ing an area of 5.37 in^2 (34.647 cm^2 ). The beam is made of 2500-lb/in^2 (17,237.5-kPa)
concrete, and the steel has a yield-point stress of 40,000 lb/in^2 (275,800 kPa). Compute
the ultimate moment this beam may resist (a) without referring to any design tables and
without applying the basic equations of ultimate-strength design except those that are
readily apparent; (b) by applying the basic equations.Calculation Procedure:- Compute the area of reinforcement for balanced design
Use the relation es =fy/Es = 40,000/29,000,000 = 0.00138. For balanced design, eld =
ej(ec + es) = 0.0037(0.003 + 0.00138) = 0.685. Solving for c by using the relation for c/d,
we find c = 13.36 in (339.344 mm). Also, a = k^c = 0.85(13.36) = 11.36 in (288.544 mm).
Then Tu = Cu = «6(0.85/; = 11.36(12)(0.85)(2500) = 290,000 Ib (1,289,920 N); A 3 =
Tulfy = 290,000/40,000 = 7.25 in^2 (46,777 cm^2 ); and Q.75AS = 5.44 in^2 (35.097 cm^2 ). In
the present instance, A 5 = 5.37 in^2 (34.647 cm^2 ). This is acceptable. - Compute the ultimate-moment capacity of this member
Thus Tn = Asfy = 5.37(40,000) = 215,000 Ib (956,320 N); Cu = 06(0.85/; = 25,50Oa =
215,000 Ib (956,320 N); a = 8.43 in (214.122 mm); Mu = </>Tu(d - a/2) = 0.90
(215,000)(19.5 - 8.43/2) = 2,960,000 in-lb (334,421 N-m). These two steps comprise the
solution to part a. The next two steps comprise the solution of part b. - Apply Eq. 10; ascertain whether the member satisfies the Code
Thus, qmax = 0.6375^(87,000)7(87,000 + fy) = 0.6375(0.85)(87/127) = 0.371; q =
[AJ(bd)]fc = [5.37/(12 x 19.5)]40/2.5 = 0.367. This is acceptable. - Compute the ultimate-moment capacity
Applying Eq. 5 yields Mn =Asfyd(l - 0.59?) = 0.90(5.37)(40,000)(19.5)(1 - 0.59 x
0.367) = 2,960,000 in-lb (334,421 N-m). This agrees exactly with the result computed in
step 2.
DESIGN OFA RECTANGULAR BEAM
A beam on a simple span of 20 ft (6.1 m) is to carry a uniformly distributed live load of
1670 Ib/lin ft (24,372 N/m) and a dead load of 470 Ib/lin ft (6859 N/m), which includes
the estimated weight of the beam. Architectural details restrict the beam width to 12 in
(304.8 mm) and require that the depth be made as small as possible. Design the section,
using/; = 3000 lb/in^2 (20,685 kPa) and/; = 40,000 lb/in^2 (275,800 kPa).
Calculation Procedure:- Compute the ultimate load for which the member
is to be designed
The beam depth is minimized by providing the maximum amount of reinforcement per-
mitted by the Code. From the previous calculation procedure, #max = 0.371.
Use the load factors given in the Code: WDL = 470 Ib/lin ft (6859 N/m); WLL = 1670
Ib/lin ft (24,372 N/m); L = 20 ft (6.1 m). Then wu = 1.5(470) + 1.8(1670) = 3710 Ib/lin ft
(54,143 N/m); Mu = % (3710)(20)
2
12 = 2,230,000 in-lb (251,945.4 N-m).