- Establish the beam size
Solve Eq. 6 for d. Thus, d^2 = Mul$bf'cq(\ - 0.590)] = 2,230,000/[0.90(12)(3000) x
(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. - Select the reinforcing bars
Using Eq. 2, we find A 8 = qbdfc'/fy = 0.371(12)(15.5)(3/40) = 5.18 in^2 (33.42.1 cm^2 ). Use
four no. 9 and two no. 7 bars, for which A 5 = 5.20 in^2 (33.550 cm^2 ). 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). - Summarize the design
Thus, the beam size is 12 x 19 in (304.8 x 482.6 mm); reinforcement, four no. 9 and two
no. 7 bars.
DESIGN OF THE REINFORCEMENT IN A
RECTANGULAR BEAM OF GIVEN SIZEA 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// = 3000 lb/in^2 (20,685 kPa) and./; = 40,000 lb/in^2 (275,800 kPa).
Calculation Procedure:
- Investigate the adequacy of the beam size
From previous calculation procedures, #max = 0.371. By Eq. 6, MMmax = 0.90 x
(9)(13.5)^2 (3)(0.371)(0.781) = 1280 in-kips (144.6 kN-m); Mu = 95(12) = 1140 in-kips
(128.8 kN-m). This is acceptable. - Apply Eq. 7 to evaluate A 3
Thus, fc = 0.85(3) = 2.55 kips/in^2 (17.582 MPa); bdfc = 9(13.5)(2.55) = 309.8 kips
(1377.99 kN); A 5 = [309.8 - (309.8^2 - 58,140)°-^5 ]/40 - 2.88 in^2 (18.582 cm^2 ).
CAPACITY OFAT BEAMDetermine the ultimate moment that may be resisted by the T beam in Fig. 4a iffcr = 3000
lb/in^2 (20,685 kPa) and/; - 40,000 lb/in^2 (275,800 kPa).
Calculation Procedure:
- Compute Tu and the resultant force that may be developed
in the flange
Thus, Tn = 8.20(40,000) = 328,000 Ib (1,458,944 N); fc = 0.85(3000) = 2550 lb/in^2
(17,582.3 kPa); Cuf= 18(6)(2550) = 275,400 Ib (1,224,979 N). Since Cuf< Tw the defi-
ciency must be supplied by the web.