Page 416 - Design of Reinforced Masonry Structures
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WALLS UNDER GRAVITY AND TRANSVERSE LOADS      6.67

           Example 6.5  Analysis of a T-Section of a Wall under Flexure: T-beam
           Analysis.
             A nominal 8-in.-wide concrete masonry wall is reinforced with No. 8 Grade 60 bars
           placed vertically at 24 in. center-to-center (Fig. E6.5A). It carries a roof dead load of
                                                                        2
           400 lb/ft and live load of 250 lb/ft. The masonry is in running bond and  ′ =f  1500 lb/in .
                                                              m
           Calculate the moment strength of the wall which is subjected to out-of-plane loads.
                                            D = 400 lb/ft
                                            Lr = 250 lb/ft
                                                 e = 4"


                                                       Roof


                                                    Ledger beam
                 Lateral loads

                     Wind or
                     seismic                      # 8 @ 24" o.c.
                                                  vertical bars
                                                                 h
                                                  Horizontal bars

                                                  8" nominal
                                                  CMU wall





                 FIGURE E6.5A

           Solution
           Estimate the area of masonry in compression. Assume that depth of compression block
           equals the face-shell thickness, t = 1.25 in.
                                  s
             The effective width, b , is the smallest of
                             e
             (a)  Center-to-center of reinforcement = 24 in. (governs)
             (b)  Six times the nominal thickness of masonry unit = 6 × 8 = 48 in.
             (c) 72 in.
                                    Use b  = 24 in.
                                        e
             Area of masonry in compression = bets = 24 (1.25) = 30 in. 2
             Compression stress resultant:
                                    ′
                         C = 0 80.  f ab = 0 80 1 5 30.  ( . )(  )  = 36 kips
                                   m
             Tensile force resultant,
                          T = A f  = 0.79(60) = 47.4 kips > 36 kips
                              s y
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