Page 486 - Design of Reinforced Masonry Structures
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7.48                      CHAPTER SEVEN

           Example 7.9  Design base shear and distribution of the base shear over
           the building height.
             Determine the base shear and its distribution over the height of a building that con-
           sists of a building frame system with special reinforced masonry shear walls (Fig. E7.9).
           The building is assigned Seismic Design Category D and importance factor I = 1.0. It
           is located in a seismic region for which spectral acceleration parameters S  and S  have
                                                                S    1
           been determined, respectively, as 0.90g and 0.45g. The site class for building site has
           been determined as C. The seismic weight assigned to various levels is as follows:








                                                            60'







                      FIGURE E7.9

                Level 1: 650 kips
                Level 2: 600 kips
                Level 3: 500 kips
                Level 4: 450 kips
                Level 5: 400 kips
           Solution
             Given:
                S  = 0.90g
                 S
                S  = 0.45g
                 1
                Seismic design category = D
                Site class = C
                I = 1.0
             1. Determine S
                         DS              2
                                     S DS  =  3 S MS

             From Table 7.2, S  = 0.98g, F  ≈ 1.0
                          S       a
                             S  = F S  = 1.0(0.90) = 0.90
                                    s
                              MS
                                  a
                                  2
                             S  = (090 ) = 060.
                                    .
                              DS
                                  3
                             S  =  2  S
                              D1    M1
                                  3
                             S M1  = F S  = (1.35)(0.45) = 0.61
                                   v
                                    1
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