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5.62                       CHAPTER FIVE

           The bending of the column occurs about its strong axis (axis parallel to the short side of
           the cross section). The design, 5 percent damped, spectral response acceleration param-
           eter at short periods, S  = 1.25g. Calculate the shear strength of this column.
                           DS
           Solution
           Given: CMU column nominal 16 × 24 in., b = 15.625 in., h = 23.625 in., d = 19.825 in.,
                             2
           d′= 3.8 in., A  = 2.41 in.  (four No. 7 bars), column height h′ = 24 ft,  ′ f  = 2000 psi,
                     st
                                                                m
           f  = 60 ksi., S  = 1.25g.
           y
                    DS
              (Note: The column described in this example is the same column as described in
           Example 5.11 for which interaction diagram was plotted. Based on that interaction dia-
           gram, it is evident that the column can support the imposed loads. These calculations are
           not repeated here. In general, for a column for which shear strength is to be determined,
           a designer must also ensure that it can support the imposed loads.)
             Check code compliance with respect to dimensional limits, h ′/t ratio and longitudi-
           nal reinforcement.
                        Nominal column width = 16 in. > 8 in.    OK
                   Nominal column depth = 24 in. < 3t = 3 × 16 = 48 in.    OK
                              h ′/t = (24)(12)/16 = 18 < 30    OK
                             A  = (15.625)(23.625) = 369 in. 2
                              n
           Longitudinal reinforcement, A  = 2.41 in. 2
                                 st
                                        .
                                 ρ =  A st  =  241  = 0 0065
                                              .
                                   A   369 0 .
                                    n
                      r   = 0.04    r   = 0.0025    r   = 0.0065
                       max         min          provided
                             0.0025 < 0.0065 < 0.04    OK
                     Lateral ties: Tie diameter = 0.375 in. > 0.25 in.    OK

                          s = 8 in. ≤ 16d  = 16 (7/8) = 14 in.    OK
                                    b
                         ≤ 48 tie diameter = 48(0.375) 18 in.    OK
                         ≤ 16 in.    OK
           Load Combinations:

             1.  U = 1.4D = 1.4(20) = 28 kips
             2.  U = 1.2D + 1.6L  = 1.2 (20) + 1.6 (20) = 56 kips
                            r
             3.  U = (1.2 + 0.2S )D + rQ  + L + 0.2S
                           DS
                                  E
                 = [1.2 + 0.2(1.25)](20) = 29 kips
                In Load Combination 3, the term rQ  represents horizontal force effects due to
                                          E
                earthquake (= 0 for gravity load effects)
                [Note: Both live load (L) and snow load (S) are zero in this problem.]
             4.  U = (0.9 – 0.2S ) D + rQ + 1.6H
                           DS
                                   E
                where H = load due to lateral pressure due to earth, ground water or bulk materials
                      = 0 in this problem
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