Page 222 - Handbook of Civil Engineering Calculations, Second Edition
P. 222

REINFORCED CONCRETE                   2.7

                              2. Establish the beam size
                                                  2
                              Solve Eq. 6 for d. Thus, d   M u / bf c 
q(1   0.59q)]   2,230,000/[0.90(12)(3000)
                              (0.371)(0.781)]; d   15.4 in. (391.16 mm).
                                Set d   15.5 in. (393.70 mm). Then the corresponding reduction in the value of q is
                              negligible.
                              3. Select the reinforcing bars
                                                                                             2
                              Using Eq. 2, we find A s   qbdf c 
/f y   0.371(12)(15.5)(3/40)   5.18 sq.in. (33.421 cm ).
                                                                                  2
                              Use four no. 9 and two no. 7 bars, for which A s   5.20 sq.in. (33.550 cm ). This group of
                              bars cannot be accommodated in the 12-in. (304.8-mm) width and must therefore be
                              placed in two rows. The overall beam depth will therefore be 19 in. (482.6 mm).
                              4. Summarize the design
                              Thus, the beam size is 12   19 in. (304.8   482.6 mm); reinforcement, four no. 9 and
                              two no. 7 bars.




                              DESIGN OF THE REINFORCEMENT IN A
                              RECTANGULAR BEAM OF GIVEN SIZE

                              A rectangular beam 9 in. (228.6 mm) wide with a 13.5-in. (342.9-mm) effective depth is
                              to sustain an ultimate moment of 95 ft·kips (128.8 kN·m). Compute the area of reinforce-
                              ment, using f c 
  3000 lb/sq.in. (20,685 kPa) and f y   40,000 lb/sq.in. (275,800 kPa).


                              Calculation Procedure:

                              1. Investigate the adequacy of the beam size
                              From previous calculation procedures,  q max   0.371. By Eq. 6,  M u,max   0.90
                                    2
                              (9)(13.5) (3)(0.371)(0.781)   1280 in·kips (144.6 kN·m); M u   95(12)   1140 in.·kips
                              (128.8 kN·m). This is acceptable.
                              2. Apply Eq. 7 to evaluate A s
                              Thus, f c   0.85(3)   2.55 kips/sq.in. (17.582 MPa); bdf c   9(13.5)(2.55)   309.8 kips
                                                        2
                                                                                        2
                                                                 0.5
                              (1377.99 kN); A s   [309.8   (309.8   58,140) ]/40   2.88 sq.in. (18.582 cm ).

                              CAPACITY OF A T BEAM

                              Determine the ultimate moment that may be resisted by the T beam in Fig. 4a if f c
                              3000 lb/sq.in. (20,685 kPa) and f y   40,000 lb/sq.in. (275,800 kPa).


                              Calculation Procedure:

                              1. Compute T u and the resultant force that may be developed
                              in the flange
                              Thus, T u   8.20(40,000)   328,000 lb (1,458,944 N); f c   0.85(3000)   2550 lb/sq.in.
                              (17,582.3 kPa);  C uf   18(6)(2550)    275,400 lb (1,224,979 N). Since  C uf <  T u , the
                              deficiency must be supplied by the web.
   217   218   219   220   221   222   223   224   225   226   227