Page 224 - Handbook of Civil Engineering Calculations, Second Edition
P. 224

REINFORCED CONCRETE                   2.9


                              C u2,max   5.42(40,000)    216,800 lb
                              (964,326.4 N); s max   216,800/[10(2550)]
                                8.50 in. (215.9 mm).
                              2. Compute the ultimate-
                              moment capacity
                              Thus, M u,max   0.90[(76,500(19.5   5/2)
                                216,800(19.5   8.50/2)]   4,145,000
                              in.·lb (468,300 N·m).
                                                               FIGURE 5
                              DESIGN OF REINFORCEMENT
                              IN A T BEAM OF GIVEN SIZE

                              The T beam in Fig. 5 is to resist an ultimate moment of 3,960,000 in.·lb (447,400.8 N·m).
                              Determine the required area of reinforcement, using f c 
  3000 lb/sq.in. (20,685 kPa) and
                              f y   40,000 lb/sq.in. (275,800 kPa).


                              Calculation Procedure:

                              1. Obtain a moment not subject to reduction
                              From the previous calculation procedure, the ultimate-moment capacity of this member is
                              4,145,000 in.·lb (468,300 N·m). To facilitate the design, divide the given ultimate moment
                              M u by the capacity-reduction factor to obtain a moment M u 
 that is not subject to reduction.
                              Thus M u 
  3,960,000/0.9   4,400,000 in.·lb (497,112 N·m).
                              2. Compute the value of s associated with the given moment
                              From step 2 in the previous calculation procedure, M u1 
  1,300,000 in.·lb (146,874 N·m).
                              Then  M u2 
  4,400,000    1,300,000    3,100,000 in.·lb (350,238 N·m). But  M u2
                              2550(10s)(19.5   s/2), so s   7.79 in. (197.866 mm).
                              3. Compute the area of reinforcement
                                               1
                              Thus, C u2   M u2 
 /(d   /2s)   (19.5   3.90)   198,700 lb (883,817.6 N). From step 1 of
                              the previous calculation procedure, C u1   76,500 lb (340,272 N); T u   76,500   198,700
                              275,200 lb (1,224,089.6 N); A s   275,200/40,000   6.88 sq.in. (174.752 mm).
                              4. Verify the solution
                              To verify the solution, compute the ultimate-moment capacity of the member. Use the
                              notational system given in earlier calculation procedures. Thus, C uf   16(5)(2550)
                              204,000 lb (907,392 N); C uw   275,200   204,000   71,200 lb (316,697.6 N); m   71,200/
                              [2550(10)]   2.79 in. (70.866 mm); M u   0.90 [204,000(19.5   2.5)   71,200(19.5   5
                              1.40)]   3,960,000 in.·lb (447,400.8 N·m). Thus, the result is verified because the com-
                              puted moment equals the given moment.


                              REINFORCEMENT AREA FOR A DOUBLY
                              REINFORCED RECTANGULAR BEAM

                              A beam that is to resist an ultimate moment of 690 ft·kips (935.6 kN·m) is restricted to a
                              14-in. (355.6-mm) width and 24-in. (609.6-mm) total depth. Using f c 
  5000 lb/sq.in. and
                              f y   50,000 lb/sq.in. (344,750 kPa), determine the area of reinforcement.
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