(a) Section (b) Effective section (c) Resultant
forces
FIGURE 4
- Compute the resultant force developed in the web and the
depth of the stress block in the web
Thus, Cuw = 328,000 - 275,400 = 52,600 Ib (233,964.8 N); m = depth of the stress block =
52,600/[2550(1O)] = 2.06 in (52.324 mm).
- Evaluate the ultimate-moment capacity
Thus, Mu = 0.90[275,400(20.5 - 3) + 52,600(20.5 - 6 - 1.03)] = 4,975,000 in-lb
(562,075.5 N-m).
- Determine if the reinforcement compiles with the Code
Let b' = width of web, in (mm); Asl = area of reinforcement needed to resist the compres-
sive force in the overhanging portion of the flange, in^2 (cm^2 ); As2 = area of reinforcement
needed to resist the compressive force in the remainder of the section, in^2 (cm^2 ). Then/? 2
- As2/(b'd); Asl = 2550(6)(18 - 10)740,000 = 3.06 in^2 (19.743 cm^2 ); As2 = 8.20 - 3.06 =
5.14 in^2 (33.163 cm^2 ). Thenp 2 - 5.147(10(20.5)] = 0.025.
A section of the ACI Code subjects the reinforcement ratio p 2 to the same restriction
as that in a rectangular beam. By Eq. 8,/?25max = 0.15pb = 0.75(0.85)(0.85)(3/40)(87/127)
= 0.0278 > 0.025. This is acceptable.
CAPACITY OFAT BEAM OF GIVEN SIZE
The T beam in Fig. 5 is made of 3000-lb/in^2 (20,685-kPa) concrete, and^ = 40,000 lb/in^2
(275,800 kPa). Determine the ultimate-moment capacity of this member if it is reinforced
in tension only.
Calculation Procedure:
- Compute CU1, Cu2finax, and smax
Let the subscript 1 refer to the overhanging portion of the flange and the subscript 2 refer
to the remainder of the compression zone. Then/c = 0.85(3000) = 2550 lb/in^2 (17,582.3
kPa); CMl ='2550(5)(16 - 10) - 76,500 Ib (340,272 N). From the previous calculation
procedure, p2max = 0.0278. Then As2max = 0.0278(10)(19.5) = 5.42 in^2 (34.970 cm^2 );
