Page 248 - Handbook of Civil Engineering Calculations, Second Edition
P. 248

REINFORCED CONCRETE                  2.33

















                              FIGURE 20


                              (275,800 kPa). By starting with c   8 in. (203.2 mm) and assigning progressively higher
                              values to c, construct the interaction diagram for this member.


                              Calculation Procedure:
                              1. Compute the value of c associated with balanced design
                              An interaction diagram, as the term is used here, is one in which every point on the curve
                              represents a set of simultaneous values of the ultimate moment and allowable ultimate ax-
                              ial load. Let 
 A and 
 B   strain of reinforcement at A and B, respectively; 
 c   strain of
                              extreme fiber of concrete; F A and F B   stress in reinforcement at A and B, respectively,
                              lb/sq.in. (kPa); F A and F B   resultant force in reinforcement at A and B, respectively; F c
                                resultant force in concrete, lb (N).
                                Compression will be considered positive and tension negative. For simplicity, disre-
                              gard the slight reduction in concrete area caused by the steel at A.
                                Referring to Fig. 20b, compute the value of c associated with balanced design. Comput-
                              ing P b and M b yields c b /d   0.003/(0.003   f y /E s )   87,000/(87,000   f y ); c b ,   10.62 in.
                              (269.748 mm). Then 
 A /
 B   (10.62   2.5)/(15.5   10.62) > 1; therefore, f A   f y ; a b
                              0.85(10.62)   9.03 in. (229.362 mm); F c   0.85(3000)(12a b )   276,300 lb (1,228,982.4 N);
                              F A    40,000(2.00)   80,000 lb (355,840 N); F B    80,000 lb ( 355,840 N); P b
                              0.70(276,300)   193,400 lb (860,243.2 N). Also,
                                                       F c (t   a)  (F A   F B )(t   2d
)
                                           M b   0.70                                      (38)
                                                                      2
                                                        2
                              Thus,  M b   0.70[276,300(18    9.03)/2    160,000(6.5)]    1,596,000 in.·lb (180,316.1
                              N·m).
                                When c > c b , the member fails by crushing of the concrete; when c < c b , it fails by
                              yielding of the reinforcement at line B.
                              2. Compute the value of c associated with incipient yielding
                              of the compression steel
                              Compute the corresponding values of P u and M u . Since 
 A and 
 B are numerically equal,
                              the neutral axis lies at N. Thus, c   9 in. (228.6 mm); a   0.85(9)   7.65 in. (194.31 mm);
                              F c   30,600(7.65)   234,100 lb (1,041,276.8 N); F A    80,000 lb (355,840 N); F B
                               80,000 lb ( 355,840 N);  P u   0.70 (234,100)    163,900 lb (729,027.2 N);  M u
                              0.70(234,100   5.18   160,000   6.5)   1,577,000 in.·lb (178,169.5 N·m).
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