4-in stud, with shearing area A sc = 0.442 in and tensile strength F u = 60 ksi,
                                  the limiting strength without reduction factors is A sc F u = 0.442 × 60 = 26.5 kips. With concrete unit
                                                 3
                                  weight w = 115 lb/ft and compressive strength f c ′= 3.0 ksi, and modulus of elasticity E c = 2136 ksi,
                                  the nominal strength given by Eq. (6.28) is
                                                        ×
                                                                 ×
                                                                          .
                                                                                      .
                                                      .
                                                 Q = 0 5 0 442 3 0 2136  = 17 7 kips  <  R R 26 5 kips
                                                          .
                                                               .
                                                                                 g
                                                                                   p
                                                   n
                                  R g = 1.0 for deck perpendicular to beam, and R p = 0.75 when e mid-ht ≥ 2 in (see Art. 5.8.6). Thus,
                                  R g R p A sc F u = 1.00 × 0.75 × 26.5 = 19.9 kips and Q n = 17.7 kips.
                                    The number of shear studs required is 2 × 119.1/17.7 = 13.5. Use 14. The total number of metal
                                  deck ribs supported on the steel beam is 30. Therefore, only one row of shear studs is required.
                                  Deflection Calculations.  Deflections are calculated based on the partial composite properties of
                                  the beam. First, the properties of the transformed full composite section (Fig. 6.7) are determined.
                                    The modular ratio E s /E n is n = 29,000/2136 = 13.6. This is used to determine the transformed con-
                                                                                           2
                                                                2
                                  crete area A 1 = 3.25 × 90/13.6 = 21.52 in . The area of the W14 × 22 is 6.49 in , and its moment of
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