Page 169 - Design of Reinforced Masonry Structures
P. 169

DESIGN OF REINFORCED MASONRY BEAMS          4.33

             Substitution of  ρ = Abd  from Eq. (4.37) in Eq. (4.60) yields
                              /
                             s
                                         ρ fd
                                     a =   y                         (4.61)
                                        080 f ′
                                         .
                                            m
             Combining Eqs. (4.5) and (4.61), we obtain
                                         ρ fd
                                   c =     y                         (4.62)
                                   b
                                      080 080  f ′)
                                           .
                                         (
                                       .
                                              m
             Equating the values of c  from Eqs. (4.59) and (4.62), we obtain
                               b
                            ⎛   ρ f   ⎞   ⎛  72 500 ⎞
                                               ,
                                  y
                            ⎜ ⎝  080 080)  f ′ ⎠ ⎟  d = ⎜ ⎝  72 500 +  f ⎠ ⎟ ⎠  d  (4.63)
                              .
                                             ,
                                  .
                                (
                                                   y
                                     m
             Substitution of r = r  in Eq. (4.63) yields
                            b
                                  ⎛  72 500  ⎞ ⎛ 064.  f ′ ⎞
                                      ,
                              ρ =  ⎜       ⎟ ⎜   m ⎟                 (4.64)
                               b
                                  ⎝ 72 500,  +  f y ⎠ ⎝  f  y  ⎠
             Equation (4.64) gives the value of the balanced steel ratio, r , for concrete masonry.
                                                        b
           Values of r  and 0.75r  for concrete masonry for various combinations of  ′ f  and f  are
                           b
                                                                 m
                   b
                                                                      y
           tabulated in Table 4.5.
         TABLE 4.5    Values of r b  and r max  for Practical Combinations of  ′ f  and f y  for Concrete Masonry
                                                     m
         (e  = 0.0025)
           mu
                        f y  = 60,000 psi         f y  = 40,000 psi
                               r max
                  r b                       r b         r max
                  Eq.          Eq.   0.75   Eq.          Eq.   0.75
            ′ f  psi  (4.64)  0.75r  (4.75)  (4.64)  0.75r  (4.75)    ′ f  psi
                                                               b
                                     b
            m              b        r / r max        b        r /r max  m
          1500   0.0088  0.0066  0.0071  0.920  0.0155  0.0116  0.0131  0.883  1500
          1800   0.0105  0.0079  0.0086  0.920  0.0186  0.0139  0.0158  0.883  1800
          2000   0.0117  0.0088  0.0095  0.920  0.0206  0.0155  0.0175  0.883  2000
          2500   0.0146  0.0109  0.0119  0.920  0.0258  0.0193  0.0219  0.883  2500
          3000   0.0175  0.0131  0.0143  0.920  0.0309  0.0232  0.0263  0.883  3000

          3500   0.0204  0.0153  0.0167  0.920  0.0361  0.0271  0.0306  0.883  3500
          4000 *  0.0233  0.0175  0.0190  0.920  0.0412  0.0309  0.0350  0.883  4000
             *
             Values for  ′ f  > 4000 psi are not listed because of the Code’s upper limit on the value of  ′ f  for concrete
                    m                                           m
          masonry.
         2.  Clay masonry: Following the above procedure, we can derive an expression for the bal-
           anced steel ratio for clay masonry as follows. For clay masonry,
                                    e  = 0.0035                      (4.65)
                                     mu
   164   165   166   167   168   169   170   171   172   173   174