Page 331 - Design of Reinforced Masonry Structures
P. 331

COLUMNS                          5.51

           The force in the tensile reinforcement is
                        T = A f  = 1.2(11.6) = 13.92 kips (compressive)
                            st T
           The force in masonry in compression is given by Eq. (5.21):
                               ′
                        C = 0 80.  f ab = 0 80 2 0.  ( . )( a 15 625)(  .  )  = 25 akips
                        m      m
           Substituting a = 0.80c = 0.80(23.625) = 18.9 in. in the above expression, we obtain
                               C  = 25 (18.9) = 472.5 kips
                                m
           For vertical equilibrium of forces, we must have ΣF  = 0. Thus,
                                                y
                                C  + C  + T – P = 0
                                  s
                                     m
                              70.08 + 472.5 + 13.92 – P = 0
           whence
                                    P = 556.5 kips

           Multiplying the above value of P by C  = 0.793 (for slenderness), we have
                                       P
                           P  = PC  = (556.5)(0.793) = 441.3 kips
                            n
                                 P
                          fP  = 0.9(441.3) = 397.2 kips
                            n
           Taking moments of C , C , and T about the centroidal axis of the column, we obtain,
                          s
                             m
                    h
                             ha
                                        h
                               −
                       d′ +
                                  −
             M =  C s( )  C m( ) ( )
                     −
                                   Td −
               n
                    2        2 2        2
                                         .
                         .
                                             −
                    .
                = 70 08 ( 23 625  − − 3 8 . ) + 472 5 .  ( 23 625 18 9 . ) −13 92 19 825.  (  .  −  23..625 )
                         2               2     2                  2
                     .5 1116 −
                =  561 +   .3 111 .5
                = 1566 .3k-in. =  130..5k-ft
            φM =  0 .(1566 . ) 1409 .7 k-in. =  117 .5 k--ft
                         3 =
                   9
               n
           Point 3: Assume that the neutral axis passes through the tension reinforcement.
             When the neutral axis passes through the tension reinforcement, the strain in this
           reinforcement is zero. Therefore, T = 0.
                         c = d = 23.625 – 3.8 = 19.825 in.
                         a = 0.8c = 0.8(19.825) = 15.86 in.

                                ′
                                                            .
                        C = 0 80.  f ab = 0 80 2 0 15 86 15 625.  ( . )(  .  )(  .  )  = 396 5kips
                         m
                                m
             The strain distribution would be as shown in Fig. E5.11C. The strain in compression
           reinforcement is given by
                      ( ) (
                                    .
                                        0 0025)
                  ε ′ = 1 −  ′ d  ε m  = 1 −  38  ) (.  = 0 00202.  < < ε y  = 0 00207.
                   s
                         h       19 825
                                   .
                          ε
           Therefore,   f s ′= ′E s  = (.0 00202 )(29 ,000 ) = 58 .58ksi
                           s
                                                    =
                                                       .
                                             0820
                       A f ′− .080
                                     . [5858
                   C = ′(  s    f ′ =) 12  .  − . ( . )] 6844kips
                                 m
                    s
                        s
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