Page 220 - Handbook of Civil Engineering Calculations, Second Edition
P. 220

REINFORCED CONCRETE                   2.5

                                To allow for material imperfections, defects in workmanship, etc., the  Code intro-
                              duces the capacity-reduction factor  . A section of the Code sets     0.90 with respect to
                              flexure and     0.85 with respect to diagonal tension, bond, and anchorage.
                                The basic equations for the ultimate-strength design of a rectangular beam reinforced
                              solely in tension are

                                                  C u   0.85abf c 
  T u   A s f y          (1)

                                                           [A s /(bd)] f y
                                                       q                                    (2)
                                                              f c
                                                                  1.18qd
                                                  a   1.18qd  c                             (3)
                                                                    k 1
                                                              
   a
                                                     M u    A s f y d                       (4)
                                                                  2
                                                    M u    A s f y d(1   0.59q)             (5)

                                                           2
                                                   M u    bd f c 
q(1   0.59q)              (6)
                                                    bdf c   [(bdf c )   2bf c M u / ] 0.5
                                                              2
                                               A s                                          (7)
                                                              f y
                                                       0.85k 1 f c 
  87,000
                                                  p b                                       (8)
                                                          f y  87,000   f y
                                                               87,000
                                                   q b   0.85k 1                            (9)
                                                            
 87,000   f y
                              In accordance with the Code,
                                                                    87,000
                                              q max   0.75q b   0.6375k 1                  (10)
                                                                 
 87,000   f y
                                Figure 3 shows the relationship between
                              M u and A s for a beam of given size. As A s in-
                              creases, the internal forces C u and T u increase
                              proportionately, but M u increases by a small-
                              er proportion because the action line of C u is
                              depressed. The  M u -A s diagram is parabolic,
                              but its curvature is small. By comparing the
                              coordinates of two points P a and P b , the fol-
                              lowing result is obtained, in which the sub-
                              scripts correspond to that of the given point:

                                           M ua
                                                 M ub
                                               >           (11)
                                           A sa  A sb
                                                                    FIGURE 3
                              where A sa < A sb
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